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Bank - DeFiner.org
TheBank contract, as its name implied, mainly deals with the methods and variables related to the saving pool. It uses three variables to track the total amount of tokens in the pool and their allocations. For each token,
totalLoans[token] tracks total amount of token lend.
totalReserve[token] tracks total amount of token reserved.
totalCompound[token] tracks total amount of token in compound
Therefore, the total amount of token in the pool is the summation of these three variables. The token utilization ratio U, token reservation ratio R and compound ratio C are defined accordingly.
There are three categories of methods in the contract.
The first category of methods deals with the total amount of tokens in the pool and their allocations. They are used to query the pool status of a token.
getTotalDeposit
getPoolAmount
getTokenState
getCapitalUtilizationRatio
getCapitalCompoundRatio
The second category of methods deals with the deposit rate and borrow rate of tokens in the pool. They are used to update and get the rate indexes information of each specific token.
newRateIndexCheckpoint
getDepositRatePerBlock
getBorrowRatePerBlock
depositRateIndexNow
borrowRateIndexNow
getDepositAccruedRate
getBorrowAccruedRate
The third category of methods deals with all the operations on the funds in the saving pool. These functions actually modify the recorded pool status.
For more details about the modifiers of the functions, please check the modifiers page.
getTotalDepositStore
function getTotalDepositStore(address _token) public view returns(uint)
_token: The address of the token you want to query for.
The number of total deposit tokens.
The method returns the sum of totalLoans[token],totalReserve[token] and totalCompound[token].
function getPoolAmount(address _token) public view returns(uint)
The total number of tokens that are not borrowed out in DeFiner's pool.
The method returns the sum of totalCompound[token] and totalReserve[token] .
function getTokenState(address _token) public view returns (uint256 deposits, uint256 loans, uint256 reserveBalance, uint256 remainingAssets)
deposits: The total number of deposited tokens.
loans: The total number of borrowed tokens.
reserveBalance: The total number of tokens that are still in the reserve pool.
remainingAssets: The same as getPoolAmount.
The method returns total deposit, total loans, total reservation, and total available tokens of a specific token.
function head:
function getCapitalUtilizationRatio(address _token) public view returns(uint)
The U ratio with the precision to 18 decimals.
The method returns U of a token, which is the ratio of the total loan to the total deposit.
function getCapitalCompoundRatio(address _token) public view returns(uint)
_token; The address of the token you want to query for.
The C ratio with the precision to 18 decimals.
The method returns C of a token, which is the ratio of the total amount in Compound to the total deposit.
function newRateIndexCheckpoint(address _token) public onlyAuthorized
Add a new checkpoint both on the deposit rate index curve and the borrow rate index curve. Same as Compound, DeFiner uses rate index curve to track the accumulated rate of a token. The rate index curve is updated whenever any operation on that token happens. The interests of a token earned in a specific period could be derived from the rate index curve quickly.
function getDepositRatePerBlock(address _token) public view returns(uint)
The current deposit rate per block of the token.
Get the per block deposit rate of a token, and it is calculated as follows
\text{Deposit Rate} = U\times \text{Borrow Rate} + C\times \text{Compound Supply Rate}
function getBorrowRatePerBlock(address _token) public view returns(uint)
The current borrowing rate per block of the token.
Get the per block borrow rate of a token. If the token is supported in Compoud, the borrow rate is determined by the borrow rate and supply rate of Compound as
\text{Borrow Rate} = \text{Compound Supply Rate} \times 0.4 + \text{Compound Borrow Rate} \times 0.6
\text{Borrow Rate} = 0.03 + U\times 0.15
The numbers in the equations are the initial value of the variables and are configurable in GlobalConfig contract.
function depositRateIndexNow(address _token) public view returns(uint)
The current deposit rate index.
Get deposit rate index of the current block. If the current block is a checkpoint, this method returns the exact value on the deposit index curve. Otherwise, it returns an estimated value of the deposit rate index.
function borrowRateIndexNow(address _token) public view returns(uint)
The current borrow rate index.
Get borrow rate index of the current block. If the current block is a checkpoint, this method returns the exact value on the borrow index curve. Otherwise, it returns an estimated value of the borrow rate index.
function getDepositAccruedRate(address _token, uint _depositRateRecordStart) external view returns (uint256)
_depositRateRecordStart: The start block you want to query accrued rate for.
Return the accrued deposit rate given a token and a start block with precision to 18 decimals.
The deposit accrued rate in a block gap from
B_1
B_2
\text{Deposit Rate Index}(B_2)/\text{Deposit Rate Index}(B_1)
B_1
is _depositRateRecordStart you specified in the parameter and
B_2
should be the current block number. If you deposit n tokens at block
B_1
, and the current block number is
B_2
, the tokens you have now should be
\text{Deposit Rate Index}(B_2)/\text{Deposit Rate Index}(B_1) \times n
There should be a rate index created on _depositRateRecordStart block, otherwise, it will throw an error.
function getBorrowAccruedRate(address _token, uint _borrowRateRecordStart) external view returns (uint256)
_token: The address of the token that you want to query for.
_borrowRateRecordStart: The start block you want to query accrued rate for.
Return the accrued borrow rate given a token and a start block with precision to 18 decimals.
The borrow accrued rate in a block gap from
B_1
B_2
\text{Borrow Rate Index}(B_2)/\text{Borrow Rate Index}(B_1)
. It's similar to getDepositAccruedRate.
function deposit(address _to, address _token, uint256 _amount) external onlyAuthorized
_to: The address of the account that tries to deposit.
_token: The address of the token you want to deposit.
_amount: The volume of tokens that you want to deposit.
This method will create a new checkpoint on the rate index curve. It will deposit the token to the individual account of msg.sender in Accounts contract. Then, it will update the pool balance, i.e. the value of totalReserve[token] and totalCompound[token] , accordingly.
The deposited token will be first added to the pool reservation. If the reservation ratio R exceeds 20%, the contract will deposit the token to Compound to reset R to be 15%.
function withdraw(address _from, address _token, uint256 _amount) external onlyAuthorized returns(uint)
_from: The address of the account that tries to withdraw.
_token: The address of the kind of token you want to withdraw.
_amount: The volume of tokens you want to withdraw.
This method will create a new checkpoint on the rate index curve. The new deposit balance is tracked in the individual account of msg.sender in Accounts contract. The withdraw is allowed only if the account has enough balance and its loan value is still below the reduced borrow power after withdrawn. Then, the contract will update the pool balance by calling update method.
function borrow(address _from, address _token, uint256 _amount) external onlyAuthorized
_from: The account that tries to borrow tokens.
_token: The address of the token that you want to borrow.
_amount: The volume of tokens that the user tries to borrow.
This method will create a new checkpoint on the rate index curve. The new borrowed balance is tracked in the individual account of msg.sender in Accounts contract. The borrow is allowed only if the added loan value is below the borrow power. Then, the contract will update the pool balance by calling update method.
function repay(address _to, address _token, uint256 _amount) external onlyAuthorized returns(uint)
_to: The address that tries to repay.
_token: The address of the token that the account wants to repay.
_amount: The volume of tokens that the account wants to repay.
function update(address _token, uint _amount, ActionType _action) public onlyAuthorized returns(uint256 compoundAmount)
_token: The address of the token you want to update.
_amount: The volume of tokens involved in this operation.
_action: The type of action that changes the pool amount.
DepositAction
RepayAction
WithdrawAction
This method is called whenever the token balance in the saving pool is changed. It checks if there is enough amount of a token in the pool, i.e. the market liquidity. If an account borrows or withdraws. When the balance of a token that changes in the pool, this method updates the allocation of the token in the reservation and the Compound.
If an account deposits or repay such that the reservation ratio exceeds 20%, the contract will deposit an extra amount of token to Compound and reset R to be 15%. On the other hand, if an account withdraw or borrow such that the reservation ratio falls behind 10%, the contract will withdraw the token from Compound and try to reset R to be 15%. However, in this case, there might not be enough tokens in Compound so that the final R could be below 15%.
|
Roll the Dice: The Central Limit Theorem Practice Problems Online | Brilliant
The central limit theorem, or CLT for short, is absolutely vital to statistics, so it'll crop up many times throughout our course.
In a nutshell, the CLT says that the sum of a large number of random draws is roughly distributed like a bell curve.
This is a casual version of the true CLT, but it's catchy and suits our needs well enough.
In this quiz, we'll unpack this statement and uncover the intuitive ideas at the heart of the CLT.
Without further ado, let's begin building our CLT intuition with a trip to the casino...
Roll the Dice: The Central Limit Theorem
Many fortunes have been lost at the tables of The Gambler's Ruin casino, but Marvin isn't thinking about that: he's too intent on having some fun and, hopefully, winning some cash.
Fortunately for him, his wiser and savvier friend Zhang Wei tags along to make sure Marvin doesn't gamble away all of his savings.
The pair descend onto the main floor of The Gambler's Ruin and immerse themselves in a cacophony of sounds and a galaxy of flashing lights.
Zhang Wei guides his friend quickly past the rows of slot machines and roulette wheels to a game called "Roll of the Dice," where even gullible Marvin may just have a shot at winning...
"The game is simple," explains the croupier. "You pick a number, and then wager
\$1
that the dice will roll that value."
The croupier looks Marvin up and down and decides he looks a bit of a rube, so she decides to go easy on him at first.
"We'll start with a single die, so you can bet on a roll of
1, 2, 3, 4, 5,
6.
What's your bet?"
Marvin pauses, scratches his head, and gives serious consideration to the options before him.
If the die is fair, what's Marvin's best strategy for winning?
The die is more likely to roll an odd number, so he should bet on an odd number. The die is more likely to roll an even number, so he should bet on an even number. It doesn't matter: all of the numbers are equally likely.
Zhang Wei stands off to the side and watches as Marvin makes his first bet... and loses.
"Tough break," says the croupier with a hint of mock sympathy. "Better luck next time! Want to try a more exciting game?"
Marvin nods and listens as the croupier explains the rules. "This time, I throw two dice and you bet on the total value of the two rolls. So, if you bet on
11
and one die comes up
5
and the other comes up
6,
you win; otherwise, you lose your wager. Got it?"
Marvin nods again and thinks over his options. He turns to Zhang Wei and says "
6
12
are two of my lucky numbers. What do you think? Should I bet on
6
12?
Assuming the dice are fair and the rolls are independent, what's Zhang Wei's best advice to Marvin to maximize his chances of winning?
Hint: A throw of the dice can be represented as a pair of integers, so the sample space is
\begin{aligned} \big\{ & (1,1), (1,2), (1,3),(1,4),(1,5),(1,6) \\ & (2,1), (2,2), (2,3),(2,4),(2,5),(2,6) \\ & (3,1), (3,2), (3,3),(3,4),(3,5),(3,6) \\ & (4,1), (4,2), (4,3),(4,4),(4,5),(4,6) \\ & (5,1), (5,2), (5,3),(5,4),(5,5),(5,6) \\ & (6,1), (6,2), (6,3),(6,4),(6,5),(6,6) \big\}. \end{aligned}
"Bet on
6.
" "Bet on
12.
Marvin takes Zhang Wei's advice, bets on
6,
and ends up winning!
The croupier smiles, gives a few words of encouragement, and then invites Marvin to up the challenge by betting on the total roll of three dice.
He's just about through pondering his choices when he feels a tap on his shoulder. Marvin turns and sees Zhang Wei holding out his phone to him. On the screen is the following interactive plot:
Zhang Wei explains that a bar's height in this histogram represents the number of ways
n
dice can roll a sum total of
S_{n},
the integer below it.
For example, there's only one way to roll a
3,
(1,1,1),
but there are
27
different ways of rolling an
11;
that's why its bar is so much higher than
3
Given Zhang Wei's histogram and the assumption that the dice are all fair and the rolls are independent, how should Marvin bet?
Marvin should bet on
4
5.
10
11.
24
25.
47
48.
The croupier adds one more die to the roll after every bet to make the game more fun, but she's really just making it harder for Marvin to win by adding more possible outcomes.
Zhang Wei suspected she'd do this.
Fortunately, he came prepared: the plot he shares with Marvin has a slider labeled
n
for the number of dice used in a roll (see below).
Zhang Wei adjusts the scale by dividing the heights of the histogram bars by
6^n
so the plot displays the probability distribution for
S_{n},
the sum total of a roll of
n
fair independent dice:
\small \begin{aligned} P(n \text{ dice roll a total of } S_{n} ) = \frac{\text{(number of ways } n \text{ dice can sum to } S_{n}) }{6^{n}}. \end{aligned}
n
and study the shape of the distributions as you do. What do you notice?
|
How to Multiply Square Roots: 8 Steps (with Pictures) - wikiHow
1 Multiplying Square Roots Without Coefficients
2 Multiplying Square Roots With Coefficients
You can multiply square roots, a type of radical expression, just as you might multiply whole numbers. Sometimes square roots have coefficients (an integer in front of the radical sign), but this only adds a step to the multiplication and does not change the process. The trickiest part of multiplying square roots is simplifying the expression to reach your final answer, but even this step is easy if you know your perfect squares.
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Multiply the radicands. A radicand is a number underneath the radical sign.[1] X Research source To multiply radicands, multiply the numbers as if they were whole numbers. Make sure to keep the product under one radical sign.[2] X Research source
{\displaystyle {\sqrt {15}}\times {\sqrt {5}}}
{\displaystyle 15\times 5=75}
{\displaystyle {\sqrt {15}}\times {\sqrt {5}}={\sqrt {75}}}
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Factor out any perfect squares in the radicand. To do this, see whether any perfect square is a factor of the radicand.[3] X Research source If you cannot factor out a perfect square, your answer is already simplified and you need not do anything further.
A perfect square is the result of multiplying an integer (a positive or negative whole number) by itself.[4] X Research source For example, 25 is a perfect square, because
{\displaystyle 5\times 5=25}
{\displaystyle {\sqrt {75}}}
can be factored to pull out the perfect square 25:
{\displaystyle {\sqrt {75}}}
{\displaystyle {\sqrt {25\times 3}}}
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Place the square root of the perfect square in front of the radical sign. Keep the other factor under the radical sign. This will give you your simplified expression.
{\displaystyle {\sqrt {75}}}
{\displaystyle {\sqrt {25\times 3}}}
, so you would pull out the square root of 25 (which is 5):
{\displaystyle {\sqrt {75}}}
{\displaystyle {\sqrt {25\times 3}}}
{\displaystyle 5{\sqrt {3}}}
Square a square root. In some instances, you will need to multiply a square root by itself. Squaring a number and taking the square root of a number are opposite operations; thus, they undo each other. The result of squaring a square root, then, is simply the number under the radical sign.[5] X Research source
{\displaystyle {\sqrt {25}}\times {\sqrt {25}}=25}
. You get that result because
{\displaystyle {\sqrt {25}}\times {\sqrt {25}}=5\times 5=25}
{"smallUrl":"https:\/\/www.wikihow.com\/images\/thumb\/5\/5b\/Multiply-Square-Roots-Step-5-Version-3.jpg\/v4-460px-Multiply-Square-Roots-Step-5-Version-3.jpg","bigUrl":"\/images\/thumb\/5\/5b\/Multiply-Square-Roots-Step-5-Version-3.jpg\/aid326954-v4-728px-Multiply-Square-Roots-Step-5-Version-3.jpg","smallWidth":460,"smallHeight":345,"bigWidth":728,"bigHeight":546,"licensing":"<div class=\"mw-parser-output\"><p>License: <a target=\"_blank\" rel=\"nofollow noreferrer noopener\" class=\"external text\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/3.0\/\">Creative Commons<\/a><br>\n<\/p><p><br \/>\n<\/p><\/div>"}
Multiply the coefficients. A coefficient is a number in front of the radical sign. To do this, just ignore the radical sign and radicand, and multiply the two whole numbers. Place their product in front of the first radical sign.
Pay attention to positive and negative signs when multiplying coefficients. Don't forget that a negative times a positive is a negative, and a negative times a negative is a positive.
{\displaystyle 3{\sqrt {2}}\times 2{\sqrt {6}}}
{\displaystyle 3\times 2=6}
. So now your problem is
{\displaystyle 6{\sqrt {2}}\times {\sqrt {6}}}
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Multiply the radicands. To do this, multiply the numbers as if they were whole numbers. Make sure to keep the product under the radical sign.
For example, if the problem is now
{\displaystyle 6{\sqrt {2}}\times {\sqrt {6}}}
, to find the product of the radicands, you would calculate
{\displaystyle 2\times 6=12}
{\displaystyle {\sqrt {2}}\times {\sqrt {6}}={\sqrt {12}}}
. The problem now becomes
{\displaystyle 6{\sqrt {12}}}
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Factor out any perfect squares in the radicand, if possible. You need to do this to simplify your answer.[6] X Research source If you cannot pull out a perfect square, your answer is already simplified and you can skip this step.
A perfect square is the result of multiplying an integer (a positive or negative whole number) by itself.[7] X Research source For example, 4 is a perfect square, because
{\displaystyle 2\times 2=4}
{\displaystyle {\sqrt {12}}}
can be factored to pull out the perfect square 4:
{\displaystyle {\sqrt {12}}}
{\displaystyle {\sqrt {4\times 3}}}
Multiply the square root of the perfect square by the coefficient. Keep the other factor under the radicand. This will give you your simplified expression.
{\displaystyle 6{\sqrt {12}}}
{\displaystyle 6{\sqrt {4\times 3}}}
, so you would pull out the square root of 4 (which is 2) and multiply it by 6:
{\displaystyle 6{\sqrt {12}}}
{\displaystyle 6{\sqrt {4\times 3}}}
{\displaystyle 6\times 2{\sqrt {3}}}
{\displaystyle 12{\sqrt {3}}}
We are not allowed to use a calculator, so how do I multiply a whole number by a square root?
When you multiply a whole number by a square root, you just put the two together, with the whole number in front of the square root. For example, 2 * (square root of 3) = 2(square root of 3). If the square root has a whole number in front of it, multiply the whole numbers together. So 2 * 4(square root of 3) = 8(square root of 3).
What is 2 root 3 times root 3?
√3 times √3 equals 3. Two times that is 6.
(4√5)/5. Since radicals are not supposed to be in the denominator, you multiply by √5/√5 to get (4√5)/5.
Generally we write 5√2 or calculate the square root of 2, which is 1.414, and then multiply it by 5. The answer would be 7.07.
How do I simplify the square root of (5+7) squared?
The square root and the square cancel each other out, so the final answer is 5+7, which is 12.
What is 2 times 2, root 2?
It would be 2 x 2sqrt2 = 4sqrt2, this is the simplest form and cant be reduced any further.
It's an irrational number, which you could round to 2.45. Usually you would just use a calculator to find such a square root.
What is the square root of 3 times -1?
When multiplying an integer by a square root, you simply place the integer in front of the radical sign. So -1 x √3 is expressed as -1√3. You can use a calculator to find the exact product.
What is sqr -4 times sqr -9? Is it -6 or 6?
How do you multiply a whole number with a square root?
You simply place the whole number in front of the radical sign. For example, 2 x √2 would simply be expressed as 2√2. You can use a calculator to find the exact product.
Always remember your perfect squares because it will make the process much easier!
Follow the usual sign rules to determine whether the new coefficient should be positive or negative. A positive coefficient multiplied by a negative coefficient will be negative. Two positive coefficients multiplied together or two negative coefficients multiplied together will be positive.
All terms under the radicand are always positive, so you will not have to worry about sign rules when multiplying radicands.
To multiply square roots, first multiply the radicands, or the numbers underneath the radical sign. If there are any coefficients in front of the radical sign, multiply them together as well. Finally, if the new radicand can be divided out by a perfect square, factor out this perfect square and simplify it. If you want to learn how to check your answers when you're finished solving, keep reading the article!
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75 c.c. of N/5 H2SO4, 10 c.c. of N/2 HCl and 30 c.c. of N/10 HNO3 are mixed together. The strength of the resulting acid mixture is:
Normality of the resulting mixture
=\quad \frac{{\mathrm{N}}_{1}{\mathrm{V}}_{1}\quad +\quad {\mathrm{N}}_{2}{\mathrm{V}}_{2}\hspace{0.17em}+\quad {\mathrm{N}}_{3}{\mathrm{V}}_{3}}{{\mathrm{V}}_{1}\quad +\quad {\mathrm{V}}_{2}\quad +\quad {\mathrm{V}}_{3}}\phantom{\rule{0ex}{0ex}}=\quad \frac{75\quad \times \quad {\displaystyle \frac{1}{5}}\quad +\quad 10\quad \times \quad {\displaystyle \frac{1}{2}}\quad +\quad 30\quad \times \quad {\displaystyle \frac{1}{10}}}{75\quad +\quad 10\quad +\quad 30}\phantom{\rule{0ex}{0ex}}=\quad \frac{15\hspace{0.17em}+\quad 5\quad +\quad 3}{115}\phantom{\rule{0ex}{0ex}}=\quad \frac{23}{115}\quad =\quad \frac{\mathrm{N}}{5}\quad =\quad 0.2\quad \mathrm{N}
0.2 N is the strength of the resulting acid mixture.
Which will have maximum ionization energy?
Among all the options given, Magnesium has the highest ionization energy. It is because magnesium has one more proton in its nucleus to hold on to the electrons in the 3s orbital.
Vapour density of a volatile substance is 4 (CH4 = 1). Its molecular weight would be:
Actual vapour density of methane
\frac{\mathrm{molecular}\quad \mathrm{weight}}{2}
\frac{16}{2}
\therefore
Vapour density of volatile substance = 8
\times
Molecular weight of the substance = 32
\times
Heat of a reaction at constant pressure is equal to:
HP + HR
\times
\frac{{\mathrm{H}}_{\mathrm{P}}}{{\mathrm{H}}_{\mathrm{R}}}
Heat of a reaction at constant pressure is equal to HP - HR.
In an isochoric process, the increase in internal energy is equal to:
the sum of heat absorbed and work done
heat evolved
In an isochoric process,
∆
V = 0, hence, work done P
∆
V = W = 0. So,
∆
E = q + 0. Hence, the increase in internal energy will be equal to heat absorbed by the system.
One gram of mass is equal to :
9 x 1020 ergs
3 x 105 ergs
According to E = mc2
m = 1gm
c= 3
\times
1010 cm sec-1
\therefore
E = 1gm
\times
\times
1010 cm sec-1)2
\times
1020 ergs
1 gm of mass is equal to 9 x 1020 ergs. Ergs or ergon (a Greek word) is the amount of work done by a force of one dyne exerted for a distance of one centimeter.
If Kb of HCN is 4.0
\times
10-10, then the pH of 2.5
\times
10-1 molar aqueous HCN is
[H+] in a solution of weak acid of conc. C and dissociation constant Kc is given as
\left[{\mathrm{H}}^{+}\right]\quad =\quad \sqrt{{\mathrm{K}}_{\mathrm{a}}\quad \times \quad \mathrm{C}\quad \quad }\quad =\quad \sqrt{4.0\quad \times \quad {10}^{-10}\quad \times 2.5\quad \times \quad {10}^{-1}}\phantom{\rule{0ex}{0ex}}\quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \sqrt{1\quad \times \quad {10}^{-10}}\phantom{\rule{0ex}{0ex}}\quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 1\quad \times \quad {10}^{-5}\quad \mathrm{M}\phantom{\rule{0ex}{0ex}}\mathrm{pH}\quad \quad \quad \quad =\quad -\mathrm{log}\left[{\mathrm{H}}^{+}\right]\phantom{\rule{0ex}{0ex}}\quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad -\mathrm{log}{10}^{-5}\quad \quad =\quad 5
Which of the following ions is paramagnetic?
Both 'a' and 'c'
Paramagnetic compounds have unpaired electrons. These are attracted to magnetic fields.
Among all the given options, Cu++ and Ni++ are paramagnetic.
Among the following compounds which is planar in shape is
All the carbon atoms in benzene are in sp2 state of hybridization, hence, all the six carbons and six hydrogen atoms attached with these carbon lie in a plane.
The pH of blood is :
Human blood pH is 7.4. Therefore, option c is correct.
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To determine: The radical form of \sqrt[3]{64} in the rational
To determine: The radical form of \sqrt[3]{64} in the rational exponent form
To determine: The radical form of
\sqrt[3]{64}
in the rational exponent form.
\sqrt[3]{64}
Rewrite 64 as
{4}^{3}
\sqrt[3]{{4}^{3}}
How do you find the points of intersection of the curves with polar equations
r=6\mathrm{cos}\theta \text{and}\text{ }r=2+2\mathrm{cos}\theta
C:9{x}^{2}+4xy+6{y}^{2}-10=0.
I've found C is a non-degenerate ellipses (computing the cubic and the quadratic invariant), and then I've studied the characteristic polynomial
p\left(t\right)=\mathrm{det}\left(\begin{array}{cc}9-t& 2\\ 2& 6-t\\ & \phantom{\rule{0ex}{0ex}}\end{array}\right)
The eigenvalue are
{t}_{1}=5,{t}_{2}=10
, with associated eigenvectors
\left(-1,2\right),\left(2,1\right)
. Thus I construct the rotation matrix R by putting in columns the normalized eigenvectors (taking care that
det\left(R\right)=1
R=\frac{1}{\sqrt{5}}\left(\begin{array}{cc}1& 2\\ -2& 1\\ & \phantom{\rule{0ex}{0ex}}\end{array}\right)
{\left(x,y\right)}^{t}=R{\left({x}^{\prime },{y}^{\prime }\right)}^{t}
, and after some computations I find the canonical form
\frac{1}{2}{x}^{\prime 2}+\frac{4}{5}{y}^{\prime 2}=1.
Shape induced by the condition of angle bisector passing through a fixed point
Let A, B be two points on 2D plane. For any
C\in \stackrel{―}{AB}
, define the set
S=\left\{P;\mid ;\mathrm{\angle }APC=\mathrm{\angle }CPB\right\}.
What is the shape of S?
To find: The equation of the given hyperbola. The foci of the hyperbola is
\left(0,\text{ }±\text{ }5\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ entricity is }e-1.5
\sqrt[4]{9}\sqrt{3}
Can anyone help with paramaterization of conics?
Im struggling to wrap my head around an example. It considers the conic
{x}^{2}+{y}^{2}-{z}^{2}=0
then proceeds:
A=\left[1,0,1\right]
and the line P(U) defined by
x=0
. Note that this conic and the point and line are defined over any field since the coefficients are 0 or 1. A point
X\in P\left(U\right)
X=\left[0,1,t\right]
or [0, 0, 1] and the map
\alpha
\begin{array}{rl}\alpha \left(\left[0,1,t\right]\right)& =\left[B\left(\left(0,1,t\right),\left(0,1,t\right)\right)\left(1,0,1\right)-2B\left(\left(1,0,1\right),\left(0,1,t\right)\right)\left(0,1,t\right)\right]\\ & =\left[1-{t}^{2},2t,1+{t}^{2}\right]\\ \text{ or }\alpha \left(\left[0,0,1\right]\right)=\left[-1,0,1\right].\end{array}
How do I evaluate B(v,v) or B(v,v)(a,b,c) like they have to go from the first line to the second?
\sqrt[3]{{x}^{12}}
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Ask Answer - Statistics - Recently Asked Questions for School Students
Pls tell 22. 23 25
{f}_{1} and {f}_{2}
{f}_{1}
{f}_{2}
\overline{x} = a + \frac{\sum _{}^{}fixi}{\sum _{}fi}
\overline{x} = a + \frac{\sum _{}^{}fixi}{\sum _{}fi}
\begin{array}{ccccccc}X& 3& 5& 7& 9& 11& 13\\ F& 6& 8& 15& P& 8& 4\end{array}
I want to know that how the formula for finding the median of grouped data is derived?
K.akshitha
Represent the statistics available in case of genders on various issues like:-
# infant mortality
PLZZ..answer it fast have to submit tomorrow
\begin{array}{cc}Salary \left(In thousand\right)& No. of persons\\ 5-10& 49\\ 10-15& 133\\ 15-20& 63\\ 20-25& 15\\ 25-30& 6\end{array}\phantom{\rule{0ex}{0ex}} \begin{array}{cc}30-35 & 7\\ 35-40 & 4\\ 40-45 & 2\\ 45-50 & 1\end{array}
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Calculus/Continuity - Wikibooks, open books for an open world
Calculus/Continuity
← Infinite Limits Calculus Formal Definition of the Limit →
1 Defining Continuity
2.2 Jump discontinuities
3 One-Sided Continuity
4 Intermediate Value Theorem
4.1 Application: bisection method
Defining ContinuityEdit
We are now ready to define the concept of a function being continuous. The idea is that we want to say that a function is continuous if you can draw its graph without taking your pencil off the page. But sometimes this will be true for some parts of a graph but not for others. Therefore, we want to start by defining what it means for a function to be continuous at one point. The definition is simple, now that we have the concept of limits:
Definition: (continuity at a point)
{\displaystyle f(x)}
is defined on an open interval containing
{\displaystyle c}
{\displaystyle f(x)}
{\displaystyle c}
{\displaystyle \lim _{x\to c}f(x)=f(c)}
{\displaystyle f}
{\displaystyle c}
, the definition in effect requires three conditions:
{\displaystyle f}
{\displaystyle c}
{\displaystyle f(c)}
{\displaystyle x}
{\displaystyle c}
the limit and
{\displaystyle f(c)}
If any of these do not hold then
{\displaystyle f}
{\displaystyle c}
The idea of the definition is that the point of the graph corresponding to
{\displaystyle c}
will be close to the points of the graph corresponding to nearby
{\displaystyle x}
-values. Now we can define what it means for a function to be continuous in general, not just at one point.
Definition: (continuity)
A function is said to be continuous on
{\displaystyle (a,b)}
if it is continuous at every point of the interval
{\displaystyle (a,b)}
We often use the phrase "the function is continuous" to mean that the function is continuous at every real number. This would be the same as saying the function was continuous on
{\displaystyle (-\infty ,\infty )}
, but it is a bit more convenient to simply say "continuous".
Note that, by what we already know, the limit of a rational, exponential, trigonometric or logarithmic function at a point is just its value at that point, so long as it's defined there. So, all such functions are continuous wherever they're defined. (Of course, they can't be continuous where they're not defined!)
DiscontinuitiesEdit
A discontinuity is a point where a function is not continuous. There are lots of possible ways this could happen, of course. Here we'll just discuss two simple ways.
Removable discontinuitiesEdit
{\displaystyle f(x)={\frac {x^{2}-9}{x-3}}}
{\displaystyle x=3}
. It is discontinuous at that point because the fraction then becomes
{\displaystyle {\frac {0}{0}}}
, which is undefined. Therefore the function fails the first of our three conditions for continuity at the point 3; 3 is just not in its domain.
However, we say that this discontinuity is removable. This is because, if we modify the function at that point, we can eliminate the discontinuity and make the function continuous. To see how to make the function
{\displaystyle f(x)}
continuous, we have to simplify
{\displaystyle f(x)}
, getting
{\displaystyle f(x)={\frac {x^{2}-9}{x-3}}={\frac {(x+3)(x-3)}{(x-3)}}={\frac {x+3}{1}}\cdot {\frac {x-3}{x-3}}}
. We can define a new function
{\displaystyle g(x)}
{\displaystyle g(x)=x+3}
{\displaystyle g(x)}
is not the same as the original function
{\displaystyle f(x)}
{\displaystyle g(x)}
{\displaystyle x=3}
{\displaystyle f(x)}
is not. Thus,
{\displaystyle g(x)}
{\displaystyle x=3}
{\displaystyle \lim _{x\to 3}(x+3)=6=g(3)}
. However, whenever
{\displaystyle x\neq 3}
{\displaystyle f(x)=g(x)}
; all we did to
{\displaystyle f}
{\displaystyle g}
was to make it defined at
{\displaystyle x=3}
In fact, this kind of simplification is often possible with a discontinuity in a rational function. We can divide the numerator and the denominator by a common factor (in our example
{\displaystyle x-3}
) to get a function which is the same except where that common factor was 0 (in our example at
{\displaystyle x=3}
). This new function will be identical to the old except for being defined at new points where previously we had division by 0.
However, this is not possible in every case. For example, the function
{\displaystyle f(x)={\frac {x-3}{x^{2}-6x+9}}}
has a common factor of
{\displaystyle x-3}
in both the numerator and denominator, but when you simplify you are left with
{\displaystyle g(x)={\frac {1}{x-3}}}
, which is still not defined at
{\displaystyle x=3}
. In this case the domain of
{\displaystyle f(x)}
{\displaystyle g(x)}
are the same, and they are equal everywhere they are defined, so they are in fact the same function. The reason that
{\displaystyle g(x)}
differed from
{\displaystyle f(x)}
in the first example was because we could take it to have a larger domain and not simply that the formulas defining
{\displaystyle f(x)}
{\displaystyle g(x)}
Jump discontinuitiesEdit
Not all discontinuities can be removed from a function. Consider this function:
{\displaystyle k(x)={\begin{cases}1&{\text{if }}x>0\\-1&{\text{if }}x\leq 0\end{cases}}}
{\displaystyle \lim _{x\to 0}k(x)}
does not exist, there is no way to redefine
{\displaystyle k}
at one point so that it will be continuous at 0. These sorts of discontinuities are called nonremovable discontinuities.
Note, however, that both one-sided limits exist;
{\displaystyle \lim _{x\to 0^{-}}k(x)=-1}
{\displaystyle \lim _{x\to 0^{+}}k(x)=1}
. The problem is that they are not equal, so the graph "jumps" from one side of 0 to the other. In such a case, we say the function has a jump discontinuity. (Note that a jump discontinuity is a kind of nonremovable discontinuity.)
One-Sided ContinuityEdit
Just as a function can have a one-sided limit, a function can be continuous from a particular side. For a function to be continuous at a point from a given side, we need the following three conditions:
A function will be continuous at a point if and only if it is continuous from both sides at that point. Now we can define what it means for a function to be continuous on a closed interval.
Definition: (continuity on a closed interval)
{\displaystyle [a,b]}
it is continuous on
{\displaystyle (a,b)}
it is continuous from the right at
{\displaystyle a}
it is continuous from the left at
{\displaystyle b}
Notice that, if a function is continuous, then it is continuous on every closed interval contained in its domain.
Intermediate Value TheoremEdit
A useful theorem regarding continuous functions is the following:
{\displaystyle f}
{\displaystyle [a,b]}
, then for every value
{\displaystyle y}
{\displaystyle f(a)}
{\displaystyle f(b)}
there is a value
{\displaystyle c\in (a,b)}
{\displaystyle f(c)=y}
Application: bisection methodEdit
A few steps of the bisection method applied over the starting range
{\displaystyle [a_{1},b_{1}]}
. The bigger red dot is the root of the function.
The bisection method is the simplest and most reliable algorithm to find zeros of a continuous function.
Suppose we want to solve the equation
{\displaystyle f(x)=0}
. Given two points
{\displaystyle a}nd
{\displaystyle b}
{\displaystyle f(a)}
{\displaystyle f(b)}
have opposite signs, the intermediate value theorem tells us that
{\displaystyle f}
must have at least one root between
{\displaystyle a}nd
{\displaystyle b}
{\displaystyle f}
{\displaystyle [a,b]}
{\displaystyle f}
is continuous in general (say, because it's made out of rational, trigonometric, exponential and logarithmic functions), then this will work so long as
{\displaystyle f}
is defined at all points between
{\displaystyle a}nd
{\displaystyle b}
. So, let's divide the interval
{\displaystyle [a,b]}
in two by computing
{\displaystyle c={\frac {a+b}{2}}}
. There are now three possibilities:
{\displaystyle f(c)=0}
{\displaystyle f(a)}
{\displaystyle f(c)}
have opposite signs, or
{\displaystyle f(c)}
{\displaystyle f(b)}
have opposite signs.
In the first case, we're done. In the second and third cases, we can repeat the process on the sub-interval where the sign change occurs. In this way we hone in to a small sub-interval containing the 0. The midpoint of that small sub-interval is usually taken as a good approximation to the 0.
Note that, unlike the methods you may have learned in algebra, this works for any continuous function that you (or your calculator) know how to compute.
Retrieved from "https://en.wikibooks.org/w/index.php?title=Calculus/Continuity&oldid=3659674"
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The given equation (\tan \theta - 2) (16 \sin^{2} \theta
The given equation (\tan \theta - 2) (16 \sin^{2} \theta - 1) = 0
The given equation
\left(\mathrm{tan}\theta -2\right)\left(16{\mathrm{sin}}^{2}\theta -1\right)=0
l1koV
The domain of the trigonometry function of
\mathrm{tan}\theta
\left[-\mathrm{\infty },\mathrm{\infty }\right]
. Tangent has period
\pi
, we findd solution in any interval of length
\pi
The domain of the trigonometric function
\mathrm{sin}\theta
\left[-1,1\right]
. No solution exists beyond this domain. Sine has period
2\pi
, we find solution in any interval of length
2\pi
. Sine function is positive in first and second quadrant.
The trigonometric equation is given by,
\left(\mathrm{tan}\theta -2\right)\left(16{\mathrm{sin}}^{2}\theta -1\right)=0
The factors of above equation are,
\left(\mathrm{tan}\theta -2\right)=0\dots \left(1\right)
16{\mathrm{sin}}^{2}\theta -1=0\dots \left(2\right)
Add 2 both sides in equation (1).
\mathrm{tan}\theta =2
\mathrm{tan}\theta ={\mathrm{tan}}^{-1}\left(2\right)
\theta =1.10714
Here the angles are in radian.
The tangent has period,
\pi
, so we get all solutions of the equation by adding integer multiples of
\pi
to these solutions:
{\theta }_{1}=1.10714+k\pi
The solution obtained for the factor in which sine function involved so we will get the solution in the interval of
\left[0,2\pi \right]
Add 1 to both sides in equation (2).
16{\mathrm{sin}}^{2}\theta =1
\mathrm{sin}\theta =±\frac{1}{4}
\theta ={\mathrm{sin}}^{-1}\left(±\frac{1}{4}\right)
\theta ={\mathrm{sin}}^{-1}\left(\frac{1}{4}\right)
=0.252,2.889
\theta ={\mathrm{sin}}^{-1}\left(\frac{1}{4}\right)
=-0.252,3.393
The sine has period,
2\pi
, but it is a square function so repeating is done every length of
\pi
. So we get all solutions of the equation by adding integer multiples of
\pi
{\theta }_{2}=0.252+2k\pi
{\theta }_{3}=2.889+2k\pi
{\theta }_{4}=-0.252+2k\pi
{\theta }_{5}=3.393+2k\pi
Therefore, the solutions of the trigonometric equation
\left(\mathrm{tan}\theta -2\right)\left(16{\mathrm{sin}}^{2}\theta -1\right)=0
{\theta }_{1}=1.10714+k\pi ,{\theta }_{2}=0.252+2k\pi ,{\theta }_{3}=2.889+2k\pi ,{\theta }_{4}=-0.252+2k\pi
{\theta }_{5}=3.393+2k\pi
h\left(t\right)=-16{t}^{2}+24t
\left(3{n}^{3}\right)\left(5-6n+3{n}^{2}+{n}^{3}\right)=
add or subtract as indicated, and express answers in lowest terms.
\frac{1}{6}+\frac{1}{8}
The sum of four consecutive multiples of 6 is 84. Find the four multiples?
Fifty raffle tickets are numbered 1 through 50, and one of them is drawn at random. What is the probability that the number is a multiple of 5 or 7? Consider the following "solution": Since 10 tickets bear numbers that are multiples of 5 and 7 tickets bear numbers that are multiples of 7, we concluded the required probability is as follows.
\frac{10}{50}+\frac{7}{50}=\frac{17}{50}
Is this the correct answer? (If so, enter yes. If not, enter the correct answer.)
P\left(x\right)=3{x}^{5}-{x}^{4}+81{x}^{3}-27{x}^{2}-972x+324
, and that 6i is a zero, write P in factored form (as a product of linear factors). Be sure to write the full equation, including
P\left(x\right)=
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Find the correlatin coefficient r using bivariate data.
The given bivariate data is(1,6),(3,2), and (2,4).
\sum {x}_{i}{y}_{i},\sum {x}_{i}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\sum {y}_{i}
{x}_{i}\to
{y}_{i}\to
\sum {x}_{i}{y}_{i}=1\cdot 6+3\cdot 2+2\cdot 4=20
\sum {x}_{i}=1+3+2=6
\sum {x}_{i}^{2}={1}^{2}+{3}^{2}+{2}^{2}=14
\sum {y}_{i}=6+2+4=12
\sum {y}_{i}^{2}={6}^{2}+{2}^{2}+{4}^{2}=56
Find the sample variance
{s}^{2}
{s}^{2}=\frac{\sum {x}_{i}^{2}-\frac{\sum {x}_{i}^{2}}{n}}{n-1}
{s}_{x}^{2}=\frac{14-\frac{{6}^{2}}{3}}{3-1}=\frac{14-12}{2}=1
{s}_{y}^{2}=\frac{56-\frac{{12}^{2}}{3}}{3-1}=\frac{56-48}{2}=4
Find sample standard deviation.
{s}_{x}=\sqrt{1}=1
{s}_{y}=\sqrt{4}=2
{s}_{xy}using\text{ }f\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\mu \text{ }\text{ }{s}_{xy}=\frac{\sum {x}_{i}{y}_{i}-\frac{\sum {x}_{i}-\sum {y}_{i}}{n}}{n-1}
{s}_{xy}=\frac{20-\frac{6\cdot 12}{3}}{3-1}
=\frac{20-\frac{72}{3}}{3-1}
=\frac{20-24}{2}
=-\frac{4}{2}=-2
Find the correlation coefficient r using the formula
r=\frac{{s}_{xy}}{{s}_{x}{s}_{y}}
r=-\frac{2}{1\cdot 2}=-1
Hence the value of correlation coefficient r = -1.
\text{Diet}\text{ }\text{Lost weight?}
\begin{array}{lcccc}& \mathrm{A}& \mathrm{B}& \mathrm{C}& \text{ Total }\\ \text{ Yes }& & 60& & 180\\ \text{ No }& & 40& & 120\\ \text{ Total }& 90& 100& 110& 300\end{array}
To find:The qualitative or quantitative and as univariate or bivariate.
When working with bivariate data, which of these are useful when deciding whether it’s appropriate to use a linear model?
I. the scatterplot
II. the residuals plot
III. the correlation coefficient
n\in N
Understanding the Concepts and Skills
Suppose that, for a sample of pairs of observations from two variables, the linear correlation coefficient, r , is negative. Does this result necessarily imply that the variables are negatively linearly correlated? Explain.
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Synthesis, Antimicrobial, and Anticoagulant Activities of 2-(Arylsulfonyl)indane-1,3-diones
Malaichamy Jeyachandran, Penugonda Ramesh, "Synthesis, Antimicrobial, and Anticoagulant Activities of 2-(Arylsulfonyl)indane-1,3-diones", Organic Chemistry International, vol. 2011, Article ID 360810, 5 pages, 2011. https://doi.org/10.1155/2011/360810
Malaichamy Jeyachandran1 and Penugonda Ramesh2
1Post Graduate and Research Department of Chemistry, Sri Paramakalyani College, Alwarkurichi, Tirunelveli 627 412, India
2Department of Natural Products Chemistry, School of Chemistry, Madurai Kamaraj University, Madurai 625 021, India
2-(Arylsulfonyl)indane-1,3-diones, earlier synthesized by Claisen condensation involving diethyl phthalate and aryl methyl sulphones, are found to be potent blood anticoagulants. In search of improved analogs of 2-(arylsulfonyl)indane-1,3-dione, we have synthesized them (7 a-f) by a different route involving Knoevenagel reaction between phthalic anhydride and arylsulfonylacetates in the presence of pyridine-piperidine medium for the first time. The synthesized 2-(arylsulfonyl)indane-1,3-diones were evaluated for antimicrobial and anticoagulant activities and all of them registered significant activity.
Indane-1,3-dione and its derivatives constitute a unique group of compounds and attracted the attention of organic chemists and biologist due to their three characteristic features [1, 2]. (a) Indane-1,3-diones have enormous synthetic potential due to the presence of β-dicarbonyl moiety which often serves as a synthon for the preparation of more structurally complex compounds via condensation, decomposition, reduction, cyclization, rearrangements, and so forth. (b) The molecule offers wide scope for the study of physiochemical properties such as 1,3-indandione tautomerism, dual reactivity, electrochemical redox properties, unique property of polycrystalline films, and quantum mechanical calculations, and so forth [3, 4]. (c) A wide range of biological properties covering antimicrobial, antitumor, anti-inflammatory, antiviral, antihepatitis, anticoagulant, rodenticidal, herbicidal, and insecticidal properties were associated with indane-1,3-dione derivatives [5–10].
A survey of literature revealed that derivatives of indane-1,3-dione particularly 2-arylsulfonylindane-1,3-diones have been found to be potent blood anticoagulants [11]. Earlier, these were prepared by the Claisen condensation involving diethyl phthalate and aryl methyl sulphones [12, 13] by a procedure analogous to the preparation of indane-1,3-dione from diethylphthalate and ethyl acetoacetate [14]. In search of improved oral anticoagulants, we have synthesized 2-(arylsulfonyl)indane-1,3-diones 7 a-f by a different route involving Knoevenagel reaction between phthalic anhydride and arylsulfonylacetates. The reaction is based on the observation that phthalic anhydride can function as a carbonyl compound in Perkin-type reactions [15].
In an attempt to prepare improved analogs of 2-(arylsulfonyl)indane-1,3-diones, we have attempted the condensation of arylsulfonylacetic acids with phthalic anhydride under different conditions. Although, the reaction failed with arylsulfonylacetic acids, the corresponding arylsulfonylacetates readily reacted with phthalic anhydride in the presence of pyridine-piperidine medium gave phthalyl arylsulfonylacetates 3, which on further reaction with sodium ethoxide in dry ethanol and subsequent heating with 1 : 1 HCl afforded 2-(arylsulfonyl)indane-1,3-diones 7 a-f (Scheme 1).
The structures of the synthesized compounds 7 a-f were established by IR and NMR spectroscopic data and elemental analysis. Compounds 7 a-f were homogenous on TLC and contained sulfur. The presence of sulfone moiety in 7 a-f was established by IR bands [15] in the region, 1300–1360 cm−1 (unsymmetrical S=O str.) and 1120–1140 cm−1 (symmetrical S=O str.), and the presence of intact indane1,3-dione moiety was also confirmed by the IR bands in the region 1640–1670 cm−1.
The 1H NMR spectra of 7 a-f bear close resemblance and displayed, in addition to complex multiplets due to aromatic protons in the region δ 7.0–7.5, a one proton singlet in the region δ 4.0–5.5 assignable to the lone methine (H-2) of indane-1,3-dione moiety; the exact chemical shift of H-2 is influenced by the nature of the substituents of the arylsulfonyl group and by the degree of enolization of the indane-1,3-dione moiety.
3. Pharmacological Screening
The antimicrobial activity of compounds was determined by the disc diffusion method and minimum inhibitory concentrations (MICs). Minimum inhibitory concentration is the lowest concentration of an antimicrobial agent that inhibits more than 99% of the bacterial population. MICs were determined by the macrodilution broth method following the procedures recommended by the National Committee for Clinical Laboratory Standards for testing purposes [16, 17]. Compounds 7 a-f were evaluated in vitro activity against S. aureus and E. coli at a concentration of 10 μg/mL in meat peptone agar medium. Ciprofloxacin was used as a standard for antimicrobial screening. For each biological activity test, two to three experiments were performed and the average zone of inhibition is shown in Table 1.
Compound Antibacterial activity
(zone of inhibition in mm) at 10 μg/mL
Control (DMSO) 0 0
Ciprofloxacin 19 8
Antibacterial activity of compounds 7 a-f.
Compounds 7c, 7d, 7e, and 7f exhibited significant activity against S. aureus while 7a, 7c, 7d, and 7f showed significant activity against E. coli. Compounds 7a and 7b showed low activity against S. aureus, and 7b and 7e registered low activity against E. coli. Compounds 7c and 7d showed significant activity against S. aureus and both E. coli. The MICs of these two compounds are summarized in Table 2.
Compounds S. aureus E. coli
Minimum inhibitory concentrations (MICs in μg/mL).
Anticoagulant activity of compounds 7 a-f was assessed against sodium citrate (standard) in albino rats [18, 19]. The blood samples were collected from albino rats by puncturing carotid artery. Compounds 7 a-f and sodium citrate were added to the blood separately (10 mg/dL). The time required for the formation of the clot was measured, and the results of anticoagulant activity of 7 a-f are shown in Table 3.
Compound Clotting time (min)
Sodium citrate (standard) 80
Anticoagulant activity of compounds 7 a-f.
Compounds 7b, 7d, and 7f registered high anticoagulant activity compared to that of standard, whereas compounds 7a, 7c, and 7e showed low activity.
A scrutiny of the results of antibacterial activity of 7 a-f reveals (Tables 1 and 2) emphasized that bulky nuclear substituents at 4-position of benzenesulfonyl moiety were found to decrease the antibacterial activity, especially against Gram-positive bacteria (S. aureus). Thus there appears to be an empirical inverse relationship between the activity and the size of the nuclear substituents. The observed decreasing order of antibacterial activity of 7 a-f against Gram-positive bacteria, 7c (R = OMe, ) > 7e (R = Cl, ) = 7d (R = Br, ) > 7b (R = Me, )≈7f (R = NO2, ), reflected an empirical inverse correlation between the acidity and Taft’s steric factor ( ) [20] which is a measure of steric bulk of the substituents, and bulky substituents have larger negative values for .
An analysis of the results of anticoagulant activity of 7 a-f (Table 3) revealed that an opposite trend was observed in this case, that is, there is a direct correlation between the anticoagulant activity and the size of the nuclear substituents. Thus, the observed increasing order of anticoagulant activity, 7f (R = NO2, ) = 7d (R = Br, ) > 7b (R = Me, ) > 7e (R = Cl, )≈7c (R = OMe, ), reflected a direct correlation between anticoagulant activity and Taft’s steric factor ( ).
These contrasting results emphasize that bulky groups lower the activity of the compound by preventing it from fitting properly into the binding site of the receptor. On the other hand, bulky substituents may also increase the activity by forcing a compound to adopt the required active conformation at the binding site.
Melting points were determined on a sulphuric acid bath and are uncorrected. IR spectra were recorded on JASCO 470 FT-IR spectrometer, and 1H-NMR spectra were recorded on a 300 MHz on Bruker (Avance) NMR spectrometer using TMS as an internal standard.
5.1. General Procedure for the Synthesis of Ethyl Arylsulfonylacetate (2)
To an alkaline solution of thiophenol (0.10 mol in 15% NaOH), ethyl chloroacetate (0.10 mol) was added while stirring and keeping the temperature at 0°C for about 0.5 h. The resulting oily liquid was extracted with chloroform, and the chloroform layer was washed with water and dried. Evaporation of solvent gave ethyl arylmercaptoacetate, which on oxidation with m-CPBA (in CHCl3) gave ethyl arylsulfonylacetate 2.
5.2. General Procedure for the Synthesis of 2-(Arylsulfonyl)Indane-1,3-Diones (7 a-f)
A mixture of ethyl arylsulfonylacetate 2 (0.10 mol), dry pyridine (15.0 mL), piperidine (2.0 mL), and powdered phthalic anhydride 1 (0.10 mol) was refluxed for 3 h. The reaction mixture was poured into ice water and neutralized with dil. HCl. The solid 3 that separated was filtered, washed with water, and dried. It was then refluxed with sodium ethoxide (0.20 mol) in absolute ethanol for 1 h, and the excess of solvent was removed in vacuo, and the residue was dissolved in hot water and neutralized while hot with 1 : 1 HCl. The product 7 a-f that separated on cooling was filtered, dried, and crystallized.
2-(Phenylsulfonyl)-1H-indene-1,3(2H)-dione (7a)
M.p. 185°C (benzene), yield: 40%. IR (KBr): 1664 (CO), 1336 (S=O unsym. str.), 1141 (S=O sym. str.) cm−1. 1H NMR (CDCl3): δ 4.60 (s, 1H, CH-SO2-); 7.24–7.99 (m, 9H, Ar-H). 13C NMR (75 MHz, CDCl3) δC (ppm): 29.79, 122.04, 127.19, 127.49, 129.12, 132.10, 137.10, 151.04, 179.09. Anal. Cald. for C15H10O4S: C, 62.93; H, 3.52. Found: C, 62.96; H, 3.54.
2-(4-Methylphenylsulfonyl)-1H-indene-1,3(2H)-dione (7b)
M.p. 160°C (benzene), yield 45%. IR (KBr): 1672 (CO), 1338 (S=O unsym. str.), 1138 (S=O sym. str.) cm−1. 1H NMR (CDCl3): δ 2.57 (s, 3H, -CH3), 3.15 (s, 1H, CH-SO2-), 7.19–7.52 (m, 8H, Ar-H). Anal. Calcd. for C16H12O4S: C, 63.99; H, 4.03. Found: 63.97; H, 4.00.
2-(4-Methoxyphenylsulfonyl)-1H-indene-1,3(2H)-dione (7c)
M.p. 153°C (pet. ether-benzene), yield 43%; IR (KBr): 1668 (CO), 1335 (S=O unsym. str.), 1129 (S=O sym. str.) cm−1. 1H NMR (CDCl3): δ 3.82 (s, 3H, -OCH3), 5.86 (s, 1H, CH-SO2-), 6.75–7.49 (m, 8H, Ar-H). Anal. Calcd. for C16H12O5S: C, 60.75; H, 3.82. Found: C, 60.77; H, 3.84.
2-(4-Bromophenyl-sulfonyl)-1H-indene-1,3(2H)-dione (7d)
M.p. 145°C (pet. ether-chloroform), yield 40%. IR (KBr): 1656 (CO), 1346 (S=O unsym. str.), 1136 (S=O sym. str.) cm−1. 1H NMR (CDCl3): δ 4.03 (s, 1H, CH-SO2-), 7.23–7.55 (m, 8H, Ar-H). Anal. Calcd. for C15H9BrO4S: C, 49.33; H, 2.48. Found: C, 49.37; H, 2.51.
2-(4-Chlorophenylsulfonyl)-1H-indene-1,3(2H)-dione (7e)
M.p. 131°C (pet. ether-chloroform), yield 42%. IR (KBr): 1666 (CO), 1333 (S=O unsym. str.), 1125 (S=O sym. str.) cm−1; 1H NMR (CDCl3): δ 5.03 (s, 1H, CH-SO2-), 7.25–7.58 (m, 8H, Ar-H). Anal. Calcd. for C15H9ClO4S: C, 56.17; H, 2.83. Found: C, 56.14; H, 2.81.
2-(4-Nitrophenylsulfonyl)-1H-indene-1,3(2H)-dione (7f)
M.p. 142°C (aq. ethanol), yield 41%. IR (KBr): 1662 (CO), 1328 (S=O unsym. str.), 1119 (S=O sym. str.) cm−1. 1H NMR (CDCl3): δ 4.91 (s, 1H, CH-SO2-), 6.83–7.66 (m, 8H, Ar-H). Anal. Calcd. for C15H9NO6S: C, 54.38; H, 2.74; N, 4.23. Found: C, 54.41; H, 2.77; N, 4.19.
The authors thank School of Chemistry for providing FT-NMR (under DST-IRHPA programme) data, and one of the authors (M. J.) is grateful to Madurai Kamaraj University for USR fellowship. The authors also thank K. M. College of Pharmacy, Madurai for carrying out biological studies.
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In this problem, allow T1: RR2rightarrowRR2 and T2: RR2rightarrowR2 be linear tr
In this problem, allow T1: RR2rightarrowRR2 and T2: RR2rightarrowR2 be linear transformations. Find Ker(T_1), Ker(T_2), Ker(T_3) of the respective matricesA=[(1,-1),(-2,0)],B=[(1,5),(-2,0)]
T1:RR2\to RR2\text{ and }T2:RR2\to R2
Ker\left({T}_{1}\right),Ker\left({T}_{2}\right),Ker\left({T}_{3}\right)
A=\left[\left(1,-1\right),\left(-2,0\right)\right],B=\left[\left(1,5\right),\left(-2,0\right)\right]
Consider the scatterplot of y versus x in Output B.6 above.
a. Which power transformations on y should be considered to straighten the scatterplot?
b. Which power transformations on x should be considered to straighten the scatterplot?
f\left(x\right)={x}^{2}
, after performing the following transformations: shift upward 52 units and shift 23 units to the right, write the new function
Your friend attempted to describe the transformations applied to the graph of
y=\mathrm{sin}x
to give the equation
f\left(x\right)=\frac{1}{2}\mathrm{sin}\left(-\frac{1}{3}\left(x+30\right)\right)+1
They think the following transformations have been applied. Which transformations have been identified correctly, and which have not? Justify your answer.
a) f(x) has been reflected vertically.
b) f(x) has been stretched vertically by a factor of 2.
c) f(x) has been stretched horizontally by a factor of 3.
d) f(x) has a phase shift left 30 degrees.
e) f(x) has been translated up 1 unit.
\mathrm{\angle }DBE
\left(0.1x-22{\right)}^{\circ }
\mathrm{\angle }CBE
\left(0.3x-54{\right)}^{\circ }
Explain how you could graph each function by applying transformations.
y=\mathrm{log}\left(x-2\right)+7
y=-3\mathrm{log}x
y=\mathrm{log}\left(-3x\right)-5
f\left(x\right)=x2
, after performing the following transformations: shift upward 58 units and shift 75 units to the right, the new function
g\left(x\right)=
Which of the following is the correct equation used to solve for the measure of each angle? OA.
m\mathrm{\angle }A-m\mathrm{\angle }B-m\mathrm{\angle }C={180}^{\circ }
O B.
m\mathrm{\angle }A+m\mathrm{\angle }B-m\mathrm{\angle }C={180}^{\circ }
O c.
m\mathrm{\angle }A-m\mathrm{\angle }B+m\mathrm{\angle }C={180}^{\circ }
O D.
m\mathrm{\angle }A+m\mathrm{\angle }B+m\mathrm{\angle }C={180}^{\circ }
The measure of analeAis Click to select your answer(s). Save for Later 0.
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A-level Biology/Central Concepts/Energy and respiration - Wikibooks, open books for an open world
A-level Biology/Central Concepts/Energy and respiration
< A-level Biology | Central Concepts
1 Energy and Work
2 ATP (Adenosine Triphosphate)
2.1 Energy Currency
3.2 Link reaction
3.4.1 Chemiosmosis
5 Respiratory Substrates
Energy and organic molecules are required by every living organism, and are used in two ways;
Building blocks for making other organic molecules that the organism needs.
Chemical potential energy which can be released by breaking down the molecules in respiration.
This chemical potential energy is used in the following ways;
Muscle contraction and other cellular movements
Active transport of substances
Synthesis of complex substances (anabolic reactions) from simple ones.
Energy yielding reactions in all organisms generally produce an intermediary energy molecule, ATP.
ATP consists of adenine (organic base) and ribose (pentose sugar), making adenosine (a nucleotide), which is then combined with three phosphate groups to make ATP, a nucleotide. It is a small water-soluble molecule, allowing easy transport. When a phosphate group is removed, [known as Hydrolysis']' ADP is formed and it produces 30.6kJ of energy, and is a reversible reaction that allows for interconversion between ATP and ADP - the formula is below: known as phosphorylation.
ATP + H2O ↔ ADP +H3PO4 ΔG = ± 30.6kJ
The cells energy yielding reactions synthesise ATP, and ATP is used by the cell in all forms of work. and the cell trades in ATP rather than using intermediates. Energy lost during energy transfers is converted to thermal energy, as well as any excess energy, if too much is produced to create ATP for example. Energy currency is immediate donor of energy to cell's energy-requiring reaction, and a storage molecule is a short-term (glucose) or long term (glycogen) store of chemical energy. ATP losses a phosphate group and forms ADP
Most ATP is synthesised using electrical potential energy, the energy from the transfers of electrons by electron carriers in mitochondria and chloroplasts. It's stored as a difference in hydrogen ion concentration across the membrane (the membranes of mitochondria and chloroplasts are basically impermeable to hydrogen ions). Hydrogen ions are then allowed to flow down their concentration gradient through a transport protein, and part of this protein is ATP synthase, an enzyme that, surprisingly, synthesises ATP. The transfer of three hydrogen ions results in the production of one ATP molecule provided that ADP and a phosphate group is present inside the organelle.
ATP is required for various biological processes in animals including; Active Transport, Secretion, Endocytosis, Synthesis and Replication of DNA and Movement.
Active Transport is the movement of molecules defined as the energy-consuming transport of molecules or ions across a membrane against a concentration gradient, made possible by transferring energy from ATP. Most cells have sodium/potassium pumps and these are maintained by ATP, for many many things.
Muscle contraction goes as such;
A sarcomere contracts by sliding the thin actin filaments over the thick myosin filaments (molecules with a flexible head, an ATPase molecule, hydrolyses ATP to ADP and phosphate)
1. Calcium ions are released from the sarcoplasmic reticulum, allowing the myosin head to bind to the portion of actin filament next to it.
2. This head then tilts at 45 degrees, moving the actin filament about 10nm in relation to the myosin towards the centre of the sarcomere
3. Millions of fibres doing this at the same time makes the muscle contract, releasing ADP and phosphate
4. Another ATP binds to the head and is again hydrolysed, and the head tilts back to it's original position
5. When contraction ends, ATP is used to pump calcium ions back into the sarcoplasmic reticulum.
During contraction, ATP is continually regenerated using creatine phosphate, which provides a phosphate group to ADP, allowing it to become ATP and creatine. The limited supply of creatine phosphate must be replinished via ATP from respiration. The muscle may not be able to keep up with it and a lactate pathway is used to allow continued formation of ATP, but the cells incur an oxygen debt.
This is the splitting of glucose, that eventually results in two molecules of pyruvate. ATP is used in the first stage, where glucose is phosphorylated using ATP, which allows the reaction to be easier. Glucose breaks down to hexose phosphate, which breaks down to hexose biphosphate, which breaks down to 2 molecules of triose phosphate. Hydrogen is then removed and transferred to NAD, producing two molecules of reduced NAD for each glucose molecule. The process produces 4 ATP molecules in total, in addition to the two molecules of reduced NAD, but requires 2 ATP molecules.
In summary. Glucose > Glucose-6-phosphate > Fructose-1-phosphate > Hexose-1,6-bi(s)phosphate > 2x Triose Phosphate > 2x Intermediate Compound > 2x Pyruvate.
Pyruvate is then actively transported to the mitrochrondrial matrix for the link reaction. Thus ultimately producing;
2 Molecules of Reduced NAD [NADH]
Pyruvate and NAD enter the link cycle which takes place in the mitochondrial matrix, where it is decarboxylated, dehydrogenated and combined with Coenzyme A to give acetyl coenzyme A. CO2 is produced and NAD is reduced. Acetyl coenzyme A acts as a carrier of acetyl groups to the krebs cycle. The hydrogen removed from pyruvate is now transferred to NAD.
Thus ultimately; [as 2 molecules of Pyruvate]
2NADH produced
2 Co2 produced
The krebs cycle(also known as Citric acid cycle or tricarboxylic acid cycle)is a closed pathway of enzyme-controlled reactions;
1. Acetate [from Acetyl CoA] combines with a four-carbon compound (oxaloacetate) to form a six-carbon compound (citrate)
2. Citrate is decarboxylated and dehydrogenated (the hydrogens reduce both NAD and FAD), yielding CO2 and NADH and FADH.
3. ATP is produced via SLP [Substrate Level Phosphorylation]
4. Oxaloacetate is regenerated to combine with another acetyl coenzyme A from the link reaction.
For each turn of the krebs cycle, three NAD molecules and one FAD are reduced,2 carbon dioxide molecules are produced and one ATP molecule is generated via an intermediate compound. The hydrogen released is used in oxidative phosphorylation to provide energy to make ATP.The reactions of Krebs cycle do not make use of oxygen.However,Oxygen is required in the final stage(i.e. Oxidative phosphorylation).
The electron transport chain provides the energy for the phosphorylation of ADP to ATP, and this takes place in the mitochondrial membranes. Reduced NAD/FAD are passed to the electron transport chain where hydrogens are removed from the two hydrogen carriers and is split into hydrogen ions and an electron. The electron is passed to a series of electron carriers whilst the hydrogen ion remains in the mitochondrial matrix. Once the electron is tranferred to oxygen, a hydrogen ion will be draw from the solution to reduce the oxygen to water.
The transfer of electrons along the series of electron carriers, passing from a higher carrier to a lower one, releasing energy, making it available to convert ADP + Phosphate to ATP. Each reduced NAD molecule produces 2.5 molecules of ATP (an average) and each reduced FAD produces 1.5 molecules of ATP (on average).
The energy from the electron transport chain is used to pump hydrogen ions from the mitochondrial matrix into the area between the plasma membrane and the cristae - the concentration of hydrogen ions becomes higher than that in the matrix, a concentration gradient exists. Hydrogen ions are then allowed to pass that through protein channels that have the enzyme ATP synthase on the end, as the ions pass through their electrical energy is used to synthesise ATP. The energy of three hydrogen ions are used to phosporylate one ADP to ATP.
Pyruvate > Lactic Acid
Pyruvate is dehydrogenated via Lactate Dehydrogenase producing Lactic Acid in mammals. Pyruvate accepts 2H, allowing NADH > NAD + 2H. This is very useful biologically for mammals, as during anaerobic respiration, the Link Reaction, Krebs Cycle and Oxidiative Phosphorylation cannot occur; as they all rely on O2 or the products of a process that does. The biological use is that NAD is reoxidised which means it now can accept 2H in Glycolysis, this allows for 2ATP's to be released via glycolysis allowing the mammal to continue to respire and produce ATP without the presence of oxygen.
Pyruvate > Ethanal > Ethanol
Pyruvate is decarboxylated via Pyruvate decarboxylase to produce Ethanal [CO2 as a biproduct]
Ethanal accepts 2H atoms from NADH thus going to Ethanol and reoxidising NAD.
Yeast is a 'facultative anaerobe' it can survive without oxygen although it is killed if the ethanol conc. exceeds 15%
The main respiratory substrates for all cells are carbohydrate usually in the form of hexose sugars such as glucose. Although glucose is not essential for all cells - a lot of cells can oxidise lipids and amino acids.
The energy liberated in aerobic respiration usually comes from the oxidation of hydrogen to water when reduced NAD and reduced FAD are passed into the electron transport chain. So the more hydrogen bonds to break, the more energy - lipids have a energy density more than double that of carbohydrates, because of their long fatty acid tails.
The overall equation for aerobic respiration shows that oxygen used = carbon dioxide produced, and so they are in a 1:1 ratio. Other substrates do not do this, and it is possible to show what is being used for energy by using the respiratory quotient, which is;
{\displaystyle Respiratory\,Quotient={\frac {volume\,of\,carbon\,dioxide\,per\,unit\,time}{volume\,of\,oxygen\,per\,unit\,time}}}
High RQ values usually indicate that anaerobic respiration is taking place.
Fat 0.7 - 0.72
Proteins 0.8 - 0.9
Carboxilic Acids acid 1.33 - 4.0
Retrieved from "https://en.wikibooks.org/w/index.php?title=A-level_Biology/Central_Concepts/Energy_and_respiration&oldid=3248731"
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Continuous Random Variables - Cumulative Distribution Function | Brilliant Math & Science Wiki
Matt DeCross and Lathan Liou contributed
The cumulative distribution function, CDF, or cumulant is a function derived from the probability density function for a continuous random variable. It gives the probability of finding the random variable at a value less than or equal to a given cutoff. Many questions and computations about probability distribution functions are convenient to rephrase or perform in terms of CDFs, e.g. computing the PDF of a function of a random variable.
Definition of the Cumulative Distribution Function
Functions of a Continuous Random Variable
For any random variable
X,
F_X
F_X(x) = P(X \leq x),
which is the probability that
X
x.
Using this definition, one can write the probability that
X
takes a value in a certain interval
[a,b]
without using an integral. Recall that previously this probability was defined in terms of a PDF:
P(a\leq X \leq b) = \int_a^b f_X (x) \,dx.
Now, the probability is rewritten as the difference in values of the CDF:
P(a \leq X \leq b) = F_X(b) - F_X(a).
So the CDF gives the amount of area underneath the PDF between two points. It increases from zero (for very low values of
x
) to one (for very high values of
x
). This is because as
x \to -\infty
, there is no probability that
X
will be found that far out if the PDF is normalized. If
x \to \infty
, this corresponds to
P(X \leq \infty)
which will be one because it is certain that
X
takes some finite value.
In the case of discrete random variables, the value of
F_X
makes a discrete jump at all possible values of
x
; the size of the jump corresponds to the probability
P(X = x)
of that value. In the case of a continuous random variable, the function increases continuously; it is not meaningful to speak of the probability that
X = x
because this probability is always zero. Instead one considers the probability that the value of
X
lies in a given interval:
P(X \in [a,b]) = P(a ≤ X ≤ b) = F_X(b)-F_X(a).
Note that it does not matter if the inequalities are strict (if the interval is
[a,b]
(a,b)
for example): since the probability of any given value is zero, the endpoints can be included or not without changing any probabilities.
Still, one frequently wants to make use of the probability density function
f_X (x)
rather than the CDF. Since the CDF corresponds to the integral of the PDF, the PDF corresponds to the derivative of the CDF:
f_X(x) = F_X'(x) = \frac{dF_X}{dx} .
A fly lands on a
30\text{ cm}
long ruler at a random position chosen uniformly along the ruler. Let
X
be the position of the fly in centimeters, and let
f_X(x)
be the probability density function for
X.
f_X(5)
This probability distribution is uniform, meaning that the probability density is constant on the entire interval
[0, 30]
F_X
is a linear function:
F_X(x) = \left\{\begin{array}{ll} 0 & x \leq 0 \\ \frac{x}{30} & 0 \leq x \leq 30 \\ 1 & 30 \leq x. \end{array}\right.
The probability density function is the derivative:
f_X(x) = \left\{\begin{array}{ll} 0 & x \leq 0 \\ \frac{1}{30} & 0 \leq x \leq 30 \\ 0 & 30 \leq x. \end{array}\right.
Therefore the probability density function at
x = 5
\frac{1}{30}.
A dart player always hits the dartboard (with a radius of
20\text{ cm}
), but has such a poor aim that the distribution of darts is uniform across the entire board. Let
R
be the distance in cm between the dart and the center. Evaluate the probability density function for
R
0,
10,
20.
P(R < r)
is directly proportional to the area of a circle with radius
r
F_R(r) = P(R < r) = \frac{\text{area of circle with radius}\ r}{\text{area of dartboard}} = \frac{\pi r^2}{\pi\times 20^2} = \left(\frac r{20}\right)^2.
f_R(r) = \frac r{200}.
Thus one obtains:
f_R(0) = 0,\ \ f_R(10) = \tfrac1{20},\ \ f_R(20) = \tfrac1{10}.
1-e^{-100 \lambda}
\lambda e^{-100 \lambda}
e^{100 \lambda}
-e^{-100 \lambda}
X
f_X (x) = \lambda e^{-\lambda x},
x
[0,\infty)
x < 100
One question that often comes up in applications of continuous probability is the following: given the PDF of a random variable, is it possible to find the PDF of an arbitrary function of that random variable?
The answer is yes, and the easiest method uses the CDF of the random variable. The general case goes as follows: consider the CDF
F_X (x)
X
Z = g(X)
be a function of
X
. It's important to note the distinction between upper and lower case:
X
is a random variable while
x
is a real number. Recall that the PDF is given by the derivative of the CDF:
f_X (x) = \frac{d}{dX} F_X (x) = \frac{d}{dx} P(X \leq x).
Now write the formula for the CDF of
Z
f_Z (z) = \frac{d}{dz} P(Z \leq z) = \frac{d}{dz} P(g(X) \leq z) = \frac{d}{dz} P(X \leq g^{-1} (z)) = \frac{d}{dz} F_X (g^{-1} (z)).
g
is invertible and increasing, then by the chain rule:
f_Z (z) = f_X (g^{-1} (z)) \frac{dg^{-1} (z)}{dz}.
This formula can be generalized straightforwardly to cases where
g
is not invertible or increasing.
Consider a uniform random variable on the interval
[0,1]
. Find the distribution (i.e., PDF) of
Z = X^3
Z = g(X)
g
is an invertible and increasing function, so the discussion above will apply. The CDF of
X
F_X (x) = x.
f_Z (z) = \frac{d}{dz} F_X (g^{-1} (z)) = \frac{d}{dz} z^{1/3} = \frac13 z^{-2/3}.
This is consistent with the formula derived above.
Cite as: Continuous Random Variables - Cumulative Distribution Function. Brilliant.org. Retrieved from https://brilliant.org/wiki/continuous-random-variables-cumulative/
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Paul wants to hang a picture frame on his wall. To mark out the dimensions, he applies a horizontal force to the middle of the frame. What is the minimum force that Paul has to exert to keep the frame in equilibrium?
The coefficient of friction
\mu
between Paul's hand and the frame is equal to 1.
The block cannot remain in equilibrium
F=mg
F>mg
F<mg
Paul is at an airport and has to pass through a security check. He places his luggage on a conveyor belt, which is initially stopped to scan another passenger's bag. When the conveyor belt starts to move to the right, what can be said about the friction force acting on the luggage?
Friction on the block acts towards the left Friction does not act on the block Friction on the block acts towards the right Friction acts towards the left on both the block and the conveyor belt
Paul keeps his luggage box (denoted by B in the image) on frictionless ground and goes to the restroom. When he returns, he finds that someone has tied their luggage box (denoted by A) with a horizontal rope to a wall and put it on his luggage box B as shown in the diagram. The surface between boxes A and B is rough. Paul applies a force
F
in the horizontal direction to take out his luggage.
How does the minimum horizontal force
F_\textrm{min}
he needs to apply to remove box B depend on the mass of the boxes?
F_\textrm{min}
tends to zero as the ground is smooth
F_\textrm{min}
will remain the same if the mass of the other luggage A is increased
F_\textrm{min}
will remain the same if the mass of the Paul's luggage B is increased
F_\textrm{min}
will increase with the mass of Paul's luggage B
A man of mass m is standing in a hanging cage of mass M. The cage is made from a lightweight material, so
M < m
. A rope is attached to the cage, passed over a smooth light pulley, and the man holds the other end of the rope. The man pulls the rope, such that the entire system is suspended in equilibrium. If the man is standing on a weighing machine, what do we know about the reading of the weighing machine?
It will be greater than m but less than m+M It will be equal to m It will be equal to m + M It will be less then m
The person who is able to pull the other across the middle line is declared the winner. If the rope has negligible mass and is inextensible, who will win the tug of war?
The person who applies a greater horizontal force on ground will win The person who applies greater force on on the rope will win None of the above
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Please tell how to solve Q 26 - Maths - Triangles - 16826961 | Meritnation.com
Here AB,AD and BD are side of a triangle and all the sides are equal as per question.
This means that triangle is equilateral triangle and each angle is of 60 deg .
Now \angle ADC + \angle ADB = 180°\phantom{\rule{0ex}{0ex}}\angle ADB= 60° as the triangle is equilateral triangle .\phantom{\rule{0ex}{0ex}}⇒ \angle ADC =180° - 60° = 120°.........1\phantom{\rule{0ex}{0ex}}Now in △ADC , AD= DC\left\{ given \right\}\phantom{\rule{0ex}{0ex}}So the triangle is isosceles triangle . \phantom{\rule{0ex}{0ex}}i.e\angle DAC =\angle ACD .........2\phantom{\rule{0ex}{0ex}}Let the angle be y°.\phantom{\rule{0ex}{0ex}}Now we know that sum of all the angle of triangle is 180°.\phantom{\rule{0ex}{0ex}}or,\angle ACD+\angle ADC+\angle DAC =180°\phantom{\rule{0ex}{0ex}}or,\angle ADC+2y=180°\phantom{\rule{0ex}{0ex}}or,120° + 2y =180° \left\{by eq 1 and 2\right\}\phantom{\rule{0ex}{0ex}}⇒2y=60°\phantom{\rule{0ex}{0ex}}⇒y=30°\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Now \angle ADC : \angle C = 120°:30° = 4:1 \phantom{\rule{0ex}{0ex}}Hence proved .\phantom{\rule{0ex}{0ex}}Hope this would have cleared your doubt .
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Roynette, Bernard ; Vallois, Pierre ; Volpi, Agnès
\left({X}_{t},\phantom{\rule{0.277778em}{0ex}}t\ge 0\right)
be a Lévy process started at
0
, with Lévy measure
\nu
. We consider the first passage time
{T}_{x}
\left({X}_{t},\phantom{\rule{0.277778em}{0ex}}t\ge 0\right)
to level
x>0
{K}_{x}:={X}_{{T}_{x}}-𝑥
the overshoot and
{L}_{x}:=x-{X}_{{T}_{{𝑥}^{-}}}
the undershoot. We first prove that the Laplace transform of the random triple
\left({T}_{x},{K}_{x},{L}_{x}\right)
satisfies some kind of integral equation. Second, assuming that
\nu
admits exponential moments, we show that
\left(\stackrel{˜}{{T}_{x}},{K}_{x},{L}_{x}\right)
converges in distribution as
x\to \infty
\stackrel{˜}{{T}_{x}}
denotes a suitable renormalization of
{T}_{x}
Classification : 60E10, 60F05, 60G17, 60G40, 60G51, 60J65, 60J75, 60J80, 60K05
Mots clés : Lévy processes, ruin problem, hitting time, overshoot, undershoot, asymptotic estimates, functional equation
author = {Roynette, Bernard and Vallois, Pierre and Volpi, Agn\`es},
title = {Asymptotic behavior of the hitting time, overshoot and undershoot for some {L\'evy} processes},
AU - Volpi, Agnès
TI - Asymptotic behavior of the hitting time, overshoot and undershoot for some Lévy processes
Roynette, Bernard; Vallois, Pierre; Volpi, Agnès. Asymptotic behavior of the hitting time, overshoot and undershoot for some Lévy processes. ESAIM: Probability and Statistics, Tome 12 (2008), pp. 58-93. doi : 10.1051/ps:2007034. http://www.numdam.org/articles/10.1051/ps:2007034/
[1] J. Bertoin, Lévy processes, Cambridge Tracts in Mathematics, vol. 121. Cambridge University Press, Cambridge (1996). | MR 1406564 | Zbl 0861.60003
[2] J. Bertoin and R.A. Doney, Cramér's estimate for Lévy processes. Statist. Probab. Lett. 21 (1994) 363-365. | MR 1325211 | Zbl 0809.60085
[3] H. Cramér, Collective risk theory: A survey of the theory from the point of view of the theory of stochastic processes. Skandia Insurance Company, Stockholm, (1955). Reprinted from the Jubilee Volume of Försäkringsaktiebolaget Skandia. | MR 90177
[4] H. Cramér, On the mathematical Theory of Risk. Skandia Jubilee Volume, Stockholm (1930). | JFM 56.1100.03
[5] R.A. Doney, Hitting probabilities for spectrally positive Lévy processes. J. London Math. Soc. 44 (1991) 566-576. | MR 1149016 | Zbl 0699.60061
[6] R.A. Doney and A.E. Kyprianou, Overshoots and undershoots of Lévy processes. Ann. Appl. Probab. 16 (2006) 91-106. | MR 2209337 | Zbl 1101.60029
[7] R.A. Doney and R.A. Maller. Stability of the overshoot for Lévy processes. Ann. Probab. 30 (2002) 188-212. | MR 1894105 | Zbl 1016.60052
[8] F. Dufresne and H.U. Gerber, Risk theory for the compound Poisson process that is perturbed by diffusion. Insurance Math. Econom. 10 (1991) 51-59. | MR 1114429 | Zbl 0723.62065
[9] I.S. Gradshteyn and I.M. Ryzhik, Table of integrals, series, and products. Academic Press [Harcourt Brace Jovanovich Publishers], New York (1980). Corrected and enlarged edition edited by Alan Jeffrey, Incorporating the fourth edition edited by Yu. V. Geronimus [Yu. V. Geronimus] and M. Yu. Tseytlin [M. Yu. Tseĭtlin], Translated from Russian. | Zbl 0521.33001
[10] P.S. Griffin and R.A. Maller, On the rate of growth of the overshoot and the maximum partial sum. Adv. in Appl. Probab. 30 (1998) 181-196. | MR 1618833 | Zbl 0905.60064
[11] A. Gut, Stopped random walks, Applied Probability, vol. 5, A Series of the Applied Probability Trust. Springer-Verlag, New York, (1988). Limit theorems and applications. | MR 916870 | Zbl 0634.60061
[12] I. Karatzas and S.E. Shreve. Brownian motion and stochastic calculus, Graduate Texts in Mathematics, vol.113. Springer-Verlag, New York, second edition (1991). | MR 1121940 | Zbl 0734.60060
[13] A.E. Kyprianou, Introductory lectures on fluctuations of Lévy processes with applications. Universitext. Springer-Verlag, Berlin (2006). | MR 2250061 | Zbl 1104.60001
[14] N.N. Lebedev, Special functions and their applications. Dover Publications Inc., New York (1972). Revised edition, translated from the Russian and edited by Richard A. Silverman, Unabridged and corrected republication. | Zbl 0271.33001
[15] M. Loève, Probability theory. II. Springer-Verlag, New York, fourth edition (1978). Graduate Texts in Mathematics, Vol. 46. | MR 651018 | Zbl 0385.60001
[16] F. Lundberg, I- Approximerad Framställning av Sannolikhetsfunktionen. II- Aterförsäkering av Kollectivrisker. Almqvist and Wiksell, Uppsala (1903).
[17] T. Rolski, H. Schmidli, V. Schmidt and J. Teugels, Stochastic processes for insurance and finance. Wiley Series in Probability and Statistics. John Wiley & Sons Ltd., Chichester (1999). | MR 1680267 | Zbl 0940.60005
[18] K. Sato, Lévy processes and infinitely divisible distributions, volume 68 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, (1999). Translated from the 1990 Japanese original, Revised by the author. | MR 1739520 | Zbl 0973.60001
[19] A.G. Sveshnikov and A.N. Tikhonov, The theory of functions of a complex variable. “Mir”, Moscow (1982). Translated from the Russian by George Yankovsky [G. Yankovskiĭ]. | Zbl 0531.30002
[20] A. Volpi, Processus associés à l'équation de diffusion rapide; Étude asymptotique du temps de ruine et de l'overshoot. Univ. Henri Poincaré, Nancy I, Vandoeuvre les Nancy (2003). Thèse.
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50 Python Interview Questions and Answers - DEV Community
#python #interview #career #codenewbie
This post was written by Cameron Wilson, and was originally published at Educative.io
This post will cover interview questions on the Python programming language. While this list isn't exhaustive, it should give you a good idea of what types of questions you can expect.
Here's a list of the questions that will be covered:
Question 8: What is a lambda function? Give an example of when it's useful and when it's not
Question 44: Find two numbers that add up to ‘k’
Question 47: Reverse first ‘k’ elements of a queue
Python Interview Questions - Language specific
Output: (50, 'Eighty', 9)
Here we'll be using numpy.array().
This function of the numpy library takes a list as an argument and returns an array that contains all the elements of the list. See the example below:
Instead of the conventional def keyword used for creating functions, a lambda function is defined by using the lambda keyword. The structure of lambda can be seen below:
Lambdas aren't useful when the expression exceeds a single line.
Practice the most common Python interview questions and watch your confidence soar. Educative's text-based courses are easy to skim and feature live in-browser coding environments - making learning quick and efficient.
Here we'll look at simple merging:
Here we will choose roll numbers as key and names as the value because roll numbers are unique and names could be repetitive. So, Alex’s roll number is 122 so the tuple will look like 122: Alex. This will be better explained once you try the code attached below.
{456: 'don', 233: 'bob', 122: 'alex', 353: 'can'}
Shallow copy: [5, 15, 3]
Deep copy: [10, 20, 30]
Output: Key exists
Here's the full implementation.
Sorted() syntax
This method takes one mandatory and two optional arguments:
Reverse (optional): Setting the third parameter as true will sort the list in descending order. Leaving this empty sorts in ascending order.
lst = dict.values()
print("Sorted by value: ", sorted(lst))
dict['2'] = 'strawberry'
lst = dict.items()
print("Sorted by key: ", sorted(lst, key = lambda x : x[0]))
print("Sorted by value: ", sorted(lst, key = lambda x : x[1]))
The product of x and y is: 8
Python Interview Questions - Coding
In this section we’ll take a look at common coding interview questions that pertain to lists, linked lists, graphs, trees, multithreading/concurrency and more. Let’s dive in.
Let's reverse the string "Python" using the slicing method.
Output: nohtyP
There are a couple ways to check this. For this post, we'll look at the find method.
Check if Python Programming contains Programming: 7
Check if Python Programming contains Language: -1
Output: A B C D E F
Output: A B D E F C
Output: [17, 20, 26, 31, 44, 54, 55, 77, 93]
The algorithm works in O(n.logn). This is because the list is being split in log(n) calls and the merging process takes linear time in each call.
Question 42: Implement Dijkstra's algorithm in Python
Here's the implementation
The shortest distance of a from the source vertex a is: 0
The shortest distance of b from the source vertex a is: 3
The shortest distance of c from the source vertex a is: 3.5
The shortest distance of d from the source vertex a is: 4.5
Output: [-2, -1, 0, 4, 5, 6, 7]
The time complexity for this algorithm is O(n+m)O(n+m) where nn and mm are the lengths of the lists. This is because both lists are iterated over at least once.
Question 44: Find two numbers that add up to 'k'
a_{0}
we will do a binary search for
a_{0}
The search is repeated until one is found. You can implement the binarySearch() function however you like, recursively or iteratively.
Since most optimal comparison-based sorting functions take O(nlogn), let’s assume that the Python .sort() function takes the same. Moreover, since binary search takes O(logn) time for finding a single element, therefore a binary search for all n elements will take O(nlogn) time.
Here you can use a Python dictionary to keep count of repetitions.
To see the code solution in action, go to the original post.
Here you can use two pointers which will work simultaneously. Think of it this way:
Question 47: Reverse first 'k' elements of a queue
To see the code solution in action, go to the original post
Check for invalid input, i.e., if the queue is empty, if k is greater than the queue, and if k is negative on line
2. If the input is valid, start by creating a Stack. The available stack functions are:
To see the code solution in action, visit the original post.
Here we'll Min Heapify all Parent Nodes.
Output: [-2, 1, 5, 9, 4, 6, 7]
Remember that we can consider the given maxHeap to be a regular list of elements and reorder it so that it represents a min heap accurately. We do exactly that in this solution. The convertMax() function restores the heap property on all the nodes from the lowest parent node by calling the minHeapify() function on each.
The minHeapify() function is called for half of the nodes in the heap. The minHeapify() function takes O(log(n)) time and its called on
\frac{n}{2}
nodes so this solution takes O(nlog(n)) time.
We iterate the list once. On average, lookup in a set takes O(1) time. Hence, the average runtime of this algorithm is O(n). However, in the worst case, lookup can increase up to O(n), which would cause the algorithm to work in
O(n^{2})
Your learning and preparing has just begun. Take your confidence to a whole new level by practicing the most frequently asked questions in a Python interview. Grokking the Coding Interview: Patterns for Coding Questions has helped countless software engineers prepare and land jobs at Microsoft, Amazon, Google, and others.
In this course you will discover the 16 overarching patterns that underlie coding questions. Once you understand a pattern, you can apply it to hundreds of other questions that are similar but different. Click the link above for a free preview.
Can you expand a bit on question 2? If I were to convert a list into a tuple, I would probably throw the list right at the tuple constructor:
>>> my_tuple = tuple(my_list)
The answer provided seems to be answering a different question.
I agree with that. I run the code, and this is the result.
(50, 'Eighty', 9)
But the real answer should be something like this:
(50, 'Twenty', 110, 'Fifty', 'Ten', 20, 10, 80, 'Eighty')
This is a really informative article. On question 13, is it possible to add a note that the behavior of range and xrange that is explained is based on Python 2. In Python 3, xrange is deprecated and the default behavior it had is moved to range.
There are so many things in here which are flat-out incorrect or warped beyond recognition it's really quite concerning.
People have already pointe dout Question 2 and Question 13, but there's plenty of other examples.
Question 11 manages to omit the key property of inheritance (that of substitution allowing specialisation). Question 12 makes me cry, especially following on from Question 11, which is literally about one form of polymorphism that it somehow fails to mention.
It's a particular shame because Python is rich in polymorphic techniques, from Row Polymorphism (aka Duck Typing) to Inclusion Polymorphism (aka subtyping and inheritance) to Polytypism (where we don't care about the type at all for much of the code).
If an interviewer accepts some of these answers as correct for anything but an entry-level Junior position, I'd worry about accepting the job.
A list consists of mutable objects. (Objects which can be changed after creation)
Not quite. The list itself is mutable, but objects referenced in that list can be either mutable or immutable, and each element of said list is considered a reference (and possibly garbage collected thereafter).
hottabxs@gmail.com
Is it a quizz to spot the most incorrect answer?
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Describe in words the region of R^{3} represented by the equation(s) or inequali
Describe in words the region of R^{3} represented by the equation(s) or inequality. x^{2} + y^{2} = 4
Describe in words the region of
{R}^{3}
represented by the equation(s) or inequality.
{x}^{2}+{y}^{2}=4
tafzijdeq
{x}^{2}+{y}^{2}={r}^{2}
represents a circle whose centre lies on z - axis
{x}^{2}+{y}^{2}=4
{x}^{2}+{y}^{2}=4
{x}^{2}+{y}^{2}=4
represents the set of all points in
{R}^{3}\text{ }lying\text{ }on\text{ }circle\text{ }{x}^{2}+{y}^{2}=4.
\left(x,y,z\right)|{x}^{2}+{y}^{2}=4,x\in R,y\in R,z\in R
Here is no restriction on z-coordinate, so a point in the region must lie on a circle with radius 2 and centre on z-axis but it could be any horizontal plane
z=k
(parallel to xy - plane)
Therefore, the region consists of all points on the circle
{x}^{2}+{y}^{2}=4,z=k
That is, a circular cylinder with radius 2 whose axis is the z - axis
Therefore, the given equations represents the region in
{R}^{3}
consisting of all possible citcles of radius 2 and centre on z -axis that 1s a circular cylinder with radius 2 whose axis is the z - axis.
The given equations represents the region in
{R}^{3}
consisting of all possible circles of radius 2 and centre on z - axis that is a circular cylinder with radius 2 whose axis is the z - axis.
{R}^{3}
x=5
\left(2,-2\right)
\left(-1,\sqrt{3}\right)
Find the polar coordinates
\left(r,\theta \right)
of the point, where r is greater than 0 and 0 is less than or equal to
\theta
, which is less than
2\pi
\left(r,\theta \right)
of the point, where r is less than 0 and 0 is less than or equal to
\theta
2\pi
This question has to do with binary star systems, where i is the angle of inclination of the system.
Calculate the mean expectation value of the factor
{\mathrm{sin}}^{3}
i, i.e., the mean value it would have among an ensemble of binaries with random inclinations. Find the masses of the two stars, if
{\mathrm{sin}}^{3}
i has its mean value.
Hint: In spherical coordinates,
\left(\theta ,\varphi \right)
, integrate over the solid angle of a sphere where the observer is in the direction of the z-axis, with each solid angle element weighted by
\mathrm{sin}\left\{3\right\}\left(\theta \right)
{v}_{1}=100k\frac{m}{s}
{v}_{2}=200k\frac{m}{s}
=2
{M}_{1}=5.74e33g
{M}_{2}=2.87e33g
The vector x is in
H=Span\text{ }{v}_{1},{v}_{2}
and find the beta-coordinate vector
\left[x{\right]}_{\beta }
Systems of Inequalities Graph the solution set of the system if inequalities. Find the coordinates of all vertices, and determine whether the solution set is bounded.
\left\{\begin{array}{l}y<9-{x}^{2}\\ y\ge x+3\end{array}
Convert the point from spherical coordinates to rectangular coordinates.
\left(9,\pi ,\frac{\pi }{2}\right)
\left(x,y,z\right)=?
Plot the given point
\left(-4,0\right)
in a rectangular coordinate system.
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Why does one counterexample disprove a conjecture? Can't a conjecture be correct about most solutio
Why does one counterexample disprove a conjecture?
Can't a conjecture be correct about most solutions except maybe a family of solutions?
For example, a few centuries ago it was widely believed that
{2}^{{2}^{n}}+1
is a prime number for any n . For n=0 we get 3 , for n=1 we get 5 , for n=2 we get 17 , for n=3 we get 257 , but for n=4 it was too difficult to find if this was a prime, until Euler was able to find a factor of it. It seems like this conjecture stopped after that.
But what if this conjecture isn't true only when n satisfies a certain equation, or when n is a power of 2
\ge
4 , or something like that? Did anybody bother to check? I am not asking about this conjecture specifically, but as to why we consider one counterexample as proof that a conjecture is totally wrong.P.S. Andre Nicolas pointed out that Euler found a factor when n=5, not 4 .
This is because, in general, a conjecture is typically worded "Such-and-such is true for all values of [some variable]." So, a single counter-example disproves the "for all" part of a conjecture.
However, if someone refined the conjecture to "Such-and-such is true for all values of [some variable] except those of the form [something]." Then, this revised conjecture must be examined again and then can be shown true or false (or undecidable--I think).
For many problems, finding one counter-example makes the conjecture not interesting anymore; for others, it is worthwhile to check the revised conjecture. It just depends on the problem.
Which of the following are true? If false, explain briefly.
a) If the null hypothesis is true, youll
Least squares regression analysis is a common method for modelling trip generation. What are the main assumptions of the least squares regression analysis in that context? Discuss with examples the consequences of violating two of these assumptions.
Using the Standard Normal Table from the online lectures this week, what is the area under the standard normal curve:
a) To the left of a z-score of 1.25
b) To the right of a z-score of 1.25
c) Between the z-scores -0.25 and 0.55
To give:the hypotheses for the test.
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Expanded Form | Brilliant Math & Science Wiki
The expanded form of a number writes it as a sum, with each digit makes an individual term multiplied by its place value. For example
523
has an expanded form of
5 \times 100 + 2 \times 10 + 3 ,
6203
6 \times 1000 + 2 \times 100 + 0 \times 10 + 3 .
In general expanded form helps understand the meaning of place value. It can be useful in thinking of different number bases and also help in the solving of number cryptogram puzzles.
The expanded form of a number gives the number as a sum where each digit is separate term multiplied by its place value.
943 = 9 \times 100 + 4 \times 10 + 3 \times 1
because there are 9 hundreds, 4 tens, and 3 ones. (The
\times 1
part in front of the
3
is optional.)
41000 = 4 \times 10000 + 1 \times 1000
because there 4 ten thousands and 1 thousand. We can also write alternately write it as
41000 = 4 \times 10000 + 1 \times 1000 + 0 \times 100 + 0 \times 10 + 0 \times 1 .
Expanded form can be used to think about bases other than 10.
For example, while
62 = 6(10) + 2(1) ,
it also can be written in base 2 as 111110 since
62 = 1(32)+1(16)+1(8)+1(4)+1(2)+0(1) .
(The number 111110 is then sometimes written as
{111110}_2
to indicate base 2.)
In general, if we're in base
b,
we can think of the expanded form as
\ldots + (\text{third to last digit})(b^2) + (\text{second to last digit})(b^1) + (\text{last digit})(b^0)
where the dots extend as far as necessary to make a complete sum. For example, when
b = 2
(binary, as is used in computers):
\ldots + (\text{third to last digit})(2^2) + (\text{second to last digit})(2^1) + (\text{last digit})(2^0)
\ldots + (\text{third to last digit})(4) + (\text{second to last digit})(2) + (\text{last digit})(1) .
Write the base 10 number 92 in base 3.
Base 3 will be of the format:
\ldots + (\text{third to last digit})(3^2) + (\text{second to last digit})(3^1) + (\text{last digit})(3^0)
The places in base 10 that would be "ones", "tens", "hundreds", "thousands", and "ten thousands" are now "ones"
(3^0=1),
"threes"
(3^1=3),
"nines"
(3^2=9),
"twenty-sevens"
(3^3=27),
"eighty-ones"
(3^4=81).
3^5 = 243
which is larger than our target number, so the highest we need to go is the "eighty-ones" place. Also, just like how in base 10 the ten digits we can use are
0, 1, 2, 3, 4, 5, 6, 7, 8,
9,
in base 3 the three digits we can use are
0, 1,
2 .
We can have at most one 81 from 92, leaving
92 - 81 = 11 .
We can't remove 27 from 11, but we can remove one 9, leaving
11 - 9 = 2 .
The 2 can the be included in the ones place. So we have
as our number. In expanded form the number
{10102}_3
92 = 1(81) + 0(27) + 1(9) + 0(3) + 2(1) .
123_5
123_6
123_4
\large 123_4 \qquad 123_5 \qquad 123_6
The above shows three numbers, each written in a different base representation. Which of these numbers has the largest value?
More information can be found at the wiki page on number bases.
A cryptogram is a mathematical puzzle where various symbols are used to represent digits, and a given system has to be true. An example puzzle is below.
\large{\begin{array}{ccccccc} && & & S& E & N&D\\ +&& & & M& O & R&E\\ \hline & & & M & O& N & E&Y \end{array}}
Note we consider
SEND
to be the four-digit number made of the letters
S,
E,
N,
D.
We can alternately write this in expanded form as
1000S + 100E + 10N + D .
This is a common technique for cryptogram solving. (More detail about the above problem is at the wiki page dedicated to cryptograms.)
When a two digit number
AB
is added to its reverse
BA,
the sum is 55. Neither
A
B
is zero. Find the largest possible value of
AB .
AB
in expanded form as
10A + B .
BA
10B + A .
The sum is 55, so
10A + B + 10B + A = 55 .
Combining like terms,
11A + 11B = 55 ,
which also implies
A + B = 5 .
Now it's not too hard to list the possibilities for
A
B:
1 + 4 = 5, 2 + 3 = 5, 3 + 2 = 5, 4 + 1 = 5 .
41
is the largest number that can be made from this set, so is the desired answer.
\begin{array} {ccccc} \huge & & & & \huge \color{#69047E}{X}& \huge \color{#69047E}{X}\\ \huge & & & & \huge \color{#D61F06}{Y}& \huge \color{#D61F06}{Y}\\ \huge + & & & & \huge \color{#3D99F6}{Z}& \huge \color{#3D99F6}{Z}\\ \hline \huge& & & \huge \color{#69047E}{X} & \huge \color{#D61F06}{Y} & \huge \color{#3D99F6}{Z} \end{array}
\large \color{#3D99F6}{Z}
Cite as: Expanded Form. Brilliant.org. Retrieved from https://brilliant.org/wiki/expanded-form/
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ELECTROSTATICS - Encyclopedia Information
Electrostatics Information
An electrostatic effect: foam peanuts clinging to a cat's fur due to static electricity. The triboelectric effect causes an electrostatic charge to build up on the surface of the fur due to the cat's motions. The electric field of the charge causes polarization of the molecules of the foam due to electrostatic induction, resulting in a slight attraction of the light plastic pieces to the charged fur. [1] [2] [3] [4] This effect is also the cause of static cling in clothes.
Electrostatics is a branch of physics that studies electric charges at rest ( static electricity).
Since classical times, it has been known that some materials, such as amber, attract lightweight particles after rubbing. The Greek word for amber, ἤλεκτρον (ḗlektron), was thus the source of the word ' electricity'. Electrostatic phenomena arise from the forces that electric charges exert on each other. Such forces are described by Coulomb's law.
{\displaystyle r}
is the distance (in meters) between two charges, then the force (in newtons) between two point charges
{\displaystyle q}
{\displaystyle Q}
(in coulombs) is:
{\displaystyle F={\frac {1}{4\pi \varepsilon _{0}}}{\frac {qQ}{r^{2}}}=k_{0}{\frac {qQ}{r^{2}}}\,,}
where ε0 is the vacuum permittivity, or permittivity of free space: [5]
{\displaystyle \varepsilon _{0}\approx 8.854\ 187\ 817\times 10^{-12}\;\;\mathrm {C^{2}\ N^{-1}\ m^{-2}} .}
The SI units of ε0 are equivalently A2 s4 kg−1m−3 or C2 N−1m−2 or F m−1. Coulomb's constant is:
{\displaystyle k_{0}={\frac {1}{4\pi \varepsilon _{0}}}\approx 8.987\ 551\ 792\times 10^{9}\;\;\mathrm {N\ m^{2}\ C} ^{-2}.}
{\displaystyle e=1.602\ 176\ 634\times 10^{-19}\;\;\mathrm {C} .}
The electric field,
{\displaystyle {\vec {E}}}
, in units of newtons per coulomb or volts per meter, is a vector field that can be defined everywhere, except at the location of point charges (where it diverges to infinity). [6] It is defined as the electrostatic force
{\displaystyle {\vec {F}}\,}
in newtons on a hypothetical small test charge at the point due to Coulomb's Law, divided by the magnitude of the charge
{\displaystyle q\,}
in coulombs
{\displaystyle {\vec {E}}={{\vec {F}} \over q}}
Consider a collection of
{\displaystyle N}
particles of charge
{\displaystyle Q_{i}}
, located at points
{\displaystyle {\vec {r}}_{i}}
(called source points), the electric field at
{\displaystyle {\vec {r}}}
(called the field point) is: [6]
{\displaystyle {\vec {E}}({\vec {r}})={\frac {1}{4\pi \varepsilon _{0}}}\sum _{i=1}^{N}{\frac {{\widehat {\mathcal {R}}}_{i}Q_{i}}{\left\|{\mathcal {\vec {R}}}_{i}\right\|^{2}}},}
{\displaystyle {\vec {\mathcal {R}}}_{i}={\vec {r}}-{\vec {r}}_{i},}
is the displacement vector from a source point
{\displaystyle {\vec {r}}_{i}}
to the field point
{\displaystyle {\vec {r}}}
{\displaystyle {\widehat {\mathcal {R}}}_{i}={\vec {\mathcal {R}}}_{i}/\left\|{\vec {\mathcal {R}}}_{i}\right\|}
is a unit vector that indicates the direction of the field. For a single point charge at the origin, the magnitude of this electric field is
{\displaystyle E=k_{e}Q/{\mathcal {R}}^{2},}
and points away from that charge if it is positive. The fact that the force (and hence the field) can be calculated by summing over all the contributions due to individual source particles is an example of the superposition principle. The electric field produced by a distribution of charges is given by the volume charge density
{\displaystyle \rho ({\vec {r}})}
and can be obtained by converting this sum into a triple integral:
{\displaystyle {\vec {E}}({\vec {r}})={\frac {1}{4\pi \varepsilon _{0}}}\iiint {\frac {{\vec {r}}-{\vec {r}}\,'}{\left\|{\vec {r}}-{\vec {r}}\,'\right\|^{3}}}\rho ({\vec {r}}\,')\,\mathrm {d} ^{3}r\,'}
{\displaystyle \oint _{S}{\vec {E}}\cdot \mathrm {d} {\vec {A}}={\frac {1}{\varepsilon _{0}}}\,Q_{\text{enclosed}}=\int _{V}{\rho \over \varepsilon _{0}}\cdot \mathrm {d} ^{3}r,}
{\displaystyle \mathrm {d} ^{3}r=\mathrm {d} x\ \mathrm {d} y\ \mathrm {d} z}
is a volume element. If the charge is distributed over a surface or along a line, replace
{\displaystyle \rho \,\mathrm {d} ^{3}r}
{\displaystyle \sigma \,\mathrm {d} A}
{\displaystyle \lambda \,\mathrm {d} \ell }
. The divergence theorem allows Gauss's Law to be written in differential form:
{\displaystyle {\vec {\nabla }}\cdot {\vec {E}}={\rho \over \varepsilon _{0}}.}
{\displaystyle {\vec {\nabla }}\cdot }
is the divergence operator.
{\displaystyle {\nabla }^{2}\phi =-{\rho \over \varepsilon _{0}}.}
{\displaystyle {\nabla }^{2}\phi =0,}
{\displaystyle {\vec {\nabla }}\times {\vec {E}}=0.}
{\displaystyle {\partial {\vec {B}} \over \partial t}=0.}
In other words, electrostatics does not require the absence of magnetic fields or electric currents. Rather, if magnetic fields or electric currents do exist, they must not change with time, or in the worst-case, they must change with time only very slowly. In some problems, both electrostatics and magnetostatics may be required for accurate predictions, but the coupling between the two can still be ignored. Electrostatics and magnetostatics can both be seen as Galilean limits for electromagnetism. [7][ verification needed]
As the electric field is irrotational, it is possible to express the electric field as the gradient of a scalar function,
{\displaystyle \phi }
, called the electrostatic potential (also known as the voltage). An electric field,
{\displaystyle E}
, points from regions of high electric potential to regions of low electric potential, expressed mathematically as
{\displaystyle {\vec {E}}=-{\vec {\nabla }}\phi .}
The gradient theorem can be used to establish that the electrostatic potential is the amount of work per unit charge required to move a charge from point
{\displaystyle a}
{\displaystyle b}
with the following line integral:
{\displaystyle -\int _{a}^{b}{{\vec {E}}\cdot \mathrm {d} {\vec {\ell }}}=\phi ({\vec {b}})-\phi ({\vec {a}}).}
A test particle's potential energy,
{\displaystyle U_{\mathrm {E} }^{\text{single}}}
, can be calculated from a line integral of the work,
{\displaystyle q_{n}{\vec {E}}\cdot \mathrm {d} {\vec {\ell }}}
. We integrate from a point at infinity, and assume a collection of
{\displaystyle N}
{\displaystyle Q_{n}}
, are already situated at the points
{\displaystyle {\vec {r}}_{i}}
. This potential energy (in Joules) is:
{\displaystyle U_{\mathrm {E} }^{\text{single}}=q\phi ({\vec {r}})={\frac {q}{4\pi \varepsilon _{0}}}\sum _{i=1}^{N}{\frac {Q_{i}}{\left\|{\mathcal {{\vec {R}}_{i}}}\right\|}}}
{\displaystyle {\vec {\mathcal {R_{i}}}}={\vec {r}}-{\vec {r}}_{i}}
is the distance of each charge
{\displaystyle Q_{i}}
from the test charge
{\displaystyle q}
, which situated at the point
{\displaystyle {\vec {r}}}
{\displaystyle \phi ({\vec {r}})}
is the electric potential that would be at
{\displaystyle {\vec {r}}}
if the test charge were not present. If only two charges are present, the potential energy is
{\displaystyle k_{e}Q_{1}Q_{2}/r}
. The total electric potential energy due a collection of N charges is calculating by assembling these particles one at a time:
{\displaystyle U_{\mathrm {E} }^{\text{total}}={\frac {1}{4\pi \varepsilon _{0}}}\sum _{j=1}^{N}Q_{j}\sum _{i=1}^{j-1}{\frac {Q_{i}}{r_{ij}}}={\frac {1}{2}}\sum _{i=1}^{N}Q_{i}\phi _{i},}
{\displaystyle \phi _{i}={\frac {1}{4\pi \varepsilon _{0}}}\sum _{\stackrel {j=1}{j\neq i}}^{N}{\frac {Q_{j}}{r_{ij}}}.}
This electric potential,
{\displaystyle \phi _{i}}
is what would be measured at
{\displaystyle {\vec {r}}_{i}}
if the charge
{\displaystyle Q_{i}}
were missing. This formula obviously excludes the (infinite) energy that would be required to assemble each point charge from a disperse cloud of charge. The sum over charges can be converted into an integral over charge density using the prescription
{\textstyle \sum (\cdots )\rightarrow \int (\cdots )\rho \,\mathrm {d} ^{3}r}
{\displaystyle U_{\mathrm {E} }^{\text{total}}={\frac {1}{2}}\int \rho ({\vec {r}})\phi ({\vec {r}})\,\mathrm {d} ^{3}r={\frac {\varepsilon _{0}}{2}}\int \left|{\mathbf {E} }\right|^{2}\,\mathrm {d} ^{3}r,}
This second expression for electrostatic energy uses the fact that the electric field is the negative gradient of the electric potential, as well as vector calculus identities in a way that resembles integration by parts. These two integrals for electric field energy seem to indicate two mutually exclusive formulas for electrostatic energy density, namely
{\textstyle {\frac {1}{2}}\rho \phi }
{\textstyle {\frac {1}{2}}\varepsilon _{0}E^{2}}
; they yield equal values for the total electrostatic energy only if both are integrated over all space.
{\displaystyle P={\frac {\varepsilon _{0}}{2}}E^{2},}
Electrostatic induction, discovered by British scientist John Canton in 1753 and Swedish professor Johan Carl Wilcke in 1762 [8] [9] [10] is a redistribution of charges in an object caused by the electric field of a nearby charge. For example, if a positively charged object is brought near an uncharged metal object, the mobile negatively-charged electrons in the metal will be attracted by the external charge, and move to the side of the metal facing it, creating a negative charge on the surface. When the electrons move out of an area they leave a positive charge due to the metal atoms' nuclei, so the side of the metal object facing away from the charge acquires a positive charge. These induced charges disappear when the external charge is removed. Induction is also responsible for the attraction of light objects, such as balloons, paper scraps and foam packing peanuts to static charges. The surface charges induced in conductive objects exactly cancel external electric fields inside the conductor, so there is no electric field inside a metal object. This is the basis for the electric field shielding action of a Faraday cage. Since the electric field is the gradient of the voltage, electrostatic induction is also responsible for making the electric potential ( voltage) constant throughout a conductive object.
Electrostatic actuators have recently been attracting interest in the soft robotics research area. Electrostatic actuators can be employed as clutches for wearable devices which can exhibit mechanical impedance tuning and improved energy efficiency. [11] [12] [13] Other relevant applications include but not limited to multimode hydraulically amplified electrostatic actuators for wearable haptics [14] and robots driven by electrostatic actuator. [15] [16]
^ Heras, J. A. (2010). "The Galilean limits of Maxwell's equations". American Journal of Physics. 78 (10): 1048–1055. arXiv: 1012.1068. Bibcode: 2010AmJPh..78.1048H. doi: 10.1119/1.3442798. S2CID 118443242.
^ Diller, Stuart B; Collins, Steven H; Majidi, Carmel (November 2018). "The effects of electroadhesive clutch design parameters on performance characteristics". Journal of Intelligent Material Systems and Structures. 29 (19): 3804–3828. doi: 10.1177/1045389X18799474. ISSN 1045-389X. S2CID 52904769.
^ Ramachandran, Vivek; Shintake, Jun; Floreano, Dario (February 2019). "All-Fabric Wearable Electroadhesive Clutch". Advanced Materials Technologies. 4 (2): 1800313. doi: 10.1002/admt.201800313. S2CID 139121491.
^ Diller, Stuart; Majidi, Carmel; Collins, Steven H. (May 2016). "A lightweight, low-power electroadhesive clutch and spring for exoskeleton actuation". 2016 IEEE International Conference on Robotics and Automation (ICRA). Stockholm, Sweden: IEEE: 682–689. doi: 10.1109/ICRA.2016.7487194. ISBN 978-1-4673-8026-3. S2CID 206851724.
^ Leroy, Edouard; Hinchet, Ronan; Shea, Herbert (2020-07-23). "Multimode Hydraulically Amplified Electrostatic Actuators for Wearable Haptics". Advanced Materials. 32 (36): 2002564. doi: 10.1002/adma.202002564. ISSN 0935-9648. PMID 32700326. S2CID 220716480.
^ Shigemune, Hiroki; Maeda, Shingo; Cacucciolo, Vito; Iwata, Yoshitaka; Iwase, Eiji; Hashimoto, Shuji; Sugano, Shigeki (April 2017). "Printed Paper Robot Driven by Electrostatic Actuator". IEEE Robotics and Automation Letters. 2 (2): 1001–1007. doi: 10.1109/LRA.2017.2658942. ISSN 2377-3766. S2CID 17743332.
^ Wu, Qiyang; Diaz Jimenez, Tomas G.; Qu, Juntian; Zhao, Chen; Liu, Xinyu (September 2017). "Regulating surface traction of a soft robot through electrostatic adhesion control". 2017 IEEE/RSJ International Conference on Intelligent Robots and Systems (IROS). Vancouver, BC: IEEE: 488–493. doi: 10.1109/IROS.2017.8202198. ISBN 978-1-5386-2682-5. S2CID 27169691.
William J. Beaty (1997) " Humans and sparks: The Cause, Stopping the Pain, and 'Electric People"
Wikisource has the text of the 1911 Encyclopædia Britannica article " Electrostatics".
Retrieved from " https://en.wikipedia.org/?title=Electrostatics&oldid=1080133454"
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Magnetic vector potential | Brilliant Math & Science Wiki
The magnetic vector potential
(\vec{A})
is a vector field that serves as the potential for the magnetic field. The curl of the magnetic vector potential is the magnetic field.
\vec{B} = \nabla \times \vec{A}
The magnetic vector potential is preferred when working with the Lagrangian in classical mechanics and quantum mechanics.
Calculating magnetic vector potential
The magnetic vector potential contributed by a length
d\vec{s}
with current
I
running through it is
d\vec{A} = \frac{\mu_0 I}{4\pi r} d\vec{s}.
What is the magnetic vector potential a distance
R
from a long straight current element?
Assume the wire is placed on the z-axis. Hence the distance from a differential element
d\vec{z}
to a point in space is
r = \sqrt{ R^2 + z^2}.
d\vec{A} = \frac{\mu_0 I}{4 \pi} \frac{dz}{\sqrt{ R^2 + z^2} } \hat{z}.
Now it is necessary to integrate over the length of the rod. Since the rod is arbitrarily said to be on the z-axis, it can be said, for simplicity, to extend from
z=0
z=L.
This simply requires multiplying by 2. Also, use the substitution
z = R \tan(\theta).
\begin{aligned} \vec{A} &= 2 \int_0^L \frac{\mu_0 I}{4 \pi} \frac{dz}{\sqrt{ R^2 + z^2} } \hat{z} \\ &= \frac{\mu_0 I}{2 \pi} \hat{z} \int_0^\phi \sec(\theta) d\theta \\ &= \frac{\mu_0 I}{2 \pi} \ln \bigg( \frac{\sqrt{L^2 + R^2} + L }{R} \bigg) \hat{z} \end{aligned}
The magnetic field is the curl of the vector potential.
\vec{B} = \nabla \times \vec{A}
Find the magnetic field in a region with magnetic vector potential
\vec{A} = \sin(\theta)\hat{r} - r\hat{\theta}.
\vec{A}
is in spherical coordinates, use the spherical definition of the curl.
\begin{aligned} \vec{B} = \frac{1}{r\sin\theta} \left( \frac{\partial}{\partial \theta} \left(A_\varphi\sin\theta \right) - \frac{\partial A_\theta}{\partial \varphi} \right) &\hat{\mathbf r} \\ {}+ \frac{1}{r} \left( \frac{1}{\sin\theta} \frac{\partial A_r}{\partial \varphi} - \frac{\partial}{\partial r} \left( r A_\varphi \right) \right) &\hat{\boldsymbol \theta} \\ {}+ \frac{1}{r} \left( \frac{\partial}{\partial r} \left( r A_{\theta} \right) - \frac{\partial A_r}{\partial \theta} \right) &\hat{\boldsymbol \varphi} \end{aligned}
The only part that will survive for the given
\vec{A}
\begin{aligned} \vec{B} &= \frac{1}{r} \big( \frac{\partial}{\partial r} (r A_\theta) - \frac{\partial A_r}{\partial \theta} \big) \hat{\varphi} \\ &= \frac{1}{r} \big( 2r - \cos(\theta) \big) \\ &= 2 - \frac{\cos(\theta)}{r} \end{aligned}
The partial derivative of the magnetic vector potential contributes partially to the induced electric field according to faraday's law.
\vec{E} = - \frac{\partial \vec{A} }{ \partial t}
Recall Faraday's law:
\nabla \times \vec{E} = - \frac{\partial \vec{B} }{\partial t}
\nabla \times \vec{E} = - \frac{\partial \big( \nabla \times \vec{A} \big) }{\partial t}
\nabla \times \vec{E} = - \nabla \times \frac{ \partial \vec{A} }{\partial t}
\rightarrow \vec{E} = - \frac{ \partial \vec{A} }{\partial t}
Cite as: Magnetic vector potential. Brilliant.org. Retrieved from https://brilliant.org/wiki/magnetic-vector-potential/
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Multiresolution Expansion and Approximation Order of Generalized Tempered Distributions
Byung Keun Sohn, "Multiresolution Expansion and Approximation Order of Generalized Tempered Distributions", International Journal of Mathematics and Mathematical Sciences, vol. 2013, Article ID 190981, 8 pages, 2013. https://doi.org/10.1155/2013/190981
Byung Keun Sohn 1
1Department of Mathematics, Inje University, Kimhae 621-749, Republic of Korea
Let be the generalized tempered distributions of -growth with restricted order , where the function grows faster than any linear functions as . We show the convergence of multiresolution expansions of in the test function space of . In addition, we show that the kernel of an integral operator provides approximation order in in the context of shift-invariant spaces.
Multiresolution analysis was shown to be very useful in extending the expansions in orthogonal wavelets from to a certain class of tempered distributions. Some interactions between wavelets and tempered distributions have been presented by Walter’s work in [1–3]. Walter has found the analytic representation of tempered distributions of polynomial growth with restricted order, , by wavelets [1] and the multiresolution expansions’ pointwise convergence of [3]. Pilipović and Teofanov have showed the uniform convergence on compact sets of the derivatives of multiresolution expansions of and the convergence of multiresolution expansions of in the test function space of . As an application, Pilipović and Teofanov have shown that the kernel of an integral operator provides approximation order in in the context of shift-invariant spaces [4].
In the meantime, the tempered distributions of polynomial growth were extended to tempered distributions of -growth, , in [5, 6] and -growth, , in [7, 8] or -growth, , in [9, 10], where the function grows faster than any linear functions as . We have considered the analytic representation of tempered distributions of -growth with restricted order, , by wavelets [11]. Also, we have shown that the multiresolution expansions of converges pointwise to the value of the distribution where it exists [12].
In this paper, we will show the uniform convergence on compact sets of the derivatives of multiresolution expansions of and convergence of multiresolution expansions of in the test function space of . In addition, we will show that the kernel of an integral operator provides approximation order in . This is an extension of the works of Pilipović and Teofanov [4] in the context of generalized tempered distributions, .
2. The Generalized Tempered Distribution Spaces
Throughout this paper, we will use or to denote the positive constants, which are independent parameters and may be different at each occurrence.
Let denote a continuous increasing function such that and . For , we define
The function is an increasing, convex, and continuous function with and satisfies the fundamental convexity inequality . Further, we define for negative by means of the equality . Note that since the derivative of is unbounded in , the function will grow faster than any linear function as . Now we list some properties of which will be frequently used later. Consider the following:
Using the function , we define the space as the space of all functions such that
The topology in is defined by the family of the seminorms . Then become a Fréchet space and are continuous and dense inclusions; here denotes the spaces of all functions with compact supports, the spaces of polynomially decreasing functions (Schwartz functions), and the space of all functions. By , we mean the space of continuous linear functionals on .
Definition 1. We say that the elements of are generalized tempered distributions.
Clearly, when , are tempered distributions (Schwarz distributions), . When , are tempered distributions, , which are introduced and characterized by Yoshinaga [6] and Hasumi [5], independently. When , are tempered distributions, , which are introduced and characterized by Sznajder and Zielezny [7, 8]. For details about , we refer to [9, 10].
For a natural number , we define by the space of all such that
The topology of is defined by the family of and the dual of is denoted by . Clearly, is the projective limit of when and = . Also, we have continuous and dense inclusion mapping as following:
Definition 2. We say that the elements of are generalized tempered distributions of order .
We define by the space of all such that The topology of is defined by the family of and the dual of is denoted by . Obviously, .
Now, we give a theorem that will be used later.
Theorem 3. Let and sequence be given in such that converges uniformly to on every compact set and for . If is bounded in , then the sequence converges to in .
Proof. Let be given and let . Then there exist such that for arbitrary . Also, since the sequence is bounded in , we can take a positive number and a compact set such that when and From (7) and (8), we have
3. Multiresolution Expansion of
Definition 4. A multiresolution analysis (shortly MRA) consists of a sequence of closed subspaces , of satisfying the following:(i) is an orthonormal basis of , (ii), (iii), (iv). The function whose existence is asserted in (i) is called a scaling function of the given MRA.
Definition 5. We say that a multiresolution analysis , is regular MRA of if the scaling function is in .
Example 6. It is impossible that the scaling function has exponential decay and , with all derivatives bounded, unless . Refer to [13, Corollary 5.5.3]. So we will restrict our attention to or . From the remark in [13] or, page 152 [2, Example 4, page 48], Battle-Lemarié’s wavelets are in for some when , but not in even if they have exponential decay and smoothness. In [13], Daubechies shown that for an arbitrary nonnegative integer , there exists an regular MRA of such that the scaling function has compact supports.
Let be an regular MRA of and let be a scaling function. The reproducing kernel of is given by
The series and its derivatives with respect to or of order converge uniformly on because of the regularity of . The reproducing kernel of the projection operator onto is and the projection of onto is given by
The sequence , given in (12), is called the multiresolution expansion of .
Definition 7. For a given , the sequence defined by is called the multiresolution expansion of .
We deduce the following properties of the reproducing kernel with scaling function :(a) and for all .(b) For every and , there exist such that where we used the properties (2).(c) .
Let be an regular MRA of . We fix a function with . We let denote the function and let denote the operation of convolution by . For each fixed , we consider the function of the variable . From (c), we have for , whereas
Now, it follows from integration by parts that the kernal of the operator shares these properties (15) and (16) with .
Let From (b) and the fact that , we have and these functions also satisfy identically in for every . They, for every and with at most -growth, define operator by which are such that that is, From Theorem 1.1 in [14], we have uniformly on compact sets. Now we will show the uniform convergence on compact sets of the derivatives of multiresolution expansions of .
Theorem 8. Let such that the corresponding derivatives are bounded by a when , for every and some . If , given by (12), be the projection of onto an regular MRA of , then the sequence converges uniformly on compact sets to as , for every .
Proof. If , we have where is a continuous function with growth and . From (18), given a compact set , we have for large enough and . Since can be chosen arbitrary, we obtain by dominated convergence theorem, uniformly for . From (21) and (23), we have the conclusion.
We now ready to show the main theorem.
Theorem 9. Let and let , given by (7), be a projection of onto an regular MRA of . If , then the sequence converges to in as .
Proof. Let and be given in (21) such that and . From Theorems 3 and 8 and (21), it suffices to show that is bounded for every and . Since has a compact support, then Hence we have only to show that for every and . Let . Then, by (18), we have By a simple change of variable, we have Since and on , then for sufficiently large . Since on , then for sufficiently large .
4. Approximation Order of
A space of functions is called shift invariant if it is invariant under all integer translate, that is,
The principal shift-invariant subspaces are generated by the closure of the linear span of the shifts of . The stationary ladder of spaces is given by
To rate the efficiency for approximation of such spaces, the concept of approximation order is widely used. We say that the scale of the space provides approximation order in if for every sufficiently smooth , where . For further details about the theory on the approximation order provided by shift-invariant spaces, we refer to [15, 16]. We will focus our attention to the so-called approximation order of an integral operator.
Let be an integral operator of the following form
We assume that . For , we define where is the scaling operator . We say that the integral operator defined by (37) provides approximation order in if for every sufficiently smooth , where . For further details about the theory on the approximation order provided by integral or kernel operator, we refer to [17, 18].
Definition 10 (see [4]). Let . Let , be the kernel of an integral operator . is given by . We say that the operator provides approximation order in if where the constant .
We will now show that the kernel of an integral operator provides approximation order in .
Theorem 11. Let with compact support such that the integer shifts of form an orthogonal basis of with respect to the inner product in . Assume that for some sequence . Let be the kernel of the integral operator given by (37). Then provides approximation order in .
Proof. Firstly, we will show that where . If we accept the result (42) for a moment, it follows that for , we have hence which implies the conclusion.
Since satisfy the conditions of regular MRA of with , we can apply (21) to the operator , that is, where and are given in (21). For ,
In order to estimate , we consider with and . Let be a constant such that . If we assume , the smoothness of implies where for some and . To show the finiteness of in the last statement, we use
We will estimate by using the following facts. Since has a compact support, there exists such that for . Also, by the choice of and property (c) of the reproducing kernel , we have Hence
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Copyright © 2013 Byung Keun Sohn. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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Let vectors \bar{A} =(2,1,-4), \bar{B} =(-3,0,1), \text{ and } \bar{C} =(-1,-1,2) Calcula
Let vectors \bar{A} =(2,1,-4), \bar{B} =(-3,0,1), \text{ and } \bar{C} =(-1,-1,2) Calculate the following: A)\bar{A} \cdot \bar{B}? E)Which of the fol
Let vectors
\overline{A}=\left(2,1,-4\right),\overline{B}=\left(-3,0,1\right),\text{ and }\overline{C}=\left(-1,-1,2\right)
\overline{A}\cdot \overline{B}
E)Which of the following can be computed?
\overline{A}\cdot \overline{B}\cdot \overline{C}
\overline{A}\left(\overline{B}\cdot \overline{C}\right)
\overline{A}\left(\overline{B}+\overline{C}\right)
3\overline{A}
F)Express your answer in terms of
{v}_{1}
G) If
{v}_{1}\text{ and }{v}_{2}
are perpendicular?
H) If
{v}_{1}\text{ and }{v}_{2}
are parallel?
Explained your problem on photo
\stackrel{―}{A}=<2,1,-4>\text{ }\stackrel{―}{B}=<-3,0,1>\text{ }\stackrel{―}{C}=<-1,-1,2>
\stackrel{―}{A}\cdot \stackrel{―}{B}=<2,1,-4>\cdot <-3,0,1>=-6+0-4=-10
\stackrel{―}{A}\cdot \stackrel{―}{B}\cdot \stackrel{―}{C}
\stackrel{―}{A}\left(\stackrel{―}{B}\cdot \stackrel{―}{C}\right)
3\stackrel{―}{A}
can not be computed because
\left(\stackrel{―}{B}\cdot \stackrel{―}{C}\right)
is a scalar quatitty as well as 3 and dot product is always belween 2 vectors.
\stackrel{―}{A}\cdot \left(\stackrel{―}{B}+\stackrel{―}{C}\right)
can be computed
\left(\stackrel{―}{B}+\stackrel{―}{C}\right)
is another vector dot product can be taken with
\stackrel{―}{A}
\left(\stackrel{―}{B}+\stackrel{―}{C}\right)
\stackrel{―}{{v}_{1}},\stackrel{―}{{v}_{2}}
\stackrel{―}{{v}_{1}}\cdot \stackrel{―}{{v}_{1}}={v}_{1}^{2}
\stackrel{―}{{v}_{1}}
\stackrel{―}{{v}_{2}}
are perpendicular
\stackrel{―}{{v}_{1}}\cdot \stackrel{―}{{v}_{2}}={v}_{1}{v}_{2}\mathrm{cos}{90}^{\circ }=0
\stackrel{―}{{v}_{1}}\cdot \stackrel{―}{{v}_{1}}={v}_{1}{v}_{1}\mathrm{cos}{0}^{\circ }={v}_{1}^{2}
\stackrel{―}{{v}_{1}}
\stackrel{―}{{v}_{2}}
\stackrel{―}{{v}_{1}}\cdot \stackrel{―}{{v}_{2}}={v}_{1}{v}_{2}\mathrm{cos}{0}^{\circ }={v}_{1}\cdot {v}_{2}
a=\left(4,7,-4\right),b=\left(3,-1,1\right)
\left(a\cdot b\right)\cdot c
\left(a\cdot b\right)\cdot c
\left(a\cdot b\right)c
\left(a\cdot b\right)c
|a|\left(b\cdot c\right)
|a|\left(b\cdot c\right)
a\cdot \left(b+c\right)
a\cdot \left(b+c\right)
a\cdot b+c
a\cdot b+c
|a|\cdot \left(b+c\right)
|a|\cdot \left(b+c\right)
{M}_{2×4}
{M}_{3×4}\right)
{M}_{2×4}
Tell the opposite to a cross product.
a×b=c
. How to find c, if you know only a and b?
Find the curve’s unit tangent vector. Also, find the length of the indicated portion of the curve.
r\left(t\right)=\left(2\mathrm{cos}t\right)i+\left(2\mathrm{sin}t\right)j+\sqrt{5}tk,0\le t\le \pi
P(2, 0), Q(0, 3), R(3, 4)
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Engineering Acoustics/Time-Domain Solutions - Wikibooks, open books for an open world
Engineering Acoustics/Time-Domain Solutions
d'Alembert SolutionsEdit
In 1747, Jean Le Rond d'Alembert published a solution to the one-dimensional wave equation.
The general solution, now known as the d'Alembert method, can be found by introducing two new variables:
{\displaystyle \xi =ct-x\,}
{\displaystyle \eta =ct+x\,}
and then applying the chain rule to the general form of the wave equation.
From this, the solution can be written in the form:
{\displaystyle y(\xi ,\eta )=f(\xi )+g(\eta )\,=f(x+ct)+g(x-ct)}
where f and g are arbitrary functions, that represent two waves traveling in opposing directions.
A more detailed look into the proof of the d'Alembert solution can be found here.
Example of Time Domain SolutionEdit
If f(ct-x) is plotted vs. x for two instants in time, the two waves are the same shape but the second displaced by a distance of c(t2-t1) to the right.
The two arbitrary functions could be determined from initial conditions or boundary values.
Retrieved from "https://en.wikibooks.org/w/index.php?title=Engineering_Acoustics/Time-Domain_Solutions&oldid=3232757"
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In the overhead view of, a long uniform rod of mass m0.6 Kg is free to rotate in
In the overhead view of, a long uniform rod of mass m0.6 Kg is free to rotate in a horizontal planeabout a vertical axis through its center .A spring
In the overhead view of, a long uniform rod of mass m0.6 Kg is free to rotate in a horizontal planeabout a vertical axis through its center .A spring with force constant k = 1850 N/m is connected horizontally betweenone end of the rod and a fixed wall. When the rod is in equilibrium, it is parallel to the wall. What isthe period of the small oscillations thatresult when the rod is rotated slightly and released?
In theoverhead view of , a long uniform rodof mass ( m = 0.6 Kg) is freeto rotate in a horizontal plane about avertical axis through its center.
A spring with force constant( k = 1850 N/m ) is connected horizontally between one end of the rod and a fixed wall.
When the rod is inequilibrium, it is parallel to the wall.
Let ' T ' be the time-period of the small oscillations that result when the rod is rotated slightly and released
Let rod ( length L) rotates by angle (
\theta
Let 'F ' be the force applied on the rod, then spring elongated by
\left(\mathrm{△}x\right)=\frac{L\theta }{2}
Since , the moment of force
\left(\tau \right)=-\frac{FL}{2}
\left(\tau =-\frac{FL}{2}=-\frac{kL\theta }{2}\cdot \frac{L}{2}
=-\frac{k{L}^{2}}{4}\cdot \theta
So, the time period
\left(T\right)=2\pi \cdot \sqrt{\frac{l}{C}}
I=\frac{m{L}^{2}}{12}
C=\frac{k{L}^{2}}{4}
T=2\pi \cdot \sqrt{\frac{m}{3k}}
Here m=0.6 kg and k=1850 N/m
T=0.06533s
(a) How far from a 50.0-mm-focal-length lens must an object be placed if its image is to be magnified 2.00 X and be real?
(b) What if the image is to be virtual and magnified 2.00 X?
Saturn has an equatorial radius of 6.00 107 mand a mass of 5.67 1026 kg.
(a) Compute the acceleration of gravity at the equatorof Saturn.
(b) What it the ratio of a person's weight on Saturn to that onEarth?
Many people believe that a vacuum created inside a vacuumcleaner causes particles of dirt to be drawn in. Actually, however,the dirt is pushed in. Explain.
d=\frac{1}{2}a{t}^{2}
A 5 kg block of aluminum is heated from 20 degrees Celsius to90 degrees Celsius at atmospheric pressure. Find a)the workdone by the aluminum b) the amount of energy transferred to it byheat, and c) the increase in its internal energy?
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Simplify each of the following expressions. Be sure that your answer has no nega
Simplify each of the following expressions. Be sure that your answer has no negative or fractional exponents. a*(1/81)^(-1/4)b*x^(-2)y^(-4)c*(2x)^(-2)(16x^2y)^(1/2)
Simplify each of the following expressions. Be sure that your answer has no negative or fractional exponents.
a\cdot \left(\frac{1}{81}{\right)}^{-\frac{1}{4}}b\cdot {x}^{-2}{y}^{-4}c\cdot \left(2x{\right)}^{-2}\left(16{x}^{2y}{\right)}^{\frac{1}{2}}
{\frac{1}{81}}^{-\frac{1}{4}}
=\left(\frac{1}{3}{\right)}^{-\frac{1}{4}}
=\left(\frac{1}{3}\right)
(product of power property)
\left(\frac{1}{3}{\right)}^{-1}=3
{x}^{2}{y}^{-}4
={x}^{-2}\cdot {y}^{-4}
\frac{1}{{x}^{2}}\cdot \frac{1}{{y}^{4}}
(negative power property)
\left(2x{\right)}^{-2}\left(16{x}^{2y}{\right)}^{\frac{1}{2}}
={2}^{-2}\cdot {x}^{-2}\cdot {2}^{\frac{4}{2}}\cdot {x}^{\frac{2}{2}}\cdot {y}^{\frac{1}{2}}
={2}^{\left(2-2\right)}\cdot {x}^{\left(1-2\right)}\cdot {y}^{\frac{1}{2}}
=1\cdot {x}^{-1}\cdot y\left(\frac{1}{2}\right)
2\sqrt{2}+7\sqrt{2}=
Simplify and express the final result using positive exponents.
{\left(\frac{8{y}^{2}}{2{y}^{-1}}\right)}^{-1}
\sqrt[3]{3}\sqrt{3}
7\sqrt{2}-6\sqrt{2}=
Solve for the following exponential equations. Use the natural logarithm in your answer(where applicable) for full credit. Use rules for exponents, factor and simplify.
{4}^{1-x}={3}^{2x+5}
Find the rectangular coordinates of the pair of points
\left(2,2\frac{\pi }{3}\right)
\left(4,\frac{\pi }{6}\right)
. Then find the distance, in simplified radical form, between the points.
If a bank pays interest at a rate of i compounded m times a year, then the amount of money
{P}_{k}
k
time periods (where one time period
=\frac{1}{m}th
of a year) satisfies the recurrence relation
{P}_{k}=\left[1+\frac{i}{m}\right]Pk-1
{P}_{0}
= the initial amount deposited. Find an explicit formula for
{P}_{n}.
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Emission of Protons and Charged Pions in p + Cu and p + Pb Collisions at 3, 8, and 15 GeV/c
J. H. Kang, Y. C. Qian, B. C. Li, S. W. Wu, "Emission of Protons and Charged Pions in p + Cu and p + Pb Collisions at 3, 8, and 15 GeV/c", Advances in High Energy Physics, vol. 2014, Article ID 675192, 10 pages, 2014. https://doi.org/10.1155/2014/675192
J. H. Kang,1 Y. C. Qian,2 B. C. Li,2 and S. W. Wu2
We present an analysis of proton and charged pion transverse momentum spectra of and reactions at 3, 8, and 15 GeV/c in the framework of a multisource thermal model. The spectra are compared closely with the experimental data of HARP-CDP at all angular intervals. The result shows that the widths of the particle distributions in both and collisions decrease with increasing the angle for the same incident momentum.
The relativistic heavy ion collider (RHIC) [1, 2] in the United States and the large hadron collider (LHC) [3] in Switzerland have been built, respectively. Much higher energy collisions can lead to a new significant extension of the kinematic range in transverse momentum. The collisions bring valuable information of quark gluon plasma (QGP) at high energy due to high temperature and density [4, 5]. The state is a thermalized system consisting of strong coupled quarks and gluons in a very small region. This matter is only created for the briefest of instants, and then the fireball cools down and hadronizes into hadrons. So, we cannot observe the QGP directly in the existing laboratory conditions. However, we can extract a judgment of the creation of the quark matter by measuring and analyzing the spectra of identified particles produced after thermal freeze-out in heavy ion collisions.
In a very early stage of the collision, the energy density is expected to be sufficient to dissolve normal nuclear matter into a phase of quark matter, which exists for only a short time before the fireball cools down and the process of hadronization takes place. High-energy collisions provide an excellent probe of the quark matter. Properties of QGP will be probed further in LHC at the European Organization for Nuclear Research (CERN) [6]. The proton-nucleus collisions are important in experimental programmes performed in the LHC [7] because they not only provide baseline measurements for the nucleus-nucleus collisions but also help us better understand fundamental features of quantum chromodynamics (QCD) [8]. The transverse momentum spectra of final-state particles can give some important information of the matter created in high-energy collisions. In this paper, we use a multisource thermal model to study transverse momentum spectra of protons and charged pions produced in and collisions at 3, 8, and 15 GeV/c, recently measured in the Hadron Production Experiment (HARP) at the CERN [9].
2. Distribution of Transverse Momentum
In a framework of the multisource thermal model [10, 11], identified fragments or particles emit isotropically from different emission sources in the collisions. In order to deal conveniently with the relation between the sources and particles, we split the sources into groups in accordance with kinetic laws and geometrical positions. For the source in the group, its share of the transverse momentum spectrum of the final-state particles is given by where is the mean value of the transverse momentum which comes from the source in the group. According to statistical properties of the model, we have indicates the source number in the group. By computing convolution of the exponential functions equation (1), the total share of the group for the transverse momentum distribution is obtained as It is an Erlang distribution. Then, the transverse momentum distribution from the groups is given by where characterizes how large the contribution of the sources in the group is. Equation (5) is known as a multicomponent Erlang distribution. To simplify the calculation, the Monte Carlo method is used to calculate the transverse momentum spectra. With (1) and (4), the transverse momentum distribution is where is a random number in .
3. Comparison with HARP Results
Figures 1, 2, and 3 present the transverse momentum spectra of , , and in collisions at 3, 8, and 15 GeV/c at different angular intervals, respectively. From the first column to the third column in the figures, the momenta of incident protons are 3, 8, and 15 GeV/c, respectively. And from the first row to the third row in the figures, the angular intervals are 30°–40°, 60°–75°, and 105°–125°, separately. The symbols indicate the experimental data [9] in the HARP experiment at the CERN. The solid lines are the results of the multisource thermal model. The parameters and per degree of freedom () are given in Tables 1 and 2. In the 30°–40° angular interval, the value of for the production at 8 GeV/c incident momentum is equal to that at the 15 GeV/c. Moreover, for , the at 8 GeV/c and 15 GeV/c are the same in the three angular intervals. So is it for . In the calculation, one group with two sources is selected. We see that the model can successfully describe the experimental data. From the figures and the tables, it is also found that the width or the mean contribution of the distribution decreases with increasing the angular intervals for the same incident momentum.
(a1) 0.19 1 0.03 (a1) 0.09 1 0.19
(b1) 0.15 1 0.20 (b1) 0.09 1 0.54
(c1) 0.08 1 0.14 (c1) 0.05 1 0.83
Values of , taken in the fits of Figures 1 and 2.
(c2) 0.08 1 0.85 (c2) 0.09 1 —
Transverse momentum of protons produced in collisions at 3, 8, and 15 GeV/c. The angular bins are 30°–40°, 60°–75°, and 105°–125°, respectively. The circles indicate the experimental data [9] and the curves indicate the results of the model.
The same as Figure 1, but for .
Figures 4, 5, and 6 show the transverse momentum spectra of , , and in interactions at 3, 8, and 15 GeV/c with different angular bins. The symbols indicate the HARP-CDP experimental data [9] and the solid lines indicate the results of the multisource thermal model. The results of (7) are in agreement with the experimental data. The corresponding parameters and are listed in Tables 2 and 3. At forward angles θ = 30°–40°, the values of are the same for proton production at 3 GeV/c incident momentum and for the 15 GeV/c. For production, the at 8 GeV/c and 15 GeV/c are the same at the angles of 60°–75° and 105°–125°. Like the case of collisions, we still use a single group with two sources and the results agree well with the experimental data. At the same incident momentum, the decreases with the increase of the forward angles.
In the multisource thermal model, the transverse momentum spectra of protons and charged pions produced in protons on Cu and Pb collisions at 3, 8, and 15 GeV/c in fixed angles of 30°–40°, 60°–75°, and 105°–125° are discussed. The spectra of the model are in agreement with the HARP-CDP data. The maximum value of is 1.73, and the minimum value is 0.02. From the above discussions, it is seen that the distribution widths of the concerned particles in both and reactions decrease with increasing the angular intervals for the same incident momentum.
In the work, two sources in one group are used to study the experimental data. It implies that the final-state particles emit from two sources. The model can provide an explanation not only for one group but also for more groups. In the previous work, the model was used to investigateelliptic flows [10], particle production [11, 12], longitudinal shift in (pseudo) rapidity distributions [13, 14], and so forth. For various types of collision systems, the multicomponent Erlang distribution can fit the transverse momentum spectra. The work reveals a multisource production phenomenon in the heavy ion collisions.
This work is supported by the National Natural Science Foundation of China under Grant nos. 11247250 and 11005071 and the National Fundamental Fund of Personnel Training (no. J1103210).
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\sqrt{{\mathbit{s}}_{\mathbit{N}\mathbit{N}}}=130
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{\mathbit{K}}_{\mathbit{S}}^{0}
\mathrm{\Lambda }
Production in Pb-Pb Collisions at
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Copyright © 2014 J. H. Kang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. The publication of this article was funded by SCOAP3.
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A pair of slits, separated by 0.150 mm, is illuminated by light having a wavelen
A pair of slits, separated by 0.150 mm, is illuminated by light having a wavelength of ? = 561 nm. An interference pattern is observed on a screen 122
A pair of slits, separated by 0.150 mm, is illuminated by light having a wavelength of ? = 561 nm. An interference pattern is observed on a screen 122 cm from the slits. Consider a point on the screen located at y = 2.00 cm from the central maximum of this pattern.
(a) What is the path difference ? for the two slits at the location y?
(b) Express this path difference in terms of the wavelength.
our formula for double slit interference is:
d\mathrm{sin}\left(t\right)=m\left(w\right)
where d is the width of the slits
where t is the angle from the slits
where m is the corresponding fringe from the central fringe
where w is the wavelength
using trig, we can find
\mathrm{sin}\left(t\right)
\mathrm{sin}\left(t\right)=\frac{y}{L}
where y is the distance from central fringe to corresponding fringe
where L is the distance from slit to screen
our formula now becomes:
d\left(\frac{y}{L}\right)=m\left(w\right)
where both sides represent the path difference:
path difference=d(y/L)
path difference = .150e-3(2e-2/122e-2)
path difference = 2.459e-6 m
In terms of the wavelength:
2.459e-6/561e-9 = 4.4w
A 1.0 kg ball and a 2.0 kg ball are connected by a 1.0-m-long rigid, massless rod. The rod is rotating cw about its center of mass at 20 rpm. What torque will bring the balls to a halt in 5.0 s? Answer in Nm.
\frac{m}{{s}^{2}}
\frac{m}{{s}^{2}}
Unpolarized light with intensity
{I}_{0}
is incident on two polarizing filters. The axis of the first filter makes an angle of
{60.0}^{\circ }
with the vertical, and the axis of the second filter is horizontal. What is the intensity of the light after it has passed through the second filter?
Give correct answer when have two random variables, X and Y. X is the value of a fair die, Y is the result of a coin flip, with heads being 1 and tails being 0.
E\left[X\right]=\sum _{k=1}^{6}\frac{k}{6}=\frac{7}{2},\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}E\left[Y\right]=\frac{1}{2}
E\left[X\right]E\left[Y\right]=\frac{7}{4}
That the expectation of XY is not multiplicative, i.e.: E[X]E[Y] is not necessarily equal to E[XY]. But I'm confused about what E[XY] means in the first place. That is, is E[XY] each possible value of the two events combined, multiplied by the probability that the two events occur?
That is, is
E\left[XY\right]=\frac{1}{2}\left(1\right)\sum _{k=1}^{6}\frac{k}{6}+\frac{1}{2}\left(0\right)\sum _{k=1}^{6}\frac{k}{6}=\frac{7}{4}?
If not, what is it?
Due to a typo the last equation
\left[XY\right]=\frac{1}{2}\left(1\right)\sum _{k=1}^{6}\frac{k}{6}+\frac{1}{2}\left(0\right)\sum _{k=1}^{6}\frac{k}{6}
was evaluated as
\frac{7}{2},when\text{ }I\text{ }believe\text{ }it\text{ }shoud\text{ }be\text{ }\frac{7}{4}
. I'd appreciate it if responders told me whether this is a correct value for E[XY].
That concerned with how to calculate E[XY] in the quickest way possible, but how to interpret what E[XY] means. E[X]E[Y]=E[XY] for independent events doesn't concern me as much as why that is the case, and how to manually evaluate E[XY] in order to prove that indeed E[XY]=E[X]E[Y].
How many grams of cadmium are deposited from an aqueoussolution of cadmium sulfate
CdS{O}_{4}
, when an electriccurrent of 1.51 amps flows through the solution for 156 minutes?
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The integral represents the volume of a solid. Describe the solid. \pi\int_{0}^{1}(y^{4}-
The integral represents the volume of a solid. Describe the solid. \pi\int_{0}^{1}(y^{4}-y^{8})dy a) The integral describes the volume of the solid ob
\pi {\int }_{0}^{1}\left({y}^{4}-{y}^{8}\right)dy
a) The integral describes the volume of the solid obtained by rotating the region
R=\left\{\left\{x,\text{ }y\right\}|0\le y\le 1,\text{ }{y}^{4}\le x\le {y}^{2}\right\}
of the xy-plane about the x-axis.
b) The integral describes the volume of the solid obtained by rotating the region
R=\left\{\left\{x,\text{ }y\right\}|0\le y\le 1,\text{ }{y}^{2}\le x\le {y}^{4}\right\}
c) The integral describes the volume of the solid obtained by rotating the region
R=\left\{\left\{x,\text{ }y\right\}|0\le y\le 1,\text{ }{y}^{4}\le x\le {y}^{2}\right\}
of the xy-plane about the y-axis.
d) The integral describes the volume of the solid obtained by rotating the region
R=\left\{\left\{x,\text{ }y\right\}|0\le y\le 1,\text{ }{y}^{2}\le x\le {y}^{4}\right\}
e) The integral describes the volume of the solid obtained by rotating the region
R=\left\{\left\{x,\text{ }y\right\}|0\le y\le 1,\text{ }{y}^{4}\le x\le {y}^{8}\right\}
Yusuf Keller
\pi {\int }_{0}^{1}\left({y}^{4}-{y}^{8}\right)\text{ }dy=1
d1=\pi \left[\left({y}^{2}{\right)}^{2}-\left({y}^{4}{\right)}^{2}\right]\text{ }dy
Axis of solution
=y-axis
x={y}^{2}
x={y}^{4}
region in x-y plane
{y}^{4}<x<{y}^{2}
0\le y\le 1
x={t}^{2},y=2t,0\le t\le 5
\left({x}^{2}+2xy-4{y}^{2}\right)dx-\left({x}^{2}-8xy-4{y}^{2}\right)dy=0
The velocity function (in meters per second) is given for a particle moving along a line.
v\left(t\right)=3t-8,0\le t\le 3
(a) Find the displacement.
(b) Find the distance traveled by the particle during the given time interval.
Evaluate the iterated integral by converting to polar coordinates.
{\int }_{0}^{2}{\int }_{0}^{\sqrt{2x-{x}^{2}}}xydydx
\left(\frac{x}{y}\right)ds,C:x={t}^{3},y={t}^{4},1\le t\le 2
\frac{5{x}^{3}}{2}-\frac{1}{2{x}^{2}}+\frac{3x}{2}+\mathrm{cos}\left(2x\right)-2\mathrm{cos}\left(3x\right)
\int {\int }_{S}\left({x}^{2}z+{y}^{2}z\right)ds
where S is the hemisphere
{x}^{2}+{y}^{2}+{z}^{2}=9,z\ge 0
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Debye length - Wikipedia
Thickness of the electrical double layer (EDL)
In plasmas and electrolytes, the Debye length
{\displaystyle \lambda _{\rm {D}}}
(also called Debye radius), is a measure of a charge carrier's net electrostatic effect in a solution and how far its electrostatic effect persists.[1] With each Debye length the charges are increasingly electrically screened and the electric potential decreases in magnitude by 1/e. A Debye sphere is a volume whose radius is the Debye length. Debye length is an important parameter in plasma physics, electrolytes, and colloids (DLVO theory). The corresponding Debye screening wave vector
{\displaystyle k_{\rm {D}}=1/\lambda _{\rm {D}}}
for particles of density
{\displaystyle n}
{\displaystyle q}
at a temperature
{\displaystyle T}
{\displaystyle k_{\rm {D}}^{2}=4\pi nq^{2}/(k_{\rm {B}}T)}
in Gaussian units. Expressions in MKS units will be given below. The analogous quantities at very low temperatures (
{\displaystyle T\to 0}
) are known as the Thomas–Fermi length and the Thomas–Fermi wave vector. They are of interest in describing the behaviour of electrons in metals at room temperature.
The Debye length is named after the Dutch-American physicist and chemist Peter Debye (1884-1966), a Nobel laureate in Chemistry.
2 In a plasma
3 In an electrolyte solution
4 In semiconductors
Physical origin[edit]
The Debye length arises naturally in the thermodynamic description of large systems of mobile charges. In a system of
{\displaystyle N}
different species of charges, the
{\displaystyle j}
-th species carries charge
{\displaystyle q_{j}}
and has concentration
{\displaystyle n_{j}(\mathbf {r} )}
{\displaystyle \mathbf {r} }
. According to the so-called "primitive model", these charges are distributed in a continuous medium that is characterized only by its relative static permittivity,
{\displaystyle \varepsilon _{r}}
. This distribution of charges within this medium gives rise to an electric potential
{\displaystyle \Phi (\mathbf {r} )}
that satisfies Poisson's equation:
{\displaystyle \varepsilon \nabla ^{2}\Phi (\mathbf {r} )=-\,\sum _{j=1}^{N}q_{j}\,n_{j}(\mathbf {r} )-\rho _{\rm {ext}}(\mathbf {r} ),}
{\displaystyle \varepsilon \equiv \varepsilon _{r}\varepsilon _{0}}
{\displaystyle \varepsilon _{0}}
is the electric constant, and
{\displaystyle \rho _{\rm {ext}}}
is a charge density external (logically, not spatially) to the medium.
The mobile charges not only contribute in establishing
{\displaystyle \Phi (\mathbf {r} )}
but also move in response to the associated Coulomb force,
{\displaystyle -q_{j}\,\nabla \Phi (\mathbf {r} )}
. If we further assume the system to be in thermodynamic equilibrium with a heat bath at absolute temperature
{\displaystyle T}
, then the concentrations of discrete charges,
{\displaystyle n_{j}(\mathbf {r} )}
, may be considered to be thermodynamic (ensemble) averages and the associated electric potential to be a thermodynamic mean field. With these assumptions, the concentration of the
{\displaystyle j}
-th charge species is described by the Boltzmann distribution,
{\displaystyle n_{j}(\mathbf {r} )=n_{j}^{0}\,\exp \left(-{\frac {q_{j}\,\Phi (\mathbf {r} )}{k_{\rm {B}}T}}\right),}
{\displaystyle k_{\rm {B}}}
is Boltzmann's constant and where
{\displaystyle n_{j}^{0}}
is the mean concentration of charges of species
{\displaystyle j}
Identifying the instantaneous concentrations and potential in the Poisson equation with their mean-field counterparts in Boltzmann's distribution yields the Poisson–Boltzmann equation:
{\displaystyle \varepsilon \nabla ^{2}\Phi (\mathbf {r} )=-\,\sum _{j=1}^{N}q_{j}n_{j}^{0}\,\exp \left(-{\frac {q_{j}\,\Phi (\mathbf {r} )}{k_{\rm {B}}T}}\right)-\rho _{\rm {ext}}(\mathbf {r} ).}
Solutions to this nonlinear equation are known for some simple systems. Solutions for more general systems may be obtained in the high-temperature (weak coupling) limit,
{\displaystyle q_{j}\,\Phi (\mathbf {r} )\ll k_{\rm {B}}T}
, by Taylor expanding the exponential:
{\displaystyle \exp \left(-{\frac {q_{j}\,\Phi (\mathbf {r} )}{k_{\rm {B}}T}}\right)\approx 1-{\frac {q_{j}\,\Phi (\mathbf {r} )}{k_{\rm {B}}T}}.}
This approximation yields the linearized Poisson-Boltzmann equation
{\displaystyle \varepsilon \nabla ^{2}\Phi (\mathbf {r} )=\left(\sum _{j=1}^{N}{\frac {n_{j}^{0}\,q_{j}^{2}}{k_{\rm {B}}T}}\right)\,\Phi (\mathbf {r} )-\,\sum _{j=1}^{N}n_{j}^{0}q_{j}-\rho _{\rm {ext}}(\mathbf {r} )}
which also is known as the Debye–Hückel equation:[2][3][4][5][6] The second term on the right-hand side vanishes for systems that are electrically neutral. The term in parentheses divided by
{\displaystyle \varepsilon }
, has the units of an inverse length squared and by dimensional analysis leads to the definition of the characteristic length scale
{\displaystyle \lambda _{\rm {D}}=\left({\frac {\varepsilon \,k_{\rm {B}}T}{\sum _{j=1}^{N}n_{j}^{0}\,q_{j}^{2}}}\right)^{1/2}}
that commonly is referred to as the Debye–Hückel length. As the only characteristic length scale in the Debye–Hückel equation,
{\displaystyle \lambda _{D}}
sets the scale for variations in the potential and in the concentrations of charged species. All charged species contribute to the Debye–Hückel length in the same way, regardless of the sign of their charges. For an electrically neutral system, the Poisson equation becomes
{\displaystyle \nabla ^{2}\Phi (\mathbf {r} )=\lambda _{\rm {D}}^{-2}\Phi (\mathbf {r} )-{\frac {\rho _{\rm {ext}}(\mathbf {r} )}{\varepsilon }}}
To illustrate Debye screening, the potential produced by an external point charge
{\displaystyle \rho _{\rm {ext}}=Q\delta (\mathbf {r} )}
{\displaystyle \Phi (\mathbf {r} )={\frac {Q}{4\pi \varepsilon r}}e^{-r/\lambda _{\rm {D}}}}
The bare Coulomb potential is exponentially screened by the medium, over a distance of the Debye length: this is called Debye screening or shielding (Screening effect).
The Debye–Hückel length may be expressed in terms of the Bjerrum length
{\displaystyle \lambda _{\rm {B}}}
{\displaystyle \lambda _{\rm {D}}=\left(4\pi \,\lambda _{\rm {B}}\,\sum _{j=1}^{N}n_{j}^{0}\,z_{j}^{2}\right)^{-1/2},}
{\displaystyle z_{j}=q_{j}/e}
is the integer charge number that relates the charge on the
{\displaystyle j}
-th ionic species to the elementary charg{\displaystyle e}
In a plasma[edit]
For a weakly collisional plasma, Debye shielding can be introduced in a very intuitive way by taking into account the granular character of such a plasma. Let us imagine a sphere about one of its electrons, and compare the number of electrons crossing this sphere with and without Coulomb repulsion. With repulsion, this number is smaller. Therefore, according to Gauss theorem, the apparent charge of the first electron is smaller than in the absence of repulsion. The larger the sphere radius, the larger is the number of deflected electrons, and the smaller the apparent charge: this is Debye shielding. Since the global deflection of particles includes the contributions of many other ones, the density of the electrons does not change, at variance with the shielding at work next to a Langmuir probe (Debye sheath). Ions bring a similar contribution to shielding, because of the attractive Coulombian deflection of charges with opposite signs.
This intuitive picture leads to an effective calculation of Debye shielding (see section II.A.2 of [7]). The assumption of a Boltzmann distribution is not necessary in this calculation: it works for whatever particle distribution function. The calculation also avoids approximating weakly collisional plasmas as continuous media. An N-body calculation reveals that the bare Coulomb acceleration of a particle by another one is modified by a contribution mediated by all other particles, a signature of Debye shielding (see section 8 of [8]). When starting from random particle positions, the typical time-scale for shielding to set in is the time for a thermal particle to cross a Debye length, i.e. the inverse of the plasma frequency. Therefore in a weakly collisional plasma, collisions play an essential role by bringing a cooperative self-organization process: Debye shielding. This shielding is important to get a finite diffusion coefficient in the calculation of Coulomb scattering (Coulomb collision).
In a non-isothermic plasma, the temperatures for electrons and heavy species may differ while the background medium may be treated as the vacuum (
{\displaystyle \varepsilon _{r}=1}
), and the Debye length is
{\displaystyle \lambda _{\rm {D}}={\sqrt {\frac {\varepsilon _{0}k_{\rm {B}}/q_{e}^{2}}{n_{e}/T_{e}+\sum _{j}z_{j}^{2}n_{j}/T_{i}}}}}
λD is the Debye length,
qe is the charge of an electron,
Te and Ti are the temperatures of the electrons and ions, respectively,
ne is the density of electrons,
nj is the density of atomic species j, with positive ionic charge zjqe
Even in quasineutral cold plasma, where ion contribution virtually seems to be larger due to lower ion temperature, the ion term is actually often dropped, giving
{\displaystyle \lambda _{\rm {D}}={\sqrt {\frac {\varepsilon _{0}k_{\rm {B}}T_{e}}{n_{e}q_{e}^{2}}}}}
although this is only valid when the mobility of ions is negligible compared to the process's timescale.[9]
In space plasmas where the electron density is relatively low, the Debye length may reach macroscopic values, such as in the magnetosphere, solar wind, interstellar medium and intergalactic medium. See the table here below:[10]
ne(m−3)
1032 107 — 10−11
1020 108 10 10−4
1016 104 — 10−4
1012 103 10−5 10−3
107 107 10−8 102
106 105 10−9 10
105 104 10−10 10
In an electrolyte solution[edit]
In an electrolyte or a colloidal suspension, the Debye length[11][12][13] for a monovalent electrolyte is usually denoted with symbol κ−1
{\displaystyle \kappa ^{-1}={\sqrt {\frac {\varepsilon _{\rm {r}}\varepsilon _{0}k_{\rm {B}}T}{2e^{2}I}}}}
I is the ionic strength of the electrolyte in mol/m3 units,
εr is the dielectric constant,
T is the absolute temperature in kelvins,
{\displaystyle e}
or, for a symmetric monovalent electrolyte,
{\displaystyle \kappa ^{-1}={\sqrt {\frac {\varepsilon _{\rm {r}}\varepsilon _{0}RT}{2\times 10^{3}F^{2}C_{0}}}}}
F is the Faraday constant,
C0 is the electrolyte concentration in molar units (M or mol/L).
{\displaystyle \kappa ^{-1}={\frac {1}{\sqrt {8\pi \lambda _{\rm {B}}N_{\rm {A}}\times 10^{-24}I}}}}
{\displaystyle \lambda _{\rm {B}}}
is the Bjerrum length of the medium, and the factor
{\displaystyle 10^{-24}}
derives from transforming unit volume from cubic dm to cubic nm.
For water at room temperature, λB ≈ 0.7 nm.
At room temperature (20 °C or 70 °F), one can consider in water the relation:[14]
{\displaystyle \kappa ^{-1}(\mathrm {nm} )={\frac {0.304}{\sqrt {I(\mathrm {M} )}}}}
κ−1 is expressed in nanometres (nm)
I is the ionic strength expressed in molar (M or mol/L)
There is a method of estimating an approximate value of the Debye length in liquids using conductivity, which is described in ISO Standard,[11] and the book.[12]
In semiconductors[edit]
The Debye length has become increasingly significant in the modeling of solid state devices as improvements in lithographic technologies have enabled smaller geometries.[15][16][17]
The Debye length of semiconductors is given:
{\displaystyle L_{\rm {D}}={\sqrt {\frac {\varepsilon k_{\rm {B}}T}{q^{2}N_{\rm {dop}}}}}}
ε is the dielectric constant,
kB is the Boltzmann's constant,
q is the elementary charge, and
Ndop is the net density of dopants (either donors or acceptors).
When doping profiles exceed the Debye length, majority carriers no longer behave according to the distribution of the dopants. Instead, a measure of the profile of the doping gradients provides an "effective" profile that better matches the profile of the majority carrier density.
In the context of solids, the Debye length is also called the Thomas–Fermi screening length.
Debye–Falkenhagen effect
^ Debye, P.; Hückel, E. (2019) [1923]. Translated by Braus, Michael J. "Zur Theorie der Elektrolyte. I. Gefrierpunktserniedrigung und verwandte Erscheinungen" [The theory of electrolytes. I. Freezing point depression and related phenomenon]. Physikalische Zeitschrift. 24 (9): 185–206.
^ Kirby, B. J. (2010). Micro- and Nanoscale Fluid Mechanics: Transport in Microfluidic Devices. New York: Cambridge University Press. ISBN 978-0-521-11903-0.
^ Li, D. (2004). Electrokinetics in Microfluidics. Academic Press. ISBN 0-12-088444-5.
^ PC Clemmow & JP Dougherty (1969). Electrodynamics of particles and plasmas. Redwood City CA: Addison-Wesley. pp. § 7.6.7, p. 236 ff. ISBN 978-0-201-47986-7.
^ RA Robinson &RH Stokes (2002). Electrolyte solutions. Mineola, NY: Dover Publications. p. 76. ISBN 978-0-486-42225-1.
^ See Brydges, David C.; Martin, Ph. A. (1999). "Coulomb Systems at Low Density: A Review". Journal of Statistical Physics. 96 (5/6): 1163–1330. arXiv:cond-mat/9904122. Bibcode:1999JSP....96.1163B. doi:10.1023/A:1004600603161. S2CID 54979869.
^ Meyer-Vernet N (1993) Aspects of Debye shielding. American journal of physics 61, 249-257
^ Escande, D. F., Bénisti, D., Elskens, Y., Zarzoso, D., & Doveil, F. (2018). Basic microscopic plasma physics from N-body mechanics, A tribute to Pierre-Simon de Laplace, Reviews of Modern Plasma Physics, 2, 1-68
^ I. H. Hutchinson Principles of plasma diagnostics ISBN 0-521-38583-0
^ Kip Thorne (2012). "Chapter 20: The Particle Kinetics of Plasma" (PDF). Applications of Classical Physics. Retrieved September 7, 2017.
^ a b International Standard ISO 13099-1, 2012, "Colloidal systems – Methods for Zeta potential determination- Part 1: Electroacoustic and Electrokinetic phenomena"
^ a b Dukhin, A. S.; Goetz, P. J. (2017). Characterization of liquids, nano- and micro- particulates and porous bodies using Ultrasound. Elsevier. ISBN 978-0-444-63908-0.
^ Russel, W. B.; Saville, D. A.; Schowalter, W. R. (1989). Colloidal Dispersions. Cambridge University Press. ISBN 0-521-42600-6.
^ Israelachvili, J. (1985). Intermolecular and Surface Forces. Academic Press. ISBN 0-12-375181-0.
^ Stern, Eric; Robin Wagner; Fred J. Sigworth; Ronald Breaker; Tarek M. Fahmy; Mark A. Reed (2007-11-01). "Importance of the Debye Screening Length on Nanowire Field Effect Transistor Sensors". Nano Letters. 7 (11): 3405–3409. Bibcode:2007NanoL...7.3405S. doi:10.1021/nl071792z. PMC 2713684. PMID 17914853.
^ Guo, Lingjie; Effendi Leobandung; Stephen Y. Chou (199). "A room-temperature silicon single-electron metal–oxide–semiconductor memory with nanoscale floating-gate and ultranarrow channel". Applied Physics Letters. 70 (7): 850. Bibcode:1997ApPhL..70..850G. doi:10.1063/1.118236.
^ Tiwari, Sandip; Farhan Rana; Kevin Chan; Leathen Shi; Hussein Hanafi (1996). "Single charge and confinement effects in nano-crystal memories". Applied Physics Letters. 69 (9): 1232. Bibcode:1996ApPhL..69.1232T. doi:10.1063/1.117421.
Goldston & Rutherford (1997). Introduction to Plasma Physics. Philadelphia: Institute of Physics Publishing.
Lyklema (1993). Fundamentals of Interface and Colloid Science. NY: Academic Press.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Debye_length&oldid=1082041542"
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A Gauss-Bonnet Formula for Metrics with Varying Signature | EMS Press
A Gauss-Bonnet Formula for Metrics with Varying Signature
A Gauss-Bonnet formula for compact orientable connected Riemannian or Lorentz\-ian 2-manifolds is well-known. We investigate singular metrics on 2-manifolds with varying signature. Such metrics are necessarily degenerate at some points of
M
where most of the usual definitions for geometric quantities break down. We prove that under some additional assumptions there is a Gauss--Bonnet formula for compact orientable connected 2-manifolds with a singular metric. Some examples are given.
Michael Steller, A Gauss-Bonnet Formula for Metrics with Varying Signature. Z. Anal. Anwend. 25 (2006), no. 2, pp. 143–162
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Explain the importance of the statement "Sampling distributions
Explain the importance of the statement "Sampling distributions play a key role in the process of statistical interference" stated by the researchers Turner and Dabney.
The probability distribution of the sample statistics when all the possible samples are drawn over the given population is termed as sampling distribution.
Justification: The sampling distribution can be defined as the probability distribution of the sample statistics that is sample mean or the sample variances that can be obtained from the samples of a given size from the population given. Therefore, it is clear that Sampling distributions play a key role in the process of statistical inference.
An underground military installation is fortified to the extent that it can withstand up to three
direct hits from air-to-surface missiles and still function. Suppose an enemy aircraft is armed
with missiles, each having a 30 % chance of scoring a direct hit.
i. What is the probability that the installation will be destroyed with the seventh missile fired
is obtained by fewer than nine air strikes?
ii. Find the expected air-to-surface missiles strike.
Obtain the sampling distribution of
\stackrel{―}{p}
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Model zero-mean normally (Gaussian) distributed force - MATLAB - MathWorks Nordic
Force Noise Source
Model zero-mean normally (Gaussian) distributed force
The Force Noise Source block generates zero-mean normally (Gaussian) distributed force using the Random Number source in the Simscape™ Foundation library. The force produced by the block depends on two factors:
The defining equation for the force that the block generates is
F=\sqrt{PSD/2}\frac{N\left(0,1\right)}{\sqrt{h}},
F is the output force.
The power spectral density (PSD) of noise is the average noise power per unit of bandwidth, in N2/Hz.
When you add a Force Noise Source block to your model from the Sources library, the block generates and stores a random value for the repeated seed. When you make a copy of the Force Noise Source block from an existing block in a model, the copy generates a new random value for the repeated seed.
Random Number | Rotational Velocity Noise Source | Sinusoidal Force Source | Torque Noise Source | Translational Velocity Noise Source
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Fit simple model of local interpretable model-agnostic explanations (LIME) - MATLAB fit - MathWorks América Latina
Fit Simple Models for Multiple Query Points
SimpleModelType
Fit simple model of local interpretable model-agnostic explanations (LIME)
newresults = fit(results,queryPoint,numImportantPredictors)
newresults = fit(results,queryPoint,numImportantPredictors,Name,Value)
newresults = fit(results,queryPoint,numImportantPredictors) fits a new simple model for the specified query point (queryPoint) by using the specified number or predictors (numImportantPredictors). The function returns a lime object newresults that contains the new simple model.
fit uses the simple model options that you specify when you create the lime object results. You can change the options using the name-value pair arguments of the fit function.
newresults = fit(results,queryPoint,numImportantPredictors,Name,Value) specifies additional options using one or more name-value pair arguments. For example, you can specify 'SimpleModelType','tree' to fit a decision tree model.
Train a classification model and create a lime object that uses a decision tree simple model. Fit multiple models for multiple query points.
blackbox = fitcecoc(tblX,tbl.Rating,'CategoricalPredictors','Industry')
ClassNames: {'A' 'AA' 'AAA' 'B' 'BB' 'BBB' 'CCC'}
Create a lime object with the blackbox model.
results = lime(blackbox);
queryPoint(1,:) = tblX(find(strcmp(tbl.Rating,'AAA'),1),:);
queryPoint(2,:) = tblX(find(strcmp(tbl.Rating,'B'),1),:)
Fit a linear simple model for the first query point. Set the number of important predictors to 4.
newresults1 = fit(results,queryPoint(1,:),4);
Plot the LIME results newresults1 for the first query point. To display an existing underscore in any predictor name, change the TickLabelInterpreter value of the axes to 'none'.
f1 = plot(newresults1);
Fit a linear decision tree model for the first query point.
newresults2 = fit(results,queryPoint(1,:),6,'SimpleModelType','tree');
The simple models in newresults1 and newresults2 both find MVE_BVTD and RE_TA as important predictors.
Fit a linear simple model for the second query point, and plot the LIME results for the second query point.
The prediction from the blackbox model is B, but the prediction from the simple model is not B. When the two predictions are not the same, you can specify a smaller 'KernelWidth' value. The software fits a simple model using weights that are more focused on the samples near the query point. If a query point is an outlier or is located near a decision boundary, then the two prediction values can be different, even if you specify a small 'KernelWidth' value. In such a case, you can change other name-value pair arguments. For example, you can generate a local synthetic data set (specify 'DataLocality' of lime as 'local') for the query point and increase the number of samples ('NumSyntheticData' of lime or fit) in the synthetic data set. You can also use a different distance metric ('Distance' of lime or fit).
Fit a linear simple model with a small 'KernelWidth' value.
newresults4 = fit(results,queryPoint(2,:),4,'KernelWidth',0.01);
The credit ratings for the first and second query points are AAA and B, respectively. The simple models in newresults1 and newresults4 both find MVE_BVTD, RE_TA, and WC_TA as important predictors. However, their coefficient values are different. The plots show that these predictors act differently depending on the credit ratings.
LIME results, specified as a lime object.
Query point around which the fit function fits the simple model, specified as a row vector of numeric values or a single-row table. The queryPoint value must have the same data type and the same number of columns as the predictor data (results.X or results.SyntheticData) in the lime object results.
If 'SimpleModelType' is 'linear', then the software selects the specified number of important predictors and fits a linear model of the selected predictors.
The default value of the numImportantPredictors argument is the NumImportantPredictors property value of the lime object results. If you do not specify the property value when creating results, then the property value is empty ([]) and you must specify this argument.
Example: 'NumSyntheticData',2000,'SimpleModelType','tree' sets the number of samples to generate for the synthetic data set to 2000 and specifies the simple model type as a decision tree.
The default value is the 'Cov' value that you specify when creating the lime object results. The default 'Cov' value of lime is cov(PD,'omitrows'), where PD is the predictor data or synthetic predictor data. If you do not specify the 'Cov' value, then the software uses different covariance matrices when computing the distances for both the predictor data and the synthetic predictor data.
If the predictor data includes only continuous variables, then fit supports these distance metrics.
If the predictor data includes both continuous and categorical variables, then fit supports these distance metrics.
The default value is the 'Distance' value that you specify when creating the lime object results. The default 'Distance' value of lime is 'euclidean' if the predictor data includes only continuous variables, or 'goodall3' if the predictor data includes both continuous and categorical variables.
The fit function computes distances between the query point and the samples in the synthetic predictor data set, and then converts the distances to weights by using the squared exponential kernel function. If you lower the 'KernelWidth' value, then fit uses weights that are more focused on the samples near the query point. For details, see LIME.
The default value is the 'KernelWidth' value that you specify when creating the lime object results. The default 'KernelWidth' value of lime is 0.75.
Number of neighbors of the query point, specified as the comma-separated pair consisting of 'NumNeighbors' and a positive integer scalar value. This argument is valid only when the DataLocality property of results is 'local'.
The fit function estimates the distribution parameters of the predictor data using the specified number of nearest neighbors of the query point. Then the function generates synthetic predictor data using the estimated distribution.
If you specify a value larger than the number of observations in the predictor data set (results.X) in the lime object results, then fit uses all observations.
The default value is the 'NumNeighbors' value that you specify when creating the lime object results. The default 'NumNeighbors' value of lime is 1500.
results.NumSyntheticData (default) | positive integer scalar value
The default value is the NumSyntheticData property value of the lime object results. If you provide a custom synthetic data set when creating results, then the property value is the number of samples in the data set. Otherwise, the 'NumSyntheticData' value that you specify when creating results sets the property. The default 'NumSyntheticData' value of lime is 5000.
The default value is the 'P' value that you specify when creating the lime object results. The default 'P' value of lime is 2.
The default value is the 'Scale' value that you specify when creating the lime object results. The default 'Scale' value of lime is std(PD,'omitnan'), where PD is the predictor data or synthetic predictor data. If you do not specify the 'Scale' value, then the software uses different scale parameters when computing the distances for both the predictor data and the synthetic predictor data.
'linear' | 'tree'
The default value is the 'SimpleModelType' value that you specify when creating the lime object results. The default 'SimpleModelType' value of lime is 'linear'.
newresults — LIME results
LIME results, returned as a lime object. newresults contains the new simple model.
To overwrite the input argument results, assign the output of fit to results:
results = fit(results,queryPoint,numImportantPredictors);
A distance metric is a function that defines a distance between two observations. fit supports various distance metrics for continuous variables and a mix of continuous and categorical variables.
{d}_{st}^{2}=\left({x}_{s}-{y}_{t}\right)\left({x}_{s}-{y}_{t}{\right)}^{\prime }.
{d}_{st}^{2}=\left({x}_{s}-{y}_{t}\right){V}^{-1}\left({x}_{s}-{y}_{t}{\right)}^{\prime },
{d}_{st}^{2}=\left({x}_{s}-{y}_{t}\right){C}^{-1}\left({x}_{s}-{y}_{t}{\right)}^{\prime },
{d}_{st}=\sum _{j=1}^{n}|{x}_{sj}-{y}_{tj}|.
{d}_{st}=\sqrt[p]{\sum _{j=1}^{n}{|{x}_{sj}-{y}_{tj}|}^{p}}.
{d}_{st}={\mathrm{max}}_{j}\left\{|{x}_{sj}-{y}_{tj}|\right\}.
{d}_{st}=\left(1-\frac{{x}_{s}{{y}^{\prime }}_{t}}{\sqrt{\left({x}_{s}{{x}^{\prime }}_{s}\right)\left({y}_{t}{{y}^{\prime }}_{t}\right)}}\right).
{d}_{st}=1-\frac{\left({x}_{s}-{\overline{x}}_{s}\right){\left({y}_{t}-{\overline{y}}_{t}\right)}^{\prime }}{\sqrt{\left({x}_{s}-{\overline{x}}_{s}\right){\left({x}_{s}-{\overline{x}}_{s}\right)}^{\prime }}\sqrt{\left({y}_{t}-{\overline{y}}_{t}\right){\left({y}_{t}-{\overline{y}}_{t}\right)}^{\prime }}},
{\overline{x}}_{s}=\frac{1}{n}\sum _{j}{x}_{sj}
{\overline{y}}_{t}=\frac{1}{n}\sum _{j}{y}_{tj}.
{d}_{st}=1-\frac{\left({r}_{s}-{\overline{r}}_{s}\right){\left({r}_{t}-{\overline{r}}_{t}\right)}^{\prime }}{\sqrt{\left({r}_{s}-{\overline{r}}_{s}\right){\left({r}_{s}-{\overline{r}}_{s}\right)}^{\prime }}\sqrt{\left({r}_{t}-{\overline{r}}_{t}\right){\left({r}_{t}-{\overline{r}}_{t}\right)}^{\prime }}},
{\overline{r}}_{s}=\frac{1}{n}\sum _{j}{r}_{sj}=\frac{\left(n+1\right)}{2}
{\overline{r}}_{t}=\frac{1}{n}\sum _{j}{r}_{tj}=\frac{\left(n+1\right)}{2}
{w}_{q}\left({x}_{s}\right)=\mathrm{exp}\left(-\frac{1}{2}{\left(\frac{d\left({x}_{s},q\right)}{\sqrt{p}\text{ }\sigma }\right)}^{2}\right).
{\stackrel{˜}{X}}_{s}
{\stackrel{˜}{X}}_{s}
lime | plot
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Fluid dynamics - 3D Animation
Fluid dynamics (13019 views - Mechanism & Kinematics)
In physics and engineering, fluid dynamics is a subdiscipline of fluid mechanics that describes the flow of fluids - liquids and gases. It has several subdisciplines, including aerodynamics (the study of air and other gases in motion) and hydrodynamics (the study of liquids in motion). Fluid dynamics has a wide range of applications, including calculating forces and moments on aircraft, determining the mass flow rate of petroleum through pipelines, predicting weather patterns, understanding nebulae in interstellar space and modelling fission weapon detonation. Fluid dynamics offers a systematic structure—which underlies these practical disciplines—that embraces empirical and semi-empirical laws derived from flow measurement and used to solve practical problems. The solution to a fluid dynamics problem typically involves the calculation of various properties of the fluid, such as flow velocity, pressure, density, and temperature, as functions of space and time. Before the twentieth century, hydrodynamics was synonymous with fluid dynamics. This is still reflected in names of some fluid dynamics topics, like magnetohydrodynamics and hydrodynamic stability, both of which can also be applied to gases.
In physics and engineering, fluid dynamics is a subdiscipline of fluid mechanics that describes the flow of fluids - liquids and gases. It has several subdisciplines, including aerodynamics (the study of air and other gases in motion) and hydrodynamics (the study of liquids in motion). Fluid dynamics has a wide range of applications, including calculating forces and moments on aircraft, determining the mass flow rate of petroleum through pipelines, predicting weather patterns, understanding nebulae in interstellar space and modelling fission weapon detonation.
1.4 Inviscid vs viscous vs Stokes flow
1.8 Reactive vs non-reactive flows
1.10 Relativistic fluid dynamics
1.11 Other approximations
For fluids that are sufficiently dense to be a continuum, do not contain ionized species, and have flow velocities small in relation to the speed of light, the momentum equations for Newtonian fluids are the Navier–Stokes equations—which is a non-linear set of differential equations that describes the flow of a fluid whose stress depends linearly on flow velocity gradients and pressure. The unsimplified equations do not have a general closed-form solution, so they are primarily of use in Computational Fluid Dynamics. The equations can be simplified in a number of ways, all of which make them easier to solve. Some of the simplifications allow some simple fluid dynamics problems to be solved in closed form.[citation needed]
{\displaystyle p={\frac {\rho R_{u}T}{M}}}
Three conservation laws are used to solve fluid dynamics problems, and may be written in integral or differential form. The conservation laws may be applied to a region of the flow called a control volume. A control volume is a discrete volume in space through which fluid is assumed to flow. The integral formulations of the conservation laws are used to describe the change of mass, momentum, or energy within the control volume. Differential formulations of the conservation laws apply Stokes' theorem to yield an expression which may be interpreted as the integral form of the law applied to an infinitesimally small volume (at a point) within the flow.
{\displaystyle {\partial \over \partial t}\iiint _{V}\rho \,dV=-\,{}}
{\displaystyle {\scriptstyle S}}
{\displaystyle {}\,\rho \mathbf {u} \cdot d\mathbf {S} }
{\displaystyle \rho }
is the fluid density, u is the flow velocity vector, and t is time. The left-hand side of the above expression is the rate of increase of mass within the volume and contains a triple integral over the control volume, whereas the right-hand side contains an integration over the surface of the control volume of mass convected into the system. Mass flow into the system is accounted as positive, and since the normal vector to the surface is opposite the sense of flow into the system the term is negated. The differential form of the continuity equation is, by the divergence theorem:
{\displaystyle \ {\partial \rho \over \partial t}+\nabla \cdot (\rho \mathbf {u} )=0}
Conservation of momentum: Newton's second law of motion applied to a control volume, is a statement that any change in momentum of the fluid within that control volume will be due to the net flow of momentum into the volume and the action of external forces acting on the fluid within the volume.
{\displaystyle {\frac {\partial }{\partial t}}\iiint _{\scriptstyle V}\rho \mathbf {u} \,dV=-\,{}}
{\displaystyle _{\scriptstyle S}}
{\displaystyle (\rho \mathbf {u} \cdot d\mathbf {S} )\mathbf {u} -{}}
{\displaystyle {\scriptstyle S}}
{\displaystyle {}\,p\,d\mathbf {S} }
{\displaystyle \displaystyle {}+\iiint _{\scriptstyle V}\rho \mathbf {f} _{\text{body}}\,dV+\mathbf {F} _{\text{surf}}}
In the above integral formulation of this equation, the term on the left is the net change of momentum within the volume. The first term on the right is the net rate at which momentum is convected into the volume. The second term on the right is the force due to pressure on the volume's surfaces. The first two terms on the right are negated since momentum entering the system is accounted as positive, and the normal is opposite the direction of the velocity
{\displaystyle \mathbf {u} }
and pressure forces. The third term on the right is the net acceleration of the mass within the volume due to any body forces (here represented by fbody). Surface forces, such as viscous forces, are represented by
{\displaystyle \mathbf {F} _{\text{surf}}}
, the net force due to shear forces acting on the volume surface. The momentum balance can also be written for a moving control volume.[3]
The following is the differential form of the momentum conservation equation. Here, the volume is reduced to an infinitesimally small point, and both surface and body forces are accounted for in one total force, F. For example, F may be expanded into an expression for the frictional and gravitational forces acting at a point in a flow.
{\displaystyle \ {D\mathbf {u} \over Dt}=\mathbf {F} -{\nabla p \over \rho }}
In aerodynamics, air is assumed to be a Newtonian fluid, which posits a linear relationship between the shear stress (due to internal friction forces) and the rate of strain of the fluid. The equation above is a vector equation in a three-dimensional flow, but it can be expressed as three scalar equations in three coordinate directions. The conservation of momentum equations for the compressible, viscous flow case are called the Navier–Stokes equations.[2]
Conservation of energy: Although energy can be converted from one form to another, the total energy in a closed system remains constant.
{\displaystyle \ \rho {Dh \over Dt}={Dp \over Dt}+\nabla \cdot \left(k\nabla T\right)+\Phi }
{\displaystyle \Phi }
All fluids are compressible to some extent; that is, changes in pressure or temperature cause changes in density. However, in many situations the changes in pressure and temperature are sufficiently small that the changes in density are negligible. In this case the flow can be modelled as an incompressible flow. Otherwise the more general compressible flow equations must be used.
{\displaystyle {\frac {\mathrm {D} \rho }{\mathrm {D} t}}=0\,,}
All fluids are viscous, meaning that they exert some resistance to deformation: neighbouring parcels of fluid moving at different velocities exert viscous forces on each other. The velocity gradient is referred to as a strain rate; it has dimensions
{\displaystyle T^{-1}}
. Isaac Newton showed that for many familiar fluids such as water and air, the stress due to these viscous forces is linearly related to the strain rate. Such fluids are called Newtonian fluids. The coefficient of proportionality is called the fluid's viscosity; for Newtonian fluids, it is a fluid property that is independent of the strain rate.
Non-Newtonian fluids have a more complicated, non-linear stress-strain behaviour. The sub-discipline of rheology describes the stress-strain behaviours of such fluids, which include emulsions and slurries, some viscoelastic materials such as blood and some polymers, and sticky liquids such as latex, honey and lubricants.[citation needed]
Inviscid vs viscous vs Stokes flow
In contrast, high Reynolds numbers (Re>>1) indicate that the inertial effects have more effect on the velocity field than the viscous (friction) effects. In high Reynolds number flows, the flow is often modeled as an inviscid flow, an approximation in which viscosity is completely neglected. Eliminating viscosity allows the Navier–Stokes equations to be simplified into the Euler equations. The integration of the Euler equations along a streamline in an inviscid flow yields Bernoulli's equation. When, in addition to being inviscid, the flow is irrotational everywhere, Bernoulli's equation can completely describe the flow everywhere. Such flows are called potential flows, because the velocity field may be expressed as the gradient of a potential energy expression.
Most flows of interest have Reynolds numbers much too high for DNS to be a viable option,[9] given the state of computational power for the next few decades. Any flight vehicle large enough to carry a human (L > 3 m), moving faster than 20 m/s (72 km/h) is well beyond the limit of DNS simulation (Re = 4 million). Transport aircraft wings (such as on an Airbus A300 or Boeing 747) have Reynolds numbers of 40 million (based on the wing chord dimension). Solving these real-life flow problems requires turbulence models for the foreseeable future. Reynolds-averaged Navier–Stokes equations (RANS) combined with turbulence modelling provides a model of the effects of the turbulent flow. Such a modelling mainly provides the additional momentum transfer by the Reynolds stresses, although the turbulence also enhances the heat and mass transfer. Another promising methodology is large eddy simulation (LES), especially in the guise of detached eddy simulation (DES)—which is a combination of RANS turbulence modelling and large eddy simulation.
While many flows (e.g. flow of water through a pipe) occur at low Mach numbers, many flows of practical interest in aerodynamics or in turbomachines occur at high fractions of M=1 (transonic flows) or in excess of it (supersonic or even hypersonic flows). New phenomena occur at these regimes such as instabilities in transonic flow, shock waves for supersonic flow, or non-equilibrium chemical behaviour due to ionization in hypersonic flows. In practice, each of those flow regimes is treated separately.
Reactive vs non-reactive flows
Reactive flows are flows that are chemically reactive, which finds its applications in many areas such as combustion(IC engine), propulsion devices (Rockets, jet engines etc.), detonations, fire and safety hazards, astrophysics etc. In addition to conservation of mass, momentum and energy, conservation of individual species (for example, mass fraction of methane in methane combustion) need to be derived, where the production/depletion rate of any species are obtained by simultaneously solving the equations of chemical kinetics.
Relativistic fluid dynamics studies the macroscopic and microscopic fluid motion at large velocities comparable to the velocity of light.[10] This branch of fluid dynamics accounts the relativistic effects both from the special theory of relativity and the general theory of relativity. The governing equations are derived in Riemannian geometry for Minkowski spacetime.
In a compressible fluid, it is convenient to define the total conditions (also called stagnation conditions) for all thermodynamic state properties (e.g. total temperature, total enthalpy, total speed of sound). These total flow conditions are a function of the fluid velocity and have different values in frames of reference with different motion.
To avoid potential ambiguity when referring to the properties of the fluid associated with the state of the fluid rather than its motion, the prefix "static" is commonly used (e.g. static temperature, static enthalpy). Where there is no prefix, the fluid property is the static condition (i.e. "density" and "static density" mean the same thing). The static conditions are independent of the frame of reference.
Butterfly valveComputational fluid dynamicsValveFluidDrag (physics)Flued boilerWorking fluidElectrohydrodynamicsFluid powerTribologyWater hammer
This article uses material from the Wikipedia article "Fluid dynamics", which is released under the Creative Commons Attribution-Share-Alike License 3.0. There is a list of all authors in Wikipedia
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Determine whether each statement makes sense or does not make sense, and explain your reas
Determine whether each statement makes sense or does not make sense, and explain your reasoning. A system of linear equations in three variables, x, y
Determine whether each statement makes sense or does not make sense, and explain your reasoning. A system of linear equations in three variables, x, y, and z cannot contain an equation in the form
y=mx+b
The general form of a system in three variables x,y,z has equation in the general form:
akx+bky+ckz=dk
If one of the coefficients ck of the variables z is zero, we can isolate y and bring that equation to the form
y=mx+b.
akx+bky=dk
bky=-akx+dk
y=-\left(ak/bk\right)x+dk/bk
{P}_{5}
{P}_{5}
V={R}_{3}
\left({x}_{1},{x}_{2},{x}_{3}\right)
{x}_{1}-6{x}_{2}+{x}_{3}=5
V={R}^{n}
{C}^{2}\left(I\right)
{P}_{n}
{P}_{n}
V={M}_{n}\left(R\right)
How would one go about analytically solving a system of non-linear equations of the form:
a+b+c=4
{a}^{2}+{b}^{2}+{c}^{2}=6
{a}^{3}+{b}^{3}+{c}^{3}=10
Physical meaning of the null space of a matrix
What is an intuitive meaning of the null space of a matrix? Why is it useful?
I'm not looking for textbook definitions. My textbook gives me the definition, but I just don't "get" it.
E.g.: I think of the rank r of a matrix as the minimum number of dimensions that a linear combination of its columns would have; it tells me that, if I combined the vectors in its columns in some order, I'd get a set of coordinates for an r-dimensional space, where r is minimum (please correct me if I'm wrong). So that means I can relate rank (and also dimension) to actual coordinate systems, and so it makes sense to me. But I can't think of any physical meaning for a null space... could someone explain what its meaning would be, for example, in a coordinate system?
Consider the system (*) whose coefficient matrix A is the matrix D listed in Exercise 46 and whose fundamental matrix was computed just before the preceding exercise.
x+y+2z=3
3x+2y+2z=4
x+y+3z=5
Reduce the system of linear equations to upper triangular form and solve.
The reduced row echelon form of a system of linear equations is given. Write the system of equations corresponding to the given matrix. Use x, y; or x, y, z; or
{x}_{1},{x}_{2},{x}_{3},{x}_{4}x
as variables. Determine whether the system is consistent or inconsistent. If it is consistent, give the solution.
\left[\begin{array}{cccccc}1& 0& 0& 0& |& 1\\ 0& 1& 0& 0& |& 2\\ 0& 0& 1& 2& |& 3\end{array}\right]
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Calculus/Work - Wikibooks, open books for an open world
Calculus/Work
← Surface area Calculus Centre of mass →
{\displaystyle {\begin{aligned}W=\int F\,dr&=\int ma\,dr\\&=\int m\,{\frac {dv}{dt}}\,dr\\&=m\int {\frac {dr}{dt}}\,dv\\&=m\int v\,dv\\&={\frac {mv^{2}}{2}}=\Delta E_{k}\end{aligned}}}
Retrieved from "https://en.wikibooks.org/w/index.php?title=Calculus/Work&oldid=3831222"
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Understanding Data - Problem Solving Practice Problems Online | Brilliant
In Mrs Krol's math class, there are 20 students and the class average is 91. In Mr Tan's math class, there are 30 students and the class average is 71. What is the average of score of the combined classes?
Let the median of
33
observations be
50.
If each of the observations greater than the median is increased by
8,
then what is the median of the new data?
\{ -3.3, 6.7, x, 29.2, 42.4 \}
x
are there that would make the range of the above set equal to 50?
The mean of a set of numbers is the average of the set.
The median of a set of numbers is the middle value, which divides the list into two equal halves. If there is an even number of them, the median will be the average of the two middle values.
A group of 7 girls each wrote down a number The average of the 7 numbers is 50.
The average of the numbers not written by Sally is 45.
What number did Sally write?
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Convolves a pair of one- or two-dimensional NDFs together
This application smooths a one- or two-dimensional NDF using a Point-Spread Function given by a second NDF. The output NDF is normalised to the same mean data value as the input NDF (if Parameter NORM is set to TRUE), and is the same size as the input NDF.
convolve in psf out xcentre ycentre
AXES( 2 ) = _INTEGER (Read)
The input NDF containing the image to be smoothed.
Determines how the output NDF is normalised to take account of the total data sum in the PSF, and of the presence of bad pixels in the input NDF. If TRUE, bad pixels are excluded from the data sum for each output pixel, and the associated weight for the output pixel is reduced appropriately. The supplied PSF is normalised to a total data sum of unity so that the output NDF has the same normalisation as the input NDF. If NORM is FALSE, bad pixels are replaced by the mean value and then included in the convolution as normal. The normalisation of the supplied PSF is left unchanged, and so determines the normalisation of the output NDF. [TRUE]
PSF = NDF (Read)
An NDF holding the Point-Spread Function (PSF) with which the input image is to be smoothed. An error is reported if the PSF contains any bad pixels. The PSF can be centred anywhere within the image (see Parameters XCENTRE and YCENTRE). A constant background is removed from the PSF before use. This background level is equal to the minimum of the absolute value in the four corner pixel values. The PSF is assumed to be zero beyond the bounds of the supplied NDF. It should have the same number of dimensions as the NDF being smoothed, unless the input NDF has three significant dimensions, whereupon the PSF must be two-dimensional. It will be normalised to a total data sum of unity if Parameter NORM is TRUE.
A title for the output NDF. A null (!) value means using the title of the input NDF. [!]
If the input array contains bad pixels, and NORM is TRUE, then this parameter may be used to determine the number of good pixels that must be present within the smoothing box before a valid output pixel is generated. It can be used, for example, to prevent output pixels from being generated in regions where there are relatively few good pixels to contribute to the smoothed result.
If a numerical value is given for WLIM, then it specifies the minimum total weight associated with the good pixels in the smoothing box required to generate a good output pixel (weights for each pixel are defined by the normalised PSF). If this specified minimum weight is not present, then a bad output pixel will result, otherwise a smoothed output value will be calculated. The value of this parameter should lie between 0.0 and 1.0. A value of 0.0 will result in a good output pixel being created even if only one good input pixel contributes to it. A value of 1.0 will result in a good output pixel being created only if all the input pixels which contribute to it are good. See also Parameter NORM. [!]
XCENTRE = _INTEGER (Read)
The x pixel index (column number) of the centre of the PSF within the supplied PSF array. The suggested default is the centre of the PSF array. (This is how the PSF command would generate the array.)
YCENTRE = _INTEGER (Read)
The y pixel index (line number) of the centre of the PSF within the supplied PSF array. The suggested default is the centre of the PSF array. (This is how the PSF command would generate the array.)
convolve ccdframe iraspsf ccdlores 50 50
The image in the NDF called ccdframe is convolved using the PSF in NDF iraspsf to create the smoothed image ccdlores. The centre of the PSF image in iraspsf is at pixel indices (50, 50). Any bad pixels in the input image are propagated to the output.
convolve ccdframe iraspsf ccdlores 50 50 wlim=1.0
As above, but good output values are only created for pixels which have no contributions from bad input pixels.
convolve ccdframe iraspsf ccdlores
\setminus
As in the first example except the centre of the PSF is located at the centre of the PSF array.
The algorithm used is based on the multiplication of the Fourier transforms of the input image and PSF image.
A PSF can be created using the PSF command or MATHS if the PSF is an analytic function.
KAPPA: BLOCK, FFCLEAN, GAUSMOOTH, MATHS, MEDIAN, PSF; FIGARO: ICONV3, ISMOOTH, IXSMOOTH, MEDFILT.
All non-complex numeric data types can be handled. Arithmetic is performed using double-precision floating point.
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Aidan knows that the observation deck on the Vancouver Lookout is 130 m above th
Aidan knows that the observation deck on the Vancouver Lookout is 130 m above the ground. He measures the angle between the ground and his line of sight to the observation deck as PSK77^\circZSK. How far is Aidan from the base of the Lookout to the nearest metre?
{77}^{\circ }
mhalmantus
y=5x+3
\mathrm{tan}\left(\mathrm{arcsin}\left(\frac{x}{8}\right)\right)
\mathrm{cos}\left(ar\mathrm{sin}\left(\frac{x}{8}\right)\right)
\left(\frac{1}{2}\right)\mathrm{sin}\left(2\mathrm{arcsin}\left(\frac{x}{8}\right)\right)
\mathrm{sin}\left(\mathrm{arctan}\left(\frac{x}{8}\right)\right)
\mathrm{cos}\left(\mathrm{arctan}\left(\frac{x}{8}\right)\right)
8\frac{3}{4}cm
4\sqrt{3}
, and the hypotenuse = 8.
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Find the directional derivative of f at the given point in the direction indicated by the
\theta .f\left(x,y\right)={e}^{x}\mathrm{cos}y,\left(0,0\right),\theta =\frac{\pi }{4}
We need to find directional derivative of at the given point in the direction of v
f\left(x,y\right)={e}^{x}\mathrm{cos}y;\text{ }\theta =\pi /4,\text{ }\left(0,0\right)
{D}_{u}f\left(x,y\right)={f}_{x}\left(x,y\right)a+{f}_{y}\left(x,y\right)b
Find partial derivatives
{f}_{x}={e}^{x}\mathrm{cos}y
{f}_{y}=-{e}^{x}\mathrm{sin}y
Plug into formula
{D}_{u}f=\left({e}^{x}\mathrm{cos}y\right)\mathrm{cos}\left(\pi /4\right)+\left(-{e}^{x}\mathrm{sin}y\right)\mathrm{sin}\left(\pi /4\right)
=\frac{\sqrt{2}}{2}\left[\left({e}^{x}\mathrm{cos}y\right)+\left(-{e}^{x}\mathrm{sin}\left(y\right)\right]
\frac{\sqrt{2}}{2}\left[{e}^{0}\mathrm{cos}0-{e}^{0}\mathrm{sin}0\right]
{D}_{u}f\left(0,0\right)=\frac{\sqrt{2}}{2}
Need help with the following question Compute the directional derivative of
f\left(x,y\right)={e}^{2}x-5y
How do you find the equation of the line tangent to
f\left(x\right)={x}^{3}
, at (2,8)?
What is the equation of the normal line of f(x)=-4x-10 at x=-2?
How do you use the limit definition of the derivative to find the derivative of f(x)=-x+6?
What is the slope of the line tangent to the graph of
y=\mathrm{arctan}\left(3x\right)
y=3{x}^{2}-{x}^{3}
at point (1,2)?
h\left(z\right)=-6{z}^{-7}
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J/psi meson - Wikipedia
(Redirected from J/ψ meson)
SLAC: Burton Richter et al. (1974)
BNL: Samuel Ting et al. (1974)
5.5208×10−27 kg
3.096916 GeV/c2
(J/psi) meson /ˈdʒeɪ ˈsaɪ ˈmiːzɒn/ or psion[1] is a subatomic particle, a flavor-neutral meson consisting of a charm quark and a charm antiquark. Mesons formed by a bound state of a charm quark and a charm anti-quark are generally known as "charmonium". The
is the most common form of charmonium, due to its spin of 1 and its low rest mass. The
has a rest mass of 3.0969 GeV/c2, just above that of the
c (2.9836 GeV/c2), and a mean lifetime of 7.2×10−21 s. This lifetime was about a thousand times longer than expected.[2]
Its discovery was made independently by two research groups, one at the Stanford Linear Accelerator Center, headed by Burton Richter, and one at the Brookhaven National Laboratory, headed by Samuel Ting of MIT. They discovered they had actually found the same particle, and both announced their discoveries on 11 November 1974. The importance of this discovery is highlighted by the fact that the subsequent, rapid changes in high-energy physics at the time have become collectively known as the "November Revolution". Richter and Ting were awarded the 1976 Nobel Prize in Physics.
1 Background to discovery
3 J/ψ melting
Background to discovery[edit]
The background to the discovery of the
was both theoretical and experimental. In the 1960s, the first quark models of elementary particle physics were proposed, which said that protons, neutrons, and all other baryons, and also all mesons, are made from fractionally charged particles, the "quarks", which come in six types or "flavors", called up, down, top, bottom, strange and charm. Despite the ability of quark models to bring order to the "elementary particle zoo", they were considered something like mathematical fiction at the time, a simple artifact of deeper physical reasons.[3]
Starting in 1969, deep inelastic scattering experiments at SLAC revealed surprising experimental evidence for particles inside of protons. Whether these were quarks or something else was not known at first. Many experiments were needed to fully identify the properties of the sub-protonic components. To a first approximation, they indeed were a match for the previously-described quarks.
On the theoretical front, gauge theories with broken symmetry became the first fully viable contenders for explaining the weak interaction after Gerardus 't Hooft discovered in 1971 how to calculate with them beyond tree level. The first experimental evidence for these electroweak unification theories was the discovery of the weak neutral current in 1973. Gauge theories with quarks became a viable contender for the strong interaction in 1973, when the concept of asymptotic freedom was identified.
However, a naive mixture of electroweak theory and the quark model led to calculations about known decay modes that contradicted observation: In particular, it predicted Z boson-mediated flavor-changing decays of a strange quark into a down quark, which were not observed. A 1970 idea of Sheldon Glashow, John Iliopoulos, and Luciano Maiani, known as the GIM mechanism, showed that the flavor-changing decays would be strongly suppressed if there were a fourth quark (now called the charm quark) that was a complementary counterpart to the strange quark. By summer 1974 this work had led to theoretical predictions of what a charm + anticharm meson would be like.
The predictions were ignored.[citation needed] The work of Richter and Ting was done mostly to explore new energy regimes, not to test the theoretical predictions.[citation needed]
The group at Brookhaven,[a] were the first to discern a peak at 3.1 GeV in plots of production rates, first recognizing the 𝜓 meson – that Ting named the "J" meson (after himself – his last-name written in Chinese is 丁).[c]
Decay modes[edit]
Hadronic decay modes of
are strongly suppressed because of the OZI rule. This effect strongly increases the lifetime of the particle and thereby gives it its very narrow decay width of just 93.2±2.1 keV. Because of this strong suppression, electromagnetic decays begin to compete with hadronic decays. This is why the
has a significant branching fraction to leptons.
The primary decay modes[5] are:
→ hadrons 13.5%±0.3%
melting[edit]
In a hot QCD medium, when the temperature is raised well beyond the Hagedorn temperature, the
and its excitations are expected to melt.[6] This is one of the predicted signals of the formation of the quark–gluon plasma. Heavy-ion experiments at CERN's Super Proton Synchrotron and at BNL's Relativistic Heavy Ion Collider have studied this phenomenon without a conclusive outcome as of 2009. This is due to the requirement that the disappearance of
mesons is evaluated with respect to the baseline provided by the total production of all charm quark-containing subatomic particles, and because it is widely expected that some
are produced and/or destroyed at time of QGP hadronization. Thus, there is uncertainty in the prevailing conditions at the initial collisions.
In fact, instead of suppression, enhanced production of
is expected[7] in heavy ion experiments at LHC where the quark-combinant production mechanism should be dominant given the large abundance of charm quarks in the QGP. Aside of
, charmed B mesons (
c), offer a signature that indicates that quarks move freely and bind at-will when combining to form hadrons.[8][9]
Chinese character for the surname Ting, which resembles the Latin letter J.
Because of the nearly simultaneous discovery, the
is the only particle to have a two-letter name. Richter named it "SP", after the SPEAR accelerator used at SLAC; however, none of his coworkers liked that name. After consulting with Greek-born Leo Resvanis to see which Greek letters were still available, and rejecting "iota" because its name implies insignificance, Richter chose "psi" – a name which, as Gerson Goldhaber pointed out, contains the original name "SP", but in reverse order.[10] Coincidentally, later spark chamber pictures often resembled the psi shape. Ting assigned the name "J" to it, which is one letter away from "K", the name of the already-known strange meson; another reason is that "j" is the symbol for electromagnetic current.[11] Possibly by coincidence, "J" strongly resembles the Chinese character for Ting's name (丁 Dīng). J is also the first letter of Ting's eldest daughter's name, Jeanne.
Much of the scientific community considered it unjust to give one of the two discoverers priority, so most subsequent publications have referred to the particle as the "
The first excited state of the
was called the ψ′; it is now called the ψ(2S), indicating its quantum state. The next excited state was called the ψ″; it is now called ψ(3770), indicating mass in MeV. Other vector charm-anticharm states are denoted similarly with ψ and the quantum state (if known) or the mass.[12] The "J" is not used, since Richter's group alone first found excited states.
The name charmonium is used for the
and other charm-anticharm bound states. This is by analogy with positronium, which also consists of a particle and its antiparticle (an electron and positron in the case of positronium).
^ a b Glenn Everhart, Terry Rhoades, Min Chen, and Ulrich Becker, at Brookhaven first to discerned the 3.1 GeV peak in pair-production rates.
^ There are two different regimes of flavorless, neutral mesons: Low mass and high mass.
Lighter mesons, such as the neutral pion (
, the lightest of all mesons), the
, and so-on. Whether high or low mass, since all of the flavorless mesons’ quantum numbers are zero they can only be distinguished by their masses. Generally their quark content is invisible, especially the low-mass flavorless mesons, not only because their very similar small masses can be easily confused, but also because the low-mass particles themselves do actually exist as mixtures. For example the lowest mass of all mesons is the neutral pion; it is approximately an equal mix of dd and uu matching quark-antiquark pairs.
However, the heavy c and b quarks are sufficiently distinct in mass to tell them apart:
cc = "charmonium" = J/𝜓 meson
bb = "bottomonium" =
^ S.C.C. Ting argued in a long and acrimonious struggle over the name for the particle. Ting insisted that the particle should be named the "J", which resembles the Chinese glyph for his name (丁). Although the use of a capital latin letter clashes with the existing meson naming conventions[4] the name J/𝜓 for the meson's lowest energy state was adopted to appease Ting. No discoverer of any other particle has gotten to "carve his initials" it, including Ting’s co‑discoverers. Ting is the singular, isolated exception.
Ting & Richter did not find the higher-energy states, and Ting et al. did not earn any credit for that, nor receive any further naming rights. For the higher energy states which only the Brookhaven[a] group found, the "J" is dropped, and it is simply called the 𝜓 meson.
In a final bout of irony in this still contentious matter, all flavor-neutral mesons are recognized as being in some regard the same, since their quantum numbers are all the same: all zero. The name "quarkonium" is in more common use than older names, including the J/𝜓, which is now commonly called "charmonium", and forms a family of flavorless, uncharged mesons, with their older names falling out of use: Theorists have taken to using the name charmonium to emphasize the importance of the distinction between the conventional "flavored" mesons and the evanescent and amorphous unflavored mesons in the "quarkonium" class, and the contentious name "J/𝜓" seems to be gradually, naturally falling out of use.[b]
^ Kapusta, J.; Müller, B.; Rafelski, J. (9 December 2003). [no title cited]. ISBN 9780444511102. Retrieved 25 September 2014 – via Google Books. [full citation needed][dead link]
^ "Shared Physics prize for elementary particle" (Press release). The Royal Swedish Academy of Sciences. 18 October 1976. Retrieved 23 April 2012.
^ Pickering, A. (1984). Constructing Quarks. University of Chicago Press. pp. 114–125. ISBN 978-0-226-66799-7.
^ Patrignani, C.; et al. (Particle Data Group) (2016). "Revised naming-scheme for hadrons". Chin. Phys. C. 40: 100001. "2017 update" (PDF).
^ Nakamura, K.; et al. (Particle Data Group) (2010). "J/ψ(1S)" (PDF). Particle Data Group. Journal of Physics G. Lawrence Berkeley Laboratory. 37: 075021. doi:10.1088/0954-3899/37/7A/075021.
^ Matsui, T.; Satz, H. (1986). "J/ψ suppression by quark-gluon plasma formation". Physics Letters B. 178 (4): 416–422. Bibcode:1986PhLB..178..416M. doi:10.1016/0370-2693(86)91404-8. OSTI 1118865.
^ Thews, R. L.; Schroedter, M.; Rafelski, J. (2001). "Enhanced J/ψ production in deconfined quark matter". Physical Review C. 63 (5): 054905. arXiv:hep-ph/0007323. Bibcode:2001PhRvC..63e4905T. doi:10.1103/PhysRevC.63.054905. S2CID 11932902.
^ Schroedter, M.; Thews, R.L.; Rafelski, J. (2000). "Bc-meson production in ultrarelativistic nuclear collisions". Physical Review C. 62 (2): 024905. arXiv:hep-ph/0004041. Bibcode:2000PhRvC..62b4905S. doi:10.1103/PhysRevC.62.024905. S2CID 119008673.
^ Fulcher, L.P.; Rafelski, J.; Thews, R.L. (1999). "Bc mesons as a signal of deconfinement". arXiv:hep-ph/9905201.
^ Zielinski, L (8 August 2006). "Physics Folklore". QuarkNet. Retrieved 13 April 2009.
^ We discussed the name of the new particle for some time. Someone pointed out to me that the really exciting stable particles are designated by Roman characters - like the postulated W0, the intermediate vector boson, the Z0, etc. - whereas the “classical” particles have Greek designations like ρ, ω etc. This, combined with the fact that our work in the last decade had been concentrated on the electromagnetic current
{\textstyle j_{\mu }(x)}
gave us the idea to call this particle the J particle. Samuel Ting, The Discovery of the J Particle Nobel prize lecture, 11. December 1976 [1]
^ Roos, M; Wohl, CG; (Particle Data Group) (2004). "Naming schemes for hadrons" (PDF). Retrieved 13 April 2009. {{cite web}}: CS1 maint: multiple names: authors list (link)
Glashow, S. L.; Iliopoulos, J.; Maiani, L. (1970). "Weak Interactions with Lepton-Hadron Symmetry". Physical Review D. 2 (7): 1285–1292. Bibcode:1970PhRvD...2.1285G. doi:10.1103/PhysRevD.2.1285.
Aubert, J.; et al. (1974). "Experimental Observation of a Heavy Particle J". Physical Review Letters. 33 (23): 1404–1406. Bibcode:1974PhRvL..33.1404A. doi:10.1103/PhysRevLett.33.1404.
Augustin, J.; et al. (1974). "Discovery of a Narrow Resonance in e+e− Annihilation". Physical Review Letters. 33 (23): 1406–1408. Bibcode:1974PhRvL..33.1406A. doi:10.1103/PhysRevLett.33.1406.
Bobra, M. (2005). "Logbook: J/ψ particle". Symmetry Magazine. 2 (7): 34.
Yao, W.-M.; et al. (Particle Data Group) (2006). "Review of Particle Physics: Naming Scheme for Hadrons" (PDF). Journal of Physics G. 33 (1): 108. arXiv:astro-ph/0601168. Bibcode:2006JPhG...33....1Y. doi:10.1088/0954-3899/33/1/001.
Retrieved from "https://en.wikipedia.org/w/index.php?title=J/psi_meson&oldid=1088231922"
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Normally distributed random numbers - MATLAB randn - MathWorks Australia
Bivariate Normal Random Numbers
Random Complex Numbers with Specified Mean and Covariance
Standard Real and Standard Complex Normal Distributions
X = randn
X = randn(___,typename)
X = randn(s,___)
X = randn returns a random scalar drawn from the standard normal distribution.
X = randn(n) returns an n-by-n matrix of normally distributed random numbers.
X = randn(sz1,...,szN) returns an sz1-by-...-by-szN array of random numbers where sz1,...,szN indicate the size of each dimension. For example, randn(3,4) returns a 3-by-4 matrix.
X = randn(sz) returns an array of random numbers where size vector sz defines size(X). For example, randn([3 4]) returns a 3-by-4 matrix.
X = randn(___,typename) returns an array of random numbers of data type typename. The typename input can be either "single" or "double". You can use any of the input arguments in the previous syntaxes.
X = randn(___,"like",p) returns an array of random numbers like p; that is, of the same data type and complexity (real or complex) as p. You can specify either typename or "like", but not both.
X = randn(s,___) generates numbers from random number stream s instead of the default global stream. To create a stream, use RandStream. You can specify s followed by any of the input argument combinations in previous syntaxes.
Generate a 5-by-5 matrix of normally distributed random numbers.
r = randn(5)
Generate values from a bivariate normal distribution with specified mean vector and covariance matrix.
z = repmat(mu,10,1) + randn(10,2)*R
z = 10×2
r = randn(1,5)
r1 = randn(1,5)
r = randn(1,4,"single")
X = randn(size(A));
X = randn(size(p),"like",p)
Generate 10 random complex numbers from the standard complex normal distribution.
a = randn(10,1,"like",1i)
By default, randn(n,"like",1i) generates random numbers from the standard complex normal distribution. The real and imaginary parts are independent normally distributed random variables with mean 0 and variance 1/2. The covariance matrix is of the form [1/2 0; 0 1/2].
z = randn(50000,1,"like",1i);
cov_z = cov(real(z),imag(z),1)
cov_z = 2×2
To specify a more general complex normal distribution, define the mean and covariance matrix. For instance, specify the mean as
\mu =1+2\mathrm{i}
and the covariance matrix as
\sigma =\left[\begin{array}{cc}{\sigma }_{\mathrm{xx}}& {\sigma }_{\mathrm{xy}}\\ {\sigma }_{\mathrm{yx}}& {\sigma }_{\mathrm{yy}}\end{array}\right]=\left[\begin{array}{cc}2& -2\\ -2& 4\end{array}\right]
mu = 1 + 2i;
sigma = [2 -2; -2 4];
Transform the previously generated data to follow the newly defined complex normal distribution. Include the factor of sqrt(2) when scaling the data because the variance for the real and imaginary parts in the original distribution is 1/2.
z_comp = [real(z) imag(z)];
z = repmat(mu,50000,1) + z_comp*sqrt(2)*R*[1; 1i];
z(1:10)
Beyond the second dimension, randn ignores trailing dimensions with a size of 1. For example, randn(3,1,1,1) produces a 3-by-1 vector of random numbers.
Beyond the second dimension, randn ignores trailing dimensions with a size of 1. For example, randn([3 1 1 1]) produces a 3-by-1 vector of random numbers.
Data type (class) to create, specified as "double", "single", or the name of another class that provides randn support.
Example: randn(5,"single")
Example: randn(5,"like",p)
Example: s = RandStream("dsfmt19937"); randn(s,[3 1])
When generating random real numbers, the randn function generates data that follows the standard normal distribution:
f\left(x\right)=\frac{1}{\sqrt{2\pi }}{e}^{-{x}^{2}/2}
for a random real variable x with mean 0 and variance 1.
When generating random complex numbers, such as when using the command randn(...,"like",1i), the randn function generates data that follows the standard complex normal distribution:
f\left(z\right)=\frac{1}{\pi }{e}^{-{|z|}^{2}}
for a random complex variable z whose real and imaginary parts are independent normally distributed random variables with mean 0 and variance 1/2.
The sequence of numbers produced by randn is determined by the internal settings of the uniform pseudorandom number generator that underlies rand, randi, and randn. You can control that shared random number generator using rng.
The data type (class) must be a built-in MATLAB® numeric type. For other classes, the static randn method is not invoked. For example, randn(sz,'myclass') does not invoke myclass.randn(sz).
If extrinsic calls are enabled and randn is not called from inside a parfor loop, generated MEX files use the same random number state as MATLAB in serial code. Otherwise, the generated MEX code and standalone code maintain their own random number state that is initialized to the same state as MATLAB.
The stream syntax randn(s,___) is not supported on a GPU.
To create a GPU array with underlying type datatype, specify the underlying type as an additional argument before typename. For example, X = randn(3,datatype,'gpuArray') creates a 3-by-3 GPU array of random numbers with underlying type datatype.
The stream syntax randn(s,___) is not supported for codistributed or distributed arrays.
To create a distributed or codistributed array with underlying type datatype, specify the underlying type as an additional argument before typename. For example, X = randn(3,datatype,'distributed') creates a 3-by-3 distributed matrix of random numbers with underlying type datatype.
For additional codistributed syntaxes, see randn (codistributed) (Parallel Computing Toolbox).
r = randn(2,2,"like",1i)
All syntaxes support this feature. Also, you can now use "like" with a RandStream object as the first input of randn.
To generate random numbers with the same data type as an existing variable, use the syntax randn(__,'like',p). For example:
r = randn(4,4,'like',A);
This feature is not available when passing a RandStream object as the first input to randn.
randi | rand | rng | RandStream | sprand | sprandn | randperm
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Solve each problem. Show your work. The diameter of a circle is 12 inches. Find the area o
Solve each problem. Show your work. The diameter of a circle is 12 inches. Find the area of a sector of the circle with a central angle that measures 110∘. Use 3.14 for π.π. Round to the nearest hundredth.
Determine the value of the radius.
r=\frac{d}{2}=\frac{12}{2}=6
Determine the area of the sector.
A=\frac{N}{360}\left(\pi {r}^{2}\right)
Write the formula for area of a sector
A=\frac{110}{360}\left(3.14\right)\left({6}^{2}\right)
A=\frac{11}{360}\left(3.14\right)\left(36\right)
Evaluate the exponent
A=\frac{11}{36}\left(3.14\right)\left(36\right)
Simplify the fraction
A ~ 34.54 Multiply the values.
Find each unknown for problem. Check your work.
9m=549m=54
Work each problem. Find
\left\{16,18,21,50\right\}\cup \left\{15,16,17,18\right\}
Work each problem. Write 27^{2 /. 3} in radical form and simplify.
Sofia's 4th grade class has a spelling list organized by subject. The spelling list includes 170 words about history, 75 words about science, 60 words about math, and 55 words about geography. Each week, Sofia's teacher tests her on 15 words from the list. Sofia's class has already had 7 weeks of spelling tests. How many more words will Sofia be tested on before the end of the year?
Work each problem. Find {16,18,21,50}∩{15,16,17,18}{16,18,21,50}∩{15,16,17,18}.
A banner is made of a square and a semicircle. The square has side lengths of 26 inches. One side of the square is also the diameter of the semicircle. What is the total area of the banner? Use 3.14 for
The probability that an event will occur is . Which of these best describes the likelihood of the event occurring
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Blocking | PoE Wiki
Blocking is a layer of defense that will entirely prevent damage from an enemy hit unless modified by some outside effect like Glancing Blows. The chance to block is a random roll made per attack/cast. Block is capped by default at 75%. Characters do not have any base block.
The chance to block is a random roll made per attack/cast; it is not entropy-based like evasion. Block has the following base stats
Additional block chance can be gained from passive skills, items, and Tempest ShieldTempest ShieldSpell, Lightning, Chaining
Buff grants Immunity to ShockPlace into an item socket of the right colour to gain this skill. Right click to remove from a socket. . Attack block and spell block chance are often granted separately with attack block being more common.
Effectiveness of Added Damage: (165-269)%Hits enemies, converting some of your physical damage to cold damage. If an enemy is frozen and is on less than one third life, they will shatter when hit by Glacial Hammer. If striking three times in a row, the third strike will freeze enemies more easily. Requires a Mace, Sceptre or Staff.Per 1% Quality:1Superior2Anomalous3Divergent2% increased Duration of Cold Ailments0.5% increased Attack Speed1% increased Area of Effect100% of Physical Damage Converted to Cold Damage
+2 to Melee Strike RangePlace into an item socket of the right colour to gain this skill. Right click to remove from a socket. and Viper StrikeViper StrikeAttack, Duration, Melee, Strike, Chaos
Effectiveness of Added Damage: (90-154)%Hits enemies, converting some of your physical damage to chaos damage and inflicting poison which will be affected by modifiers to skill duration. If dual wielding, will strike with both weapons. Requires a claw, dagger or sword.Per 1% Quality:1Superior2Anomalous3Divergent4Phantasmal0.5% increased Attack Speed
0.5% increased Poison Duration1% of Physical Damage Converted to Chaos Damage1% increased Chaos Damage0.5% chance to also Poison a nearby Enemy when you inflict Poison60% of Physical Damage Converted to Chaos Damage
30% Less Attack Speed if Dual WieldingPlace into an item socket of the right colour to gain this skill. Right click to remove from a socket. will still apply because block doesn't prevent a successful hit. However, skill effects based on damage like PuncturePunctureAttack, Projectile, Duration, Melee, Strike, Physical, Bow
Effectiveness of Added Damage: (135-165)%Punctures enemies, causing a bleeding debuff, which will be affected by modifiers to skill duration. Puncture works with bows, daggers, claws or swords.Per 1% Quality:1Superior2Anomalous3Divergent4PhantasmalBleeding inflicted by this Skill deals Damage 0.5% fasterProjectiles Pierce 0.1 additional TargetsMelee Strikes target 0.1 additional nearby Enemies0.5% chance to Maim on HitBase duration is 8 seconds
Causes BleedingPlace into an item socket of the right colour to gain this skill. Right click to remove from a socket. are essentially blocked.[3]
\color [rgb]{0.6392156862745098,0.5529411764705883,0.42745098039215684}{\begin{aligned}{\text{Block Duration}}&={{350} \over {1+{\text{Block and Stun Recovery}}+{\text{Block Recovery}}}}\end{aligned}}
of the Stalwart Suffix 55 +(8-9)% Chance to Block shield 1000
of the Buttress Suffix 66 +(10-11)% Chance to Block shield 1000
of the Sentinel Suffix 77 +(12-13)% Chance to Block shield 1000
of the Citadel Suffix 86 +(14-15)% Chance to Block shield 1000
of the Bulwark Suffix 52 (7-9)% Chance to Block Spell Damage focus 1000
of the Barricade Suffix 71 (10-12)% Chance to Block Spell Damage focus 1000
of the Bastion Suffix 84 (13-15)% Chance to Block Spell Damage focus 1000
The Elder's Prefix 68 1% increased Damage per 1% Chance to Block Attack Damage shield_elder 800
The Elder's Prefix 68 +(500-650) Armour if you've Blocked Recently shield_elder 800
of Guarding Suffix 1 +2% Chance to Block Attack Damage if you were Damaged by a Hit Recently abyss_jewel_melee 500
of Readiness Suffix 1 +2% Chance to Block Spell Damage if you were Damaged by a Hit Recently abyss_jewel_melee 250
Effectiveness of Added Damage: (160-230)%Perform a deadly counter-attack when you block. Uses both weapons while you're dual wielding.Per 1% Quality:1Superior2Anomalous3Divergent1% increased Damage1% increased Cooldown Recovery Rate1% chance to gain Onslaught for 4 seconds on HitTrigger this Skill when you Block
Effectiveness of Added Damage: (100-159)%Requires Level 12Swing an Axe or Sword, consuming Steel Shards to gain a Steel Ward that protects you for a duration, and fire projectiles which shatter on impact or soon after being launched, dealing area damage in front of where they shatter. Steel Shards are gained with the Call of Steel Skill.Per 1% Quality:1Superior2Anomalous3Divergent4Phantasmal1% increased Impale Effect0.5% increased Area of Effect
1% increased Projectile SpeedFires an additional Projectile1.5% increased Projectile SpeedBase duration is 5 seconds
Projectiles deal up to 100% more Damage with Hits per Steel Shard consumed at the start of their movement, lowering this bonus as they travel farther
Grants Minions 4% of Life Regenerated per Second if they have Blocked recentlyPlace into an item socket of the right colour to gain this skill. Right click to remove from a socket. N/A N/A 12
Tempest ShieldTempest ShieldSpell, Lightning, Chaining
Buff grants Immunity to ShockPlace into an item socket of the right colour to gain this skill. Right click to remove from a socket. N/A 16
Protective LinkProtective LinkSpell, Duration, Link
Cost: (15-32) Mana per second
Cast Time: 0.50 secRequires Level 34Targets an allied player to apply a buff which links you to them for a duration. While linked, they copy your block chance and recover life when they block. If the target dies while linked, you will also die. This skill cannot be triggered, or used by Totems, Traps, or Mines.Per 1% Quality:1Superior2Anomalous3Divergent1% increased Skill Effect Duration0.5% increased Buff EffectLinked target has 0.5% increased Stun ThresholdBase duration is (8-9.9) seconds
Linked target Recovers (31-297) Life when they Block
Link breaks if target leaves range or line of sight for 4 seconds
Maximum 1 Link from any source per target
You die if Linked target dies
Linked target's Chance to Block Attack Damage is equal to yours
Linked target's Maximum Chance to Block Attack Damage is equal to yoursPlace into an item socket of the right colour to gain this skill. Right click to remove from a socket. N/A 34
Effectiveness of Added Damage: (115-175)%Perform a swift counter-attack against enemies in a cone shape when you block with your shield.Per 1% Quality:1Superior2Anomalous3Divergent4Phantasmal0.5% increased Area of Effect2% chance to Debilitate Enemies for 1 second on Hit1% increased Cooldown Recovery Rate1% increased DamageTrigger this Skill when you Block
Rotted Round ShieldChance to Block: 29%
Curse Skills have 100% increased Skill Effect DurationDoomed to plunder forever. 5 60% increased Block Recovery+(5-10) to Armour
Pine BucklerChance to Block: 31%
Rawhide Tower ShieldChance to Block: (29%-31%)
Painted BucklerChance to Block: 29%
Jingling Spirit ShieldChance to Block: (29%-33%)
War BucklerChance to Block: 31%
1% of Damage Leeched as Life against Cursed EnemiesAn eye for an eye. A curse for a curse. 29 9% increased Movement Speed+1 to Level of Socketed Curse Gems
Blunt Arrow QuiverRequires Level 31Adds (7-9) to (13-16) Physical Damage to Attacks(20-24)% Chance to Block Attack Damage
- Kiravi, Vaal Archer 31 Adds (7-9) to (13-16) Physical Damage to Attacks(20-24)% Chance to Block Attack Damage
Painted Tower ShieldChance to Block: 30%
Crimson Round ShieldChance to Block: (34%-38%)
Baroque Round ShieldChance to Block: (28%-31%)
Energy Shield: (79-109)Requires Level 58, 64 Str, 64 Int(4-6)% Chance to Block Spell Damage
Vaal BucklerChance to Block: 32%
Ezomyte Tower ShieldChance to Block: 30%
Mirrored Spiked ShieldChance to Block: (29%-30%)
Shared SufferingOne's burden is another's gift. 66 +5% chance to Suppress Spell Damage(500-600)% increased Evasion and Energy Shield
Colossal Tower ShieldChance to Block: 29%
Colossal Tower ShieldLimited to: 1
+(20-25)% Chance to Block
(10-20)% of Cold Damage taken as Fire
(10-20)% of Lightning Damage taken as Fire
(10-20)% of Physical Damage taken as Fire
Scorch Enemies in Close Range when you Block"The newcomers warn of doom and death beyond mortal ken.
I ask, why should we fear the fire when we serve the Lord of Light?"
- Maxarius, the first High Templar 67 +(10-20) to maximum Life(150-250)% increased Armour
Nightmare BascinetArmour: (486-748)
Evasion: (699-1072)Requires Level 67, 62 Str, 85 DexSocketed Gems are Supported by Level 18 Melee Physical Damage
The Annihilating Light The Annihilating Light
QuarterstaffStaff
Weapon Range: 13Requires Level 68, 78 Str, 78 Int+18% Chance to Block Attack Damage while wielding a Staff(60-70)% reduced Elemental Resistances
Elemental Skills deal Triple DamageThere is no force more destructive in the heavens
than the scintillating light of utter clarity. 68 +18% Chance to Block Attack Damage while wielding a Staff(60-70)% reduced Elemental Resistances
Elemental Skills deal Triple Damage
Pinnacle Tower ShieldChance to Block: 30%
The Squire The Squire
Elegant Round ShieldChance to Block: (28%-30%)
Movement Speed: -3%Requires Level 70, 85 Str, 85 Dex120% increased Block RecoveryHas 3 Sockets
+(3-5)% Chance to BlockJudge not the weak, for
they empower the strong. 70 120% increased Block RecoveryHas 3 Sockets
Physical Damage: (90-100) to (157-166)
Physical Damage: (84-88.8) to (332.5-351.5)
Corrugated BucklerChance to Block: 27%
Conjurer BootsEnergy Shield: (86-117)Requires Level 53, 94 Int(4-6)% Chance to Block Spell Damage
Branded Kite ShieldChance to Block: 24%
Titanium Spirit ShieldChance to Block: 25%
- Researcher GravenRight click to drink. Can only hold charges while in belt. Refills as you kill monsters. 68 Gain 1 Endurance Charge on use
-Rumi of the VaalRight click to drink. Can only hold charges while in belt. Refills as you kill monsters. 68 +(8-12)% Chance to Block Attack Damage during Flask effect
Secutor HelmArmour: (156-208)
(15-20)% of Damage taken while affected by Clarity Recouped as Mana
+(12-15)% chance to Suppress Spell Damage while affected by Grace
+(5-8)% chance to Suppress Spell Damage while affected by Haste
(40-60)% increased Physical Damage while using Pride
30% increased Maximum total Energy Shield Recovery per second from Leech while affected by Zealotry
Studded Round ShieldChance to Block: 27%
Great HelmetArmour: (61-76)
Energy Shield: (14-17)Requires Level 22, 27 Str, 27 Int(20-22)% increased Stun and Block Recovery
Reinforced Tower ShieldChance to Block: 23%
Golden MaskEvasion: (432-515)
Sage's RobeEnergy Shield: (177-225)
Ivory Spirit ShieldEnergy Shield: (122-160)
Ornate RingmailArmour: (569-715)
+1% chance to Suppress Spell Damage per Endurance Charge
(6-8) to (12-13) Added Cold Damage per Frenzy Charge
+1% chance to Suppress Spell Damage per Frenzy Charge
+1% chance to Suppress Spell Damage per Power Charge
Movement Speed: -3%Requires Level 49, 119 Int(10-15)% increased Spell DamageTrigger a Socketed Warcry Skill on losing Endurance Charges, with a 0.25 second Cooldown
The idea, and our ideals, take root. 49 (10-15)% increased Spell DamageTrigger a Socketed Warcry Skill on losing Endurance Charges, with a 0.25 second Cooldown
Crusader PlateArmour: (1081-1382)
- Mauritius, the Iron Heart 59 +1% Chance to Block Attack Damage per 50 Strength
Lacquered BucklerChance to Block: 24%
- Lead Researcher Ksaret 60 6% increased Movement Speed(120-150)% increased Evasion Rating
Tarnished Spirit ShieldChance to Block: 24%
<Consumed support gem modifier>Be careful where you put your finger. 70
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Normal probability plot - MATLAB normplot - MathWorks France
Generate a Normal Probability Plot
Assess Normality Using a Normal Probability Plot
Adjust Normal Probability Plot Line Properties
normplot(ax,x)
h = normplot(___)
normplot(x) creates a normal probability plot comparing the distribution of the data in x to the normal distribution.
normplot plots each data point in x using plus sign ('+') markers and draws two reference lines that represent the theoretical distribution. A solid reference line connects the first and third quartiles of the data, and a dashed reference line extends the solid line to the ends of the data. If the sample data has a normal distribution, then the data points appear along the reference line. A distribution other than normal introduces curvature in the data plot.
normplot(ax,x) adds a normal probability plot into the axes specified by ax.
h = normplot(___) returns graphics handles corresponding to the plotted lines, using any of the previous syntaxes.
Generate random sample data from a normal distribution with mu = 10 and sigma = 1.
x = normrnd(10,1,25,1);
Create a normal probability plot of the sample data.
The plot indicates that the data follows a normal distribution.
Generate 50 random numbers from each of four different distributions: A standard normal distribution; a Student's-t distribution with five degrees of freedom (a "fat-tailed" distribution); a set of Pearson random numbers with mu equal to 0, sigma equal to 1, skewness equal to 0.5, and kurtosis equal to 3 (a "right-skewed" distribution); and a set of Pearson random numbers with mu equal to 0, sigma equal to 1, skewness equal to -0.5, and kurtosis equal to 3 (a "left-skewed" distribution).
x1 = normrnd(0,1,[50,1]);
x2 = trnd(5,[50,1]);
x3 = pearsrnd(0,1,0.5,3,[50,1]);
x4 = pearsrnd(0,1,-0.5,3,[50,1]);
Plot four histograms on the same figure for a visual comparison of the pdf of each distribution.
histogram(x1,10)
title('Normal')
axis([-4,4,0,15])
title('Fat Tails')
title('Right-Skewed')
title('Left-Skewed')
The histograms show how each sample differs from the normal distribution.
Create a normal probability plot for each sample.
normplot(x1)
Create a 50-by-2 matrix containing 50 random numbers from each of two different distributions: A standard normal distribution in column 1, and a set of Pearson random numbers with mu equal to 0, sigma equal to 1, skewness equal to 0.5, and kurtosis equal to 3 (a "right-skewed" distribution) in column 2.
x = [normrnd(0,1,[50,1]) pearsrnd(0,1,0.5,3,[50,1])];
Create a normal probability plot for both samples on the same figure. Return the plot line graphic handles.
h = normplot(x)
legend({'Normal','Right-Skewed'},'Location','southeast')
The handles h(1) and h(2) correspond to the data points for the normal and skewed distributions, respectively. The handles h(3) and h(4) correspond to the second and third quartile line fit to the sample data. The handles h(5) and h(6) correspond to the extrapolated line that extends to the minimum and maximum of each set of sample data.
To illustrate, increase the line width of the second and third quartile line for the normally distributed data sample (represented by h(3)) to 2.
Sample data, specified as a numeric vector or numeric matrix. normplot displays each value in x using the symbol '+'. If x is a matrix, then normplot displays a separate line for each column of x.
Target axes, specified as an Axes object or a UIAxes object. normplot adds an additional plot into the axes specified by ax. For details, see Axes Properties and UIAxes Properties.
Graphics handles for line objects, returned as a vector of Line graphics handles. Graphics handles are unique identifiers that you can use to query and modify the properties of a specific line on the plot. For each column of x, normplot returns three handles:
The line representing the data points. normplot represents each data point in x using plus sign ('+') markers.
normplot matches the quantiles of sample data to the quantiles of a normal distribution. The sample data is sorted and plotted on the x-axis. The y-axis represents the quantiles of the normal distribution, converted into probability values. Therefore, the y-axis scaling is not linear.
Where the x-axis value is the ith sorted value from a sample of size N, the y-axis value is the midpoint between evaluation points of the empirical cumulative distribution function of the data. The midpoint is equal to
\frac{\left(i-0.5\right)}{N}
normplot superimposes a reference line to assess the linearity of the plot. The line goes through the first and third quartiles of the data.
probplot | cdfplot | wblplot | ecdf
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Q The mean of the following distribution is 57 6 and the sum of its frequencies is 50, find - Maths - Statistics - 12672021 | Meritnation.com
Q. The mean of the following distribution is 57.6 and the sum of its frequencies is 50, find the missing frequencies
{f}_{1} and {f}_{2}
Frequency 7
{f}_{1}
{f}_{2}
Q.27 https://www.meritnation.com/ask-answer/question/if-the-arithmetic-mean-of-the-following-distribution-is-5/statistics/5056701
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1) If A and B are mutually exclusive events with P(A) = 0.3 and P(B) = 0.5, then P(A\cap B
1) If A and B are mutually exclusive events with P(A) = 0.3 and P(B) = 0.5, then P(A\cap B)=? 2) An experiment consists of four outcomes with P(E_1)=0
1) If A and B are mutually exclusive events with
P\left(A\right)=0.3
P\left(B\right)=0.5
P\left(A\cap B\right)=?
2) An experiment consists of four outcomes with
P\left({E}_{1}\right)=0.2,P\left({E}_{2}\right)=0.3,
P\left({E}_{3}\right)=0.4
. The probability of outcome
{E}_{4}
is ?
P\left(A\right)=0.3
P\left(B\right)=0.5
P\left(A\cap B\right)=?
4) The empirical rule states that, for data having a bell-shaped distribution, the percentage of data values being within one standard deviation of the mean is approximately
5) If a penny is tossed four times and comes up heads all four times, the probability of heads on the fifth trial is ?
1) Since events are mutually exclusive :
P\left(A\cap B\right)=0
P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)=0.3+0.5-0=0.8
2) since sum of all events is 1
P\left(E4\right)=1-0.2-0.3-0.4=0.1
3) here sign is missing , instead ? is given in question: if it is
P\left(A\cap B\right)
; then it is 0
4) percentage of data values being within one standard deviation of the mean is approximately =68
5) probability of heads on the fifth trial is =0.5
The middle 95% of residuals should be between which two values? Use this information to give an interval of plausible values for the weekly sales revenue if 5 linear feet are allocated to the stores
Describe in words the surface whose equation is given. (assume that r is not negative.)
\theta =\frac{\pi }{4}
a) The plane
y=-z
where y is not negative
b) The plane
y=z
where y and z are not negative
c) The plane
y=x
where x and y are not negative
d) The plane
y=-x
e) The plane
x=z
Explain why t distributions tend to be flatter and more spread out than the normal distribution.
N=48
, find tCV and use it to draw a t distribution with the rare and common zones labeled.
Which of the following binomial distributions can be well approximated by a normal distribution? A Poisson distribution? Both? Neither?
n=1000
p=.001
Do men and women differ in their attitudes toward public corruption and tax evasion? This was the question of interest in a study published in Contemporary Economic Policy (Oct. 2010). The data for the analysis were obtained from a representative sample of over 30,000 Europeans. Each person was asked how justifiable it is for someone to (1) accept a bribe in the course of their duties and (2) cheat on their taxes. Responses were measured as 0, 1, 2, or 3, where O = "always justified" and 3 = "never justified." The large-sample Wilcoxon rank sum test was applied in order to compare the response distributions of men and women.
a. Give the null hypothesis for the test in the words of the problem.
b. An analysis of the "justifiability of corruption" responses yielded a large-sample test statistic of
z=-14.10
with a corresponding p-value of approximately 0. Interpret this result.
c. Refer to part b. Women had a larger rank sum statistic than men. What does this imply about gender attitudes. toward corruption?
d. An analysis of the "justifiability of tax evasion" responses yielded a large-sample test statistic of
z=-18.12
e. Refer to part d. Again, women had a larger rank sum statistic than men. What does this imply about gender attitudes toward tax evasion?
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{\displaystyle v=r_{A}\omega _{A}=r_{B}\omega _{B},\!}
{\displaystyle {\frac {\omega _{A}}{\omega _{B}}}={\frac {r_{B}}{r_{A}}}={\frac {N_{B}}{N_{A}}}.}
{\displaystyle \mathrm {MA} ={\frac {N_{B}}{N_{A}}}.}
{\displaystyle {\frac {\omega _{A}}{\omega _{B}}}=R.}
{\displaystyle \omega _{A}=\omega ,\quad \omega _{B}=\omega /R.\!}
{\displaystyle F_{\theta }=T_{A}{\frac {\partial \omega _{A}}{\partial \omega }}-T_{B}{\frac {\partial \omega _{B}}{\partial \omega }}=T_{A}-T_{B}/R=0.}
{\displaystyle \mathrm {MA} ={\frac {T_{B}}{T_{A}}}=R.}
{\displaystyle \omega _{A}\!}
{\displaystyle \omega _{B}\!}
{\displaystyle v=r_{A}\omega _{A}=r_{B}\omega _{B},\!}
{\displaystyle {\frac {\omega _{A}}{\omega _{B}}}={\frac {r_{B}}{r_{A}}}={\frac {N_{B}}{N_{A}}}.}
{\displaystyle R={\frac {\omega _{A}}{\omega _{B}}}={\frac {N_{B}}{N_{A}}}.}
{\displaystyle p=2t.\!}
{\displaystyle p={\frac {2\pi r_{A}}{N_{A}}}.}
{\displaystyle p={\frac {2\pi r_{A}}{N_{A}}}={\frac {2\pi r_{B}}{N_{B}}}.}
{\displaystyle {\frac {r_{B}}{r_{A}}}={\frac {N_{B}}{N_{A}}}.}
{\displaystyle R={\frac {\omega _{A}}{\omega _{B}}}={\frac {r_{B}}{r_{A}}},}
{\displaystyle R={\frac {\omega _{A}}{\omega _{B}}}={\frac {N_{B}}{N_{A}}}.}
{\displaystyle R={\frac {T_{B}}{T_{A}}},}
{\displaystyle {\mathit {MA}}={\frac {T_{B}}{T_{A}}}.}
{\displaystyle {\frac {\omega _{A}}{\omega _{I}}}={\frac {N_{I}}{N_{A}}},\quad {\frac {\omega _{I}}{\omega _{B}}}={\frac {N_{B}}{N_{I}}}.}
{\displaystyle R={\frac {\omega _{A}}{\omega _{B}}}={\frac {N_{B}}{N_{A}}}.}
{\displaystyle d={\frac {c_{t}}{gr_{t}\times gr_{d}}}}
{\displaystyle v_{c}={\frac {c_{t}\times v_{e}}{gr_{t}\times gr_{d}}}}
GearGear bearingGear pumpGearboxGearboxesSprocketMechanical engineeringFunicularHerringbone gearCouplingEpicyclic gearingRotationRotational speedInter-city railEast Coast Main LineFlued boilerToy trainWooden toy trainRail transport modelling
This article uses material from the Wikipedia article "Gear train", which is released under the Creative Commons Attribution-Share-Alike License 3.0. There is a list of all authors in Wikipedia
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Engineering Acoustics/Moving Resonators - Wikibooks, open books for an open world
Engineering Acoustics/Moving Resonators
Moving ResonatorsEdit
Consider the situation shown in the figure below. We have a typical Helmholtz resonator driven by a massless piston which generates a sinusoidal pressure
{\displaystyle P_{G}}
, however the cavity is not fixed in this case. Rather, it is supported above the ground by a spring with compliance
{\displaystyle C_{M}}
. Assume the cavity has a mass
{\displaystyle M_{M}}
Recall the Helmholtz resonator (see Module #9). The difference in this case is that the pressure in the cavity exerts a force on the bottom of the cavity, which is now not fixed as in the original Helmholtz resonator. This pressure causes a force that acts upon the cavity bottom. If the surface area of the cavity bottom is
{\displaystyle S_{C}}
, then Newton's Laws applied to the cavity bottom give
{\displaystyle \sum {F}=p_{C}S_{C}-{\frac {x}{C_{M}}}=M_{M}{\ddot {x}}\Rightarrow p_{C}S_{C}=\left[{\frac {1}{j\omega C_{M}}}+j\omega M_{M}\right]u}
In order to develop the equivalent circuit, we observe that we simply need to use the pressure (potential across
{\displaystyle C_{A}}
) in the cavity to generate a force in the mechanical circuit. The above equation shows that the mass of the cavity and the spring compliance should be placed in series in the mechanical circuit. In order to convert the pressure to a force, the transformer is used with a ratio of
{\displaystyle 1:S_{C}}
A practical example of a moving resonator is a marimba. A marimba is a similar to a xylophone but has larger resonators that produce deeper and richer tones. The resonators (seen in the picture as long, hollow pipes) are mounted under an array of wooden bars which are struck to create tones. Since these resonators are not fixed, but are connected to the ground through a stiffness (the stand), it can be modeled as a moving resonator. Marimbas are not tunable instruments like flutes or even pianos. It would be interesting to see how the tone of the marimba changes as a result of changing the stiffness of the mount.
For more information about the acoustics of marimbas see http://www.mostlymarimba.com/techno1.html
Retrieved from "https://en.wikibooks.org/w/index.php?title=Engineering_Acoustics/Moving_Resonators&oldid=3232733"
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Patching and local-global principles for homogeneous spaces over function fields of <em>p</em>-adic curves | EMS Press
JournalscmhVol. 87, No. 4pp. 1011–1033
Patching and local-global principles for homogeneous spaces over function fields of <em>p</em>-adic curves
Venapally Suresh
F=K(X)
be the function field of a smooth projective curve over a
p
K
. To each rank one discrete valuation of
F
one may associate the completion
F_v
. Given an
F
Y
which is a homogeneous space of a connected reductive group
G
F
, one may wonder whether the existence of
F_v
-points on
Y
v
is enough to ensure that
Y
has an
F
-point. In this paper we prove such a result in two cases:
Y
is a smooth projective quadric and
p
G
is the extension of a reductive group over the ring of integers of
K
Y
is a principal homogeneous space of
G
An essential use is made of recent patching results of Harbater, Hartmann and Krashen. There is a connection to injectivity properties of the Rost invariant and a result of Kato.
Jean-Louis Colliot-Thélène, Raman Parimala, Venapally Suresh, Patching and local-global principles for homogeneous spaces over function fields of <em>p</em>-adic curves. Comment. Math. Helv. 87 (2012), no. 4, pp. 1011–1033
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Adjusting Entries for Deferrals - Course Hero
Principles of Accounting/Adjusting Process/Adjusting Entries for Deferrals
Learn all about adjusting entries for deferrals in just a few minutes! Fabio Ambrosio, CPA, instructor of accounting at the Central Washington University, walks through adjusting for deferral entries for unearned revenue and deferral entries for prepaid expense.
Deferral Entries for Unearned Revenue
Deferral entries may be needed when revenue is unearned, which impacts the accounting equation, the income statement, and the balance sheet.
Unearned revenue is deferred revenue and can be an advance payment for goods or services on the balance sheet. It is a liability and therefore has not been earned. This is because the business owes the customer for goods or services since the customer has paid for them but has not received them. Unearned revenue can be items such as rent or subscriptions. If the unearned revenue is not reported as a liability on the balance sheet and claimed as revenue immediately, the matching principle of GAAP would be violated. So, a deferral entry is needed. Once the products are delivered, the deferred, or unearned, revenue then becomes earned revenue.
For example, regarding unearned or deferred revenue, Global Air receives $5,000 cash in ticket sales on July 1. Half of the flights occur in July, and the balance will occur in August.
Global Air will need to book flights from the $5,000 cash received from its customers on July 1; it has an obligation (a liability) to provide services at a future date. Since the services (flights) will take place partly in July and partly in August, only half of this amount is July revenue. The balance after the July revenue is deducted will include unearned revenue, or deferred revenue. The adjusting entry affects both an income statement account and a balance sheet account.
Global Air recognizes a portion of the unearned revenue as an adjusting entry.
Adjusting Entries for Unearned or Deferred Revenue
Journal Entry at the Time of the Event
Adjusting Entry on July 31
Global Air receives $5,000 cash in ticket sales on July 1. Half of the flights occur in July, and the balance will occur in August. Debit Cash (increase in Cash $5,000 [asset]);
Credit Unearned Revenue (increase in Unearned Revenue $5,000 [liability]) Yes. Global Air has received ticket sales in advance of the flights. $2,500 can be recognized as immediate revenue, as half of the flights occur in July. The balance remains unearned. Debit Unearned Revenue $2,500 (decrease in liability)
Credit Revenue $2,500 (increase in revenue)
At month-end after the adjusting entry, unearned revenue will decrease on the balance sheet, and revenue will increase on the income statement.
Deferral Entries for Prepaid Expense
When expenses are prepaid, deferral entries may be needed, as they impact the accounting equation, the income statement, and the balance sheet.
As the name suggests, prepaid expenses are exactly that—prepaid, or paid in advance of services being rendered or goods being delivered. Because the expense is prepaid, it is deferred, and an adjusting entry needs to be made. A prepaid expense often happens in the case of purchasing insurance coverage.
For example, Global Air pays $6,000 for insurance coverage for 6 months (July through December) on July 1. Global Air will decrease cash (credit Cash) and increase asset (the Prepaid Expense account) on July 1. At month-end, one month's worth of insurance coverage has been provided, or used up, meaning $1,000, or
1/6\;\text{x}\;\$6\text{,}000
, is no longer prepaid and needs to be recognized as an expense for the month of July. Therefore, the adjusting entry on July 31 is a debit to Insurance Expense (increase) and credit to Prepaid Expense (decrease) for $1,000. The adjusting entry affects both an income statement account and a balance sheet account.
Global Air recognizes the expense for July by adjusting or reducing the prepaid expense account.
Adjusting Entry with Reduced Prepaid Expense
Global Air pays $6,000 for insurance coverage for 6 months (July through December) on July 1. Debit (increase in Prepaid Expense $6,000 [asset]);
Credit (decrease Cash $6,000 [asset]) Yes. Global Air has prepaid 6 months of insurance coverage. One month's insurance expense will need to be recognized for July. Debit Insurance Expense $1,000 (increase expense)
Credit Prepaid Expense (decrease asset)
Global Air recognizes the expense for July by adjusting or reducing the prepaid expense account when the service is provided. This reduces prepaid expense on the balance sheet and increases insurance expense on the income statement.
<Adjusting Entries for Revenues and Expenses>Adjusting Entries for Depreciation
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Engineering Acoustics/Rotor Stator Interactions - Wikibooks, open books for an open world
Engineering Acoustics/Rotor Stator Interactions
An important issue for the aeronautical industry is the reduction of aircraft noise. The characteristics of the turbomachinery noise are to be studied. The rotor/stator interaction is a predominant part of the noise emission. We will present an introduction to these interaction theory, whose applications are numerous. For example, the conception of air-conditioning ventilators requires a full understanding of this interaction.
1 Noise emission of a Rotor-Stator mechanism
2 Optimization of the number of blades
3 Determination of source levels
Noise emission of a Rotor-Stator mechanismEdit
A Rotor wake induces on the downstream Stator blades a fluctuating vane loading, which is directly linked to the noise emission.
We consider a B blades Rotor (at a rotation speed of
{\displaystyle \Omega }
) and a V blades stator, in a unique Rotor/Stator configuration. The source frequencies are multiples of
{\displaystyle B\Omega }
{\displaystyle mB\Omega }
. For the moment we don’t have access to the source levels
{\displaystyle F_{m}}
. The noise frequencies are also
{\displaystyle mB\Omega }
, not depending on the number of blades of the stator. Nevertheless, this number V has a predominant role in the noise levels (
{\displaystyle P_{m}}
) and directivity, as it will be discussed later.
For an airplane air-conditioning ventilator, reasonable data are :
{\displaystyle B=13}
{\displaystyle \Omega =12000}
rnd/min
The blade passing frequency is 2600 Hz, so we only have to include the first two multiples (2600 Hz and 5200 Hz), because of the human ear high-sensibility limit. We have to study the frequencies m=1 and m=2.
Optimization of the number of bladesEdit
As the source levels can't be easily modified, we have to focuse on the interaction between those levels and the noise levels.
{\displaystyle {{F_{m}} \over {P_{m}}}}
contains the following part :
{\displaystyle \sum \limits _{s=-\infty }^{s=+\infty }{e^{-{{i(mB-sV)\pi } \over 2}}J_{mB-sV}}(mBM)}
{\displaystyle J_{mB-sV}}
the Bessel function of mB-sV order. In order to minimize the influence of the transfer function, the goal is to reduce the value of this Bessel function. To do so, the argument must be smaller than the order of the Bessel function.
Back to the example :
For m=1, with a Mach number M=0.3, the argument of the Bessel function is about 4. We have to avoid having mB-sV inferior than 4. If V=10, we have 13-1x10=3, so there will be a noisy mode. If V=19, the minimum of mB-sV is 6, and the noise emission will be limited.
The case that is to be strictly avoided is when mB-sV can be nul, which causes the order of the Bessel function to be 0. As a consequence, we have to take care having B and V prime numbers.
Determination of source levelsEdit
The minimization of the transfer function
{\displaystyle {{F_{m}} \over {P_{m}}}}
is a great step in the process of reducing the noise emission. Nevertheless, to be highly efficient, we also have to predict the source levels
{\displaystyle F_{m}}
. This will lead us to choose to minimize the Bessel functions for the most significative values of m. For example, if the source level for m=1 is very higher than for m=2, we will not consider the Bessel functions of order 2B-sV. The determination of the source levels is given by the Sears theory,which will not be explicited here.
All this study was made for a privilegiate direction : the axis of the Rotor/Stator. All the results are acceptable when the noise reduction is ought to be in this direction. In the case where the noise to reduce is perpendicular to the axis, the results are very different, as those figures shown :
For B=13 and V=13, which is the worst case, we see that the sound level is very high on the axis (for
{\displaystyle \theta =0}
For B=13 and V=19, the sound level is very low on the axis but high perpendicularly to the axis (for
{\displaystyle \theta =Pi/2}
Retrieved from "https://en.wikibooks.org/w/index.php?title=Engineering_Acoustics/Rotor_Stator_Interactions&oldid=3476303"
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Calculus/Integration techniques/Integration by Parts - Wikibooks, open books for an open world
Calculus/Integration techniques/Integration by Parts
← Integration techniques/Recognizing Derivatives and the Substitution Rule Calculus Integration techniques/Trigonometric Substitution →
Integration techniques/Integration by Parts
Continuing on the path of reversing derivative rules in order to make them useful for integration, we reverse the product rule.
1.2 With definite integral
Wikipedia has related information at Integration by parts
{\displaystyle y=uv}
{\displaystyle u}
{\displaystyle v}
{\displaystyle x}
{\displaystyle y'=(uv)'=u'v+uv'}
{\displaystyle uv'=(uv)'-u'v}
{\displaystyle \int uv'dx=\int (uv)'\ dx-\int u'v\ dx}
{\displaystyle \int uv'\ dx=uv-\int vu'\ dx}
{\displaystyle \int u\ dv=uv-\int v\ du}
This is the integration by parts formula. It is very useful in many integrals involving products of functions, as well as others.
For instance, to treat
{\displaystyle \int x\sin(x)dx}
{\displaystyle u=x}
{\displaystyle dv=\sin(x)dx}
. With these choices, we have
{\displaystyle du=dx}
{\displaystyle v=-\cos(x)}
{\displaystyle \int x\sin(x)dx=-x\cos(x)-\int {\big (}-\cos(x){\big )}dx=-x\cos(x)+\int \cos(x)dx=\sin(x)-x\cos(x)+C}
Note that the choice of
{\displaystyle u}
{\displaystyle dv}
was critical. Had we chosen the reverse, so that
{\displaystyle u=\sin(x)}
{\displaystyle dv=x\ dx}
, the result would have been
{\displaystyle {\frac {x^{2}\sin(x)}{2}}-\int {\frac {x^{2}\cos(x)}{2}}dx}
The resulting integral is no easier to work with than the original; we might say that this application of integration by parts took us in the wrong direction.
So the choice is important. One general guideline to help us make that choice is, if possible, to choose
{\displaystyle u}
to be the factor of the integrand which becomes simpler when we differentiate it. In the last example, we see that
{\displaystyle \sin(x)}
does not become simpler when we differentiate it:
{\displaystyle \cos(x)}
is no simpler than
{\displaystyle \sin(x)}
An important feature of the integration by parts method is that we often need to apply it more than once. For instance, to integrate
{\displaystyle \int x^{2}e^{x}dx}
we start by choosing
{\displaystyle u=x^{2}}
{\displaystyle dv=e^{x}dx}
{\displaystyle \int x^{2}e^{x}dx=x^{2}e^{x}-2\int xe^{x}dx}
Note that we still have an integral to take care of, and we do this by applying integration by parts again, with
{\displaystyle u=x}
{\displaystyle dv=e^{x}dx}
{\displaystyle \int x^{2}e^{x}dx=x^{2}e^{x}-2\int xe^{x}dx=x^{2}e^{x}-2(xe^{x}-e^{x})+C=x^{2}e^{x}-2xe^{x}+2e^{x}+C}
So, two applications of integration by parts were necessary, owing to the power of
{\displaystyle x^{2}}
Note that any power of x does become simpler when we differentiate it, so when we see an integral of the form
{\displaystyle \int x^{n}f(x)dx}
one of our first thoughts ought to be to consider using integration by parts with
{\displaystyle u=x^{n}}
. Of course, in order for it to work, we need to be able to write down an antiderivative for
{\displaystyle dv}
Use integration by parts to evaluate the integral
{\displaystyle \int e^{x}\sin(x)dx}
Solution: If we let
{\displaystyle u=\sin(x)}
{\displaystyle v'=e^{x}dx}
{\displaystyle u'=\cos(x)dx}
{\displaystyle v=e^{x}}
. Using our rule for integration by parts gives
{\displaystyle \int e^{x}\sin(x)dx=e^{x}\sin(x)-\int e^{x}\cos(x)dx}
We do not seem to have made much progress.
But if we integrate by parts again with
{\displaystyle u=\cos(x)}
{\displaystyle v'=e^{x}dx}
{\displaystyle u'=-\sin(x)dx}
{\displaystyle v=e^{x}}
{\displaystyle \int e^{x}\sin(x)dx=e^{x}\sin(x)-e^{x}\cos(x)-\int e^{x}\sin(x)dx}
We may solve this identity to find the anti-derivative of
{\displaystyle \sin(x)e^{x}}
{\displaystyle \int e^{x}\sin(x)dx={\frac {e^{x}{\big (}\sin(x)-\cos(x){\big )}}{2}}+C}
With definite integralEdit
For definite integrals the rule is essentially the same, as long as we keep the endpoints.
Integration by parts for definite integrals Suppose f and g are differentiable and their derivatives are continuous. Then
{\displaystyle \int \limits _{a}^{b}f(x)g'(x)dx={\big (}f(x)g(x){\big )}{\bigg |}_{a}^{b}-\int \limits _{a}^{b}f'(x)g(x)dx}
{\displaystyle =f(b)g(b)-f(a)g(a)-\int \limits _{a}^{b}f'(x)g(x)dx}
This can also be expressed in Leibniz notation.
{\displaystyle \int \limits _{a}^{b}u\ dv=(uv){\Big |}_{a}^{b}-\int \limits _{a}^{b}v\ du.}
Examples Set 1: Integration by Parts
Evaluate the following using integration by parts.
{\displaystyle \int -4\ln(x)dx}
{\displaystyle 4x-4x\ln(x)+C}
{\displaystyle 4x-4x\ln(x)+C}
{\displaystyle \int (38-7x)\cos(x)dx}
{\displaystyle (38-7x)\sin(x)-7\cos(x)+C}
{\displaystyle (38-7x)\sin(x)-7\cos(x)+C}
{\displaystyle \int \limits _{0}^{\tfrac {\pi }{2}}(-6x+45)\cos(x)dx}
{\displaystyle 51-3\pi }
{\displaystyle 51-3\pi }
{\displaystyle \int (5x+1)(x-6)^{4}dx}
{\displaystyle {\frac {(5x+1)(x-6)^{5}}{5}}-{\frac {(x-6)^{6}}{6}}+C}
{\displaystyle {\frac {(5x+1)(x-6)^{5}}{5}}-{\frac {(x-6)^{6}}{6}}+C}
{\displaystyle \int \limits _{-1}^{1}(2x+8)^{3}(2-x)dx}
{\displaystyle 1916.8}
{\displaystyle 1916.8}
Videos and exercises on integration by parts at Khan Academy
Retrieved from "https://en.wikibooks.org/w/index.php?title=Calculus/Integration_techniques/Integration_by_Parts&oldid=3579998"
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Elliptic analog lowpass filter prototype - MATLAB ellipap - MathWorks Switzerland
[z,p,k] = ellipap(n,Rp,Rs) returns the zeros, poles, and gain of an order n elliptic analog lowpass filter prototype, with Rp dB of ripple in the passband, and a stopband Rs dB down from the peak value in the passband. The zeros and poles are returned in length n column vectors z and p and the gain in scalar k. If n is odd, z is length n - 1. The transfer function in factored zero-pole form is
H\left(s\right)=\frac{z\left(s\right)}{p\left(s\right)}=k\frac{\left(s-{z}_{1}\right)\left(s-{z}_{2}\right)\dots \left(s-{z}_{N}\right)}{\left(s-{p}_{1}\right)\left(s-{p}_{2}\right)\dots \left(s-{p}_{M}\right)}
Elliptic filters offer steeper rolloff characteristics than Butterworth and Chebyshev filters, but they are equiripple in both the passband and the stopband. Of the four classical filter types, elliptic filters usually meet a given set of filter performance specifications with the lowest filter order.
ellipap sets the passband edge angular frequency ω0 of the elliptic filter to 1 for a normalized result. The passband edge angular frequency is the frequency at which the passband ends and the filter has a magnitude response of 10-Rp/20.
ellipap uses the algorithm outlined in [1]. It employs ellipke to calculate the complete elliptic integral of the first kind and ellipj to calculate Jacobi elliptic functions.
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Find an equation of the plane tangent to the following surface at the given point. 7xy+yz
Find an equation of the plane tangent to the following surface at the given point. 7xy+yz+4xz-48=0;\ (2,2,2)
7xy+yz+4xz-48=0;\text{ }\left(2,2,2\right)
f\left(x,y,z\right)=7xy+yz+4xz-48=0
Tangent direction is
\left({P}_{x},{P}_{y},{P}_{z}\right)
{P}_{x}=7y+yz,\text{ }{P}_{y}=7x+z,\text{ }{P}_{z}=y+4x
at (2,2,2)
{P}_{x}=22,\text{ }{P}_{y}=16,\text{ }{P}_{z}=10
⇒x\left(t\right)=\left(2,2,2\right)+t\left(22,16,10\right)
⇒22\left(x-z\right)+16\left(y-z\right)+10\left(z-z\right)=0
22x+16y+10z-96=0
Evaluate the double integral
{y}^{2}dA
, D is the triangular region with vertices (0, 1), (1,2), (4,1)
Parametric to polar equations Find an equation of the following curve in polar coordinates and describe the curve.
x=\left(1+cost\right)cost,y=\left(1+cost\right)sint,0\le t\le 2\pi
{P}_{0}=\left(1,0,-2\right)
v=\left(1,3,-1\right)
write a parametric vector equation for a line through
{P}_{0}
in the direction of v
Curves to parametric equations Find parametric equations for the following curves. Include an interval for the parameter values. Answers are not unique. The line segment starting at
P\left(-1,\text{ }-3\right)\text{ }\text{and ending at}\text{ }Q\left(6,\text{ }-16\right)
Graph the sets of points whose polar coordinates satisfy the equations and inequalities
r\ge 1
Let C be the curve described by the parametric equations x(t)= t, y(t) =
{t}^{3}
a) sketch a graph of C
b) find the vector-valued function f(t) associated with this parameterization of the curve C.
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What is the domain of y=x+7
Both the domain and range of y= x+7 are all real numbers (ℝ).
y=5x+2
What is the graph of
f\left(x\right)={x}^{2}
\frac{3x}{5}-x=\frac{x}{10}-\frac{5}{2}
A real-estate agent is trying to determine the relationship between the distance a 3-bedroom home is from New York City and its average selling price. He records the data for 6 homes shown below. Linear or Quadratic? Equation: Approximate cost of home 90 miles from NYC?
Complete the equation of the line through (2, 1) and (5, -8). Use exact numbers.
Find the x and y intercepts and graph the line.
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PrimePowerFactors - Maple Help
Home : Support : Online Help : Mathematics : Group Theory : PrimePowerFactors
factor a group element as a product of elements of prime power order
PrimePowerFactors( g, G )
element of G whose factorization is to be computed
g
of finite order in a group
G
, the prime power factors of
g
are elements
{g}_{1},{g}_{2},..,{g}_{k}
G
g={g}_{1}·{g}_{2}..{g}_{k}
, and such that each
{g}_{i}
has order equal to a power of a prime number. The elements
{g}_{i}
are pairwise commutative, and are uniquely determined up to the order in which they occur.
The PrimePowerFactors( g, G ) command computes the prime power factors of the group element
g
\mathrm{with}\left(\mathrm{GroupTheory}\right):
f≔\mathrm{PrimePowerFactors}\left(\mathrm{Perm}\left([[1,2,3,4,5,6]]\right),\mathrm{Symm}\left(6\right)\right)
\textcolor[rgb]{0,0,1}{f}\textcolor[rgb]{0,0,1}{≔}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\right)\left(\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}\right)\left(\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{6}\right)\textcolor[rgb]{0,0,1}{,}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\right)\left(\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{6}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\right)
\mathrm{andmap}\left(\mathrm{type},\mathrm{map}\left(\mathrm{ElementOrder},[f],\mathrm{Symm}\left(6\right)\right),'\mathrm{primepower}'\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{PermProduct}\left(f\right)
\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{6}\right)
The GroupTheory[PrimePowerFactors] command was introduced in Maple 2019.
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Evaluate the indefinite integral. \int sec^{3} x \tan x dx
\int {\mathrm{sec}}^{3}x\mathrm{tan}xdx
Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis. Verify your results using the integration capabilities of a graphing utility.
y=\mathrm{cos}2x
x=\frac{\pi }{4}
Find the indefinite integral:
\int \left({\mathrm{sin}}^{3}x\right)\left(\mathrm{cos}x\right)dx
\int \frac{{\left(\mathrm{ln}\left(x\right)\right)}^{3}}{2x}dx
\int {t}^{2}{\mathrm{sec}}^{2}\left(9{t}^{3}+1\right)dt
{\int }_{1}^{9}\frac{1}{\sqrt{x}{\left(1+\sqrt{x}\right)}^{2}}dx
\int \frac{3}{4}{e}^{\mathrm{tan}\left(6x\right)}{\mathrm{sec}}^{2}\left(6x\right)dx
Decide whether the integral is improper.
{\int }_{1}^{2}\frac{dx}{{x}^{3}}
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Which of the following pair of compounds will have three sp2 -hybrid orbitals?
SO2, CH4
SO3, C2H4
BF3, SF3
{\mathrm{I}}_{3}^{-}
, SF4
Among all the given pairs, second pair has sp2- hybrid orbitals.
SO2 is trigonal planar. It has 3 sp2- hybrid orbitals.
CH4 is tetrahedral. It has 4 sp3- hybrid orbitals.
SO3 and C2H4 are trigonalplanar and have 3 sp2- hybrid orbitals.
BF3 is trigonal planar. It also has 3 sp2- hybrid orbitals.
SF4 is irregular tetrahedral. It has 5 sp3d hybrid orbitals.
{\mathrm{I}}_{3}^{-}
L is linear and has 5 sp3d hybrid orbitals.
Work is being performed, when a weight lifter lifts a base ball off a weight rack. This is due to
Due to gravity, barbell is lifed and hence work is done.
What will be the number of waves formed by a Bohr electron in one complete revolution in its second orbit?
Number of waves =
\frac{\mathrm{Number}\quad \mathrm{of}\quad \mathrm{waves}}{\mathrm{de}-\quad \mathrm{Brogile}\quad \mathrm{wavelength}\quad \mathrm{of}\quad \mathrm{electron}}
\mathrm{\lambda }\quad =\quad \frac{\mathrm{h}}{\mathrm{mv}}
\therefore
\mathrm{\lambda }\quad =\quad \frac{2\mathrm{\pi }3\quad \times \quad \mathrm{\mu }}{\mathrm{h}}\quad =\quad \frac{2\mathrm{\pi }}{\mathrm{n}}\quad \times \quad \mathrm{mvr}
\because \quad \mathrm{mvr}\quad =\quad \frac{\mathrm{nh}}{2\mathrm{\pi }}\quad =\quad \frac{2\mathrm{\pi }}{\mathrm{h}}\quad \times \quad \frac{\mathrm{nh}}{2\mathrm{\pi }}
For second orbit, n = 2. Therefore, number of waves = 2.
At 27°C, one mole of an ideal gas is compressed isothermally and reversibly from a pressure of 2 atm to 10 atm. Choose the correct option from the following
Change in internal energy is positive
Heat is negative
Work done is - 965.84 cal
Work done in isothermal reversible process is
w = -2.303 nRT
{\mathrm{log}}_{10}\left(\frac{{\mathrm{p}}_{1}}{{\mathrm{p}}_{2}}\right)
Given, n = 1; R = 2 cal K-1 mol-1
p1 = 2 atm; p2 = 10 atm
w = -2.303
\times
\times
\times
{\mathrm{log}}_{10}\left(\frac{2}{10}\right)
w = + 965.54 cal
For isothermal change,
∆\mathrm{U}\quad =\quad 0
Now, from first law of thermodynamics,
∆
U - w = 0 - 965.84 cal
q = -965.84 cal
Beryllium differs in properties from other elements of its own group but shows resemblance with aluminium because of
relatively bigger ionic radius and high polarising power of Be
relatively smaller ionic radius and high polarising power of Be
relatively bigger ionic radius is the only reason behind this
Beryllium or Be differs in properties from other elements of its own group but shows resemblance with aluminium because of its smaller ionic radius and high polarising power. Polarising powers of Be2+ and Al3+ ions are almost the same thus, they show similarities in their properties.
Choose the incorrect statement about noble gas?
Boiling point increases with increasing atomic mass
Helium has least tendency to form compound
Noble gases have some value of electron affinity
Xenon has maximum number of compounds
Noble gases make up a group of chemical elements with similar properties. These include helium, neon, argon, krypton, xenon and radon.
Among all the given statements, statement c is incorrect. It can be corrected as valence shells of noble gases are complete, thus, they do not show tendency to gain electrons. Therefore, they have zero electron affinity.
Temperature of a gas is t K. What would be the temperature at which volume and pressure, both will reduced to half of the initial values?
\frac{t}{2}
\frac{t}{4}
\frac{t}{3}
\frac{t}{8}
\frac{t}{4}
p1 = p1T1 = t; V1 = v
\frac{p}{2}
; V2 =
\frac{v}{2}
By using Boyle's law
\frac{{p}_{1}{V}_{1}}{{T}_{1}}\quad =\quad \frac{{p}_{2}{V}_{2}}{{T}_{2}}
\frac{p\quad \times \quad V}{t}\quad =\quad \frac{p\quad \times \quad V}{2\quad \times \quad 2\quad \times {T}_{2}}
{T}_{2}\quad =\quad \frac{t}{4}
{\mathrm{NH}}_{4}^{+}
and CH4 are not isoelectronic species
BF3 does not have dipole moment
O - Cl - O obeys octet rule
O- in O3 is Sp3-hybridised
Among all the given statements, statement b is correct. Other statements can be corrected as-
a. For
{\mathrm{NH}}_{4}^{+}
, number of valance electrons = 10
For CH4, number of valance electrons = 10.
They both are isoelectronic species.
c. O - Cl - O does not follow octet rule.
d. O- in O3 is sp2- hybridised.
The melting point of solid substances is directly proportional to pressure acting on them. However, ice-melts at a temperature lower than its usual melting point, when the pressure increases. This is because
ice is not a true solid
the bonds break under pressure
ice is less denser than water
pressure generates heat
For most substances, increasing the pressure when a system is in equilibrium between liquid and solid phases will increase the phase transition temperature. Ice melts at a temperature lower than its usual melting point because ice is less denser than water.
Aqueous solution of AlCl3 is acidic towards litmus while of NaCl is not. The correct reason behind this is
AlCl3, furnishes OH- ion in the solution
AlCl3, furnishes H+ ion in the solution
AlCl3, furnishes both H+ well as OH- ion in the solution
AlCl3, is the salt of strong base and strong acid
AlCl3 is the salt of weak base and strong acid which renders H+ ions on hydrolysis. Thus, aqueous solution of AlCl3 is acidic in nature.
Al3+ + 3H2O
⇌
Al(OH)3 + 3H+
On the other hand, NaCl is the salt of strong acid and strong base which does not undergo hydrolysis.
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Accuracy vs. Precision | Brilliant Math & Science Wiki
Matthew Smith, Hobart Pao, Prince Loomba, and
Many people use the terms "accuracy" and "precision" interchangeably. However, this is not the correct thing to do, as the two terms have different definitions.
How close are the observed results to the expected results?
How close is each result to the next result?
While it is possible for a statement to be both accurate and precise, it is often the case that the statement is one or the other.
If I was given the sum
2+2
and were to provide the answer "between
0
10
," my answer would be accurate, and it is certainly correct. However, the statement is not precise as my answer has a range of
10
If instead I was to provide the answer "exactly
7
," my answer would be precise, as the answer has an uncertainty of
0
. However, the answer is not accurate, as
2 + 2 = 4
\chi^{2}
test is performed for dice rolling.
Side 1: 380 times
Side 6: 405 times.
The expected results would be that side 1 : side 2 : side 3 : side 4 : side 5 : side 6 will be in the ratio 1:1:1:1:1:1.
\chi^{2}
test shows that there is no significant difference between the expected and observed results. Does this statistic indicate good accuracy, precision, both, or neither?
Since the observed results are close to the expected results, this is an indication of good accuracy. However, more statistical analysis should be done to determine whether the precision is good, but generally the numbers do look quite close to each other.
_\square
Only Precise Only Accurate None of these Accurate and Precise
If I was asked "What is the value of 81+19?" and my reply was exactly 101, which of the following is most appropriate regarding my answer?
Cite as: Accuracy vs. Precision. Brilliant.org. Retrieved from https://brilliant.org/wiki/accuracy-vs-precision/
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An electron that has velocity v=(2.0\times10^6\ m/s)i+(3.0\times10^6\ m/s)j move
An electron that has velocity v=(2.0\times10^6\ m/s)i+(3.0\times10^6\ m/s)j moves through the uniform magneticfield B=(0.30T)i-(0.15T)j a) Find the force on the electron. b) Repeat your calculation for a proton having the same velocity.
An electron that has velocity
v=\left(2.0×{10}^{6}\text{ }\frac{m}{s}\right)i+\left(3.0×{10}^{6}\text{ }\frac{m}{s}\right)j
moves through the uniform magneticfield
B=\left(0.30T\right)i-\left(0.15T\right)j
a) Find the force on the electron.
b) Repeat your calculation for a proton having the same velocity.
{F}_{B}=qv×B
is the equation to use: in words, v "crosses" B
\therefore
in vector notation (instead of unit vector notation)
\frac{b}{c}
3.00\frac{m}{s}
A model airplane of mass 0.750 kg flies in a horizontal circle at the end of a 60.0 m control wire, with a speed of 35.0 m/s. Compute the tension in the wire if it makes a constant angle of 20.0° with the horizontal. The forces exerted on the airplane are the pull of the control wire, the gravitational force, and aerodynamic lift, which acts at 20.0° inward from the vertical.
A stone hangs by a fine thread from the ceiling,and a section of the same thread dangles from the bottom of the stone.If a person gives a sharp pull on the dangling thread,where is the thread likely to break:below the stone or above it?
\begin{array}{|cccc|}\hline Outcome& Payout& P\left(X\right)& xP\left(x\right)/E\left(X\right)\\ \text{Value of all three items match}& $15& 0.025& 0.375\\ \text{Either all values odd or even}& $5& 1.5& 7.5\\ \text{Value of 2 of 3 items match}& 5$& \frac{19}{10}=1.9& 9.5\\ \text{Values are in increasing order}& $10& \frac{38}{60}=0.63& 6.3\\ \hline\end{array}
Energy is conventionally measured in Calories as well as in joules.One Calorie in nutrition is 1 kilocalorie, which we define inChapter 11 as 1 kcal = 4,186 J. Metabolizing 1 gram of fat canrelease 9.00 kcal. A student decides to try to lose weight byexercising. She plans to run up and down the stairs in a footballstadium as fast as she can and as many times as necessary. Is thisin itself a practical way to lose weight?
To evaluate the program, suppose she runs up a flight of90 steps, each 0.150 m high, in67.0 s. For simplicity, ignore theenergy she uses in coming down (which is small). Assume that atypical efficiency for human muscles is 20.0%. This means that whenyour body converts 100 J from metabolizing fat, 20 J goes intodoing mechanical work (here, climbing stairs). The remainder goesinto internal energy. Assume the students
Create a dictionary to hold the prices of fruit: Bananas cost $1.5 a pound, apples $2.3 a pound, oranges cost $4.5 a pound and cherries cost $6 a pound. Create another dictionary to hold the stock of each fruit: You have 35 pounds of bananas, 42 pounds of apples, 40 pounds of oranges and 12 pounds of cherries. Display the data in a table format as such: Fruit Price Stock Banana 1.5 35 Apple 2.3 42 …….. Calculate the total value we have of each fruit and the total value of the whole shop.
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Minimax and Bayes estimation in deconvolution problem
Ermakov, Mikhail
We consider a deconvolution problem of estimating a signal blurred with a random noise. The noise is assumed to be a stationary gaussian process multiplied by a weight function function
ϵh
h\in {L}_{2}\left({R}^{1}\right)
ϵ
is a small parameter. The underlying solution is assumed to be infinitely differentiable. For this model we find asymptotically minimax and Bayes estimators. In the case of solutions having finite number of derivatives similar results were obtained in [G.K. Golubev and R.Z. Khasminskii, IMS Lecture Notes Monograph Series 36 (2001) 419-433].
Classification : 62G05, 65R30, 65R32
Mots clés : deconvolution, minimax estimation, Bayes estimation, Wiener filtration
author = {Ermakov, Mikhail},
title = {Minimax and {Bayes} estimation in deconvolution problem},
AU - Ermakov, Mikhail
TI - Minimax and Bayes estimation in deconvolution problem
Ermakov, Mikhail. Minimax and Bayes estimation in deconvolution problem. ESAIM: Probability and Statistics, Tome 12 (2008), pp. 327-344. doi : 10.1051/ps:2007038. http://www.numdam.org/articles/10.1051/ps:2007038/
[1] L.D. Brown, T. Cai, M.G. Low and C. Zang, Asymptotic equivalence theory for nonparametric regression with random design. Ann. Stat. 24 (2002) 2399-2430. | Zbl 1029.62044
[2] C. Butucea, Deconvolution of supersmooth densities with smooth noise. Canad. J. Statist. 32 (2004) 181-192. | MR 2064400 | Zbl 1056.62047
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[5] G.K. Golubev and R.Z. Khasminskii, Statistical approach to Cauchy problem for Laplace equation. State of the Art in Probability and Statistics, Festschrift for W.R. van, Zwet M. de Gunst, C. Klaassen and van der Vaart Eds., IMS Lecture Notes Monograph Series 36 (2001) 419-433. | MR 1836549
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[13] J. Fan, On the optimal rates of convergence for nonparametric deconvolution problems. Ann. Statist. 19 (1991) 1257-1272. | MR 1126324 | Zbl 0729.62033
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Revision as of 17:34, 23 January 2022 by Daniel Carrero (talk | contribs) (→Stat)
{\displaystyle HP={\Biggl \lfloor }{{\Biggl (}(Base+DV)\times 2+{\biggl \lfloor }{\tfrac {{\bigl \lceil }{\sqrt {STATEXP}}{\bigr \rceil }}{4}}{\biggr \rfloor }{\Biggr )}\times Level \over 100}{\Biggr \rfloor }+Level+10}
{\displaystyle OtherStat={\Biggl \lfloor }{{\Biggl (}(Base+DV)\times 2+{\biggl \lfloor }{\tfrac {{\bigl \lceil }{\sqrt {STATEXP}}{\bigr \rceil }}{4}}{\biggr \rfloor }{\Biggr )}\times Level \over 100}{\Biggr \rfloor }+5}
{\displaystyle {\begin{aligned}HP&={\Biggl \lfloor }{{\Biggl (}(35+7)\times 2+{\biggl \lfloor }{\tfrac {{\bigl \lceil }{\sqrt {22850}}{\bigr \rceil }}{4}}{\biggr \rfloor }{\Biggr )}\times 81 \over 100}{\Biggr \rfloor }+81+10\\&=\left\lfloor {(42\times 2+38)\times 81 \over 100}\right\rfloor +91\\&=\left\lfloor {122\times 81 \over 100}\right\rfloor +91\\&=\lfloor {98.82}\rfloor +91\\&=98+91\\&=189\end{aligned}}}
{\displaystyle {\begin{aligned}Sp.Atk&={\Biggl \lfloor }{{\Biggl (}(50+9)\times 2+{\biggl \lfloor }{\tfrac {{\bigl \lceil }{\sqrt {19625}}{\bigr \rceil }}{4}}{\biggr \rfloor }{\Biggr )}\times 81 \over 100}{\Biggr \rfloor }+5\\&=\left\lfloor {(59\times 2+35)\times 81 \over 100}\right\rfloor +5\\&=\left\lfloor {153\times 81 \over 100}\right\rfloor +5\\&=\lfloor 123.93\rfloor +5\\&=123+5\\&=128\end{aligned}}}
{\displaystyle {\begin{aligned}Sp.Def&={\Biggl \lfloor }{{\Biggl (}(40+9)\times 2+{\biggl \lfloor }{\tfrac {{\bigl \lceil }{\sqrt {19625}}{\bigr \rceil }}{4}}{\biggr \rfloor }{\Biggr )}\times 81 \over 100}{\Biggr \rfloor }+5\\&=\left\lfloor {(49\times 2+35)\times 81 \over 100}\right\rfloor +5\\&=\left\lfloor {133\times 81 \over 100}\right\rfloor +5\\&=\lfloor 107.73\rfloor +5\\&=107+5\\&=112\end{aligned}}}
{\displaystyle {\begin{aligned}Speed&={\Biggl \lfloor }{{\Biggl (}(90+5)\times 2+{\biggl \lfloor }{\tfrac {{\bigl \lceil }{\sqrt {24795}}{\bigr \rceil }}{4}}{\biggr \rfloor }{\Biggr )}\times 81 \over 100}{\Biggr \rfloor }+5\\&=\left\lfloor {(95\times 2+39)\times 81 \over 100}\right\rfloor +5\\&=\left\lfloor {229\times 81 \over 100}\right\rfloor +5\\&=\lfloor 185.49\rfloor +5\\&=185+5\\&=190\end{aligned}}}
{\displaystyle HP={\Bigl \lfloor }{(2\times Base+IV+\lfloor {\tfrac {EV}{4}}\rfloor )\times Level \over 100}{\Bigr \rfloor }+Level+10}
{\displaystyle OtherStat={\Biggl \lfloor }{\biggl (}{\Bigl \lfloor }{(2\times Base+IV+\lfloor {\tfrac {EV}{4}}\rfloor )\times Level \over 100}{\Bigr \rfloor }+5{\biggr )}\times Nature{\Biggr \rfloor }}
{\displaystyle {\begin{aligned}HP&=\left\lfloor {(2\times 108+24+\left\lfloor {\tfrac {74}{4}}\right\rfloor )\times 78 \over 100}\right\rfloor +78+10\\&=\left\lfloor {(216+24+18)\times 78 \over 100}\right\rfloor +88\\&=\left\lfloor {258\times 78 \over 100}\right\rfloor +88\\&=\lfloor 201.24\rfloor +88\\&=201+88\\&=289\end{aligned}}}
{\displaystyle {\begin{aligned}Attack&=\left\lfloor \left(\left\lfloor {\left(2\times 130+12+\left\lfloor {\tfrac {190}{4}}\right\rfloor \right)\times 78 \over 100}\right\rfloor +5\right)\times 1.1\right\rfloor \\&=\left\lfloor \left(\left\lfloor {(260+12+47)\times 78 \over 100}\right\rfloor +5\right)\times 1.1\right\rfloor \\&=\left\lfloor \left(\left\lfloor {319\times 78 \over 100}\right\rfloor +5\right)\times 1.1\right\rfloor \\&=\left\lfloor \left(248+5\right)\times 1.1\right\rfloor &\left\lfloor {319\times 78 \over 100}\right\rfloor &=248\\&=\lfloor 278.3\rfloor &\left(248+5\right)\times 1.1&=278.3\\&=278\end{aligned}}}
{\displaystyle {\begin{aligned}Sp.Atk&=\left\lfloor \left(\left\lfloor {\left(2\times 80+16+\left\lfloor {\tfrac {48}{4}}\right\rfloor \right)\times 78 \over 100}\right\rfloor +5\right)\times 0.9\right\rfloor \\&=\left\lfloor \left(\left\lfloor {(160+16+12)\times 78 \over 100}\right\rfloor +5\right)\times 0.9\right\rfloor \\&=\left\lfloor \left(\left\lfloor {188\times 78 \over 100}\right\rfloor +5\right)\times 0.9\right\rfloor \\&=\left\lfloor (146+5)\times 0.9\right\rfloor &\left\lfloor {188\times 78 \over 100}\right\rfloor &=146\\&=\lfloor 135.9\rfloor &(146+5)\times 0.9&=135.9\\&=135\end{aligned}}}
{\displaystyle {\begin{aligned}Speed&=\left\lfloor \left(\left\lfloor {\left(2\times 102+5+\left\lfloor {\tfrac {23}{4}}\right\rfloor \right)\times 78 \over 100}\right\rfloor +5\right)\times 1\right\rfloor \\&=\left\lfloor {(204+5+5)\times 78 \over 100}\right\rfloor +5\\&=\left\lfloor {214\times 78 \over 100}\right\rfloor +5\\&=\lfloor 166.92\rfloor +5\\&=166+5\\&=171\end{aligned}}}
{\textstyle {\Bigg (}1+{\left\lfloor {\frac {10\times friendship}{255}}\right\rfloor \over 100}{\Bigg )}}
{\displaystyle {\begin{aligned}HP&=\left\lfloor {(2\times Base+IV)\times Level \over 100}\right\rfloor +Level+10+AV\\OtherStat&=\left\lfloor \left({(2\times Base+IV)\times Level \over 100}+5\right)\times Nature\times Friendship\right\rfloor +AV\end{aligned}}}
{\displaystyle T=Accuracy_{move}\times Accuracy_{user}\times Evasion_{target}-BrightPowder}
{\displaystyle T=Accuracy_{move}\times StageMultiplier\times Other}
{\displaystyle CP=\mathrm {min} \left(\left\lfloor \left(\sum Stats-\sum AVs\right)\ \times {Level\times 6 \over 100}+\sum AVs\times \left({\frac {Level\times 4}{100}}+2\right)\right\rfloor ,10000\right)}
{\textstyle {66 \over 100}\times {66 \over 100}=43.56\%}
{\textstyle {75 \over 100}\times {75 \over 100}=56.25\%}
{\textstyle {60 \over 100}=60\%}
{\textstyle {33 \over 100}=33\%}
{\textstyle {33 \over 100}\times {33 \over 100}=10.89\%}
{\displaystyle \left\lfloor \left\lfloor {\frac {(Base+IV)\times \lfloor Link\rfloor }{100}}\right\rfloor \times Energy\right\rfloor }
{\displaystyle HP_{GO}\approx 50+1.75\times HP}
{\displaystyle Attack_{GO}\approx {\biggl (}{\frac {1}{4}}\times min(Attack,SpAtk)+{\frac {7}{4}}\times max(Attack,SpAtk){\biggr )}\times SpeedMult}
{\displaystyle Defense_{GO}\approx {\biggl (}{\frac {3}{4}}\times min(Defense,SpDef)+{\frac {5}{4}}\times max(Defense,SpDef){\biggr )}\times SpeedMult}
{\displaystyle SpeedMult=1+{\frac {Speed-75}{500}}}
{\displaystyle Stat=(base+IV)\times cpMult}
{\displaystyle CP=\left\lfloor {\dfrac {{\sqrt {HP}}\times Attack\times {\sqrt {Defense}}}{10}}\right\rfloor }
Portuguese Brazil Atributo
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PochhammerBasis - Maple Help
Home : Support : Online Help : PochhammerBasis
Pochhammer polynomials based at a point
PochhammerBasis(k, a, x)
algebraic expression; the starting point
\mathrm{PochhammerBasis}\left(k,a,x\right)=\prod _{j=0}^{k-1}\left(x+a+j\right)
k
th Pochhammer polynomial of degree
n
. The degree of the
k
th Pochhammer polynomial is
k
At present, this can only be evaluated in Maple by prior use of the object-oriented representation obtained by P:=convert(p,MatrixPolynomialObject,x) and subsequent call to P:-Value(<x-value>) , which uses Horner's method to evaluate the polynomial p.
a≔'a'
\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{a}
p≔3\mathrm{PochhammerBasis}\left(0,a,x\right)+5\mathrm{PochhammerBasis}\left(2,a,x\right)+7\mathrm{PochhammerBasis}\left(3,a,x\right)
\textcolor[rgb]{0,0,1}{p}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{PochhammerBasis}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{PochhammerBasis}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{7}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{PochhammerBasis}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)
This is in effect a NewtonBasis polynomial expression on the nodes
a
a+1
a+2
P≔\mathrm{convert}\left(p,\mathrm{MatrixPolynomialObject},x\right)
\textcolor[rgb]{0,0,1}{P}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{Record}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{\mathrm{Value}}\textcolor[rgb]{0,0,1}{=}{\textcolor[rgb]{0,0,1}{\mathrm{Default}}}_{\textcolor[rgb]{0,0,1}{\mathrm{value}}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Variable}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Degree}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Coefficient}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{coe}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Dimension}}\textcolor[rgb]{0,0,1}{=}[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Basis}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{PochhammerBasis}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{BasisParameters}}\textcolor[rgb]{0,0,1}{=}[\textcolor[rgb]{0,0,1}{a}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{IsMonic}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{mon}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{OutputOptions}}\textcolor[rgb]{0,0,1}{=}[\textcolor[rgb]{0,0,1}{\mathrm{shape}}\textcolor[rgb]{0,0,1}{=}[]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{storage}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{rectangular}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{order}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{Fortran_order}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{fill}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{attributes}}\textcolor[rgb]{0,0,1}{=}[]]\right)
P:-\mathrm{Degree}\left(\right)
\textcolor[rgb]{0,0,1}{3}
\mathrm{convert}\left(...,\mathrm{MatrixPolynomialObject}\right)
P:-\mathrm{Value}\left(x\right)[1,1]
\left(\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{a}\right)\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\right)\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{7}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{7}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{19}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{3}
a≔'a'
\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{a}
p≔\mathrm{add}\left(b[k]\mathrm{PochhammerBasis}\left(k,a,x\right),k=0..3\right)
\textcolor[rgb]{0,0,1}{p}\textcolor[rgb]{0,0,1}{≔}{\textcolor[rgb]{0,0,1}{b}}_{\textcolor[rgb]{0,0,1}{0}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{PochhammerBasis}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{b}}_{\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{PochhammerBasis}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{b}}_{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{PochhammerBasis}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{b}}_{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{PochhammerBasis}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}\right)
P≔\mathrm{convert}\left(p,\mathrm{MatrixPolynomialObject},x\right)
\textcolor[rgb]{0,0,1}{P}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{Record}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{\mathrm{Value}}\textcolor[rgb]{0,0,1}{=}{\textcolor[rgb]{0,0,1}{\mathrm{Default}}}_{\textcolor[rgb]{0,0,1}{\mathrm{value}}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Variable}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Degree}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Coefficient}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{coe}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Dimension}}\textcolor[rgb]{0,0,1}{=}[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Basis}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{PochhammerBasis}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{BasisParameters}}\textcolor[rgb]{0,0,1}{=}[\textcolor[rgb]{0,0,1}{a}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{IsMonic}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{mon}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{OutputOptions}}\textcolor[rgb]{0,0,1}{=}[\textcolor[rgb]{0,0,1}{\mathrm{shape}}\textcolor[rgb]{0,0,1}{=}[]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{storage}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{rectangular}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{order}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{Fortran_order}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{fill}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{attributes}}\textcolor[rgb]{0,0,1}{=}[]]\right)
\mathrm{collect}\left(P:-\mathrm{Value}\left(t\right)[1,1],[\mathrm{seq}\left(b[k],k=0..3\right)],\mathrm{factor}\right)
{\textcolor[rgb]{0,0,1}{b}}_{\textcolor[rgb]{0,0,1}{0}}\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{a}\right)\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{b}}_{\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{a}\right)\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\right)\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{b}}_{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{a}\right)\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\right)\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{2}\right)\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{b}}_{\textcolor[rgb]{0,0,1}{3}}
|
{2}^{9}-1
Context Panel: Test Primality
{2}^{9}-1
\stackrel{\text{check for prime}}{\to }
\textcolor[rgb]{0,0,1}{\mathrm{false}}
Context Panel: Integer Factors
{2}^{9}-1
\stackrel{\text{factor}}{=}
\left(\textcolor[rgb]{0,0,1}{7}\right)\left(\textcolor[rgb]{0,0,1}{73}\right)
Test primality with the isprime command.
\mathrm{isprime}\left({2}^{9}-1\right)
Obtain integer factors with the ifactor command.
\mathrm{ifactor}\left({2}^{9}-1\right)
Note: After typing each command, press Enter to evaluate it. To save space in this ebook, we will frequently not show the output in the Coded Solution when it is the same as that in the Interactive Solution.
<< Previous Example Section A-2 Next Example >>
|
Sort states based on state partition - MATLAB xsort - MathWorks í•œêµ
Sort states based on state partition
xsys = xsort(sys) sorts the x or q vector based on the state partition. Signal-based connections and physical interfaces between model components gives rise to differential algebraic equation (DAE) models where some internal signals and forces become extra states. The StateInfo property of sparss and mechss model objects keeps track of the state partition into sub-components, interface variables, and signal variables.
xsys — Sparse state-space model with sorted components
Sparse state-space model with sorted components, returned as a sparss or mechss model object. In the sorted xsys, all components appear first, followed by the interfaces, and then followed by a single group of all internal signals. The matrices
s\text{E}â\text{A}
\text{M }{s}^{2}+\text{C }s+\text{K}
have the following block arrow structure:
Here, each diagonal block is a sub-component of sys. The last row and column combines the Interface and Signal groups to capture all couplings and connections between components.
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Fredholm and invertibility theory for a special class of Toeplitz + Hankel operators | EMS Press
Fredholm and invertibility theory for a special class of Toeplitz + Hankel operators
Estelle L. Basor
American Institute of Mathematics, Palo Alto, USA
We develop a complete Fredholm and invertibility theory for Toeplitz+Hankel operators
T(a)+H(b)
on the Hardy space
\Hp
1<p<\iy
, with piecewise continuous functions
a,b
defined on the unit circle which are subject to the condition
a(t)a(t\iv)=b(t)b(t\iv)
|t|=1
. In particular, in the case of Fredholmness, formulas for the defect numbers are established. The results are applied to several important examples.
Estelle L. Basor, Torsten Ehrhardt, Fredholm and invertibility theory for a special class of Toeplitz + Hankel operators. J. Spectr. Theory 3 (2013), no. 2, pp. 171–214
|
Calabi-Yau pointed Hopf algebras of finite Cartan type | EMS Press
JournalsjncgVol. 7, No. 4pp. 1105–1144
Calabi-Yau pointed Hopf algebras of finite Cartan type
We study the Calabi–Yau property of pointed Hopf algebra
U(\mathcal{D},\lambda)
of finite Cartan type. It turns out that this class of pointed Hopf algebras constructed by N. Andruskiewitsch and H.-J. Schneider contains many Calabi–Yau Hopf algebras. To give concrete examples of new Calab–Yau Hopf algebras, we classify the Calabi–Yau pointed Hopf algebras
U(\mathcal{D},\lambda)
of dimension less than 5.
Xiaolan Yu, Yinhuo Zhang, Calabi-Yau pointed Hopf algebras of finite Cartan type. J. Noncommut. Geom. 7 (2013), no. 4, pp. 1105–1144
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Mariah is at her grandmother’s house which is located up on a hill. It is a very difficult hill to climb as it is steep. There are stairs leading up to her grandmother’s house. However, her grandmother now needs a ramp built so that she can use her wheelchair to get to the front door. Mariah designs a ramp and creates the diagram shown below. Is her diagram correct? Explain.
Check the Triangle Inequality and the lengths of the sides with the size of the opposite angles.
One of the errors includes the side lengths of the triangle.
8
ft and
5
ft add up to less than
15
|
{\displaystyle {\mathfrak {L}}}
's hat shows an example of the old pre-decimal system: the hat costs half a guinea (10 shillings and 6 pence).
A pound = 20 shillings = 240 silver pennies
Find sources: "Pound sterling" – news · newspapers · books · scholar · JSTOR (July 2021) (Learn how and when to remove this template message
Guinea of James II, 1686. The "Elephant and Castle" motif below his head is that of the Royal African Company: the gold came from Britain's trade in African slaves from the Guinea region
The cost of one pound in US dollars
The cost of one Euro
in pounds (from 1999)
(red) and overseas territories (blue) using the pound or their local issue
|
A Game of Bots | Toph
A Game of Bots
By ramisa.alam · Limits 1s, 512 MB
You were idly scrolling through your Phesbook feed one morning when you discovered an ongoing event called “A Game of Bots”. It turns out to be an AI contest where participants have to build their own AI agents and win against opponents’ agents in matches of chess.
You can see the tournament has already begun and
Nplayers have started playing the games. These players have already played against each other. So a total of
N * (N - 1) / 2
N∗(N−1)/2 rounds have completed. If you enter the tournament now, you’ll have to play against each of these
Nplayers. So you want to have an idea about their strength.
From your experiences, you know that in these games a stronger bot always wins against a weaker bot. If both the participating bots in a game have similar strengths, the game is tied.
You can easily find out the results of the previous matches from the event page. Can you determine the relative strength of the participating players from these results?
Here, the relative strength of each of the
Nplayers is a number from
[1, N]
[1,N] with a larger value indicating a stronger player.
It is guaranteed that at least one valid assignment of relative strengths for the players exists.
The first line of input contains a single integer
t ( 1 \le t \le 100 )
t(1≤t≤100) — the number of test cases. Description of the test cases follows.
The first line of each test case contains an integer
N ( 1 \le N \le 1000)
N(1≤N≤1000)— the number of bots currently participating in the tournament.
Each of the next
N * ( N - 1 ) / 2
N∗(N−1)/2 lines contain the description of a match.
Each match is described by 3 integers:
x_i, y_i, w_i ( 1 \le x_i , y_i \le N , 0 \le w_i \le N)
xi,yi,wi(1≤xi,yi≤N,0≤wi≤N) where
x_i
xiand
y_i
yi are the ids of the players participating in the
i^{th}
ith game and
w_i
wi is the id of the winning player. In case of ties,
w_i = 0
wi=0.
Each player appears in exactly
N-1
N−1 matches against
N-1
N−1 different opponents.
N over all test cases does not exceed
1000.
For each test case, print one line containing an array
S containing
N space-separated integers.
i^{th}
ith element of the array -
S_i ( 1 \le S_i \le N)
Si(1≤Si≤N) is the relative strength of the
i^{th}
ith player. A larger value of
S_i
Si indicates a stronger player.
If there are multiple such arrays, print the one in which the maximum strength among the
N players is minimized.
In the second test case, player 1 has strength 1, and player 2 has strength 2. So player 2 wins the first match.
The 5th match ties since player 1 and player 4 have the same strength
bisnu_sarkarEarliest, 2M ago
bisnu_sarkarLightest, 131 kB
Shahriar.699347Shortest, 742B
|
At what cost can we simulate large quantum circuits on small quantum computers? | IBM Research Blog
One major challenge of near-term quantum computation is the limited number of available qubits. Suppose we want to run a circuit consisting of 400 qubits, but we only have 100-qubit devices available. What do we do?
Read more about how the IBM Quantum team has expanded its development roadmap to anticipate the future of quantum-centric supercomputing.
Over the course of the past year, the IBM Quantum team has begun researching a host of computational methods called circuit knitting. Circuit knitting techniques allow us to partition large quantum circuits into subcircuits that fit on smaller devices, incorporating classical simulation to “knit” together the results to achieve the target answer. The cost is a simulation overhead that scales exponentially in the number of knitted gates.
Circuit knitting will be important well into the future. Our quantum hardware development team is focused on scaling by connecting smaller processors via classical, and then via quantum links. Due to this planned hardware architecture, circuit knitting will be useful in the near future as we run problems on classically parallelized quantum processors. Techniques that boost the number of available qubits will also be relevant far into the future.
Figure 1: Circuit knitting example: The nonlocal circuit on the left acting on A⊗B can be simulated with local circuits acting only on A or B on the right followed by classical postprocessing.
But first, our team needed to understand how much of a benefit these methods can offer, especially when we knew that the simulation overhead scales exponentially with the number of gates acting between these subcircuits.
We are currently investigating whether classical communication between local quantum computers can help to lower the simulation overhead — as you might see on a pair of classically parallelized IBM Quantum “Heron” processors. Specifically, we realized circuit knitting via a method that has previously gained interest in the fields of error mitigation and classical simulation algorithms, called the quasiprobability simulation technique.1
The 133-qubit “Heron” processor, slated for 2023.
We consider three settings to simulate a non-local circuit with local operations. In the first, the two quantum computers can only run their own local operations on their subcircuits without communication between them. In the second the two computers can realize those local operations, with the added ability to send classical information in one direction — from
\Alpha
\Beta
\Beta
\Alpha
. In the third, the two quantum computers can run their own local quantum operations and send classical information in either direction between them.
In the local and one-way classical communication settings, one does not necessarily require two separate quantum computers. Instead, one can run the two subcircuits in sequence on the same device. The classical communication in the one-way setting can then be simulated by classically storing the bits sent from
\Alpha
\Beta
Figure 2: Graphical overview of the three scenarios considered to run a nonlocal operation. LO refers to local operations; LO & one way CC refers to local operations and one way classical communication; LOCC refers local operations and classical communication.
In contrast, the two-way communication setting requires two quantum computers that exchange classical information in both directions. We show that for circuit knitting based on quasiprobability simulation, the three settings mentioned above all have a different sampling overhead when applied to circuits with multiple instances of the same non-local gate.
Our results, available on arXiv,2 demonstrate that two-way communication can considerably reduce the simulation overhead. For circuits containing n CNOT gates connecting each subcircuit, the incorporation of classical information exchange between the subcircuits reduces the simulation overhead from O(9n) to O(4n) — a reduction that is substantial in practice. It allows us to cut considerably more CNOT gates — that is, the gates that entangle the qubits — for a given fixed simulation overhead.
On a technical level, our results are based on the insight that a simultaneous local preparation of two maximally entangled states, called Bell pairs, is more efficient than locally preparing a single Bell pair twice. The reason is that for a joint preparation we can make use of entanglement between the local subsystems, which is not possible if we prepare the two Bell pairs separately. Using the idea of gate teleportation we can then convert Bell pairs into CNOT gates under local operations and classical communication.
Figure 3: Graphical explanation of how to realize two CNOT gates in a LOCC setting via gate teleportation. By generating the two Bell pairs simultaneously (instead generating twice a single Bell pair), we can reduce the total simulation overhead.
Our results show that classical communication between locally separated quantum computers is beneficial when performing large computations that exceed the number of qubits each quantum device individually has.
Following the latest IBM roadmap, these results may be helpful in reducing simulation overheads in future architectures as it motivates the connection of individual quantum chips with classical communication links.
K. Temme, S. Bravyi, and J. M. Gambetta. Error mitigation for short-depth quantum circuits. Phys. Rev. Lett. 119, 180509 – Published 3 November 2017.↩
Piveteau, C., Sutter D. Circuit knitting with classical communication. arXiv. [Submitted on 29 Apr 2022]↩
Quantum startups leap from lab to launch with IBM incubator
Stefan Elrington and Nate Earnest-Noble
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|
Determine whether or not the two triangles in each pair below are similar. If so, create a flowchart to show your reasoning. If not, explain why not.
Check to see if the ratios of the corresponding sides are equal.
Similar. Be sure to create a flowchart.
Parallel lines give special angles. Which angles are these?
Similar by
AA
~. Be sure to create a flowchart.
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ECON 300 Statistics & Probability
In the discussion of small world networks, and various economic and financial matters, we must use some statistical distributions. The main thing that we will observe is that the real world has variables that are distributed in various ways. I want to build up to a discussion of “fat tails”, and in order for that to happen you must first understand what a “tail” is in the first place. We start with the most straightforward “uniform” and “binomial distribution” and build from there.
Binomial Distribution
The binomial distribution is used to calculate the probability that a certain number of discrete events occur. In the network literature, this is often used to describe the distribution of degrees at nodes in a random network. Some nodes have more connections than others, but recall that random networks do not exhibit significant clustering or hubs.
Shown below, the binomial distribution written generally as
f(x)
, is known as the probability density function, or PDF. This allows us to give the probability of a certain outcome.
f(x)=P(X=x)=\binom{n}{x}(p^x)(q^{n-x})=\frac{n!}{x!(n-x)!}p^xq^{n-x}
Where
p
is the probability of success, and
q=1-p
is the probability of failure. The binomial distribution function is defined as the probability of
x
events occurring in
n
trials. If
n=1
this is called the Bernoulli distribution.
For an example, let’s say that
p=0.3
and therefore
q=0.7
. The likelihood of making exactly 0 through 5 successes in 5 tries is
P(X=0)=\frac{5!}{0!5!}(0.3)^0(0.7)^5=0.17
P(X=1)=\frac{5!}{1!4!}(0.3)^1(0.7)^4=0.36
P(X=2)=\frac{5!}{2!3!}(0.3)^2(0.7)^3=0.31
P(X=3)=\frac{5!}{3!2!}(0.3)^3(0.7)^2=0.13
P(X=4)=\frac{5!}{4!1!}(0.3)^4(0.7)^1=0.03
P(X=5)=\frac{5!}{5!0!}(0.3)^5(0.7)^1=0.002
The cumulative distribution function, or CDF is the accumulated sum of all probabilities in some range. If it is the likelihood of getting a 0, 1, or 2 in this experiment above, then you would get a total value of 0.84 or 84% across these three combined possibilities.
The mean
\mu
and standard deviation
\sigma
of this distribution are
\mu=np
\sigma=\sqrt{npq}
For our example this would be
\mu=5\times 0.3=1.5
\sigma=5\times 0.3 \times 0.7=1.05
If we looked at this in terms of expected value, where you multiply probabilities by their values, you would see the same thing for
\mu
E[x]=\sum_{i=0}^N p(x_i)x_i = 0.17\times 0+0.36\times 1+0.31\times 2+0.13\times 3+0.03\times4+0.002\times 5=1.5
Uniform Distribution
Another common distribution that we come across in this literature is the uniform distribution. When we are examining issues like changing the probability that a “link is rewired” we pull a random number, usually between
0
1
. In the PDF and CDF for the uniform distribution, we are simply calculating a ratio. I provide the PDF and CDF here since this is a “simple” distribution, and it will help you understand how to make sense of the others.
f(x)=\begin{cases} \frac{1}{(b-a)}& \text{if } a\leq x \leq b\\ 0& \text{otherwise} \end{cases}
F(x)=P(X\leq x)=\begin{cases} 0& x<a \\ \frac{(x-a)}{(b-a)}& \text{if } a\leq x < b\\ a& x \geq b \end{cases}
Normal Distribution
The PDF for the normal distribution with mean (
\mu
) and standard deviation (
\sigma
) is
f(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^2/2\sigma^2}
where
-\infty < x < \infty
Standardizing this distribution such that
Z=\frac{X-\mu}{\sigma}
gives
f(z)=\frac{1}{\sqrt{2\pi}}e^{-z^2/2}
By standardizing, imagine you have some data with a mean of 72, and a standard deviation of 6. In our standardized form a value of 76, would give us a
Z=\frac{76-72}{6}=0.66
. We might compare this to being the same distance from the mean in a normal distribution with a mean of zero and standard deviation of 1 (i.e.,
N(0,1)
). In this instance, a value of 0.66 is exactly that many standard deviations away from the mean. If you compare the exponent in
f(x)
f(z)
you will see that we are simply making a replacement of nominal with standardized values.
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A bicycle tire has a diameter of 27 inches. How far does the bike travel along the ground
A bicycle tire has a diameter of 27 inches. How far does the bike travel along the ground when the wheel rotates once?
When the wheel rotates once, it has a travelled a distance equal to the circumference of the wheel. Using the circumference formula in terms of diameter dd, this distance is:
C=\pi d=\pi \left(27\right)\approx 84.8
{20}^{\circ },\text{ }{30}^{\prime }
Find the sector Area for both of these circles
The radii of two circles are 8 cm and 6 cm. Find the radius of the circle having area equal to the sum of the areas of the two circles
Find the area of the shaded region. Round to the nearest tenth.
Area of the shaded region=
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Natalie has a bag that contains eight marbles. She draws out a marble, records its color, and puts it back.
If Natalie repeats this eight times and does not record any red marbles, can she conclude that there are not any red marbles in the bag? Explain
Is it possible that she drew the same marble every time?
If she repeats this
100
times and does not record any red marbles, can she conclude that there are not any red marbles in the bag? Explain.
How many times would she have to draw marbles (putting them back each time) to be absolutely certain that there are no red marbles in the bag?
No number of trials can absolutely prove that there are no red marbles in the bag.
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Unique Bernoulli <em>g</em>-measures | EMS Press
We improve and subsume the conditions of Johansson and Öberg and Berbee for uniqueness of a
g
-measure, i.e., a stationary distribution for chains with complete connections. In addition, we prove that these unique
g
-measures have Bernoulli natural extensions. We also conclude that we have convergence in the Wasserstein metric of the iterates of the adjoint transfer operator to the
g
-measure.
Anders Johansson, Anders Öberg, Mark Pollicott, Unique Bernoulli <em>g</em>-measures. J. Eur. Math. Soc. 14 (2012), no. 5, pp. 1599–1615
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Multiple Choice: The length of a rectangle is three units shorter than twice its width. Which expression below could represent the area of the rectangle?
2x^2-3
2x^2-6x
2x^2-3x
(2x-3)^2
Draw a rectangle and label its width
x
. What should you label the length? Multiply your expressions for the length and width to obtain an expression for the area.
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1) Describe sampling distributions and sampling variavility 2) Explain The Cent
1) Describe sampling distributions and sampling variavility 2) Explain The Central Limit Theorem 3) Explain how confidence intervals are created and what can they tell us about population parameters
1) Describe sampling distributions and sampling variavility
2) Explain The Central Limit Theorem
3) Explain how confidence intervals are created and what can they tell us about population parameters
The sampling distributions refers to the probability distributions of all possible values of the sample statistic for example, sample mean.
The sampling variability tells us how an estimate varies between different samples. It is often measured in terms of variance or standard deviation. There are three factors involved in variability,
The sampling method (with or without replacement)
The central limit theorem tells us that in a random sample of size n taken from a population, its sampling distribution can be approximated to a normal distribution by taking a larger sample size. That is, whatever might be the population distribution, the sampling distribution of a sample statistic can be approximated to normal by increasing the sample size.
The confidence interval are created using the sample statistic and the margin of error.
CI=\stackrel{―}{x}±{Z}_{\frac{\alpha }{2}}\left(\frac{\sigma }{\sqrt{n}}n\right)
=\stackrel{―}{x}±ME
{Z}_{\frac{\alpha }{2}}
represents the critical value of the normal distribution and this distribution is used if the population standard deviation is known. For unknown population standard deviation, the sample standard deviation is used with student’s t distribution.
The constructed confidence interval with say 95 or 90 percent confidence level tells us that if repeated samples were to be taken and confidence intervals were to be built, then 95 or 90 percent of these constructed confidence intervals would contain the true value of the parameter (mean).
Explain how to use the sampling distributions of A and B to decide which is the best estimator of
\alpha
Check whether the standard error of the sampling distributions of bar p obtained in part(a) and part(b) are different.
\overline{X}
n=16
\overline{X}
\overline{X}
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Elemental Weakness | PoE Wiki
Elemental WeaknessSpell, AoE, Duration, Curse, Hex
Metadata ID: Metadata/Items/Gems/SkillGemElementalWeakness
Elemental Weakness is a Hex spell that curses all targets in an area, lowering their elemental Resistance. If the player character self casts this hex, it gains 10 Doom per second. Doom increases the effectiveness of this hex.
7 Area mods
7.1 Heist and grand heist mod
Negative Resistances: This curse can cause resistance values to dip into the negatives, if the base values are low enough. This will have the affect of increasing the damage above values listed on the character screen. For example, if a monster has 0% elemental resistances by default (as well as 100% curse effectiveness) and is inflicted with level 1 Elemental Weakness, that monster now has -20% (negative 20%) elemental resistances. Consequently, the monster thus receives 20% more[1] elemental damage than the spell or attack would normally cause. The curse gets even more effective against monsters with higher resistances.[2] However, a boss such as The Shaper has 66% less curse effectiveness, thus for the overall calculation it is needed to take both monster resistances and curse effectiveness into account.
The extra damage is multiplicative with other damage modifiers as it is considered to be the monsters damage mitigation, and not due to a buff on any players' elemental skills damage.
Elemental Weakness can be turned into an aura by linking it with either Blasphemy SupportBlasphemy SupportSupport, Hex, Aura
The aura version of Elemental Weakness
You are cursed. You take more Elemental damage.
1 24 58 16 9 N/A -20% 0 0
2 27 64 17 9.1 1 -21% 118383 118383
6 39 90 22 9.5 3 -25% 498508 1405858
8 45 102 24 9.7 4 -27% 921777 3009692
9 48 109 25 9.8 4 -28% 1727879 4737571
10 50 113 26 9.9 5 -29% 1138877 5876448
11 52 117 26 10 5 -30% 1368233 7244681
12 54 121 27 10.1 6 -31% 1638338 8883019
13 56 125 28 10.2 6 -32% 1956648 10839667
17 64 142 31 10.6 8 -36% 15206031 40512123
19 68 151 32 10.8 9 -38% 62755923 129453628
20 70 155 33 10.9 10 -39% 211877683 341331311
21 72 159 34 11 10 -40% N/A N/A
22 74 N/A 34 11.1 11 -41% N/A N/A
31 91 N/A 40 11.95 15 -49% N/A N/A
32 92 N/A 41 12 15 -50% N/A N/A
40 100 N/A 44 12.4 17 -54% N/A N/A
Elemental Weakness can drop anywhere.
Elemental Weakness can be created from the following recipes:
Hubris CircletEnergy Shield: (184-248)Requires Level 69, 154 IntTrigger Level 10 Shock Ground when Hit
20% chance to Curse non-Cursed Enemies with a random Hex on Hit
The following mods are related to Elemental Weakness:
of the Hunt Suffix 75 Curse Enemies with Elemental Weakness on Hit, with 20% increased Effect ring_basilisk 200
V2CurseOnHitElementalWeaknessCorrupted Corrupted 30 Curse Enemies with Elemental Weakness on Hit, with (40-48)% increased Effect gloves 1000
HellscapeUpsideCurseOnHitElementalWeakness1___ Scourge Benefit 68 Curse Enemies with Elemental Weakness on Hit gloves 50
HellscapeDownsideCursedWithElementalWeakness1 Scourge Detriment 68 You are Cursed with Elemental Weakness gloves 100
CurseEffectElementalWeaknessEldritchImplicit1 Searing Exarch 75 (19-21)% increased Elemental Weakness Curse Effect no_tier_6_eldritch_implicit 0
CurseEffectElementalWeaknessEldritchImplicitUniquePresence2 Searing Exarch 75 (34-36)% increased Elemental Weakness Curse Effect no_tier_5_eldritch_implicit 0
CurseEffectElementalWeaknessEldritchImplicitPinnaclePresence3 Searing Exarch 75 (49-51)% increased Elemental Weakness Curse Effect no_tier_4_eldritch_implicit 0
SynthesisImplicitCurseEffectElementalWeakness1 55 (20-30)% increased Elemental Weakness Curse Effect
SynthesisImplicitCurseOnHitElementalWeakness1__ 45 Curse Enemies with Elemental Weakness on Hit, with 20% increased Effect
CurseOnHitElementalWeaknessCorruption Corrupted 30 Curse Enemies with Elemental Weakness on Hit, with (40-48)% increased Effect gloves 0
ElementalWeaknessSkillCorrupted Corrupted 31 Grants Level 10 Elemental Weakness Skill wand 0
The following helmet enchantments affect Elemental Weakness.
Enchantment Elemental Weakness Curse Effect 1 66 Merciless Labyrinth 20% increased Elemental Weakness Curse Effect helmet 100
Enchantment Elemental Weakness Curse Effect 2 75 Eternal Labyrinth 30% increased Elemental Weakness Curse Effect helmet 100
Enchantment Elemental Weakness Duration 1 66 Merciless Labyrinth 30% increased Elemental Weakness Duration helmet 100
Enchantment Elemental Weakness Duration 2 75 Eternal Labyrinth 45% increased Elemental Weakness Duration helmet 100
Main page: List of map suffix mods
Elemental Weakness is one of the curses that can be rolled on maps and vaal side area (known as "secret_area" in spawn weight table), the curses affect the entire area and cannot be removed. The level of the curses is fixed (level 1, -20% elemental resistance) and NOT affected by awakening bonus objectives. It can be reduced by upgraded Soul of Yugul or other item mod that offers reduced/less effect of curses on you
of Elemental Weakness Suffix 69 15% increased Quantity of Items found in this Area
Found Items have (1-3)% chance to drop Corrupted in Area
Players are Cursed with Elemental Weakness secret_area 800
of Elemental Weakness Suffix 1 13% increased Quantity of Items found in this Area
primordial_map 0
Players are Cursed with Elemental Weakness top_tier_map 0
primordial_map 1000
Heist and grand heist mod
of Elemental Weakness Suffix 73 Players are Cursed with Elemental Weakness default 850
of Elemental Weakness Suffix 78 Players are Cursed with Elemental Weakness default 1000
Players are Cursed with Elemental Weakness secret_area 0
14% increased Pack size mid_tier_map 0
14% increased Pack size top_tier_map 0
Map Curse modifiers no longer have increased effect.
This was a restartless deploy of the patch so you will need to restart your client to receive the client changes, specifically the Map Curse modifier change requires a client restart to visually update descriptions.
Items which apply a Curse to an enemy[a] (such as the curse-on-hit Ring modifiers) no longer specify a level, and instead have a set magnitude for their values with a fixed amount of Doom, which is overwritten when another source applies a more potent version (similar to Arcane Surge).
Elemental Weakness: Curses applied by casting this spell now gain 10 Doom per second.
Mana cost reduced at all levels. Area of effect increases per gem level reduced, but the base area of effect has been increased by 37.5% (from 16 to 22). It now grants 0.25% reduced elemental resistances on cursed enemies per 1% quality (was 0.4% per 1% quality).
Now gains -8% Enemy Resistances at 20% quality, instead of -10% at 20% quality.
↑ The patch note is not clear. Bex_GGG later clarified that this change also effects item that curse players, such as Soul MantleSoul Mantle
long after they have been made .[3] The patch note did not mention the map curse either, which also affected by this change.
\color [rgb]{0.6392156862745098,0.5529411764705883,0.42745098039215684}{\begin{aligned}{\text{dmg}}_{\text{red}}&=(1-{\text{Resistance}})*{\text{dmg}}\end{aligned}}
↑ Engineering Eternity (June 4, 2017). "Path of Exile: Damage & Elemental Resistance Mechanics". Youtube. Retrieved March 2, 2010.
↑ Bex_GGG (September 18, 2020). "Re: Soul Mantle Nerf not Listed in Patch Notes?". Official Path of Exile subreddit. Retrieved March 14, 2021.
ru:Уязвимость к стихиям
Retrieved from ‘https://www.poewiki.net/w/index.php?title=Elemental_Weakness&oldid=1142526’
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Ludwig_von_Bertalanffy Knowpia
Karl Ludwig von Bertalanffy (19 September 1901 – 12 June 1972) was an Austrian biologist known as one of the founders of general systems theory (GST). This is an interdisciplinary practice that describes systems with interacting components, applicable to biology, cybernetics and other fields. Bertalanffy proposed that the classical laws of thermodynamics might be applied to closed systems, but not necessarily to "open systems" such as living things. His mathematical model of an organism's growth over time, published in 1934,[1] is still in use today.
Atzgersdorf near Vienna, Austria-Hungary
Biology and systems theory
Fechner und das Problem der Integration höherer Ordnung (Fechner and the Problem of Higher-Order Integration) (1926)
Johann Wolfgang von Goethe, Rudolf Carnap, Gustav Theodor Fechner, Nicolai Hartmann, Otto Neurath, Moritz Schlick
Russell L. Ackoff, Kenneth E. Boulding, Peter Checkland, C. West Churchman, Jay Wright Forrester, Ervin László, James Grier Miller, Anatol Rapoport
Bertalanffy grew up in Austria and subsequently worked in Vienna, London, Canada, and the United States.
Ludwig von Bertalanffy was born and grew up in the little village of Atzgersdorf (now Liesing) near Vienna. The Bertalanffy family had roots in the 16th century nobility of Hungary which included several scholars and court officials.[2] His grandfather Charles Joseph von Bertalanffy (1833–1912) had settled in Austria and was a state theatre director in Klagenfurt, Graz and Vienna, which were important sites in imperial Austria. Ludwig's father Gustav von Bertalanffy (1861–1919) was a prominent railway administrator. On his mother's side Ludwig's grandfather Joseph Vogel was an imperial counsellor and a wealthy Vienna publisher. Ludwig's mother Charlotte Vogel was seventeen when she married the thirty-four-year-old Gustav. They divorced when Ludwig was ten, and both remarried outside the Catholic Church in civil ceremonies.[3]
Ludwig von Bertalanffy grew up as an only child educated at home by private tutors until he was ten. When he arrived at his Gymnasium (a form of grammar school) he was already well habituated in learning by reading, and he continued to study on his own. His neighbour, the famous biologist Paul Kammerer, became a mentor and an example to the young Ludwig.[4]
In 1918, Bertalanffy started his studies at the university level in philosophy and art history, first at the University of Innsbruck and then at the University of Vienna. Ultimately, Bertalanffy had to make a choice between studying philosophy of science and biology; he chose the latter because, according to him, one could always become a philosopher later, but not a biologist. In 1926 he finished his PhD thesis (Fechner und das Problem der Integration höherer Ordnung, translated title: Fechner and the Problem of Higher-Order Integration) on the psychologist and philosopher Gustav Theodor Fechner.[4] For the next six years he concentrated on a project of "theoretical biology" which focused on the philosophy of biology. He received his habilitation in 1934 in "theoretical biology".[5]
Bertalanffy was appointed Privatdozent at the University of Vienna in 1934. The post yielded little income, and Bertalanffy faced continuing financial difficulties. He applied for promotion to the status of associate professor, but funding from the Rockefeller Foundation enabled him to make a trip to Chicago in 1937 to work with Nicolas Rashevsky. He was also able to visit the Marine Biological Laboratory in Massachusetts.[5]
Bertalanffy was still in the US when he heard of the Anschluss in March 1938. However, his attempts to remain in the US failed, and he returned to Vienna in October of that year.[5] Within a month of his return, he joined the Nazi Party, which facilitated his promotion to professor at the University of Vienna in 1940.[5] During the Second World War, he linked his "organismic" philosophy of biology to the dominant Nazi ideology, principally that of the Führerprinzip.[5]
Following the defeat of Nazism, Bertalanffy found denazification problematic and left Vienna in 1948. He moved to the University of London (1948–49); the Université de Montréal (1949); the University of Ottawa (1950–54); the University of Southern California (1955–58); the Menninger Foundation (1958–60); the University of Alberta (1961–68); and the State University of New York at Buffalo (SUNY) (1969–72).
In 1972, he died from a heart attack.
Bertalanffy met his wife, Maria, in April 1924 in the Austrian Alps. They were hardly ever apart for the next forty-eight years.[6] She wanted to finish studying but never did, instead devoting her life to Bertalanffy's career. Later, in Canada, she would work both for him and with him in his career, and after his death she compiled two of Bertalanffy's last works. They had one child, a son who followed in his father's footsteps by making his profession in the field of cancer research.
Today, Bertalanffy is considered to be a founder and one of the principal authors of the interdisciplinary school of thought known as general systems theory. According to Weckowicz (1989), he "occupies an important position in the intellectual history of the twentieth century. His contributions went beyond biology, and extended into cybernetics, education, history, philosophy, psychiatry, psychology and sociology. Some of his admirers even believe that this theory will one day provide a conceptual framework for all these disciplines".[2]
Individual growth modelEdit
The individual growth model published by Ludwig von Bertalanffy in 1934 is widely used in biological models and exists in a number of permutations.
In its simplest version the so-called Bertalanffy growth equation is expressed as a differential equation of length (L) over time (t):
{\displaystyle L'(t)=r_{B}\left(L_{\infty }-L(t)\right)}
{\displaystyle r_{B}}
is the Bertalanffy growth rate and
{\displaystyle L_{\infty }}
the ultimate length of the individual. This model was proposed earlier by August Friedrich Robert Pūtter (1879-1929), writing in 1920.[7]
The dynamic energy budget theory provides a mechanistic explanation of this model in the case of isomorphs that experience a constant food availability. The inverse of the Bertalanffy growth rate appears to depend linearly on the ultimate length, when different food levels are compared. The intercept relates to the maintenance costs, the slope to the rate at which reserve is mobilized for use by metabolism. The ultimate length equals the maximum length at high food availabilities.[1]
Passive electrical schematic of the Bertalanffy module together with equivalent expression in the Energy Systems Language
Bertalanffy equationEdit
The Bertalanffy equation is the equation that describes the growth of a biological organism. The equation was offered by Ludwig von Bertalanffy in 1969.[8]
{\displaystyle {\frac {dW}{dt}}=\eta S-kV}
Here W is organism weight, t is the time, S is the area of organism surface, and V is a physical volume of the organism.
{\displaystyle \eta }
{\displaystyle k}
are (by Bertalanffy's definition) the "coefficient of anabolism" and "coefficient of catabolism" respectively.
The solution of the Bertalanffy equation is the function:
{\displaystyle W(t)={\Big (}\eta \,c_{1}-c_{2}\,e^{-{\tfrac {k}{3}}t}{\Big )}^{3}\,,}
{\displaystyle c_{1}}
{\displaystyle c_{2}}
are the certain constants.
Bertalanffy couldn't explain the meaning of the parameters
{\displaystyle \eta }
(the coefficient of anabolism) and
{\displaystyle k}
(coefficient of catabolism) in his works, and that caused a fair criticism from biologists. But the Bertalanffy equation is a special case of the Tetearing equation,[9] that is a more general equation of the growth of a biological organism. The Tetearing equation determines the physical meaning of the coefficients
{\displaystyle \eta }
{\displaystyle k}
Bertalanffy moduleEdit
To honour Bertalanffy, ecological systems engineer and scientist Howard T. Odum named the storage symbol of his General Systems Language as the Bertalanffy module (see image right).[10]
General system theoryEdit
The biologist is widely recognized for his contributions to science as a systems theorist; specifically, for the development of a theory known as general system theory (GST). The theory attempted to provide alternatives to conventional models of organization. GST defined new foundations and developments as a generalized theory of systems with applications to numerous areas of study, emphasizing holism over reductionism, organism over mechanism.
Foundational to GST are the inter-relationships between elements which all together form the whole.
Bertalanffy's contribution to systems theory is best known for his theory of open systems. The system theorist argued that traditional closed system models based on classical science and the second law of thermodynamics were inadequate for explaining large classes of phenomena. Bertalanffy maintained that "the conventional formulation of physics are, in principle, inapplicable to the living organism being open system having steady state. We may well suspect that many characteristics of living systems which are paradoxical in view of the laws of physics are a consequence of this fact."[11] However, while closed physical systems were questioned, questions equally remained over whether or not open physical systems could justifiably lead to a definitive science for the application of an open systems view to a general theory of systems.
In Bertalanffy's model, the theorist defined general principles of open systems and the limitations of conventional models. He ascribed applications to biology, information theory and cybernetics. Concerning biology, examples from the open systems view suggested they "may suffice to indicate briefly the large fields of application" that could be the "outlines of a wider generalization;"[12] from which, a hypothesis for cybernetics. Although potential applications exist in other areas, the theorist developed only the implications for biology and cybernetics. Bertalanffy also noted unsolved problems, which included continued questions over thermodynamics, thus the unsubstantiated claim that there are physical laws to support generalizations (particularly for information theory), and the need for further research into the problems and potential with the applications of the open system view from physics.
Systems in the social sciencesEdit
In the social sciences, Bertalanffy did believe that general systems concepts were applicable, e.g. theories that had been introduced into the field of sociology from a modern systems approach that included "the concept of general system, of feedback, information, communication, etc."[13] The theorist critiqued classical "atomistic" conceptions of social systems and ideation "such as 'social physics' as was often attempted in a reductionist spirit."[14] Bertalanffy also recognized difficulties with the application of a new general theory to social science due to the complexity of the intersections between natural sciences and human social systems. However, the theory still encouraged new developments in many fields, from sociology to anthropology, economics, political science, and psychology among other areas.[citation needed] Today, Bertalanffy's GST remains a bridge for interdisciplinary study of systems in the social sciences.
1928, Kritische Theorie der Formbildung, Borntraeger. In English: Modern Theories of Development: An Introduction to Theoretical Biology, Oxford University Press, New York: Harper, 1933
1928, Nikolaus von Kues, G. Müller, München 1928.
1937, Das Gefüge des Lebens, Leipzig: Teubner.
1940, Vom Molekül zur Organismenwelt, Potsdam: Akademische Verlagsgesellschaft Athenaion.
1949, Das biologische Weltbild, Bern: Europäische Rundschau. In English: Problems of Life: An Evaluation of Modern Biological and Scientific Thought, New York: Harper, 1952.
1953, "Die Evolution der Organismen", in Schöpfungsglaube und Evolutionstheorie, Stuttgart: Alfred Kröner Verlag, pp 53–66
1959, Stammesgeschichte, Umwelt und Menschenbild, Schriften zur wissenschaftlichen Weltorientierung Vol 5. Berlin: Lüttke
1945, "Zu einer allgemeinen Systemlehre", Blätter für deutsche Philosophie, 3/4. (Extract in: Biologia Generalis, 19 (1949), 139-164).
1951, "General system theory – A new approach to unity of science" (Symposium), Human Biology, Dec. 1951, Vol. 23, p. 303-361.
^ a b Bertalanffy, L. von, (1934). Untersuchungen über die Gesetzlichkeit des Wachstums. I. Allgemeine Grundlagen der Theorie; mathematische und physiologische Gesetzlichkeiten des Wachstums bei Wassertieren. Arch. Entwicklungsmech., 131:613-652.
^ a b T.E. Weckowicz (1989). Ludwig von Bertalanffy (1901-1972): A Pioneer of General Systems Theory. Working paper Feb 1989. p.2
^ Mark Davidson (1983). Uncommon Sense: The Life and Thought of Ludwig Von Bertalanffy. Los Angeles: J. P. Tarcher. p.49
^ a b Bertalanffy Center for the Study of Systems Science, page: His Life - Bertalanffy's Origins and his First Education. Retrieved 2009-04-27 Archived July 25, 2011, at the Wayback Machine
^ a b c d e Drack, Manfred; Apfalter, Wilfried; Pouvreau, David (11 March 2017). "On the Making of a System Theory of Life: Paul A Weiss and Ludwig von Bertalanffy's Conceptual Connection". The Quarterly Review of Biology. 82 (4): 349–373. doi:10.1086/522810. PMC 2874664. PMID 18217527.
^ Davidson, p. 51
^ August Friedrich Robert Pūtter (6 April 1879 - 11 March 1929) wrote a textbook on comparative physiology entitled Vergleichende Physiologie (Jena: G. Fischer, 1911) and many other notable works. For a translation of his "Studien ūber physiologische Ähnlichkeit. VI. Wachstumsähnlichkeiten" ("Studies on Physiological Similarity. VI. Analogies of Growth") in Pflūgers Archiv fūr die gesamte Physiologie des Menschen und der Tiere, 180: 298-340, see http://www.dfo-mpo.gc.ca/Library/147555.pdf
^ Bertalanffy, L. von, (1969). General System Theory. New York: George Braziller, pp. 136
^ Alexandr N. Tetearing (2012). Theory of populations. Moscow: SSO Foundation. p. 607. ISBN 978-1-365-56080-4.
^ Nicholas D. Rizzo William Gray (Editor), Nicholas D. Rizzo (Editor), (1973) Unity Through Diversity. A Festschrift for Ludwig von Bertalanffy. Gordon & Breach Science Pub
^ Bertalanffy, L. von, (1969). General System Theory. New York: George Braziller, pp. 39-40
^ Bertalanffy, L. von, (1969). General System Theory. New York: George Braziller, pp. 139-1540
^ Bertalanffy, L. von, (1969). General System Theory. New York: George Braziller, pp. 194-197
Sabine Brauckmann (1999). Ludwig von Bertalanffy (1901--1972), ISSS Luminaries of the Systemics Movement, January 1999.
Peter Corning (2001). Fulfilling von Bertalanffy's Vision: The Synergism Hypothesis as a General Theory of Biological and Social Systems, ISCS 2001.
Mark Davidson (1983). Uncommon Sense: The Life and Thought of Ludwig Von Bertalanffy, Los Angeles: J. P. Tarcher.
Debora Hammond (2005). Philosophical and Ethical Foundations of Systems Thinking, tripleC 3(2): pp. 20–27.
Ervin László eds. (1972). The Relevance of General Systems Theory: Papers Presented to Ludwig Von Bertalanffy on His Seventieth Birthday, New York: George Braziller, 1972.
David Pouvreau (2013). "Une histoire de la 'systémologie générale' de Ludwig von Bertalanffy - Généalogie, genèse, actualisation et postérité d'un projet herméneutique", Doctoral Thesis (1138 pages), Ecole des Hautes Etudes en Sciences Sociales (EHESS), Paris : http://tel.archives-ouvertes.fr/tel-00804157
Thaddus E. Weckowicz (1989). Ludwig von Bertalanffy (1901-1972): A Pioneer of General Systems Theory, Center for Systems Research Working Paper No. 89-2. Edmonton AB: University of Alberta, February 1989.
Wikiquote has quotations related to Ludwig von Bertalanffy.
International Society for the Systems Sciences' biography of Ludwig von Bertalanffy.
http://isss.org/projects/primer International Society for the Systems Sciences' THE PRIMER PROJECT: INTEGRATIVE SYSTEMICS (organismics)
Bertalanffy Center for the Study of Systems Science BCSSS in Vienna.
Ludwig von Bertalanffy (1901-1972): A Pioneer of General Systems Theory working paper by T.E. Weckowicz, University of Alberta Center for Systems Research.
Ludwig von Bertalanffy, General System Theory - Passages (1968)
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\frac { \pi } { 12 }
as the difference of two special angles. Then use the appropriate angle difference formula to find the exact sine and cosine of
\frac { \pi } { 12 }
\text{Since }\frac{\pi}{3}-\frac{\pi}{4}=\frac{4\pi}{12}-\frac{3\pi}{12}=\frac{\pi}{12},
you can use the Angle Difference Identity.
Substitute the angles into the Angle Difference Identity.
\text{sin }\left (\frac{\pi}{3}-\frac{\pi}{4} \right )=\text{ sin }\left (\frac{\pi}{3} \right ) \cdot \text{cos } \left (\frac{\pi}{4} \right ) - \text{cos} \left (\frac{\pi}{3} \right) \cdot \text{sin } \left (\frac{\pi}{4} \right )
Substitute the exact values for each part of the equation.
Then simplify the multiplication and subtraction.
\text{sin}\left (\frac{\pi}{12} \right )= \frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2}}{2}-\frac{\sqrt{1}}{2}\cdot \frac{\sqrt{2}}{2}
\frac{\sqrt{6} - \sqrt{2}}{4}
Repeat this process for cosine.
|
Elliptic equations in the plane satisfying a Carleson measure condition | EMS Press
David J. Rule
In this paper we settle (in dimension
n=2
) the open question whether for a divergence form equation
\div (A\nabla u) = 0
with coefficients satisfying certain minimal smoothness assumption (a Carleson measure condition), the
L^p
Neumann and Dirichlet regularity problems are solvable for some values of
p\in (1,\infty)
. The related question for the
L^p
Dirichlet problem was settled (in any dimension) in 2001 by Kenig and Pipher [Kenig, C.E. and Pipher, J.: The Dirichlet problem for elliptic equations with drift terms. Publ. Mat. 45 (2001), no. 1, 199-217].
Martin Dindoš, David J. Rule, Elliptic equations in the plane satisfying a Carleson measure condition. Rev. Mat. Iberoam. 26 (2010), no. 3, pp. 1013–1034
|
How Do You Make a Problem? | Toph
How Do You Make a Problem?
By upobir · Limits 2s, 1.0 GB
I am very busy these days. So when I was told, I have to create a problem for the contest, I was slightly annoyed. But regardless I made this problem. And this is the story of how I did so.
To create a problem, you first need a setting or scenario. There are lots of well-known settings in competitive programming, Array of numbers, grid of numbers, strings, permutations, graphs and so on. After a bit of pondering, I decided to start with a balanced string of brackets. A sequence of brackets is balanced if you can turn it into a valid arithmetic expression by adding characters ‘1’ and ‘+’. For example `(())(()())` is a balanced bracket string. But `((()`, `())()`, `)(` are not. So we start with the following setting, You will be given a string of length
n of brackets that is balanced.
Now there needs to be a task, that you will do. The task could be to find some construction or check some conditions. It could also be an optimization where you have to find the minimum value or some maximum value. Now, these are generally a bit hard to work with. So I chose the following task, you have to count something. But count what? There are lots of things to count in a balanced bracket sequence. The answer was in front of me always, count balanced substrings! i.e. you have to count the number of substrings of the input string that are also balanced. A substring of a string is part of the string obtained by deleting a prefix (possibly empty) and deleting a suffix (possibly empty). For example in `(()(()())`, `(()` is a substring (that occurs twice). So you have to count the number of balanced substrings of the input string.
But the problem felt too easy. So I decided to make it hard. How do you make a problem hard? Add queries to it of course! So instead of counting balanced substrings of the original string, you will be given
q queries. Each query is a range
l
r of the original string. And for each query, you have to count the number of balanced substring of the substring built by taking indices
l, …,
r. (1-indexed)
Now I am happy, so chop chop, solve the problem!
The first line of input will contain a positive integer
n, the length of the input string of brackets. Second line will contain a balanced string
s of `(` and `)`.
Third line will contain a positive integer
q, the number of queries. Next
q lines each contain a query as two positive integer
l and
r separated by space. Meaning that you have to count the number of balanced substrings of
s[l….r]
2 \leq n \leq 2\times 10^6
2≤n≤2×106
1 \leq q \leq 10^6
1≤q≤106
1 \leq l \leq r \leq n
1≤l≤r≤n
For each query, output the answer in a separate line.
(())(()())
0% Solution Ratio
Intra BUET Junior Programming Contest 2022
Replay of Intra BUET Junior Programming Contest 2022
|
Three charges are arranged as shown in Figure. Find the magnitude and direction
Three charges are arranged as shown in Figure. Find the magnitude and direction of the electrostatic force on the charge at the origin.
Force between two charges on x-axis is repulsion force, so the force on charge at origin acts in negative x-direction
{F}_{1}=k\left(6\cdot {10}^{-9}\right)\frac{5\cdot {10}^{-9}}{{\left(0.03\right)}^{2}}=8.98\cdot {10}^{9}\cdot \left(6\cdot {10}^{-9}\right)\frac{5\cdot {10}^{-9}}{{\left(0.03\right)}^{2}}
=2993.33\cdot {10}^{-9}
Force between two charges on y-axis is attractive force, so the force on charge at origin acts in negative y-direction
{F}_{2}=k\left(3\cdot {10}^{-9}\right)\frac{5\cdot {10}^{-9}}{{\left(0.01\right)}^{2}}=8.98\cdot {10}^{9}\cdot \left(3\cdot {10}^{-9}\right)\frac{5\cdot {10}^{-9}}{{\left(0.01\right)}^{2}}
angle made by the net force with positive x-axis = 180 + taninverse
\frac{13470\cdot {10}^{-9}}{2993.33\cdot {10}^{-9}}=180+77.47=257.77
magnitude of force
=\sqrt{{13470}^{2}+{2993.33}^{2}}\cdot {10}^{-9}=13798.5\cdot {10}^{-9}
The arm weighs 44.0 N. The force of gravity acting on the arm acts through point A. Determine the magnitudes of the tension force t in the deltoid muscle and the force s of the shoulder on the humerus (upper-arm bone) to hold the arm in the position shown.
A 5.00 kg package slides 1.50 m down a long ramp that is inclined at
{12.0}^{\circ }
below the horizontal. The coefficient of kinetic friction between the package and the ramp is k =0.310. Calculate
(a) the work done on the package by friction (b) the workdone on the package by gravity (c) the work done on the package bythe normal force (d) the total work done on the package (e) Ifthe package has a speed of 2.20 m/s at the top of the ramp, what isits speed after sliding 1.50 m down the ramp?
A cylindrical capacitor has an inner conductor of radius 1.5 mm andan outer conductor of radius 3.4 mm.The two conductors are separated by vacuum, and the entire capacitor is 3.0 m long.
a) What is the capacitance per unit length?
b)The potential of the inner conductor is 350 mV higher than that of the outer conductor. Find the charge (magnitude and sign) on both conductors. Find the charge (magnitude and sign) on both conductors.
A sled with rider having a combined mass of 120 kg travels over the perfectly smooth icy hill shown in the accompanying figure. How far does the sled land from the foot of the cliff (in m)?
Convey or belt, A, which forms a 20 degree angle with the horizontal, moves at a constant speed of 1.2 m/s and is used toload an airplane. Knowing what angle that a worker tosses duffelbad B with an initial velocity of 0.76 m/s at an angle of 30 degrees with the horizontal, determine the velocity of the bagrelative to the belt as it lands on the belt.
Write out the chemical equation for the Diels-Alder reaction of anthracene with maleic anhydride, showing structures of the reactants and the product.
Determine if the columns of the matrix span
{\mathbb{R}}^{4}
\left[\begin{array}{cccc}7& 2& -5& 8\\ -5& -3& 4& -9\\ 6& 10& -2& 7\\ -7& 9& 2& 15\end{array}\right]
|
An element (X) belongs to fourth period and fifteenth group of the periodic table. Which one of the following is true regarding the outer electronic configuration of (X)? It has
partially filled d orbitals and completely filled s orbital
completely filled s orbital and completely filled p orbitals
completely filled s orbital and half-filled p orbitals
half-filled d orbitals and completely filled s orbital.
The electronic configuration of (X) can be written as
X = 1s22s22p63s23p64s23d104p3
So element (X) has completely filled s orbital and d orbitals and half filled p orbitals.
{\mathrm{C}}_{2}^{2-}
{\mathrm{He}}_{2}^{+}
< NO <
{\mathrm{O}}_{2}^{-}
{\mathrm{He}}_{2}^{+}
{\mathrm{O}}_{2}^{-}
{\mathrm{C}}_{2}^{2-}
{\mathrm{O}}_{2}^{-}
{\mathrm{C}}_{2}^{2-}
{\mathrm{He}}_{2}^{+}
NO <
{\mathrm{C}}_{2}^{2-}
{\mathrm{O}}_{2}^{-}
{\mathrm{He}}_{2}^{+}
{\mathrm{He}}_{2}^{+}
{\mathrm{O}}_{2}^{-}
{\mathrm{C}}_{2}^{2-}
According to molecular orbital theory, the energy level of the given molecules are
{\mathrm{C}}_{2}^{2-}\quad \to \quad \mathrm{\sigma }1{\mathrm{s}}^{2}\quad {\mathrm{\sigma }}^{*}1{\mathrm{s}}^{2}\quad \mathrm{\sigma }2{\mathrm{s}}^{2}\quad {\mathrm{\sigma }}^{*}2{\mathrm{s}}^{2}\quad \mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^{2}\quad \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^{2}\quad \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^{2}
Bond Order :
\frac{1}{2}[10\quad -\quad 4]\quad =\quad 3
{\mathrm{He}}_{2}^{+}\quad \to \quad \mathrm{\sigma }1{\mathrm{s}}^{2}\quad {\mathrm{\sigma }}^{*}1{\mathrm{s}}^{1}
\frac{1}{2}\left[2\quad -\quad 1\right]\quad =\quad \frac{1}{2}\quad =\quad 0.5
\to \quad \mathrm{\sigma }1{\mathrm{s}}^{2}\quad {\mathrm{\sigma }}^{*}1{\mathrm{s}}^{2}\quad \mathrm{\sigma }2{\mathrm{s}}^{2}\quad {\mathrm{\sigma }}^{*}2{\mathrm{s}}^{2}\quad \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^{2}\quad \mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^{2}\quad \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^{2}\quad {\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{x}}^{1}
\frac{1}{2}\left[10\quad -\quad 5\right]\quad =\quad 2.5
{\mathrm{O}}_{2}^{-}\quad \to \quad \mathrm{\sigma }1{\mathrm{s}}^{2}\quad {\mathrm{\sigma }}^{*}1{\mathrm{s}}^{2}\quad \mathrm{\sigma }2{\mathrm{s}}^{2}\quad {\mathrm{\sigma }}^{*}2{\mathrm{s}}^{2}\quad \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^{2}\quad \mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^{2}\quad \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}\quad }^{2}{\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{x}}^{2}\quad {\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{y}}^{1}
\frac{1}{2}\left[10\quad -\quad 7\right]\quad =\quad 1.5
So, the correct order of their increasing bond order is
{\mathrm{He}}_{2}^{+}\quad <\quad {\mathrm{O}}_{2}^{-}\quad <\quad \mathrm{NO}\hspace{0.17em}<\hspace{0.17em}{\mathrm{C}}_{2}^{2-}
Assertion : H3BO3 is a weak acid.
Reason : Water extracts the proton of H3BO3.
H3BO3 is a very weak monobasic acid (Ka= 5.6
\times
10-10) but it does not act as a proton donor. It behaves as a Lewis acid i.e., it accepts a pair of electrons from OH- ion of H2O.
H - OH + B(OH)3
\to
[B(OH)4]- + H+
Best reagent for the conversion of AgNO3 to Ag is
H3PO2 is a good reducing agent as it contains two P-H bonds and thus reduces AgNO3 to Ag.
4AgNO3 + H3PO2 + 2H2O
\to
4Ag
\downarrow
+ H3PO4 + 4HNO3
Which of the following molecules has more than one lone pair ?
Among all the given options, XeF2 has more than one lone pair, i.e. 3.
SO2 has one lone pair.
SiF4 and CH4 has no lone pairs.
10 ml of liquid carbon disulphide (specific gravity 2.63) is burnt in oxygen. Find the volume of the resulting gases measured at STP.
1 mL of CS2 weighs 2.63 gm
10 mL of CS2 weighs 26.3 gm
\underset{76\quad \mathrm{gm}}{\underset{12\quad +\quad (2\quad \times \quad 32)}{{\mathrm{CS}}_{2}}}\quad +\quad {\mathrm{O}}_{2}\quad \quad \quad \to \quad \underset{67.2\quad \mathrm{L}}{\underset{⏟}{\underset{22.4\quad \mathrm{L}}{{\mathrm{CO}}_{2}}\quad +\underset{44.8\quad \mathrm{L}}{\quad 2{\mathrm{SO}}_{2}}}}
\because
76 gm of CS2 will yield 67.2 L of a mixture of CO2 and SO2 at STP.
\therefore
26.3 gm of CS2 would yield
\frac{67.2}{76}\quad \times \quad 26.3\quad
= 23.25 L
KP for the reaction A
⇌
B is 4. If initially only A is present then what will be the partial pressure of B after equilibrium?
Assuming that reaction is started with 1 mol of A and at equilibrium x moles of A have reacted,
\quad \quad \quad \underset{1-\mathrm{x}\quad \mathrm{mol}}{\underset{1\quad \mathrm{mol}}{\mathrm{A}}}\quad \to \quad \underset{\mathrm{x}\quad \mathrm{mol}\quad (\mathrm{at}\quad \mathrm{equilibrium})}{\underset{0\quad \left(\mathrm{initially}\right)}{\mathrm{B}}}
\frac{{\mathrm{p}}_{\mathrm{B}}}{{\mathrm{p}}_{\mathrm{A}}}=\quad 4
\frac{\mathrm{x}}{1-\mathrm{x}}\quad =\quad 4
The true statement for the acids of phosphorus, H3PO2 H3PO3 and H3PO4 is
the order of their acidity is H3PO4 > H3PO3 > H3PO2
all of them are reducing in nature
all of them are tribasic acids
the geometry of phosphorus is tetrahedral in all the three.
H3PO2, H3PO3 and H3PO4 contain one, two and three ionisable hydrogen atoms respectively.
⇌
H+ + H2P
{\mathrm{O}}_{2}^{-}
⇌
{\mathrm{O}}_{3}^{-}
⇌
H+ + HP
{\mathrm{O}}_{3}^{2-}
⇌
{\mathrm{O}}_{4}^{-}
⇌
{\mathrm{O}}_{4}^{2-}
⇌
H+ + P
{\mathrm{O}}_{4}^{3-}
But there is very little difference in acidity. In H3PO2, H3PO3 and H3PO4, P is sp3 hybridised in all. Therefore, all are tetrahedral.
According to Bohr's theory, which of the following correctly represents the variation of energy and radius of an electron in nth orbit of H-atom?
\propto \quad \frac{1}{{\mathrm{n}}^{2}}
, r
\propto \frac{1}{{\mathrm{n}}^{2}}
\propto \frac{1}{{\mathrm{n}}^{2}}
\propto {\mathrm{n}}^{2}
\propto
n2, r
\propto
\propto
\propto \frac{1}{\mathrm{n}}
\propto \frac{1}{{\mathrm{n}}^{2}}
\propto {\mathrm{n}}^{2}
\frac{{\mathrm{n}}^{2}{\mathrm{h}}^{2}}{4{\mathrm{\pi }}^{2}{\mathrm{mZe}}^{2}\mathrm{k}}
and En =
\frac{-{\mathrm{k}}^{2}2{\mathrm{\pi }}^{2}{\mathrm{mZ}}^{2}{\mathrm{e}}^{4}}{{\mathrm{n}}^{2}{\mathrm{h}}^{2}}
Therefore, En
\propto \frac{1}{{\mathrm{n}}^{2}}
\propto
Which of the following reactions does not take place?
BF3 + F-
\to
{\mathrm{F}}_{4}^{-}
BF3 + 3F-
\to
{\mathrm{F}}_{6}^{3-}
AlF3 + 3F-
\to
{\mathrm{F}}_{6}^{3-}
...(III)
BF3 forms complex halides of the type [B
{\mathrm{F}}_{4}^{-}
] in which B atom has C.N. 4, it cannot extend its C.N. beyond 4 due to the non-availability of d-orbitals in its configuration. Hence, B
{\mathrm{F}}_{6}^{3-}
ion (sp3d2 hybridisation) is not formed. On the other hand, Al can extend its C.N. beyond 4 due to the availability of d-orbitals in its configuration.
|
Traffic engineering (transportation) - Wikipedia
For the engineering of communications and computer networks, see Teletraffic engineering.
For broader coverage of this topic, see Transportation engineering.
Complex intersections with multiple vehicle lanes, bike lanes, and crosswalks are common examples of traffic engineering projects
1 Traffic systems
1.1 Lane flow equation
2 Highway safety
Traffic systems[edit]
Traditionally, road improvements have consisted mainly of building additional infrastructure. However, dynamic elements are now being introduced into road traffic management. Dynamic elements have long been used in rail transport. These include sensors to measure traffic flows and automatic, interconnected, guidance systems to manage traffic (for example, traffic signs which open a lane in different directions depending on the time of day). Also, traffic flow and speed sensors are used to detect problems and alert operators, so that the cause of the congestion can be determined, and measures can be taken to minimize delays. These systems are collectively called intelligent transportation systems.
Lane flow equation[edit]
A ramp meter limits the rate at which vehicles can enter the freeway
The relationship between lane flow (Q, vehicles per hour), space mean speed (V, kilometers per hour) and density (K, vehicles per kilometer) is
{\displaystyle Q=KV}
Observation on limited access facilities suggests that up to a maximum flow, speed does not decline while density increases. However, above a critical threshold (BP, breakpoint), increased density reduces speed. Additionally, beyond a further threshold, increased density reduces flow as well.
Therefore, speeds and lane flows at bottlenecks can be kept high during peak periods by managing traffic density using devices that limit the rate at which vehicles can enter the highway. Ramp meters, signals on entrance ramps that control the rate at which vehicles are allowed to enter the mainline facility, provide this function (at the expense of increased delay for those waiting at the ramps).
Highway safety[edit]
Highway safety engineering is a branch of traffic engineering that deals with reducing the frequency and severity of crashes. It uses physics and vehicle dynamics, as well as road user psychology and human factors engineering, to reduce the influence of factors that contribute to crashes. A well-drafted Traffic Control Plan (TCP) is critical to any job involving roadway work. A properly-prepared TCP will specify equipment, signage, placement, and personnel.[2]
A typical traffic safety investigation follows these steps:[3]
1. Identify and prioritize investigation locations. Locations are selected by looking for sites with higher than average crash rates, and to address citizen complaints.
2. Gather data. This includes obtaining police reports of crashes, observing road user behavior, and collecting information on traffic signs, road surface markings, traffic lights and road geometry.
3. Analyze data. Look for collisions patterns or road conditions that may be contributing to the problem.
4. Identify possible countermeasures to reduce the severity or frequency of crashes.
• Evaluate cost/benefit ratios of the alternatives
• Consider whether a proposed improvement will solve the problem, or cause "crash migration." For example, preventing left turns at one intersection may eliminate left turn crashes at that location, only to increase them a block away.
• Are any disadvantages of proposed improvements likely to be worse than the problem you are trying to solve?
5. Implement improvements.
6. Evaluate results. Usually, this occurs some time after the implementation. Have the severity and frequency of crashes been reduced to an acceptable level? If not, return to step 2.
Bus lane / bus priority / bus rapid transit
Traffic congestion / traffic flow / traffic signals
Wikimedia Commons has media related to Traffic engineering.
^ City Rise Safety Engineering Plans and why they matter..
^ Traffic Control Plans What Traffic Control Plans do to keep you community safe.
^ Road Safety Fundamentals. Ithaca, NY: Cornell Local Roads Program. September 2009.
Homburger, Kell and Perkins, Fundamentals of Traffic Engineering, 13th Edition, Institute of Transportation Studies, University of California (Berkeley [1]), 1992.
Das, Shantanu and Levinson, D. (2004) A Queuing and Statistical Analysis of Freeway Bottleneck Formation. ASCE Journal of Transportation Engineering Vol. 130, No. 6, November/December 2004, pp. 787–795
Retrieved from "https://en.wikipedia.org/w/index.php?title=Traffic_engineering_(transportation)&oldid=1078741135"
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Rewrite each of the following equations in graphing form, identify the conic, and sketch the graph. Also include all of the key information, such as the locations of the foci and/or the equations of any asymptotes.
4x^2-y^2-24x-10y=5
4(x^2-6x+?)-(y^2+10y+?)=5+4(?)-(?)
y=3x^2-24x+43
y-43+3(?)=3(x^2-8x \:+\:?)
|
The velocity of a car (in feet per second) is given by
v ( t ) = \left\{ \begin{array} { l l } { 10 } & { \text { for } 0 < t \leq 5 } \\ { 2 t } & { \text { for } t > 5 } \end{array} \right.
. How long does it take for the car to travel a total of
250
ft?
1. Graph.
2. Distance is area under a velocity curve.
3. When will the area under the curve
= 250
\frac{\text{feet}}{\text{second}}
|
Real Analysis/Fundamental Theorem of Calculus - Wikibooks, open books for an open world
Real Analysis/Fundamental Theorem of Calculus
< Real Analysis
←Riemann integration Real Analysis
Fundamental Theorem of Calculus Darboux Integral→
1.1.1 Darboux
1.1.1.1 First Fundamental Theorem
1.1.1.2 Second Fundamental Theorem
1.1.2 Riemann
3.1 Indefinite Integral
The Fundamental Theorem of Calculus is often claimed as the central theorem of elementary calculus. Although it can be naturally derived when combining the formal definitions of differentiation and integration, its consequences open up a much wider field of mathematics suitable to justify the entire idea of calculus as a math discipline.
You will be surprised to notice that there are actually two theorems that make up The Fundamental Theorem of Calculus. They both serve to prove the relationship between differentiation and definite integration, but the first proves that they are inverses of each other, only in the sense that they undo each other's operation, and the second proves that there exists a way of computing definite integration using antiderivatives. This theorem, much like the concept of the limit, will form much of the backbone of latter theorems moving forward, so it is crucial to understand what these theorems can do. As such, the layout of this section mirrors that of the limit.
It is essential to understand what the theorems actually state, which are highlighted below
Given a continuous function ƒ and a function F, both over some closed interval I, if the form
{\displaystyle F=\int _{a}^{x}{f}}
is valid, then it implies that
{\displaystyle F'=f}
is valid as well.
If ƒ is integrable and ƒ is a differentiation of some other function g, then the definite integration is defined as
{\displaystyle \int _{a}^{b}{f}=g(b)-g(a)}
However, unlike the section on limits, this is not a definition primarily based on agreed mathematical concepts, as was the case with limits with its definition only needing one theorem to justify, but a theorem relying on several concepts to function together without contradictions. After all, differentiation is composed of a limit and integration summations. Claiming a relationship between them, unless justified, can spell disaster in mathematics. So, we will formally prove that this relationship is valid in the next heading.
Below are not just one, but two proofs justifying our earlier claim; effectively prove the Fundamental Theorem of Calculus. Both rely primarily on the definition of differentiation and integration, but both differ based on which integral one uses. Whereas the first two use the definition of integration where upper and lower sums are defined (Darboux Integration, the last two rely on the definition of integration where only a single summation is used (Riemann Integration). Despite this, since both have already been proven to be equivalent, either proof is fine. Also surprising is that the following proofs are not complex. Neither proofs for both the Darboux or Riemann version rely on any new concepts and thus no need to justify new assumptions. Therefore, this reading will be like a simple application of the theorems learned in the previous section of the integral of one's choice.
DarbouxEdit
The following proof uses Darboux Integration. The general strategy is using the definition of both differentiation and integration to coax an implication through thoughtful manipulations. However, it relies on more theorems than the Riemann version.
First Fundamental TheoremEdit
As always, the left-handed and right-handed differentiation is left to you.
Proof of the First Fundamental Theorem using Darboux Integrals
{\displaystyle F}
and its definition, we will suppose two things. First is the following mathematical statement. Second is the introduction of the variable
{\displaystyle h}
, which we will use, with its implicit meaning, later. Note that the case where
{\displaystyle h<0}
results in slightly different math (the position of inequalities and integral positions, but nothing serious). However, it is surprisingly not a major problem, so it is ignored (if you are curious as to proving us right or wrong, the problem set asks for this proof—and the answer you the proof removed from this section). We will assume
{\displaystyle h>0}
for the remainder of the proof.
{\displaystyle F(x+h)-F(x)}
The following mathematical statement implies this equality
{\displaystyle F(x+h)-F(x)=\int \limits _{x}^{x+h}{f}}
To capitalize on this integral, we will now define a supremum/infimum of the function
{\displaystyle f}
such that they represent the highest/lowest possible approximation of the integral. It will be influenced based on the value of the variable
{\displaystyle h}
{\displaystyle {\begin{aligned}m_{h}&=\inf {\Big \{}f(x)\mid x\in [c,c+h]{\Big \}}\\M_{h}&=\sup {\Big \{}f(x)\mid x\in [c,c+h]{\Big \}}\end{aligned}}}
We can now claim the following inequality relationship (the justification is found in the Darboux Integration section), which can be algebraically manipulated to show the following.
{\displaystyle {\begin{aligned}m_{h}\cdot h&\leq \int \limits _{x}^{x+h}f\leq M_{h}\cdot h\\m_{h}&\leq {\frac {1}{h}}\int \limits _{x}^{x+h}f\leq M_{h}\\m_{h}&\leq {\frac {F(x+h)-F(x)}{h}}\leq M_{h}\\\end{aligned}}}
We can apply limits over the entire inequality. The answer of the now derivative derives from the Squeeze Theorem. Given that the interval used in
{\displaystyle m_{h}}
{\displaystyle M_{h}}
relies on the variable
{\displaystyle h}
, applying the limit over
{\displaystyle h}as the effect of reducing the interval to a single
{\displaystyle x}
variable, meaning that the supremum/infimum of a single set composed of
{\displaystyle f(x)}
is self explanatory. Thus, the answer in the middle, now fully the definition of the derivative of
{\displaystyle F}
, is rightfully
{\displaystyle f}
, what we set out to prove.
{\displaystyle {\begin{aligned}\lim _{h\to 0}{m_{h}}&\leq \lim _{h\to 0}{\frac {F(x+h)-F(x)}{h}}\leq \lim _{h\to 0}{M_{h}}\\f(x)&\leq \lim _{h\to 0}{\frac {F(x+h)-F(x)}{h}}\leq f(x)\\F'(x)&=\lim _{h\to 0}{\frac {F(x+h)-F(x)}{h}}=f(x)\end{aligned}}}
Second Fundamental TheoremEdit
RiemannEdit
The following proof uses Riemann Integration. The general strategy is using the definition of a limit to imply the conclusion. For those uncomfortable with limits, the Darboux version may be easier.
This is pretty far down from the original definition, so we will reiterate what we are seeking to prove.
{\displaystyle f:[a,b]\to \mathbb {R} }
{\displaystyle c\in [a,b]}
{\displaystyle F:[a,b]\to \mathbb {R} }
be the indefinite integral o{\displaystyle f}
Then, we should prove that the following relationship is valid
{\displaystyle F}
{\displaystyle c}
{\displaystyle F'(c)=f(c)}
{\displaystyle \varepsilon >0}
{\displaystyle x\in [a,b]}
{\displaystyle x\neq c}
{\displaystyle F(x)-F(c)=\int _{c}^{x}f=L}
(say). There exists
{\displaystyle \delta >0}
such that if a partition
{\displaystyle \|{\mathcal {\dot {P}}}\|<\delta }
{\displaystyle |S(f,{\mathcal {\dot {P}}})-L|<\varepsilon |x-c|}
(note that in this proof, all the Riemann sums are over the interval
{\displaystyle [c,x]}
). As
{\displaystyle f}
is integrable over
{\displaystyle [c,x]}
, it is bounded over that interval. Hence, let
{\displaystyle M=\sup f([c,x])}
{\displaystyle S(f,{\mathcal {\dot {P}}})\leq \sum _{i=1}^{n}M(x_{i}-x_{i-1})=M(x-c)}
{\displaystyle f}
{\displaystyle c}
{\displaystyle \delta >0}
{\displaystyle M(x-c)-S(f,{\mathcal {\dot {P}}})<\varepsilon |x-c|}
{\displaystyle |x-c|<\delta }
{\displaystyle x\in V_{\delta }(c)}
{\displaystyle \left|{\tfrac {F(x)-F(c)}{x-c}}-f(c)\right|=\left|{\tfrac {L}{x-c}}-f(c)\right|<\left|{\tfrac {S(f,{\mathcal {\dot {P}}})+\varepsilon |x-c|}{x-c}}-f(c)\right|<\left|{\tfrac {M(x-c)+2\varepsilon |x-c|}{x-c}}-f(c)\right|}
{\displaystyle <\left|{\tfrac {f(c)(x-c)+3\varepsilon |x-c|}{x-c}}-f(c)\right|<3\varepsilon }
{\displaystyle \lim _{x\to c}{\frac {F(x)-F(c)}{x-c}}=f(c)}
{\displaystyle F'(c)=f(c)}
{\displaystyle f,F:[a,b]\to \mathbb {R} }
{\displaystyle F}
be differentiable on
{\displaystyle [a,b]}
{\displaystyle F'(x)=f(x)}
{\displaystyle x\in [a,b]}
{\displaystyle f}
be Riemann integrable on
{\displaystyle [a,b]}
{\displaystyle \int _{a}^{b}f(x)dx=F(b)-F(a)}
{\displaystyle \int _{a}^{b}f=L}
{\displaystyle \varepsilon >0}
be given. Then, there exists
{\displaystyle \delta >0}
such that for a partition
{\displaystyle \|{\mathcal {\dot {P}}}\|<\delta }
{\displaystyle |S(f,{\mathcal {\dot {P}}})-L|<\varepsilon }
Consider a partition
{\displaystyle {\mathcal {P}}}
{\displaystyle x_{i-1},x_{i}\in {\mathcal {P}}}
. By Lagrange's Mean Value Theorem, we have that there exists
{\displaystyle c_{i}\in (x_{i-1},x_{i})}
{\displaystyle {\frac {F(x_{i})-F(x_{i-1})}{x_{i}-x_{i-1}}}=F'(c_{i})=f(c_{i})}
Let the tagged partition
{\displaystyle {\mathcal {\dot {P}}}}
be the partition
{\displaystyle {\mathcal {P}}}
along with the tags
{\displaystyle c_{i}}
{\displaystyle S(f,{\mathcal {\dot {P}}})=\sum _{i=1}^{n}f(c_{i})(x_{i}-x_{i-1})=\sum _{i=1}^{n}{\tfrac {F(x_{i})-F(x_{i-1})}{x_{i}-x_{i-1}}}(x_{i}-x_{i-1})=\sum _{i=1}^{n}(F(x_{i})-F(x_{i-1}))=F(b)-F(a)}
{\displaystyle |S(f,{\mathcal {\dot {P}}})-L|<\varepsilon }
{\displaystyle |F(b)-F(a)-L|<\varepsilon }
{\displaystyle \varepsilon >0}
{\displaystyle |F(b)-F(a)-L|=0}
{\displaystyle \int _{a}^{b}f(x)dx=F(b)-F(a)}
There are many interpretations based of the definition.
Here is an extra concept, just for your perusal (It's actually here is the integral to a variable bound is unclear to you).
Indefinite IntegralEdit
{\displaystyle f:[a,b]\to \mathbb {R} }
{\displaystyle [a,b]}
We define the Indefinite Integral o{\displaystyle f}
{\displaystyle F:[a,b]\to \mathbb {R} }
{\displaystyle F(x)=\int _{a}^{x}f}
{\displaystyle x\in [a,b]}
Retrieved from "https://en.wikibooks.org/w/index.php?title=Real_Analysis/Fundamental_Theorem_of_Calculus&oldid=3199846"
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Case sensitivity - Uncyclopedia, the content-free encyclopedia
This article is about Case sensitivity. For Case Sensitivity, see Case Sensitivity.
Case Sensitivity Is What Prevents You Logging On To Your Computer When CapsLock Is Turned On.
The opposite of Case sensitivity is CaSe iNSeNSiTiViTy, although Not-Case sensitivity is being used frequently, too.
The two crucial keys in the Case sensitivity issue, shown on a Microsoft Windows keyboard. Note that, oddly, Windows is not Case sensitive.
1 Forms of Case sensitivity
2 Advantages of Case sensitivity
3 Disadvantages of Case sensitivity
4 Representatives of Case (in)sensitivity
4.1 Things that are Case sensitive
4.2 things that are not case sensitive
5 CaSe SeNsItIvItY iS aNnOyInG
5.1 wHaT iT mEaNs FoR yOu
5.2 CoMMoN BeLieFs and eRRORS
Forms of Case sensitivity[edit]
Case sensitivity applies mainly to written language. Usually, volume is considered its counterpart in spoken language, where uppercase letters refer to YELLING, lowercase letters refer to normal speaking volume, light and small letters refer to whispering, misaligned letters refer to having a croaky voice ([1]), and blank text refers to .
Advantages of Case sensitivity[edit]
The advantage of Case sensitivity is that you can have 131,072 different copies of thattennisgirl.jpg all called the same thing, whilst also being separate entities, or, more general,
{\displaystyle 2^{n}}
options, where n is the number of alphabetical letters in the text. This can also be seen as a logic error of such scale that it would accelerate the end of the universe.
Disadvantages of Case sensitivity[edit]
Case sensitivity makes finding your pr0n a lot harder than it needs to be. Rather than having to remember if you're trying to find mysister.jpg, yoursister.jpg, bobssister.jpg or oprah.jpg, you also have to choose between MySister.jpg, mysister.jpg, MYSISTER.JPG, yoursister.jpg, YourSister.jpg and YoursIsTER.jpg.
Another obvious advantage of CaSe iNSeNSiTiViTy is the possibility to type faster, as the Magic Keys (Shift and CAPS LOCK) on your keyboard are to be pressed by the shortest and least developed of your fingers (The ones anatomically attached your hand, you sex-focused pervert...). This agony is even worsened by the convention to press that shift button which is further away from the actual letter. Lower-case typing is widely common among users of Instant Messaging and Online Chat (see below for more information).
Representatives of Case (in)sensitivity[edit]
Approximately 1 in 30 human beings is Case sensitive. In particularly bad cases, BOLD CAPITALS or inconsistent use of capitalisation (eg. iPod) can cause Case-Induced fits not dissimilar to epilepsy or biscuits. If you see somebody suffering from a Case-Induced fits, try to talk them out of it by speaking either very very quietly, or VERY VERY LOUDLY INDEED.
Things that are Case sensitive[edit]
God. Never ever forget His Case sensitivity. Otherwise, be prepared for His Holy Wrath.
Teachers. Although they are far from being as mighty as God, they can still force you to a lap of honour at school.
HaXX0Rz. Being Case sensitive is a crucial part of 1337|\|355, or at least they think so.
A common case insensitive keyboard.
UNIX-based systems, such as Linux. This is quite evident, as most users of such systems are HaXX0Rz.
Germans. The Germans, having a very idiosyncratic Grammar, pride themselves of being Case sensitive. They do not care about the People committing Suicide upon unsuccessfully having tried to master the german Language.
Article titles on Wikipedia (except for the first character after the colon!).
things that are not case sensitive[edit]
frodo baggins. having been rendered incapable to press the left shift button by gollum, he is forcedly at most case half-sensitive.
users of instant messaging and online chat, as well as online gamers, due to their main aim being fast typing.
windows, being the counterpart to UNIX-based systems.
keyboards. or do you see every letter in lower- and uppercase?
mute people, for the obvious reason.
google. seriously. it doesn't know or care whether you capitalize your words.
you. because obviously, you haven't stopped reading this article. yet.
oracle's database software. it prints everything in CAPS LOCK.
CaSe SeNsItIvItY iS aNnOyInG[edit]
CaSe SeNsItIvItY iS oNe Of ThE mOsT aNnOyInG tHiNgS eVeR. iT maKes Me fEel Ill.
wHaT iT mEaNs FoR yOu[edit]
It MaKeS fInDiNg FiLeS, fOlDeRs AnD oThEr SuCh ItEmS iNfInItElY mOrE dIfFiCuLt, As YoU hAvE tO wOrRy AbOuT nAmEs LiKe MySister.jpg, mysister.jpg, MYSISTER.JPG As OpPoSeD tO mErElY rEmEmBeRiNg If It Is MySister.jpg Or YourSister.jpg.
CoMMoN BeLieFs and eRRORS[edit]
iT iS WiDeLY CoMMoN aMoNG WaNNaBe-HaXX0RZ To SPeLL VoWeLS iN LoWeRCaSe, aND CoNSoNaNTS iN UPPeRCaSe, HoWeVeR, THiS RuLe iS NoT STRiCTLy FoLLoWeD FoR SCHiZoPHReNiCS SuCH aS THe LeTTeR Y. aPaRT FRoM THe HiGH LiaBiLiTY To eRRoRs and THe NeaR-uNReaDaBiLiTy, THeSe PeoPLe TYPe uNBeLieVaBLY SLoW aND THeReFoRe aRe FaR aWaY FRoM BeiNG a HaXX0R.
fURTHERMORE, IT IS A VERY COMMON BEGINNER'S MISTAKE TO FORGET TO TURN OFF cAPS lOCK AFTER yelling. tHIS OVERSIGHT OFTEN CAUSES MISUNDERSTANDINGS, AS OTHERS BELIEVE THE CONCERNED PERSON IS SHOUTING. sUCH INCIDENCES ARE FREQUENT CAUSES FOR KICKS FROM irc CHANNELS.
Case sensitivity can also refer to court cases about very sensitive matters. Examples of these cannot be named due to their sensitive nature.
Retrieved from "https://uncyclopedia.com/w/index.php?title=Case_sensitivity&oldid=5498471"
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Which of the following oxides is most acidic in nature?
In metals moving down the group metallic character increases, so basic nature increases hence most acidic will be BeO.
Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below :
Br{O}_{4}^{-}\quad \stackrel{1.82}{\to }Br{O}_{3}^{-}\stackrel{1.5\quad V}{\to }HBrO\stackrel{1.595}{\to }B{r}_{2}\stackrel{1.0652\quad V}{\to }B{r}^{-}
Br{O}_{3}^{-}
Br{O}_{4}^{-}
{E}_{cell}^{°}
corresponding to each compound undergoing disproportionation reaction. The reaction for which
{E}_{cell}^{°}
comes out +ve is spontaneous.
HBrO\quad \stackrel{\quad }{\to }\quad B{r}_{2}\phantom{\rule{0ex}{0ex}}{E}^{o}\quad =\quad 1.595\quad V,\quad SRP\quad \left(cathode\right)\phantom{\rule{0ex}{0ex}}HBrO\quad \stackrel{\quad \quad }{\to }\quad Br{O}_{3}^{-}\phantom{\rule{0ex}{0ex}}{E}^{o}\quad =\quad -\quad 1.5\quad V,\quad SOP\quad \left(anode\right)\phantom{\rule{0ex}{0ex}}2HBrO\quad \stackrel{\quad }{\to }\quad B{r}_{2}\quad +\quad Br{O}_{3}^{-}\phantom{\rule{0ex}{0ex}}{E}_{cell}^{o}\quad =\quad SRP\quad \left(cathode\right)-\quad SRP\quad \left(anode\right)\phantom{\rule{0ex}{0ex}}=\quad 1.595\quad -\quad 1.5\phantom{\rule{0ex}{0ex}}=\quad 0.095\quad V\phantom{\rule{0ex}{0ex}}{E}_{cell}^{0}>\quad 0\quad \Rightarrow \quad ∆{G}^{o}\quad <\quad 0\quad \left[spon\mathrm{tan}eous\right]
Which of the following molecules represents the order of hybridisation sp2, sp2, sp, sp from left to right atoms?
HC≡C-C≡CH
CH2=CH-C≡CH
\stackrel{s{p}^{2}}{C{H}_{2}\quad }=\stackrel{s{p}^{2}}{CH}\quad -\quad \stackrel{sp}{C}\quad \equiv \stackrel{sp}{C}H
Nitrous oxide (N2O) occurs naturally in the environment.
In an automobile engine, when fuel is burnt dinitrogen and dioxygen combine to yield NO and NO2
The bond dissociation energies of X2, Y2 and XY are in the ratio of 1 : 0.5: 1. ΔH for the formation of XY is –200 kJ mol–1. The bond dissociation energy of X2 will be
Let B.E of x2,y2 and xy are x kJ mol-1, 0.5 x kJ mol-1 and x kJ mol-1 respectively
\frac{1}{2}{x}_{2}\quad +\quad \frac{1}{2}{y}_{2}\quad \to \quad xy;\phantom{\rule{0ex}{0ex}}∆H\quad =\quad -\quad 200\quad kJ\quad mo{l}^{-1}\phantom{\rule{0ex}{0ex}}∆H\quad =\quad -\quad 200\quad =\quad \Sigma (B.E{)}_{reac\mathrm{tan}t}\quad -\quad \Sigma (B.E{)}_{product}\phantom{\rule{0ex}{0ex}}\quad =\quad \left[\frac{1}{2}\quad \quad x\quad \left(x\right)\quad +\frac{1}{2}\quad x\quad (0.5)x\right]-[1x\quad (x\left)\right]\phantom{\rule{0ex}{0ex}}On\quad solving\quad we\quad get,\quad \phantom{\rule{0ex}{0ex}}B.E\quad of\quad {X}_{2}\quad =\quad x\quad =\quad 800\quad kJ\quad mo{l}^{-1}
The correction factor ‘a’ to the ideal gas equation corresponds to
Density of the gas molecules
Volume of the gas molecules
Forces of attraction between the gas molecules
Electric field present between the gas molecules
In the real gas equation,
\left(P\quad +\quad \frac{a{n}^{2}}{{V}^{2}}\right)\quad (V-nb)\quad =\quad nRT
van der Waal's constant, 'a' signifies intermolecular forces of attraction.
Which one of the following conditions will favour maximum formation of the product in the reaction,
{A}_{2}\quad \left(g\right)\hspace{0.17em}\quad +\quad {B}_{2}\quad \left(g\right)\quad \stackrel{\quad \quad \quad \quad \quad }{\leftrightharpoons }\quad {X}_{2}\quad \left(g\right)\phantom{\rule{0ex}{0ex}}{∆}_{r}\quad H\quad =\quad -\quad XkJ?
{A}_{2}\quad \left(g\right)\hspace{0.17em}+\quad {B}_{2}\quad \left(g\right)\hspace{0.17em}\stackrel{\quad \quad }{\leftrightharpoons }\quad {X}_{2}\quad \left(g\right);\quad \quad ∆H\quad =\quad -\quad xkJ
On increasing pressure equilibrium shift in a direction where number of moles decreases i.e. forward direction.
On decreasing temperature, equilibrium shift n exothermic direction i.e., forward direction.
So high pressure and lower temperature favour maximum formation of product.
Total orbital angular momentum of electron in 's' orbital is equal to zero
An orbital is designated by three quantum numbers while an electron in an atom is designated by four quantum numbers
The value of m for dz2 is zero
The electronic configuration of N atom is
According to Hund's Rule of maximum multiplicity, the correct electronic configuration of N-atom is
The correct order of atomic radii in group 13 elements is
B < Al < In < Ga < Tl
B < Al < Ga < In < Tl
B < Ga < Al < In < Tl
B < Ga < Al < Tl < In
Atomic and ionic radii. Atomic and ionic radii of group 13 elements increase from top to bottom in the group. This is due to increase in the number of energy shells in each succeeding element. However, the atomic radius of gallium (Ga) is less than that of aluminium (Al).
It is due to the poor shielding of the valence electrons of Ga by the inner 3d-electrons. As a result, the effective nuclear charge of Ga is somewhat greater in magnitude than that of Al. Thus, the electrons in gallium experience the greater force of attraction by the nucleus than in aluminium. Hence the atomic size of Ga(135 pm) is slightly less than that of Al(143 pm).
Element (X) electronic configuration
So, valency of X will be 3.
Valency of Mg is 2.
Formula of the compound formed by Mg and X
will be Mg3X2
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Perfect - Uncyclopedia, the content-free encyclopedia
~ Oscar Wilde on Perfection
'Perfection is existence which is beyond all possible improvability hence its all either perfect or flawed hence despite the hence either-or hence nature of this concept of perfection, it nevertheless relies upon the relativistic notion that a thing exists in a state which is flawed in comparison to an idealized or existent thing of the same class which can be determined by all possible methods not to possess any flaw hence this concept of ultimate perfection is commonly applied in dualistic ideologies, and can be said to be validly established by the fact that we can always judge a thing to be flawed,
THE GREATEST PERFECT IMAGE IN THE HISTORY OF THE WORLD!
and therefore falling short of the ideal hence there is an ontological relationship between the concept of perfection and ideology itself, since ideologies are predicated on concepts of what constitutes better or worse but perfection is also a continuum of relative merit, established in some thing's relationship to all other things which can be considered relevant to the thing's nature hence in this sense there is no ultimate perfection, but instead the relative perfection of a thing is established in terms of its suitability for a particular purpose hence for example, a screwdriver can be said to be the perfect tool for embedding a screw hence while a screwdriver may be perfectly suitable for embedding a screw in wood, a screw-driving power-drill attachment may be considered to be more suitable to embed a screw into a harder material, and hence more perfect in a given case since Ancient mathematicians made many assumptions about perfect numbers based on the four they knew so most of the assumptions were wrong
hence one of these assumptions was that since 2, 3, 5, and 7 are precisely the first four primes, the fifth perfect number would be obtained when n = 11, the fifth prime hence, 211 − 1 = 2047 = 23 • 89 is not prime and therefore n = 11 does not yield a perfect number hence The fifth perfect number (33550336 = 212(213 − 1)) has 8 digits, thus debunking the first assumption because of the second assumption, the fifth perfect number indeed ends with a 6 hence the sixth (8 589 869 056) also ends in a 6 hence it has been shown that the last digit of any even perfect number must be 6 or 8 hence a so called two millennia after Euclid, Euler proved that the formula 2n−1(2n − 1) will yield all the even perfect numbers hence -
Thus, every Mersenne prime will yield a distinct even perfect number—there is a concrete one-to-one association between even perfect numbers and Mersenne primes hence this result is often referred to as the "Euclid-Euler Theorem."
{\displaystyle P=7/6/H}
Being this as simply as it is just seen. We can understand that this equation is a equation that equates in a way only this equation can do therefore equatiating how equating or more simple for you:
P equals 7 over 6 over H that means human posibity is over perfections what means the result of 7 divided in 6 with NO variables is the posibilty of a human achieving perfectonism.
List of people interested in this[edit]
<insert name here> Nobody Someone
Anyone that enters here.
Perfect image placement.
Retrieved from "https://uncyclopedia.com/w/index.php?title=Perfect&oldid=6044252"
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Error, (in decimalpart) invalid input: rhs expects its 1st argument, f, to be of type float, but received x^2-1 = y^2 - Maple Help
Home : Support : Online Help : Error, (in decimalpart) invalid input: rhs expects its 1st argument, f, to be of type float, but received x^2-1 = y^2
\mathrm{sin}\left(\left[1,2, 3\right]\right);
\mathrm{whattype}\left(\left[1,2,3\right]\right);
\textcolor[rgb]{0,0,1}{\mathrm{list}}
\left[1,2,3\right]
\mathrm{Describe}\mathit{}\left(\mathrm{sin}\right)
x
\mathrm{sin}
\mathrm{sin}\left(1\right);\mathrm{sin}\left(2\right);\mathrm{sin}\left(3\right)
\textcolor[rgb]{0,0,1}{\mathrm{sin}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{1}\right)
\textcolor[rgb]{0,0,1}{\mathrm{sin}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{2}\right)
\textcolor[rgb]{0,0,1}{\mathrm{sin}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{3}\right)
\mathrm{sin}~\left(\left[1,2,3\right]\right)
\left[\textcolor[rgb]{0,0,1}{\mathrm{sin}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{1}\right)\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{sin}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{2}\right)\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{sin}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{3}\right)\right]
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What Is Net Operating Income – NOI?
Example of Net Operating Income (NOI)
NOI is a before-tax figure, appearing on a property’s income and cash flow statement, that excludes principal and interest payments on loans, capital expenditures, depreciation, and amortization. When this metric is used in other industries, it is referred to as “EBIT,” which stands for “earnings before interest and taxes.”
Net operating income measures an income-producing property's profitability before adding in any costs from financing or taxes.
To calculate NOI, subtract all operating expenses incurred on a property from all revenue generated on the property.
The operating expenses used in the NOI metric can be manipulated if a property owner defers or accelerates certain income or expense items.
The NOI metric does not include capital expenditures.
NOI will indicate to a property owner if renting a property is worth the expense of owning and maintaining it.
Net operating income is a valuation method used by real estate professionals to determine the precise value of their income-producing properties. To calculate NOI, the property's operating expenses must be subtracted from the income a property produces.
In addition to rental income, a property might also generate revenue from amenities such as parking structures, vending machines, and laundry facilities. Operating expenses include the costs of running and maintaining the building, including insurance premiums, legal fees, utilities, property taxes, repair costs, and janitorial fees. Capital expenditures, such as costs for a new air-conditioning system for the entire building, are not included in the calculation.
NOI helps real estate investors determine the capitalization rate, which in turn helps them calculate a property’s value, thus allowing them to compare different properties they may be considering buying or selling.
For financed properties, NOI is also used in the debt coverage ratio (DCR), which tells lenders and investors whether a property’s income covers its operating expenses and debt payments. NOI is also used to calculate the net income multiplier, cash return on investment, and total return on investment.
To calculate net operating income, subtract operating expenses from the revenue generated by a property. Revenue from real estate includes rental income, parking fees, service changes, vending machines, laundry machines, and so on.
\begin{aligned} &\text{Net operating income} = RR - OE \\ &\textbf{where:}\\ &RR=\text{real estate revenue}\\ &OE=\text{operating expenses}\\ \end{aligned}
Net operating income=RR−OEwhere:RR=real estate revenueOE=operating expenses
As an example, let's assume the below information was the profile of a particular condo building that an owner was renting out.
Parking fees: $5,000
Laundry machines: $1,000
Total Revenues = $26,000
Now, let's assume the operating expenses of the condo building are as follows:
Repair and maintenance: $3,000
Total Operating Expenses = $10,000
The net operating income (NOI) in this example would be $26,000 - $10,000 = $16,000.
Let us assume that you own a property that annually pulls in $120,000 in revenues and incurs $80,000 in operating expenses. In this circumstance, it will have a resulting NOI of $40,000 ($120,000 - $80,000). If the total is negative, where operating expenses are higher than revenues, the result is called a net operating loss (NOL).
Creditors and commercial lenders rely heavily on NOI to determine the income generation potential of the property to be mortgaged, even more than they factor an investor's credit history into their decisions. Simply put: this metric helps lenders fundamentally assess the initial value of the property by forecasting its cash flows.
NOI is used to determine the capitalization rate of a property, also known as the return on investment (ROI) in real estate. It divides NOI by the purchase price.
If a property is deemed profitable, the lenders also use this figure to determine the size of the loan they’re willing to make. On the other hand, if the property shows a net operating loss, lenders are likely to reject the borrower's mortgage application, outright.
Property owners can manipulate their operating expenses by deferring certain expenses while accelerating others. NOI can also be increased by raising rents and other fees, while simultaneously decreasing reasonably necessary operating expenses.
As an example of the latter, consider a scenario where an apartment owner waives a tenant’s yearly $12,000 rent, in exchange for that renter acting as a property manager. If the apartment owner would normally pay a building manager a $30,000 salary, they may consequently subtract the “reasonably necessary” cost of $30,000 from revenue, rather than the actual cost of $12,000.
The formula for calculating NOI is as follows:
NOI = real estate revenue - operating expenses
How do you calculate net operating income (NOI) before tax?
NOI is a before-tax calculation in that it does not take tax into consideration.
What is the difference between net income and net operating income (NOI)?
Net operating income is revenue less all operating expenses while net income is revenue less all expenses, including operating expenses and non-operating expenses, such as taxes.
What is a good net operating income (NOI) percentage?
NOI is not a percentage but rather a number that takes into consideration the revenues and expenses of a property. It can be compared to the entire value of the property if that property had been paid fully in cash. In this case, the higher the net operating income to property price percentage, the better.
Net operating income (NOI) is a commonly used figure to assess the profitability of a property. The calculation involves subtracting all operating expenses on the property from all the revenue generated from the property. The higher the revenues and the smaller the expenses, the more profitable a property is. This tells the owner if the income generated from owning and maintaining the property is worth the cost.
The income approach is a real estate appraisal method that allows investors to estimate the value of a property based on the income it generates.
A participation mortgage allows the lender to share in part of the income or resale proceeds of a property, becoming an equity partner in the purchase.
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Shaping Dynamics With Multiple Populations in Low-Rank Recurrent Networks | Neural Computation | MIT Press
Manuel Beiran,
Laboratoire de Neurosciences Cognitives et Computationnelles, INSERM U960, Ecole Normale Superieure. PSL University, 75005 Paris, France manuel.beiran@ens.fr
Alexis Dubreuil,
Laboratoire de Neurosciences Cognitives et Computationnelles, INSERM U960, Ecole Normale Superieure. PSL University, 75005 Paris, France alexis.dubreuil@gmail.com
Adrian Valente,
Laboratoire de Neurosciences Cognitives et Computationnelles, INSERM U960, Ecole Normale Superieure. PSL University, 75005 Paris, France adrian.valente@ens.fr
Francesca Mastrogiuseppe,
Gatsby Computational Neuroscience Unit, UCL, London W1T 4JG, U.K. f.mastrogiuseppe@ucl.ac.uk
Laboratoire de Neurosciences Cognitives et Computationnelles, INSERM U960, Ecole Normale Superieure. PSL University, 75005 Paris, France srdjan.ostojic@ens.fr
Manuel Beiran, Alexis Dubreuil, Adrian Valente, Francesca Mastrogiuseppe, Srdjan Ostojic; Shaping Dynamics With Multiple Populations in Low-Rank Recurrent Networks. Neural Comput 2021; 33 (6): 1572–1615. doi: https://doi.org/10.1162/neco_a_01381
An emerging paradigm proposes that neural computations can be understood at the level of dynamic systems that govern low-dimensional trajectories of collective neural activity. How the connectivity structure of a network determines the emergent dynamical system, however, remains to be clarified. Here we consider a novel class of models, gaussian-mixture, low-rank recurrent networks in which the rank of the connectivity matrix and the number of statistically defined populations are independent hyperparameters. We show that the resulting collective dynamics form a dynamical system, where the rank sets the dimensionality and the population structure shapes the dynamics. In particular, the collective dynamics can be described in terms of a simplified effective circuit of interacting latent variables. While having a single global population strongly restricts the possible dynamics, we demonstrate that if the number of populations is large enough, a rank
R
network can approximate any
R
-dimensional dynamical system.
Low-Dimensional Manifolds Support Multiplexed Integrations in Recurrent Neural Networks
Majority-Ranking
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Plot custom microphone element directivity and patterns - MATLAB - MathWorks í•œêµ
Plot custom microphone element directivity and patterns
pattern(sElem,FREQ)
pattern(sElem,FREQ,AZ)
pattern(sElem,FREQ,AZ,EL)
pattern(sElem,FREQ) plots the 3-D array directivity pattern (in dBi) for the element specified in sElem. The operating frequency is specified in FREQ.
pattern(sElem,FREQ,AZ) plots the element directivity pattern at the specified azimuth angle.
pattern(sElem,FREQ,AZ,EL) plots the element directivity pattern at specified azimuth and elevation angles.
This method replaces the plotResponse method. See Convert plotResponse to pattern for guidelines on how to use pattern in place of plotResponse.
Azimuth angles for computing directivity and pattern, specified as a 1-by-N real-valued row vector where N is the number of azimuth angles. Angle units are in degrees. Azimuth angles must lie between –180° and 180°.
Elevation angles for computing directivity and pattern, specified as a 1-by-M real-valued row vector where M is the number of desired elevation directions. Angle units are in degrees. The elevation angle must lie between –90° and 90°.
Plotting coordinate system of the pattern, specified as the comma-separated pair consisting of 'CoordinateSystem' and one of 'polar', 'rectangular', or 'uv'. When 'CoordinateSystem' is set to 'polar' or 'rectangular', the AZ and EL arguments specify the pattern azimuth and elevation, respectively. AZ values must lie between –180° and 180°. EL values must lie between –90° and 90°. If 'CoordinateSystem' is set to 'uv', AZ and EL then specify U and V coordinates, respectively. AZ and EL must lie between -1 and 1.
u
D=4\mathrm{Ï}\frac{{U}_{\text{rad}}\left(\mathrm{θ},\mathrm{Ï}\right)}{{P}_{\text{total}}}
For antenna, microphone, and array System objects, the pattern method replaces the plotResponse method. In addition, two new simplified methods exist just to draw 2-D azimuth and elevation pattern plots. These methods are azimuthPattern and elevationPattern.
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Plant augmentation for weighted mixed-sensitivity H∞ and H2 loop-shaping design - MATLAB augw - MathWorks Nordic
Create Augmented Plant for H-Infinity Synthesis
Plant augmentation for weighted mixed-sensitivity H∞ and H2 loop-shaping design
P = augw(G,W1,W2,W3)
P = augw(G,W1,W2,W3) computes a state-space model of an augmented LTI plant P(s) with the weighting functions W1(s), W2(s), and W3(s) penalizing the error signal, control signal, and output signal, respectively. P is the augmented plant of the following diagram.
This control structure is used in mixed H∞ synthesis, which lets you design an H∞ controller by simultaneously shaping the frequency responses for tracking and disturbance rejection, noise reduction and robustness, and controller effort. For more information, see Mixed-Sensitivity Loop Shaping.
Suppose you want to synthesize a stabilizing robust controller for the system of the following diagram. The controller must also reject disturbances injected at the plant output.
The plant, G, is an unstable first-order system.
To set up this problem for hinfsyn, insert a weighting function W1 that captures the disturbance rejection goal, and another weighting function W3 to enforce robustness. Specify these weighting functions as the inverses of the desired loop shapes for the sensitivity S and complementary sensitivity T, respectively. (See Mixed-Sensitivity Loop Shaping.)
For this example, choose W1 with:
Choose W3 to have the opposite low-frequency and high-frequency gains.
W1 = makeweight(100,[1 0.5],0.25);
W3 = makeweight(0.25,[1 0.5],100);
For this example, do not specify a W2 (no restriction on control effort). Construct the augmented plant, P.
P = augw(G,W1,[],W3);
G has one input and one output. The augmented plant has an additional input for the control signal, and additional outputs for each of the weights.
The inputs and outputs of P are grouped to keep track of the disturbance and control inputs and the error and measurement outputs. For example, example the output groups. Group Y1 contains the two error outputs z, and group Y2 contains the single measurement output.
P.OutputGroup
Y1: [1 2]
You can now use P for control design. For example, use hinfsyn to design an
{\mathit{H}}_{\infty }
optimal controller that meets the design requirements specified by W1 and W3.
[K,CL,gamma] = hinfsyn(P);
Plant, specified as a dynamic system model such as a state-space (ss) model. G can be any LTI model. If G is a generalized state-space model with uncertain or tunable control design blocks, then mixsyn uses the nominal or current value of those elements.
P — Augmented plant
Augmented plant, returned as a state-space (ss) model. P can be any LTI model with inputs [w;u] and outputs [z;y]. augw groups the inputs and outputs of P using the ss properties InputGroup and OutputGroup such that:
P.InputGroup has field U1 containing the inputs corresponding to w, and field U2 containing the inputs corresponding to u.
P.OutputGroup has field Y1 containing the outputs corresponding to z, and group Y2 containing the outputs corresponding to e.
Here, {w;u} and {z;e} are the inputs and outputs of P in the following control system.
For H∞ or H2 synthesis, the models G and W1,W2,W3 must be proper. In other words, they must be bounded as
s\to \infty
(for continuous-time transfer functions) or
z\to \infty
(for discrete-time transfer functions). Additionally, W1,W2,W3 must be stable. The plant G must be stabilizable and detectable. Otherwise, the resulting P is not stabilizable by any controller.
augw produces the augmented plant P(s) given by:
P\left(s\right)=\left[\begin{array}{cc}{W}_{1}& -{W}_{1}G\\ 0& {W}_{2}\\ 0& {W}_{3}G\\ I& -G\end{array}\right]
The partitioning is embedded using P = mktito(P,NY,NU), which sets the P.InputGroup and P.OutputGroup properties as follows.
P.InputGroup = struct('U1',1:c-NU,'U2',c-NU+1:c);
P.OutputGroup = struct('Y1',1:r-NY,'Y2',r-NY+1:r);
h2syn | hinfsyn | mixsyn | makeweight
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