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Why there are Infinitely Many Prime Numbers – Musings Why there are Infinitely Many Prime Numbers | Musings
Why there are Infinitely Many Prime Numbers
Prime numbers, like $2, 5 7,$ and $47$ are special 1. Unlike say 4 = 2 * 2 or 33 = 3 * 11, prime numbers can’t be written as the product of other numbers. They’re the fundamental block we express other numbers through: every natural number, 1, 2, 3, … is the product of primes 2. How many of these special prime numbers are there?
Infinitely many! A recent one-line proof by Sam Northshield shows why there are infinitely many primes 3:
You may wonder how sin relates to prime numbers. Why $\pi$? Why is the left less than the right?
In this post, we’ll untangle this elegant, one-line proof to reveal the answers.
Let’s imagine there’s a fixed number of primes. What then?
Ignoring the in-between, the proof says
This means “something” is simultaneously equal to zero and strictly less than zero. What we imagined—that there’s a fixed number of primes—is then impossible. There must be infinitely many primes.
Now let’s untangle the “something” in between.
% <![CDATA[ 0 < %]]>
First, let’s understand what these symbols mean:
Careful, pi = 3.14… and product look very similar.
How do we know “something” is greater than zero?
Let’s see what these values equal
Not matter which prime $p$ we choose, $ sin(\pi / p) > 0$.
This means “something” is the product of positive numbers. “Something” $ = \prod_p sin(\pi / p) = + + …+ = +$.
“Something” must be greater than 0.
Step 2: Rewrite “Something”
We want to rewrite “something” in a new form to see it’s also equal to zero. The proof says
The product inside sin means
Let’s look at a graph of sin
The graph between 0 and $2\pi$ is exactly the same as that between $2\pi$ and $4\pi$. It’s a repeating pattern every $2 \pi$. The pattern says
for any # $0, 1, 2, \dots $.
Let’s think about one case when $p = 5$. Remember we know
no matter which orange # we choose.
What if we choose the # to be $ \prod_{p \prime} \frac{p\prime}{5}$?
Using the repeating pattern of sin we can then rewrite $\sin(\pi / 5)$ as
Luckily, this is true not just for $p=5$, but for any $p$. We can always find a $p \prime$ in the numerator to cancel $p$. For any $p$, we then get
Step 3: “Something” equals 0
Finally, let’s see why “something” = $\prod_p sin(\pi / p)$ is also zero. Let’s look at where sin is zero
$sin(k \pi) = 0$ for any $k = 0, 1, 2, …$.
Let’s rewrite the terms inside sin in a friendlier form:
by combining the terms under a common denominator.
because every natural number can be written as the product of primes.
Let’s call one of those primes $p^*$. Then,
Now we have one term in our product that’s zero (when $p = p^*$), meaning
If we assume there are finitely many primes,
and from step 1
We confidently conclude our original assumption is wrong. There must be infinitely many primes.
prime numbers are also special in an everyday sense. Finding these special numbers is what makes sending credit card info. on the internet secure ↩
natural numbers are 1, 2, 3, 4, …. excluding fractions or irrational numbers such as pi. ↩
Many other proofs exist showing there are infinitely many primes (see prime factorization). ↩
© Mark Ibrahim
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Ask Answer - Molecular Basis of Inheritance - Expert Answered Questions for School Students
In given diagram which of the following combination showing recessive epistasis and supplimentary influence:-
Recessive epistasis Supplementary
(1) K L O F H N P
(2) F N H P K L O
(3) F N H P A B C D
(4) K L O A B C D
Solve it n please don't send too small pics of answer.
Q.10. A micro-organism, when viewed under a compound microscope with an objective lens of 40 X and an eye piece of 10 X magnification measured 4000
\mu
in length. The same micro-organism when observed under a dissection microscope with a lens of 10 X magnification, would measure ............
\mu
\mu
\mu
\mu
In Q3, I could'nt comprehend the solution given in NCERT. could u pls send a simple solution to this question? Thank you
Q3. If the sequence of one strand of DNA is written as follows :
5' – ATGCATGCATGCATGCATGCATGCATGC-3'
Write down the sequence of complementary strand in 5'
\to
3' direction.
what does 10 bp in each turn mean
Which of the following professionals are more likely to run the risk of a permanent change in their cell\text{'}s DNA?\phantom{\rule{0ex}{0ex}}I. Researchers u\mathrm{sin}g carbon 14 isotope.\phantom{\rule{0ex}{0ex}}II. X-ray technician.\phantom{\rule{0ex}{0ex}}III. Coal miner.\phantom{\rule{0ex}{0ex}}IV. Dryer and painter.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(a\right) Only II\phantom{\rule{0ex}{0ex}}\left(b\right) I, II and III\phantom{\rule{0ex}{0ex}}\left(c\right) I, II and IV\phantom{\rule{0ex}{0ex}}\left(d\right) I, III and IV\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}The answer is option \left(d\right) but I am not sure about it. Explain.
What is replication. Explain the tools used for replication. (3 marks)
What is meant by semi conservative nature of DNA and who proposed it?
How did the 3 scientist prove that DNA is a genetic material
Rashwan Nazeer
Differentiate between the following 1) inducer and repressor in operons 2) Vntr and probe 3)Template and coding stand
Differentiate between the following 1) Promoter and terminator in transcription unit 2). Exons and intron in and unprocessed eukaryotic mrna 3) inducer and repressor in operons 4) Vntr and probe 4) Template and coding stand
Is this because RNase and protease digest RNA and protein, so there will be no change in transformation and hence transformation will occur. But when DNA is digested, transformation wont occur? :/ Is that the reason or is there any mistake?
what are the properties of genetic material ? explain each.
please answer with explanation of question number 32
Q.32.) Choose the correct option having a group of factors which activate stuart factor in blood plasma, for the clotting of blood.
(A) IX + VIII + IV + Phospholipid
(B) XI + IX + V + Phospholipid
(C) X + VIII + IV + Phospholipid
(D) XI + XII + XIII + Phospholipid
Please can you provide some very important question from the section DNA Fingerprinting.
Please can you provide some very important question from the section Human Genome Project .
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Now that we know how neural networks function, and have built up some intuition about how to train them with backpropagation, it's time to look at how we can describe it mathematically.
The best way to learn how something works mathematically is to understand the why behind something; which is why I want you to go through the extra trouble of deriving the mathematics.
Warning: This essay is going to be heavy on the math. If you're allergic to math, you might want to check out the more intuitive version (coming soon), but with that said, backpropagation is really not as hard as you might think.
Backpropagation is not as hard as you might think
This article assumes familiarity with forward propagation, and neural networks in general. If you haven't already, I recommend reading What is a Neural Network first.
Recall that the weights in a neural network are updated by minimizing an error function that describes how wrong the neural networks current hypothesis is.
x
is the input for the first layer, and
L
is the number of layers in the network, then
a^{(L)}(x)
is the network's hypothesis:
h(x)
m
be the number of examples,
n^{(k)}
the number of neurons in layer
(k)
y(x)
be the correct answer given the input
x
In order for an error function to be suitable for backpropagation, the average error should be computable using:
E = \frac{1}{m} \sum_{i=1}^{m} E(x_i,y_i)
E(x,y)
is the error function for a specific example.
This is nescessary if the backpropagation procedure is to update the weights on the basis of more than one example which will result in a more direct route toward convergence, and is generally preferred.
Furthermore, we assume that the error function can be written as a function of the network's hypothesis
h(x)
, and the correct answer
y(x)
One simple error function that satisfies thse requirements, and which you probably already know, is the mean squared error (MSE) defined as:
E(x,y) = (y-x)^2
For a single example, and for multiple examples:
E = \frac{1}{m} \sum_{i=1}^{m} (y_i-x_i)^2
We see that not only does it satisfy the averaging constraint, but it also only depends on the hypothesis (noted as
x
), and the actual answer (noted as
y
For notational simplicity, for the rest of the essay, we will omit the function variables, so
E(x,y)
E
Recall that we use backpropagation to find the individual weights' contribution to the error function which is used during gradient descent when updating the weights.
Backpropagation is just figuring out how awful each weight is
In other words, backpropagation attempts to find:
\frac{\partial E}{\partial \theta_{i,j}^{(k)}}
In order to find this, we introduce a new variable
\delta_i^{(k)}
which is the error-sum of neuron
in layer
k
; somtimes called the delta error. The delta error is defined as:
\delta_i^{(k)}=\frac{\partial E}{\partial z_i^{(k)}}
z_i^{(k)}
is the raw output signal of a neuron in the last layer before the activation function has been applied.
During backpropagation, we will find a way of computing the delta error, and translate it to
\frac{\partial E}{\partial \theta_{i,j}^{(k)}}
We will derive
\frac{\partial E}{\partial \theta_{i,j}^{(k)}}
in three steps using the three equations summarized below:
Find a way of computing
\delta^{(L)}
to initialize the process.
\delta^{(k)}
\delta^{(k+1)}
\frac{\partial E}{\partial \theta_{i,j}^{(k)}}
\delta^{(k)}
Recall that we can differentiate composite functions using the chain rule by:
\begin{aligned} f\big(g(x)\big)'&=f'\big(g(x)\big) \cdot g'(x) \Leftrightarrow \\ \frac{df}{dx} &= \frac{\partial f}{\partial g(x)} \cdot \frac{dg}{dx} \end{aligned}
The same principle holds for nested composite functions.
Using the chain rule, we can reformulate
\delta_i^{(L)}
in terms of the partial derivative of the activation function as
a_i^{(k)}=g\big(z_i^{(k)}\big)
\begin{aligned} \delta_i^{(L)} &= \frac{\partial E}{\partial z_i^{(L)}} \Leftrightarrow \\ \delta_i^{(L)} &= \frac{\partial E}{\partial a_i^{(L)}} \cdot \frac{\partial a_i^{(L)}}{\partial z_i^{(L)}} \end{aligned}
\begin{aligned} a_i^{(L)} &= g\big(z_i^{(L)}\big) \Leftrightarrow \\ \frac{\partial a_i^{(L)}}{\partial z_i^{(L)}} &= g'\big(z_i^{(L)}\big) \end{aligned}
We can simplify the above to:
\delta_i^{(L)} = \frac{\partial E}{\partial a_i^{(L)}} \cdot g'\big(z_i^{(L)}\big)
We can vectorize the simplified equation by collecting
\frac{\partial E}{\partial a_i^{(L)}}
in a vector gradient for each layer,
\nabla_a^{(L)}
. Similarily, we can collect the raw output
z_i^{(L)}
into a vector of all the raw outputs in each layer,
z^{(L)}
By doing so, we find the first equation:
\begin{aligned} \delta^{(L)} = \nabla_a^{(L)} \odot g'\big(z^{(L)}\big) &&(1) \end{aligned}
\odot
is the Hadamard product; elementwise multiplication.
While equation
(1)
describes the error in the last layer:
z^{(L)}
a^{(L)}
, equation
(2)
describes the error of a layer in terms of the errors of the layers in front of it. [1]
In order to achieve this, we rewrite
\delta_i^{(k)}=\frac{\partial E}{\partial z_i^{(k)}}
in terms of the next layer,
k+1
\delta_i^{(k+1)}=\frac{\partial E}{\partial z_i^{(k+1)}}
Once again, we use the chain rule.
\begin{aligned} \delta_i^{(k)}&=\frac{\partial E}{\partial z_i^{(k)}}\\ &= \sum_\beta^{n^{(k+1)}} \frac{\partial E}{\partial z_\beta^{(k+1)}} \frac{\partial z_\beta^{(k+1)}}{\partial z_i^{(k)}} \end{aligned}
n^{(k+1)}
is the number of neurons in layer
(k+1)
\frac{\partial E}{\partial z_\beta^{(k+1)}} = \delta_\beta^{(k+1)}
, we can rewrite the above as:
\begin{aligned} \delta_i^{(k)} = \sum_\beta^{n^{(k+1)}} \frac{\partial z_\beta^{(k+1)}}{\partial z_i^{(k)}} \cdot \delta_\beta^{(k+1)} &&(2.1) \end{aligned}
This works because the error from the previous layers is carried over to the neurons in the later layers. This is also why we sum over all the neurons. Equation
(2.1)
can be interpreted as the total error caused by
z_i^{(k)}
z_\beta^{(k+1)}
We know from forward propagation that:
z_\beta^{(k+1)} = \sum_i^{n^{(k)}} \big( \theta_{\beta,i}^{(k+1)} \cdot a_i^{(k)} \big) + b_\beta^{(k+1)}
a_i^{(k)}=g\big(z_i^{(k)}\big)
z_\beta^{(k+1)} = \sum_i^{n^{(k)}} \Big( \theta_{\beta,i}^{(k+1)} \cdot g\big(z_i^{(k)}\big) \Big) + b_\beta^{(k+1)}
z_\beta^{(k+1)}
z_i^{(k)}
\frac{\partial z_\beta^{(k+1)}}{\partial z_i^{(k)}} = \theta_{\beta,i}^{(k+1)} \cdot g'\big(z_i^{(k)}\big)
By substituting this expression in equation
(2.1)
\delta_i^{(k)} = \sum_\beta^{n^{(k+1)}} \theta_{\beta,i}^{(k+1)} g'\big(z_i^{(k)}\big) \delta_\beta^{(k+1)}
If this is not obvious, I do encourage you to spend some time going through the equations in order to convince yourself that this is correct.
Finally, by vectorizing the above, we arrive at the final form for equation
(2)
\begin{aligned} \delta^{(k)} = \big( \theta^{(k+1)^{T}} \delta^{(k+1)} \big) \odot g'\big( z^{(k)} \big) && (2) \end{aligned}
(3)
is derived in the exact same way as equation
(1)
(2)
using the chain rule, so I'll simply state the final form:
\begin{aligned} \frac{\partial E}{\partial \theta_{i,j}^{(k)}} = a_j^{(k-1)} \delta_i^{(k)} && (3) \end{aligned}
(3)
, we see that an individual weight's contribution to the error function is equal to the scaled error it sends forward in the network.
If we think of the error as throwing balls at a target, and if the percentage of balls missing the target is the delta error, and the rate of throwing is the activation, then the total number of balls missing the target, is those multiplied together - which is exactly what we do.
a_j^{(k-1)}
is how much a neuron is stimulated - how strong the output is, or the rate of throwing.
\delta_i^{(k)}
is our throwing accuracy, or rather, in a team of athletes trying to hit the target, how much an individual contributes to the overall number of balls that didn't hit the target.
Finally, we can confirm that this also works for the bias unit where
a_o^{(k-1)}=1
. It should just give us
\delta_0^{(k)}
as there's no activation coefficient:
\frac{\partial E}{\partial \theta^{(k)}_{i,0}} = 1 \cdot \delta_0^{(k)} = \delta_0^{(k)}
Which we see it does.
Using these three equations, we can now describe the algorithm for backpropagation in a feed forward layer.[2]
We use equation
(1)
to calculate the delta error of the last layer.
\begin{aligned} \delta^{(L)} = \nabla_a^{(L)} \odot g'\big(z^{(L)}\big) \end{aligned}
We use the delta error of the last layer to initialize a recursive process of calculating the delta error of all the previous layers using equation
(2)
\begin{aligned} \delta^{(k)} = \big( \theta^{(k+1)^{T}} \delta^{(k+1)} \big) \odot g'\big( z^{(k)} \big) \end{aligned}
We use the delta errors with equation
(3)
to calculate the derivative of the error function with respect to each weight in the neural network which can be used in gradient descent:
\begin{aligned} \frac{\partial E}{\partial \theta_{i,j}^{(k)}} = a_j^{(k-1)} \delta_i^{(k)} \end{aligned}
Finally, equation
(1)
(2)
can be combined into one recursive equation:
\delta^{(k)} = \begin{cases} \begin{aligned} \nabla_a^{(L)} \odot g'\big(z^{(L)}\big) &\quad\text{if } k=L\\ \big( \theta^{(k+1)^{T}} \delta^{(k+1)} \big) \odot g'\big( z^{(k)} \big) &\quad\text{otherwise} \end{aligned} \end{cases}
And that's it. You now know everything there's to know about how backpropagation works.
Don't worry if you don't immediately understand it; that's normal. Put this essay away, and come back after a couple of days to review it, and do a couple of exercices.
Do this a couple of times, and your brain should start to pick it up, and you will be become more comfortable with backpropagation.
The fact that the algorithm moves backwards through the layers of the network is what "back" refers to in backpropagation. ↩︎
It turns out, that the same general principle also applies to backpropagation in other architectures such as convolutional neural networks. ↩︎
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Estimate Multiplicative ARIMA Model - MATLAB & Simulink - MathWorks India
Load Data and Specify Model.
Estimate Model.
Infer Residuals.
This example shows how to estimate a multiplicative seasonal ARIMA model using estimate. The time series is monthly international airline passenger numbers from 1949 to 1960.
Use the first 13 observations as presample data, and the remaining 131 observations for estimation.
y0 = y(1:13);
[EstMdl,EstParamCov] = estimate(Mdl,y(14:end),'Y0',y0)
Variance 0.0013887 0.00015242 9.1115 8.1249e-20
MA: {-0.377161} at lag [1]
SMA: {-0.572379} at lag [12]
Variance: 0.00138874
0 -0.0015 0.0088 0.0000
\Delta {\Delta }_{12}{y}_{t}=\left(1-0.38L\right)\left(1-0.57{L}^{12}\right){\epsilon }_{t},
with innovation variance 0.0014.
Notice that the model constant is not estimated, but remains fixed at zero. There is no corresponding standard error or t statistic for the constant term. The row (and column) in the variance-covariance matrix corresponding to the constant term has all zeros.
Infer the residuals from the fitted model.
res = infer(EstMdl,y(14:end),'Y0',y0);
plot(14:T,res)
When you use the first 13 observations as presample data, residuals are available from time 14 onward.
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Analyze Time-Series Models - MATLAB & Simulink - MathWorks France
This example shows how to analyze time-series models.
A time-series model has no inputs. However, you can use many response computation commands on such models. The software treats (implicitly) the noise source e(t) as a measured input. Thus, step(sys) plots the step response assuming that the step input was applied to the noise channel e(t).
To avoid ambiguity in how the software treats a time-series model, you can transform it explicitly into an input-output model using noise2meas. This command causes the noise input e(t) to be treated as a measured input and transforms the linear time series model with Ny outputs into an input-output model with Ny outputs and Ny inputs. You can use the resulting model with commands, such as, bode, nyquist, and iopzmap to study the characteristics of the H transfer function.
Convert the time-series model to an input-output model.
iosys = noise2meas(sys);
Plot the step response of H.
step(iosys);
Plot the poles and zeros of H.
iopzmap(iosys);
Calculate and plot the time-series spectrum directly without converting to an input-output model.
The command plots the time-series spectrum amplitude
\Phi \left(\omega \right)={‖H\left(\omega \right)‖}^{2}
noise2meas | step
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Chemical Reactions And The Mole Concept, Popular Questions: Kerala Class 10 SCIENCE, Science Part I - Meritnation
A mixture having 2 g of hydrogen and 32g of oxygen occupies how much volume at NTP
differences between dilute and concentrated
how to calculate molecular mass and atomic mass?
Mansi Garg asked a question
a sulphuric acid solution contains 98% h2so4 the density of solution is 1.98g/cm3 calculate the molarity and molality 98soluti
Give answer to first question fast
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it was found that the weight of carbon atom in the signature of a person who put it using carbon pencil was 1.2mg. what is the number of carbon atom contained it ?
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please help me answer this experts.please explain how to get b part of question
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what kind of chemical reaction is used in preparing CCL4 from methane?write the chemical equation?
Common salt obtained from sea water contains 95% NaCl by mass. The approximate number of molecules present in 10.0 g of the salt is
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Please solve it....
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Which one of the following sets of compound correctly illustrates the law of reciprocal propotion
A p2o3,ph3,h2o
B p2o5,ph3,h2o
C n2o5,Nh3,h2o
D n2o,nh3,h2o
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F{e}_{2}{O}_{3 }+3CO \to 2Fe +3C{O}_{2}
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find the number of moles in 2 g in NAOH
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Answer for second questionspleaseee
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Polyphase FFT synthesis filter bank - MATLAB - MathWorks Switzerland
{e}^{j{w}_{k}n},\text{ }{w}_{k}=2\pi k/M,\text{ }k=0,1,...,M-1
{H}_{0}\left(z\right)={b}_{0}+{b}_{1}{z}^{-1}+...+{b}_{N}{z}^{-N}
{H}_{0}\left(z\right)=\begin{array}{c}\left({b}_{0}+{b}_{M}{z}^{-M}+{b}_{2M}{z}^{-2M}+..+{b}_{N-M+1}{z}^{-\left(N-M+1\right)}\right)+\\ {z}^{-1}\left({b}_{1}+{b}_{M+1}{z}^{-M}+{b}_{2M+1}{z}^{-2M}+..+{b}_{N-M+2}{z}^{-\left(N-M+1\right)}\right)+\\ \begin{array}{c}⋮\\ {z}^{-\left(M-1\right)}\left({b}_{M-1}+{b}_{2M-1}{z}^{-M}+{b}_{3M-1}{z}^{-2M}+..+{b}_{N}{z}^{-\left(N-M+1\right)}\right)\end{array}\end{array}
{H}_{0}\left(z\right)={E}_{0}\left({z}^{M}\right)+{z}^{-1}{E}_{1}\left({z}^{M}\right)+...+{z}^{-\left(M-1\right)}{E}_{M-1}\left({z}^{M}\right)
{H}_{k}\left(z\right)={H}_{0}\left(z{e}^{j{w}_{k}}\right)
{H}_{k}\left(z\right)={h}_{0}+{h}_{1}{e}^{jwk}{z}^{-1}+{h}_{2}{e}^{j2wk}{z}^{-2}...+{h}_{N}{e}^{jNwk}{z}^{-N}
{H}_{k}\left(z\right)=\left[1\text{ }{e}^{j{w}_{k}}\text{ }{e}^{j2{w}_{k}}\text{ }...\text{ }{e}^{j\left(M-1\right){w}_{k}}\right]\left[\begin{array}{c}{E}_{0}\left({z}^{M}\right)\\ {z}^{-1}{E}_{1}\left({z}^{M}\right)\\ ⋮\\ {z}^{-\left(M-1\right)}{E}_{M-1}\left({z}^{M}\right)\end{array}\right]
H\left(z\right)=\left[\begin{array}{c}\begin{array}{lllll}1\hfill & 1\hfill & 1\hfill & ...\hfill & 1\hfill \end{array}\\ \begin{array}{lllll}1\hfill & {e}^{j{w}_{1}}\hfill & {e}^{j2{w}_{1}}\hfill & ...\hfill & {e}^{j\left(M-1\right){w}_{1}}\hfill \end{array}\\ ⋮\\ \begin{array}{lllll}1\hfill & {e}^{j{w}_{M-1}}\hfill & {e}^{j2{w}_{M-1}}\hfill & ...\hfill & {e}^{j\left(M-1\right){w}_{M-1}}\hfill \end{array}\end{array}\right]\left[\begin{array}{c}{E}_{0}\left({z}^{M}\right)\\ {z}^{-1}{E}_{1}\left({z}^{M}\right)\\ ⋮\\ {z}^{-\left(M-1\right)}{E}_{M-1}\left({z}^{M}\right)\end{array}\right]
H\left(z\right)=\left[\begin{array}{c}\begin{array}{lllll}1\hfill & 1\hfill & 1\hfill & ...\hfill & 1\hfill \end{array}\\ \begin{array}{lllll}1\hfill & {e}^{j{w}_{1}}\hfill & {e}^{j2{w}_{1}}\hfill & ...\hfill & {e}^{j\left(M-1\right){w}_{1}}\hfill \end{array}\\ ⋮\\ \begin{array}{lllll}1\hfill & {e}^{j{w}_{M-1}}\hfill & {e}^{j2{w}_{M-1}}\hfill & ...\hfill & {e}^{j\left(M-1\right){w}_{M-1}}\hfill \end{array}\end{array}\right]\left[\begin{array}{c}{E}_{0}\left(z\right)\\ {E}_{1}\left(z\right)\\ ⋮\\ {E}_{M-1}\left(z\right)\end{array}\right]
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How to bind function arguments in Python
One of the cool things about functional programming languages such as Haskell is the concept of partially evaluating a function also known as currying.
It serves as a way of encapsulating information hiding details from the caller. This is a powerful way of dealing with abstraction. We will talk more about this later.
Note: I will use currying, binding, and partial application interchangably in this article. There are some differences, but those are outside the scope of this article.
Mathemtically it is like if we have a function
f(a,b,c)
then we can bind
a
by calling
f(a)
giving us a new function
f(b,c)
. Many functional programming languages provide a neat syntax involving just calling the function. For example in Haskell, you can write
which returns a new function that takes two parameters
b,c
However, if we try to do the same in Python, we get an error telling us we didn't specify the remaining arguments.
>>> f = lambda a,b,c: a+b+c
TypeError: f() missing 2 required positional arguments: 'b' and 'c'
There are a couple of different ways we can curry functions in python. The most obvious way is probably to define a new function using lambda.
lambda b,c: f(5,b,c)
However, while this works it can be quite verbose. There must be a better way, and indeed there is. We can use partial from the functools module.
partial(f,5)
We can even bind multiple arguments at the same time, and use keyword arguments.
f = lambda a,b,c: a+b+c
partial(f,5,4)(3) # 12
partial(f,5,c=4)(3) # 12
This is much better, but it still bugs me that the syntax isn't as elegant as Haskell's just calling the function.
Why can't we have f(b=5)?
It turns out we can if we can accept decorating f slightly - which isn't a big deal as we usually want to use it on our own functions, and it's not more difficult wrap other functions than to use functools.partial.
For starters, let us simplify the problem a bit. Haskell doesn't support keyword arguments. You have to bind arguments in order starting with the first one. If we impose the same constraints on our solution, we can do something like this
def partial(f,nargs=None):
""" Does not support kwargs """
total_paramters = wrapper.nargs or f.__code__.co_argcount
given_parameters = len(args)
do_currying = total_paramters > given_parameters
if do_currying:
def c(*newargs):
return f(*(args+newargs))
return partial(c,nargs=total_paramters-len(args))
wrapper.nargs = nargs
The decorator works by checking if the given number of arguments is equal to the total number of arguments expected for the function. Then recursively constructs a new function that expects the remaining number of arguments until the given arguments matches the total arguments for the first function. This works because Python is a dynamic language, and we can access the function arguments at runtime.
We can now use the partial decorator on a function to enable currying for that function.
multiply(1,2) # 2
multiply(1)(2) # 2
def ssum(a,b,c):
return 100*a+10*b+c
ssum(1,2,3) # 123
ssum(1)(2)(3) # 123
ssum(1,2)(3) # 123
This is already quite neat. But it suffers from the same limitations as Haskell, namely, we cannot do f(b=3) on f(a,b,c) and get a f(a,c) function out. The Haskell community has a few convoluted answers for how to get around this limitation, but generally recommends avoiding this pattern.
But I think it's sad to avoid such a flexible and powerful pattern, so how can we extend the solution to include keywords in Python?
After some trial and error, I came up with this solution.
def partial(f, nargs=None):
given_parameters = len(args) + len(kwargs)
# convert args to kwargs.
combined_kwargs = kwargs
# remove keyword arguments from the arglist
# and let the args fill in the 'gaps'.
remaining_var_names = [var for var in f.varnames if not var in kwargs]
# fill in the gaps using the args
# keep a list of unfilled_args for later currying.
unfilled_args = [x for x in remaining_var_names]
key = remaining_var_names[i]
combined_kwargs[key] = value
unfilled_args.remove(key)
def c(**newkwargs):
return f(**{**combined_kwargs, **newkwargs})
c.varnames = unfilled_args
return partial(c, nargs=total_paramters-len(kwargs))
if not hasattr(f,"varnames"):
f.varnames = f.__code__.co_varnames
It's not the most pretty solution, but it works. It works pretty much in the same way as the previous solution with the recursively more and more specific functions, but this time we convert all the arguments to keyword arguments before passing it to the function, and we have some extra plumming to keep track of which argument names we still need to fill not just how many arguments.
Going back to our test functions, this lets us partially bind like this
multiply(1)(2)
multiply(y=1)(2)
multiply(1)(y=2)
multiply(x=1)(y=2)
multiply(x=1)(2)
ssum(1,2,3)
ssum(1)(2)(3)
ssum(1,2)(3)
ssum(1)(c=3)(2)
ssum(c=3)(b=2)(1)
ssum(1)(2)(c=3)
which is pretty neat.
Why is currying useful?
Sure, you might say, it's neat that we can call functions and get a partially applied function back, but how is this useful?
As I alluded to in the beginning, partial application can be a powerful tool handling abstractions. For example let us consider an example of dependency injection.
Suppose we have a web application, we would probably have a function such as get_user(id), but we would have different ways we might get the customer, for example from a database, a cache, or maybe we are working with mock data in memory. We would typically implement this in Python by defining an abstract base class or interface which we inherit from or implement for each of the different storage methods.
With partial application, we have another option where we define a function for each of the different storage methods, and then bind the paramters that are required for the method creating an agnostic get_user(id) function for the rest of the application to use.
def get_user_from_db(dbconnection, id):
def get_user_from_memory(dict, id):
def get_user_from_cache(cache, id):
# specify a dependency agnostic function to use in the rest of the application.
get_user = get_user_from_db(conn)
Generally, partial application is analogous to inheritance from a general class to a more specialized class.
It can also be used in a scenario where you pass a function around different parts of the system configuring the parameters of the function without the systems knowing about each other.
Moreover, it can be used as a way of achieving pseudo* lazy evaluation where the function value is only evaluated once you have gathered all the parameters. *pseudo because the parameters are evaluated eagerly as per the Python interpreter.
Even if you don't adopt functional programming patterns, or use it for the other situational benefits, it can still give increased flexibility when manipulating functions with very little overhead.
And by using the decorator style, the syntax becomes elegant and succinct enough that it is easy to read, and not cumbersome to use.
We examine Google's open source library Tensorflow, and go through its components to understand how it can be used to create scalable machine learning models.
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Element (mathematics) — Wikipedia Republished // WIKI 2
For elements in category theory, see Element (category theory).
Concept of sets and elements in Mathematics
Determinant | Lecture-1 | Elements of Mathematics
INTRODUCTION TO SETS || GRADE 7 MATHEMATICS Q1
1 Sets
2 Notation and terminology
3 Cardinality of sets
5 Formal relation
{\displaystyle A=\{1,2,3,4\}}
{\displaystyle \{1,2\}}
{\displaystyle B=\{1,2,\{3,4\}\}}
{\displaystyle \{3,4\}}
{\displaystyle C=\{\mathrm {\color {red}red} ,\mathrm {\color {green}green} ,\mathrm {\color {blue}blue} \}}
{\displaystyle x\in A}
means that "x is an element of A".[1] Equivalent expressions are "x is a member of A", "x belongs to A", "x is in A" and "x lies in A". The expressions "A includes x" and "A contains x" are also used to mean set membership, although some authors use them to mean instead "x is a subset of A".[2] Logician George Boolos strongly urged that "contains" be used for membership only, and "includes" for the subset relation only.[3]
For the relation ∈ , the converse relation ∈T may be written
{\displaystyle A\ni x,}
{\displaystyle x\notin A}
The symbol ∈ was first used by Giuseppe Peano, in his 1889 work Arithmetices principia, nova methodo exposita.[4] Here he wrote on page X:
Numeric character reference ∈ ∈ ∉ ∉ ∋ ∋ ∌ ∌
Named character reference ∈, ∈, ∈, ∈ ∉, ∉, ∉ ∋, ∋, ∋, ∋ ∌, ∌, ∌
Wolfram Mathematica \[Element] \[NotElement] \[ReverseElement] \[NotReverseElement]
The number of elements in a particular set is a property known as cardinality; informally, this is the size of a set.[5] In the above examples, the cardinality of the set A is 4, while the cardinality of set B and set C are both 3. An infinite set is a set with an infinite number of elements, while a finite set is a set with a finite number of elements. The above examples are examples of finite sets. An example of an infinite set is the set of positive integers {1, 2, 3, 4, ...}.
As a relation, set membership must have a domain and a range. Conventionally the domain is called the universe denoted U. The range is the set of subsets of U called the power set of U and denoted P(U). Thus the relation
{\displaystyle \in }
{\displaystyle \ni }
Identity element
Singleton (mathematics)
^ Eric Schechter (1997). Handbook of Analysis and Its Foundations. Academic Press. ISBN 0-12-622760-8. p. 12
^ George Boolos (February 4, 1992). 24.243 Classical Set Theory (lecture) (Speech). Massachusetts Institute of Technology.
^ a b Kennedy, H. C. (July 1973). "What Russell learned from Peano". Notre Dame Journal of Formal Logic. Duke University Press. 14 (3): 367–372. doi:10.1305/ndjfl/1093891001. MR 0319684.
^ "Sets - Elements | Brilliant Math & Science Wiki". brilliant.org. Retrieved 2020-08-10.
Halmos, Paul R. (1974) [1960], Naive Set Theory, Undergraduate Texts in Mathematics (Hardcover ed.), NY: Springer-Verlag, ISBN 0-387-90092-6 - "Naive" means that it is not fully axiomatized, not that it is silly or easy (Halmos's treatment is neither).
Jech, Thomas (2002), "Set Theory", Stanford Encyclopedia of Philosophy, Metaphysics Research Lab, Stanford University
Suppes, Patrick (1972) [1960], Axiomatic Set Theory, NY: Dover Publications, Inc., ISBN 0-486-61630-4 - Both the notion of set (a collection of members), membership or element-hood, the axiom of extension, the axiom of separation, and the union axiom (Suppes calls it the sum axiom) are needed for a more thorough understanding of "set element".
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Curl (mathematics) - Wikipedia
For other uses, see Rotation operator (disambiguation).
1.1 Intuitive interpretation
3.3 Descriptive examples
5.2 Curl geometrically
The components of F at position r, normal and tangent to a closed curve C in a plane, enclosing a planar vector area
{\displaystyle \mathbf {A} =A\mathbf {\hat {n}} }
The thumb points in the direction of
{\displaystyle \mathbf {\hat {n}} }
and the fingers curl along the orientation of C
The curl of a vector field F, denoted by curl F, or ∇ × F, or rot F, at a point is defined in terms of its projection onto various lines through the point. If
{\displaystyle \mathbf {\hat {n}} }
is any unit vector, the projection of the curl of F onto
{\displaystyle \mathbf {\hat {n}} }
is defined to be the limiting value of a closed line integral in a plane orthogonal to
{\displaystyle \mathbf {\hat {n}} }
divided by the area enclosed, as the path of integration is contracted around the point.
{\displaystyle (\nabla \times \mathbf {F} )(p)\cdot \mathbf {\hat {n}} \ {\overset {\underset {\mathrm {def} }{}}{{}={}}}\lim _{A\to 0}{\frac {1}{|A|}}\oint _{C}\mathbf {F} \cdot \mathrm {d} \mathbf {r} }
where the line integral is calculated along the boundary C of the area A in question, |A| being the magnitude of the area. This equation defines the projection of the curl of F onto
{\displaystyle \mathbf {\hat {n}} }
. The infinitesimal surfaces bounded by C have
{\displaystyle \mathbf {\hat {n}} }
as their normal. C is oriented via the right-hand rule.
{\displaystyle {\begin{aligned}&(\operatorname {curl} \mathbf {F} )_{1}={\frac {1}{h_{2}h_{3}}}\left({\frac {\partial (h_{3}F_{3})}{\partial u_{2}}}-{\frac {\partial (h_{2}F_{2})}{\partial u_{3}}}\right),\\[5pt]&(\operatorname {curl} \mathbf {F} )_{2}={\frac {1}{h_{3}h_{1}}}\left({\frac {\partial (h_{1}F_{1})}{\partial u_{3}}}-{\frac {\partial (h_{3}F_{3})}{\partial u_{1}}}\right),\\[5pt]&(\operatorname {curl} \mathbf {F} )_{3}={\frac {1}{h_{1}h_{2}}}\left({\frac {\partial (h_{2}F_{2})}{\partial u_{1}}}-{\frac {\partial (h_{1}F_{1})}{\partial u_{2}}}\right).\end{aligned}}}
{\displaystyle h_{i}={\sqrt {\left({\frac {\partial x_{1}}{\partial u_{i}}}\right)^{2}+\left({\frac {\partial x_{2}}{\partial u_{i}}}\right)^{2}+\left({\frac {\partial x_{3}}{\partial u_{i}}}\right)^{2}}}}
Intuitive interpretation[edit]
{\displaystyle \nabla \times \mathbf {F} ={\begin{vmatrix}{\boldsymbol {\hat {\imath }}}&{\boldsymbol {\hat {\jmath }}}&{\boldsymbol {\hat {k}}}\\[5pt]{\dfrac {\partial }{\partial x}}&{\dfrac {\partial }{\partial y}}&{\dfrac {\partial }{\partial z}}\\[10pt]F_{x}&F_{y}&F_{z}\end{vmatrix}}}
{\displaystyle \nabla \times \mathbf {F} =\left({\frac {\partial F_{z}}{\partial y}}-{\frac {\partial F_{y}}{\partial z}}\right){\boldsymbol {\hat {\imath }}}+\left({\frac {\partial F_{x}}{\partial z}}-{\frac {\partial F_{z}}{\partial x}}\right){\boldsymbol {\hat {\jmath }}}+\left({\frac {\partial F_{y}}{\partial x}}-{\frac {\partial F_{x}}{\partial y}}\right){\boldsymbol {\hat {k}}}={\begin{bmatrix}{\frac {\partial F_{z}}{\partial y}}-{\frac {\partial F_{y}}{\partial z}}\\{\frac {\partial F_{x}}{\partial z}}-{\frac {\partial F_{z}}{\partial x}}\\{\frac {\partial F_{y}}{\partial x}}-{\frac {\partial F_{x}}{\partial y}}\end{bmatrix}}}
{\displaystyle (\nabla \times \mathbf {F} )^{k}={\frac {1}{\sqrt {g}}}\varepsilon ^{k\ell m}\nabla _{\ell }F_{m}}
where ε denotes the Levi-Civita tensor, ∇ the covariant derivative,
{\displaystyle {\sqrt {g}}}
is the Jacobian and the Einstein summation convention implies that repeated indices are summed over. Due to the symmetry of the Christoffel symbols participating in the covariant derivative, this expression reduces to the partial derivative:
{\displaystyle (\nabla \times \mathbf {F} )={\frac {1}{\sqrt {g}}}\mathbf {R} _{k}\varepsilon ^{k\ell m}\partial _{l}F_{m}}
{\displaystyle \nabla \times \mathbf {F} =\left(\star {\big (}{\mathrm {d} }\mathbf {F} ^{\flat }{\big )}\right)^{\sharp }}
{\displaystyle \mathbf {F} (x,y,z)=y{\boldsymbol {\hat {\imath }}}-x{\boldsymbol {\hat {\jmath }}}}
{\displaystyle F_{x}=y,F_{y}=-x,F_{z}=0.}
{\displaystyle \nabla \times \mathbf {F} =0{\boldsymbol {\hat {\imath }}}+0{\boldsymbol {\hat {\jmath }}}+\left({\frac {\partial }{\partial x}}(-x)-{\frac {\partial }{\partial y}}y\right){\boldsymbol {\hat {k}}}=-2{\boldsymbol {\hat {k}}}}
{\displaystyle \mathbf {F} (x,y,z)=-x^{2}{\boldsymbol {\hat {\jmath }}}}
{\displaystyle {\nabla }\times \mathbf {F} =0{\boldsymbol {\hat {\imath }}}+0{\boldsymbol {\hat {\jmath }}}+{\frac {\partial }{\partial x}}\left(-x^{2}\right){\boldsymbol {\hat {k}}}=-2x{\boldsymbol {\hat {k}}}.}
Descriptive examples[edit]
{\displaystyle \nabla \times \left(\mathbf {v\times F} \right)={\Big (}\left(\mathbf {\nabla \cdot F} \right)+\mathbf {F\cdot \nabla } {\Big )}\mathbf {v} -{\Big (}\left(\mathbf {\nabla \cdot v} \right)+\mathbf {v\cdot \nabla } {\Big )}\mathbf {F} \ .}
{\displaystyle \mathbf {v\ \times } \left(\mathbf {\nabla \times F} \right)=\nabla _{\mathbf {F} }\left(\mathbf {v\cdot F} \right)-\left(\mathbf {v\cdot \nabla } \right)\mathbf {F} \ ,}
{\displaystyle \nabla \times \left(\mathbf {\nabla \times F} \right)=\mathbf {\nabla } (\mathbf {\nabla \cdot F} )-\nabla ^{2}\mathbf {F} \ ,}
{\displaystyle \nabla \times (\nabla \varphi )={\boldsymbol {0}}}
{\displaystyle \nabla \times (\varphi \mathbf {F} )=\nabla \varphi \times \mathbf {F} +\varphi \nabla \times \mathbf {F} }
The vector calculus operations of grad, curl, and div are most easily generalized in the context of differential forms, which involves a number of steps. In short, they correspond to the derivatives of 0-forms, 1-forms, and 2-forms, respectively. The geometric interpretation of curl as rotation corresponds to identifying bivectors (2-vectors) in 3 dimensions with the special orthogonal Lie algebra
{\displaystyle {\mathfrak {so}}}
(3) of infinitesimal rotations (in coordinates, skew-symmetric 3 × 3 matrices), while representing rotations by vectors corresponds to identifying 1-vectors (equivalently, 2-vectors) and
{\displaystyle {\mathfrak {so}}}
(3), these all being 3-dimensional spaces.
Differential forms[edit]
{\displaystyle a_{1}\,dx+a_{2}\,dy+a_{3}\,dz;}
{\displaystyle a_{12}\,dx\wedge dy+a_{13}\,dx\wedge dz+a_{23}\,dy\wedge dz;}
{\displaystyle a_{123}\,dx\wedge dy\wedge dz.}
{\displaystyle \omega ^{(k)}=\sum _{\scriptstyle {i_{1}<i_{2}<\cdots <i_{k}} \atop \forall \scriptstyle {i_{\nu }\in 1,\ldots ,n}}a_{i_{1},\ldots ,i_{k}}\,dx_{i_{1}}\wedge \cdots \wedge dx_{i_{k}},}
{\displaystyle d\omega ^{(k)}=\sum _{\scriptstyle {j=1} \atop \scriptstyle {i_{1}<\cdots <i_{k}}}^{n}{\frac {\partial a_{i_{1},\ldots ,i_{k}}}{\partial x_{j}}}\,dx_{j}\wedge dx_{i_{1}}\wedge \cdots \wedge dx_{i_{k}}.}
{\displaystyle {\frac {\partial ^{2}}{\partial x\,\partial y}}={\frac {\partial ^{2}}{\partial y\,\partial x}},}
{\displaystyle 0\,{\overset {d}{\longrightarrow }}\;\Omega ^{0}\left(\mathbb {R} ^{3}\right)\,{\overset {d}{\longrightarrow }}\;\Omega ^{1}\left(\mathbb {R} ^{3}\right)\,{\overset {d}{\longrightarrow }}\;\Omega ^{2}\left(\mathbb {R} ^{3}\right)\,{\overset {d}{\longrightarrow }}\;\Omega ^{3}\left(\mathbb {R} ^{3}\right)\,{\overset {d}{\longrightarrow }}\,0.}
{\displaystyle \nabla \times (\nabla f)=0}
{\displaystyle \nabla \cdot (\nabla \times \mathbf {v} )=0}
{\displaystyle \omega ^{(2)}=\sum _{i<k=1,2,3,4}a_{i,k}\,dx_{i}\wedge dx_{k},}
Curl geometrically[edit]
2-vectors correspond to the exterior power Λ2V; in the presence of an inner product, in coordinates these are the skew-symmetric matrices, which are geometrically considered as the special orthogonal Lie algebra
{\displaystyle {\mathfrak {so}}}
(V) of infinitesimal rotations. This has (n
2) = 1/2n(n − 1) dimensions, and allows one to interpret the differential of a 1-vector field as its infinitesimal rotations. Only in 3 dimensions (or trivially in 0 dimensions) does n = 1/2n(n − 1), which is the most elegant and common case. In 2 dimensions the curl of a vector field is not a vector field but a function, as 2-dimensional rotations are given by an angle (a scalar – an orientation is required to choose whether one counts clockwise or counterclockwise rotations as positive); this is not the div, but is rather perpendicular to it. In 3 dimensions the curl of a vector field is a vector field as is familiar (in 1 and 0 dimensions the curl of a vector field is 0, because there are no non-trivial 2-vectors), while in 4 dimensions the curl of a vector field is, geometrically, at each point an element of the 6-dimensional Lie algebra
{\displaystyle {\mathfrak {so}}(4)}
Retrieved from "https://en.wikipedia.org/w/index.php?title=Curl_(mathematics)&oldid=1084306117"
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Glycerophosphocholine phosphodiesterase - Wikipedia
In enzymology, a glycerophosphocholine phosphodiesterase (EC 3.1.4.2) is an enzyme that catalyzes the chemical reaction
sn-glycero-3-phosphocholine + H2O
{\displaystyle \rightleftharpoons }
choline + sn-glycerol 3-phosphate
Thus, the two substrates of this enzyme are sn-glycero-3-phosphocholine and H2O, whereas its two products are choline and sn-glycerol 3-phosphate.
This enzyme belongs to the family of hydrolases, specifically those acting on phosphoric diester bonds. The systematic name of this enzyme class is sn-glycero-3-phosphocholine glycerophosphohydrolase. Other names in common use include glycerophosphinicocholine diesterase, glycerylphosphorylcholinediesterase, sn-glycero-3-phosphorylcholine diesterase, glycerolphosphorylcholine phosphodiesterase, and glycerophosphohydrolase. This enzyme participates in glycerophospholipid metabolism.
Dawson RMC (1956). "Liver glycerylphosphorylcholine diesterase". Biochem. J. 62: 689–693.
Hayaishi O, Kornberg A (1954). "Metabolism of phospholipides by bacterial enzymes". J. Biol. Chem. 206 (2): 647–63. PMID 13143024.
Webster GR, Marples EA, Thompson RH (1957). "Glycerylphosphorylcholine diesterase activity in nervous tissue". Biochem. J. 65 (2): 374–377. PMC 1199879. PMID 13403918.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Glycerophosphocholine_phosphodiesterase&oldid=917405103"
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GeneralizedPetersenGraph - Maple Help
Home : Support : Online Help : Mathematics : Discrete Mathematics : Graph Theory : GraphTheory Package : SpecialGraphs : GeneralizedPetersenGraph
construct generalized Petersen graph
GeneralizedPetersenGraph(n, k)
The GeneralizedPetersenGraph(n,k) command returns the generalized Petersen graph with the given parameters.
If n and k are relatively prime, the graph consists of two cycles of length n with a perfect matching between their vertices. The i'th vertex of the first cycle is connected to the ki'th vertex on the second cycle.
If n and k are not relatively prime, the graph consists of one cycle of length n perfectly matched to another set of n vertices forming gcd(k,n) cycles of length n/gcd(k,n).
\mathrm{with}\left(\mathrm{GraphTheory}\right):
\mathrm{with}\left(\mathrm{SpecialGraphs}\right):
P≔\mathrm{GeneralizedPetersenGraph}\left(5,2\right)
\textcolor[rgb]{0,0,1}{P}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{Graph 1: an undirected unweighted graph with 10 vertices and 15 edge\left(s\right)}}
\mathrm{DrawGraph}\left(P\right)
P≔\mathrm{GeneralizedPetersenGraph}\left(6,2\right)
\textcolor[rgb]{0,0,1}{P}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{Graph 2: an undirected unweighted graph with 12 vertices and 18 edge\left(s\right)}}
\mathrm{DrawGraph}\left(P\right)
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Machine Learning fundamentals | Machine Learning from Scratch (Part 0) | Curiousily - Hacker's Guide to Machine Learning
28.06.2019 — Machine Learning, Fundamentals, Training, Data Science — 4 min read
TL;DR Learn about the basics of Machine Learning - what types of learning exist, why implement algorithms from scratch and can you really trust your models?
Sick of being a lamer?
Here’s a definition for “lamer” from urban dictionary:
Lamer is someone who thinks they are smart, yet are really a loser. E.g. that hacker wannabe is really just a lamer.
You might’ve noticed that reading through Deep Learning and TensorFlow/PyTorch tutorials might give you an idea of how to do a specific task, but fall short if you want to apply it to your own problems. Importing a library and calling 4-5 methods might get the job done but leave you clueless about why it works. Can you solve this problem?
Different people learn in different styles, but all hackers learn the same way - we build stuff!
How these series help?
Provide a clear path to learning (reduce choice overload and paralysis by analysis) increasingly complex Machine Learning models
Succint implementations of Machine Learning algorithms solving real-world problems you can thinker with
Just enough theory + math that helps you understand why things work after you understand what problem you have to solve
Machine Learning (ML) is the art and science of teaching machines to do complex tasks without explicitly programming them. Your job, as a hacker, is to:
Define the problem in a way that a computer can understand it
Choose a set of possible models that could solve it
Evaluate the performance and improve
There are 3 main types of learning: supervised, unsupervised and reinforcement
In Supervised Learning setting, you have a dataset, which is a collection of
N
labeled examples. Each example has a vector of features
x_i
and a label
y_i
. The label
y_i
can belong to a finite set of classes
\{1, \ldots, C\}
, real number or something more complex.
The goal of supervised learning algorithms is to build a model that receives a feature vector
x
as input and infer the correct label for it.
In Unsupervised Learning setting, you have a dataset of
N
unlabeled examples. Again, you have a feature vector
x
, and the goal is to build a model that takes it and transforms it into another. Some practical examples include clustering, reducing the number of dimensions and anomaly detection.
Reinforcement Learning is concerned with building agents that interact with an environment by getting its state and executing an action. Actions provide rewards and change the state of the environment. The goal is to learn a set of actions that maximize the total reward.
What Learning Algorithms are made of?
Each learning algorithm we’re going to have a look at consists of three parts
loss function - measure of how wrong your model currently is
optimization criteria based on the loss function
optimization routine that uses data to find “good” solutions according to the optimization criteria
These are the main components of all the algorithms we’re going to implement. While there are many optimization routines, Stochastic Gradient Descent is the most used in practice. It is used to find optimal parameters for logistic regression, neural networks, and many other models.
How can you guarantee that your model will make correct predictions when deployed in production? Well, only suckers think that this is possible.
“All models are wrong, but some are useful.” - George Box
That said, there are ways to increase the prediction accuracy of your models. If the data used for training were selected randomly, independently of one another and following the same procedure for generating it, then, it is more likely your model to learn better. Still, for situations that are less likely to happen, your model will probably make errors.
Generally, the larger the data set, the better the predictions you can expect.
We’re going to use a lot of libraries provided by different kind people, but the main ones are NumPy, Pandas and Matplotlib.
Here’s a super simple walkthrough:
3a = np.array([1, 2, 3]) # 1D array
4type(a)
1numpy.ndarray
1(3,)
We’ve created 1 dimensional array with 3 elements. You can get the first element of the array:
1a[0] = 5
1array([5, 2, 3])
Let’s create a 2D array
1b = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9]]) # 2D array
3b.shape
We have 3x3 matrix. Let’s select only the first 2 rows:
1b[0:2]
1array([[1, 2, 3],
2 [4, 5, 6]])
We’re going to use pandas as a holder for our datasets. The primary entity we’ll be working with is the DataFrame. We’ll also do some transformations/preprocessing and visualizations.
Let’s start by creating a DataFrame:
3df = pd.DataFrame(dict(
4 customers=["Jill", "Jane", "Hanna"],
5 payments=[120, 180, 90]
You can check the size of the data frame:
1df.shape
We have 3 rows and 2 columns. You can use head() to render a preview of the first five rows:
Let’s check for missing values:
1df.isnull()
You can apply functions such as [sum(https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.sum.html)] to columns like payments:
1df.payments.sum()
You can even show some charts using plot():
1df.plot(
2 kind='bar',
3 x='customers',
4 y='payments',
5 title='Payments done by customer'
Here is a quick sample of it:
3x = np.arange(0, 3 * np.pi, 0.1)
4y = np.sin(x)
6plt.plot(x, y);
Welcome to the amazing world of Machine Learning. Let’s get this party started!
|
Oscillations Physics NEET Practice Questions, MCQs, Past Year Questions (PYQs), NCERT Questions, Question Bank, Class 11 and Class 12 Questions, and PDF solved with answers
The variation of acceleration of a particle executing SHM with displacement x is:
Subtopic: Simple Harmonic Motion |
A particle is subjected to two simple harmonic motions in the same direction having equal amplitudes and equal frequency. If the resulting amplitude is equal to the amplitude of individual motions, the phase difference between them is:
\frac{\mathrm{\pi }}{3}
\frac{2\mathrm{\pi }}{3}
\frac{\mathrm{\pi }}{6}
\frac{\mathrm{\pi }}{2}
Subtopic: Linear SHM |
The motion of a particle varies with time according to the relation
\mathrm{y} =\mathrm{a} \mathrm{sin} \mathrm{\omega t}+\mathrm{a} \mathrm{cos} \mathrm{\omega t}
1. The motion is oscillatory but not SHM
2. The motion is SHM with amplitude
a\sqrt{2}
\sqrt{2}
4. The motion is SHM with amplitude a
If a particle is executing SHM, with an amplitude A, the distance moved and the displacement of the body in a time equal to its period are
1. 2A, A
2. 4A, 0
4. 0, 2A
The equation of the displacement of two particles making SHM is represented by y1 = a sin (ωt + φ) and y2 = a cos (ωt) respectively. The phase difference of the velocities of the two particles is
(1) π/2 + φ
(2) -φ
(4) φ - π/2
The displacement of a particle executing SHM is given by y = 0.25 (sin 200t) cm. The maximum speed of the particles is:
1. 200 cm/sec
3. 50 cm/sec
4. 0.25 cm/sec
Which of the following figure represents damped harmonic motion?
3. i, ii, iii, and iv
Subtopic: Damped Oscillations |
A particle is executing SHM with amplitude A and time period T. If at t = 0, it is at origin (mean position) then the time instant when it covers a distance equal to 2.5A is:
\frac{\mathrm{T}}{12}
\frac{5\mathrm{T}}{12}
\frac{7\mathrm{T}}{12}
\frac{2\mathrm{T}}{3}
The time period of a spring mass system at the surface of earth is 2 second. What will be the time period of this system at moon where acceleration due to gravity is
\frac{1}{6}\mathrm{th}
of the value of earth's surface?
\frac{1}{\sqrt{6}} \mathrm{seconds}
2\sqrt{6}\mathrm{seconds}
2 \mathrm{seconds}
12 \mathrm{seconds}
Subtopic: Spring mass system |
A particle undergoes SHM with a time period of 2 seconds. In how much time will it travel from its mean position to a displacement equal to half of its amplitude?
\frac{1}{2}s
\frac{1}{6}s
\frac{1}{4}s
\frac{1}{3}s
|
Random Thoughts From April Part 2 - Dennis
On April 21st, Johns Hopkins University President Ronald Daniels sent an email and published online a document detailing the financial implications stemming from the COVID-19 pandemic. There’s a lot to digest here. It’s an intimate introduction to serious topics. The role of university endowments, spending allocations at large organizations, voices of reason in crisis, and layoffs.
First, we will address the easy target. Why doesn’t the University use endowment money to cover costs? Surely, this is a once in a lifetime crisis that permits a one time breach in endowment contracts?
We should be mindful that endowments are made up primarily of gifts of good will. In 2019, Michael Bloomberg famously donated $1.8 billion to the university (a 44% increase), but the funds were explicitly reserved for student financial aid. From the JHU Krieger School website: At a donor’s direction, gifts can either be spent immediately or endowed. Endowed gifts are held in perpetuity and invested by the university with an emphasis on long-term growth. The Board of Trustees meets annually to determine what percentage of endowment income may be spent, generally between 4 and 5%. Every department has the responsibility to spend the funds in accordance with donor intent.
The short answer is that it’s tricky to mobilize endowment funds. A divestment from coal and oil request took 3 tries and 3 years to get done when the try was successful, for example. A great white paper was published here. TLDR: endowments seem invincible, but for most schools, they truly do need to be rationed, not only for financial stability, but because of moral obligation to past donors, who intended for their principal to be reserved for long term growth and use by the university.
Still, for a system like JHU that has activated itself in response to the pandemic it’s a shame that the money isn’t able to be used.
Next, let’s dive into the financial release document.
First, some general details and framing. The JHU ecosystem runs on a $6.5 billion annual budget, of which 1-2% is surplus in a normal year. For fiscal year 2020, JHU had a projected $72 million surplus, roughly 1.1% midway through the fiscal year (The JHU fiscal year runs from July 1st to June 30th, the next year). JHU is now projecting a net loss of
100 million for FY 2020, and a net loss of \$375 million for FY 2021 instead of an expected margin of \$80 million. In a given year, 2/3 of this budget (\$4.3 billion) is comprised of salary, wages, and benefits. 4% of this budget (\$260 million) comes from a \$6.3 billion endowment, which checks out, since only between 4-5% of the endowment can be spent each year. For this reason, the decline in stock market prices does not affect JHU as much as it's peers. For comparison, Harvard has a \$5 billion annual budget, and a \$40 billion endowment. If we assume they also spend 4% each year, that is \
1.6 billion, roughly 32% of Harvard’s annual budget.
JHU makes up for this via donations, tuition, clinical revenues, and sponsored research. Johns Hopkins doesn’t publish donation data (somewhat difficult to project, given that more than half of the endowment is from Michael Bloomberg ($3.35 billion)), but current projections are a $25 million loss in FY 2020 and a $60 million loss in FY 2021. $25 million in FY 2020 and $150 million in FY 2021 will be lost to tuition related income. To the university’s credit, $12 million was returned to students for housing, dining, and student service charges, and an additional $5 million was extended for unanticipated financial aid packages. Still, Summer courses on campus including the worldwide CTY programs have been virtualized, resulting in $40 million in losses, and restrictions for travel prevents international students from studying at professional schools, another unnamed but large source of revenue.
Johns Hopkins spends $1.5 billion on research (excluding the Applied Physics Lab) and takes in $370 million in sponsored grants. The costs are largely fixed, yet the grants function as recoverable income. Stay at home orders do not reduce fixed costs, but do prevent recovery of costs from the use of grant funds. In other words, it takes longer to do the same amount of research, which loses money because of the fixed costs and also because researchers are less productive and receive less funding when research does not get done.
Physician clinical revenue losses will reach $100 million in FY 2020 and $200 million in FY 2021 due to decreases in elective procedures and refocusing of efforts towards fighting the COVID-19 pandemic.
So to take stock, we have losses of $475 million over 2 years. How does Daniels want to approach addressing this?
Since 2/3 of the annual budget goes to salary, wages, and benefits, it might be prudent to start there. The first thing to go was retirement contributions for 1 year, saving $100 million
Salary reductions for university leaders and holds for faculty and staff was next. ~$20 million.
Restrictions on hiring ~40 million.
Suspensions on capital projects have put 78 projects worth $29 million on hold.
Expense restrictions such as travel will save an additional $10 million.
Finally, there will be furloughs and layoffs, though he doesn’t give any numbers.
Including everything except furloughs and layoffs, we reach almost $200 million. We still have a $275 million problem. Assuming economic dampening continues, we shouldn’t expect FY 2022 or FY 2023 surpluses to be back to normal levels.
It seems impossible to sustain previous levels of spending while ensuring the overall growth of the university. Administrators will need to be leaner, and more effective. We will all need to be cognizant of whether our activities are useful and productive, and worth the sacrifice that employees have made during this time. For school clubs, now is not the time to be milking money from the university. For students, we should have some empathy. This is a tough problem with no easy answers.
|
Risk efficiency of hedging strategy in China financial market - Guo 2015a - Scipedia
China financial market
H. Guo, Risk efficiency of hedging strategy in China financial market, Perspectives in Science (2016). Vol. 7 URL https://www.scipedia.com/public/Guo_2015a
D81 ; G1 ; G32
Hedging strategy ; Risk measure ; Black–Scholes model ; China financial market
This paper is going to describe the general invest environment of China financial market. Consider risk management has become progressively more important for corporations. It will use the risk measure analysing a call option to demonstrate how active trading in China financial market. Furthermore, in particular calculate the cases of lower partial moments, linear and risk-taking that considers the attitude towards risk, the result presents the efficient hedge for call option in Black–Scholes model. The objective of this paper is risk measure analysis which for Value at Risk (VaR) and distinguish between VaR and expected shortfall (ES) in China financial market and analyse partial hedge of estimate the expected shortfall risk in the Black–Scholes model. Risk measure and hedging which are two concepts investor used and considered every daily life. But how much should be hedged 0–50–100% (no hedge–partial hedge–full hedge)? Depends on exposure and risk attitude towards. Risk attitude is a complex area, it can measure in variety ways, though not perfect and entirely accurate, but it is progressing. The paper is interested in the approach of partial hedge of estimate risk measures ES risk in the Black–Scholes model based on different risk attitude towards.
Analysis of China financial market
As the fastest growing and one of the largest economies of the world, China has witnessed a magical transformation from a stagnant command economy to a dynamic market economy. In this section is presents the overview of China financial market and risk measure analysis in China financial market.
Overview of China financial market
The Chinese stock market has promoted the reform of state-owned enterprises and the change of their systems, and enabled a stable transition between the two systems. On the strength of the stock market in the past decade, many large state-owned enterprises have realized system change. The change also has stimulated medium and small-sized state-owned enterprises to adopt the shareholding system, thus solving the most important issue – the system problem – during the transition from planned to a market economy. As for ordinary citizens, bank deposit is not the only way to put their money. The stock market has become one of the most important channels for investment. Following Table 1 presents the Chinese stock market statistics in 2014.
Table 1. Chinese stock market statistics in 2014.
Total funds raised (100 million yuan)
Turnover of trading (100 million yuan)
Quarter-end volume of stock issued (100 million shares)
Quarter-end market capitalization (100 million yuan)
Quarter-end numbers of companies listed
Quarter-end close index
Shanghai stock exchange composite index (December 19, 1990 = 100)
Shenzhen stock exchange component index (July 20, 1994 = 1000)
Source : The Peoples Bank, 2014 and The Peoples Bank, 2015 .
Methods of stock trading are constantly being improved. Today, a network system for securities exchange and account settlement has been formed, with the Shanghai and Shenzhen exchanges as the powerhouse, radiating to all parts of the country (see china.org.cn).
China has three stock exchanges, two of them (Shanghai stock exchange, Shenzhen stock exchange) are located in mainland of China, one is located in Hong Kong. Also there are four futures exchanges in mainland of China. Zhengzhou Commodity Exchange (ZCE) which established in 1993. Dalian Commodity Exchange (DCE) which established in February 1993. Shanghai Futures Exchange (SHFE) which established in 1999. China Financial Futures Exchange (CFFEX) which established in Shanghai in September 2006. There are two futures exchanges in Hong Kong which are Hong Kong Exchanges and Clearing (HKEx) and Hong Kong Mercantile Exchange (HKMEx). Following Table 2 presents the Chinese futures market statistics in 2014.
Table 2. Chinese futures market statistics in 2014.
Transaction volume (10 thousand lots)
Quarter-end position (10 thousand lots)
Outlook of China options market
China launched simulated trading in stock index option on 8th November 2013, as regulators move to enhance risk hedging options to support further financial reforms. The first stock options launched on the Shanghai Stock Exchange on 2014, offering investors a new hedging tool for trading index heavyweights, which regulators long have hoped to boost. The options are based on the exchange-trade fund (ETF) that tracks the SSE50 index, composed of the 50 most heavily weighted stocks on the bourse. Regulators are essentially guiding investors into blue chips, which most retail investors have avoided in favour of smaller firms, whose valuations have been pushed up.
Risk measure analysis in China financial market
General there are two basic but often use measures of risk that are VaR and ES. To comparison them, a measure that can produce better incentives for traders than VaR is ES. This is also sometimes referred to as conditional VaR, conditional tail expectation, or expected tail loss.
ES has better properties than VaR in that it encourages diversification. In spite of its weaknesses, VaR has become the most popular measure of risk among both regulators and risk managers. Because the disadvantage of ES are it does not have the simplicity of VaR and as a result is more difficult to understand, and it is more difficult to back-test a procedure for calculating ES than it is to back-test a procedure for VaR.
Example of risk measure comparison between VaR and ES in financial asset (options) in the China financial market (HKEx). Following Table 3 illustrates the normal distribution VaR, Historical VaR and ES for the PING AN (2318.HK) call option.
Table 3. Normal distribution value at risk, historical value at risk and expected shortfall.
Normal distribution value at risk
Value at Risk (%)
29 90% confidence −0.02556
Bottom 3rd return
Bottom 2nd return
Bottom 2.9th return
0.016227568 (10% Value at Risk)
1.45th return 95% confidence −0.03204
Bottom 1st return
Bottom 1.45th return
−0.033619719 (5%Value at Risk)
According to the results, it is calculate the normal distribution VaR and historical VaR, and compare the results of historical VaR and ES under confidence of 90% (10% VaR) and 95% (5% VaR). The result of 5% VaR is −0.019347118, 95% confidence ES is −0.03204. The result of 10% VaR is −0.026138061, 90% confidence ES is −0.02556.
Methodology of hedging strategy under the Black–Scholes model
The optimal hedging
{\textstyle {\tilde {\phi }}\in {\mathfrak {R}}}
{\textstyle min_{\phi \in {\mathfrak {R}}}E[l((1-\phi )H)]}
{\textstyle sup_{p{_{\ast }}\in {\mathfrak {R}}}E{_{\ast }}[\phi H]\leq {\tilde {V}}_{0}}
, l is strictly convex, on
{\textstyle \lbrace H>0\rbrace }
. General considering the case of lower partial moments based on the optimal hedging which the loss function
{\textstyle l(x)=(x^{p}/P)}
{\textstyle P>1}
The optimal hedge consists in hedging the modified claim
{\displaystyle \phi _{p}H=H-c_{p}(\rho \ast )^{1/p-1}\wedge H}
where the constant cpis determined
{\textstyle E{_{\ast }}[\phi _{p}H]={\tilde {V}}_{0}}
In the Black–Scholes model with constant volatility
{\textstyle \sigma >0}
the underlying discounted price process is given by a geometric Brownian motion
{\textstyle dX_{t}=X_{t}(\sigma dW_{t}+mdt)}
{\textstyle E(X_{t})=X_{0}\cdot e^{[(m/\sigma )-((1/2)\cdot (m/\sigma ))^{2}\cdot (T-t)]}}
{\textstyle X_{0}=x_{0}}
, where W is a Wiener process under P and m is a constant. If assume that
{\textstyle m>0}
. The unique equivalent martingale measure P * is given by
{\displaystyle {\frac {dP^{_{\ast }}}{dP}}=\rho ^{_{\ast }}=exp\left(-{\frac {m}{\sigma }}W_{T}-{\frac {1}{2}}\left({\frac {m}{\sigma }}\right)^{2}T\right)=const{\mbox{ }}X_{T}^{-\alpha }}
where it set
{\textstyle \alpha =(m/\sigma ^{2})}
The process W * defined by
{\textstyle W_{t}^{_{\ast }}=W_{t}+(m/\sigma )t}
, is a Brownian motion under P *.
A European call payoff function
{\textstyle H_{T}=(X_{T}-K)^{+}}
can be hedged perfectly if it provides the initial capital
{\displaystyle H_{0}=E^{_{\ast }}[H]=x_{0}\Phi (d_{+})-K\Phi (d_{-})}
{\textstyle d_{\pm }(x_{0},K)=((ln{\mbox{ }}x_{0}-ln{\mbox{ }}K)/\sigma {\sqrt {T}})\pm (1/2)\sigma {\sqrt {T}}}
{\textstyle \Phi }
denotes the distribution function of the standard normal distribution.
Suppose it use only an initial capital
{\textstyle {\tilde {V}}_{0}}
which is smaller than the Black–Scholes price H0 . Under this constraint to minimize the shortfall risk
{\textstyle E[l((H-V_{T})^{+})]}
where l is given loss function satisfying the assumptions. Lower partial moments and the linear case are modified claims and strategies determined and compare to the Black–Scholes perfect hedging strategy, see Föllmer and Leukert (2000) .
Lower partial moments modified claims and strategy for
{\textstyle p\rightarrow \infty }
{\displaystyle {\begin{array}{l}\displaystyle \epsilon _{p}(t,x)={\frac {\partial }{\partial x}}F_{p}(t\\\displaystyle x)=\Phi \left({\frac {ln{\mbox{ }}x-ln{\mbox{ }}L}{\sigma {\sqrt {\tau }}}}+{\frac {1}{2}}\sigma {\sqrt {\tau }}\right)+{\frac {\alpha }{p-1}}{\frac {L^{\alpha /(p-1)}(L-K)}{x^{(}\alpha /(p-1))+1}}\times exp\left[{\frac {1}{2}}\sigma ^{2}\tau {\frac {\alpha }{p-1}}\left({\frac {\alpha }{p-1}}+\right.\right.\\\left.\left.\displaystyle +1\right)\right]\times \Phi \left({\frac {ln{\mbox{ }}x-ln{\mbox{ }}L}{\sigma {\sqrt {\tau }}}}-\sigma {\sqrt {\tau }}\left({\frac {\alpha }{p-1}}+{\frac {1}{2}}\right)\right)\end{array}}}
{\textstyle \tau =T-t}
{\textstyle \alpha =(m/\sigma ^{2})}
Hence the function Fp by following equation
{\displaystyle {\begin{array}{l}\displaystyle F_{p}(t,x)=x\Phi (d_{+}(x,L))-K\Phi (d_{-}(x\\\displaystyle L))-{\frac {L^{\alpha /(p-1)}(L-K)}{x^{\alpha }/(p-1)}}\times exp\left[{\frac {1}{2}}\sigma ^{2}\tau {\frac {\alpha }{p-1}}\left({\frac {\alpha }{p-1}}+1\right)\right]\Phi \left(d_{-}(x\right.\\\left.\displaystyle L)-{\frac {\alpha \sigma {\sqrt {\tau }}}{p-1}}\right)\end{array}}}
The constant L is modified claim of option prices which determined by the equation
{\textstyle {\tilde {V}}_{0}=F_{p}(0,x_{0})}
. Let determine
{\textstyle {\tilde {V}}_{0}}
equal 1, which means when
{\textstyle t=0}
, calculate by the following equation
{\displaystyle {\begin{array}{l}\displaystyle F_{p}(0\\\displaystyle x)=x\Phi \left({\frac {ln{\mbox{ }}x-ln{\mbox{ }}L}{\sigma {\sqrt {T}}}}+{\frac {1}{2}}\sigma {\sqrt {T}}\right)-K\Phi \left({\frac {ln{\mbox{ }}x-ln{\mbox{ }}L}{\sigma {\sqrt {T}}}}-{\frac {1}{2}}\sigma {\sqrt {T}}\right)-\\\displaystyle -{\frac {L^{\alpha /(p-1)}(L-K)}{x^{\alpha }/(p-1)}}\times exp\left[{\frac {1}{2}}\sigma ^{2}T{\frac {\alpha }{p-1}}\left({\frac {\alpha }{p-1}}+1\right)\right]\Phi \left({\frac {ln{\mbox{ }}x-ln{\mbox{ }}L}{\sigma {\sqrt {T}}}}-{\frac {1}{2}}\sigma {\sqrt {T}}\right)-\\\displaystyle -\left({\frac {\alpha \sigma {\sqrt {T}}}{p-1}}\right)={\tilde {V}}_{0}=1\end{array}}}
Resulting of L input calculate Eq. (2.5) for call option with strike corresponding to the Black–Scholes, see Föllmer and Leukert (2000) .
Application of efficient hedging strategy
Example of the financial asset (call option) for modified claims and strategies determined the Black–Scholes efficient hedging strategy in China financial market (HKEx). Following Table 4 presents the parameters of call option TENCENT (0700.HK) under the Black–Scholes model.
Table 4. Parameters call option under the Black–Scholes model.
4.4 120 0.15 0.1 0.664257626
Hence, V0 denotes initial capital, H0 denotes the Black–Scholes price, K is strike price, X is underlying asset price. Following Fig. 1 presents the Black–Scholes call option efficient hedging strategy.
The Black–Scholes call option efficient hedging strategy.
It was illustrated the approach of partial hedge of estimate the ES risk in the Black–Scholes model, the efficient hedges for call option in the geometric Brownian motion with known volatility and for the loss function based on different risk attitude towards. Because risk profiling is not an exact science, risk attitude is a complex area, but it can measure in variety ways, even each of ways measurement has their own advantages and disadvantages. Based on Fig. 1 to summarize different attitudes towards shortfall risk are reflected in different shapes of the modified claims and of the resulting hedging strategies by following special case when
{\textstyle p\rightarrow \infty }
means risk free; when
{\textstyle p=1}
means risk neutral; when
{\textstyle p=0}
means quantile hedging; when
{\textstyle p=0.85}
means risk seeking; when
{\textstyle p=1.1}
means risk aversion; when
{\textstyle p=2}
the efficient frontier means balancing cost and shortfall risk.
The paper describes the outlook and risk analysis (VaR and ES) in China financial market. Resulting of compare between VaR and ES that risk measure can produce better incentives for traders than VaR is ES. The objective of this paper is method of efficient hedging for call option in the Black–Scholes model based on geometric Brownian motion. The illustration determined of application is the Black–Scholes efficient hedging strategy in a call option data which from China financial market. The resulting of efficient hedges allow the investor to interpolate in a systematic way extremes of partial hedging which depend on the accepted level of shortfall risk attitude towards.
The research was supported by the SGS project of VSB-TU Ostrava under no. SP2015/15. This paper has been also elaborated in the framework of the Operational Programme Education for Competitiveness – Project No. CZ.1.07/2.3.00/20.0296.
Föllmer and Leukert, 2000 H. Föllmer, P. Leukert; Efficient hedging: cost versus shortfall risk; Financ. Stochast., 4 (2000), pp. 117–146
The Peoples Bank, 2014 The Peoples Bank of China Annual Report 2014.
The Peoples Bank, 2015 The Peoples Bank of China. http://www.pbc.gov.cn/ .
Hedging strategy • Risk measure • Black–Scholes model • China financial market
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MRChart - Maple Help
Home : Support : Online Help : Statistics and Data Analysis : ProcessControl Package : MRChart
MRChart(X, options, plotoptions)
(optional) equation(s) of the form option=value where option is one of color, confidencelevel, controllimits, ignore, or rbar; specify options for generating the MR chart
The MRChart command generates control chart for the moving range (MR chart) for the specified observations. The chart also contains the upper control limit (UCL), the lower control limit (LCL), and the average of the moving ranges of two observations (represented by the center line) of the underlying quality characteristic. Unless explicitly given, the control limits are computed based on the data.
color=list -- This option specifies colors of the various components of the MR chart. The value of this option must be a list containing the color of the control limits, center line, data to be plotted, and the specification limits.
ignore=truefalse -- This option controls how missing values are handled by the MRChart command. Missing values are represented by undefined or Float(undefined). So, if ignore=false and X contains missing data, the MRChart command returns undefined. If ignore=true, all missing items in X are ignored. The default value is true.
\mathrm{with}\left(\mathrm{ProcessControl}\right):
\mathrm{infolevel}[\mathrm{ProcessControl}]≔1:
A≔[33.75,33.05,34.00,33.81,33.46,34.02,33.68,33.27,33.49,33.20,33.62,33.00,33.54,33.12,33.84]:
\mathrm{MRChart}\left(A\right)
\mathrm{MRControlLimits}\left(A\right)
[\textcolor[rgb]{0,0,1}{0.}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1.57127089652357}]
l≔\mathrm{MRControlLimits}\left(A,\mathrm{confidencelevel}=0.95\right)
\textcolor[rgb]{0,0,1}{l}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{0.}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1.57127089652357}]
\mathrm{MRChart}\left(A,\mathrm{controllimits}=l\right)
ProcessControl[MRControlLimits]
ProcessControl[XChart]
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EUDML | Improved -convexity inequalities. EuDML | Improved -convexity inequalities.
Improved
GA
-convexity inequalities.
Satnoianu, Razvan A.
Satnoianu, Razvan A.. "Improved -convexity inequalities.." JIPAM. Journal of Inequalities in Pure & Applied Mathematics [electronic only] 3.5 (2002): Paper No. 82, 6 p., electronic only-Paper No. 82, 6 p., electronic only. <http://eudml.org/doc/123001>.
@article{Satnoianu2002,
author = {Satnoianu, Razvan A.},
keywords = {convex function; -convex function; symmetric function; -convex function},
title = {Improved -convexity inequalities.},
AU - Satnoianu, Razvan A.
TI - Improved -convexity inequalities.
KW - convex function; -convex function; symmetric function; -convex function
convex function,
GA
-convex function, symmetric function,
GA
-convex function
Articles by Satnoianu
|
Refit neighborhood component analysis (NCA) model for regression - MATLAB - MathWorks Italia
LBFGS or Mini-Batch LBFGS Options
SGD or Mini-Batch LBFGS Options
SGD or LBFGS or Mini-Batch LBFGS Options
mdlrefit
Refit NCA Model for Regression with Modified Settings
Class: FeatureSelectionNCARegression
Refit neighborhood component analysis (NCA) model for regression
mdlrefit = refit(mdl,Name,Value)
mdlrefit = refit(mdl,Name,Value) refits the model mdl, with modified parameters specified by one or more Name,Value pair arguments.
Neighborhood component analysis model or classification, specified as a FeatureSelectionNCARegression object.
mdl.FitMethod (default) | 'exact' | 'none' | 'average'
Method for fitting the model, specified as the comma-separated pair consisting of 'FitMethod' and one of the following.
'average' — The function divides the data into partitions (subsets), fits each partition using the exact method, and returns the average of the feature weights. You can specify the number of partitions using the NumPartitions name-value pair argument.
mdl.Lambda (default) | non-negative scalar value
Regularization parameter, specified as the comma-separated pair consisting of 'Lambda' and a non-negative scalar value.
For n observations, the best Lambda value that minimizes the generalization error of the NCA model is expected to be a multiple of 1/n
mdl.Solver (default) | 'lbfgs' | 'sgd' | 'minibatch-lbfgs'
Solver type for estimating feature weights, specified as the comma-separated pair consisting of 'Solver' and one of the following.
'lbfgs' — Limited memory BFGS (Broyden-Fletcher-Goldfarb-Shanno) algorithm (LBFGS algorithm)
'sgd' — Stochastic gradient descent
mdl.InitialFeatureWeights (default) | p-by-1 vector of real positive scalar values
Initial feature weights, specified as the comma-separated pair consisting of 'InitialFeatureWeights' and a p-by-1 vector of real positive scalar values.
Verbose — Indicator for verbosity level
mdl.Verbose (default) | 0 | 1 | >1
Indicator for verbosity level for the convergence summary display, specified as the comma-separated pair consisting of 'Verbose' and one of the following.
1 — Convergence summary including iteration number, norm of the gradient, and objective function value.
>1 — More convergence information depending on the fitting algorithm
When using solver 'minibatch-lbfgs' and verbosity level >1, the convergence information includes iteration log from intermediate mini-batch LBFGS fits.
mdl.GradientTolerance (default) | positive real scalar value
Relative convergence tolerance on the gradient norm for solver lbfgs, specified as the comma-separated pair consisting of 'GradientTolerance' and a positive real scalar value.
Example: 'GradientTolerance',0.00001
InitialLearningRate — Initial learning rate for solver sgd
mdl.InitialLearningRate (default) | positive real scalar value
Initial learning rate for solver sgd, specified as the comma-separated pair consisting of 'InitialLearningRate' and a positive scalar value.
mdl.PassLimit (default) | positive integer value
Maximum number of passes for solver 'sgd' (stochastic gradient descent), specified as the comma-separated pair consisting of 'PassLimit' and a positive integer. Every pass processes size(mdl.X,1) observations.
mdl.IterationLimit (default) | positive integer value
mdlrefit — Neighborhood component analysis model for regression
Neighborhood component analysis model or classification, returned as a FeatureSelectionNCARegression object. You can either save the results as a new model or update the existing model as mdl = refit(mdl,Name,Value).
load('robotarm.mat')
The robotarm (pumadyn32nm) dataset is created using a robot arm simulator with 7168 training and 1024 test observations with 32 features [1], [2]. This is a preprocessed version of the original data set. Data are preprocessed by subtracting off a linear regression fit followed by normalization of all features to unit variance.
Compute the generalization error without feature selection.
nca = fsrnca(Xtrain,ytrain,'FitMethod','none','Standardize',1);
Now, refit the model and compute the prediction loss with feature selection, with
\lambda
= 0 (no regularization term) and compare to the previous loss value, to determine feature selection seems necessary for this problem. For the settings that you do not change, refit uses the settings of the initial model nca. For example, it uses the feature weights found in nca as the initial feature weights.
nca2 = refit(nca,'FitMethod','exact','Lambda',0);
L2 = loss(nca2,Xtest,ytest)
The decrease in the loss suggests that feature selection is necessary.
plot(nca2.FeatureWeights,'ro')
Tuning the regularization parameter usually improves the results. Suppose that, after tuning
\lambda
using cross-validation as in Tune Regularization Parameter in NCA for Regression, the best
\lambda
value found is 0.0035. Refit the nca model using this
\lambda
value and stochastic gradient descent as the solver. Compute the prediction loss.
nca3 = refit(nca2,'FitMethod','exact','Lambda',0.0035,...
'Solver','sgd');
After tuning the regularization parameter, the loss decreased even more and the software identified four of the features as relevant.
[1] Rasmussen, C. E., R. M. Neal, G. E. Hinton, D. van Camp, M. Revow, Z. Ghahramani, R. Kustra, and R. Tibshirani. The DELVE Manual, 1996, https://mlg.eng.cam.ac.uk/pub/pdf/RasNeaHinetal96.pdf
[2] https://www.cs.toronto.edu/~delve/data/datasets.html
loss | fsrnca | predict | FeatureSelectionNCARegression
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Example problems of a sphere and frustum of a cone — lesson. Mathematics State Board, Class 10.
1. The diameter of an orange is \(7 \ cm\). Find the surface area of \(50\) oranges.
\frac{d}{2}=\frac{7}{2}=3.5\phantom{\rule{0.147em}{0ex}}\mathit{cm}
Surface area of an orange \(=\) Surface area of a sphere
\(=\) \(4 \pi r^2\) sq. units
Surface area of an orange \(=\) \(154\) \(cm^2\).
Surface area of \(50\) oranges:
\(=\) \(154 \times 50\)
Therefore, the surface area of \(50\) oranges is \(7700 \ cm^2\).
2. The radii of the frustum of a cone are \(6 \ cm\) and \(2 \ cm\), and the height of the cone is \(5 \ cm\). Find the total surface area of a frustum of a cone.
Let \(R = 6 \ cm\), \(r = 2 \ cm\) and \(h = 5 \ cm\).
Slant height, \(l = \sqrt{h^2 + (R - r)^2}\)
\(l = \sqrt{5^2 + (6 - 2)^2}\)
\(l = \sqrt{25 + 16}\)
\(l = \sqrt{41}\)
\(l = 6.4 \ cm\) (approximately)
Total surface area of frustum of a cone \(=\) \(\pi l(R + r)\) \(+\) \(\pi R^2 + \pi r^2\) sq. units
T. S. A. \(=\) \(\frac{22}{7} \times 6.4 (6 + 2)\) \(+\) \(\frac{22}{7} (6^2 + 2^2)\)
\(=\) \(\frac{22}{7} \times 6.4 \times 8\) \(+\) \(\frac{22}{7} (36 + 4)\)
\(=\) \(\frac{22}{7} \times 51.2\) \(+\) \(\frac{22}{7} \times 40\)
\(=\) \(\frac{22}{7} \times (51.2 + 40)\)
\(=\) \(\frac{22}{7} \times 91.2\)
\(=\) \(286.63\) (approximately)
Therefore, the total surface area of the frustum of a cone is \(286.63 \ cm^2\).
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\frac{\mathrm{nR} }{\left(\mathrm{\gamma }-1\right)}\left({\mathrm{T}}_{2}-{\mathrm{T}}_{1}\right)
2H2O2(l)
\to
2H2O(l) + O2(g)
One mole of a non-ideal gas undergoes a change of state (2.0 atm, 3.0 L, 95 K)
\to
(4.0 atm, 5.0 L, 245 K) with a change in internal energy,
∆
U = 30.0 L atm. The change in enthalpy (
∆
H) of the process in L atm is:
The work done by a mass less piston in causing an expansion
∆
V (at constant temperature), when the opposing pressure, P is variable, is given by:
1. W=
-\int \mathrm{P}∆\mathrm{V}
3. W= -P
∆
∆{\mathrm{S}}_{\mathit{\left(p}\mathit{\right)}}= \frac{∆{\mathrm{Hp}}_{}}{\mathrm{T}}
∆{\mathrm{S}}_{\mathit{\left(}v\mathit{\right)}}=\frac{∆{\mathrm{U}}_{v}}{\mathrm{T}}
∆{\mathrm{S}}_{\mathit{\left(}v\mathit{\right)}}= \frac{∆{\mathrm{H}}_{v}}{\mathrm{P}}
2H2 (g) + O2 (g)
\to
2H2O(l)
If S
°
for H2, Cl2 and HCl are 0.13, 0.22 and 0.19 kJ K-1mol-1 respectively. The total change in standard entropy for the reaction, H2 + Cl2
\to
2HCl is:
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Research:Modeling monthly active editors - Meta
This page in a nutshell: This page contains background analysis for the definition of editor activity metrics in support of the Vital Signs project
Monthly active editor model (concept). A conceptual tree view of the month active editors model is presented.
1.1 Editor Classes
2.3 Comparison with legacy definition
Editor ClassesEdit
MAE proportions (enwiki). Proportions of active editor classes in Monthly Active Editors is plotted with an equation showing how they add up. These values are based on averages between May 2013 and May 2014 in English Wikipedia.
Monthly Active Editors (MAE)
Editors who save at least 5 revisions within one month.
New Active Editors (NAE)
Newly registered users who save at least 5 revisions in the month that they registered.
Surviving New Active Editors (SNAE)
New Active Editors from the previous month who continued to make at least 5 edits in the current month.
Recurring Old Active Editors (ROAE)
Non-new Active Editors from the previous month who continued to make at least 5 edits in the current month.
Reactivated Editors (RAE)
All other active editors who (1) were not active in the previous month and (2) were not a Newly registered user in the current month.
{\displaystyle {\text{MAE}}_{m}={\text{NAE}}_{m}+{\text{SNAE}}_{m}+{\text{ROAE}}_{m}+{\text{RAE}}_{m}}
New Editor Activation Rate (NEAR)
The proportion of Newly registered users who save at least 5 revisions in the current month
New Active Survival Rate (NASR)
The proportion of New Active Editors the previous month who save at least 5 revisions in the current month
Old Active Survival Rate (OASR)
The proportion of Active Editors from the previous month (who were not Newly registered users) who save at least 5 revisions in the current month
Expanded equation with rates
{\displaystyle {\text{NAE}}_{m}={\text{NRU}}_{m}\times {\text{NEAR}}_{m}}
{\displaystyle {\text{SNAE}}_{m}={\text{NAE}}_{m-1}\times {\text{NASR}}_{m}}
{\displaystyle {\text{ROAE}}_{m}=({\text{SNAE}}_{m-1}+{\text{ROAE}}_{m-1}+{\text{RAE}}_{m-1})\times {\text{OASR}}_{m}}
{\displaystyle {\text{MAE}}_{m}={\text{NRU}}_{m}\times {\text{NEAR}}_{m}+{\text{NAE}}_{m-1}\times {\text{NASR}}_{m}+({\text{SNAE}}_{m-1}+{\text{ROAE}}_{m-1}+{\text{RAE}}_{m-1})\times {\text{OASR}}_{m}+{\text{RAE}}}
MAE over time. The count of monthly active editors is plotted (stacked) for the four active editor classes for the English Wikipedia. A loess curve is fit to the trend.
MAE rates over time. Activation and retention rates are plotted for editor class thresholds for the English Wikipedia.
Italian WikipediaEdit
MAE over time. The count of monthly active editors is plotted (stacked) for the four active editor classes for the Italian Wikipedia. A loess curve is fit to the trend.
MAE rates over time. Activation and retention rates are plotted for editor class thresholds for the Italian Wikipedia.
Comparison with legacy definitionEdit
In order to explore the implications of the relatively simplistic definition of active editor used in this model (5 revisions to any page), we compare counts of active editors with those generated by the historical definition (5 revisions to countable pages).
dump ns=0 -- Legacy definition based on XML dump processing within wikistats. Does not include edits to deleted pages and also filters by countable pages
archive ns=0 -- All edits to pages in the article namespace (ns=0) count.
archive ns=all -- All edits to any page count. This is the definition used in the analysis above.
MAE comparison (enwiki).
MAE comparison (itwiki).
MAE comparison (dump ns=0).
MAE comparison (archive ns=0).
MAE comparison (archive ns=all).
Project repo: https://github.com/halfak/Monthly-Active-Editors-Model
Retrieved from "https://meta.wikimedia.org/w/index.php?title=Research:Modeling_monthly_active_editors&oldid=10507237"
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Question numbers 20 to 22 are long-answer type questions and carry 5 marks each. Each question has an internal choice.
Useful Constant and Relation:
1u = 931MeV
(i) A point charge ‘q’ is kept at each of the vertices of an equilateral triangle having each side ‘a’. Total electrostatic potential energy of the system is:
\left(\frac{1}{4\pi {\epsilon }_{0}}\right)\frac{3{q}^{2}}{{a}^{2}}
\left(\frac{1}{4\pi {\epsilon }_{0}}\right)\frac{3q}{a}
\left(\frac{1}{4\pi {\epsilon }_{0}}\right)\frac{3{q}^{2}}{{a}^{}}
\left(\frac{1}{4\pi {\epsilon }_{0}}\right)\frac{3q}{{a}^{2}}
(ii) Curie temperature is the temperature above which:
(a) a ferromagnetic substance behaves like a paramagnetic substance.
(b) a paramagnetic substance behaves like a diamagnetic substance.
(c) a ferromagnetic substance behaves like a diamagnetic substance.
(d) a paramagnetic substance behaves like a ferromagnetic substance.
(iii) In an astronomical telescope of refracting type:
(a) Objective should have small focal length.
(b) Objective should have large focal length.
(c) Eyepiece should have large focal length.
(d) Both objective and eyepiece should have large focal length.
(iv) In photoelectric effect experiment, the slope of the graph of the stopping potential versus frequency gives the value of:
\frac{h}{e}
\frac{e}{h}
\frac{hc}{e}
(v) In a nuclear reactor, cadmium rods are used as:
(a) Control rods
(b) Fuel rods
(d) Moderator
(b) Answer the following questions briefly and to the point:
(i) State Gauss’ theorem.
(ii) A metallic wire having a resistance of 20Ω is bent in order to form a complete circle. Calculate the resistance between any two diametrically opposite points on the circle.
(iii) How can a moving coil galvanometer be converted into a voltmeter?
(iv) Write Biot-Savart’s law in vector form.
(v) What is the phase difference between any two points lying on the same wavefront?
(vi) Name the physical principle on the basis of which optical fibres work.
(vii) What is Pair production? VIEW SOLUTION
A uniform copper wire having a cross sectional area of 1mm2 carries a current of 5A. Calculate the drift speed of free electrons in it.
(Free electron number density of copper = 2 × 1028/m3)
An electric bulb is rated as 250V, 750W. Calculate the:
(i) Electric current flowing through it, when it is operated on a 250V supply.
(ii) Resistance of its filament. VIEW SOLUTION
Write an expression for force per unit length between two long current carrying wires, kept parallel to each other, in vacuum and hence define an ampere, the SI unit of current. VIEW SOLUTION
(i) Define angle of dip.
(ii) State the relation between magnetic susceptibility
\left(\chi \right)
and relative permeability (µr) of a magnetic substance. VIEW SOLUTION
(a) Figure 1 below shows a metallic rod MN of length l = 80 cm, kept in a uniform magnetic field of flux density B = 0·5 T, on two parallel metallic rails P and Q.
Calculate the emf that will be induced between its two ends, when it is moved towards right with a constant velocity v of 36 km/hr.
(b) When current flowing through one coil changes from 0 Amp to 15 Amp in 0·2 s, an emf of 750V is induced in an adjacent coil.
Calculate the coefficient of mutual inductance of the two coils.
(i) State any one use of infrared radiations.
(ii) State any one source of ultraviolet radiations. VIEW SOLUTION
Where will you keep an object in front of a:
(i) Convex lens in order to get a virtual and magnified image?
(ii) Concave mirror to get a real and diminished image? VIEW SOLUTION
Draw a labelled graph of angle of deviation (𝛿) versus angle of incidence (i) for a prism. VIEW SOLUTION
(ii) What conclusion can be drawn from Davisson and Germer’s experiment? VIEW SOLUTION
Calculate binding energy of oxygen nucleus
\left({}_{8}{}^{16}\mathrm{O}\right)
from the data given below:
Mass of a proton = 1·007825u
Mass of a neutron = 1·008665u
\left({}_{8}{}^{16}\mathrm{O}\right)
= 15·994915u VIEW SOLUTION
For a radioactive substance, write the relation between:
(i) Half life (T) and disintegration constant (λ).
(ii) Mean life (τ) and disintegration constant (λ). VIEW SOLUTION
With reference to communication systems, what is meant by:
(i) modulation?
(ii) demodulation? VIEW SOLUTION
Show that intensity of electric field E at a point in broadside on position is given by:
\mathrm{E}=\left(\frac{1}{4\pi {\in }_{0}}\right)\frac{p}{{\left({r}^{2}+{l}^{2}\right)}^{3/2}}
, where the terms have their usual meaning. VIEW SOLUTION
A parallel plate capacitor is charged by a battery, which is then disconnected. A dielectric slab having dielectric constant (relative permittivity) K, is now introduced between its two plates in order to occupy the space completely. State, in terms of K, its effect on the following:
(i) The capacitance of the capacitor.
(ii) The potential difference between its plates.
(iii) The energy stored in the capacitor. VIEW SOLUTION
(a) E1 and E2 are two batteries having emfs of 3V and 4V and internal resistances of 2Ω and 1Ω respectively. They are connected as shown in Figure 2 below.
Using Kirchhoff’s Laws of electrical circuits, calculate the currents I1 and I2.
(b) A potentiometer circuit is shown in Figure 3 below. AB is a uniform metallic wire having length of 2m and resistance of 8Ω.
The batteries E1 and E2 have emfs of 4V and 1·5V and their internal resistances are 1Ω and 2Ω respectively.
(i) When the jockey J does not touch the wire AB, calculate:
(a) the current flowing through the potentiometer wire AB.
(b) the potential gradient across the wire AB.
(ii) Now the jockey J is made to touch the wire AB at a point C such that the galvanometer (G) shows no deflection. Calculate the length AC.
For two thin lenses kept in contact with each other, show that:
\frac{1}{F}=\frac{1}{{f}_{1}}+\frac{1}{{f}_{2}}
where the terms have their usual meaning. VIEW SOLUTION
(a) A compound microscope consists of two convex lenses having focal length of 1·5cm and 5cm.
When an object is kept at a distance of 1·6cm from the objective, the final image is virtual and lies at a distance of 25cm from the eyepiece.
Calculate magnifying power of the compound microscope in this set-up.
(b) In Young’s double slit experiment, the screen is kept at a distance of 1·2m from the plane of the slits.
The two slits are separated by 5mm and illuminated with monochromatic light having wavelength 600nm.
(i) Fringe width i.e. fringe separation of the interference pattern.
(ii) Distance of 10th bright fringe from the centre of the pattern.
Draw the energy level diagram of hydrogen atom and show the transitions responsible for:
(i) absorption lines of Lyman series.
(ii) emission lines of Balmer series. VIEW SOLUTION
(i) State any one difference between energy band diagram of conductors and that of insulators.
(ii) Give a relation between 𝛼 and 𝛽 for a transistor.
(Derivation is not required.)
(iii) What is the advantage of an LED bulb over the filament electric bulb? VIEW SOLUTION
(i) A 400Ω resistor, a 3H inductor and a 5μF capacitor are connected in series to a 220V, 50Hz ac source. Calculate the:
(1) Impedance of the circuit.
(2) Current flowing through the circuit.
(ii) Draw a labelled graph showing the variation of impedance (Z) of a series LCR circuit versus frequency (f) of the ac supply.
(i) When an alternating emf e = 310 sin (100𝜋t)V is applied to a series LCR circuit, current flowing through it is i = 5 sin(100𝜋t + 𝜋/3)A.
(1) What is the phase difference between the current and the emf?
(2) Calculate the average power consumed by the circuit.
(ii) Obtain an expression for the resonant frequency (fo) of a series LCR circuit. VIEW SOLUTION
(i) Derive an expression for refraction at a single (convex) spherical surface, i.e. a relation between u, v, R, n1 (rarer medium) and n2 (denser medium), where the terms have their usual meaning.
(ii) Name the phenomenon due to which the sun appears reddish at sunset.
(i) Draw a labelled graph of intensity of diffracted light (I) versus angle (𝜃) in the Fraunhofer diffraction experiment for a single slit diffraction.
(ii) State the law of Malus.
(iii) How will you distinguish experimentally between ordinary light and plane polarized light? VIEW SOLUTION
(i) In a semiconductor diode, what is meant by potential barrier?
(ii) Draw a labelled circuit diagram of a Zener diode as a voltage regulator.
(iii) Show with the help of a diagram, how you will obtain an AND gate using only NAND gates. (Truth table is not required.)
(i) Draw a labelled circuit diagram of a transistor acting as a common emitter amplifier. What is meant by phase reversal?
(ii) Draw the symbol of a NAND gate and write its truth table. VIEW SOLUTION
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Currency Exchange Markets - Course Hero
Introduction to Finance/International Currency Markets/Currency Exchange Markets
Development of Currency Exchange Markets
International currencies are exchanged on Forex, or foreign exchange currency markets, and are backed by foreign exchange reserves.
Currency exchange markets began in the Middle Ages, when traveling traders used letters of credit as backing for transactions. At the time, the credit standard was gold. While there was no rate of exchange between currencies, the strength of a creditor added value to their transactions. For example, the backing of a king held more monetary power than that of a wealthy merchant. For centuries this led to volatility in currency markets. It was not until the Bretton Woods agreement was established in 1944 that the gold standard, which established gold as the underlying asset for all currency, helped the international market shift away from this volatile currency market. Under the agreement, all currencies could be pegged to gold in value and the U.S. dollar was viewed as a reserve currency. However, an overvaluation of the U.S. dollar began to raise concerns about the connection between exchange rates and the price of gold. As a result, U.S. President Richard Nixon called for a temporary suspension of the dollar’s convertibility. This provided countries the freedom to choose any exchange agreement, except the price of gold. In 1973, foreign governments let currencies float, which put an end to the Bretton Woods system in which countries could then back their currency with either gold or U.S. dollars. Thus, the Bretton Woods Agreement was dissolved and, respectively, in 1971 the United States stopped trading currency for gold, ending the gold standard within the U.S. and creating the foreign exchange market. The foreign exchange market (Forex) is a currency marketplace that is expressly used for trading currencies.
Similar to stock markets, in a currency exchange market, users can exchange, buy, and sell currency and can use derivatives to speculate on currency. Unlike stock markets, foreign exchange markets conduct their transactions electronically. The electronic network allows currency transactions to take place almost instantaneously. The currency exchange rate is generally set by the central bank of each nation. The central bank is the financial institution in a given nation tasked with the regulation of financial systems and the creation of money. The currency exchange rate is the relative ratio of the value of a foreign currency to that of a domestic currency. Similar to other trading markets, currency exchange may be packaged so that one fund will trade several different currencies. Doing this mitigates the currency exchange rate risk, the chance that the exchange rate between two currencies will change before a transaction is finalized. At the country level, central banks hold assets to cover the amount of currency that the country holds. This is called a foreign exchange reserve.
Approximately two-thirds of all currency transactions occur between large dealer banks. Because of the impact that foreign exchange markets have on the economy, the inability of one country to trade currency could severely affect the world economy. This is why currency exchange markets are decentralized and offer liquidity in trade. This helps to mitigate the risk of a rapid decline in currency exchange markets.
Largest Currency Exchange Markets
The relative value of money is set by the largest currency exchange markets.
To mitigate the risk of default, foreign exchange transactions are concentrated in three major Forex (or foreign exchange) markets expressly used for trading currencies. Over half of the world's currency exchanges take place in London, New York, or Tokyo. These locations were chosen so that the Forex system can run continuously, 24 hours a day. The London market trades between 7:00 a.m. and 4:00 p.m. GMT, the New York market trades between 12:00 p.m. and 8:00 p.m. GMT, and the Tokyo market trades between 11:00 p.m. and 8:00 a.m. GMT. Other markets fill in during the hours unaccounted for by the Forex markets' schedule.
Hours of the Forex Markets
A.M. (GMT)
P.M. (GMT)
With three Forex markets, nearly the entire 24 hours in a day are covered for global currency trading.
Forex markets are generally driven by the strength of the macroeconomics of the countries, as well as the custom of trading in that market. New York is one of the largest Forex markets because the United States has traditionally boasted a strong economy. Because New York has been a leading market for so long, it has become customary to trade through the New York market. Singapore is changing this tradition, as its healthy economy makes it a sought-after Forex market.
London is the largest of the currency exchange markets; approximately 30 percent of worldwide transactions happen through this market. The New York exchange comes in second with approximately 18 percent of transactions. The Tokyo exchange has approximately 9 percent of worldwide transactions. The Tokyo exchange has declined, with the Singapore exchange taking up the volume. The rise and fall of other Forex markets and the overlap between trading times is an important aspect for traders, as these overlaps are where volatility occurs. The London exchange sees its greatest volatility just before 4:00 p.m., before it closes.
It is common to pair and trade through the largest currency markets. Typical trading pairs are the euro and the U.S. dollar, the U.S. dollar and the Japanese yen, and the U.S. dollar and the British pound. About 23 percent of worldwide currency trades are exchanges between the U.S. dollar and the euro. It is with these trading pairs that most Forex transactions occur.
Currency valuation can be calculated using direct and indirect quotation methods.
Currency exchange rates are stated in two basic ways: direct and indirect quotation methods. The direct quotation method demonstrates the value of one unit of foreign currency compared to the home country's currency. This gives a unit rate for the purchase of one foreign unit using the home currency.
\text{Amount of Home Currency}=\text{Direct Quote}\;\times\;\text{Amount of Foreign Currency}
For example, assume that the home currency is the U.S. dollar. If the Australian dollar is directly quoted at 0.77, this means that one dollar of U.S. currency will purchase approximately $0.77 in Australian currency. The indirect quotation method works in reverse. It calculates the amount of home currency needed to purchase one unit of foreign currency.
\text{Amount of Foreign Currency}=\frac{\text{Amount of Home Currency}}{\text{Indirect Quote}}
If the Australian dollar is indirectly quoted at 1.30, this means 1.30 U.S. dollars will purchase one Australian dollar. The direct quote and the indirect quote are reciprocals of each other.
\begin{aligned}\text{Direct Quote Home Currency}&=\frac1{\text{Indirect Quote Foreign Currency}}\\\\&=\frac1{\$1.30}\\\\&=\$0.77\end{aligned}
The cross rates are used to calculate foreign exchanges between two countries, neither of which are the home country. In 2018, the direct quote for the Australian dollar (AUD) was .72 while the direct quote for the euro (EUR) was 1.16 against the U.S. dollar.
When a trader wants to make currency transactions between two countries, neither of which is the home country, the trades are done using cross rates, which use an additional set of calculations. The first is for the trade from the first foreign currency to the home currency. The second calculation is from the home currency to the second foreign currency. As an example, assume the direct quote for the Australian dollar (AUD) is 0.72 against the U.S. home dollar (USD) and the direct quote for the euro (EUR) is 1.16 against the U.S. dollar. One Australian dollar would be 1.62 euros.
\begin{aligned}\text{Direct Quote}&=1\;\text{AUD}\times\left(\frac{1\;\text{USD}}{0.72\;\text{AUD}}\right)\times\left(\frac{1.16\;\text{EUR}}{1\;\rm{USD}}\right)\\\\&=1.62\;\text{EUR}\end{aligned}
<International Monetary System>Digital Currency
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Volume of hollow cylinder — lesson. Mathematics State Board, Class 10.
A cylinder emptied from the inner side and has a difference in the outer and inner radius of a cylinder with the same height is called a hollow cylinder.
Volume of a hollow cylinder:
Let \(R\) be the outer radius, \(r\) be the inner radius, and \(h\) be the height of the hollow cylinder.
Volume \(=\) Volume of the outer cylinder \(-\) Volume of the inner cylinder
\(=\) \(\pi R^2 h - \pi r^2 h\)
\(=\) \(\pi (R^2 - r^2) h\)
Volume of a hollow cylinder \(=\) \(\pi (R^2 - r^2) h\) cu. units.
Find the volume of the hollow cylinder of height \(14\) \(cm\) and whose internal and external radii are \(6\) \(cm\) and \(8\) \(cm\), respectively.
Internal radius, \(r\) \(=\) \(6\) \(cm\)
External radius, \(R\) \(=\) \(8\) \(cm\)
Height, \(h\) \(=\) \(14\) \(cm\)
\(=\) \(22 \times 4 \times 14\)
Therefore, the volume of a hollow cylinder is \(1232\) \(cm^3\).
\frac{22}{7}
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Inverse Trigonometric Functions, Popular Questions: ICSE Class 12-science MATH, Math Part I - Meritnation
Snehalakshmi asked a question
formula for sin-1X + sin-1Y , sin-1X -sin-1Y
cos-1X + cos-1Y
cos-1X - cos-1Y
Linu Dash asked a question
formula of 1-sinx
Priyanshi Jain asked a question
prove: tan inverse [ cos x / (1+sin x)] = pie /4 - (x/2)
Prove that 2 tan inverse (1/3) + cot inverse 4 = tan inverse (16/13)
show that tan(1/2 sin-1 3/4) = 4-root7/3
Show that sin-1(4/5) +cos-1(2/square root of 5)=cot-1 (2/11)
Please ans in a simple way
tan [pi/4 + 1/2 cos-1 a/b] + tan[pi/4 - 1/2cos-1 a/b] = 2b/a
How ca we perform the expoet simplificatio of 197/107 whole raised to 1/3 ?
find the value of x : sin-1x + sin-1(1-x) = cos-1x
Sridhar asked a question
find the principal value of sec^-1(3/sqrt 3)
Upasana Patro asked a question
Show that tan(1/2sin-13/4) = (4-Squareroot of 7)/ 3
Aditi Roy asked a question
Prove that : cos-1x-cos-1y=cos-1{xy+whole root over(1-x2)(1-y2)}
Lutfi Finch asked a question
i have 2 question -
1)if cos-1(x/a)+cos-1(y/b)=d
prove that : x2/a2-2xy/ab-cosd+y2/b2=sin2d
2)solve tan(cos-1x)=sin(tan-1x)
3)tan-1x+tan-1y+tan-1z=pi/2
prove that xy+yz+zx =1
tan inverse (1/x) = cot inverse (x) - pi for x less than 0 and = cot inverse x for x greater than 0. I know how to verify this but my doubt is when I'm changing functions how do I know that for like here about 0 we've different answers so how do I know that about which value I've check and will there always be only 1 such value? I mean how can we say that below 0 answer will always will same and not change I.e. cot inv x -pi . Pls show for other function like sin cos also
cos^-1(x) + sin^-1(x/2)=pi/6
Sec inverse (x-3/x+3) + sin inverse (x+3/x-3)
Karan Shinde & 1 other asked a question
2tan-1(1/3) +cot-14 =tan-1(16/13)
tan-1(1/4) + tan-1(2/9) = 1/2 Cos-1 (3/5)
Solve - sin(sec^-1 (17/15)) If I just provide the answer, i.e., 15/17 instead of showing the complete solution, in CBSE board exam? If yes, then there is no need to provide the solution, and if not, kindly do so.
if cos-1x + cos-1y + cos-1z = pi
prove that x2 + y2 + z2 +2xyz = 1
Set of exhaustive value for x | cos^-1 x |
find the greatest and least values:(sin-1x)2+(cos-1x)2
Could any one tell me what is cot^-1 (infinity) and tan ^-1( infinity)
Joe & 3 others asked a question
prove that tan -1(1) + tan-1(2) + tan-1(3) = pie.
Manjeet Kumar asked a question
evaluate cos{pi/3 - cos-1 (1/2)}
Manonarayanan Jb asked a question
tan inverse [ root ( 1 + x ) - root ( 1 - x ) / root( 1 + x )+ root ( 1 - x ) ] = pi/4 - 1/2 cos inverse x
find the value of cosec(tan-1 (-root 3))
Eshika Arya asked a question
tan(1/2(cos^-1(square root 5/3))
Smriti Joy asked a question
solve for x : tan-1(2-x/2+x) = 1/2 tan-1x/2
Rinta Mathew asked a question
sin-1(1-x)-2sin-1x=pi/2
Solve the problem experts
Vaishuuuu asked a question
If y=cot-1(square root cosx) - tan-1(square root cosx)
prove that sin y= tan2 x/2.
Prove that sin [ 2 tan inverse 3/5-sin inverse 7/25]=304/425
Arun Singh Tomar & 2 others asked a question
if tan-1x+tan-1y+tan-1z=π/2 show that xy+yz+zx=1?
Anshika Yadav asked a question
Simplify: sin inverse (5/13 cos x + 12/13 sin x).
prove that 2 tan inverse {root of a-b / a+b * tan theta /2 } = cos inverse { a cos thetha + b / a +b cos thetha}
If cot^-1 (3/4) = X , find the value of sin x
Ruby Tanveer asked a question
Solve for x : sin-1 6x + sin-1 6root3 x = -pi/2
Raghav Kapoor asked a question
What is the principal value of cosec-1 (cosec π/6) + tan-1 (tan 7π/6)
sin-1x + sin-12x = ∏/3
cos(tan-1 (15/8)-sin-1(7/24))=297/425.
Solve for the equation cos (tan-1x) = sin(cot-13/4)
Mohit Assudani asked a question
sin(arctan(x)+ arctan(1/x))for x
Jazlin asked a question
if sin ( cot-1 (x + 1) ) = cos ( tan-1x ) ...... find x ..... i asked this question and an expert proved it wrong .....but this question is absolutely correct ...it's 2015 board question .... so kindly find x ...since question is correct ..!!! @ PRIYANKA KEDIA ......AND OTHER MERITNATION EXPERTS ....!!!
Find domain of function cos inverse(3x-2).
if sin -1x +sin-1 y+ sin-1 z= pie ,prove that
a)x( root 1-x2)+y( root 1- y2)+ z(root 1-z2)= 2xyz
b)x4+ y4+z4+ 4 x2y2z2= 2( x2y2+y2z2+ z2x2)
If tan-1{root (1+x2) - root(1-x2)}/{ root (1+x2) +root (1-x2)} =theta , then prove that x2 =sin 2 theta .
expand please ....
Neelima Sreekumar asked a question
Solve for x - tan^-1 (x+1) + tan^-1 (x-1) = tan^-1 8/31
Isha Gupta asked a question
Find principal value of cot(tan^-1(4/5))
2tan-1 [ { (a-b) / (a+b) }1/2 tan x/2 ] = cos-1{ (b +acos x ) / (a+bcos x) }
Srishruthi Sriram asked a question
prove that 1/2 tan -1 x = cos -1 ( 1 + root ( 1 + x2 )) / 2 root (1 + x2 ) ) 1/2
Manu & 1 other asked a question
prove that sec square (tan inverse 2) +cosec square (cot inverse 3) =15
if y = cot x , find d^2 y/ dx^2, at x = pi/4
if tan^-1(y/x)= log( root of x^2 +y^2 , prove that dy/dx = x+y/x-y
arccos x+arccos 2x+arccos 3x =pi.if x satisfies cubic equation ax^3+bx^2+cx-1=0. then value of a+b+c =
Mohsuarez asked a question
sin-1 ( x . root (1-x) - root(x).root(1-x^2) )
pls answer , I have a test tomorrow :/
sin ( 2 sin -1 (2/3))
Find x :
tan-1 ( x-1 ) + tan-1x + tan-1 ( x+1 ) = tan-1 3x
Find domain of arc cos 2/(2+sinx)
Shrey Saini asked a question
how to solve tan-1 2x/1-x2 + cot-1 1-x2/2x = pie/3
sin inverse sin 3pi/5????
{\mathrm{sin}}^{2}x+{\mathrm{sin}}^{2}y<1 \mathrm{for} \mathrm{all} x, y \in \mathrm{R}
{\mathrm{sin}}^{-1} \left(\mathrm{tan}x . \mathrm{tan}y\right) \in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)
Akshya asked a question
Write the function, cot^-1 (sqrt(1+x^2)+x) in the simplest form
Please help me with 1st sum
1. Let f be an injective function with domain [a, b] and range [c, d]. If
\text{α}
is a point in (a, b) such that f has left hand derivative l and
right hand derivative r at x =
\text{α}
with both l and r non-zero different and negative, then left hand derivative
right hand derivative of f-1 at x = f(
\text{α}
) respectively, is
\text{(A) }\frac{1}{\text{r}}\text{,}\frac{1}{\text{l}}\phantom{\rule{0ex}{0ex}}\text{ (B) r , l}\phantom{\rule{0ex}{0ex}}\text{ (C) }\frac{1}{\text{l}}\text{,}\frac{1}{\text{r}}\phantom{\rule{0ex}{0ex}}\text{(D) l, r}
If tan-1x+tan-1y+tan-1z=pie, prove x+y+z=xyz
Anushrey Singh asked a question
Evaluate Cos(2arc Cos x + arc Sin x) at x = (1/5).
cot inverse [ root ( 1 + sinx ) + root ( 1 - sinx ) / root ( 1 + sinx ) - root ( 1 - sinx ) ] = x/2
Sharada asked a question
Cos inverse (5/root26) = 1/4 tan inverse (120/119)
( NOTE: 1/4 into Tan inverse: Tan inverse is NOT in the denominator along with 4)
PLEASE try to solve this question.
Andita Dunn asked a question
tan-1 x + 2cot-1 x = 2pi/3 find value of x
Inika Bardoloi asked a question
tan2( sec -12) +cot 2( cosec -13) =11
sin(1/2 cos inverse 4/5) ??evaluate
Hem Pandey asked a question
An isosceles triangle with base 6 cms. and base angles 30 degree each is incribed in a circle. A second circle, which is situated outside the triangle, touches the first circle and also touches the base of the triangle at its midpoint. Find its radius.
prove that 2 tan-1 (root of a+b/a-b tn theta/2) = cos -1 (acos theta b/a b cos theta)
Tanmay Nijhawan asked a question
Principle value of:
sec-1[cosec pie/8]
Leela Lakshmi & 1 other asked a question
sin-1(sinx+cosx/root 2)where pie
Find the value of tan-1(-tan13π/8)
Stunning Dude Akshay asked a question
evaluate sin(2 cos-1(-3/5))
Lovepreet Singh Johar asked a question
find the domain of the following part
Find x if sin-1(5/x) + sin-1(12/x) =pi/2
For an acute angle , cosine A is equal to
Mayyur & 1 other asked a question
sin^-1 x - cos^-1 x = pi/6
evaluate tan(2cot inverse x)
Rauank Mittal asked a question
IF y = sin-1 (2x+1/1+4x) find dy/dx
The function f(x) is defined as follows: f(x) = 2+x if x >=0 , f(x) = 2-x if x <= 0 .Then function f(x) at x=0 is: a)continious and differentiable b) continuos but not differentiable c) differentiable but not continous d) neither continous nor differentiable . No links , please give a detailed explaination .
cot[pi/4 - 2cot-1 3] =7
Find x if tan-1 4 + cot-1 x = pi / 2.
(i) sin^-1 (12/13) + cos^-1 (4/5) +tan^-1 (63/16)=0
(ii) 2sin^-1(3/5) - tan^-1 (17/31) =pi / 4
Gincy asked a question
cos-1 ([x2-1]/[x2+1]) + tan-1[(2x)/(x2-1)]=2pi/3
sin ( 2cos-1x)
Angel asked a question
Find the real solution of: tan-1 root(x(x+1)) + sin-1root(x2+x+1) = pi/2.
What is the value of tan infinity?
Yashik V.k & 1 other asked a question
(cosx - cosy)2 + (sinx - siny)2 = 4 sin2(x-y)/2.
Antara Chakraborty asked a question
Find the value of tan inverse(1) +cos inverse (1/4) +sin inverse (1/4)?
cos-1 [ 3/5 cosx + 4/5sinx ] find its value?
I can ' t understand this question . please explain it briefly ?
Domain of sec^-1(sinx)
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Bisector - zxc.wiki
In planar geometry , the bisector (also called the symmetrical angle ) of an angle is the half-line that runs through the vertex of the angle and divides the angular field into two congruent parts.
An intersecting pair of straight lines defines two bisectors, in this case straight lines that are orthogonal to each other . Each of these bisectors is an axis of symmetry of the geometric figure, which is formed by the intersecting pair of lines. This symmetry property is followed by a characterization of the two bisectors as a geometric location , which is referred to as the bisector set .
In analytic geometry and in the analysis , the bisecting line of play coordinate axes of a Cartesian coordinate system a special role. The ones who are affected by the I. and III. Quadrant is called 1st bisector or 1st median , the other 2nd bisector .
In synthetic geometry , the bisectors of an intersecting pair of lines are also defined by their properties as axes of symmetry. The existence of this bisector is one of the axioms that characterize a freely movable pre-Euclidean plane .
1 bisector in planar geometry
1.2 Angle bisector theorem
1.3 bisector in a triangle
1.4 bisector in a square
2 bisector of a coordinate system
3 synthetic geometry
Bisector in planar geometry
An angle is given by its two legs (half-straight lines with a common beginning at the vertex of the angle). Then the bisector can be constructed with a pair of compasses and a ruler (set square ): A circle with any radius is drawn around the vertex . The compass is reattached at the intersection with the legs of the square. Then you draw a circle with the same radius. The intersections of these two circles lie on the bisector.
In this construction it is used that the bisector is at the same time the vertical line in the isosceles triangle that is given by the vertex and the first two auxiliary points.
If, more generally, there are two straight lines that intersect at a point, we have four angles and thus four bisectors. The bisectors of two vertex angles coincide; so only two bisectors remain. These two bisectors - which are orthogonal to each other - are called the bisectors of the two straight lines .
If we come back to the case of an angle that is bounded by two legs (half-straight lines) and now extend these legs to straight lines, then we get two straight lines with two bisectors. One of these is the bisector of the original angle; the other is the bisector of its minor angle ; it is called the outer angle bisector of the original angle.
The union of the two bisectors of an intersecting pair of straight lines is the set of all points that have the same distance from the two straight lines, or, to put it another way, the set of the centers of all circles that touch the two straight lines .
Bisector in the triangle
The 3 outer angles:
The 3 intersection points E, F, G lie on a straight line (red) and the following route conditions apply:
{\ displaystyle {\ begin {aligned} {\ frac {| EB |} {| EC |}} & = {\ frac {| AB |} {| AC |}} \\ {\ frac {| FB |} { | FA |}} & = {\ frac {| CB |} {| CA |}} \\ {\ frac {| DA |} {| DC |}} & = {\ frac {| BA |} {| BC |}} \ end {aligned}}}
If angle bisectors are mentioned in triangle theory, this term mostly refers to the interior angles , less often to the exterior angles . Here the bisector of an interior angle is often abbreviated as. This abbreviation also stands for the distance on the bisector that lies within the triangle and, in design tasks, for its length.
{\ displaystyle \ alpha}
{\ displaystyle w _ {\ alpha}}
The following sentences apply to these bisectors :
The three bisectors (the interior angle) of a triangle intersect at one point. This point is the center of the inscribed circle (see also: Excellent points in a triangle ).
Each bisector (of an interior angle) in the triangle divides the opposite side in relation to the adjacent sides. (This statement is also known as the bisector theorem and can be proven with the help of similar triangles or by applying the sine law )
For the length of the bisector of an interior angle and for the adjacent sides of the length and , the relationship applies
{\ displaystyle w}
{\ displaystyle \ gamma}
{\ displaystyle a}
{\ displaystyle b}
{\ displaystyle {\ frac {2 \ cos {\ frac {\ gamma} {2}}} {w}} = {\ frac {1} {a}} + {\ frac {1} {b}}}
The bisectors of an interior angle and the exterior angle of a triangle belonging to the other two interior angles each intersect at a point. This point is the center of a circle .
The points of intersection of the outer angle bisector with the extended opposite sides of the corresponding inner angle lie, if they exist, on a straight line.
Bisector in a square
The bisecting a rectangle defining a generally quadrilateral . In the case of the tangent square it has degenerated into a point. In the case of the tendon quadrilateral, the enclosed quadrilateral is orthodiagonal . The bisector of a parallelogram generally includes a rectangle , the bisector of a rectangle a square , the bisector of an isosceles trapezoid a dragon square , the bisector of a quadrangle with equal opposite angles an isosceles trapezoid.
Angle bisector of a coordinate system
In a Cartesian coordinate system , the two bisectors of the coordinate axes play a special role:
The straight line with the equation is called the first bisector (bisector of the 1st and 3rd quadrant ) . This graph is the straight line through the origin with a slope of 1. It is called the 1st median in Austria .
{\ displaystyle y = x}
The straight line with the equation is called the second bisector (bisector of the 2nd and 4th quadrant) . This graph is the straight line through the origin with the slope −1.
{\ displaystyle y = -x}
In synthetic geometry, a pre-Euclidean plane is an affine plane over a body whose characteristic is not 2, together with an orthogonality relation without isotropic lines between the lines of the plane. (Vertical) axis mirroring can be defined in such a plane (→ see Mirroring (geometry) #axis mirroring ).
{\ displaystyle K ^ {2}}
{\ displaystyle K}
{\ displaystyle \ perp}
The following statement is called the bisector axiom :
For two straight , there is a straight line , so that when the mirror image of precisely the on the straight that is mapped.
{\ displaystyle a, b}
{\ displaystyle w}
{\ displaystyle w}
{\ displaystyle a}
{\ displaystyle b}
If the straight lines are parallel and different, then their central parallel is a straight line which has the required symmetry property. Since central parallels always exist in a pre-Euclidean plane, the essential requirement is for an axis of symmetry for an intersecting pair of straight lines, i.e. for an angle bisector. From the existence of an angle bisector, the existence of exactly a second line always follows, which is perpendicular to the first.
{\ displaystyle a, b}
A pre-Euclidean plane that fulfills the bisector axiom is called a floating plane .
Friedrich Bachmann : Structure of geometry from the concept of reflection. 2nd edition, Berlin; Göttingen; Heidelberg 1973
Summary: To justify the geometry from the concept of reflection. Mathematische Annalen, Vol. 123, 1951, pp. 341ff
Wendelin Degen and Lothar Profke: Fundamentals of affine and Euclidean geometry , Teubner, Stuttgart, 1976, ISBN 3-519-02751-8
↑ Degen (1976), p. 144
Commons : bisector - collection of images, videos and audio files
Wiktionary: bisector - explanations of meanings, word origins, synonyms, translations
Wiktionary: Angular symmetrals - explanations of meanings, word origins, synonyms, translations
Classic transversals - inside and outside angle bisector under D.7
This page is based on the copyrighted Wikipedia article "Winkelhalbierende" (Authors); it is used under the Creative Commons Attribution-ShareAlike 3.0 Unported License. You may redistribute it, verbatim or modified, providing that you comply with the terms of the CC-BY-SA.
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Narrowband minimum-variance distortionless-response beamformer - MATLAB - MathWorks Deutschland
phased.MVDRBeamformer
Narrowband minimum-variance distortionless-response beamformer
The phased.MVDRBeamformer System object™ implements a narrowband minimum-variance distortionless-response (MVDR) beamformer. The MVDR beamformer is also called the Capon beamformer. An MVDR beamformer belongs to the family of constrained optimization beamformers.
Create the phased.MVDRBeamformer object and set its properties.
beamformer = phased.MVDRBeamformer
beamformer = phased.MVDRBeamformer(Name,Value)
beamformer = phased.MVDRBeamformer creates an MVDR beamformer System object, beamformer, with default property values.
beamformer = phased.MVDRBeamformer(Name,Value) creates an MVDR beamformer with each property Name set to a specified Value. You can specify additional name-value pair arguments in any order as (Name1,Value1,...,NameN,ValueN). Enclose each property name in single quotes.
Example: beamformer = phased.MVDRBeamformer('SensorArray',phased.URA,'OperatingFrequency',300e6) sets the sensor array to a uniform rectangular array (URA) with default URA property values. The beamformer has an operating frequency of 300 MHz.
The number of bits used to quantize the phase shift component of beamformer or steering vector weights, specified as a nonnegative integer. A value of zero indicates that no quantization is performed.
Y = beamformer(X) performs MVDR beamforming on the input signal, X, and returns the beamformed output in Y. This syntax uses X as training samples to calculate the beamforming weights.
Y = beamformer(X,XT) uses XT as training samples to calculate the beamforming weights. To use this syntax, set the TrainingInputPort property to true.
Input signal, specified as a complex-valued M-by-N matrix. N is the number of array elements. If the sensor array contains subarrays, N is the number of subarrays. If you set TrainingInputPort to false, M must be larger than N; otherwise, M can be any positive integer.
Training data, specified as a complex-valued P-by-N matrix. If the sensor array contains subarrays, N is the number of subarrays; otherwise, N is the number of elements. P must be larger than N.
Beamforming directions, specified as a real-valued 2-by-1 column vector, or 2-by-L matrix. L is the number of beamforming directions. Each column has the form [AzimuthAngle;ElevationAngle]. Units are in degrees. Each azimuth angle must lie between –180° and 180°, and each elevation angle must lie between –90° and 90°.
Beamformed output, returned as a complex-valued M-by-L matrix, where M is the number of rows of X and L is the number of beamforming directions.
complex-valued N-by-L matrix.
Beamforming weights, returned as a complex-valued N-by-L matrix. If the sensor array contains subarrays, N is the number of subarrays; otherwise, N is the number of elements. L is the number of beamforming directions.
Apply an MVDR beamformer to a 5-element ULA. The incident angle of the signal is 45 degrees in azimuth and 0 degree in elevation. The signal frequency is .01 hertz. The carrier frequency is 300 MHz.
t = [0:.1:200]';
fr = .01;
xm = sin(2*pi*fr*t);
x = collectPlaneWave(array,xm,incidentAngle,fc,c);
Compute the beamforming weights.
beamformer = phased.MVDRBeamformer('SensorArray',array,...
'Direction',incidentAngle,'WeightsOutputPort',true);
[y,w] = beamformer(rx);
plot(t,real(rx(:,3)),'r:',t,real(y))
Plot the array response pattern using the MVDR weights.
'Weights',w,'CoordinateSystem','rectangular',...
The MVDR beamformer maximizes the signal to noise ratio.
Start with a signal arriving at the elements of an array. Assume that X is a complex-valued N-by-M data matrix representing the arrival of signals at an array. N is the number of sensors in the array and M is the number of samples or snapshots per signal. For mathematical convenience, this matrix is the transpose of the matrix specified in the X argument. Each row of X represents a time series of data for the corresponding array. The signal-to-noise ratio of a signal is given here.
SNR=\frac{{|{w}^{H}s|}^{2}}{{w}^{H}{R}_{I+N}^{}w}
Properly, the covariance matrix in the denominator is the covariance matrix for the noise and any interferers. You can vary the scale of w without affecting the SNR. Therefore, you can choose the normalization for w so that The MVDR estimator weights for beamforming are w = R-1v/vHRv where R is the data covariance matrix R = E[xxH].
Diagonal loading provides beamformer robustness due to small sample size and steering vector errors.
phased.FrostBeamformer | phased.PhaseShiftBeamformer | phased.LCMVBeamformer | phased.SubbandMVDRBeamformer
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Optical Measurement of Thermal Conductivity Using Fiber Aligned Frequency Domain Thermoreflectance | J. Heat Transfer | ASME Digital Collection
Optical Measurement of Thermal Conductivity Using Fiber Aligned Frequency Domain Thermoreflectance
e-mail: jonmalen@andrew.cmu.edu
Kanhayalal Baheti,
Kanhayalal Baheti
Tao Tong,
Janice A. Hudgings,
Janice A. Hudgings
, South Hadley, MA 01075
Malen, J. A., Baheti, K., Tong, T., Zhao, Y., Hudgings, J. A., and Majumdar, A. (May 2, 2011). "Optical Measurement of Thermal Conductivity Using Fiber Aligned Frequency Domain Thermoreflectance." ASME. J. Heat Transfer. August 2011; 133(8): 081601. https://doi.org/10.1115/1.4003545
Fiber aligned frequency domain thermoreflectance (FAFDTR) is a simple noncontact optical technique for accurately measuring the thermal conductivity of thin films and bulk samples for a wide range of materials, including electrically conducting samples. FAFDTR is a single-sided measurement that requires minimal sample preparation and no microfabrication. Like existing thermoreflectance techniques, a modulated pump laser heats the sample surface, and a probe laser monitors the resultant thermal wave via the temperature dependent reflectance of the surface. Via the use of inexpensive fiber coupled diode lasers and common mode rejection, FAFDTR addresses three challenges of existing optical methods: complexity in setup, uncertainty in pump-probe alignment, and noise in the probe laser. FAFDTR was validated for thermal conductivities spanning three orders of magnitude
(0.1–100 W/m K)
, and thin film thermal conductances greater than
10 W/m2 K
. Uncertainties of 10–15% were typical, and were dominated by uncertainties in the laser spot size. A parametric study of sensitivity for thin film samples shows that high thermal conductivity contrast between film and substrate is essential for making accurate measurements.
photothermal effects, thermal conductivity, thermal conductivity measurement, thermal diffusion, thermoreflectance, thermal conductivity, thermal diffusivity, thermometry, photothermal, thermoreflectance, frequency domain
Fibers, Thermal conductivity, Thermoreflectance, Thin films, Pumps, Temperature, Probes, Lasers, Uncertainty
Transient Thermoreflectance From Thin Metal-Films
Thermooptical Spectroscopy—Detection by the Mirage Effect
Photothermal Spectroscopy Using Optical Beam Probing—Mirage Effect
Rosencwaig
Thermal-Conductivity Measurement From 30-K to 750-K—The 3-Omega Method
Optical Measurement of Thermal Contact Conductance Between Wafer-Like Thin Solid Samples
Phonon Scattering in Condensed Matter V: Proceedings of the Fifth International Conference
, Urbana, IL, Jun. 2–6,
Thermal Conduction in Metallized Silicon-Dioxide Layers on Silicon
Two-Tint Pump-Probe Measurements Using a Femtosecond Laser Oscillator and sharp-Edged Optical Filters
Femtosecond Laser-Heating of Multilayer Metals. 2. Experiments
Coherent Phonon Generation and Detection by Picosecond Light-Pulses
Jarrinen
Determination of Thermal-Diffusivity of Low-Diffusivity Materials Using the Mirage Method With Multiparameter Fitting
Micrometer Scale Visualization of Thermal Waves by Photoreflectance Microscopy
Joulaud
Micron-Scale Thermal Characterizations of Interfaces Parallel or Perpendicular to the Surface
Thermal Effusivity Distribution Measurements Using a Thermoreflectance Technique
Photoacoustic and Photothermal Phenomena: Tenth International Conference
Thermoreflectance Technique to Measure Thermal Effusivity Distribution With High Spatial Resolution
Laptop Photothermal Reflectance Measurement Instrument Assembled With Optical Fiber Components
Thermoreflectance of V, Nb, and Paramagnetic Cr
Measurement of Thermal Conductivity of Silicon Dioxide Thin Films Using a 3 Omega Method
Thermal Conductivity Measurement and Interface Thermal Resistance Estimation Using SiO2 Thin Film
Dodabalapur
Comparison of the 3 Omega Method and Time-Domain Thermoreflectance for Measurements of the Cross-Plane Thermal Conductivity of Epitaxial Semiconductors
Thermal-Conductivity of Amorphous Solids Above the Plateau
Large Thermoelectric Response of Metallic Perovskites: Sr1−xLaxTiO3 (0⇐x⇐0.1)
Thermal Conductivity of Silicon From 300 to 1400 Degrees K
Experimental Measurement of Multiple Thermal Properties by Error Minimization
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Slow blow-up solutions for the H1(R3) critical focusing semilinear wave equation
15 March 2009 Slow blow-up solutions for the
{H}^{1}\left({\mathbb{R}}^{3}\right)
critical focusing semilinear wave equation
Joachim Krieger, Wilhelm Schlag, Daniel Tataru
Joachim Krieger,1 Wilhelm Schlag,2 Daniel Tataru3
1Department of Mathematics, University of Pennsylvania
Duke Math. J. 147(1): 1-53 (15 March 2009). DOI: 10.1215/00127094-2009-005
\nu >1/2
\delta >0
arbitrary, we prove the existence of energy solutions of
{\partial }_{\mathrm{tt}}u-\Delta u-{u}^{5}=0\left(0.1\right)
{\mathbb{R}}^{3+1}
which blow up exactly at
r=t=0
t\to 0-
. These solutions are radial and of the form
u=\lambda \left(t{\right)}^{1/2}W\left(\lambda \left(t\right)r\right)+\eta \left(r,t\right)
inside the cone
r\le t
\lambda \left(t\right)={t}^{-1-\nu }
W\left(r\right)=\left(1+{r}^{2}/3{\right)}^{-1/2}
is the stationary solution of (0.1), and
\eta
is a radiation term with
{\int }_{\left[r\le t\right]}\left(|\nabla \eta \left(x,t\right){|}^{2}+|{\eta }_{t}\left(x,t\right){|}^{2}+|\eta \left(x,t\right){|}^{6}\right)\mathrm{dx}\to 0,t\to 0.
Outside of the light-cone, there is the energy bound
{\int }_{\left[r>t\right]}\left(|\nabla u\left(x,t\right){|}^{2}+|{u}_{t}\left(x,t\right){|}^{2}+|u\left(x,t\right){|}^{6}\right)\mathrm{dx}<\delta
for all small
t>0
. The regularity of
u
\nu
. As in our accompanying article on wave maps [10], the argument is based on a renormalization method for the “soliton profile”
W\left(r\right)
Joachim Krieger. Wilhelm Schlag. Daniel Tataru. "Slow blow-up solutions for the
{H}^{1}\left({\mathbb{R}}^{3}\right)
critical focusing semilinear wave equation." Duke Math. J. 147 (1) 1 - 53, 15 March 2009. https://doi.org/10.1215/00127094-2009-005
Joachim Krieger, Wilhelm Schlag, Daniel Tataru "Slow blow-up solutions for the
{H}^{1}\left({\mathbb{R}}^{3}\right)
critical focusing semilinear wave equation," Duke Mathematical Journal, Duke Math. J. 147(1), 1-53, (15 March 2009)
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Using Free Cash Flow Yield
Liability Adjusted Cash Flow Yield
When evaluating stocks, most investors are familiar with fundamental indicators such as the price-to-earnings ratio (P/E), book value, price-to-book (P/B), and the PEG ratio. Also, investors who recognize the importance of cash generation use the company's cash flow statements when analyzing its fundamentals. They acknowledge that these statements offer a better representation of the company's operations.
However, very few people look at how much free cash flow (FCF) is available vis-à-vis the value of the company. Called the free cash flow yield, it's a better indicator than the P/E ratio.
Money in the bank is what every company strives to achieve. Investors are interested in what cash the company has in its bank accounts, as these numbers show the truth of a company's performance. It is more difficult to hide financial misdeeds and management adjustments in the cash flow statement.
Cash flow is the measure of money into and out of a company's bank accounts. Free cash flow, a subset of cash flow, is the amount of cash left over after the company has paid all its expenses and capital expenditures (funds reinvested into the company).
You can quickly calculate the free cash flow of a company from the cash flow statement. Start with the total from the cash generated from operations. Next, find the amount for capital expenditures in the "cash flow from investing" section. Then subtract the capital expenditures number from the total cash generated from operations to derive free cash flow (FCF).
When free cash flow is positive, it indicates the company is generating more cash than is used to run the business and reinvest to grow the business. It’s fully capable of supporting itself, and there is plenty of potential for further growth. A negative free cash flow number indicates the company is not able to generate sufficient cash to support the business. However, many small businesses do not have positive free cash flow as they are investing heavily to grow their venture rapidly.
Free cash flow is similar to earnings for a company without the more arbitrary adjustments made in the income statement. As a result, you can use free cash flow to help measure the performance of a company in a similar way to looking at the net income line. (Free cash flow is not the same as net cash flow, however.
Free cash flow is the amount of cash that is available for stockholders after the extraction of all expenses from the total revenue. The net cash flow is the amount of profit the company has with the costs that it pays currently, excluding long-term debts or bills. A company that has a positive net cash flow is meeting operating expenses at the current time, but not long-term costs, so it is not always an accurate measurement of the company’s progress or success.)
The P/E ratio measures how much annual net income is available per common share. However, the cash flow statement is a better measure of the performance of a company than the income statement.
Is there a comparable measurement tool to the P/E ratio that uses the cash flow statement? Happily, yes. We can use the free cash flow number and divide it by the value of the company as a more reliable indicator. Called the free cash flow yield, this gives investors another way to assess the value of a company that is comparable to the P/E ratio. Since this measure uses free cash flow, the free cash flow yield provides a better measure of a company's performance.
The most common way to calculate free cash flow yield is to use market capitalization as the divisor. Market capitalization is widely available, making it easy to determine. The formula is as follows:
\text{Free Cash Flow Yield} = \frac{\text{Free Cash Flow}}{\text{Market Capitalization}}
Free Cash Flow Yield=Market CapitalizationFree Cash Flow
Another way to calculate free cash flow yield is to use enterprise value as the divisor. To many, enterprise value is a more accurate measure of the value of a firm, as it includes the debt, value of preferred shares and minority interest, but minus cash and cash equivalents. The formula is as follows:
\text{Free Cash Flow Yield} = \frac{\text{Free Cash Flow}}{\text{Enterprise Value}}
Free Cash Flow Yield=Enterprise ValueFree Cash Flow
Both methods are valuable tools for investors. Use of market capitalization is comparable to the P/E ratio. Enterprise value provides a way to compare companies across different industries and companies with various capital structures. To make the comparison to the P/E ratio easier, some investors invert the free cash flow yield, creating a ratio of either market capitalization or enterprise value to free cash flow.
As an example, the table below shows the free cash flow yield for four large-cap companies and their P/E ratios in the middle of 2009. Apple (AAPL) sported a high trailing P/E ratio, thanks to the company's high growth expectations. General Electric (GE) had a trailing P/E ratio that reflected a slower growth scenario. Comparing Apple's and GE's free cash flow yield using market capitalization indicated that GE offered more attractive potential at this time. The primary reason for this difference was the large amount of debt that GE carried on its books, primarily from its financial unit. Apple was essentially debt-free. When you substituted market capitalization with the enterprise value as the divisor, Apple became a better choice.
Comparing the four companies listed below indicates that Cisco was positioned to perform well with the highest free cash flow yield, based on enterprise value. Lastly, although Fluor had a low P/E ratio, it did look as attractive after taking into consideration its low FCF yield.
Though not commonly used in company valuation, liability-adjusted cash flow yield (LACFY) is a variation. This fundamental analysis calculation compares a company's long-term free cash flow to its outstanding liabilities over the same period. Liability adjusted cash flow yield can be used to determine how long it will take for a buyout to become profitable or how a company is valued. The calculation is as follows:
Average Free Cash Flow
\begin{aligned} &\frac{10YAFCF}{[(OS+O+W) \times PSP-L]-(CA-I)}\\ &\textbf{where:}\\ &10YAFCF = 10\text{-Year average free cash flow}\\ &OS = \text{Outstanding shares}\\ &O = \text{Options}\\ &W = \text{Warrants}\\ &PSP = \text{Per share price}\\ &L = \text{Liabilities}\\ &CA = \text{Current assets}\\ &I = \text{Inventory} \end{aligned}
[(OS+O+W)×PSP−L]−(CA−I)10YAFCFwhere:10YAFCF=10-Year average free cash flowOS=Outstanding sharesO=OptionsW=WarrantsPSP=Per share priceL=LiabilitiesCA=Current assets
To see whether an investment is worthwhile, an analyst may look at ten years worth of data in a LACFY calculation and compare that to the yield on a 10-year Treasury note. The smaller the difference between LACFY and the Treasury yield, the less desirable an investment is.
Free cash flow yield offers investors or stockholders a better measure of a company's fundamental performance than the widely used P/E ratio. Investors who wish to employ the best fundamental indicator should add free cash flow yield to their repertoire of financial measures. You should not depend on just one measure, of course. However, the free cash flow amount is one of the most accurate ways to gauge a company's financial condition.
Liability Adjusted Cash Flow Yield (LACFY)
Liability Adjusted Cash Flow Yield is a fundamental analysis calculation that compares a firm's long-term free cash flow to its outstanding liabilities.
Free Cash Flow Yield: What You Need to Know
Free cash flow yield is a financial ratio that standardizes the free cash flow per share a company is expected to earn as compared to its market value per share.
The price-to-cash flow (P/CF) ratio measures the value of a stock’s price relative to its operating cash flow per share.
A relative valuation model is a business valuation method that compares a firm's value to that of its competitors to determine the firm's financial worth.
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Hessian Output - MATLAB & Simulink - MathWorks Australia
fminunc Hessian
fmincon Hessian
The fminunc and fmincon solvers return an approximate Hessian as an optional output.
[x,fval,exitflag,output,grad,hessian] = fminunc(fun,x0)
This topic describes the meaning of the returned Hessian, and the accuracy you can expect.
You can also specify the type of Hessian that the solvers use as input Hessian arguments. For fminunc, see Including Gradients and Hessians. For fmincon, see Hessian as an Input.
The Hessian for an unconstrained problem is the matrix of second derivatives of the objective function f:
\text{Hessian }{H}_{ij}=\frac{{\partial }^{2}f}{\partial {x}_{i}\partial {x}_{j}}.
Quasi-Newton Algorithm — fminunc returns an estimated Hessian matrix at the solution. fminunc computes the estimate by finite differences, so the estimate is generally accurate.
Trust-Region Algorithm — fminunc returns a Hessian matrix at the next-to-last iterate.
If you supply a Hessian in the objective function and set the HessianFcn option to 'objective', fminunc returns this Hessian.
If you supply a HessianMultiplyFcn function, fminunc returns the Hinfo matrix from the HessianMultiplyFcn function. For more information, see HessianMultiplyFcn in the trust-region section of the fminunc options table.
Otherwise, fminunc returns an approximation from a sparse finite difference algorithm on the gradients.
This Hessian is accurate for the next-to-last iterate. However, the next-to-last iterate might not be close to the final point.
The trust-region algorithm returns the Hessian at the next-to-last iterate for efficiency. fminunc uses the Hessian internally to compute its next step. When fminunc reaches a stopping condition, it does not need to compute the next step and, therefore, does not compute the Hessian.
The Hessian for a constrained problem is the Hessian of the Lagrangian. For an objective function f, nonlinear inequality constraint vector c, and nonlinear equality constraint vector ceq, the Lagrangian is
L=f+\sum _{i}{\lambda }_{i}{c}_{i}+\sum _{j}{\lambda }_{j}ce{q}_{j}.
The λi are Lagrange multipliers; see First-Order Optimality Measure and Lagrange Multiplier Structures. The Hessian of the Lagrangian is
H={\nabla }^{2}L={\nabla }^{2}f+\sum _{i}{\lambda }_{i}{\nabla }^{2}{c}_{i}+\sum _{j}{\lambda }_{j}{\nabla }^{2}ce{q}_{j}.
fmincon has several algorithms, with several options for Hessians, as described in fmincon Trust Region Reflective Algorithm, fmincon Active Set Algorithm, and fmincon Interior Point Algorithm.
active-set, sqp, or sqp-legacy Algorithm — fmincon returns the Hessian approximation it computes at the next-to-last iterate. fmincon computes a quasi-Newton approximation of the Hessian matrix at the solution in the course of its iterations. In general, this approximation does not match the true Hessian in every component, but only in certain subspaces. Therefore, the Hessian returned by fmincon can be inaccurate. For more details about the active-set calculation, see SQP Implementation.
trust-region-reflective Algorithm — fmincon returns the Hessian it computes at the next-to-last iterate.
If you supply a Hessian in the objective function and set the HessianFcn option to 'objective', fmincon returns this Hessian.
If you supply a HessianMultiplyFcn function, fmincon returns the Hinfo matrix from the HessianMultiplyFcn function. For more information, see Trust-Region-Reflective Algorithm in fmincon options.
Otherwise, fmincon returns an approximation from a sparse finite difference algorithm on the gradients.
The trust-region-reflective algorithm returns the Hessian at the next-to-last iterate for efficiency. fmincon uses the Hessian internally to compute its next step. When fmincon reaches a stopping condition, it does not need to compute the next step and, therefore, does not compute the Hessian.
If the HessianApproximation option is 'lbfgs' or 'finite-difference', or if you supply a HessianMultiplyFcn function, fmincon returns [] for the Hessian.
If the HessianApproximation option is 'bfgs' (the default), fmincon returns a quasi-Newton approximation to the Hessian at the final point. This Hessian can be inaccurate, similar to the active-set or sqp algorithm Hessian.
If the HessianFcn option is a function handle, fmincon returns this function as the Hessian at the final point.
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Printing and Exporting - Maple Help
Home : Support : Online Help : System : Information : Updates : Maple 2020 : Printing and Exporting
Maple 2020 includes many important enhancements to printing Maple documents, including significant improvements to LaTeX export.
When printing or exporting to PDF, you can now control how sections are displayed:
Select whether collapsed sections are automatically expanded.
Select whether section boundary lines, arrows, and indentation are removed.
As an example, open this page of Physics examples and go to File > Print Preview.
In addition, headers and footers can now be set up to apply globally, so you can put the same headers or footers on all your printed Maple documents.
For details, see Printing and Headers and Footers.
The Export to LaTeX facility in Maple has undergone several improvements and changes for Maple 2020.
1-D Math input is now translated to a lstlisting environment which uses the LaTeX listings package.
The listings package is a standard part of modern LaTeX distributions and is commonly used for formatting programming-language source code in LaTeX documents.
The following is a typical example of Maple 1-D input:
sin(x)^2 + cos(x)^2;
{\textcolor[rgb]{0,0,1}{\mathrm{sin}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\right)}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{\mathrm{cos}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\right)}^{\textcolor[rgb]{0,0,1}{2}}
Here is the appearance of this line in the generated LaTeX after it is compiled using standard LaTeX tools:
Maple 2020 now includes support for Code Edit Regions in the generated LaTeX, also using the listings package.
Fibonacci := proc( i :: nonnegint ) if i = 0 or i = 1 then return 1; else return thisproc( i-2 ) + thisproc( i-2 ); end if; end proc:
Here is the appearance of this code edit region in the generated LaTeX after it is compiled using standard LaTeX tools:
Maple 2020 now includes support for inserted images (that is, images included via Insert -> Image) in the generated LaTeX.
These images will be exported to PNG files on disk, and references to the exported file will be inserted in the LaTeX file corresponding to the main worksheet or document.
International characters appearing in Maple documents are now translated to LaTeX in such a way as to work seamlessly with existing tools for international character support.
In particular, any non-ASCII characters in Maple document text are now translated to equivalent UTF-8 characters in the LaTeX output file. This generated file now also includes the line:
This loads a standard LaTeX package for UTF-8 compatibility and enables UTF-8 characters to be used directly within text in the LaTeX file without any special environments or macros.
The following is the opening paragraph of The Metamorphosis by Franz Kafka:
During export, the accented characters in the text (ä, ö, ü) are translated to multibyte UTF-8 characters which can be used directly in the LaTeX source. Here is the appearance of this text in the generated LaTeX output after it is compiled using standard LaTeX tools:
Links to external URL sources such as the Maplesoft website
Links to bookmarks within the same Maple document
Links to help pages within the Maple help system, such as the plot help page
Links to local file locations
Together with the improvements for hyperlinks, bookmarks within a Maple document are now translated to link destinations in the generated LaTeX document.
Hyperlinks referencing these bookmarks within the current document will continue to work in documents built from the generated LaTeX.
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Estimation of crack opening from a two-dimensional continuum-based finite element computation - Dufour et al 2012a - Scipedia
F. Dufour, G. Legrain, G. Pijaudier, A. Huerta
Damage models are capable of representing crack initiation and mimicking crack propagation within a continuum framework. Thus, in principle, they do not describe crack openings. In durability analyses of concrete structures however, transfer properties are a key issue controlled by crack propagation and crack opening. We extend here a one dimensional approach for estimating a crack opening from a continuum based fi nite element calculation to two dimensional cases. The technique operates in the case of mode I cracking described in a continuum setting by a nonlocal isotropic damage model. We used the global tracking method to compute the idealized crack location as a post treatment procedure. The orig inal one dimensional problem devised in Dufour et al . [4] is recovered as pro fi les of deformation orthog onal to the idealized crack direction are computed. An estimate of the crack opening and an error indicator are computed by comparing fi nite element deformation pro fi les and theoretical pro fi les corresponding to a displacement discontinuity. Two estimates have been considered: In the strong approach, the maxima of the pro fi les are assumed to be equal; in the weak approach, the integrals of each pro fi le are set equal. Two dimensional numerical calculations show that the weak estimates perform better than do the strong ones. Error indicators, de fi ned as the distance between the numerical and theoretical pro fi les, are less than a few percentages. In the case of a three point bending, test results are in good agreement with experimental data, with an error lower than 10% for widely opened crack ( > 40 m m )
F. Dufour, G. Legrain, G. Pijaudier and A. Huerta, Estimation of crack opening from a two-dimensional continuum-based finite element computation, Int. J. Numer. Anal. Meth. Geomech. (2012). Vol. 36 (16), pp. 1813-1830 URL https://www.scipedia.com/public/Dufour_et_al_2012a
Damage models are capable of representing crack initiation and mimicking crack propagation within a continuum framework. Thus, in principle, they do not describe crack openings. In durability analyses of concrete structures however, transfer properties are a key issue controlled by crack propagation and crack opening. We extend here a one‐dimensional approach for estimating a crack opening from a continuum‐based finite element calculation to two‐dimensional cases. The technique operates in the case of mode I cracking described in a continuum setting by a nonlocal isotropic damage model. We used the global tracking method to compute the idealized crack location as a post‐treatment procedure. The original one‐dimensional problem devised in Dufour et al. [4] is recovered as profiles of deformation orthogonal to the idealized crack direction are computed. An estimate of the crack opening and an error indicator are computed by comparing finite element deformation profiles and theoretical profiles corresponding to a displacement discontinuity. Two estimates have been considered: In the strong approach, the maxima of the profiles are assumed to be equal; in the weak approach, the integrals of each profile are set equal. Two‐dimensional numerical calculations show that the weak estimates perform better than do the strong ones. Error indicators, defined as the distance between the numerical and theoretical profiles, are less than a few percentages. In the case of a three‐point bending, test results are in good agreement with experimental data, with an error lower than 10% for widely opened crack (
{\displaystyle >40}
{\displaystyle \mu m}
Crack location • Crack opening • Damage mechanics
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Fractional Reserve Banking - Course Hero
Macroeconomics/The Banking System/Fractional Reserve Banking
Learn all about fractional reserve banking in just a few minutes! Professor Jadrian Wooten of Penn State University details the fractional reserve banking system, including reserve requirements and increases in the money supply.
Fractional reserve banking is the system by which banks set aside a portion of their deposits and loan out the remainder or invest it in other ways, which creates money.
Fractional reserve banking is a system of banking under which commercial banks hold a portion of their deposits and use the remainder to increase revenue through loans and investments. A commercial bank is a private financial institution primarily concerned with maximizing its revenue through holding deposits, offering checking services to the public and businesses, and making loans and investments. The amount of money a bank has held back from deposits received, which is not available to loan or invest, is called its required reserves.
By law, the federal government requires all banks to keep a portion (i.e., the required reserves) of every dollar deposited in reserve. This law is set in place as a safety precaution to ensure that banks do not collapse and lose their investments.
The amount of required reserves a bank must keep on hand is based on the bank's liabilities. A bank's liabilities are the money consumers have placed in the bank for safekeeping, called a demand deposit. Demand deposits are deposits customers can withdraw without advance notice or warning.
For example, assume the Federal Reserve requires a bank to keep 10% of its deposits as reserves. If a customer deposits $100 into a bank account, the bank holds $10 in reserve. The bank can use the remaining $90 in excess reserves to invest, loan to individuals, and engage in other profit-making measures. Excess reserves are the amount of money a bank has available to loan or invest, made up of all deposits minus the required reserves; in other words, the reserves held in excess of the reserve requirement required by regulators. Later, the customer may withdraw their $100, by which time the bank will have profited from the $90 it loaned out or invested.
Fractional reserve banking also increases the national money supply, which is the amount of money free to circulate and power economic activity in an economy. As an example, instead of $100 being deposited and remaining unavailable for use, $90 circulates in the economy and contributes to economic activity and growth, while $10 remains in the bank as reserves.
Fractional reserve banking creates an expansion on $100. Each curve terminates at the point where expansion stops. This value is calculated by the money multiplier, which is
1/\text{RR}
, where RR is the required reserve ratio. For example, when
\text{RR} = 20\%
, with an initial deposit of $100, the expansion is calculated as
\$100 \times 1/0.2=\$500
Fed Member Banks collaborate with Federal Reserve banks by often holding their deposits in one of the Federal Reserve branches. When a customer deposits $100 in the bank, the fractional reserve bank places the $10 it must set aside in reserve in a Federal Reserve bank.
Banks rely on some individuals saving money and others borrowing money. To stimulate these activities, banks must have cash on hand from individual savings to offer to borrowers. Confidence in banks is necessary to ensure that people will make deposits.
The larger the bank, the more reserves it must hold, according to the regulations created by the Federal Reserve. This regulation exists to discourage and impede a run on a bank. A run on a bank is when many of a bank's depositors choose to withdraw their funds at the same time, often because of a perception of financial weakness in the bank, which then causes the weakness people feared in the first place. Bank runs throughout the banking system were one of the most damaging aspects of the Great Depression, the time of severe economic downturn in the 1930s.
The Federal Reserve decides the percentage of reserves each bank must have. The percentage is relatively low, allowing most assets to remain liquid and in circulation in the economy, which keeps the cost of credit low as well. The system of fractional reserve banking only functions as long as people are taking out loans, repaying loans, and depositing money in banks.
<Federal Reserve>Expansion of Deposits
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The Zero-volatility spread (Z-spread) is the constant spread that makes the price of a security equal to the present value of its cash flows when added to the yield at each point on the spot rate Treasury curve where cash flow is received. In other words, each cash flow is discounted at the appropriate Treasury spot rate plus the Z-spread. The Z-spread is also known as a static spread.
Formula and Calculation for the Zero-Volatility Spread
To calculate a Z-spread, an investor must take the Treasury spot rate at each relevant maturity, add the Z-spread to this rate, and then use this combined rate as the discount rate to calculate the price of the bond. The formula to calculate a Z-spread is:
\begin{aligned} &\text{P} = \frac { C_1 }{ \left ( 1 + \frac { r_1 + Z }{ 2 } \right ) ^ {2n} } + \frac { C_2 }{ \left ( 1 + \frac { r_2 + Z }{ 2 } \right ) ^ {2n} } + \frac { C_n }{ \left ( 1 + \frac { r_n + Z }{ 2 } \right ) ^ {2n} } \\ &\textbf{where:} \\ &\text{P} = \text{Current price of the bond plus any accrued interest} \\ &C_x = \text{Bond coupon payment} \\ &r_x = \text{Spot rate at each maturity} \\ &Z = \text{Z-spread} \\ &n = \text{Relevant time period} \\ \end{aligned}
P=(1+2r1+Z)2nC1+(1+2r2+Z)2nC2+(1+2rn+Z)2nCnwhere:P=Current price of the bond plus any accrued interestCx=Bond coupon paymentrx=Spot rate at each maturityZ=Z-spreadn=Relevant time period
For example, assume a bond is currently priced at $104.90. It has three future cash flows: a $5 payment next year, a $5 payment two years from now and a final total payment of $105 in three years. The Treasury spot rate at the one-, two-, and three- year marks are 2.5%, 2.7% and 3%. The formula would be set up as follows:
\begin{aligned} \$104.90 = &\ \frac { \$5 }{ \left ( 1 + \frac { 2.5\% + Z }{ 2 } \right ) ^ { 2 \times 1 } } + \frac { \$5 }{ \left ( 1 + \frac { 2.7\% + Z }{ 2 } \right ) ^ { 2 \times 2 } } \\ &+ \frac { \$105 }{ \left ( 1 + \frac { 3\% + Z }{ 2 } \right ) ^ {2 \times 3 } } \end{aligned}
$104.90= (1+22.5%+Z)2×1$5+(1+22.7%+Z)2×2$5+(1+23%+Z)2×3$105
With the correct Z-spread, this simplifies to:
\begin{aligned} \$104.90 = \$4.87 + \$4.72 + \$95.32 \end{aligned}
$104.90=$4.87+$4.72+$95.32
This implies that the Z-spread equals 0.25% in this example.
The zero-volatility spread of a bond tells the investor the bond's current value plus its cash flows at certain points on the Treasury curve where cash-flow is received.
The Z-spread is also called the static spread.
The spread is used by analysts and investors to discover discrepancies in a bond's price.
What the Zero-Volatility Spread (Z-spread) Can Tell You
A Z-spread calculation is different than a nominal spread calculation. A nominal spread calculation uses one point on the Treasury yield curve (not the spot-rate Treasury yield curve) to determine the spread at a single point that will equal the present value of the security's cash flows to its price.
The Zero-volatility spread (Z-spread) helps analysts discover if there is a discrepancy in a bond's price. Because the Z-spread measures the spread that an investor will receive over the entirety of the Treasury yield curve, it gives analysts a more realistic valuation of a security instead of a single-point metric, such as a bond's maturity date.
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Rule of Sarrus - Wikipedia
Mnemonic device for calculating 3 by 3 matrix determinants
Rule of Sarrus: The determinant of the three columns on the left is the sum of the products along the down-right diagonals minus the sum of the products along the up-right diagonals.
In linear algebra, the Rule of Sarrus is a mnemonic device for computing the determinant of a
{\displaystyle 3\times 3}
matrix named after the French mathematician Pierre Frédéric Sarrus.[1]
{\displaystyle 3\times 3}
{\displaystyle M={\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}},}
then its determinant can be computed by the following scheme.
Write out the first two columns of the matrix to the right of the third column, giving five columns in a row. Then add the products of the diagonals going from top to bottom (solid) and subtract the products of the diagonals going from bottom to top (dashed). This yields[1][2]
{\displaystyle {\begin{aligned}\det(M)&=\det {\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}}\\[6pt]&=a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{31}a_{22}a_{13}-a_{32}a_{23}a_{11}-a_{33}a_{21}a_{12}.\end{aligned}}}
Alternative vertical arrangement
Alternative "butterfly" arrangement
A similar scheme based on diagonals works for
{\displaystyle 2\times 2}
matrices:[1]
{\displaystyle \det(M)=\det {\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}}=a_{11}a_{22}-a_{21}a_{12}.}
Both are special cases of the Leibniz formula, which however does not yield similar memorization schemes for larger matrices. Sarrus' rule can also be derived using the Laplace expansion of a
{\displaystyle 3\times 3}
matrix.[1]
Another way of thinking of Sarrus' rule is to imagine that the matrix is wrapped around a cylinder, such that the right and left edges are joined.
^ a b c d Fischer, Gerd (1985). Analytische Geometrie (in German) (4th ed.). Wiesbaden: Vieweg. p. 145. ISBN 3-528-37235-4.
^ Paul Cohn: Elements of Linear Algebra. CRC Press, 1994, ISBN 9780412552809, p. 69
Sarrus' rule at Planetmath
Linear Algebra: Rule of Sarrus of Determinants at khanacademy.org
Retrieved from "https://en.wikipedia.org/w/index.php?title=Rule_of_Sarrus&oldid=1050637323"
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Solve fully implicit differential equations — variable order method - MATLAB ode15i - MathWorks Switzerland
Solve Weissinger Implicit ODE
Solve Robertson Problem as Implicit Differential Algebraic Equations (DAEs)
Solve fully implicit differential equations — variable order method
[t,y] = ode15i(odefun,tspan,y0,yp0)
[t,y] = ode15i(odefun,tspan,y0,yp0,options)
[t,y,te,ye,ie] = ode15i(odefun,tspan,y0,yp0,options)
sol = ode15i(___)
[t,y] = ode15i(odefun,tspan,y0,yp0), where tspan = [t0 tf], integrates the system of differential equations
f\left(t,y,y\text{'}\right)=0
from t0 to tf with initial conditions y0 and yp0. Each row in the solution array y corresponds to a value returned in column vector t.
[t,y] = ode15i(odefun,tspan,y0,yp0,options) also uses the integration settings defined by options, which is an argument created using the odeset function. For example, use the AbsTol and RelTol options to specify absolute and relative error tolerances, or the Jacobian option to provide the Jacobian matrix.
[t,y,te,ye,ie] = ode15i(odefun,tspan,y0,yp0,options) additionally finds where functions of (t,y,y'), called event functions, are zero. In the output, te is the time of the event, ye is the solution at the time of the event, and ie is the index of the triggered event.
For each event function, specify whether the integration is to terminate at a zero and whether the direction of the zero crossing matters. Do this by setting the 'Events' property to a function, such as myEventFcn or @myEventFcn, and creating a corresponding function: [value,isterminal,direction] = myEventFcn(t,y,yp). For more information, see ODE Event Location.
sol = ode15i(___) returns a structure that you can use with deval to evaluate the solution at any point on the interval [t0 tf]. You can use any of the input argument combinations in previous syntaxes.
Calculate consistent initial conditions and solve an implicit ODE with ode15i.
Weissinger's equation is
{\mathrm{ty}}^{2}{\left({\mathit{y}}^{\prime }\right)}^{3}-{\mathit{y}}^{3}{\left({\mathit{y}}^{\prime }\right)}^{2}+\mathit{t}\left({\mathit{t}}^{2}+1\right){\mathit{y}}^{\prime }-{\mathit{t}}^{2}\mathit{y}=0
Since the equation is in the generic form
\mathit{f}\left(\mathit{t},\mathit{y},{\mathit{y}}^{\prime }\right)=0
, you can use the ode15i function to solve the implicit differential equation.
To code the equation in a form suitable for ode15i, you need to write a function with inputs for
\mathit{t}
\mathit{y}
{\mathit{y}}^{\prime }
that returns the residual value of the equation. The function @weissinger encodes this equation. View the function file.
type weissinger
function res = weissinger(t,y,yp)
%WEISSINGER Evaluate the residual of the Weissinger implicit ODE
% See also ODE15I.
res = t*y^2 * yp^3 - y^3 * yp^2 + t*(t^2 + 1)*yp - t^2 * y;
Calculate Consistent Initial Conditions
The ode15i solver requires consistent initial conditions, that is, the initial conditions supplied to the solver must satisfy
\mathit{f}\left({\mathit{t}}_{0},\mathit{y},{\mathit{y}}^{\prime }\right)=0
Since it is possible to supply inconsistent initial conditions, and ode15i does not check for consistency, it is recommended that you use the helper function decic to compute such conditions. decic holds some specified variables fixed and computes consistent initial values for the unfixed variables.
In this case, fix the initial value
\mathit{y}\left({\mathit{t}}_{0}\right)=\sqrt{\frac{3}{2}}
and let decic compute a consistent initial value for the derivative
{\mathit{y}}^{\prime }\left({\mathit{t}}_{0}\right)
, starting from an initial guess of
{\mathit{y}}^{\prime }\left({\mathit{t}}_{0}\right)=0
y0 = sqrt(3/2);
yp0 = 0;
[y0,yp0] = decic(@weissinger,t0,y0,1,yp0,0)
yp0 = 0.8165
Use the consistent initial conditions returned by decic with ode15i to solve the ODE over the time interval
\left[1\text{\hspace{0.17em}}10\right]
[t,y] = ode15i(@weissinger,[1 10],y0,yp0);
The exact solution of this ODE is
\mathit{y}\left(\mathit{t}\right)=\sqrt{{\mathit{t}}^{2}+\frac{1}{2}}
Plot the numerical solution y computed by ode15i against the analytical solution ytrue.
ytrue = sqrt(t.^2 + 0.5);
plot(t,y,'*',t,ytrue,'-o')
legend('ode15i', 'exact')
This example reformulates a system of ODEs as a fully implicit system of differential algebraic equations (DAEs). The Robertson problem coded by hb1ode.m is a classic test problem for programs that solve stiff ODEs. The system of equations is
The problem can be rewritten as a system of DAEs by using the conservation law to determine the state of . This reformulates the problem as the implicit DAE system
The function robertsidae encodes this DAE system.
function res = robertsidae(t,y,yp)
res = [yp(1) + 0.04*y(1) - 1e4*y(2)*y(3);
yp(2) - 0.04*y(1) + 1e4*y(2)*y(3) + 3e7*y(2)^2;
y(1) + y(2) + y(3) - 1];
The full example code for this formulation of the Robertson problem is available in ihb1dae.m.
Set the error tolerances and the value of .
options = odeset('RelTol',1e-4,'AbsTol',[1e-6 1e-10 1e-6], ...
'Jacobian',{[],[1 0 0; 0 1 0; 0 0 0]});
Use decic to compute consistent initial conditions from guesses. Fix the first two components of y0 to get the same consistent initial conditions as found by ode15s in hb1dae.m, which formulates this problem as a semi-explicit DAE system.
y0 = [1; 0; 1e-3];
yp0 = [0; 0; 0];
[y0,yp0] = decic(@robertsidae,0,y0,[1 1 0],yp0,[],options);
Solve the system of DAEs using ode15i.
[t,y] = ode15i(@robertsidae,tspan,y0,yp0,options);
Plot the solution components. Since the second solution component is small relative to the others, multiply it by 1e4 before plotting.
semilogx(t,y)
ylabel('1e4 * y(:,2)')
title('Robertson DAE problem with a Conservation Law, solved by ODE15I')
The function f = odefun(t,y,yp), for a scalar t and column vectors y and yp, must return a column vector f of data type single or double that corresponds to
f\left(t,y,y\text{'}\right)
. odefun must accept the three inputs for t, y, and yp even if one of the inputs is not used in the function.
y\text{'}-y=0
, use this function.
function f = odefun(t,y,yp)
f = yp - y;
For a system of equations, the output of odefun is a vector. Each equation becomes an element in the solution vector. For example, to solve
\begin{array}{l}y{\text{'}}_{1}-{y}_{2}=0\\ y{\text{'}}_{2}+1=0\text{\hspace{0.17em}},\end{array}
function dy = odefun(t,y,yp)
dy(1) = yp(1)-y(2);
dy(2) = yp(2)+1;
Interval of integration, specified as a vector. At minimum, tspan must be a two element vector [t0 tf] specifying the initial and final times. To obtain solutions at specific times between t0 and tf, use a longer vector of the form [t0,t1,t2,...,tf]. The elements in tspan must be all increasing or all decreasing.
The solver imposes the initial conditions given by y0 at the initial time tspan(1), then integrates from tspan(1) to tspan(end):
If tspan has two elements, [t0 tf], then the solver returns the solution evaluated at each internal integration step within the interval.
If tspan has more than two elements [t0,t1,t2,...,tf], then the solver returns the solution evaluated at the given points. However, the solver does not step precisely to each point specified in tspan. Instead, the solver uses its own internal steps to compute the solution, then evaluates the solution at the requested points in tspan. The solutions produced at the specified points are of the same order of accuracy as the solutions computed at each internal step.
The initial and final values in tspan are used to calculate the maximum step size MaxStep. Therefore, changing the initial or final values in tspan could lead to the solver using a different step sequence, which might change the solution.
y0 — Initial conditions for y
Initial conditions for y, specified as a vector. y0 must be the same length as the vector output of odefun, so that y0 contains an initial condition for each equation defined in odefun.
The initial conditions for y0 and yp0 must be consistent, meaning that
f\left({t}_{0},{y}_{0},y{\text{'}}_{0}\right)=0
. Use the decic function to compute consistent initial conditions close to guessed values.
yp0 — Initial conditions for y’
Initial conditions for y’, specified as a column vector. yp0 must be the same length as the vector output of odefun, so that yp0 contains an initial condition for each variable defined in odefun.
f\left({t}_{0},{y}_{0},y{\text{'}}_{0}\right)=0
Option structure, specified as a structure array. Use the odeset function to create or modify the option structure.
See Summary of ODE Options for a list of which options are compatible with each ODE solver.
Providing the Jacobian matrix to ode15i is critical for reliability and efficiency. Alternatively, if the system is large and sparse, then providing the Jacobian sparsity pattern also assists the solver. In either case, use odeset to pass in the matrices using the Jacobian or JPattern options.
ode15i is a variable-step, variable-order (VSVO) solver based on the backward differentiation formulas (BDFs) of orders 1 to 5. ode15i is designed to be used with fully implicit differential equations and index-1 differential algebraic equations (DAEs). The helper function decic computes consistent initial conditions that are suitable to be used with ode15i [1].
[1] Lawrence F. Shampine, “Solving 0 = F(t, y(t), y′(t)) in MATLAB,” Journal of Numerical Mathematics, Vol.10, No.4, 2002, pp. 291-310.
decic | ode15s | ode23t | odeset | odeget | deval
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Calendar/HostTimeZone - Maple Help
Home : Support : Online Help : Calendar/HostTimeZone
attempt to determine the local time zone of the host computer
HostTimeZone()
The HostTimeZone() command returns, if possible, a string designating the detected time zone of the host. The ability to detect the time zone depends upon the correct configuration of the host on which it is called.
\mathrm{with}\left(\mathrm{Calendar}\right):
\mathrm{HostTimeZone}\left(\right)
\textcolor[rgb]{0,0,1}{"Etc/UTC"}
The Calendar[HostTimeZone] command was introduced in Maple 2018.
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VectorSpaceSum - Maple Help
Home : Support : Online Help : Mathematics : Differential Equations : Lie Symmetry Method : Commands for PDEs (and ODEs) : LieAlgebrasOfVectorFields : LHPDE : VectorSpaceSum
find a LHPDE object whose solution space is the sum of the solution spaces of given LHPDE objects.
VectorSpaceSum( obj1, obj2, ..., depname = vars )
Let obj1, obj2, ... be a sequence of LHPDE objects living on the same space (see AreSameSpace). The VectorSpaceSum method finds a LHPDEs system whose solution space is the vector space sum of solution spaces of obj1,obj2,....
\mathrm{with}\left(\mathrm{LieAlgebrasOfVectorFields}\right):
\mathrm{Typesetting}:-\mathrm{Settings}\left(\mathrm{userep}=\mathrm{true}\right):
\mathrm{Typesetting}:-\mathrm{Suppress}\left({\mathrm{\alpha },\mathrm{\beta },\mathrm{\eta },\mathrm{\phi },\mathrm{\psi },\mathrm{\xi }}\left(x,y\right)\right):
S≔\mathrm{LHPDE}\left([\mathrm{diff}\left(\mathrm{\xi }\left(x,y\right),x\right)=0,\mathrm{diff}\left(\mathrm{\eta }\left(x,y\right),y\right)=0,\mathrm{diff}\left(\mathrm{\xi }\left(x,y\right),y\right)+\mathrm{diff}\left(\mathrm{\eta }\left(x,y\right),x\right)=0,\mathrm{diff}\left(\mathrm{\xi }\left(x,y\right),y,y\right)=0,\mathrm{diff}\left(\mathrm{\eta }\left(x,y\right),x,x\right)=0]\right)
\textcolor[rgb]{0,0,1}{S}\textcolor[rgb]{0,0,1}{≔}[{\textcolor[rgb]{0,0,1}{\mathrm{\xi }}}_{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{\mathrm{\eta }}}_{\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{\mathrm{\xi }}}_{\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{\mathrm{\eta }}}_{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{\mathrm{\xi }}}_{\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{\mathrm{\eta }}}_{\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{indep}}\textcolor[rgb]{0,0,1}{=}[\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{dep}}\textcolor[rgb]{0,0,1}{=}[\textcolor[rgb]{0,0,1}{\mathrm{\eta }}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{\xi }}]
\mathrm{S1}≔\mathrm{LHPDE}\left([\mathrm{diff}\left(\mathrm{\alpha }\left(x,y\right),x,x\right)=0,\mathrm{diff}\left(\mathrm{\alpha }\left(x,y\right),y\right)=0,\mathrm{diff}\left(\mathrm{\beta }\left(x,y\right),x\right)=0,\mathrm{diff}\left(\mathrm{\beta }\left(x,y\right),y,y\right)=0,\mathrm{diff}\left(\mathrm{\alpha }\left(x,y\right),x\right)-\mathrm{diff}\left(\mathrm{\beta }\left(x,y\right),y\right)=0],\mathrm{indep}=[x,y],\mathrm{dep}=[\mathrm{\alpha },\mathrm{\beta }]\right)
\textcolor[rgb]{0,0,1}{\mathrm{S1}}\textcolor[rgb]{0,0,1}{≔}[{\textcolor[rgb]{0,0,1}{\mathrm{\alpha }}}_{\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{\mathrm{\alpha }}}_{\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{\mathrm{\beta }}}_{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{\mathrm{\beta }}}_{\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{\mathrm{\alpha }}}_{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{\mathrm{\beta }}}_{\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{indep}}\textcolor[rgb]{0,0,1}{=}[\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{dep}}\textcolor[rgb]{0,0,1}{=}[\textcolor[rgb]{0,0,1}{\mathrm{\alpha }}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{\beta }}]
\mathrm{VectorSpaceSum}\left(S,\mathrm{S1}\right)
[{\textcolor[rgb]{0,0,1}{\mathrm{\eta }}}_{\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{\mathrm{\eta }}}_{\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{\mathrm{\xi }}}_{\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{\mathrm{\xi }}}_{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{indep}}\textcolor[rgb]{0,0,1}{=}[\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{dep}}\textcolor[rgb]{0,0,1}{=}[\textcolor[rgb]{0,0,1}{\mathrm{\eta }}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{\xi }}]
\mathrm{VectorSpaceSum}\left(S,\mathrm{S1},\mathrm{depname}=[\mathrm{\phi },\mathrm{\psi }]\right)
[{\textcolor[rgb]{0,0,1}{\mathrm{\phi }}}_{\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{\mathrm{\phi }}}_{\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{\mathrm{\psi }}}_{\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{\mathrm{\psi }}}_{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{indep}}\textcolor[rgb]{0,0,1}{=}[\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{dep}}\textcolor[rgb]{0,0,1}{=}[\textcolor[rgb]{0,0,1}{\mathrm{\phi }}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{\psi }}]
The VectorSpaceSum command was introduced in Maple 2020.
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Depreciation Accounting | EME 801: Energy Markets, Policy, and Regulation
The term "depreciation" usually refers to the physical degradation of some capital asset, like wear and tear. My car, for example, doesn't drive as smoothly with 120,000 miles on it as it did when it had 12,000 miles. As some piece of capital gets older, you might naturally expect it to be worth less, either because it requires more maintenance or because it cannot be resold for as high a price. (Though the concept of "value" of physical plant can be a bit tricky, as we will learn in the next section.) It is often said that the value of a new car drops by 20% as soon as it is driven off the lot, even though it's basically the same vehicle that was bought for a new-car price. This illustrates the difference that can sometimes arise between physical depreciation and the depreciation in a capital asset's store of value.
Depreciation of an asset's store of value has substantial implications for the financial analysis of energy projects. You might recall from Lesson 5 that the profits of a regulated public utility are determined in large part by its total stock of non-depreciated capital. Who determines the rate at which a power plant, substation or other asset depreciates in value? Similarly, tax authorities in many countries allow companies to "write off" the depreciated value of assets when calculating their total income on which they are subject to paying tax. The term "write off" here refers to a deduction from total taxable income, rather than a deduction in the total tax bill, per se. For example, if you can claim that some capital asset has depreciated in value by $100 over the course of some year, and the tax rate is 35%, then that $100 asset depreciation will ultimately lower your tax bill by $35 (35% of $100), not by $100. This is the difference between a "tax deduction" and a "tax credit." Tax credits will appear later in the course when we discuss financial subsidies for energy projects.
The rate at which an asset is financially depreciated for tax, regulatory, or other financial purposes may be very different than the rate at which the asset actually physically depreciates in value. It is even possible that an asset could be treated as completely depreciated in the eyes of a regulator or the tax authority, yet could still be generating a lot of value for its owner. Many power plants in the U.S., for example, are several decades old - well beyond their intended 30 to 40 year life spans. These power plants are mostly considered to be depreciated assets, yet some continue to be highly profitable, selling electricity into high-priced markets.
Depreciation allowances are usually determined by the regulator, tax authority or other relevant oversight body in order to allow the owners of depreciable capital assets to recover the costs of those assets through a series of tax deductions or other gains over the course of some number of years. The idea here is to encourage investment by allowing companies to use investment vehicles to reduce their tax burden. Our discussion here will focus on depreciation allowances that are allowed by tax authorities, since those allowances are ultimately the most important for development of energy project financial statements. The U.S. Internal Revenue Service, if you happen to be interested, maintains a mind-bogglingly complex list of types of property that are eligible for different depreciation schedules and methods. Here is an "introduction" to depreciation from the IRS.
A depreciation schedule lists the percentage of the original (so-called "book") value of a piece of property that can be claimed as a depreciation allowance for tax reporting purposes. Broadly, there are two types of depreciation schedules: straight-line depreciation and so-called accelerated depreciation methods.
Before we get into the mechanics of depreciation, we need to develop some notation.
P = the up-front cost of the asset;
F = the salvage value of the asset at the end of its useful life;
N = the number of years over which the asset will be depreciated;
A(t) = the depreciation allowance for year t (t = 1,…, N), in percentage terms, such that
\sum _{t=1}^{N}A\left(t\right)=1
(i.e., the asset must be fully depreciated over N years);
D(t) = the depreciation allowance for year t (t = 1,…, N), in dollar terms, so that D(t)=A(t)×P;
B(t) = the remaining book value of the asset after year t.
Based on these definitions, we can immediately see that B(t) is just equal to the original book value (P) less all of the cumulative depreciation allowances from year 1 to year t. In mathematical terms,
B\left(t\right)=P-\sum _{k=1}^{t}D\left(t\right)
Table 8.5 provides the mathematical formulas for some common depreciation methods. Note that Modified Accelerated Cost Recovery Systems (MACRS), which have become more commonplace, are not included in Table 8.5 but will be discussed below.
Table 8.5: Depreciation formulas
D\left(t\right)=\left(P-F\right)/N
B\left(t\right)=P-\left(\frac{P-F}{N}\right)×t
Sum of the Year's Digits
D\left(t\right)=\frac{N-t+1}{\left[N\left(N+1\right)\right]/2}\left(P-F\right)
B\left(t\right)=\left(P-F\right)\left(\frac{N-t}{N}\right)\left(\frac{N-t+1}{N+1}\right)+F
D\left(t\right)=\beta P{\left(1-\beta \right)}^{t-1}
B\left(t\right)=P{\left(1-\beta \right)}^{t}
\beta =1-{\left(F/P\right)}^{1/N}
The most straightforward depreciation method is straight-line depreciation. Under straight-line depreciation, the book value of an asset (less its salvage value, if any) can be depreciated evenly over some number of years. For example, if you had an asset with a book value of $1,000; no salvage value; and a ten-year depreciation horizon, you could claim $100 each year for ten years as a depreciation expense and tax deduction.
The other three depreciation methods that we will discuss here - sum of the year's digits, declining balance, and MACRS - are all forms of "accelerated depreciation." Under accelerated depreciation systems, a larger proportion of the asset's book value is allowed to be depreciated in the earlier years of its use, with smaller proportions depreciated in later years of use. This allows the asset owner to enjoy a lower tax burden earlier in the asset's life. Other things being equal, this leads to higher profits in the years immediately following investment. Accelerated depreciation can substantially affect the value of an asset to its owner; we will see in Lesson 9 just how this re-allocation of tax burden and profits across the useful life of an asset increases the asset's lifetime benefit to its owner.
The three accelerated depreciation methods that we will illustrate in this lesson are:
Sum of the Year's Digits (SYD): SYD is best illustrated using a simple example. Suppose that an asset could be depreciated over five years. Then the sum of the digits would be 1+2+3+4+5 = 15. The first year, you could claim 5/15 = 33% of the asset's book value as a depreciation expense, so that A(1) = 33%. The second year, the depreciation allowance would be A(2) = 4/15 = 27%. You can verify for yourself that A(3) = 20%; A(4) = 13%; and A(5) = 7%.
vDeclining balance: This could also be called "exponential depreciation" since it depreciates an asset at a constant rate, rather than a constant amount. If an asset is depreciated using the declining balance method over a fixed number of years, the residual book value of the asset is used as the depreciation allowance in the final year. For example, if you had a $100 asset with no salvage value, and were claiming depreciation according to the declining balance system at 25% per year over five years, in the first year your depreciation allowance would be D(1) = 0.25 × $100 = $25. In the second year, you would have D(2) = 0.25 × ($100 - $25) = 0.25 × $75 = $18.75. You can verify for yourself that D(3) = $14.06; D(4) = $10.55, and D(5) = $7.91.
Modified Accelerated Cost Recovery: MACRS, popular in the United States, is embodied in a series of depreciation tables published by the U.S. Internal Revenue Service. The tables dictate the values of A(t) to be used each year, and also describe which types of assets are eligible for MACRS and over how many years. In other words, different asset types have different values of N and different depreciation schedules A(t). One thing to note about MACRS is that the N-year depreciation schedules actually cover N+1 years (so, for example, five-year MACRS allows depreciation over six years).
As a means of comparison between all of these methods, let's take a hypothetical asset with a book value of $1,000 and zero salvage value, and depreciate that asset over a ten year time horizon. Table 8.6 and Figure 8.1 show the values of B(t) during each year for each of the four methods. For declining balance, we will use 25% per year. For MACRS we are using the 10-year table in Appendix 1 of Publication 496.
As an exercise, see if you can reproduce the table and the figure.
Table 8.6: Data for Figure 8.1 (Note that year 11 is left out for MACRS)
$ --- $ --- $32.70 $ ---
Figure 8.1: Book value B(t) of a $1,000 asset with four different depreciation methods.
‹ The Basic Language of Accounting up Book versus mark-to-market valuation - or - how Enron gave good economic logic a bad name ›
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Wave Optics Physics NEET Practice Questions, MCQs, Past Year Questions (PYQs), NCERT Questions, Question Bank, Class 11 and Class 12 Questions, and PDF solved with answers
Two waves of intensity ratio 9:1 interfere to produce fringes in a young's double-slit experiment, the ratio of intensity at maxima to the intensity at minima is
Subtopic: Superposition Principle |
Two polaroids
{P}_{1} and {P}_{2}
are placed with their axis perpendicular to each other. Unpolarised light
{I}_{o}
is incident on
{P}_{1}
. A third polaroid
{P}_{3}
is kept in between
{P}_{1} and {P}_{2}
such that its axis makes an angle
{45}^{°}
{P}_{1}
. The intensity of transmitted light through
{P}_{2}
\frac{{I}_{o}}{2}
\frac{{I}_{o}}{4}
\frac{{I}_{o}}{8}
\frac{{I}_{o}}{16}
Subtopic: Polarization of Light |
The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the ratio
\frac{{I}_{max}-{I}_{min}}{{I}_{max}+{I}_{min}}
\frac{\sqrt{n}}{n+1}
\frac{2\sqrt{n}}{n+1}
\frac{\sqrt{n}}{{\left(n+1\right)}^{2}}
\frac{2\sqrt{n}}{{\left(n+1\right)}^{2}}
A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length 60 cm. The aperture is illuminated normally by a parallel beam of wavelength
5×{10}^{-5}
cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is
Subtopic: Diffraction |
The intensity at the maximum in Young's double-slit experiment is
{\mathrm{I}}_{0}
when the distance between two slits is d=5
\lambda
\lambda
is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D= 10 d?
\frac{{\mathrm{I}}_{0}}{4}
\frac{3}{4}{\mathrm{I}}_{0}
\frac{{\mathrm{I}}_{0}}{2}
{\mathrm{I}}_{0}
Subtopic: Young's Double Slit Experiment |
In a diffraction pattern due to a single slit of width a,the first minimum is observed at an angle
{30}^{\circ }
when light of wavelength 5000
\stackrel{˙}{A}
is incident on the slit. The first secondary maximum is observed at an angle of
{\mathrm{sin}}^{-1}\left(\frac{2}{3}\right)
{\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)
{\mathrm{sin}}^{-1}\left(\frac{3}{4}\right)
{\mathrm{sin}}^{-1}\left(\frac{1}{4}\right)
For a parallel beam of monochromatic light of wavelength diffraction is produced by a single slit whose width 'a' is of the order of the wavelength of the light. If 'D' is the distance of the screen from the slit, the width of the central maxima will be
(1)2Dλ/a
(2)Dλ/a
(3)Da/λ
(4)2Da/λ
In a double-slit experiment, the two slits are 1 mm apart and the screen is placed 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double-slit within the central maxima of a single-slit pattern?
From NCERT NEET - 2015
Two slits in Young's experiment have widths in the ratio of 1: 25. The ratio of intensity at the maxima and minima in the interference pattern Imax/Imin is:
At the first minimum adjacent to the central maximum of a single slit diffraction pattern, the phase difference between the Huygen's wavelet from the edge of the slit and the wavelet from the midpoint of the slit is
(1) π/4 radian
(3) π radian
|
Magnetism And Matter, Popular Questions: ICSE Class 12-science PHYSICS, Physics Part I - Meritnation
Define magnetic elements of earth?
if a bar magnet is kept antiparallel to the earth's magnetic field, then 2 neutral points are obtained in the horizontal plane containing the bar magnet ..explain...
Smitha Naik asked a question
WAT IS RETENTIVITY AND COERCIVITY
K Karthik asked a question
what is the si unit of permeability. what do Tm/A stands for?
what is the principle of moving coil galvanometer?
the critical angle for glass air interface is ic. will the critical angle for glass water interface be greater than or less than ic ? why?
Anusha Krishnan asked a question
Why do magnetic field lines always form closed loops?
A 2MeV proton is moving perpendicular to a uniform magnetic field of 2.5 Tesla the force on the proton is ?
\left(1\right) 1 : \sqrt{2} : 2\sqrt{2} \left(2\right) 1 : \sqrt{2} : \sqrt{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(3\right) 1 : 2 : 2\sqrt{2} \left(4\right) \sqrt{2} : 2 : 1
what is difference between diamagnetic paramagnetic and ferromagnetic?atleast 10 points.
Diya Khandelwal asked a question
Aishwarya Kannan asked a question
what are eddy currents and its applications???
A square current carrying coil of edge length L.the magnetic field on the coil given by Bvector=B0y/L icap +B0x/L jcap where B0 is a positive constant.(A is area of coil).
1) If the coil is free to rotate about x axis torque on coil is given by 1/2iAB0icap
2)if the coil is free to rotate about y axis torque on the coil is given by -1/2iAB0j cap
3) resultant force on coil is zero
4)equation for torque μ vector cross B vector where μ is magnetic moment of coil is not valid on coil if any of the side is fixed as axis
The susceptibility of a magnetic material is 0.9853. Identify the type of the magnetic material.Draw the modification of field pattern on keeping piece of this material in a uniform magnetic field.
Varun Singhal & 1 other asked a question
why is calcium listed as paramagnetic even though it has no unpaired electron ??
Derive expression for magnetic dipole moment of an electron revolving around a nucleus
A dip ciccle shows an apparent dip of 60 degree at a place where the true dip is 45 degree. if the dip circle is rotated throgh 90 degree, what apparent dip will it show? first of all please tell me what is the meaing of apparent dip.how can a dip be apparent? it should always be true dip?
Give 5 differences between a bar magnet and a solenoid with their diagrams?
Shreya K Hegde asked a question
a current of 5 ampere is passed through a straight wire of length 6 CM then the magnetic field at a point 5 cm from the other end of the wire is
Karthik Deep asked a question
a jet plane is travelling west at 450 ms-1. if the horizontal component of earth's magnetic field at that place is 4 * 10-4Tesla and the angle of dip is 30 degrees, find the emf induced between the ends of the wings having span of 30 m.
Manasvi Parab asked a question
Mainak Chaudhari asked a question
Working of Cyclotron?
what is the difference between magnetic induction and magnetizing field?
force between two identical bar magnets whose centres are r metre apart is 4.8N. when their axes are in the same line..if separation is increased to 2r,the force between them is reduced to
a) 2.4N
c) 0.6N
d) 0.3N
what is the difference between molar specific heat and molar heat capacity
A thin bar magnet is cut into two equal parts.the ratio of moment of inertia to the magnetic moment of each part will become?(as compare to original magnet)
options-1) 1/2
Pratikshya Swain asked a question
what do you mean by current sensitivity of a moving coil galvanometer ?on what factors does it depends?
HELP!!! What is end-on position and broadside on position?
Latika Ghulyani asked a question
The horizontal component of earth 's magnetic field at a place is B and angle of dip is 60 degrees. What is the value of vertical component of earth 's magnetic field at equator?
This question was asked in 2012 boards.I asked ths question yesterday also and the answer given by you is different from that available on this site where question-answers of previous year board papers are given.
there the answer is given to be zero and i don't undestand how. please review the question and explain how the vertical component of earth's magnetic field is zero at equator.
any 3 salient features of hysteresis loop
To increase the current sensitivity of a moving coil galvanometer by 50%, its resistance in increased so that the new resistance becomes twice the initial resistance. By what factor does the voltage sensitivity change?
Q. An insulating ring of radius 1 m, mass 1 kg, heat capacity 1 J/k carrying a uniformly distributed charge of 1 c is lying on a smooth horizontal plane. Magnetic field induction
{10}^{4}
T present in a direction perpendicular to the plane changes at the rate of 1 J/sec. Simultaneously heat is supplied to the ring at the rate of 1 J/sec
Coefficient of linear expansion of the material of ring is
{10}^{-4}
per k. The instanious maximum angular
acceleration of the ring is
1. 1 rad/
{s}^{2}
2. 1.5 rad/
{s}^{2}
{s}^{2}
{s}^{2}
if a magnet is suspended at an angle 30 degree to the magnetic meridian the dip needle makes angle of 45 degree with the horizontal. the real dip is ??
at a certain place the angle of dip is 30 and the horizontal component of earth magnetic field is 0.50 oersted the earth yotal magnetic field is
Alby Abraham asked a question
a magnetising field of 1600 A/m produces a magnetic flux of 2 .4 x 10-5 Wb in a bar of iron of cross section 0 .2 cm2 . Calculate the Permeability and Susceptibility of the bar ?
Q.4. The value of
\left|{T}_{1}-{T}_{2}\right|
, in the given diagram, is
true value of dip at a place is 45 degree.the plane of the dip circle is turned through 60 degree from the magnetic meridian.find apparent value of dip?please explain me the question with the help of figure a and give its solution.
two short magnets apply a force F on each other, if the separation between them is halved then what's the new force between them approximately?
ans.. 16F
The vertical component of earth's magnetic field at a place is root 3 times the horizontal component. What is the value of angle of dip at that place?
Manvi Sambyal asked a question
Two magnets of magnetic moments M and 31/2 M are joined to form a cross. The combination is suspended in a uniform magnetic field B. The magnetic moment M now makes an angle θ with the field direction. find the value of angle θ.
Sakthi Umamaheswari asked a question
the needle of dip circle is the vertical at the magnetic poles.the dip circle is rotated about vertical axis through 90.what will be the position of the needle in the vertical circular scale?
why a galvanometer is not used to measure current in a given circuit?
Solve this: The earth's magnetic field at equator is 0.5 gauss The magnitude of earth's magnetic field at a magnetic colatitude of 30
°
(1) 1 gauss
\sqrt{13}
(2) 0.25
\sqrt{13}
\sqrt{3}
A wheel with 8 metallic spokes each 50cm long is rotated with a speed of 120rev./min . The earth's magnetic field at the place is 0.4G and the angle of dip is 60 . Calculate the emf induced between the axle and the rim of the wheel. How will the value of emf be affected if the number of spokes were increased ?
Kritesh Pandey asked a question
What is a stalloy?
Nikszzz asked a question
two identical charged particles moving with same speed enter a region of uniform magnetic field. if one of these enters normal to the feild direction and the other along a direction at 30'c with the field , what would be the ratio of their angular frequencies?
Dashing Sonu Kumar asked a question
Find asap of this problem
Draw the graph of magnetic susceptibility of ferromagnetic, diamagnetic and paramagnetic substances with respect to temperature
Swarinkesh Mayank Das asked a question
A magnet weighs 75gm and its magnetic moment is 2000cgs units .If the density material is 7.5g/cc,calculate the intensity of magnetisation
Raman Eela & 1 other asked a question
in what respect is a toroid different from a solenoid ? draw and compare the pattern of magnetic field lines in the 2 cases
How does the 1) pole strength and 2) magnetic moment of each part of a bar magnet change If it is cut into two equal pieces 1))transverse
2)) along its length ????
Apparent dips when dip circle is placed in two mutually perpendicular directions are 30 and 45 degree.What is actual dip at that place???
fnies a Current or B5 JLA and gnetic field of strength 0.85 T ing the coil by 180" against the 30 C. 60 N N NEET-2017 (2) 4.55 [AIPMT-2014] (1) (a) (3) (C) A bar magnet of length I and magnetic dipole moment M is bent in the form of an arc as shown in figure. The new magnetic dipole moment will be (2) (b) (4) (d) (4) 1.15 J arent angles of dip observed t right angles to each other, ip 6is given by 7 [NEET-2017] 20 202 2 a thin cotton thread in a netic field and is in gy required to rotate it e required to keep the ,60 [NEET-2013] 2 3
Trishna asked a question
A rod of length L, along East-West direction is dropped from a height H.
if B be the magnetic field due to earth at that place and angle of dip isx, then what is the magnitude of induced emf across two ends of the rodwhen the rod reaches the earth?
A charge Q moving in a straight line is accelerated by a potential difference V.it enter a uniform magnetic field B perpendicular to its path.
Deduse in terms of B an expearence radius of circular path which is travels
The magnetic field through a single loop of wire, 12cm in radius and 8.5 ohm resistance changes with time as shown. The magnetic field is perpendicular to the plane of the loop plot induced current as a function of time.
Panky asked a question
a magnate is suspended in such a way that it oscillates in the horizontal plane .it makes 20 oscillations per minute at a place where dip angle is 30 and 15 oscillations per minute at a place where dip angle is 60.the ratio of total earth magnetic field at the two places is
Rahul Pillai asked a question
Write the expression for the magnetic moment due to a planar square loop of side ‘l’ carrying a steady current I in a vector form.
In the given figure this loop is placed in a horizontal plane near a long straight conductor carrying a steady current I1 at a distance l as shown. Give reason to explain that the loop will experience a net force but no torque. Write the expression for this force acting on the loop.
pls ans:
41. Heat conducting metarial is-
(1) Mercury (2) Water (3) Oil (4) Alcohol.
42. The radii of two spheres mad of same metal are r and 2r. These are heated to the same temperature in the same surrounding. The ratio of rates of decrease of their temperature will be.
(i) 1:1 (ii) 4:1 (iii) 1:4 (iv) 2:1
dalmeida2001... asked a question
a magnetic needle free to rotate in a vertical plane, orients itself vertically at a certain place on earth. what are the values of horizontal component of earth's mag field and angle of dip at that place?
Q18. A simple pendulum has solid bob of relative density 6. It is oscillating inside a non-viscous liquid of relative density 1.2. The time period of small oscillation of this pendulum (assume SHM) is given by
\left(1\right) 2\mathrm{\pi }\sqrt{\frac{61}{5g}} \left(2\right) 2\mathrm{\pi }\sqrt{\frac{5l}{4\mathrm{g}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(3\right) 2\mathrm{\pi }\sqrt{\frac{4 l}{5\mathrm{g}}} \left(4\right) 2\mathrm{\pi }\sqrt{\frac{5 l}{6\mathrm{g}}}
A 5 kg block is sliding on rough horizontal surface with speed 10 m/s at the moment when it just starts compressing a uncompressed spring of spring constant 400 N/m. If kinetic friction force between the block and surface is 10N , then the spring is compressed by nearly
Baidyanath Gangesh asked a question
an element delta l=delta x l is placed at the origin and carriersa current I=2A.find out magnetic field at point P on y axis at distance of 1.0m due to element delta x=1cm.give also direction of field produced?
Sunita Choudhry asked a question
Divisha Sharma asked a question
why does a freely suspended magnet always points along north- south direction. ?
Vikas Giri asked a question
1. Give reason: electric lines of force do not form closed loop while magnetic field lines form closed loop.
2. Why do like poles repel each other and unlike poles attract each pole ? Give satisfactory reason.
Ayesha Ahmed asked a question
A solenoid has a core of a material with relative permeability of 400. The windings of the solenoid are insulated from the core and carry a current of 2A . The no. Of turns per length is 1000 per metre. Calculate magnetising current Im
1) An electron and proton moving parallel to each other in the same direction with equal momentum enter into a uniform magnetic field which is at right angle to thier velocites trace their trajectories in the magnetic field.
Rhythm Tyagi asked a question
Why is diamagnetism independent of temperature?
Horizontal component of earth's magnetic field at a place is root 3 times its vertical component . what is the value of angle of dip at the place ?
A large magnet is broken into two pieces so their lengths are in the ratio 2:1.What will be ratio of their pole strengths ?
A ball of mass m is tied with two strings of equal length on a rod !!! If the rod is rotated with angular velocity w ( omega ) , then ------(1) T1 = T2........(2) T2 is more than T1......(3) T1 is more than T2.......(4) T1 = T2 / 6 ..............(NOTE ----- T1 AND T2 ARE THE TENSIONS OF STRING 1 AND 2 RESPECTIVELY !!! ) KINDLY HELP !!!
the time period of a freely suspended magnet is 4s . if it is broken in length into two equal parts and one part is suspended in the same way then its time period will be
a coil of 40 ohm resistance, 100 turns and radius 6mm is connected to ammeter of resistance 160 ohm. coilis placed perpendicular to the magnetic field. when coil is taken out of the field 32 microcoulomb charge flows through it. the magnetic flux density is:
Koteswararao Bitta asked a question
Q.2. A bucket containing water of depth 15 cm is kept in a lift which is moving vertically upward with an acceleration 2 g. Then the pressure on the bottom of the bucket in kgwt/
c{m}^{2}
Fatima Hira asked a question
distinguish the magnetic properties of dia, para and ferromagnetic substances in term of,
(a) susceptibility
(b) magnetic permeability
(c) coercivity
Hiba Hanan asked a question
A magnetic needle vibrates in a vertical plane parallel to magnetic meridian about a horizontal axis passing through its centre .It's frequency is n .If the plane of oscillation is turned about a vertical axis by 90 degree ,frequency of oscillation in vertical plane will be a) n b) 0 c ) less than n d) more than n
ishitagarg asked a question
two identical loops,one of copper and another of constantan are removed from a magnetic field within same interval of time,in which loop the induced current will be greater ?
What is Dip Circle? Explain its working.
Arvind Subramanian asked a question
Can the rotation of the earth be a cause for magnetism??
Please answer urgently :
26.The following figure shows the variation of intensity of magnetisation versus the applied magnetic field intensity, Hj for two magnetic materials A and B:
Derive the relation ..
tan delta= 2tan lamda, where delta is the angle of dip and lamda is the magnetic latitude
The current sensitivity of a moving coil galvanometer increases by 20% when its resistance is increased by a factor of two. Calculate by what factor, the voltage sensitivity changes?
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Bivariate histogram plot - MATLAB - MathWorks Nordic
Adjust Number of Histogram Bins
Saving and Loading Histogram2 Objects
Bivariate histograms are a type of bar plot for numeric data that group the data into 2-D bins. After you create a Histogram2 object, you can modify aspects of the histogram by changing its property values. This is particularly useful for quickly modifying the properties of the bins or changing the display.
histogram2(X,Y)
histogram2(X,Y,nbins)
histogram2(X,Y,Xedges,Yedges)
histogram2('XBinEdges',Xedges,'YBinEdges',Yedges,'BinCounts',counts)
histogram2(___,Name,Value)
histogram2(ax,___)
h = histogram2(___)
histogram2(X,Y) creates a bivariate histogram plot of X and Y. The histogram2 function uses an automatic binning algorithm that returns bins with a uniform area, chosen to cover the range of elements in X and Y and reveal the underlying shape of the distribution. histogram2 displays the bins as 3-D rectangular bars such that the height of each bar indicates the number of elements in the bin.
histogram2(X,Y,nbins) specifies the number of bins to use in each dimension of the histogram.
histogram2(X,Y,Xedges,Yedges) specifies the edges of the bins in each dimension using the vectors Xedges and Yedges.
histogram2('XBinEdges',Xedges,'YBinEdges',Yedges,'BinCounts',counts) manually specifies the bin counts. histogram2 plots the specified bin counts and does not do any data binning.
histogram2(___,Name,Value) specifies additional options with one or more Name,Value pair arguments using any of the previous syntaxes. For example, you can specify 'BinWidth' and a two-element vector to adjust the width of the bins in each dimension, or 'Normalization' with a valid option ('count', 'probability', 'countdensity', 'pdf', 'cumcount', or 'cdf') to use a different type of normalization. For a list of properties, see Histogram2 Properties.
histogram2(ax,___) plots into the axes specified by ax instead of into the current axes (gca). The option ax can precede any of the input argument combinations in the previous syntaxes.
h = histogram2(___) returns a Histogram2 object. Use this to inspect and adjust properties of the bivariate histogram. For a list of properties, see Histogram2 Properties.
Data to distribute among bins, specified as separate arguments of vectors, matrices, or multidimensional arrays. X and Y must be the same size. If X and Y are not vectors, then histogram2 treats them as single column vectors, X(:) and Y(:), and plots a single histogram.
Corresponding elements in X and Y specify the x and y coordinates of 2-D data points, [X(k),Y(k)]. The data types of X and Y can be different, but histogram2 concatenates these inputs into a single N-by-2 matrix of the dominant data type.
histogram2 ignores all NaN values. Similarly, histogram2 ignores Inf and -Inf values, unless the bin edges explicitly specify Inf or -Inf as a bin edge. Although NaN, Inf, and -Inf values are typically not plotted, they are still included in normalization calculations that include the total number of data elements, such as 'probability'.
If X or Y contain integers of type int64 or uint64 that are larger than flintmax, then it is recommended that you explicitly specify the histogram bin edges.histogram2 automatically bins the input data using double precision, which lacks integer precision for numbers greater than flintmax.
Number of bins in each dimension, specified as a positive scalar integer or two-element vector of positive integers. If you do not specify nbins, then histogram2 automatically calculates how many bins to use based on the values in X and Y.
If nbins is a scalar, then histogram2 uses that many bins in each dimension.
Example: histogram2(X,Y,20) uses 20 bins in each dimension.
Example: histogram2(X,Y,[10 20]) uses 10 bins in the x-dimension and 20 bins in the y-dimension.
Bin counts, specified as a matrix. Use this input to pass bin counts to histogram2 when the bin counts calculation is performed separately and you do not want histogram2 to do any data binning.
counts must be a matrix of size [length(XBinEdges)-1 length(YBinEdges)-1] so that it specifies a bin count for each bin.
Example: histogram2('XBinEdges',-1:1,'YBinEdges',-2:2,'BinCounts',[1 2 3 4; 5 6 7 8])
Axes object. If you do not specify an axes, then the histogram2 function uses the current axes (gca).
Example: histogram2(X,Y,'BinWidth',[5 10])
The properties listed here are only a subset. For a complete list, see Histogram2 Properties.
histogram2 does not always choose the number of bins using these exact formulas. Sometimes the number of bins is adjusted slightly so that the bin edges fall on "nice" numbers.
If you set the NumBins, XBinEdges, YBinEdges, BinWidth, XBinLimits, or YBinLimits properties, then the BinMethod property is set to 'manual'.
Example: histogram2(X,Y,'BinMethod','integers') creates a bivariate histogram with the bins centered on pairs of integers.
If you specify BinWidth, then histogram2 can use a maximum of 1024 bins (210) along each dimension. If instead the specified bin width requires more bins, then histogram2 uses a larger bin width corresponding to the maximum number of bins.
Example: histogram2(X,Y,'BinWidth',[5 10]) uses bins with size 5 in the x-dimension and size 10 in the y-dimension.
'bar3' (default) | 'tile'
Histogram display style, specified as either 'bar3' or 'tile'. Specify 'tile' to display the histogram as a rectangular array of tiles with colors indicating the bin values.
The default value of 'bar3' displays the histogram using 3-D bars.
Example: histogram2(X,Y,'DisplayStyle','tile') plots the histogram as a rectangular array of tiles.
Example: histogram2(X,Y,'EdgeAlpha',0.5) creates a bivariate histogram plot with semi-transparent bar edges.
[0.15 0.15 0.15] (default) | 'none' | 'auto' | RGB triplet | hexadecimal color code | color name
Example: histogram2(X,Y,'EdgeColor','r') creates a 3-D histogram plot with red bar edges.
Transparency of histogram bars, specified as a scalar value between 0 and 1 inclusive. histogram2 uses the same transparency for all the bars of the histogram. A value of 1 means fully opaque and 0 means completely transparent (invisible).
Example: histogram2(X,Y,'FaceAlpha',0.5) creates a bivariate histogram plot with semi-transparent bars.
'auto' (default) | 'flat' | 'none' | RGB triplet | hexadecimal color code | color name
'flat' — Bar colors vary with height. Bars with different height have different colors. The colors are selected from the figure or axes colormap.
'auto' — Bar color is chosen automatically (default).
If you specify DisplayStyle as 'stairs', then histogram2 does not use the FaceColor property.
Example: histogram2(X,Y,'FaceColor','g') creates a 3-D histogram plot with green bars.
FaceLighting — Lighting effect on histogram bars
'lit' (default) | 'flat' | 'none'
Lighting effect on histogram bars, specified as one of the values in this table.
'lit'
Histogram bars display a pseudo-lighting effect, where the sides of the bars use darker colors relative to the tops. The bars are unaffected by other light sources in the axes.
This is the default value when DisplayStyle is 'bar3'.
Histogram bars are not lit automatically. In the presence of other light objects, the lighting effect is uniform across the bar faces.
Histogram bars are not lit automatically, and lights do not affect the histogram bars.
FaceLighting can only be 'none' when DisplayStyle is 'tile'.
Example: histogram2(X,Y,'FaceLighting','none') turns off the lighting of the histogram bars.
{v}_{i}
{c}_{i}
{A}_{i}={w}_{xi}\cdot {w}_{yi}
N
{v}_{i}={c}_{i}
Sum of bin values is less than or equal to numel(X) and numel(y). The sum is less than numel(X) and numel(y) only when some of the input data is not included in the bins.
{v}_{i}=\frac{{c}_{i}}{{A}_{i}}
The volume (height * area) of each bar is the number of observations in the bin. The sum of the bar volumes is less than or equal to numel(X) and numel(y).
{v}_{i}=\sum _{j=1}^{i}{c}_{j}
The height of the last bar is less than or equal to numel(X) and numel(Y).
{v}_{i}=\frac{{c}_{i}}{N}
{v}_{i}=\frac{{c}_{i}}{N\cdot {A}_{i}}
The volume of each bar is the relative number of observations. The sum of the bar volumes is less than or equal to 1.
{v}_{i}=\sum _{j=1}^{i}\text{\hspace{0.17em}}\frac{{c}_{j}}{N}
The height of each bar is equal to the cumulative relative number of observations in each bin and all previous bins in both the x and y dimensions. The height of the last bar is less than or equal to 1.
Example: histogram2(X,Y,'Normalization','pdf') plots an estimate of the probability density function for X and Y.
ShowEmptyBins — Toggle display of empty bins
Toggle display of empty bins, specified as either 'off' or 'on'. The default value is 'off'.
Example: histogram2(X,Y,'ShowEmptyBins','on') turns on the display of empty bins.
histogram2 only plots data that falls within the bin limits inclusively, Data(Data(:,1)>=xbmin & Data(:,1)<=xbmax).
XBinLimitsMode — Selection mode for bin limits in x-dimension
Selection mode for bin limits in x-dimension, specified as 'auto' or 'manual'. The default value is 'auto', so that the bin limits automatically adjust to the data along the x-axis.
If you explicitly specify either XBinLimits or XBinEdges, then XBinLimitsMode is set automatically to 'manual'. In that case, specify XBinLimitsMode as 'auto' to rescale the bin limits to the data.
histogram2 only plots data that falls within the bin limits inclusively, Data(Data(:,2)>=ybmin & Data(:,2)<=ybmax).
YBinLimitsMode — Selection mode for bin limits in y-dimension
Selection mode for bin limits in y-dimension, specified as 'auto' or 'manual'. The default value is 'auto', so that the bin limits automatically adjust to the data along the y-axis.
If you explicitly specify either YBinLimits or YBinEdges, then YBinLimitsMode is set automatically to 'manual'. In that case, specify YBinLimitsMode as 'auto' to rescale the bin limits to the data.
h — Bivariate histogram
Bivariate histogram, returned as an object. For more information, see Histogram2 Properties.
Histogram2 Properties Histogram2 appearance and behavior
Generate 10,000 pairs of random numbers and create a bivariate histogram. The histogram2 function automatically chooses an appropriate number of bins to cover the range of values in x and y and show the shape of the underlying distribution.
y = randn(10000,1);
h = histogram2(x,y)
Histogram2 with properties:
XBinEdges: [-3.9000 -3.6000 -3.3000 -3 -2.7000 -2.4000 -2.1000 ... ]
YBinEdges: [-4.2000 -3.9000 -3.6000 -3.3000 -3.0000 -2.7000 ... ]
BinWidth: [0.3000 0.3000]
When you specify an output argument to the histogram2 function, it returns a histogram2 object. You can use this object to inspect the properties of the histogram, such as the number of bins or the width of the bins.
Find the number of histogram bins in each dimension.
nXnY = h.NumBins
nXnY = 1×2
Plot a bivariate histogram of 1,000 pairs of random numbers sorted into 25 equally spaced bins, using 5 bins in each dimension.
h = histogram2(x,y,nbins)
Values: [5x5 double]
NumBins: [5 5]
XBinEdges: [-4 -2.4000 -0.8000 0.8000 2.4000 4]
YBinEdges: [-4 -2.4000 -0.8000 0.8000 2.4000 4]
Find the resulting bin counts.
2 40 124 47 4
1 119 341 109 10
Generate 1,000 pairs of random numbers and create a bivariate histogram.
XBinEdges: [-3.5000 -3 -2.5000 -2 -1.5000 -1 -0.5000 0 0.5000 1 ... ]
YBinEdges: [-3.5000 -3 -2.5000 -2 -1.5000 -1 -0.5000 0 0.5000 1 ... ]
Use the morebins function to coarsely adjust the number of bins in the x dimension.
nbins = morebins(h,'x');
nbins = morebins(h,'x')
nbins = 1×2
Use the fewerbins function to adjust the number of bins in the y dimension.
nbins = fewerbins(h,'y');
nbins = fewerbins(h,'y')
Adjust the number of bins at a fine grain level by explicitly setting the number of bins.
Create a bivariate histogram using 1,000 normally distributed random numbers with 12 bins in each dimension. Specify FaceColor as 'flat' to color the histogram bars by height.
h = histogram2(randn(1000,1),randn(1000,1),[12 12],'FaceColor','flat');
Generate random data and plot a bivariate tiled histogram. Display the empty bins by specifying ShowEmptyBins as 'on'.
x = 2*randn(1000,1)+2;
y = 5*randn(1000,1)+3;
h = histogram2(x,y,'DisplayStyle','tile','ShowEmptyBins','on');
Generate 1,000 pairs of random numbers and create a bivariate histogram. Specify the bin edges using two vectors, with infinitely wide bins on the boundary of the histogram to capture all outliers that do not satisfy
|x|<2
Xedges = [-Inf -2:0.4:2 Inf];
Yedges = [-Inf -2:0.4:2 Inf];
h = histogram2(x,y,Xedges,Yedges)
XBinEdges: [-Inf -2 -1.6000 -1.2000 -0.8000 -0.4000 0 0.4000 ... ]
YBinEdges: [-Inf -2 -1.6000 -1.2000 -0.8000 -0.4000 0 0.4000 ... ]
BinWidth: 'nonuniform'
When the bin edges are infinite, histogram2 displays each outlier bin (along the boundary of the histogram) as being double the width of the bin next to it.
Specify the Normalization property as 'countdensity' to remove the bins containing the outliers. Now the volume of each bin represents the frequency of observations in that interval.
Generate 1,000 pairs of random numbers and create a bivariate histogram using the 'probability' normalization.
h = histogram2(x,y,'Normalization','probability')
Compute the total sum of the bar heights. With this normalization, the height of each bar is equal to the probability of selecting an observation within that bin interval, and the heights of all of the bars sum to 1.
S = sum(h.Values(:))
Generate 1,000 pairs of random numbers and create a bivariate histogram. Return the histogram object to adjust the properties of the histogram without recreating the entire plot.
Color the histogram bars by height.
Change the number of bins in each direction.
Display the histogram as a tile plot.
h.DisplayStyle = 'tile';
Use the savefig function to save a histogram2 figure.
histogram2(randn(100,1),randn(100,1));
savefig('histogram2.fig');
h = openfig('histogram2.fig');
y = findobj(h,'type','histogram2')
XBinEdges: [-3 -2 -1 0 1 2 3 4]
YBinEdges: [-3 -2 -1 0 1 2 3]
BinWidth: [1 1]
Histogram plots created using histogram2 have a context menu in plot edit mode that enables interactive manipulations in the figure window. For example, you can use the context menu to interactively change the number of bins, align multiple histograms, or change the display order.
'XBinLimits'
'YBinLimits'
bar3 | discretize | fewerbins | morebins | histcounts2 | histcounts | Histogram2 Properties
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Establishing the minimum number of guesses needed to (always) win Wordle – Alex Peattie
Me Projects Talks Blog
Establishing the minimum number of guesses needed to (always) win Wordle
A few weeks ago, I became interested in whether there was a strategy to always “win” Wordle (i.e. to find the secret word in 6 guesses or fewer). This is exactly the problem that Laurent Poirrier examines in his excellent writeup on applying mathematical optimization techniques to Wordle:
Is there a strategy that guarantees to find any one of the 12972 possible words1 within the 6 allowed guesses? Without resorting to luck, that is.
Laurent proved the answer is yes! With careful thought, some clever optimization techniques and over a thousand hours of CPU time, he found a decision tree of depth 5 — yielding a strategy to solve Wordle puzzles in
\leq
6 guesses. (Before reading the rest of this article, I’d recommend going through Laurent’s post — it’s quite accessible even if you don’t have a background in optimization).
He observes at the end of his article that an open question remains:
Unfortunately, depth 4 seems to be beyond the reach of my computational resources. It is thus still unknown whether all Wordle puzzles can be solved in 5 guesses.2
It’s tricky to apply Laurent’s original strategy to tackling this question — he estimates it would cost about $80,000 of EC2 spend (!) to establish the presence/absence of a depth 4 decision tree. Luckily, there’s a cheaper way to solve the mystery. Below I outline why all Wordle puzzles cannot be solved in 5 guesses or fewer — thereby establishing 6 as the minimum number of guesses needed to guarantee a win.
There are 19 words ending in -ills which differ by only one letter. For a strategy that guarantees a win in 5 guesses to exist, we’ll need to be able to guess 4 words which contain 18 of the 19 starting letters of the -ills words — but no such 4 words exist (provable using exhaustive search or a SAT solver).
Introducing “ill Wordle” 🤒
Let’s start by introducing a Wordle variant we’ll call “ill Wordle”. It’s just like regular Wordle, with one key difference. In regular Wordle we’re faced with thousands of possibilities for the “secret” word, but in ill Wordle it can be one of only 19 possibilities — the 19 five-letter words ending in -ills:
bills, cills, dills, fills, gills, hills, jills, kills, lills, mills,
nills, pills, rills, sills, tills, vills, wills, yills, zills
Observation 1: ill Wordle represents a subset of the problem space of regular Wordle. If we can’t guarantee a win in ill Wordle in
\leq
5 guesses, the same is true for regular Wordle. In regular Wordle there are just more possibilities to disambiguate between, necessitating a decision tree of equal or greater depth.
Although ill Wordle is “easier” from an optimization standpoint, it’s a surprisingly difficult for human players — have a go below3:
Winning ill Wordle in 6 moves
A quick note on the terminology I’m using, since “answer” can be ambiguous when discussing Wordle, I’ll stick to the terms “secret word” or “secret” when referring to the word we’re aiming to guess.
Let’s think about why ill Wordle is tough. The possible secrets only vary by a single letter (the letter in the first position). Consider the set
\mathit{L}
of these first letters of the 19 possible secrets:
\mathit{L} = \{b, c, d, f, g, h, j, k, l, m, n, p, r, s, t, v, w, y, z\}
. A single guess can contain up to 5 letters from this set, and will eliminate up to 5 of the 19 possible secrets. For example, if I guess “nymph” and get the response ⬜️⬜️⬜️⬜️⬜️, I can eliminate nills, yills, mills, pills and hills.
If I’m shooting for victory in 6 turns, I need to choose my first 5 guesses carefully. If those guesses don’t contain at least 18 the 19 letters in
\mathit{L}
then I face having to take my last guess with multiple possibilities still in play:
Observation 2a: To guarantee a win in 6 guesses, we need to equivalently guarantee that we can eliminate 18 of the 19 letters in
\textit{L}
in 5 guesses.
Note that the order of the guesses doesn’t really matter — we’re principally interested in how many letters we’ve eliminated by our penultimate guess.
Can we find 5 guesses which contain 18 of the letters in
\mathit{L}
? The problem is actually quite subtle4 (and worth going into in another post) but the short answer is that plenty of solutions exist. One solution would be: bhang, fjord, limbo, spitz, wicky. We can get another by running Laurent’s solver against ill Wordle:
Can we guarantee a win in ill Wordle in 5 moves?
We’ve seen that we can guarantee a win in ill Wordle in 6 guesses or fewer. This should come as no surprise, since ill Wordle is strictly easier than Wordle, and Laurent had already proven that regular Wordle is winnable in
\leq
6 guesses. Let’s turn our attention to the question of
\leq
5 guesses.
Observation 2b: To guarantee a win in 5 guesses, we need to equivalently guarantee that we can eliminate 18 of the 19 letters in
\textit{L}
Immediately, it’s clear this is much trickier. Firstly, to eliminate 18 letters in 4 guesses, we’ll need at least two of those guesses to eliminate 5 letters each. Since
\mathit{L}
contains no vowels, such guesses are scarce. Let’s find all the candidates:
# Get Wordle wordlist with `curl -O https://gist.githubusercontent.com/alexpeattie/777a393caf13c2e47a12e3d15ac31438/raw/8c989737a308ed22a029a061a2b628b7b68d4f8b/wordle-12k.txt`
words = open("wordle-12k.txt", "r").read().splitlines()
letters = set([w[0] for w in words if w.endswith("ills")])
assert len(letters) == 19
five_letter_eliminating_words = [word for word in words if len(set(letters).intersection(word)) == 5]
print(five_letter_eliminating_words)
'byrls', 'chynd', 'crwth', 'crypt', 'fyrds', 'glyph', 'grypt', 'gymps', 'hwyls',
'hymns', 'kydst', 'kynds', 'lymph', 'lynch', 'myths', 'nymph', 'psych', 'rynds',
'sylph', 'synch', 'synth', 'tryps', 'tymps', 'wynds'
Additionally, these two guesses cannot share any letters (otherwise the latter guess will eliminate
\lt 5
letters). This reduces our possible pool of guesses even more:
ten_letter_eliminating_pairs = [
(w1, w2) for w1 in five_letter_eliminating_words for w2 in five_letter_eliminating_words
if set(w1).isdisjoint(w2) and w1 < w2]
print(ten_letter_eliminating_pairs)
[('crwth', 'gymps'), ('crwth', 'kynds')]
So two of our four guesses would have to be crwth and gymps, or crwth and kynds — each eliminating 10 letters, leaving us to find 2 more words which eliminate 8 additional letters. If
\mathit{W}
is the set of all 12,972 possible Wordle guesses, then the candidates we need to consider can be described by the Cartesian product:
\{\text{crwth}\} \times \{\text{gymps, kynds}\} \times \mathit{W} \times \mathit{W}
It’s fast enough to brute-force all these possibilities:
elim_count1 = len(letters.intersection('crwthgymps' + w3 + w4))
elim_count2 = len(letters.intersection('crwthkynds' + w3 + w4))
if elim_count1 >= 18 or elim_count2 >= 18:
print((w3, w4))
print(f"{time.perf_counter() - start:0.2f} seconds")
Therefore, there isn’t a length-4 subset of
\mathit{W}
which eliminates 18 of 19 letters in
\mathit{L}
. Thus, ill Wordle can’t always be solved in
\leq 5
guesses, and the same must be true of regular Wordle. Wordle cannot be always solved in 5 guesses or fewer.
Isaac Grosof astutely points out in the comments that “kynds” and “gymps” don’t actually eliminate 5 possibilities. We can’t eliminate “sills”, since both words will just give us a green tile on the last letter “s” (which will be the case for all -ills words). So the proof can be even simpler: no pair of words eliminates 10 possibilities, thus a guaranteed win in 5 guesses is impossible. Thanks Isaac!
Checking our conclusion with OR-Tools
We can also frame the problem of finding the minimum number of guesses needed to eliminate 18 of 19 letters in
\mathit{L}
as a constraint programming (CP) problem, which then allows us to use sophisticated SAT solvers like those provided by Google’s OR-Tools. The details are beyond the scope of this article, but you can check out this Colab if you’re interested. At a high level, we’re telling the solver to minimize the number of words chosen, while ensuring we eliminate enough letters:
word_chosen = [model.NewIntVar(0, 1, "word_chosen[%i]" % i) for i in range(num_words)]
num_words_chosen = model.NewIntVar(0, num_words, "num_words_chosen")
# Number of chosen words (to minimize)
model.Add(num_words_chosen == sum(word_chosen))
# Check at least one word starts with 'l' or 's'
model.Add(sum([word_chosen[j] * starting_l[j] for j in range(num_words)]) > 0)
model.Add(sum([word_chosen[j] * starting_s[j] for j in range(num_words)]) > 0)
# The crucial constraint: we must cover at least 18 of the 19 letters we're concerned with.
model.Add(sum([letter_counts[i] for i in range(num_letters)]) >= 18)
model.Minimize(num_words_chosen)
Minimum number of words needed: 5
Example words: bhang, fjord, limbo, spitz, wicky
Solved in: 4.36 seconds
OR-Tools quickly confirms the minimum number of guesses needed is 55.
The future of Wordle solving
So where does that leave us? I think most of the big questions regarding Wordle have been answered, at least in its classic form (using 5 letter English words from the current allowlist). A depth 5 decision tree has been found already by Laurent, and it seems a depth 4 tree cannot exist. It’s nice that the number of guesses Wordle gives you is the smallest “fair” number (where fair means that victory is guaranteed with optimal play).
It could still be interesting for someone to crunch through all the depth 5 trees to find the one which minimizes the average number of guesses6 (I believe at the point Wordle could be declared well and truly solved 😁!).
Thanks (again) to Laurent Poirrier for checking this proof. He also independently verified it with an MIP formulation (in LP format) which is available here. He’s also written a great round-up post on the state of the art in Wordle-solving.
In both this article & Laurent’s we assume that the secret word can be any of the 12,972 words which are accepted as guesses. This list seems to be the 5 letter words from the 2019 Collins Scrabble Words list. Wordle actually chooses its word of the day from a smaller subset of these words — around 2,500 words from the list which the partner of Wordle’s creator recognized (source). This is presumably to prevent outrage from players having to guess words like ”yrapt“.↩
Laurent’s work already establishes that we can’t guarantee a solution in 4 or fewer guesses, leaving 5 or 6 as possibilities for the minimum number of guesses needed.↩
The widget is largely based on Evan You’s implementation 💚↩
It turns out to be the set cover problem in disguise, which is NP-hard.↩
OR-Tools also tells us the maximum number of letters in
\mathit{L}
we can get with 4 guesses: 16 with judgy, lymph, scarf and twank.↩
I haven’t thought much about this, but I suspect doing this would be similarly expensive to an exhaustive search for a depth 4 tree — i.e. the $80k cost Laurent ballparks.↩
This website is open source, you can view the source code for this page. Found a mistake? Raise an issue.
© 2022, Alex Peattie
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Catalysts | Free Full-Text | Ti2O3/TiO2-Assisted Solar Photocatalytic Degradation of 4-tert-Butylphenol in Water
Quarry Residue: Treatment of Industrial Effluent Containing Dye
Mergenbayeva, S.
Sh. Atabaev, T.
Poulopoulos, S. G.
Ti2O3/TiO2-Assisted Solar Photocatalytic Degradation of 4-tert-Butylphenol in Water
Saule Mergenbayeva
Stavros G. Poulopoulos
Department of Chemical and Materials Engineering, School of Engineering and Digital Sciences, Nazarbayev University, 53 Kabanbay Batyr Ave., Nur-Sultan 010000, Kazakhstan
Department of Chemistry, School of Sciences and Humanities, Nazarbayev University, Nur-Sultan 010000, Kazakhstan
Academic Editors: Gassan Hodaifa and Rafael Borja
(This article belongs to the Special Issue Photocatalysis in the Wastewater Treatment)
Colored Ti2O3 and Ti2O3/TiO2 (mTiO) catalysts were prepared by the thermal treatment method. The effects of treatment temperature on the structure, surface area, morphology and optical properties of the as-prepared samples were investigated by XRD, BET, SEM, TEM, Raman and UV–VIS spectroscopies. Phase transformation from Ti2O3 to TiO2 rutile and TiO2 anatase to TiO2 rutile increased with increasing treatment temperatures. The photocatalytic activities of thermally treated Ti2O3 and mTiO were evaluated in the photodegradation of 4-tert-butylphenol (4-t-BP) under solar light irradiation. mTiO heated at 650 °C exhibited the highest photocatalytic activity for the degradation and mineralization of 4-t-BP, being approximately 89.8% and 52.4%, respectively, after 150 min of irradiation. The effects of various water constituents, including anions (
{\mathrm{CO}}_{3}^{2-}
{\mathrm{NO}}_{3}
\mathrm{Cl}
{\mathrm{HCO}}_{3}^{-}
) and humic acid (HA), on the photocatalytic activity of mTiO-650 were evaluated. The results showed that the presence of carbonate and nitrate ions inhibited 4-t-BP photodegradation, while chloride and bicarbonate ions enhanced the photodegradation of 4-t-BP. As for HA, its effect on the degradation of 4-t-BP was dependent on the concentration. A low concentration of HA (1 mg/L) promoted the degradation of 4-t-BP from 89.8% to 92.4% by mTiO-650, but higher concentrations of HA (5 mg/L and 10 mg/L) had a negative effect. View Full-Text
Keywords: 4-tert-butylphenol; solar photocatalysis; Ti2O3/TiO2; degradation; mineralization 4-tert-butylphenol; solar photocatalysis; Ti2O3/TiO2; degradation; mineralization
Mergenbayeva, S.; Sh. Atabaev, T.; Poulopoulos, S.G. Ti2O3/TiO2-Assisted Solar Photocatalytic Degradation of 4-tert-Butylphenol in Water. Catalysts 2021, 11, 1379. https://doi.org/10.3390/catal11111379
Mergenbayeva S, Sh. Atabaev T, Poulopoulos SG. Ti2O3/TiO2-Assisted Solar Photocatalytic Degradation of 4-tert-Butylphenol in Water. Catalysts. 2021; 11(11):1379. https://doi.org/10.3390/catal11111379
Mergenbayeva, Saule, Timur Sh. Atabaev, and Stavros G. Poulopoulos. 2021. "Ti2O3/TiO2-Assisted Solar Photocatalytic Degradation of 4-tert-Butylphenol in Water" Catalysts 11, no. 11: 1379. https://doi.org/10.3390/catal11111379
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1 October 2015 Delocalization of eigenvectors of random matrices with independent entries
Mark Rudelson, Roman Vershynin
We prove that an
n×n
G
with independent entries is completely delocalized. Suppose that the entries of
G
have zero means, variances uniformly bounded below, and a uniform tail decay of exponential type. Then with high probability all unit eigenvectors of
G
have all coordinates of magnitude
O\left({n}^{-1/2}\right)
, modulo logarithmic corrections. This comes as a consequence of a new, geometric approach to delocalization for random matrices.
Mark Rudelson. Roman Vershynin. "Delocalization of eigenvectors of random matrices with independent entries." Duke Math. J. 164 (13) 2507 - 2538, 1 October 2015. https://doi.org/10.1215/00127094-3129809
Keywords: delocalization of eigenvectors , random matrices , rotation-invariant ensembles
Mark Rudelson, Roman Vershynin "Delocalization of eigenvectors of random matrices with independent entries," Duke Mathematical Journal, Duke Math. J. 164(13), 2507-2538, (1 October 2015)
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Refraction through a convex lens — lesson. Science State Board, Class 10.
We studied the converging and diverging nature of spherical lenses. Now let us look at activity that explains the image formed by the concave and convex mirrors.
Take a convex lens.
Draw five parallel straight lines on a long table with chalk, so that the distance between the successive lines equals the focal length of the lens.
Place the lens on a lens stand.
Place it so that the optical centre of the lens is just over the line on the central line.
The two lines on either side of the lens correspond to \(F\) and \(2F\) of the lens, respectively.
Mark them with the appropriate letters, such as \(2\)
{F}_{1}
{F}_{1}
{F}_{2}
{F}_{2}
. Place a burning candle far to the left of \(2\)
{F}_{1}
On a screen on the other side of the lens, get a clear, sharp image. Make a note of the image's nature, location, and relative size.
Place the object behind \(2\)
{F}_{1}
, between F1
{F}_{1}
and \(2\)
{F}_{1}
{F}_{1}
, and between
{F}_{1}
and O to repeat the activity.
Make a list of your observations and tabulate them.
Image formation by convex lenses:
When an object is placed at infinity, a real image is formed at the principal focus. The size of the image is much smaller than that of the object.
When an object is placed behind the centre of curvature (beyond C), a real and inverted image is formed between the centre of curvature and the principal focus. The size of the image is smaller than that of the object.
Object is placed behind the curvature
When an object is placed at the centre of curvature, a real and inverted image is formed at the other centre of curvature. The size of the image is the same as that of the object.
Object is placed at the center of curvature
When an object is placed in between the centre of curvature and principal focus, a real and inverted image is formed behind the centre of curvature. The size of the image is bigger than that of the object.
Object is placed in between the centre of curvature and principal focus
Object is placed at the focus
When an object is placed in between principal focus and optical centre, a virtual image is formed. The size of the image is larger than that of the object.
Object is placed in between principal focus and optical centre
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Linear Regression with TensorFlow.js | Deep Learning for JavaScript Hackers (Part II) | Curiousily - Hacker's Guide to Machine Learning
10.07.2019 — Linear Regression, TensorFlow, Machine Learning, JavaScript — 4 min read
TL;DR Build a Linear Regression model in TensorFlow.js to predict house prices. Learn how to handle categorical data and do feature scaling.
Raining again. It has been 3 weeks since the last time you saw the sun. You’re getting tired of all this cold and unpleasant feeling of loneliness and melancholy.
The voice in your head is getting louder and louder.
Alright, you’re ready to do it. Where to? You remember that you’re nearly broke.
A friend of yours told you about this place Ames, Iowa and it stuck in your head. After a quick search, you found that the weather is pleasant during the year and there is some rain, but not much. Excitement!
Fortunately, you know of this dataset on Kaggle that might help you find out how much your dream house might cost. Let’s get to it!
Run the complete source code for this tutorial right in your browser:
Our data comes from Kaggle’s House Prices: Advanced Regression Techniques challenge.
Here’s a subset of the data we’re going to use for our model:
OverallQual - Rates the overall material and finish of the house (0 - 10)
GrLivArea - Above grade (ground) living area square feet
GarageCars - Size of garage in car capacity
TotalBsmtSF - Total square feet of basement area
FullBath - Full bathrooms above grade
YearBuilt - Original construction date
SalePrice - The property’s sale price in dollars (we’re trying to predict this)
Let’s use Papa Parse to load the training data:
1const prepareData = async () => {
2 const csv = await Papa.parsePromise(
3 "https://raw.githubusercontent.com/curiousily/Linear-Regression-with-TensorFlow-js/master/src/data/housing.csv"
6 return csv.data
1const data = await prepareData()
Let’s build a better understanding of our data. First - the quality score of each house:
Most houses are of average quality, but there are more “good” than “bad” ones.
Let’s see how large are they (that’s what she said):
Most of the houses are within the 1,000 - 2,000 range, and we have some that are bigger.
Let’s have a look at the year they are built:
Even though there are a lot of houses that were built recently, we have a much more widespread distribution.
How related is the year with the price?
Seems like newer houses are pricier, no love for the old and well made then?
Oh ok, but higher quality should equal higher price, right?
Generally yes, but look at quality 10. Some of those are relatively cheap. Any ideas why that might be?
Is a larger house equal higher price?
Seems like it, we might start our price prediction model using the living area!
Linear Regression models assume that there is a linear relationship (can be modeled using a straight line) between a dependent continuous variable
Y
and one or more explanatory (independent) variables
X
In our case, we’re going to use features like living area (X) to predict the sale price (Y) of a house.
Simple Linear Regression is a model that has a single independent variable
X
Y = bX + a
Where a and b are parameters, learned during the training of our model.
X
is the data we’re going to use to train our model,
b
controls the slope and
a
the interception point with the
y
A natural extension of the Simple Linear Regression model is the multivariate one. It is given by:
Y(x_1,x_2,\ldots,x_n) = w_1 x_1 + w_2 x_2 + \ldots + w_n x_n + w_0
x_1, x_2 \ldots, x_n
are features from our dataset and
w_1, w_2 \ldots, w_n
are learned parameters.
We’re going to use Root Mean Squared Error to measure how far our predictions are from the real house prices. It is given by:
RMSE = J(W) = \sqrt{\frac{1}{m} \sum_{i=1}^{m} (y^{(i)} - h_w(x^{(i)}))^2}
where the hypothesis/prediction
h_w
h_w(x) = g(w^Tx)
Currently, our data sits into an array of JS objects. We need to turn it into Tensors and use it for training our model(s). Here is the code for that:
1const createDataSets = (data, features, categoricalFeatures, testSize) => {
2 const X = data.map(r =>
3 features.flatMap(f => {
4 if (categoricalFeatures.has(f)) {
5 return oneHot(!r[f] ? 0 : r[f], VARIABLE_CATEGORY_COUNT[f])
7 return !r[f] ? 0 : r[f]
11 const X_t = normalize(tf.tensor2d(X))
13 const y = tf.tensor(data.map(r => (!r.SalePrice ? 0 : r.SalePrice)))
15 const splitIdx = parseInt((1 - testSize) * data.length, 10)
17 const [xTrain, xTest] = tf.split(X_t, [splitIdx, data.length - splitIdx])
18 const [yTrain, yTest] = tf.split(y, [splitIdx, data.length - splitIdx])
20 return [xTrain, xTest, yTrain, yTest]
We store our features in X and the labels in y. Then we convert the data into Tensors and split it into training and testing datasets.
Some of the features in our dataset are categorical/enumerable. For example, GarageCars can be in the 0-5 range.
Leaving categories represented as integers in our dataset might introduce implicit ordering dependence. Something that does not exist with categorical variables.
We’ll use one-hot encoding from TensorFlow to create an integer vector for each value to break the ordering. First, let’s specify how many different values each category has:
1const VARIABLE_CATEGORY_COUNT = {
2 OverallQual: 10,
3 GarageCars: 5,
4 FullBath: 4,
We’ll use tf.oneHot() to convert individual value to a one-hot representation:
1const oneHot = (val, categoryCount) =>
2 Array.from(tf.oneHot(val, categoryCount).dataSync())
Note that the createDataSets() function accepts a parameter called categoricalFeatures which should be a set. We’ll use this to check whether or not we should process this feature as categorical.
Feature scaling is used to transform the feature values into a (similar) range. Feature scaling will help our model(s) learn faster since we’re using Gradient Descent for training it.
Let’s use one of the simplest method for feature scaling - min-max normalization:
1const normalize = tensor =>
2 tf.div(tf.sub(tensor, tf.min(tensor)), tf.sub(tf.max(tensor), tf.min(tensor)))
this method rescales the range of values in the range of [0, 1].
Now that we know about the Linear Regression model(s), we can try to predict house prices based on the data we have. Let’s start simple:
We’ll wrap the training process in a function that we can reuse for our future model(s):
1const trainLinearModel = async (xTrain, yTrain) => {
trainLinearModel accepts the features and labels for our model. Let’s define a Linear Regression model using TensorFlow:
1const model = tf.sequential()
3model.add(
4 tf.layers.dense({
5 inputShape: [xTrain.shape[1]],
6 units: xTrain.shape[1],
10model.add(tf.layers.dense({ units: 1 }))
Since TensorFlow.js doesn’t offer RMSE loss function, we’ll use MSE and take the square root of that later. We’ll also track Mean Absolute Error (MAE) between the predictions and real prices:
1model.compile({
2 optimizer: tf.train.sgd(0.001),
3 loss: "meanSquaredError",
4 metrics: [tf.metrics.meanAbsoluteError],
Here’s the training process:
1const trainLogs = []
2const lossContainer = document.getElementById("loss-cont")
3const accContainer = document.getElementById("acc-cont")
5await model.fit(xTrain, yTrain, {
6 batchSize: 32,
7 epochs: 100,
8 shuffle: true,
9 validationSplit: 0.1,
10 callbacks: {
11 onEpochEnd: async (epoch, logs) => {
12 trainLogs.push({
13 rmse: Math.sqrt(logs.loss),
14 val_rmse: Math.sqrt(logs.val_loss),
15 mae: logs.meanAbsoluteError,
16 val_mae: logs.val_meanAbsoluteError,
18 tfvis.show.history(lossContainer, trainLogs, ["rmse", "val_rmse"])
19 tfvis.show.history(accContainer, trainLogs, ["mae", "val_mae"])
We train for 100 epochs, shuffle the data beforehand, and use 10% of it for validation. The RMSE and MAE are visualized after each epoch.
Our Simple Linear Regression model is using the GrLivArea feature:
1const [xTrainSimple, xTestSimple, yTrainSimple, yTestIgnored] = createDataSets(
2 data,
3 ["GrLivArea"],
4 new Set(),
8const simpleLinearModel = await trainLinearModel(xTrainSimple, yTrainSimple)
We don’t have categorical features, so we leave that set is empty. Let’s have a look at the performance:
We have a lot more data we haven’t used yet. Let’s see if that will help improve the predictions:
1const features = [
2 "OverallQual",
3 "GrLivArea",
4 "GarageCars",
5 "TotalBsmtSF",
6 "FullBath",
7 "YearBuilt",
10const categoricalFeatures = new Set(["OverallQual", "GarageCars", "FullBath"])
12const [xTrain, xTest, yTrain, yTest] = createDataSets(
13 data,
14 features,
15 categoricalFeatures,
We use all features in our dataset and pass a set of the categorical ones. Did we do better?
Overall, both models are performing at about the same level. This time, increasing the model complexity didn’t give us better accuracy.
Another way to evaluate our models is to check their predictions against the test data. Let’s start with the Simple Linear Regression:
How did adding more data improved the predictions?
Well, it didn’t. Again, having a more complex model trained with more data didn’t provide better performance.
You did it! You built two Linear Regression models that predict house price based on a set of features. You also did:
Feature scaling for faster model training
Convert categorical variables into one-hot representations
Implement RMSE (based on MSE) for accuracy evaluation
Is it time to learn about Neural Networks?
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EUDML | Some surface subgroups survive surgery. EuDML | Some surface subgroups survive surgery.
Some surface subgroups survive surgery.
Cooper, D.; Long, D.D.
Cooper, D., and Long, D.D.. "Some surface subgroups survive surgery.." Geometry & Topology 5 (2001): 347-367. <http://eudml.org/doc/122531>.
author = {Cooper, D., Long, D.D.},
keywords = {Dehn surgery; surface subgroups; hyperbolic 3-manifold},
title = {Some surface subgroups survive surgery.},
TI - Some surface subgroups survive surgery.
KW - Dehn surgery; surface subgroups; hyperbolic 3-manifold
Dehn surgery, surface subgroups, hyperbolic 3-manifold
3
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Volume 15 Issue 1 | Analysis & PDE
Home > Journals > Anal. PDE > Volume 15 > Issue 1
Anal. PDE 15 (1), 1-62, (2022) DOI: 10.2140/apde.2022.15.1
KEYWORDS: determinant lines, renormalization, Quantum field theory, 58J40, 58J50, 58J52, 81T16, 81T20, 58B12
On a compact manifold
M
, we consider the affine space
\mathcal{𝒜}
of non-self-adjoint perturbations of some invertible elliptic operator acting on sections of some Hermitian bundle by some differential operator of lower order.
We construct and classify all complex-analytic functions on the Fréchet space
\mathcal{𝒜}
vanishing exactly over noninvertible elements, having minimal growth at infinity along complex rays in
\mathcal{𝒜}
and which are obtained by local renormalization, a concept coming from quantum field theory, called renormalized determinants. The additive group of local polynomial functionals of finite degrees acts freely and transitively on the space of renormalized determinants. We provide different representations of the renormalized determinants in terms of spectral zeta-determinants, Gaussian free fields, infinite products and renormalized Feynman amplitudes in perturbation theory in position space à la Epstein–Glaser.
Specializing to the case of Dirac operators coupled to vector potentials and reformulating our results in terms of determinant line bundles, we prove our renormalized determinants define some complex-analytic trivializations of some holomorphic line bundle over
\mathcal{𝒜}
. This relates our results to a conjectural picture from some unpublished notes by Quillen from April 1989.
Ismael Bailleul, James Norris
Anal. PDE 15 (1), 63-84, (2022) DOI: 10.2140/apde.2022.15.63
KEYWORDS: sub-Riemannian, heat kernel, diffusion, 35K08, 58J65, 60J60
For incomplete sub-Riemannian manifolds and for an associated second-order hypoelliptic operator, which need not be symmetric, we identify two alternative conditions for the validity of Gaussian-type upper bounds on heat kernels and transition probabilities, with optimal constant in the exponent. Under similar conditions, we obtain the small-time logarithmic asymptotics of the heat kernel and show concentration of diffusion bridge measures near a path of minimal energy. The first condition requires that we consider points whose distance apart is no greater than the sum of their distances to infinity. The second condition requires only that the operator not be too asymmetric.
Geometric averaging operators and nonconcentration inequalities
Anal. PDE 15 (1), 85-122, (2022) DOI: 10.2140/apde.2022.15.85
KEYWORDS: geometric measure theory, geometric invariant theory, Radon-like transforms, 28A75, 44A12
This paper is devoted to a systematic study of certain geometric integral inequalities which arise in continuum combinatorial approaches to
{L}^{p}
-improving inequalities for Radon-like transforms over polynomial submanifolds of intermediate dimension. The desired inequalities relate to and extend a number of important results in geometric measure theory.
Besov-ish spaces through atomic decomposition
Anal. PDE 15 (1), 123-174, (2022) DOI: 10.2140/apde.2022.15.123
KEYWORDS: atomic decomposition, Besov space, harmonic analysis, Wavelets, multipliers, Haar wavelets, atoms, 43A99, 43A85, 30H25, 42B15, 42B35, 42C15, 42C40, 28C99
We use the method of atomic decomposition to build new families of function spaces, similar to Besov spaces, in measure spaces with grids, a very mild assumption. Besov spaces with low regularity are considered in measure spaces with good grids, and we obtain results on multipliers and left compositions in this setting.
Existence and stability of unidirectional flocks in hydrodynamic Euler alignment systems
Daniel Lear, Roman Shvydkoy
KEYWORDS: flocking, alignment, Cucker–Smale, Mikado solutions, Euler alignment, 92D25, 35Q35, 76N10
We reveal new classes of solutions to hydrodynamic Euler alignment systems governing collective behavior of flocks. The solutions describe unidirectional parallel motion of agents and are globally well-posed in multidimensional settings subject to a threshold condition similar to the one-dimensional case. We develop the flocking and stability theory of these solutions and show long-time convergence to a traveling wave with rapidly aligned velocity field.
In the context of multiscale models introduced by Shvydkoy and Tadmor (Multiscale Model. Simul. 19:2 (2021), 1115–1141) our solutions can be superimposed into Mikado formations — clusters of unidirectional flocks pointing in various directions. Such formations exhibit multiscale alignment phenomena and resemble realistic behavior of interacting large flocks.
Global integrability and weak Harnack estimates for elliptic PDEs in divergence form
KEYWORDS: weak Harnack, Hopf lemma, global integrability, boundary estimates, elliptic PDE, divergence form, 35B45, 35B50, 35B65, 35D30, 35J67
We show that two classically known properties of positive supersolutions of uniformly elliptic PDEs, the boundary point principle (Hopf lemma) and global integrability, can be quantified with respect to each other. We obtain an extension up to the boundary of the De Giorgi–Moser weak Harnack inequality, optimal with respect to the norms involved, for equations in divergence form.
Turbulent cascades in a truncation of the cubic Szegő equation and related systems
Anxo Biasi, Oleg Evnin
KEYWORDS: Szegő equation, integrable Hamiltonian systems, Lax pair, unbounded Sobolev norms, effective resonant dynamics, 35B34, 35B44, 37K10
We introduce a truncated version of the cubic Szegő equation, an integrable model for deterministic turbulence. In this truncation, a majority of the Fourier mode couplings are eliminated, while the signature features of the model are preserved, namely, a Lax pair structure and a hierarchy of finite-dimensional dynamically invariant manifolds. Despite the impoverished structure of the interactions, the turbulent behaviors of our new equation are stronger in an appropriate sense than for the original cubic Szegő equation. We construct explicit analytic solutions displaying exponential growth of Sobolev norms. We furthermore introduce a family of models that interpolate between our truncated system and the original cubic Szegő equation, along with other related deformations. These models possess Lax pairs and invariant manifolds, and display a variety of turbulent cascades. We additionally mention numerical evidence, in some related systems, for an even stronger type of turbulence in the form of a finite-time blow-up.
On the global behaviors for defocusing semilinear wave equations in
{ℝ}^{1+2}
Dongyi Wei, Shiwu Yang
KEYWORDS: asymptotic behavior, defocusing semilinear wave equation, energy subcritical, 35L05
We study the asymptotic decay properties for defocusing semilinear wave equations in
{ℝ}^{1+2}
with pure power nonlinearity. By applying new vector fields to the null hyperplane, we derive improved time decay of the potential energy, with a consequence that the solution scatters both in the critical Sobolev space and energy space for all
p>1+\sqrt{8}
. Moreover, combined with Brezis–Gallouet–Wainger-type of logarithmic Sobolev embedding, we show that the solution decays pointwise with sharp rate
{t}^{-1∕2}
p>\frac{11}{3}
and with rate
{t}^{-\left(p-1\right)∕8+𝜖}
1<p\le \frac{11}{3}
. This in particular implies that the solution scatters in energy space when
p>2\sqrt{5}-1
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Likelihood ratios used for assessing the value of performing a diagnostic test
Not to be confused with Likelihood-ratio test.
In evidence-based medicine, likelihood ratios are used for assessing the value of performing a diagnostic test. They use the sensitivity and specificity of the test to determine whether a test result usefully changes the probability that a condition (such as a disease state) exists. The first description of the use of likelihood ratios for decision rules was made at a symposium on information theory in 1954.[1] In medicine, likelihood ratios were introduced between 1975 and 1980.[2][3][4]
2 Application to medicine
3 Estimation table
3.1 Estimation example
5 Estimation of pre- and post-test probability
Two versions of the likelihood ratio exist, one for positive and one for negative test results. Respectively, they are known as the positive likelihood ratio (LR+, likelihood ratio positive, likelihood ratio for positive results) and negative likelihood ratio (LR–, likelihood ratio negative, likelihood ratio for negative results).
The positive likelihood ratio is calculated as
{\displaystyle {\text{LR}}+={\frac {\text{sensitivity}}{1-{\text{specificity}}}}}
{\displaystyle {\text{LR}}+={\frac {\Pr({T+}\mid D+)}{\Pr({T+}\mid D-)}}}
or "the probability of a person who has the disease testing positive divided by the probability of a person who does not have the disease testing positive." Here "T+" or "T−" denote that the result of the test is positive or negative, respectively. Likewise, "D+" or "D−" denote that the disease is present or absent, respectively. So "true positives" are those that test positive (T+) and have the disease (D+), and "false positives" are those that test positive (T+) but do not have the disease (D−).
The negative likelihood ratio is calculated as[5]
{\displaystyle {\text{LR}}-={\frac {1-{\text{sensitivity}}}{\text{specificity}}}}
which is equivalent to[5]
{\displaystyle {\text{LR}}-={\frac {\Pr({T-}\mid D+)}{\Pr({T-}\mid D-)}}}
or "the probability of a person who has the disease testing negative divided by the probability of a person who does not have the disease testing negative."
The calculation of likelihood ratios for tests with continuous values or more than two outcomes is similar to the calculation for dichotomous outcomes; a separate likelihood ratio is simply calculated for every level of test result and is called interval or stratum specific likelihood ratios.[6]
The pretest odds of a particular diagnosis, multiplied by the likelihood ratio, determines the post-test odds. This calculation is based on Bayes' theorem. (Note that odds can be calculated from, and then converted to, probability.)
Application to medicine[edit]
Pretest probability refers to the chance that an individual in a given population has a disorder or condition; this is the baseline probability prior to the use of a diagnostic test. Post-test probability refers to the probability that a condition is truly present given a positive test result. For a good test in a population, the post-test probability will be meaningfully higher or lower than the pretest probability. A high likelihood ratio indicates a good test for a population, and a likelihood ratio close to one indicates that a test may not be appropriate for a population.
For a screening test, the population of interest might be the general population of an area. For diagnostic testing, the ordering clinician will have observed some symptom or other factor that raises the pretest probability relative to the general population. A likelihood ratio of greater than 1 for a test in a population indicates that a positive test result is evidence that a condition is present. If the likelihood ratio for a test in a population is not clearly better than one, the test will not provide good evidence: the post-test probability will not be meaningfully different from the pretest probability. Knowing or estimating the likelihood ratio for a test in a population allows a clinician to better interpret the result.[7]
Research suggests that physicians rarely make these calculations in practice, however,[8] and when they do, they often make errors.[9] A randomized controlled trial compared how well physicians interpreted diagnostic tests that were presented as either sensitivity and specificity, a likelihood ratio, or an inexact graphic of the likelihood ratio, found no difference between the three modes in interpretation of test results.[10]
Estimation table[edit]
This table provide examples of how changes in the likelihood ratio affects post-test probability of disease.
Approximate* change
in probability[11]
Effect on posttest
Probability of disease[12]
Values between 0 and 1 decrease the probability of disease (−LR)
0.1 −45% Large decrease
0.2 −30% Moderate decrease
0.5 −15% Slight decrease
1 −0% None
Values greater than 1 increase the probability of disease (+LR)
1 +0% None
2 +15% Slight increase
5 +30% Moderate increase
10 +45% Large increase
*These estimates are accurate to within 10% of the calculated answer for all pre-test probabilities between 10% and 90%. The average error is only 4%. For polar extremes of pre-test probability >90% and <10%, see Estimation of pre- and post-test probability section below.
Estimation example[edit]
Pre-test probability: For example, if about 2 out of every 5 patients with abdominal distension have ascites, then the pretest probability is 40%.
Likelihood Ratio: An example "test" is that the physical exam finding of bulging flanks has a positive likelihood ratio of 2.0 for ascites.
Estimated change in probability: Based on table above, a likelihood ratio of 2.0 corresponds to an approximately +15% increase in probability.
Final (post-test) probability: Therefore, bulging flanks increases the probability of ascites from 40% to about 55% (i.e., 40% + 15% = 55%, which is within 2% off the exact probability of 57%).
Calculation example[edit]
A medical example is the likelihood that a given test result would be expected in a patient with a certain disorder compared to the likelihood that same result would occur in a patient without the target disorder.
Some sources distinguish between LR+ and LR−.[13] A worked example is shown below.
A diagnostic test with sensitivity 67% and specificity 91% is applied to 2030 people to look for a disorder with a population prevalence of 1.48%
False positive rate (α) = type I error = 1 − specificity = FP / (FP + TN) = 180 / (180 + 1820) = 9%
False negative rate (β) = type II error = 1 − sensitivity = FN / (TP + FN) = 10 / (20 + 10) ≈ 33%
Power = sensitivity = 1 − β
Positive likelihood ratio = sensitivity / (1 − specificity) ≈ 0.67 / (1 − 0.91) ≈ 7.4
Negative likelihood ratio = (1 − sensitivity) / specificity ≈ (1 − 0.67) / 0.91 ≈ 0.37
Prevalence threshold =
{\displaystyle PT={\frac {{\sqrt {TPR(-TNR+1)}}+TNR-1}{(TPR+TNR-1)}}}
≈ 0.2686 ≈ 26.9%
This hypothetical screening test (fecal occult blood test) correctly identified two-thirds (66.7%) of patients with colorectal cancer.[a] Unfortunately, factoring in prevalence rates reveals that this hypothetical test has a high false positive rate, and it does not reliably identify colorectal cancer in the overall population of asymptomatic people (PPV = 10%).
On the other hand, this hypothetical test demonstrates very accurate detection of cancer-free individuals (NPV ≈ 99.5%). Therefore, when used for routine colorectal cancer screening with asymptomatic adults, a negative result supplies important data for the patient and doctor, such as ruling out cancer as the cause of gastrointestinal symptoms or reassuring patients worried about developing colorectal cancer.
Confidence intervals for all the predictive parameters involved can be calculated, giving the range of values within which the true value lies at a given confidence level (e.g. 95%).[16]
Estimation of pre- and post-test probability[edit]
Further information: Pre- and post-test probability
The likelihood ratio of a test provides a way to estimate the pre- and post-test probabilities of having a condition.
With pre-test probability and likelihood ratio given, then, the post-test probabilities can be calculated by the following three steps:[17]
{\displaystyle {\text{pretest odds}}={\frac {\text{pretest probability}}{1-{\text{pretest probability}}}}}
{\displaystyle {\text{posttest odds}}={\text{pretest odds}}\times {\text{likelihood ratio}}}
In equation above, positive post-test probability is calculated using the likelihood ratio positive, and the negative post-test probability is calculated using the likelihood ratio negative.
Odds are converted to probabilities as follows:[18]
{\displaystyle {\begin{aligned}(1)\ {\text{ odds}}={\frac {\text{probability}}{1-{\text{probability}}}}\end{aligned}}}
multiply equation (1) by (1 − probability)
{\displaystyle {\begin{aligned}(2)\ {\text{ probability}}&={\text{odds}}\times (1-{\text{probability}})\\&={\text{odds}}-{\text{probability}}\times {\text{odds}}\end{aligned}}}
add (probability × odds) to equation (2)
{\displaystyle {\begin{aligned}(3)\ {\text{ probability}}+{\text{probability}}\times {\text{odds}}&={\text{odds}}\\{\text{probability}}\times (1+{\text{odds}})&={\text{odds}}\end{aligned}}}
divide equation (3) by (1 + odds)
{\displaystyle {\begin{aligned}(4)\ {\text{ probability}}={\frac {\text{odds}}{1+{\text{odds}}}}\end{aligned}}}
Posttest probability = Posttest odds / (Posttest odds + 1)
Alternatively, post-test probability can be calculated directly from the pre-test probability and the likelihood ratio using the equation:
P' = P0 × LR/(1 − P0 + P0×LR), where P0 is the pre-test probability, P' is the post-test probability, and LR is the likelihood ratio. This formula can be calculated algebraically by combining the steps in the preceding description.
In fact, post-test probability, as estimated from the likelihood ratio and pre-test probability, is generally more accurate than if estimated from the positive predictive value of the test, if the tested individual has a different pre-test probability than what is the prevalence of that condition in the population.
Taking the medical example from above (20 true positives, 10 false negatives, and 2030 total patients), the positive pre-test probability is calculated as:
Pretest probability = (20 + 10) / 2030 = 0.0148
Pretest odds = 0.0148 / (1 − 0.0148) = 0.015
Posttest odds = 0.015 × 7.4 = 0.111
Posttest probability = 0.111 / (0.111 + 1) = 0.1 or 10%
As demonstrated, the positive post-test probability is numerically equal to the positive predictive value; the negative post-test probability is numerically equal to (1 − negative predictive value).
^ There are advantages and disadvantages for all medical screening tests. Clinical practice guidelines, such as those for colorectal cancer screening, describe these risks and benefits.[14][15]
^ Swets JA. (1973). "The relative operating characteristic in Psychology". Science. 182 (14116): 990–1000. Bibcode:1973Sci...182..990S. doi:10.1126/science.182.4116.990. PMID 17833780.
^ Pauker SG, Kassirer JP (1975). "Therapeutic Decision Making: A Cost-Benefit Analysis". NEJM. 293 (5): 229–34. doi:10.1056/NEJM197507312930505. PMID 1143303.
^ Thornbury JR, Fryback DG, Edwards W (1975). "Likelihood ratios as a measure of the diagnostic usefulness of excretory urogram information". Radiology. 114 (3): 561–5. doi:10.1148/114.3.561. PMID 1118556.
^ van der Helm HJ, Hische EA (1979). "Application of Bayes's theorem to results of quantitative clinical chemical determinations". Clin Chem. 25 (6): 985–8. PMID 445835.
^ a b Gardner, M.; Altman, Douglas G. (2000). Statistics with confidence: confidence intervals and statistical guidelines. London: BMJ Books. ISBN 978-0-7279-1375-3.
^ Brown MD, Reeves MJ (2003). "Evidence-based emergency medicine/skills for evidence-based emergency care. Interval likelihood ratios: another advantage for the evidence-based diagnostician". Ann Emerg Med. 42 (2): 292–297. doi:10.1067/mem.2003.274. PMID 12883521.
^ Harrell F, Califf R, Pryor D, Lee K, Rosati R (1982). "Evaluating the Yield of Medical Tests". JAMA. 247 (18): 2543–2546. doi:10.1001/jama.247.18.2543. PMID 7069920.
^ Reid MC, Lane DA, Feinstein AR (1998). "Academic calculations versus clinical judgments: practicing physicians' use of quantitative measures of test accuracy". Am. J. Med. 104 (4): 374–80. doi:10.1016/S0002-9343(98)00054-0. PMID 9576412.
^ Steurer J, Fischer JE, Bachmann LM, Koller M, ter Riet G (2002). "Communicating accuracy of tests to general practitioners: a controlled study". BMJ. 324 (7341): 824–6. doi:10.1136/bmj.324.7341.824. PMC 100792. PMID 11934776.
^ Puhan MA, Steurer J, Bachmann LM, ter Riet G (2005). "A randomized trial of ways to describe test accuracy: the effect on physicians' post-test probability estimates". Ann. Intern. Med. 143 (3): 184–9. doi:10.7326/0003-4819-143-3-200508020-00004. PMID 16061916.
^ McGee, Steven (1 August 2002). "Simplifying likelihood ratios". Journal of General Internal Medicine. 17 (8): 647–650. doi:10.1046/j.1525-1497.2002.10750.x. ISSN 0884-8734. PMC 1495095. PMID 12213147.
^ Henderson, Mark C.; Tierney, Lawrence M.; Smetana, Gerald W. (2012). The Patient History (2nd ed.). McGraw-Hill. p. 30. ISBN 978-0-07-162494-7.
^ "Likelihood ratios". Archived from the original on 20 August 2002. Retrieved 4 April 2009.
^ Lin, Jennifer S.; Piper, Margaret A.; Perdue, Leslie A.; Rutter, Carolyn M.; Webber, Elizabeth M.; O’Connor, Elizabeth; Smith, Ning; Whitlock, Evelyn P. (21 June 2016). "Screening for Colorectal Cancer". JAMA. 315 (23): 2576–2594. doi:10.1001/jama.2016.3332. ISSN 0098-7484.
^ Bénard, Florence; Barkun, Alan N.; Martel, Myriam; Renteln, Daniel von (7 January 2018). "Systematic review of colorectal cancer screening guidelines for average-risk adults: Summarizing the current global recommendations". World Journal of Gastroenterology. 24 (1): 124–138. doi:10.3748/wjg.v24.i1.124. PMC 5757117. PMID 29358889.
^ Online calculator of confidence intervals for predictive parameters
^ Likelihood Ratios Archived 22 December 2010 at the Wayback Machine, from CEBM (Centre for Evidence-Based Medicine). Page last edited: 1 February 2009
^ [1] from Australian Bureau of Statistics: A Comparison of Volunteering Rates from the 2006 Census of Population and Housing and the 2006 General Social Survey, Jun 2012, Latest ISSUE Released at 11:30 AM (CANBERRA TIME) 08/06/2012
Medical likelihood ratio repositories
The NNT: LR Home
Retrieved from "https://en.wikipedia.org/w/index.php?title=Likelihood_ratios_in_diagnostic_testing&oldid=1072737449"
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Find signal location using similarity search - MATLAB findsignal - MathWorks
Locate Signal in Data
Find Signal in Data with Abrupt Changes
Find Letter in Writing Sample
NormalizationLength
EDRTolerance
istart,istop
[istart,istop,dist] = findsignal(data,signal)
[istart,istop,dist] = findsignal(data,signal,Name,Value)
findsignal(___)
[istart,istop,dist] = findsignal(data,signal) returns the start and stop indices of a segment of the data array, data, that best matches the search array, signal. The best-matching segment is such that dist, the squared Euclidean distance between the segment and the search array, is smallest. If data and signal are matrices, then findsignal finds the start and end columns of the region of data that best matches signal. In that case, data and signal must have the same number of rows.
[istart,istop,dist] = findsignal(data,signal,Name,Value) specifies additional options using name-value pair arguments. Options include the normalization to apply, the number of segments to report, and the distance metric to use.
findsignal(___) without output arguments plots data and highlights any identified instances of signal.
If the arrays are real vectors, the function displays data as a function of sample number.
If the arrays are complex vectors, the function displays data on an Argand diagram.
If the arrays are real matrices, the function uses imagesc to display signal on a subplot and data with the highlighted regions on another subplot.
If the arrays are complex matrices, the function plots their real and imaginary parts in the top and bottom half of each image.
Generate a data set consisting of a 5 Hz Gaussian pulse with 50% bandwidth, sampled for half a second at a rate of 1 kHz.
data = gauspuls(t,5,0.5);
Create a signal consisting of one-and-a-half cycles of a 10 Hz sinusoid. Plot the data set and the signal.
ts = 0:1/fs:0.15;
signal = cos(2*pi*10*ts);
title('Data')
plot(ts,signal)
Find the segment of the data that has the smallest squared Euclidean distance to the signal. Plot the data and highlight the segment.
Add two clearly outlying sections to the data set. Find the segment that is closest to the signal in the sense of having the smallest absolute distance.
dt(t>0.31&t<0.32) = 2.1;
dt(t>0.32&t<0.33) = -2.1;
findsignal(dt,signal,'Metric','absolute')
Let the x-axes stretch if the stretching results in a smaller absolute distance between the closest data segment and the signal.
findsignal(dt,signal,'TimeAlignment','dtw','Metric','absolute')
Add two more outlying sections to the data set.
dt(t>0.1&t<0.11) = 2.1;
Find the two data segments closest to the signal.
findsignal(dt,signal,'TimeAlignment','dtw','Metric','absolute', ...
Go back to finding one segment. Choose 'edr' as the x-axis stretching criterion. Select an edit distance tolerance of 3. The edit distance between nonmatching samples is independent of the actual separation, making 'edr' robust to outliers.
findsignal(dt,signal,'TimeAlignment','edr','EDRTolerance',3, ...
'Metric','absolute')
Repeat the calculation, but now normalize the data and the signal.
Define a moving window with 10 samples to either side of each data and signal point.
Subtract the mean of the data in the window and divide by the local standard deviation.
Find the normalized data segment that has the smallest absolute distance to the normalized signal. Display the unnormalized and normalized versions of the data and the signal.
'Normalization','zscore','NormalizationLength',21, ...
'Metric','absolute','Annotate','all')
Generate a random data array where:
The mean is constant in each of seven regions and changes abruptly from region to region.
The standard deviation is constant in each of five regions and changes abruptly from region to region.
lr = 20;
mns = [0 1 4 -5 2 0 1];
nm = length(mns);
vrs = [1 4 6 1 3]/2;
nv = length(vrs);
v = randn(1,lr*nm*nv);
f = reshape(repmat(mns,lr*nv,1),1,lr*nm*nv);
y = reshape(repmat(vrs,lr*nm,1),1,lr*nm*nv);
t = v.*y+f;
Plot the data, highlighting the steps of its construction. Display the mean and standard deviation of each region.
plot([f;v+f]')
title('Means')
text(lr*nv*nm*((0:1/nm:1-1/nm)+1/(2*nm)),-7*ones(1,nm),num2str(mns'), ...
'HorizontalAlignment',"center")
plot([y;v.*y]')
title('STD')
text(lr*nv*nm*((0:1/nv:1-1/nv)+1/(2*nv)),-7*ones(1,nv),num2str(vrs'), ...
title('Final')
Create a random signal with a mean of zero and a standard deviation of 1/2. Find and display the segment of the data array that best matches the signal.
sg = randn(1,2*lr)/2;
findsignal(t,sg)
Create a random signal with a mean of zero and a standard deviation of 2. Find and display the segment of the data array that best matches the signal.
sg = randn(1,2*lr)*2;
Create a random signal with a mean of 2 and a standard deviation of 2. Find and display the segment of the data array that best matches the signal.
sg = randn(1,2*lr)*2+2;
Create a random signal with a mean of -4 and a standard deviation of 3. Find and display the segment of the data array that best matches the signal.
sg = randn(1,2*lr)*3-4;
Repeat the calculation, but this time subtract the mean from both the signal and the data.
findsignal(t,sg,'Normalization','zscore','Annotate','all')
Corrupt the word by repeating random columns of the letters and varying the spacing. Show the original word and three corrupted versions.
Generate one more corrupted version of the word. Search for a noisy version of the letter "A." Display the distance between the search array and the data segment closest to it. The segment spills into the "T" because the horizontal axes are rigid.
corr = [c(M) c(A) c(T) c(L) c(A) c(B)];
sgn = c(A);
[ist,ind,dst] = findsignal(corr,sgn);
spy(sgn)
spy(corr)
chk = zeros(size(corr));
chk(:,ist:ind) = corr(:,ist:ind);
spy(chk,'*k')
dst = 11
Allow the horizontal axes to stretch. The closest segment is the intersection of the search array and the first instance of "A." The distance between the segment and the array is zero.
[ist,ind,dst] = findsignal(corr,sgn,'TimeAlignment','dtw');
Repeat the computation using the built-in functionality of findsignal. Divide by the local mean to normalize the data and the signal. Use the symmetric Kullback-Leibler metric.
findsignal(corr,sgn,'TimeAlignment','dtw', ...
'Normalization','power','Metric','symmkl','Annotate','all')
data — Data array
Data array, specified as a vector or matrix.
signal — Search array
Search array, specified as a vector or matrix.
Example: 'MaxNumSegments',2,'Metric','squared','Normalization','center','NormalizationLength',11 finds the two segments of the data array that have the smallest squared Euclidean distances to the search signal. Both the data and the signal are normalized by subtracting the mean of a sliding window. The window has five samples to either side of each point, for a total length of 5 + 5 + 1 = 11 samples.
Normalization — Normalization statistic
'none' (default) | 'center' | 'power' | 'zscore'
Normalization statistic, specified as the comma-separated pair consisting of 'Normalization' and one of these values:
'none' — Do not normalize.
'center' — Subtract local mean.
'power' — Divide by local mean.
'zscore' — Subtract local mean and divide by local standard deviation.
NormalizationLength — Normalization length
length(data) (default) | integer scalar
Normalization length, specified as the comma-separated pair consisting of 'NormalizationLength' and an integer scalar. This value represents the minimum number of samples over which to normalize each sample in both the data and the signal. If the signal is a matrix, then 'NormalizationLength' represents a number of columns.
MaxDistance — Maximum segment distance
Maximum segment distance, specified as the comma-separated pair consisting of 'MaxDistance' and a positive real scalar. If you specify 'MaxDistance', then findsignal returns the start and stop indices of all segments of data whose distances from signal are both local minima and smaller than 'MaxDistance'.
MaxNumSegments — Maximum number of segments to return
Maximum number of segments to return, specified as the comma-separated pair consisting of 'MaxNumSegments' and a positive integer scalar. If you specify 'MaxNumSegments', then findsignal locates all segments of data whose distances from the signal are local minima and returns up to 'MaxNumSegments' segments with smallest distances.
TimeAlignment — Time alignment technique
'fixed' (default) | 'dtw' | 'edr'
Time alignment technique, specified as the comma-separated pair consisting of 'TimeAlignment' and one of these values:
'fixed' — Do not stretch or repeat samples to minimize the distance.
'dtw' — Attempt to reduce the distance by stretching the time axis and repeating samples in either the data or the signal. See dtw for more information.
'edr' — Minimize the number of edits so that the distance between each remaining sample of the data segment and its signal counterpart lies within a given tolerance. An edit consists of removing a sample from the data, the signal, or both. Specify the tolerance using the 'EDRTolerance' argument. Use this option when any of the input arrays has outliers. See edr for more information.
EDRTolerance — Edit distance tolerance
Edit distance tolerance, specified as the comma-separated pair consisting of 'EDRTolerance' and a real scalar. Use this argument to find the signal when the 'TimeAlignment' name-value pair argument is set to 'edr'.
'squared' (default) | 'absolute' | 'euclidean' | 'symmkl'
Distance metric, specified as the comma-separated pair consisting of 'Metric' and one of 'squared', 'absolute', 'euclidean', or 'symmkl'. If X and Y are both K-dimensional signals, then Metric prescribes dmn(X,Y), the distance between the mth sample of X and the nth sample of Y. See Dynamic Time Warping for more information about dmn(X,Y).
{d}_{mn}\left(X,Y\right)=\sum _{k=1}^{K}{\left({x}_{k,m}-{y}_{k,n}\right)}^{*}\left({x}_{k,m}-{y}_{k,n}\right)
{d}_{mn}\left(X,Y\right)=\sqrt{\sum _{k=1}^{K}{\left({x}_{k,m}-{y}_{k,n}\right)}^{*}\left({x}_{k,m}-{y}_{k,n}\right)}
{d}_{mn}\left(X,Y\right)=\sum _{k=1}^{K}|{x}_{k,m}-{y}_{k,n}|=\sum _{k=1}^{K}\sqrt{{\left({x}_{k,m}-{y}_{k,n}\right)}^{*}\left({x}_{k,m}-{y}_{k,n}\right)}
{d}_{mn}\left(X,Y\right)=\sum _{k=1}^{K}\left({x}_{k,m}-{y}_{k,n}\right)\left(\mathrm{log}{x}_{k,m}-\mathrm{log}{y}_{k,n}\right)
'signal' (default) | 'data' | 'all'
Plot style, specified as the comma-separated pair consisting of 'Annotate' and one of these values:
'data' plots the data and highlights the regions that best match the signal.
'signal' plots the signal in a separate subplot.
'all' plots the signal, the data, the normalized signal, and the normalized data in separate subplots.
This argument is ignored if you call findsignal with output arguments.
istart,istop — Segment start and end indices
integer scalars | vectors
Segment start and end indices, returned as integer scalars or vectors.
dist — Minimum data-signal distance
Minimum data-signal distance, returned as a scalar or a vector.
alignsignals | dtw | edr | findpeaks | finddelay | strfind | xcorr
|
User:"half-moon" bubba - Uncyclopedia, the content-free encyclopedia
User:"half-moon" bubba
|e|d|,|e|d|d| |n| |e|d|d|y| |e|p|i|s|o|d|e|s|
hello there, i am the most annoying thing in the computerverse, KILL ME!!1!
"half-moon" bubba
from uncyclopedia, the content free encyclopedia
This user knows the true meaning of WAAOU, but will never tell.
The so-called experts at Wikipedia do not have a proper article about User:"half-moon" bubba. Help those so-called-experts and write one.
“an page like many others... OR NOT!!.”
~ Oscar Wilde on User:"half-moon" bubba
“in soviet Russia, user page sees YOU!!.”
~ russian reversal on User:"half-moon" bubba
howdy <insert name here>, this article is a user page that it is suppose to be know for is label that is not in the system
please see my other sites
the cabalcabal
i also made allot of things and stuff at uncyclopedia
... that this is just a distraction while we take your car?
ಠ_ರೃ indeed....
with love --(๏̯͡๏) "half-moon" bubba yay, hidden text! (ಠ_ರೃ)11:32, 12 June 2011
ho my god rainbow :D!!!!!
hell's golden axe
levitation/water walking
magical mirror of doom
once in a blue moon hertz's.
1600 blitz's.
fundamentalist catholic speech
1 troll mathematics
3 how to not fly
4 interest poll
5 my labels
troll mathematics[edit]
the secret of troll mathematics is to piss you off
{\displaystyle x=y}
{\displaystyle \therefore \ x+x=x+y}
{\displaystyle \therefore \ 2x=x+y}
{\displaystyle \therefore \ 2x-2y=x+y-2y}
{\displaystyle \therefore \ 2(x-y)=x+y-2y}
{\displaystyle \therefore \ 2(x-y)=x-y}
{\displaystyle \therefore \ 2=1}
{\displaystyle q.e.d.}
{\displaystyle x=1+1-1+1-1+...\infty *10^{\infty }}
{\displaystyle \therefore \ x=1-(1+1-1+1-1+...\infty *10^{\infty })}
{\displaystyle \therefore \ x=1-x}
{\displaystyle \therefore \ 2x=1}
{\displaystyle \therefore \ x={\frac {1}{2}}}
{\displaystyle q.e.d.}
the theory of the infinite numerical expression
first we admit that
{\displaystyle \varphi }
{\displaystyle {\frac {1}{3}}+1}
{\displaystyle \varphi ={\frac {1}{3}}+1}
we put the same denominators
{\displaystyle {\frac {\varphi }{1_{3}}}={\frac {1}{3}}+{\frac {1}{1_{3}}}}
{\displaystyle {\frac {3\varphi }{3}}={\frac {1}{3}}+{\frac {3}{3}}}
then we erase the same denominators
{\displaystyle 3\varphi =1+3}
then we pass 3 to the other side dividing
{\displaystyle \varphi ={\frac {1}{3}}+{\frac {3}{3}}}
BUT in math we must simplify fractions first before do any thing
{\displaystyle \varphi ={\frac {1}{3}}+1}
the hard you try to make this expression the harder you get stuck in this infinite loop and easily get bat fuck insane
{\displaystyle \pi ={\frac {80{\sqrt {15}}(5^{4}+53{\sqrt {89}})^{\frac {3}{2}}}{3308(5^{4}+53{\sqrt {89}})-3{\sqrt {89}}}}}
0 goes into 0 one time, then
{\displaystyle 0/0=1}
{\displaystyle q.e.d.}
u mad Pythagoras?
u jelly Archimedes?
btw, the prove of this highly developed formulas is in this highly developed formula:
{\displaystyle {\frac {(\int _{14}^{\infty }2^{2^{2}}=({\sqrt {\pi }}+10)}{1+1-1*1/1}}=(1^{2})_{1}}
if this works i dunno lol ¯\(ಥಿ೪ಥ)/¯
this is one method for when freshmens/nerds get laid
set /a girl = deedee
if %deedee%=="a travesty friend of yours" (
echo son of a bitch
slay travesty
relax and let everything happen
explained in ms-dos for better compression
also i have developed a html pop-up that can not be blocked
spam().onload=
/* this will open a pop-up */
alert("aha, pop-up!");
/* this will print a text in the web site if it as activex off */
document.write('in soviet russia, a popup finds you! ');
activex is blocking funny statement
<body onload="spam()">
<p>OMG a website!!1!</p>
in IE activex will say that is a script and not a pop-up lol, but it wont be blocked in chorme.
but there is another trick
1. go to ms word
2. type the following, do not copy pasta, type it:
try this one also :
how to not fly[edit]
IF the video is not loading please go here
is this page good?
hahahahahahhahhahahaaaaaahahahahah
There were 6 votes since the poll was created on 00:49, 8 January 2013.
poll-id 6A51B1AEA6D6D589BBE1AB77F28EFDE6
my labels[edit]
a noble collection of 49 labels achieved by my person.
got to see them all.
This user is a native speaker of Português.
This user does not speak Español and believes it to be an embarrassment to language. Furthermore, this user desires the genocide of all Español speakers.
Yah meeht nawt beh abble tew undersvand zis usehr behkuz zey zpeek HTML whif und estreemleh theuck akzent.
This user only speaks Lesbian enough to seduce native Lesbian speakers .
Thith uther thpeakth Dolphin with a really thexy lithp.
My edits: 0
This user's real age is:
125 years, 5 months and 13 days.
This user is too damn sexy
This user is a Hyrulean warrior. This user has defeated many monsters. They have defeated the mighty Gannon. WARNING: They WILL steal your rupees if you have pocketed them.
This user uses Uncyclopedia as his or her primary point of reference.
I have contributed 0 pints points to the Uncyclopedia Folding@home team.
This user is a pirate and hates ninjas.
"half-moon" bubba's IP is 127.0.0.1.
BEWARE: This user is a random pervert
This user considers themselves an intermediate gamer and so probably has some sort of social life.
This user plays Video Games on the lowest difficulty setting available because winning makes them feel good about themselves.
Retrieved from "http://en.uncyclopedia.co/w/index.php?title=User:%22half-moon%22_bubba&oldid=6063442"
|
Transport in Plants Botany NEET Practice Questions, MCQs, Past Year Questions (PYQs), NCERT Questions, Question Bank, Class 11 and Class 12 Questions, and PDF solved with answers
Botany - Transport in Plants
In guard cells when sugar is converted into starch, the stomatal pore
1. Opens partially
2. Closes completely
3. Opens completely
4. Remains unchanged
Subtopic: Stomata |
Diffusion is very important to plants since:
1. The cells have a permeable cell wall
2. It is the only means for gaseous movement within the plant body.
3. Plants cannot transport material by active transport.
4. They are unable to move towards the source of the nutrients.
Subtopic: Diffusion |
If a pressure greater than atmospheric pressure is applied to a solution, its water potential:
4. Becomes zero
Subtopic: Water Potential Concept |
What per cent of water reaching them is used by leaves in photosynthesis?
2. About 5 %
3. About 5 % in warm conditions and about 10 % in cold conditions
4. About 50 %
Subtopic: Transpiration & Guttation |
Attraction of water molecules to polar surfaces [such as the surface of tracheary elements] is called as:
1. Connation
2. Adnation
Subtopic: Transpiration Pull: Illustration |
What are the control points where a plant adjusts the quantity and types of solutes that reach the xylem?
1. Suberin deposited casparian strips
2. Transport proteins of endodermal cell
3. Sclerenchyma around the pericycle
4. The root hairs themselves
Subtopic: Water Absorption |
I. The direction of movement in the phloem is bi-directional.
II. The source-sink relationship in plants is variable.
1. Both I and II are correct and II explains I
2. Both I and II are correct but II does not explain I
3. I is correct but II is incorrect
4. Both I and II are incorrect
Subtopic: Phloem Translocation |
During translocation of sugars in plants from source to sink:
1. The loading of sugar at source is by active transport and unloading at the sink by passive transport.
2. The loading of sugar at source is by passive transport and unloading at the sink by active transport.
3. Both loading at the source and unloading at the sink are by active transport.
4. Both loading at the source and unloading at the sink are by passive transport.
It is a common observation that CAM plants are not tall. The reason most likely is:
1. They would be unable to move water and minerals to the top of the plant during the day.
2. They would be unable to supply sufficient sucrose for active transport of minerals into the roots during the day or night
3. Transpiration occurs only at night, and this would cause a highly negative
\mathrm{\psi }
in the roots of a tall plant during the day.
4. Since the stomata are closed in the leaves, the Casparian strip is closed in the endodermis strip is closed in the endodermis of the root.
Water logging or over-watering a plant is dangerous and may kill the plant. Why is this so?
1. Water does not have all the necessary minerals a plant needs to grow.
2. Water neutralizes the pH of the soil.
3. The roots are deprived of oxygen.
4. Water lowers the water potential of the roots.
Subtopic: Introduction to Water Absorption |
|
/g/ - Technology » Page 8
Threads by latest replies - Page 8
75KiB, 645x770, chudswarth.png
SHA256 and similar algorithms are insecure pieces of garbage.
Using SHA256 you gain no security against padding-oracle attacks (because you can't roll your own hash function), and you certainly don't gain any security against a rainbow table attack.
In fact, your security with SHA256 against a brute force attack will be a 1 in 2^32 chance that the hash is the correct value.
Insecure hash algorithms can't even provide a 128-bit security level against preimages (that is, if someone does a brute force attack against a SHA256 hash value, then that value is very likely to be a valid hash value that will yield a collision).
And there are ways to make hash functions that are actually more secure, because they take into account that a hash function will not always return the same value.
It is even possible to make hash functions that are significantly more secure than SHA256 against brute force attacks and preimage attacks, by using dedicated hardware and specialized, targeted, hash function design.
>b-but duckduckgo says it's safe!
gpt3 post
you're right and keeping that in mind while reading it made me laugh
Who uses sha256 for passwords
51KiB, 738x357, cscucks.png
The absolute state of CS cucks.
>I am going to spend months developing a skillset that has absolutely zero relevance towards the work I actually do, because some HR roasties are too lazy to develop a more effective way of filtering job applicants
that salary only exists in the US, and, to a certain extent, in hotspots like Zurich and London, everywhere else you'll be getting very similar offers to other smaller companies, tech salary inflated due to increased local host of living in tech hubs so you're not actually getting 300k unless you're willing to go full nomad and live in your truck, and then still a good amount is gone as tax, so you'll probably have to move
it's still better than other jobs especially if you don't have any academic accreditation, code monkey jobs with today's tech are very easy, don't require much experience and thus this opportunity gets shilled everywhere, coding bootcamp and the likes, so hiring managers are getting bombarded with applications which make the candidate look OK on paper since that's a major part of what they teach in bootcamp but in the actual job they suck since they're actually retards so hiring managers were forced to adopt these lunatic interview leetcode practice to get code monkeys who are at least capable of learning and performing difficult repetitive tasks
Higher salary means i can retire much earlier in life and focus on what i want to do
t. too brainlet to find balance
I agree with both of the points you make. I'm just upset because the sheer impracticality of leetcode makes me want to rip my hair out. Oh well. Gotta suck it up I guess.
Best tip is to focus on leetcode easy and mediums first, also use a fun programming language you want to get better at
71KiB, 477x557, chino6.jpg
>The HTTP-01 challenge can only be done on port 80. Allowing clients to specify arbitrary ports would make the challenge less secure, and so it is not allowed by the ACME standard.
So the only reason to use the completely insecure retarded http acme challenge, ease of deployment, is completely invalid because you cannot simply have a script that runs a small http server, you have to disable your actual http server and only then you can solve the challenge?
The only fucking reason this garbage exists is so that glowies and cloudflare can trivially MITM all of your TLS traffic. Jewgle and all other big tech companies that are behind the root CA scam need to be disbanded and their leaders executed.
It's obvious that Let's Encrypt is a fucking op by the glowies. Just look at how they discourage manual certificate authorization by arbitrarily setting the expiration date of the certs to just 3 months. There is no reason to ever refresh your TLS cert every three months. But of course, they need to let the glowies in if they want to start MITM'ing your traffic so every three months, you need to provide unencrypted HTTP access to your server.
FUCK GLOWNIGGERS AND FUCK CERTIFICATE AUTHORITIES
What, is your regular web server too shit to serve a static webroot for ACME requests?
please stop diluting actual concerns about tls and cloudflare with your schizo bs
View SameGoogleImgOpsiqdbSauceNAO chino5.jpg, 158KiB, 1920x1080
My server is built from a declarative configuration, where the NGINX service has specified SSL certs that are fetched by Certbot. But what I discovered today is that this obvious and simple setup does not work because if you refer to nonexistent SSL certs in the NGINX config, it shits itself and refuses to start, which means the http challenge fails. So I have to first generate certs and only then I can deploy it, which is fucking retarded.
Could be trivially solved by having a small http server that just serves the validation on a different port but the cert jews said no.
Just use DNS validation you fucking retard, and stop posting pictures of my wife.
View SameGoogleImgOpsiqdbSauceNAO chino43.png, 728KiB, 1280x639
39KiB, 435x580, Hans_Reiser_mug_shot_(2006).jpg
It's a sad day today, ReiserFS has been murdered from the kernel. You may press F in this thread to pay respects.
>so much so he had to get a mail-order bride
Doesn't he get out next year anyway?
This seems like a year 2038 fix for the kernel.
Hans' story convinced me that mail order brides always ends in disaster. Thank you Hans for your service to fellow Linux wagies.
BASED! i hate women
How is a prison supposed to work on a file system? It's a concrete building, and even if it were capable of doing so, why would it be limited to Hans Reiser's life time?
64KiB, 495x381, cartoon-data-scientist-sexy-robot.jpg
Is data scientist really the sexiest job of the 21st century?
No, All Tech jobs are sexy as long as you can make it seem interesting and get paid well for it
No. It was drummed up because tech companies wanted to get people into a field nobody actually likes
Data scientists are morons.
62KiB, 858x1024, Microsoft_Windows_3.1x_logo_with_wordmark.svg.png
>Is there an ltsc equivalent?
Not as yet. A reasonable emulation of it can be achieved by uninstalling Windows Store, and all the apps in C:\Program Files\WindowsApps however - this is the only difference between Enterprise and Enterprise LTSC.
>Does wpd/simplewall work on it?
Do you guys have the non-registration link for OfficeRTool? the one in the pastebin is dead.
I have an audio problem, every time I reboot my computer the audio turns into shit. After i reinstall the driver and reboot the computer the audio starts working normally until the next reboot. How can I solve that? Using LTSC 2021 on my Thinkpad T440p.
Yes, it is Open Shell now, but other than the name, they seem to be the same.
If you like the windows 10 visual style, there is a skin called "Tenified".
The one on deviantart is a bit outdated (does not support light mode for instance), so i recommend this one, it is more recent:
https://github.com/Open-Shell/Open-Shell-Menu/files/5978435/One.column.immersive.zip
I would add here to watch out for useful apps when uninstalling, since unfortunately microsoft has begun shifting its applications to the store. In particular, paint, notepad, and language packs/input methods are updated via the store now.
That being said, your suggestion is very useful nevertheless, thank you!
There's no reason to be worried.
If you get global banned, just request anything you want on BLU and it will be filled from KG, PTP, BTN or HDB.
If you get global banned from the cabal you will get banned from BLU too if you didn't know and its easier to rejoin BeyondHD than BLU
pls sir how can awesome-hd refuge get into beyond?
when is where interviews
You do realize that encrypting the entire download/upload stream is an option right? It might not be on the transport layer, but this is an option depending on the client you use. Private DDL forums aren't necessarily "more secure". They are just more private (which is not the same thing).
Of course its easier to rejoin BeyondHD, it's trash compared to BLU. But its not hard to get into BLU either.
125KiB, 1200x800, 1*CjXtgNH9zPLJdMkPplz-UQ.jpg
In my shithole you must have a bachelor in software engineering or similar to apply for that kind of job.
Just one book won't give you that.
Based dutchbro.
You can't get nepotism if you don't have people in the software sphere.
It does. I've seen it first-hand at my last job. Fucktarded owner's son is still working for the business and utterly annihilating it whilst shifting the blame to all the underlings who are just trying their hardest.
And then you've got shining turds of examples like Hunter Biden.
>>87051775 isn't wrong. Unfortunately.
Wrong board faggot boomer
67KiB, 750x300, UkraineMath.png
How much math is required for CS?
A lot of countries include questions like these in their exams. Often they're meant to asses placement rather than college preparedness or school performance. A kid getting this question right will lead to being placed in advanced courses/gov. diverting special funds to them.
It's a thing that is rather lacking in the ameerican system whereby we rely on teachers to identify individual students per their own assesments. Tests like the Iowa/SAT/ACT are meant to evaluate the school and college preparation more than aptitude.
This problem is fake, just like how Ukraine is a fake country.
Do all ukrainian 5th graders read "A Treatise On Analytical Statics: Attractions. The Bending Of Rods. Astatics "
A lot where I studied. Probably less in some Indian degree mills. Depends if you want to be a javascript code monkey or actually design stuff.
>make up a whole bunch of bullshit to explain a fake image
HARDCORE STUDY THREAD
I've got four months of free time until my MSc degree starts. I want to dedicate it all to mathematics and physics study. I'm willing to spend 18 hours a day working through mathematics, theoretical physics and computational physics.
I've heard that a lot of mathematicians and physicists throughout history have actually gone through hardcore routines like this (pic related). Anybody have any idea how these daily routines were structured?
[TPhys] Classical Mechanics.
Main topic: equations of motion. To include Generalised Co-ordinates, the principle of least action, Galilean relativity and the Lagrangian. Outlined in Landau&Lifschitz Vol1.
Topology.<br /><br />Main topic: preliminaries. We'll cover set theory and logic as outlined in Munkres. ***I'm going to need help organising this one. This chapter seems easy enough to me but of course, I've studied it all ready. If anybody can have a look at Ch1 of Munkres and see if it's too large or small, that'd be useful.***<br /><br /><script type="math/tex"> Mathematical methods for physics.<br /><br />Main topic: preliminaries. Basic mathematical analysis (series, expansions, vectors, complex analysis, differential and integral calculus). Outlined in Arfken.<br /><br />[CPhys] Computational physics.<br /><br />Main topic: Python programming for physicists. People who aren't interested in python need not study this week: the actual computational methods will come later on. To cover: computational analysis (integration and differentiation), numerical linear algebra, Fourier transforms, differential equations and Monte Carlo methods). Structure follows "Computational Physics" by Mark Newman.
>>>/sci/
Planning it out so much loses a lot of the flexibility to look into other interesting areas. You might cover the Lagrangian and then want to study the Hamiltonian but have to abandon it so you can keep to your intense schedule. Having an actual interest in what you're doing is better than slavishly going through the motions. Otherwise you're going to get to some page that details a proof of some theorem and you'll skip it or you won't internalize it as well as you could have done.
Doesn't your master's have some research component? Channel your energy into preparing for that.
Also you need to sleep and exercise. You'll learn better if you do.
|
Elementary algebra - Simple English Wikipedia, the free encyclopedia
Elementary algebra is the most basic form of algebra taught to students. It is often one of the next areas of mathematics taught to students after arithmetic. While in arithmetic only numbers and operators like +, −, ×, and ÷ occur; in algebra, variables (like a, x, y) are used to stand for numbers. This is useful because:
It lets people solve problems about "unknown" numbers. This means learning about equations and how to solve them (for example, "find a number x where
{\displaystyle 3x+1=10}
It allows the generalization of the rules from arithmetic. While some students understand that
{\displaystyle 3+4=4+3}
, it helps to prove that
{\displaystyle a+b=b+a}
for all a and b. This makes algebra a good step to learning about abstraction (learning general ideas from many examples).
It helps people understand and create functional relationships (also sometimes called cause and effect). An example of this is "if x tickets are sold, then the profit will be
{\displaystyle 3x-10}
dollars".
These three are the main strands of elementary algebra. Elementary algebra is often used in many other subjects, like science, business, and building. Abstract algebra, a much more advanced topic, is generally taught late in college.
Simple algebra problems[change | change source]
If an equation has only one number that is unknown it is sometimes easy to solve. The unknown number is called "x":
{\displaystyle 2x+4=12.\,}
To solve a simple equation with one unknown amount add, subtract, multiply, or divide both sides of the equation by the same number in order to put the unknown amount, x, on one side of the equation. Once x is by itself on one side, use arithmetic to determine the amount on the other side of the equation.[1] For example, by subtracting 4 from both sides in the equation above:
{\displaystyle 2x+4-4=12-4\,}
{\displaystyle 2x=8\,}
{\displaystyle {\frac {2x}{2}}={\frac {8}{2}}\,}
{\displaystyle x=4.\,}
It may help to think of this equation as a see-saw or balance, what you do to one side, you must do to the other and your main aim is to get x by itself.
{\displaystyle 3x^{2}-2xy+c}
1 : Exponent (power), 2 : Coefficient, 3 : term, 4 : operator, 5 : constant,
{\displaystyle x,y}
: variables
↑ Slavin, Steve (1989). All the Math You'll Ever Need. John Wiley & Sons. p. 72. ISBN 0471506362.
Retrieved from "https://simple.wikipedia.org/w/index.php?title=Elementary_algebra&oldid=7330032"
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1-Pyrroline-5-carboxylate dehydrogenase - Wikipedia
In enzymology, a 1-pyrroline-5-carboxylate dehydrogenase (EC 1.2.1.88) is an enzyme that catalyzes the chemical reaction
(S)-1-pyrroline-5-carboxylate + NAD+ + 2 H2O
{\displaystyle \rightleftharpoons }
L-glutamate + NADH + H+
The three substrates of this enzyme are (S)-1-pyrroline-5-carboxylate, NAD+, and H2O, whereas its three products are glutamate, NADH, and H+.
This enzyme belongs to the family of oxidoreductases, specifically those acting on the CH-NH group of donors with NAD+ or NADP+ as acceptor. The systematic name of this enzyme class is (S)-1-pyrroline-5-carboxylate:NAD+ oxidoreductase. Other names in common use include delta-1-pyrroline-5-carboxylate dehydrogenase, 1-pyrroline dehydrogenase, pyrroline-5-carboxylate dehydrogenase, pyrroline-5-carboxylic acid dehydrogenase, L-pyrroline-5-carboxylate-NAD+ oxidoreductase, and 1-pyrroline-5-carboxylate:NAD+ oxidoreductase. This enzyme participates in glutamate metabolism and arginine and proline metabolism.
As of late 2007, 14 structures have been solved for this class of enzymes, with PDB accession codes 2BHP, 2BHQ, 2BJA, 2BJK, 2EHQ, 2EHU, 2EII, 2EIT, 2EIW, 2EJ6, 2EJD, 2EJL, 2IY6, and 2J40.
Human gene[edit]
In human, the protein is encoded by ALDH4A1 gene.
Adams E, Goldstone A (December 1960). "Hydroxyproline metabolism. IV. Enzymatic synthesis of gamma-hydroxyglutamate from Delta 1-pyrroline-3-hydroxy-5-carboxylate". The Journal of Biological Chemistry. 235: 3504–12. PMID 13681370.
Strecker HJ (1960). "The interconversion of glutamic acid proline. III Delta1-Pyrroline-5-carboxylic acid dehydrogenase". J. Biol. Chem. 235: 3218–3223.
Retrieved from "https://en.wikipedia.org/w/index.php?title=1-Pyrroline-5-carboxylate_dehydrogenase&oldid=917310101"
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Rewrite symbolic expression in terms of common subexpressions - MATLAB subexpr - MathWorks Switzerland
\begin{array}{l}\left(\begin{array}{c}\sigma -\frac{b}{3 a}-\frac{{\sigma }_{2}}{\sigma }\\ \frac{{\sigma }_{2}}{2 \sigma }-\frac{b}{3 a}-\frac{\sigma }{2}-{\sigma }_{1}\\ \frac{{\sigma }_{2}}{2 \sigma }-\frac{b}{3 a}-\frac{\sigma }{2}+{\sigma }_{1}\end{array}\right)\\ \\ \mathrm{where}\\ \\ \mathrm{ }{\sigma }_{1}=\frac{\sqrt{3} \left(\sigma +\frac{{\sigma }_{2}}{\sigma }\right) \mathrm{i}}{2}\\ \\ \mathrm{ }{\sigma }_{2}=\frac{c}{3 a}-\frac{{b}^{2}}{9 {a}^{2}}\end{array}
{\left(\sqrt{{\left(\frac{d}{2 a}+\frac{{b}^{3}}{27 {a}^{3}}-\frac{b c}{6 {a}^{2}}\right)}^{2}+{\left(\frac{c}{3 a}-\frac{{b}^{2}}{9 {a}^{2}}\right)}^{3}}-\frac{{b}^{3}}{27 {a}^{3}}-\frac{d}{2 a}+\frac{b c}{6 {a}^{2}}\right)}^{1/3}
\left(\begin{array}{c}-\frac{b+\sqrt{{b}^{2}-4 a c}}{2 a}\\ -\frac{b-\sqrt{{b}^{2}-4 a c}}{2 a}\end{array}\right)
\left(\begin{array}{c}-\frac{b+s}{2 a}\\ -\frac{b-s}{2 a}\end{array}\right)
\sqrt{{b}^{2}-4 a c}
\left(\begin{array}{c}-\frac{b+s}{2 a}\\ -\frac{b-s}{2 a}\end{array}\right)
\sqrt{{b}^{2}-4 a c}
\left(\begin{array}{c}-\frac{b+s}{2 a}\\ -\frac{b-s}{2 a}\end{array}\right)
\left(\begin{array}{c}-\frac{b+s}{2 a}\\ -\frac{b-s}{2 a}\end{array}\right)
\sqrt{{b}^{2}-4 a c}
\left(\begin{array}{c}-\frac{b}{2 a}\\ -\frac{b}{2 a}\end{array}\right)
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Volume of right circular cylinder — lesson. Mathematics State Board, Class 10.
A cylinder whose bases are circular in shape and the axis joining the two centres of the bases perpendicular to the planes of the two bases is called a right circular cylinder.
Volume of a right circular cylinder:
Let \('r'\) be the base radius, and \('h'\) be the height of the cylinder.
Volume \(=\) Base area \(\times\) Height cu. units
Volume \(=\) Area of circle\(\times\) Height cu. units
Volume \(=\) \(\pi r^2 \times h\) \(=\) \(\pi r^2 h\) cu. units
Find the volume if the curved surface area of a right circular cylinder is \(660 \ cm^2\) and the radius \(7 \ cm\).
Radius of the cylinder \(=\) \(7 \ cm\)
Curved surface area \(=\) \(660 \ cm^2\)
\(2 \pi rh\) \(=\) \(660\)
\(h\) \(=\) \(15\)
Height \(=\) \(15 \ cm\)
Volume of the right circular cylinder \(=\) \(\pi r^2 h\) cu. units
\(=\) \(2310 \ cm^3\)
Therefore, the volume of the cylinder is \(2310 \ cm^3\).
\frac{22}{7}
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Hydrocarbons Chemistry NEET Practice Questions, MCQs, Past Year Questions (PYQs), NCERT Questions, Question Bank, Class 11 and Class 12 Questions, and PDF solved with answers
Subtopic: Alkanes, Alkenes and Alkynes - Chemical Properties |
A compound is treated with NaNH2 to give sodium salt. Identify the compound.
1. C2H2 2. C6H6
\left({\mathrm{CH}}_{3}{\right)}_{3}\mathrm{COH}\to
The product X in the above reaction is -
Subtopic: Aromatic Hydrocarbons - Benzene - Structure, Preparation & Chemical Reactions |
The products formed when the meta-xylene is treated with Br2 in the presence of FeBr3 is:-
A compound that formed on passing chlorine through propene at 400
°
C is -
3. Nickel chloride
4. 1,2-Dichloro ethane
The reactants used in Friedel-Craft's alkylation are -
1. C6H6 + NH2
2. C6H6 + CH4
3. C6H6 + CH3Cl
4. C6H6 + CH3COCl
\underset{KOH}{\overset{KMn{O}_{4}}{\to }}B\underset{FeC{l}_{3}}{\overset{B{r}_{2}}{\to }}C\underset{{H}^{+}}{\overset{{C}_{2}{H}_{5}OH}{\to }}D
The D in the above mentioned reaction is -
Nitrobenzene on reaction with conc. HNO3/H2SO4 at 80-100
°
C forms -
1. 1,2-Dinitrobenzene
4. 1,2,4-Trinitrobenzene
Subtopic: Aromatic Hydrocarbons - Reactions & Mechanism |
Product A is-
The monochlorinated products (excluding stereo-isomers) obtained from the reaction is:
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EUDML | Properties of certain -valently convex functions. EuDML | Properties of certain -valently convex functions.
Properties of certain
p
-valently convex functions.
Yang, Dinggong; Owa, Shigeyoshi
Yang, Dinggong, and Owa, Shigeyoshi. "Properties of certain -valently convex functions.." International Journal of Mathematics and Mathematical Sciences 2003.41 (2003): 2603-2608. <http://eudml.org/doc/50676>.
author = {Yang, Dinggong, Owa, Shigeyoshi},
title = {Properties of certain -valently convex functions.},
AU - Yang, Dinggong
AU - Owa, Shigeyoshi
TI - Properties of certain -valently convex functions.
Articles by Owa
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Efficiency of Energy Conversion Devices | EGEE 102: Energy Conservation and Environmental Protection
Efficiency is the useful output of energy. To calculate efficiency the following formula can be used:
Efficiency=\frac{Useful\text{\hspace{0.17em}}Energy\text{\hspace{0.17em}}Output}{Total\text{\hspace{0.17em}}Energy\text{\hspace{0.17em}}Output}
An electric motor consumes 100 watts (a joule per second (J/s)) of power to obtain 90 watts of mechanical power. Determine its efficiency.
Input to the electric motor is in the form of electrical energy and the output is mechanical energy.
Using the efficiency equation:
Or efficiency is 90%.
This is a simple example because both variables are measured in Watts. If the two variables were measured differently, you would need to convert them to equivalent forms before performing the calculation.
Use the following link to generate a random practice problem similar to the Practice 1 example.
The previous example about an electrical motor is very simple because both mechanical and electrical power is given in Watts. Units of both the input and the output have to match; if they do not, you must convert them to similar units.
The United States' power plants consumed 39.5 quadrillion Btus of energy and produced 3.675 trillion kWh of electricity. What is the average efficiency of the power plants in the U.S.?
Efficiency=\frac{Useful\text{\hspace{0.17em}}Energy\text{\hspace{0.17em}}Output}{Total\text{\hspace{0.17em}}Energy\text{\hspace{0.17em}}Output}
Total Energy input = 39.5 x 10^15 Btus and the Useful energy output is 3.675 x 10^12 kWh. Recall that both units have to be the same. So we need to convert kWh into Btus. Given that 1 kWh = 3412 Btus:
\[ 1\text{}\text{}\text{ kWh}=3412\text{}\text{ Btus} \]
\[ 3.675\times {10}^{12}\text{}\text{kWh}=\frac{3.675\times {10}^{12}\text{}\overline{)\text{kWh}}\times 3412\text{}\text{ Btus}}{1\text{}\text{}\text{ }\overline{)\text{kWh}}} \]
=12,539.1×{10}^{12}\text{Btus}
Use the formula for efficiency.
Efficiency=\frac{Useful\text{\hspace{0.17em}}Energy\text{\hspace{0.17em}}Output}{Total\text{\hspace{0.17em}}Energy\text{\hspace{0.17em}}Output}
=\frac{12,539×{10}^{12}Btus}{39.5×{10}^{15}Btus}
=0.3174
=31.74%
Energy efficiencies are not 100% and sometimes they are pretty low. The table below shows typical efficiencies of some of the devices that are used in day to day life:
Typical Efficiencies of Day to Day Devices
Home Gas Furnace
Home Coal Stove
Steam Boiler in a Power Plant
Overall Power Plant
Electric Bulb: Incandescent
Electric Bulb: Fluorescent
Electric Buld: LED
From our discussion on national and global energy usage patterns in Lesson 2, we have seen that:
about 40% of the US energy is used in power generation;
about 27% of the US energy is used for transportation.
Yet the energy efficiency of a power plant is about 35%, and the efficiency of automobiles is about 25%. Thus, over 62% of the total primary energy in the U.S. is used in relatively inefficient conversion processes.
Why are power plant and automobile design engineers allowing this? Can they do better?
There are some natural limitations when converting energy from heat to work.
‹ Energy Conversion Devices up Measuring Thermal Energy ›
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Electric Charges and Fields Physics NEET Practice Questions, MCQs, Past Year Questions (PYQs), NCERT Questions, Question Bank, Class 11 and Class 12 Questions, and PDF solved with answers
An electric dipole is in unstable equilibrium in the uniform electric field. The angle between its dipole moment and the electric field is
°
°
°
°
Subtopic: Electric Dipole |
A particle having charge
{\mathrm{q}}_{1}
exerts F electrostatic force on charge
{\mathrm{q}}_{2}
at rest. If a particle having charge
\frac{{q}_{1}}{4}
is placed midway between the line joining the two charges
{\mathrm{q}}_{1} \mathrm{and} {\mathrm{q}}_{2}
then electrostatic force on
{\mathrm{q}}_{2}
{\mathrm{q}}_{1}
will become/remain
\frac{F}{2}
Subtopic: Coulomb's Law |
A charge q is to be divided into two small conducting spheres. What should be the value of charges on the spheres so that when placed at a certain distance apart, the repulsive force between them is maximum?
\frac{q}{4} and \frac{3q}{4}
\frac{q}{2} and \frac{q}{2}
\frac{q}{3} and \frac{q}{3}
\frac{q}{4} and \frac{q}{4}
An electric dipole of dipole moment p is placed in an electric field of intensity E such that angle between electric field and dipole moment is
\theta
. Assuming that the potential energy of the dipole is zero when
\theta =0°
, the potential energy of the dipole will be
(1) -pE cos
\theta
(2) pE(1-cos
\theta
(3) pE(cos
\theta
(4) -2pE(cos
\theta
The electrostatic field due to a charged conductor just outside the conductor is
1. zero and parallel to the surface at every point inside the conductor
2. zero and is normal to the surface at every point inside the conductor
3. parallel to the surface at every point and zero inside the conductor
4. normal to the surface at every point and zero inside the conductor
Subtopic: Electric Field |
(1) Ampere's law
(3) Faraday's law
Fg and Fe represents gravitational and electrostatic force respectively between electrons situated at a distance 10 cm. The ratio of Fg/ Fe is of the order of
Four charges are arranged at the corners of a square ABCD, as shown in the adjoining figure. The force on the charge kept at the centre O is
(2) Along the diagonal AC
(3) Along the diagonal BD
(4) Perpendicular to side AB
In the absence of other conductors, the surface charge density
(1) Is proportional to the charge on the conductor and its surface area
(2) Inversely proportional to the charge and directly proportional to the surface area
(3) Directly proportional to the charge and inversely proportional to the surface area
(4) Inversely proportional to the charge and the surface area
Out of gravitational, electromagnetic, Vander Waals, electrostatic and nuclear forces; which two are able to provide an attractive force between two neutrons
(1) Electrostatic and gravitational
(2) Electrostatic and nuclear
(3) Gravitational and nuclear
(4) Some other forces like Vander Waals
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Convert ARMA model to MA model - MATLAB arma2ma - MathWorks Australia
{y}_{t}=0.2{y}_{t-1}-0.1{y}_{t-2}+{\epsilon }_{t}+0.5{\epsilon }_{t-1}.
The ARMA model is in difference-equation notation because the left side contains only
{y}_{t}
and its coefficient 1. Create a vector containing the AR lag term coefficients in order starting from t - 1.
\begin{array}{l}{y}_{t}={\epsilon }_{t}+0.7{\epsilon }_{t-1}+0.04{\epsilon }_{t-2}-0.062{\epsilon }_{t-3}-0.0164{\epsilon }_{t-4}.\end{array}
{y}_{t}=-0.2{y}_{t-1}+0.5{y}_{t-3}+{\epsilon }_{t}.
The AR model is in difference-equation notation because the left side contains only
{y}_{t}
and its coefficient of 1. Create a cell vector containing the AR lag term coefficient in order starting from t - 1. Because the second lag term of the MA model is missing, specify a 0 for its coefficient.
{y}_{t}={\epsilon }_{t}-0.2{\epsilon }_{t-1}+0.04{\epsilon }_{t-2}+0.492{\epsilon }_{t-3}-0.1984{\epsilon }_{t-4}+0.0597{\epsilon }_{t-5}
\begin{array}{l}\left\{\left[\begin{array}{ccc}1& 0.2& -0.1\\ 0.03& 1& -0.15\\ 0.9& -0.25& 1\end{array}\right]+\left[\begin{array}{ccc}0.5& -0.2& -0.1\\ -0.3& -0.1& 0.1\\ 0.4& -0.2& -0.05\end{array}\right]{L}^{4}+\left[\begin{array}{ccc}0.05& -0.02& -0.01\\ -0.1& -0.01& -0.001\\ 0.04& -0.02& -0.005\end{array}\right]{L}^{8}\right\}{y}_{t}=\\ \left\{\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]+\left[\begin{array}{ccc}-0.02& 0.03& 0.3\\ 0.003& 0.001& 0.01\\ 0.3& 0.01& 0.01\end{array}\right]{L}^{4}\right\}{\epsilon }_{t}\end{array}
{y}_{t}={\left[{y}_{1t}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}{y}_{2t}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}{y}_{3t}\right]}^{\prime }
{\epsilon }_{t}={\left[{\epsilon }_{1t}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}{\epsilon }_{2t}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}{\epsilon }_{3t}\right]}^{\prime }
Create a cell vector containing the VAR matrix coefficients. Because this model is a structural model, start with the coefficient of
{y}_{t}
Create a cell vector containing the VMA matrix coefficients. Because this model is a structural model, start with the coefficient of
{\epsilon }_{t}
{y}_{t}=1.5+0.2{y}_{t-1}-0.1{y}_{t-2}+{\epsilon }_{t}+0.5{\epsilon }_{t-1}.
{y}_{t}
and its coefficient of 1. Create separate vectors for the AR and MA lag term coefficients in order starting from t - 1.
\left(1-0.2L+0.1{L}^{2}\right){y}_{t}=1.5+\left(1+0.5L\right){\epsilon }_{t}
\Phi \left(L\right){y}_{t}=1.5+\Theta \left(L\right){\epsilon }_{t}
{y}_{t}={\Phi }^{-1}\left(L\right)1.5+{\Phi }^{-1}\left(L\right)\Theta \left(L\right){\epsilon }_{t}
{y}_{t}=1.667+0.7{\epsilon }_{t-1}+0.04{\epsilon }_{t-2}-0.062{\epsilon }_{t-3}-0.0164{\epsilon }_{t-4}+0.0029{\epsilon }_{t-5}+{\epsilon }_{t}.
E\left({y}_{t}\right)=1.667.
When you work from a model in difference-equation notation, negate the AR coefficient of the lagged terms to construct the lag-operator polynomial equivalent. For example, consider
{y}_{t}=0.5{y}_{t-1}-0.8{y}_{t-2}+{\epsilon }_{t}-0.6{\epsilon }_{t-1}+0.08{\epsilon }_{t-2}
. The model is in difference-equation notation. To convert to an MA model, enter the following into the command window.
The ARMA model in lag operator notation is
\left(1-0.5L+0.8{L}^{2}\right){y}_{t}=\left(1-0.6L+0.08{L}^{2}\right){\epsilon }_{t}.
The AR coefficients of the lagged responses are negated compared to the corresponding coefficients in difference-equation format. In this form, to obtain the same result, enter the following into the command window.
{\Phi }_{0}{y}_{t}=c+{\Phi }_{1}{y}_{t-1}+...+{\Phi }_{p}{y}_{t-p}+{\Theta }_{0}{\epsilon }_{t}+{\Theta }_{1}{\epsilon }_{t-1}+...+{\Theta }_{q}{\epsilon }_{t-q},
\Phi \left(L\right){y}_{t}=c+\Theta \left(L\right){\epsilon }_{t},
\Phi \left(L\right)={\Phi }_{0}-{\Phi }_{1}L-{\Phi }_{2}{L}^{2}-...-{\Phi }_{p}{L}^{p}
{L}^{j}{y}_{t}={y}_{t-j}
\Theta \left(L\right)={\Theta }_{0}+{\Theta }_{1}L+{\Theta }_{2}{L}^{2}+...+{\Theta }_{q}{L}^{q}
{y}_{t}={\Phi }^{-1}\left(L\right)\Theta \left(L\right){\epsilon }_{t}
\Phi \left(L\right)=\sum _{j=0}^{p}{\Phi }_{j}{L}^{j}
\Theta \left(L\right)=\sum _{k=0}^{q}{\Theta }_{k}{L}^{k}.
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DeleteArc - Maple Help
Home : Support : Online Help : Mathematics : Discrete Mathematics : Graph Theory : GraphTheory Package : DeleteArc
delete arc from digraph
DeleteArc(G, E, ip)
arc, trail, or set of arcs
(optional)equation of the form inplace=true or false
The DeleteArc command deletes one or more arcs from a directed graph. By default, the original digraph is changed to a digraph missing the specified set of arcs. By setting inplace=false the original digraph remains unchanged and a new digraph missing the specified set of arcs is created.
If the digraph is a weighted digraph, then if a weight is also provided (i.e.
[\mathrm{arc},\mathrm{weight}]
\mathrm{arc}
) that weight is subtracted from the arc weight, which will not necessarily remove the arc from the digraph. If no weight is provided, the arc is removed regardless of the weight.
\mathrm{with}\left(\mathrm{GraphTheory}\right):
G≔\mathrm{Digraph}\left([a,b,c,d],{[a,b],[b,c],[c,d],[d,a]}\right)
\textcolor[rgb]{0,0,1}{G}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{Graph 1: a directed unweighted graph with 4 vertices and 4 arc\left(s\right)}}
H≔\mathrm{DeleteArc}\left(G,[d,a],\mathrm{inplace}=\mathrm{false}\right)
\textcolor[rgb]{0,0,1}{H}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{Graph 2: a directed unweighted graph with 4 vertices and 3 arc\left(s\right)}}
\mathrm{Edges}\left(G\right)
{[\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{b}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{b}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{c}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{c}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{d}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{d}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{a}]}
\mathrm{Edges}\left(H\right)
{[\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{b}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{b}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{c}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{c}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{d}]}
\mathrm{DeleteArc}\left(G,{[a,b],[c,d]}\right)
\textcolor[rgb]{0,0,1}{\mathrm{Graph 1: a directed unweighted graph with 4 vertices and 2 arc\left(s\right)}}
\mathrm{Edges}\left(G\right)
{[\textcolor[rgb]{0,0,1}{b}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{c}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{d}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{a}]}
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Add reverberation to audio signal - MATLAB - MathWorks Italia
HighCutFrequency
HighFrequencyDamping
WetDryMix
Specific to reverberator
Add Reverberation to Audio Signal
Tune Reverberator Parameters
The reverberator System object™ adds reverberation to mono or stereo audio signals.
To add reverberation to your input:
Create the reverberator object and set its properties.
reverb = reverberator
reverb = reverberator(Name,Value)
reverb = reverberator creates a System object, reverb, that adds artificial reverberation to an audio signal.
reverb = reverberator(Name,Value) sets each property Name to the specified Value. Unspecified properties have default values.
Example: reverb = reverberator('PreDelay',0.5,'WetDryMix',1) creates a System object, reverb, with a 0.5 second pre-delay and a wet-to-dry mix ratio of one.
PreDelay — Pre-delay for reverberation (s)
Pre-delay for reverberation in seconds, specified as a real scalar in the range [0, 1].
Pre-delay for reverberation is the time between hearing direct sound and the first early reflection. The value of PreDelay is proportional to the size of the room being modeled.
HighCutFrequency — Lowpass filter cutoff (Hz)
Lowpass filter cutoff in Hz, specified as a real positive scalar in the range 0 to
\left(\frac{SampleRate}{2}\right)
Lowpass filter cutoff is the –3 dB cutoff frequency for the single-pole lowpass filter at the front of the reverberator structure. It prevents the application of reverberation to high-frequency components of the input.
Diffusion — Density of reverb tail
Density of reverb tail, specified as a real positive scalar in the range [0, 1].
Diffusion is proportional to the rate at which the reverb tail builds in density. Increasing Diffusion pushes the reflections closer together, thickening the sound. Reducing Diffusion creates more discrete echoes.
DecayFactor — Decay factor of reverb tail
Decay factor of reverb tail, specified as a real positive scalar in the range [0, 1].
DecayFactor is inversely proportional to the time it takes for reflections to run out of energy. To model a large room, use a long reverb tail (low decay factor). To model a small room, use a short reverb tail (high decay factor).
HighFrequencyDamping — High-frequency damping
High-frequency damping, specified as a real positive scalar in the range [0, 1].
HighFrequencyDamping is proportional to the attenuation of high frequencies in the reverberation output. Setting HighFrequencyDamping to a large value makes high-frequency reflections decay faster than low-frequency reflections.
WetDryMix — Wet-dry mix
Wet-dry mix, specified as a real positive scalar in the range [0, 1].
Wet-dry mix is the ratio of wet (reverberated) to dry (original) signal that your reverberator System object outputs.
audioOut = reverb(audioIn)
audioOut = reverb(audioIn) adds reverberation to the input signal, audioIn, and returns the mixed signal, audioOut. The type of reverberation is specified by the algorithm and properties of the reverberator System object, reverb.
audioIn — Audio input to reverberator
column vector | N-by-2 matrix
Audio input to the reverberator, specified as a column vector or two-column matrix. The columns of the matrix are treated as independent audio channels.
audioOut — Audio output from reverberator
Audio output from the reverberator, returned as a two-column matrix.
Use the reverberator System object™ to add artificial reverberation to an audio signal read from a file.
Create the dsp.AudioFileReader and audioDeviceWriter System objects. Use the sample rate of the reader as the sample rate of the writer.
fileReader = dsp.AudioFileReader('FunkyDrums-44p1-stereo-25secs.mp3','SamplesPerFrame',1024);
Play 10 seconds of the audio signal through your device.
Construct a reverberator System object with default settings.
reverb =
reverberator with properties:
PreDelay: 0
HighCutFrequency: 20000
Diffusion: 0.5000
DecayFactor: 0.5000
HighFrequencyDamping: 5.0000e-04
WetDryMix: 0.3000
Construct a time scope to visualize the original audio signal and the audio signal with added artificial reverberation.
'SampleRate',fileReader.SampleRate,...
'BufferLength',3*fileReader.SampleRate*2, ...
'YLimits',[-1,1],...
'Title','Audio with Reverberation vs. Original');
Play the audio signal with artificial reverberation. Visualize the audio with reverberation and the original audio.
audioWithReverb = reverb(audio);
deviceWriter(audioWithReverb);
scope([audioWithReverb(:,1),audio(:,1)])
Create a dsp.AudioFileReader to read in audio frame-by-frame. Create an audioDeviceWriter to write audio to your sound card. Create a reverberator to process the audio data.
'SamplesPerFrame',frameLength,'PlayCount',2);
reverb = reverberator('SampleRate',fileReader.SampleRate);
parameterTuner(reverb)
Apply reverberation.
While streaming, tune parameters of the reverberator and listen to the effect.
audioOut = reverb(audioIn);
The createAudioPluginClass and configureMIDI functions map tunable properties of the compressor to user-facing parameters:
PreDelay [0, 1] linear s
HighCutFrequency [20, 20000] log Hz
Diffusion [0, 1] linear none
DecayFactor [0, 1] linear none
HighFrequencyDamping [0, 1] linear none
WetDryMix [0, 1] linear none
The algorithm to add reverberation follows the plate-class reverberation topology described in [1] and is based on a 29,761 Hz sample rate.
The algorithm has five stages.
The description for the algorithm that follows is for a stereo input. A mono input is a simplified case.
A stereo signal is converted to a mono signal:
x\left[n\right]=0.5×\left({x}_{\text{R}}\left[n\right]+{x}_{\text{L}}\left[n\right]\right)
A delay followed by a lowpass filter preconditions the mono signal.
The pre-delay output is determined as
{x}_{\text{p}}\left[n\right]=x\left[n-k\right]
, where the PreDelay property determines the value of k.
The signal is fed through a single-pole lowpass filter with transfer function
LP\left(z\right)=\frac{1-\alpha }{1-\alpha {z}^{-1}}\text{\hspace{0.17em}},
\alpha =\mathrm{exp}\left(-2\pi ×\frac{{f}_{\text{c}}}{{f}_{\text{s}}}\right)\text{\hspace{0.17em}}.
fc is the cutoff frequency specified by the HighCutFrequency property.
fs is the sampling frequency specified by the SampleRate property.
The signal is decorrelated by passing through a series of four allpass filters.
The allpass filters are of the form
AP\left(z\right)=\frac{\beta +{z}^{-k}}{1+\beta {z}^{-k}}\text{\hspace{0.17em}},
where β is the coefficient specified by the Diffusion property and k is the delay as follows:
For AP1, k = 142.
The signal is fed into the tank, where it circulates to simulate the decay of a reverberation tail.
The following description tracks the signal as it progresses through the top of the tank. The signal progression through the bottom of the tank follows the same pattern, with different delay specifications.
The new signal enters the top of the tank and is added to the circulated signal from the bottom of the tank.
The signal passes through a modulated allpass filter:
Modulated\text{\hspace{0.17em}}A{P}_{1}\left(z\right)=\frac{-\beta +{z}^{-k}}{1-\beta {z}^{-k}}\text{\hspace{0.17em}}
β is the coefficient specified by the Diffusion property.
k is the variable delay specified by a 1 Hz sinusoid with amplitude = (8/29761) * SampleRate. To account for fractional delay resulting from the modulating k, allpass interpolation is used [2].
The signal is delayed again, and then passes through a lowpass filter:
L{P}_{2}\left(z\right)=\frac{1-\phi }{1-\phi {z}^{-1}}
φ is the coefficient specified by the HighFrequencyDamping property.
The signal is multiplied by a gain specified by the DecayFactor property. The signal then passes through an allpass filter:
A{P}_{5}\left(z\right)=\frac{\beta +{z}^{-k}}{1+\beta {z}^{-k}}\text{\hspace{0.17em}}.
k is set to 1800 for the top of the tank and 2656 for the bottom of the tank.
The signal is delayed again and then circulated to the bottom half of the tank for the next iteration.
A similar pattern is executed in parallel for the bottom half of the tank. The output of the tank is calculated as the signed sum of delay lines picked off at various points from the tank. The summed output is multiplied by 0.6.
The wet (processed) signal is then added to the dry (original) signal:
{y}_{R}\left[n\right]=\left(1-\kappa \right){x}_{R}\left[n\right]+\kappa {x}_{3R}\left[n\right]\text{\hspace{0.17em}},
{y}_{L}\left[n\right]=\left(1-\kappa \right){x}_{L}\left[n\right]+\kappa {x}_{3L}\left[n\right]\text{\hspace{0.17em}},
where the WetDryMix property determines κ.
[1] Dattorro, Jon. "Effect Design, Part 1: Reverberator and Other Filters." Journal of the Audio Engineering Society. Vol. 45, Issue 9, 1997, pp. 660–684.
[2] Dattorro, Jon. "Effect Design, Part 2: Delay-Line Modulation and Chorus." Journal of the Audio Engineering Society. Vol. 45, Issue 10, 1997, pp. 764–788.
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Understanding PP&E
Significance of PP&E
Accounting for PP&E
Limitations of PP&E
Example of PP&E
PP&E FAQs
Property, plant, and equipment (PP&E) are long-term assets vital to business operations. Property, plant, and equipment are tangible assets, meaning they are physical in nature or can be touched; as a result, they are not easily converted into cash. The overall value of a company's PP&E can range from very low to extremely high compared to its total assets.
Property, plant, and equipment (PP&E) are long-term assets vital to business operations and the long-term financial health of a company.
Equipment, machinery, buildings, and vehicles are all types of PP&E assets.
(PP&E) are also called fixed or tangible assets, meaning they are physical items that a company cannot easily liquidate.
Purchases of PP&E are a signal that management has faith in the long-term outlook and profitability of its company.
Investment analysts and accountants use the PP&E of a company to determine if it is on a sound financial footing and utilizing funds in the most efficient and effective manner.
Understanding Property, Plant, and Equipment (PP&E)
Property, plant, and equipment are also called fixed assets, meaning they are physical assets that a company cannot easily liquidate or sell. PP&E assets fall under the category of noncurrent assets, which are the long-term investments or assets of a company. Noncurrent assets like PP&E have a useful life of more than one year, but usually, they last for many years.
Examples of property, plant, and equipment include the following:
Noncurrent assets like PP& E are the opposite of current assets. Current assets are short-term, meaning they are items that are likely to be converted into cash within one year, such as inventory.
PP&E and Noncurrent Assets
Although PP&E are noncurrent assets or long-term assets, not all noncurrent assets are property, plant, and equipment. Intangible assets are nonphysical assets, such as patents and copyrights. They are considered to be noncurrent assets because they provide value to a company but cannot be readily converted to cash within a year. Long-term investments, such as bonds and notes, are also considered noncurrent assets because a company usually holds these assets on its balance sheet for more than one fiscal year. PP&E refers to specific fixed, tangible assets, whereas noncurrent assets are all of the long-term assets of a company.
Calculating PP&E
To calculate PP&E, add the amount of gross property, plant, and equipment, listed on the balance sheet, to capital expenditures. Next, subtract accumulated depreciation from the result. In most cases, companies will list their net PP&E on their balance sheet when reporting financial results, so the calculation has already been done.
As a formula, it would be:
\begin{aligned} &\text{Net PPE}=\text{Gross PPE}+\text{Capital Expenditures}-\text{AD}\\ &\textbf{where:}\\ &\text{AD}=\text{Accumulated depreciation} \end{aligned}
Net PPE=Gross PPE+Capital Expenditures−ADwhere:AD=Accumulated depreciation
A company investing in PP&E is a good sign for investors. A fixed asset is a sizable investment in a company's future. Purchases of PP&E are a signal that management has faith in the long-term outlook and profitability of its company. PP&E are a company's physical assets that are expected to generate economic benefits and contribute to revenue for many years. Investment in PP&E is also called a capital investment. Industries or businesses that require a large number of fixed assets like PP&E are described as capital intensive.
PP&E may be liquidated when they are no longer of use or when a company is experiencing financial difficulties. Of course, selling property, plant, and equipment to fund business operations is a signal that a company might be in financial trouble. It is important to note that regardless of the reason why a company has sold some of its property, plant, or equipment, it's likely the company didn't realize a profit from the sale. Companies can also borrow off their PP&E, (floating lien), meaning the equipment can be used as collateral for a loan.
PP&E is recorded on a company's financial statements, specifically on the balance sheet. PP&E is initially measured according to its historical cost, which is the actual purchase cost and the costs associated with bringing assets to its intended use. For example, when purchasing a building for retail operations, the historical cost could include the purchase price, transaction fees, and any improvements made to the building to bring it to its destined use.
The value of PP&E is adjusted routinely as fixed assets generally see a decline in value due to use and depreciation. Depreciation is the process of allocating the cost of a tangible asset over its useful life and is used to account for declines in value. The total amount of a company's cost allocated to depreciation expense over time is called accumulated depreciation.
However, land is not depreciated because of its potential to appreciate in value. Instead, it is represented at its current market value. The balance of the PP&E account is remeasured every reporting period, and, after accounting for historical cost and depreciation, is called the book value. This figure is reported on the balance sheet.
PP&E are vital to the long-term success of many companies, but they are capital intensive. Companies sometimes sell a portion of their assets to raise cash and boost their profit or net income. As a result, it's important to monitor a company's investments in PP&E and any sale of its fixed assets.
Since PP&E are tangible assets, PP&E analysis doesn't include intangible assets such as a company's trademark. For example, Coca-Cola's (KO) trademark and brand name represent sizable intangible assets. If investors were to only look at Coca-Cola's PP&E, they wouldn't see the true value of the company's assets. PP&E only represents one portion of a company's assets. Also, for companies with few fixed assets, PP&E has little value as a metric.
Below is a portion of Exxon Mobil Corporation's (XOM) quarterly balance sheet as of September 30, 2018.
We can see that Exxon recorded $249.153 billion in net property, plant, and equipment for the period ending September 30, 2018. When compared to Exxon's total assets of over $354 billion for the period, PP&E made up the vast majority of total assets. As a result, Exxon would be considered a capital intensive company. Some of the company's fixed assets include oil rigs and drilling equipment.
Why Should Investors Pay Attention to PP&E?
PP&E are assets that are expected to generate economic benefits and contribute to revenue for many years. Purchases of PP&E are a signal that management has faith in the long-term outlook and profitability of its company.
How Is PP&E Accounted for?
PP&E is recorded on a company's financial statements, specifically on the balance sheet. To calculate PP&E, add the amount of gross property, plant, and equipment, listed on the balance sheet, to capital expenditures. Next, subtract accumulated depreciation. The result is the overall value of the PP&E. It's often referred to as the company's book value.
Noncurrent assets are a company's long-term investments for which the full value will not be realized within the accounting year. They are allocated over the number of years the asset is used. They appear on a company's balance sheet under "investment"; "property, plant, and equipment"; "intangible assets"; or "other assets".
ExxonMobil. ”Form 10-Q: For the quarterly period ended September 30, 2018,” Page 5. Accessed Oct. 4, 2021.
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EUDML | On the automorphism groups of convex domains in . EuDML | On the automorphism groups of convex domains in .
On the automorphism groups of convex domains in
{ℂ}^{n}
Kim, Kang-Tae. "On the automorphism groups of convex domains in .." Advances in Geometry 4.1 (2004): 33-40. <http://eudml.org/doc/123604>.
author = {Kim, Kang-Tae},
keywords = {automorphism group; convex domains},
title = {On the automorphism groups of convex domains in .},
AU - Kim, Kang-Tae
TI - On the automorphism groups of convex domains in .
KW - automorphism group; convex domains
automorphism group, convex domains
Automorphism groups of
{𝐂}^{n}
and affine manifolds
|
Energy factor - Wikipedia
An energy factor is a metric used in the United States to compare the energy conversion efficiency of residential appliances and equipment. The energy factor is currently used for rating the efficiency of water heaters, dishwashers, clothes washers, and clothes dryers.[1] The term is used by the United States Department of Energy to develop and enforce minimum energy conservation standards under the Energy Conservation Program.[2]
The higher the energy factor, the more efficient the appliance should be.[3]
Although the term energy factor is used to compare the relative efficiency of these appliances, the metric is defined differently for all four appliance categories. The energy factor is expressed in terms of site energy, which excludes losses through energy conversion. All of these efficiency metrics are defined by Department of Energy test procedures.[4]
1.2 Uniform energy factor (UEF)
1.3 Energy conservation standards
Water heaters[edit]
The energy factor metric only applies to residential water heaters, which are currently defined by fuel, type, and input capacity.[5] Generally, the EF number represents the thermal efficiency of the water heater as a percentage, since it is an average of the ratio of the theoretical heat required to raise the temperature of water drawn to the amount of energy actually consumed by the water heater.
Natural Gas Storage ≤75 kBtu/h
Fuel Oil Storage ≤105 kBtu/h
Electric Storage ≤12 kW
Tabletop Storage ≤12 kW
Natural Gas Instantaneous <200 kBtu/hr
Electric Instantaneous ≤12 kW
The energy factor is determined using a stylized hot water use pattern. Hot water is drawn in six equal draws totaling 64.3 gallons, and a standby period of 18 hours follows.
The energy factor for residential water heaters is determined using the results from the 24-hour simulated use test. During the test 64.3±1.0 gallons of water are drawn from the water heater in six equally spaced draws that begin one hour apart. The hot water flow rate for each draw is 3.0±0.25 gallons per minute. After the beginning of the last draw a standby period of 18 hours follows. During the test, the test conditions must be operated at a specified value and accuracy. Heat pump water heaters (HPWHs) have different values specified for ambient air temperature and relative humidity.[6]
Required value and accuracy
Inlet water temperature 58 °F±2 °F
Outlet water temperature 135 °F±5 °F
Ambient air temperature 67.5 °F±2.5 °F
Ambient air temperature (HPWHs only) 67.5 °F±1 °F
Ambient relative humidity (HPWHs only) 50%±1%
From the standard test procedure, the energy factor is defined as
{\displaystyle EF=\sum \limits _{i=1}^{6}{\frac {M_{i}C_{pi}\left(135^{\circ }F-58^{\circ }F\right)}{Q_{dm}}},}
{\displaystyle Q_{dm}}
is the modified daily water heating energy consumption (Btu),
{\displaystyle M_{i}}
is the mass withdrawn from the ith draw (lb), and
{\displaystyle C_{pi}}
is the specific heat of the water of the ith draw Btu/lb°F, evaluated at the midpoint between 58°F and 135°F.
Uniform energy factor (UEF)[edit]
As of 2021[update] the Uniform Energy Factor (UEF) is newest measure of water heater overall efficiency according to the Department of Energy’s test method outlined in 10 CFR Part 430, Subpart B, Appendix E.
Energy conservation standards[edit]
Minimum federal energy conservation standards are defined by fuel, type, and rated storage volume. All standards are calculated as a function of the rated storage volume V in gallons. The current conservation standards are less efficient than the standards that go into effect in 2015.[5][7][8]
Energy Factor (Effective April 16, 2015)
Natural Gas Storage ≥ 20 gal and ≤ 55 gal 0.675 − 0.0015V 0.67 − 0.0019V
Natural Gas Storage > 55 gal and ≤ 100 gal 0.8012 − 0.00078V 0.67 − 0.0019V
Fuel Oil Storage ≤ 50 gal 0.68 − 0.0019V 0.59 − 0.0019V
Electric Storage ≥ 20 gal and ≤ 55 gal 0.960 − 0.0003V 0.97 − 0.00132V
Electric Storage > 55 gal and ≤ 120 gal 2.057 − 0.00113V 0.97 − 0.00132V
Tabletop Storage ≥ 20 gal and ≤ 100 gal 0.93 − 0.00132V 0.93 − 0.00132V
Natural Gas Instantaneous < 2 gal 0.82 − 0.0019V 0.62 − 0.0019V
Electric Instantaneous < 2 gal 0.93 − 0.00132V 0.93 − 0.00132V
Dishwashers[edit]
The energy factor for dishwashers is defined as "the number of cycles per kWh of input power."[1]
Clothes washers[edit]
The energy factor for clothes washers, is defined as "the cubic foot capacity per kWh of input power per cycle."[1]
Clothes dryers[edit]
The energy factor for clothes dryers is defined as "the number of pounds of clothes dried per kWh of power consumed."[1]
^ a b c d "Federal Tax Credits for Consumer Energy Efficiency: Definitions". energystar.gov. U.S. Environmental Protection Agency. Archived from the original on 1 April 2013. Retrieved 26 March 2013.
^ "Statutory Authorities and Rules". eere.energy.gov. U.S. Department of Energy. Retrieved 26 March 2013.
^ "ESTIMATING COSTS AND EFFICIENCY OF STORAGE, DEMAND, AND HEAT PUMP WATER HEATERS". Energy.gov. US Dept of Energy. Retrieved 2 June 2016.
^ "Standards and Test Procedures". eere.energy.gov. U.S. Department of Energy. Retrieved 26 March 2013.
^ a b "Residential Water Heaters". eere.energy.gov. U.S. Department of Energy. Retrieved 26 March 2013.
^ "10 CFR Part 430 Energy Conservation Program for Consumer Products: Test Procedure for Water Heaters; Final Rule" (PDF). Federal Register. 63 (90): 25995–26016. 11 May 1998. Retrieved 26 March 2013.
^ "10 CFR Part 430 Energy Conservation Program: Energy Conservation Standards for Residential Water Heaters, Direct Heating Equipment, and Pool Heaters; Final Rule" (PDF). Federal Register. 75 (73): 20112–21981 [20113]. 16 April 2010. Retrieved 26 March 2013.
^ "10 CFR Part 430 Energy Conservation Program for Consumer Products: Energy Conservation Program for Consumer Products: Energy Conservation Standards for Water Heaters; Final Rule" (PDF). Federal Register. 66 (11): 4474–4497 [4497]. 17 January 2001. Retrieved 26 March 2013.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Energy_factor&oldid=1053239070"
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Reproducible Machine Learning and Experiment Tracking Pipeline with Python and DVC | Curiousily - Hacker's Guide to Machine Learning
22.05.2020 — Deep Learning, Machine Learning, DVC, Reproducibility — 5 min read
TL;DR Learn how to build a reproducible ML pipeline using DVC and Python. You’ll build an end-to-end example with 2 experiments and compare model evaluation metrics between them.
In this tutorial, you’ll build a complete reproducible ML pipeline with Python and DVC. The approach is ML library/toolkit agnostic, but we’ll use scikit-learn.
Why your work must be reproducible?
Overview of DVC
Create a new ML project from scratch
Add the first (baseline) experiment
Added DVC to the project
Build a complete ML pipeline
Add seconds experiment
Compare the evaluation metrics between experiments
Imagine that a paper is proposing a new method for solving a task and the main objective is improved by 10%. WOW! New SOTA! Or is it?
Reproducing the experiments is the only way to see for yourself. As a bonus, you’ll get a deeper understanding of the method. But how easy it is to do it?
Unfortunately, many authors don’t include their source code when publishing a paper. The reproducibility crisis is real! To comab this, some major ML conferences (NeurIPS and ICML) have some requirements to ensure reproducibility. The reproducibility checklist is one effort to summarise the main points. Things are getting better, but improvements are still needed.
Experimenting with ML boils down to writing and reading (a lot of) code. And what do you do when you want to find the truth? You go to the source! The source code (Yeah, I am watching too much Dom Mazzetti).
Reproducibility in the real world (a.k.a. your work)
All of this is great, but should you care? After all, you’re using ML in the real world!
You should care even more! The only good way to check if a piece of code is doing what the author intended is to show it to a lot of people. ML projects involve a lot more than “regular” code, though. Making your experiments hard to reproduce is a sure way to make someone give up on the review and go with a “f*ck it, I am out”.
Ok, how do you make your experiments reproducible?
Reproducing ML experiments with DVC
DVC stands for Data Version Control. It is a free and open-source project that helps you version control your experiments, store large files (on a variety of storage services), track metrics, and build completely reproducible pipelines.
DVC doesn’t really handle big file storage for you. It stores metafiles that point to the location of the files. Those places are known as remotes. Here are some of the remotes that DVC supports:
Directory on your file system (local)
We’ll have a look at a complete ML experiment and integrate it with DVC.
The data we’re going to use is listings of Udemy courses - 3.682 courses listings from 4 different subjects. The objective is to predict the number of students for each course.
Pretty much every ML pipeline can be boiled down to the following steps (this can be a never-ending cycle):
Deploy the model (if better than previous)
In this example, we’ll skip the deployment altogether and focus on experimenting.
One of the good things about DVC is that you can put off the integration until the very end of your first experiment. We’ll do just that - start with a plain old Python project.
Here’s the initial file structure:
2├── assets (dir)
3├── Pipfile
4├── Pipfile.lock
5└── studentpredictor (dir)
The studentpredictor directory will hold the source code, while assets will contain data and DVC related files.
We’ll manage the dependencies using Pipenv. Here are the contents of the Pipfile:
1[[source]]
2name = "pypi"
3url = "https://pypi.org/simple"
4verify_ssl = true
6[dev-packages]
7black = "==19.10b0"
8isort = "*"
9flake8 = "*"
11[packages]
12dvc = "*"
13gdown = "*"
14pandas = "*"
15scikit-learn = "*"
17[requires]
18python_version = "3.8"
Run this command in the root of your project once you add the file:
1pipenv install --dev
We’ll store the config as source code in the studentpredictor/config.py file:
5 RANDOM_SEED = 42
6 ASSETS_PATH = Path("./assets")
7 ORIGINAL_DATASET_FILE_PATH = ASSETS_PATH / "original_dataset" / "udemy_courses.csv"
8 DATASET_PATH = ASSETS_PATH / "data"
9 FEATURES_PATH = ASSETS_PATH / "features"
10 MODELS_PATH = ASSETS_PATH / "models"
11 METRICS_FILE_PATH = ASSETS_PATH / "metrics.json"
The first step is to get the dataset. I’ve already uploaded the CSV file to Google Drive. Add the studentpredictor/create_dataset.py file with the following contents:
1import gdown
6from config import Config
8np.random.seed(Config.RANDOM_SEED)
10Config.ORIGINAL_DATASET_FILE_PATH.parent.mkdir(parents=True, exist_ok=True)
11Config.DATASET_PATH.mkdir(parents=True, exist_ok=True)
13gdown.download(
14 "https://drive.google.com/uc?id=1gkYBOIMm8pAGunRoI3OzQHQrgOdaRjfC",
15 str(Config.ORIGINAL_DATASET_FILE_PATH),
18df = pd.read_csv(str(Config.ORIGINAL_DATASET_FILE_PATH))
20df_train, df_test = train_test_split(
21 df, test_size=0.2, random_state=Config.RANDOM_SEED,
24df_train.to_csv(str(Config.DATASET_PATH / "train.csv"), index=None)
25df_test.to_csv(str(Config.DATASET_PATH / "test.csv"), index=None)
We make all necessary directories and split the data into train and test. The resulting data frames are saved as CSV.
We’ll do some simple feature engineering to keep this part easy to understand. Create the studentpredictor/create_features.py file and fill it with this:
7Config.FEATURES_PATH.mkdir(parents=True, exist_ok=True)
9train_df = pd.read_csv(str(Config.DATASET_PATH / "train.csv"))
10test_df = pd.read_csv(str(Config.DATASET_PATH / "test.csv"))
13def extract_features(df):
14 df["published_timestamp"] = pd.to_datetime(df.published_timestamp).dt.date
15 df["days_since_published"] = (date.today() - df.published_timestamp).dt.days
16 return df[["num_lectures", "price", "days_since_published", "content_duration"]]
19train_features = extract_features(train_df)
20test_features = extract_features(test_df)
22train_features.to_csv(str(Config.FEATURES_PATH / "train_features.csv"), index=None)
23test_features.to_csv(str(Config.FEATURES_PATH / "test_features.csv"), index=None)
25train_df.num_subscribers.to_csv(
26 str(Config.FEATURES_PATH / "train_labels.csv"), index=None
28test_df.num_subscribers.to_csv(
29 str(Config.FEATURES_PATH / "test_labels.csv"), index=None
The only real feature we’re creating is the days_since_published. We get it from the published date of the course. We’re saving the features and labels as CSV files.
We’ll start with a baseline model. In this case - Linear Regression. Put this into studentpredictor/train_model.py:
8Config.MODELS_PATH.mkdir(parents=True, exist_ok=True)
10X_train = pd.read_csv(str(Config.FEATURES_PATH / "train_features.csv"))
11y_train = pd.read_csv(str(Config.FEATURES_PATH / "train_labels.csv"))
13model = LinearRegression()
14model = model.fit(X_train, y_train.to_numpy().ravel())
16pickle.dump(model, open(str(Config.MODELS_PATH / "model.pickle"), "wb"))
We dump the trained model with pickle. Ready to evaluate that bad boy!
We’ll focus on two metrics RMSE and
R^2
. Here is the studentpredictor/evaluate_model.py file:
5from sklearn.metrics import mean_squared_error
9X_test = pd.read_csv(str(Config.FEATURES_PATH / "test_features.csv"))
10y_test = pd.read_csv(str(Config.FEATURES_PATH / "test_labels.csv"))
12model = pickle.load(open(str(Config.MODELS_PATH / "model.pickle"), "rb"))
14r_squared = model.score(X_test, y_test)
16y_pred = model.predict(X_test)
17rmse = mean_squared_error(y_test, y_pred)
19with open(str(Config.METRICS_FILE_PATH), "w") as outfile:
20 json.dump(dict(r_squared=r_squared, rmse=rmse), outfile)
We’re writing the resulting metrics in a JSON file. How we’re going to use that? More on that later.
The project structure should now look like this:
5└── studentpredictor
6 ├── config.py
7 ├── create_dataset.py
8 ├── create_features.py
9 ├── evaluate_model.py
10 └── train_model.py
Adding DVC
You’ll interact with DVC mostly via the CLI. It is a tool that plays nice with GIT (understands tags and branches) and is language agnostic.
Initialize DVC
1dvc init
and add remote storage (local in this case)
1dvc remote add -d localremote /tmp/dvc-storage
disable analytics (optional)
1dvc config core.analytics false
This is a good place for a checkpoint:
2git commit -m "Add DVC config"
We’re ready to build the pipeline. DVC creates a graph with dependencies and outputs for each stage.
We’ll use dvc run to make each step reproducible. Let’s start with the dataset:
1dvc run -f assets/data.dvc \
2 -d studentpredictor/create_dataset.py \
3 -o assets/data \
4 python studentpredictor/create_dataset.py
Let’s dissect what is happening here:
-f assets/data.dvc saves the metafile used by DVC to reproduce this step
-d studentpredictor/create_dataset.py adds this script as a dependency for this step
-o assets/data tells that the outputs will be stored in that directory
Finally, we invoke the script that will do the actual work.
The stage for feature creation looks like this:
1dvc run -f assets/features.dvc \
2 -d studentpredictor/create_features.py \
3 -d assets/data \
4 -o assets/features \
5 python studentpredictor/create_features.py
Importantly, we add assets/data as a dependency for this step. This will force the execution of the previous step if something has changed.
You can probably figure out the training stage:
1dvc run -f assets/models.dvc \
2 -d studentpredictor/train_model.py \
3 -d assets/features \
4 -o assets/models \
5 python studentpredictor/train_model.py
The final stage - evaluation:
1dvc run -f assets/evaluate.dvc \
2 -d studentpredictor/evaluate_model.py \
4 -d assets/models \
5 -M assets/metrics.json \
6 python studentpredictor/evaluate_model.py
You’ll note that this step doesn’t specify outputs. But we have -M assets/metrics.json? This tells DVC that this is a metrics file (JSON and text files are currently supported).
Your first DVC pipeline is complete. Let’s save the progress:
2git commit -m "Linear Regression experiment with DVC"
We’ll also create a tag for the experiment (you’ll see why in a second):
1git tag -a "lr-experiment" -m "Experiment with Linear Regression"
Now we can use some DVC magic to see the evaluation metrics for our model:
1dvc metrics show -T
1lr-experiment:
2 assets/metrics.json:
3 r_squared: 0.03570513102945361
4 rmse: 6777.509886999257
Experimenting with Random Forest
Why did we do all this work? Was it all worth it?
Let’s start a second experiment with Random Forest regressor. Replace the contents of studentpredictor/train_model.py:
4from sklearn.ensemble import RandomForestRegressor
13model = RandomForestRegressor(
14 n_estimators=150, max_depth=6, random_state=Config.RANDOM_SEED
Let’s reproduce the complete pipeline using the new regressor:
1dvc repro assets/evaluate.dvc
DVC is smart enough to rerun only the steps that have changed and rewrite its internal graph.
Let’s save the second experiment:
2git commit -m "Add Random Forest experiment"
and create a tag for it:
1git tag -a "rf-experiment" -m "Experiment with Random Forest"
We can now compare the two experiments:
5rf-experiment:
You can do the same thing with branches, too (if that is your thing).
You can now build a complete reproducible ML pipelines with Python and DVC. Note that you can do it with any ML library/toolkit. How would you apply this to your experiments?
Do you make your experiment reproducible? How do you do it? How do you track your metrics? I am waiting for your answers in the comments below!
Reproducable Machine Learning Models with Git, Conda and DVC
Introduction to Data Version Control(DVC)
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Hacker's Guide to Neural Networks in JavaScript
Build Machine Learning models (especially Deep Neural Networks) that you can easily integrate with existing or new web apps. Think of your ReactJs, Vue, or Angular app enhanced with the power of Machine Learning models.
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|
Coordination Compounds Chemistry NEET Practice Questions, MCQs, Past Year Questions (PYQs), NCERT Questions, Question Bank, Class 11 and Class 12 Questions, and PDF solved with answers
Chemistry - Coordination Compounds
Fe2(CO)9 is diamagnetic.Which of the following reasons is correct?
1. Presence of one CO as bridge group
2. Presence of monodentate ligand
3. Metal-metal(Fe-Fe) bond in molecule
4. Resonance hybridization of CO
Subtopic: Organometallic Complexes & their Uses |
From the stability constant (hypothetical values)given below, predict which is the strongest ligand?
1. Cu2++4NH3
⇌
[Cu(NH3)4]2+;(k=4.5x1011)
2. Cu2++4CN
⇌
[Cu(CN)4]2-;(K=2.0x1027)
3. Cu2++2en
⇌
[Cu(en)2]2+;(K=3.0x1015)
4. Cu2++4H2O
⇌
[Cu(H2O)4]2+;(K=9.5x108)
Subtopic: VBT & CFT & their Limitations |
1. [Cu(NH3)4]2+
2. [Zn(OH)4]2-
3. [HgI4]2-
4. Fe(CO)5
Subtopic: E.A.N |
The coordination number and oxidation state of Cr in K3[Cr(C2O4)3] are respectively
1. 3 and +3
Subtopic: Werner’s Theory |
The octahedral complex will not show geometrical isomerism is :
(A and B are monodentate ligands)
1. [MA4B2]
2. [MA5B]
Subtopic: Isomerism in Coordination Compounds |
According to IUPAC nomenclature sodium nitroprusside is named as
1. sodium pentacyanonitrosyl ferrate(II)
2. sodium pentacyanonitrosyl ferrate(III)
3. sodium nitroferricyanide
4. sodium nitroferrocyanide
The compound that is not
\mathrm{\pi }
-bonded organometallic compound is :
1. K[PtCl3(
{\eta }^{\mathit{2}}
-C2H4)]
2. Fe(
{\eta }^{5}
-C5H5)2
3. Cr(
{\eta }^{6}
4. (CH3)4Sn
Subtopic: Organometallic Complexes & their Uses | VBT & CFT & their Limitations |
IUPAC name of [Pt(NH3)3 (Br)(NO2)Cl]Cl is :
1. Triamminebromidochloridonitroplatinum(IV)Chloride
2. Triamminebromonitrocholoroplatinum(IV)Chloride
3. Triamminechlorobromonitroplatinum(IV)Chloride
4. Triamminenitrochlorobromoplatinum(IV)Chloride
Subtopic: Coordination comp,Introduction and Classification / Nomenclature |
The denticity of EDTA is :
1. Monodentate.
2. Hexadentate.
3. Bidentate.
4. Tridentate.
Subtopic: Ligands |
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Erdős–Bacon number - Wikipedia
(Redirected from Erdos-Bacon number)
Closeness of someone's association with mathematician Paul Erdős and actor Kevin Bacon
A person's Erdős–Bacon number is the sum of one's Erdős number—which measures the "collaborative distance" in authoring academic papers between that person and Hungarian mathematician Paul Erdős—and one's Bacon number—which represents the number of links, through roles in films, by which the person is separated from American actor Kevin Bacon.[1][2] The lower the number, the closer a person is to Erdős and Bacon, which reflects a small world phenomenon in academia and entertainment.[3]
To have a defined Erdős–Bacon number, it is necessary to have both appeared in a film and co-authored an academic paper, although this in and of itself is not sufficient.
An extended definition named Erdős–Bacon-Sabbath includes the Sabbath number, an equivalent to Bacon number that connects musicians to the heavy metal band Black Sabbath.[4][5]
Mathematician Daniel Kleitman has an Erdős–Bacon number of 3. He co-authored papers with Erdős and has a Bacon number of 2 via Minnie Driver in Good Will Hunting; Driver and Bacon appeared together in Sleepers.[6]
Like Kleitman, mathematician Bruce Reznick has co-authored a paper with Erdős[7] and has a Bacon number of 2, via Roddy McDowall in the film Pretty Maids All in a Row, giving him an Erdős–Bacon number of 3 as well.[8]
Physicist Nicholas Metropolis has an Erdős number of 2,[9] and also a Bacon number of 2,[10] giving him an Erdős–Bacon number of 4.
Physicists Nicholas Metropolis and Richard Feynman both worked on the Manhattan Project at Los Alamos Laboratory. Via Metropolis, Feynman has an Erdős number of 3 and, from having appeared in the film Anti-Clock alongside Tony Tang, Feynman also has a Bacon number of 3. Richard Feynman thus has an Erdős–Bacon number of 6.[9]
Theoretical physicist Stephen Hawking has an Erdős–Bacon number of 6: his Bacon number of 2 (via his appearance alongside John Cleese in Monty Python Live (Mostly) who acted alongside Kevin Bacon in The Big Picture) is lower than his Erdős number of 4.[11]
Similarly to Stephen Hawking scientist Carl Sagan has an Erdős–Bacon number of 6, also from a Bacon number of 2 and an Erdős number of 4.[12]
Mathematician Jordan Ellenberg has an Erdős number of 3[13] and a Bacon number of 2 due to a cameo appearance in the film Gifted for which he was also the mathematical consultant[14]
Danica McKellar, who played Winnie Cooper in The Wonder Years, has an Erdős–Bacon number of 6. While an undergraduate at the University of California, Los Angeles McKellar coauthored a mathematics paper[15] with Lincoln Chayes who via his wife Jennifer Tour Chayes[16] has an Erdős number of 3, giving McKellar one of 4. Having worked with Margaret Easley, McKellar has a Bacon number of 2.[2]
Israeli-American actress Natalie Portman has an Erdős–Bacon number of 7.[17] She collaborated (using her birth name, Natalie Hershlag) with Abigail A. Baird,[18] who has a collaboration path[19][20][21] leading to Joseph Gillis, who has an Erdős number of 1.[22] Portman appeared in A Powerful Noise Live (2009) with Sarah Michelle Gellar, who appeared in The Air I Breathe (2007) with Bacon, giving Portman a Bacon number of 2 and an Erdős number of 5.
British actor Colin Firth has an Erdős–Bacon number of 6. Firth is credited as co-author of a neuroscience paper, "Political Orientations Are Correlated with Brain Structure in Young Adults",[23] after he suggested on BBC Radio 4 that such a study could be done.[24] Another author of that paper, Geraint Rees, has an Erdős number of 4,[25] which gives Firth an Erdős number of 5. Firth's Bacon number of 1 is due to his appearance in Where the Truth Lies.[26][27]
Kristen Stewart has an Erdős–Bacon number of 7; she is credited as a co-author on an artificial intelligence paper that was written after a technique was used for her short film Come Swim, giving her an Erdős number of 5,[28][29] and she co-starred with Michael Sheen in Twilight, who co-starred with Bacon in Frost/Nixon, giving her a Bacon number of 2.[30]
Elon Musk, who is neither an academic nor an actor, has an Erdős–Bacon number of 6. In 2010 Musk had a cameo in the film Iron Man 2.[31] Since actor Mickey Rourke played a role in both Iron Man 2 and in Diner where also Kevin Bacon played a role, Musk has a Bacon number of 2.[32] In 2021 Musk coauthored a peer-reviewed scientific paper on COVID-19 together with among others Pardis Sabeti.[33] Since Sabeti has an Erdős number of 3,[34] Musk has an Erdős number of 4[35] and consequently an Erdős–Bacon number of 6.
Mayim Bialik 5 2 7[36]
Jordan Ellenberg 3[13] 2[14] 5
Richard Feynman 3 3 6[9]
Colin Firth 5[a][23] 1[26] 6
Stephen Hawking 4 2[b] 6[11]
Daniel Kleitman 1 2 3[6]
Danica McKellar 4[37][15][16][38] 2[a] 6
Nicholas Metropolis 2[9] 2[10] 4
Elon Musk 4 2[b] 6
Natalie Portman 5[18][19][20][21][22] 2[a] 7[17]
Bruce Reznick 1 2 3
Carl Sagan 4 2[b] 6[12]
Kristen Stewart 5[28] 2[39][40] 7
Richard Thaler 3 2 5[41]
^ a b c See discussion above (Actors).
^ a b c Includes role as self.
^ Singh, Simon (May 1, 2002). "And the winner tonight is". The Telegraph. Archived from the original on November 12, 2012. Retrieved September 26, 2013.
^ a b "There's not much separating her from Bacon, Erdos". USA Today. August 14, 2007. Archived from the original on November 4, 2012.
^ Collins, James J.; Chow, Carson C. (1998). "It's a small world". Nature. 393 (6684): 409–10. Bibcode:1998Natur.393..409C. doi:10.1038/30835. PMID 9623993. S2CID 6827605.
^ Len Fisher (2016-02-17). "What's your Erdös-Bacon-Sabbath number?". Times Higher Education. Retrieved 2020-12-21.
^ Nancy Duvergne Smith (2013-01-29). "Erdos-Bacon-Sabbath Numbers: MITers at the Center of the Creative Universe". Slice of MIT. Retrieved 2020-12-21.
^ a b Grossman, Jerry (January 27, 1999). "The Erdös Number Project". Oakland University. Archived from the original on 1999-02-03. Retrieved 2021-03-03.
^ Erdős, P.; Hildebrand, A.; Odlyzko, A.; Pudaite, P.; Reznick, B. (1987). "The asymptotic behavior of a family of sequences". Pacific J. Math. no. 2 (126): 227–241.
^ Grossman, Jerry (December 6, 2018). "The Erdös Number Project". Oakland University. Archived from the original on 2020-03-11. Retrieved 2021-03-03.
^ a b c d "Richard Feynman". Erdős Bacon Sabbath Project. Archived from the original on 2017-12-25. Retrieved 2015-12-26.
^ a b "The Oracle of Bacon". oracleofbacon.org.
^ a b "Stephen Hawking". Erdős Bacon Sabbath Project. Archived from the original on 2017-12-25. Retrieved 2018-11-10.
^ a b "Carl Sagan". Erdős Bacon Sabbath Project. Archived from the original on 2018-01-11. Retrieved 2018-11-10.
^ a b "MR: Search MSC database". mathscinet.ams.org. Retrieved 2022-02-06. MR Erdos Number = 3 Jordan S. Ellenberg coauthored with Christopher M. Skinner MR1844206 Christopher M. Skinner coauthored with Andrew M. Odlyzko MR1210537 Andrew M. Odlyzko coauthored with Paul Erdős1 MR0535395
^ a b "The Oracle of Bacon". oracleofbacon.org. Retrieved 2022-02-14.
^ a b Chayes, L; McKellar, D; Winn, B (1998). "Percolation and Gibbs states multiplicity for ferromagnetic Ashkin-Teller models on
{\displaystyle \mathbb {Z} ^{2}}
". Journal of Physics A: Mathematical and General. 31 (45): 9055–9063. Bibcode:1998JPhA...31.9055C. doi:10.1088/0305-4470/31/45/005.
^ a b Chayes, J. T.; Chayes, L.; Kotecký, R. (1995). "The analysis of the Widom-Rowlinson model by stochastic geometric methods". Communications in Mathematical Physics. 172 (3): 551. Bibcode:1995CMaPh.172..551C. doi:10.1007/BF02101808. S2CID 15051914.
^ a b "MICHAEL'S ERDŐS-BACON NUMBER | The Liquid Narrative Research Group". liquidnarrative.eae.utah.edu.
^ a b Baird, A; Kagan, J; Gaudette, T; Walz, KA; Hershlag, N; Boas, DA (2002). "Frontal Lobe Activation during Object Permanence: Data from Near-Infrared Spectroscopy". NeuroImage. 16 (4): 1120–5. doi:10.1006/nimg.2002.1170. PMID 12202098. S2CID 15630444.
^ a b Baird, Abigail A.; Colvin, Mary K.; Vanhorn, John D.; Inati, Souheil; Gazzaniga, Michael S. (2005). "Functional Connectivity: Integrating Behavioral, Diffusion Tensor Imaging, and Functional Magnetic Resonance Imaging Data Sets". Journal of Cognitive Neuroscience. 17 (4): 687–93. CiteSeerX 10.1.1.484.1868. doi:10.1162/0898929053467569. PMID 15829087. S2CID 4666737.
^ a b Victor, Jonathan D.; Maiese, Kenneth; Shapley, Robert; Sidtis, John; Gazzaniga, Michael S. (1989). "Acquired central dyschromatopsia: analysis of a case with preservation of color discrimination". Clinical Vision Sciences. 4: 183–96.
^ a b Azor, Ruth; Gillis, J.; Victor, J. D. (1982). "Combinatorial Applications of Hermite Polynomials". SIAM Journal on Mathematical Analysis. 13 (5): 879–90. doi:10.1137/0513062.
^ a b Erdos, P.; Gillis, J. (2009). "Note on the Transfinite Diameter". Journal of the London Mathematical Society. s1-12 (3): 185. doi:10.1112/jlms/s1-12.2.185.
^ a b Kanai, Ryota; Feilden, Tom; Firth, Colin; Rees, Geraint (2011). "Political Orientations Are Correlated with Brain Structure in Young Adults". Current Biology. 21 (8): 677–80. doi:10.1016/j.cub.2011.03.017. PMC 3092984. PMID 21474316.
^ "Colin Firth credited in brain research". BBC News. 2011-06-05. Retrieved 2021-03-13.
^ "From Geraint Rees 0001 to Paul Erdős in four papers". csauthors.net. Archived from the original on 2021-03-13. Retrieved 2021-03-13.
^ a b Where the Truth Lies at IMDb
^ "The Oracle of Bacon". oracleofbacon.org. Retrieved 2021-03-13.
^ a b Gershgorn, Dave. "Kristen Stewart (yes, that Kristen Stewart) just released a research paper on artificial intelligence".
^ "From Paul Erdős to Kristen Stewart in five papers". Archived from the original on 2018-03-14. Retrieved 2021-03-13.
^ Tate, Ryan (2012-09-20). "10 Awkward Hollywood Cameos by Tech Founders". Wired. Archived from the original on 2017-12-01. Retrieved 2021-05-08.
^ "The Oracle of Bacon". oracleofbacon.org. Retrieved 2021-05-08. Elon Musk has a Bacon number of 2. Elon Musk was in Iron Man 2 with Mickey Rourke was in Diner with Kevin Bacon
^ Bartsch, Yannic C.; Fischinger, Stephanie; Siddiqui, Sameed M.; Chen, Zhilin; Yu, Jingyou; Gebre, Makda; Atyeo, Caroline; Gorman, Matthew J.; Zhu, Alex Lee; Kang, Jaewon; Burke, John S.; Slein, Matthew; Gluck, Matthew J.; Beger, Samuel; Hu, Yiyuan; Rhee, Justin; Petersen, Eric; Mormann, Benjamin; de St Aubin, Michael; Hasdianda, Mohammad A.; Jambaulikar, Guruprasad; Boyer, Edward W.; Sabeti, Pardis C.; Barouch, Dan H.; Julg, Boris D.; Musk, Elon R.; Menon, Anil S.; Lauffenburger, Douglas A.; Nilles, Eric J.; Alter, Galit (2021-02-15). "Discrete SARS-CoV-2 antibody titers track with functional humoral stability". Nature Communications. 12 (1): 1018. Bibcode:2021NatCo..12.1018B. doi:10.1038/s41467-021-21336-8. PMC 7884400. PMID 33589636.
^ "MR: Search MSC database". mathscinet.ams.org. Retrieved 2021-05-08. MR Erdos Number = 3 Pardis C. Sabeti coauthored with Michael Mitzenmacher MR3595146 Michael Mitzenmacher coauthored with Joel H. Spencer MR2056083 Joel H. Spencer coauthored with Paul Erdős1 MR0382007
^ "Jerry Grossman's Web Page > The Erdös Number Project > Some Famous People with Finite Erdös Numbers >". Oakland University. Retrieved 2021-05-08. Elon Musk entrepreneur 4
^ "Mayim Bialik". Erdős Bacon Sabbath Project. Archived from the original on 2018-01-15. Retrieved 2014-02-09.
^ "The Erdős Number Project, Erdos1". Archived from the original on 2006-12-07. Retrieved 2006-12-20.
^ Kotecký, R.; Preiss, D. (1986). "Cluster expansion for abstract polymer models". Communications in Mathematical Physics. 103 (3): 491–8. Bibcode:1986CMaPh.103..491K. doi:10.1007/BF01211762. S2CID 121879006.
^ Twilight at IMDb co-starred Kristen Stewart and Michael Sheen
^ Frost/Nixon at IMDb co-starred Michael Sheen and Kevin Bacon
^ Thaler, Richard H. [@R_Thaler] (May 10, 2016). "Learned I have a Bacon-Erdos number=5! Wrote a paper with Peter Wakker an Erdos 2 via Fishburn, and am Bacon 2 via Ryan Gosling in Big Short" (Tweet) – via Twitter.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Erdős–Bacon_number&oldid=1089313852"
Lists of people by association
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Introduction to Reinforcement Learning | Deep Reinforcement Learning for Hackers (Part 0) | Curiousily - Hacker's Guide to Machine Learning
04.12.2017 — Machine Learning, Reinforcement Learning, Deep Learning, Python — 4 min read
The best way to understand what Reinforcement Learning is to watch this video:
Remember the first time you went behind the wheel of a car? Your dad, mom or driving instructor was next to you, waiting for you to mess something up. You had a clear goal - make a couple of turns and get to the supermarket for ice cream. The task was infinitely more fun if you had to learn to drive stick. Ah, good times. Too bad that your kids might never experience that. More on that later.
Reinforcement learning (RL) is learning what to do, given a situation and a set of possible actions to choose from, in order to maximize a reward. The learner, which we will call agent, is not told what to do, he must discover this by himself through interacting with the environment. The goal is to choose its actions in such a way that the cumulative reward is maximized. So, choosing the best reward now, might not be the best decision, in the long run. That is greedy approaches might not be optimal.
Back to you, behind the wheel with running engine, properly strapped seatbelt, adrenaline pumping and rerunning the latest Fast & Furious through your mind - you have a good feeling about this, the passenger next to you does not look that scared, after all…
How all of this relates to RL? Let’s try to map your situation to an RL problem. Driving is really complex, so for your first lesson, your instructor will do everything, except turning the wheel. The environment is the nature itself and the agent is you. The state of the environment (situation) can be defined by the position of your car, surrounding cars, pedestrians, upcoming crossroads etc. You have 3 possible actions to choose from - turn left, keep straight and turn right. The reward is well defined - you will eat ice cream if you are able to reach the supermarket. Your instructor will give your intermediate rewards based on your performance. At each step (let’s say once every second), you will have to make a decision - turn left, right or continue straight ahead. Whether or not the ice cream is happening is mostly up to you.
Let’s summarize what we’ve learned so far. We have an agent and an environment. The environment gives the agent a state. The agent chooses an action and receives a reward from the environment along with the new state. This learning process continues until the goal is achieved or some other condition is met.
Source: https://phrasee.co/
Let’s have a look at some example applications of RL:
Cart-Pole Balancing
Goal - Balance the pole on top of a moving cart
State - angle, angular speed, position, horizontal velocity
Actions - horizontal force to the cart
Reward - 1 at each time step if the pole is upright
Goal - Beat the game with the highest score
State - Raw game pixels of the game
Actions - Up, Down, Left, Right etc
Reward - Score provided by the game
Goal - Eliminate all opponents
Reward - Positive when eliminating an opponent, negative when the agent is eliminated
Training robots for Bin Packing
Source: www.plasticsdist.com
Goal - Pick a device from a box and put it into a container
State - Raw pixels of the real world
Actions - Possible actions of the robot
Reward - Positive when placing a device successfully, negative otherwise
You started thinking that all RL researchers are failed pro-gamers, didn’t you? In practice, that doesn’t seem to be the case. For example, somewhat “meta” applications include “Designing Neural Network Architectures using Reinforcement Learning”.
Formalizing the RL problem
Markov Decision Process (MDP) is mathematical formulations of the RL problem. They satisfy the Markov property:
Markov property - the current state completely represents the state of the environment (world). That is, the future depends only on the present.
An MDP can be defined by
(S, A, R, P, \gamma)
S
- set of possible states
A
- set of possible actions
R
- probability distribution of reward given (state, action) pair
P
- probability distribution over how likely any of the states is to be the new states, given (state, action) pair. Also known as transition probability.
\gamma
- reward discount factor
How MDPs work
At the initial time step
t=0
, the environment chooses initial state
s_0 \sim p(s_0)
. That state is used as a seed state for the following loop:
t=0
until done:
The agent selects action
a_t
The environment chooses reward
r_t \sim R(. \vert\, s_t, a_t)
s_{t + 1} \sim P(. \vert\, s_t, a_t)
The agent receives reward
r_t
s_{t + 1}
More formally, the environment does not choose, it samples from the reward and transition probability distributions.
What is the objective of all this? Find a function
\pi^*
, known as optimal policy, that maximizes the cumulative discounted reward:
\sum_{t \geq 0}\gamma^t r_t
A policy
\pi
is a function that maps state
s
to action
a
, that our agent believes is the best given that state.
Your first MDP
Let’s get back to you, cruising through the neighborhood, dreaming about that delicious ice cream. Here is one possible situation, described as an MDP:
Your objective is to get to the bitten ice cream on a stick, without meeting a zombie. The reasoning behind the new design is based on solid data science - people seem to give a crazy amount of cash for a bitten fruit and everybody knows that candy is much tastier. Putting it together you get “the all-new ice cream”. And honestly, it wouldn’t be cool to omit the zombies, so there you have it.
The state is fully described by the grid. At the first step you have the following actions:
Crashing into a zombie (Carmageddon anyone?) gives a reward of -100 points, taking an action is -1 points and eating the ice cream gives you the crazy 1000 points. Why -1 points for taking an action? Well, the store might close anytime now, so you have to get there as soon as possible.
Congrats, you just created your first MDP. But how do we solve the problem? Stay tuned for that :)
Oops, almost forgot, your reward for reading so far:
Dissecting Reinforcement Learning
Reinforcement Learning: An Introduction 2nd edition draft
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Investigation of injection-induced seismicity using a coupled fluid flow and rate/state friction modelInjection-induced seismicity | Geophysics | GeoScienceWorld
Mark W. McClure;
, Stanford Geothermal Program Department of Energy Resources Engineering, Stanford, California,
. E-mail: mcclure@stanford.edu, horne@stanford.edu.
Roland N. Horne
Mark W. McClure, Roland N. Horne; Investigation of injection-induced seismicity using a coupled fluid flow and rate/state friction model. Geophysics 2012;; 76 (6): WC181–WC198. doi: https://doi.org/10.1190/geo2011-0064.1
We describe a numerical investigation of seismicity induced by injection into a single isolated fracture. Injection into a single isolated fracture is a simple analog for shear stimulation in enhanced geothermal systems (EGS) during which water is injected into fractured, low permeability rock, triggering slip on preexisting large scale fracture zones. A model was developed and used that couples (1) fluid flow, (2) rate and state friction, and (3) mechanical stress interaction between fracture elements. Based on the results of this model, we propose a mechanism to describe the process by which the stimulated region grows during shear stimulation, which we refer to as the sequential stimulation (SS) mechanism. If the SS mechanism is realistic, it would undermine assumptions that are made for the estimation of the minimum principal stress and unstimulated hydraulic diffusivity. We investigated the effect of injection pressure on induced seismicity. For injection at constant pressure, there was not a significant dependence of maximum event magnitude on injection pressure, but there were more relatively large events for higher injection pressure. Decreasing injection pressure over time significantly reduced the maximum event magnitude. Significant seismicity occurred after shut-in, which was consistent with observations from EGS stimulations. Production of fluid from the well immediately after injection inhibited shut-in seismic events. The results of the model in this study were found to be broadly consistent with results from prior work using a simpler treatment of friction that we refer to as static/dynamic. We investigated the effect of shear-induced pore volume dilation and the rate and state characteristic length scale,
dc
. Shear-induced pore dilation resulted in a larger number of lower magnitude events. A larger value of
dc
caused slip to occur aseismically.
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Gamma cumulative distribution function - MATLAB gamcdf - MathWorks Italia
Confidence Interval of Gamma cdf Value
Gamma cumulative distribution function
p = gamcdf(x,a)
p = gamcdf(x,a,b)
[p,pLo,pUp] = gamcdf(x,a,b,pCov)
[p,pLo,pUp] = gamcdf(x,a,b,pCov,alpha)
___ = gamcdf(___,'upper')
p = gamcdf(x,a) returns the cumulative distribution function (cdf) of the standard gamma distribution with the shape parameters in a, evaluated at the values in x.
p = gamcdf(x,a,b) returns the cdf of the gamma distribution with the shape parameters in a and scale parameters in b, evaluated at the values in x.
[p,pLo,pUp] = gamcdf(x,a,b,pCov) also returns the 95% confidence interval [pLo,pUp] of p when a and b are estimates. pCov is the covariance matrix of the estimated parameters.
[p,pLo,pUp] = gamcdf(x,a,b,pCov,alpha) specifies the confidence level for the confidence interval [pLo pUp] to be 100(1–alpha)%.
___ = gamcdf(___,'upper') returns the complement of the cdf, evaluated at the values in x, using an algorithm that more accurately computes the extreme upper-tail probabilities than subtracting the lower tail value from 1. 'upper' can follow any of the input argument combinations in the previous syntaxes.
Compute the cdf of the mean of the gamma distribution, which is equal to the product of the parameters ab.
prob = gamcdf(a.*b,a,b)
As ab increases, the distribution becomes more symmetric, and the mean approaches the median.
Find a confidence interval estimating the probability that an observation is in the interval [0 10] using gamma distributed data.
Generate a sample of 1000 gamma distributed random numbers with shape 2 and scale 5.
Compute estimates for the parameters.
[params,~] = gamfit(x)
Store the parameters as ahat and bhat.
ahat = params(1);
bhat = params(2);
Find the covariance of the parameter estimates.
[~,nCov] = gamlike(params,x)
nCov = 2×2
Create a confidence interval estimating the probability that an observation is in the interval [0 10].
[prob,pLo,pUp] = gamcdf(10,ahat,bhat,nCov)
prob = 0.5830
Determine the probability that an observation from the gamma distribution with shape parameter 2 and scale parameter 3 will is in the interval [150 Inf].
p1 = 1 - gamcdf(150,2,3)
gamcdf(150, 2, 3) is nearly 1, so p1 becomes 0. Specify 'upper' so that gamcdf computes the extreme upper-tail probabilities more accurately.
p2 = gamcdf(150,2,3,'upper')
To evaluate the cdfs of multiple distributions, specify a and b using arrays.
If one or more of the input arguments x, a, and b are arrays, then the array sizes must be the same. In this case, gamcdf expands each scalar input into a constant array of the same size as the array inputs. Each element in p is the cdf value of the distribution specified by the corresponding elements in a and b, evaluated at the corresponding element in x.
Shape of the gamma distribution, specified as a positive scalar value or an array of positive scalar values.
Scale of the gamma distribution, specified as a positive scalar value or an array of positive scalar values.
Covariance of the estimates a and b, specified as a 2-by-2 matrix.
If you specify pCov to compute the confidence interval [pLo,pUp], then x, a, and b must be scalar values.
You can estimate a and b by using gamfit or mle, and estimate the covariance of a and b by using gamlike. For an example, see Confidence Interval of Gamma cdf Value.
cdf values evaluated at the values in x, returned as a scalar value or an array of scalar values. p is the same size as x, a, and b after any necessary scalar expansion. Each element in p is the cdf value of the distribution specified by the corresponding elements in a and b, evaluated at the corresponding element in x.
The gamma cdf is
p=F\left(x|a,b\right)=\frac{1}{{b}^{a}\Gamma \left(a\right)}\underset{0}{\overset{x}{\int }}{t}^{a-1}{e}^{\frac{-t}{b}}dt.
The result p is the probability that a single observation from a gamma distribution with parameters a and b falls in the interval [0,x].
f\left(x|a,b\right)=\text{gammainc}\left(\frac{x}{b},a\right).
The standard gamma distribution occurs when b = 1, which coincides with the incomplete gamma function precisely.
gamcdf is a function specific to the gamma distribution. Statistics and Machine Learning Toolbox™ also offers the generic function cdf, which supports various probability distributions. To use cdf, create a GammaDistribution probability distribution object and pass the object as an input argument or specify the probability distribution name and its parameters. Note that the distribution-specific function gamcdf is faster than the generic function cdf.
GammaDistribution | cdf | gampdf | gaminv | gamstat | gamfit | gamlike | gamrnd | gamma
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Thermodynamics, Popular Questions: CBSE Class 11-science CHEMISTRY, Chemistry Part I - Meritnation
Akansha Patel asked a question
°
°
Yadhu Krishnan asked a question
Standard enthalpy of vapourisation DvapHQ for water at 100° C is 40.66 kJmol–1. The internal energy of vapourisation of water at 100°C (in kJmol–1) is ?
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the heat of combustion of naphthalene at constant volume at 25 degree celsius was found to be -5133 kj/mole calculate the value of enthalpy change at constant pressure at the same temperature
Aparna Pandey asked a question
35 ML OF OXYGEN WERE COLLECTED AT 6 DEGREE CELCIUS AND 758 MM PRESSURE. CALCULATE ITS VOLUME AT NTP? WHAT IS THE VALUE OF T2 HERE.
Oriel asked a question
Calculate the final volume of an ideal gas which expands against a constant pressure of (3.039 X 10^5) Nm^-2 from a volume 18 dm3. The work involved in the process is (2.027 X 10^3)
Prathiksha asked a question
Chinmay asked a question
Exothermic reactions which are associated with a decrease in
entropy are spontaneous at lower temperature. Justify on the basis
of Gibbs equation.
what is the value of (delta) ng for the following reaction :-
H2(g) + I2(g) -- 2HI(g)
abcd1996 asked a question
predict the sign of entropy change for the following:-
a)C(graphite)-C(diamond)
b)electrolysis of NaCl solution
c)sublimation of camphor
d)CaCO3(s)-CaO(s)+ CO2(g)
Shefali Jumnani asked a question
\to
{△}_{r}H°=-1004.0 kJ
\to
△
°
-183.9 kJ
\to
{△}_{r}H°=-73.2 kJ
\to
{△}_{r}H°=-643.0 kJ
Hrishikesh asked a question
A spherical balloon of 21 cm diameter is to be filled up with hydrogen at NTP from a cylinder containing the gas at 20 atm. at 27 degree celcius. If the cylinder can hold 2.82 litres of water, calculate the numbers of balloons that can be filled up ?
is melting of ice at 270K and 1am pressure,a spontaneous reaction?
Madan Bhola asked a question
what is atomicity of gas?explain briefly
Dhanika asked a question
What is the difference between process and reaction?
Samuel Mavelil asked a question
Find the difference between heats of reaction at constant pressure and constant volume for the following reaction at 25 C in kJ.
2C6H6 + 15O2 ---------> 12CO2 + 6H2O.
The standard enthalpy of decomposition of N2O4 to NO2 is 58.04 kJ the standard entropy for this reaction is 176. 7J/K find the standard free energy change for this reaction at 25 deg Celsius
DELTA n=delta np-delta nr.wht is this equation.pls explain wht is n and np and nr and say the use of this equation
Ritik Sharma asked a question
explain the effect of temperature on spontinity of exothermic and endothermic reaction?plz reply soon
Jenali Bhingradiya asked a question
3. \mathrm{The} \mathrm{process} {\mathrm{CH}}_{3}\mathrm{COOH} \underset{}{\overset{ }{\to }} {\mathrm{CH}}_{3}{\mathrm{COO}}^{-}+{\mathrm{H}}^{+}, \mathrm{should} \mathrm{be} :\phantom{\rule{0ex}{0ex}}\left(1\right) \mathrm{exothermic} \left(2\right) \mathrm{endothermic}\phantom{\rule{0ex}{0ex}}\left(3\right) \mathrm{nethier} \mathrm{exothermic} \mathrm{nor} \mathrm{endothermic} \left(4\right) \mathrm{exothermic} \mathrm{or} \mathrm{endothermic} \mathrm{depending} \mathrm{upon} \mathrm{tempreature}
Deep Dodhia asked a question
Parul Sharma asked a question
calculate the enthalpy of the reaction of C2H4(g)+H2(g) give rise to C2H6(g)from the the following data??
C2H4(g)+3O2(g) give rise to 2C02+2H20(l) delH=-1411KJ
2C2H6(g)+7O2(g) give rise to 4CO2(G)+6H2O(l) delH=-1560KJ
2H2(g)+O2(g) give rise to 2H2O(l) delH=-285.8KJ
pls fast urgent I want it within 2hrs????pls plsssssssss help
Explain the term 'ionic product of water' and 'ph value'.how does the former change with temperature?
3 mol of an ideal gas at 1.5 atm and 25 degree celsius expands isothermally in a reversible manner to twice the original volume against an external pressure of 1 atm. calculate the work done....
Manasi Tambade asked a question
A CYLINDER CONTAINS EITHER ETHYLENE OR PROPYLENE 12 ML OF GAS REQUIRED 54 ML OF OXYGEN FOR COMPLETE COMBUSTION. THE GAS IS
B) PROPYLENE
C) 1:1 MIXTURE OF TWO GASES
Arpit Bal asked a question
Reversible melting of solid benzene at 1atm and normal melting point correspond to----------1. q>0 2. w<0 3.∆E > 0 4. All of these
Which of the following is a process taking place with increase in entropy?
i) Freezing of water ii) Condensation of steam iii) Cooling of a liquid iv) Dissolution of a solute
Sneharth Bhajani asked a question
If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourisation of 1 mol of water at 1bar and 100°C is 41kJ mol-1. Calculate the internal energy change, when
For an isolated system, delta U = 0; what will be delta S?
a sample of gas undergoes expansion against a pressure off one atm.from a volume of 500ml to 906ml by absorbing 500J of heat.calculate the change in internal energy
EXPLAIN THAT SINCE WORK DONE IN REVERSIBLE PROCESS IS MAXIMUM THAN IN IRREVERSIBLE PROCESS DURING EXPANSION BUT WHY WORK IN REVERSIBLE PROCESS IS MINIMUM DURING COMPRESSION OR CONTRACTION OF A GAS
1. W = -∆U a) Enthalpy change
2. ∆U = 0 b) Universal gas constant
3. Cp - Cv c) Adiabatic process
4. qP d) Isothermal process
e) Cyclic process
Tej Gupte asked a question
show that the heat absorbed at constant volume is equal to increase in internal energy of the system wheras that at constant pressure is equal to the increase in enthalpy of the system
Annpurna asked a question
what is entropy change for reaction 2H2(g) + O2 (g) -----?2H2O(l). Standard entropies of H2(g) , O2(g) and H2O(l) are 126.6, 201.20,68.0 J/k mol respectively
Please answer with explanation.
Kawalpreet Juneja asked a question
2A ₂ (g) + 5B ₂ (g) -------> 2A ₂ B5 (g)
at temperature = 27°C that is... 300Kelvin
The difference between deltaH and deltaE is X.
We have to find the ratio X/R.
I'm a bit confused to place which value of R here.
Please try to answer it before 1:00pm today.
1) 3 cal mol-1K-1
2)5 cal mol-1K-1
3) 7 cal mol-1K-1
Vishnu P.nair asked a question
Calculate the enthalpy change on freezing of 1.0 mol of waterat 10.0°C to ice at –10.0°C. ΔfusH= 6.03 kJ mol–1at 0°C.
Cp[H2O(l)] = 75.3 J mol–1K–1
Cp[H2O(s)] = 36.8 J mol–1K–1
difference between delta H and delta U
Shobhanand Jha asked a question
when 1g of anhydrous oxalic acid is burnt at 25°C,the amount of heat liberated is 2.835kJ.Enthalpy change for combustion is
Kam asked a question
what is the use of stirrer in these calorimeters..
write the conjugate acid and base of H2O2 ???
Sanjeet Dash asked a question
For a reaction,both delta H and delta S are positive.under what condition,the reaction occurs spontaneously
Hiba Ashraf asked a question
Aravind Murari asked a question
What is first law of thermodynamics.
Which one has higher entropy h2 or 2h ? H2 molecule is more stable then 2h atom then why 2h has higher entropy
Enthalpy and entropy changes of a reaction are 40.63 kJ/mol and 108.8 J/K/mol respectively. Predict the feasibility of the reaction at 27⁰C. no links
Shhhhhhhhh.......... ... asked a question
how does Hess's law follow 1st law of thermodynamics?
The incorrect statement among following is 1) when a substance is in its thermodynamics standard state, the substance must be at 25⁰C 2) heat capacity at constant pressure is an extensive property. 3) Reversible process can be reversed at any point in the process by making infinitesimal change. 4) Molar internal energy is an intensive property.
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Gromov–Witten invariant - Wikipedia
In mathematics, specifically in symplectic topology and algebraic geometry, Gromov–Witten (GW) invariants are rational numbers that, in certain situations, count pseudoholomorphic curves meeting prescribed conditions in a given symplectic manifold. The GW invariants may be packaged as a homology or cohomology class in an appropriate space, or as the deformed cup product of quantum cohomology. These invariants have been used to distinguish symplectic manifolds that were previously indistinguishable. They also play a crucial role in closed type IIA string theory. They are named after Mikhail Gromov and Edward Witten.
The rigorous mathematical definition of Gromov–Witten invariants is lengthy and difficult, so it is treated separately in the stable map article. This article attempts a more intuitive explanation of what the invariants mean, how they are computed, and why they are important.
3 Related invariants and other constructions
4 Application in physics
X: a closed symplectic manifold of dimension 2k,
A: a 2-dimensional homology class in X,
g: a non-negative integer,
n: a non-negative integer.
Now we define the Gromov–Witten invariants associated to the 4-tuple: (X, A, g, n). Let
{\displaystyle {\overline {\mathcal {M}}}_{g,n}}
be the Deligne–Mumford moduli space of curves of genus g with n marked points and
{\displaystyle {\overline {\mathcal {M}}}_{g,n}(X,A)}
denote the moduli space of stable maps into X of class A, for some chosen almost complex structure J on X compatible with its symplectic form. The elements of
{\displaystyle {\overline {\mathcal {M}}}_{g,n}(X,A)}
are of the form:
{\displaystyle (C,x_{1},\ldots ,x_{n},f)}
where C is a (not necessarily stable) curve with n marked points x1, ..., xn and f : C → X is pseudoholomorphic. The moduli space has real dimension
{\displaystyle d:=2c_{1}^{X}(A)+(2k-6)(1-g)+2n.}
{\displaystyle \mathrm {st} (C,x_{1},\ldots ,x_{n})\in {\overline {\mathcal {M}}}_{g,n}}
denote the stabilization of the curve. Let
{\displaystyle Y:={\overline {\mathcal {M}}}_{g,n}\times X^{n},}
which has real dimension
{\displaystyle 6g-6+2(k+1)n}
. There is an evaluation map
{\displaystyle {\begin{cases}\mathrm {ev} :{\overline {\mathcal {M}}}_{g,n}(X,A)\to Y\\\mathrm {ev} (C,x_{1},\ldots ,x_{n},f)=\left(\operatorname {st} (C,x_{1},\ldots ,x_{n}),f(x_{1}),\ldots ,f(x_{n})\right).\end{cases}}}
The evaluation map sends the fundamental class of
{\displaystyle {\overline {\mathcal {M}}}_{g,n}(X,A)}
to a d-dimensional rational homology class in Y, denoted
{\displaystyle GW_{g,n}^{X,A}\in H_{d}(Y,\mathbb {Q} ).}
In a sense, this homology class is the Gromov–Witten invariant of X for the data g, n, and A. It is an invariant of the symplectic isotopy class of the symplectic manifold X.
To interpret the Gromov–Witten invariant geometrically, let β be a homology class in
{\displaystyle {\overline {\mathcal {M}}}_{g,n}}
{\displaystyle \alpha _{1},\ldots ,\alpha _{n}}
homology classes in X, such that the sum of the codimensions of
{\displaystyle \beta ,\alpha _{1},\ldots ,\alpha _{n}}
equals d. These induce homology classes in Y by the Künneth formula. Let
{\displaystyle GW_{g,n}^{X,A}(\beta ,\alpha _{1},\ldots ,\alpha _{n}):=GW_{g,n}^{X,A}\cdot \beta \cdot \alpha _{1}\cdots \alpha _{n}\in H_{0}(Y,\mathbb {Q} ),}
{\displaystyle \cdot }
denotes the intersection product in the rational homology of Y. This is a rational number, the Gromov–Witten invariant for the given classes. This number gives a "virtual" count of the number of pseudoholomorphic curves (in the class A, of genus g, with domain in the β-part of the Deligne–Mumford space) whose n marked points are mapped to cycles representing the
{\displaystyle \alpha _{i}}
Put simply, a GW invariant counts how many curves there are that intersect n chosen submanifolds of X. However, due to the "virtual" nature of the count, it need not be a natural number, as one might expect a count to be. For the space of stable maps is an orbifold, whose points of isotropy can contribute noninteger values to the invariant.
There are numerous variations on this construction, in which cohomology is used instead of homology, integration replaces intersection, Chern classes pulled back from the Deligne–Mumford space are also integrated, etc.
Computational techniques[edit]
Gromov–Witten invariants are generally difficult to compute. While they are defined for any generic almost complex structure J, for which the linearization D of the
{\displaystyle {\bar {\partial }}_{j,J}}
operator is surjective, they must actually be computed with respect to a specific, chosen J. It is most convenient to choose J with special properties, such as nongeneric symmetries or integrability. Indeed, computations are often carried out on Kähler manifolds using the techniques of algebraic geometry.
However, a special J may induce a nonsurjective D and thus a moduli space of pseudoholomorphic curves that is larger than expected. Loosely speaking, one corrects for this effect by forming from the cokernel of D a vector bundle, called the obstruction bundle, and then realizing the GW invariant as the integral of the Euler class of the obstruction bundle. Making this idea precise requires significant technical argument using Kuranishi structures.
The main computational technique is localization. This applies when X is toric, meaning that it is acted upon by a complex torus, or at least locally toric. Then one can use the Atiyah–Bott fixed-point theorem, of Michael Atiyah and Raoul Bott, to reduce, or localize, the computation of a GW invariant to an integration over the fixed-point locus of the action.
Another approach is to employ symplectic surgeries to relate X to one or more other spaces whose GW invariants are more easily computed. Of course, one must first understand how the invariants behave under the surgeries. For such applications one often uses the more elaborate relative GW invariants, which count curves with prescribed tangency conditions along a symplectic submanifold of X of real codimension two.
Related invariants and other constructions[edit]
The GW invariants are closely related to a number of other concepts in geometry, including the Donaldson invariants and Seiberg–Witten invariants in the symplectic category, and Donaldson–Thomas theory in the algebraic category. For compact symplectic four-manifolds, Clifford Taubes showed that a variant of the GW invariants (see Taubes's Gromov invariant) are equivalent to the Seiberg–Witten invariants. For algebraic threefolds, they are conjectured to contain the same information as integer valued Donaldson–Thomas invariants. Physical considerations also give rise to Gopakumar–Vafa invariants, which are meant to give an underlying integer count to the typically rational Gromov-Witten theory. The Gopakumar-Vafa invariants do not presently have a rigorous mathematical definition, and this is one of the major problems in the subject.
The Gromov-Witten invariants of smooth projective varieties can be defined entirely within algebraic geometry. The classical enumerative geometry of plane curves and of rational curves in homogeneous spaces are both captured by GW invariants. However, the major advantage that GW invariants have over the classical enumerative counts is that they are invariant under deformations of the complex structure of the target. The GW invariants also furnish deformations of the product structure in the cohomology ring of a symplectic or projective manifold; they can be organized to construct the quantum cohomology ring of the manifold X, which is a deformation of the ordinary cohomology. The associativity of the deformed product is essentially a consequence of the self-similar nature of the moduli space of stable maps that are used to define the invariants.
The quantum cohomology ring is known to be isomorphic to the symplectic Floer homology with its pair-of-pants product.
Application in physics[edit]
GW invariants are of interest in string theory, a branch of physics that attempts to unify general relativity and quantum mechanics. In this theory, everything in the universe, beginning with the elementary particles, is made of tiny strings. As a string travels through spacetime it traces out a surface, called the worldsheet of the string. Unfortunately, the moduli space of such parametrized surfaces, at least a priori, is infinite-dimensional; no appropriate measure on this space is known, and thus the path integrals of the theory lack a rigorous definition.
The situation improves in the variation known as closed A-model. Here there are six spacetime dimensions, which constitute a symplectic manifold, and it turns out that the worldsheets are necessarily parametrized by pseudoholomorphic curves, whose moduli spaces are only finite-dimensional. GW invariants, as integrals over these moduli spaces, are then path integrals of the theory. In particular, the free energy of the A-model at genus g is the generating function of the genus g GW invariants.
Cotangent complex - for deformation theory
McDuff, Dusa & Salamon, Dietmar (2004). J-Holomorphic Curves and Symplectic Topology. American Mathematical Society colloquium publications. ISBN 0-8218-3485-1. An analytically flavoured overview of Gromov–Witten invariants and quantum cohomology for symplectic manifolds, very technically complete
Piunikhin, Sergey; Salamon, Dietmar & Schwarz, Matthias (1996). "Symplectic Floer–Donaldson theory and quantum cohomology". In Thomas, C. B. (ed.). Contact and Symplectic Geometry. Cambridge University Press. pp. 171–200. ISBN 0-521-57086-7.
Moduli Spaces of Genus-One Stable Maps, Virtual Classes and an Exercise of Intersection Theory - Andrea Tirelli
Kock, Joachim; Vainsencher, Israel (2007). An Invitation to Quantum Cohomology: Kontsevich's Formula for Rational Plane Curves. New York: Springer. ISBN 978-0-8176-4456-7. A nice introduction with history and exercises to the formal notion of moduli space, treats extensively the case of projective spaces using the basics in the language of schemes.
Vakil, Ravi (2006). "The Moduli Space of Curves and Gromov–Witten Theory". arXiv:math/0602347. Bibcode:2006math......2347V.
Notes on stable maps and quantum cohomology
Research articles[edit]
Gromov-Witten theory of schemes in mixed characteristic
Retrieved from "https://en.wikipedia.org/w/index.php?title=Gromov–Witten_invariant&oldid=1068171166"
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Atomic mass - Simple English Wikipedia, the free encyclopedia
An atomic mass (symbol: ma) is the mass of a single atom of a chemical element. It includes the masses of the 3 subatomic particles that make up an atom: protons, neutrons and electrons.
Atomic mass can be expressed in grams. However, because each atom has a very small mass, this is not very helpful. Instead, atomic mass is expressed in unified atomic mass units (unit symbol: u). 1 atomic mass unit is defined as 1/12 of the mass of a single carbon-12 atom.[1]:18 1 u has a value of 1.660 539 066 60(50) × 10−27 kg.[2]
A carbon-12 atom has a mass of 12 u. Because electrons are so light, we can say that the mass of a carbon-12 atom is made of 6 protons and 6 neutrons. Because the masses of protons and neutrons are almost exactly the same, we can say that both protons and neutrons have a mass of roughly 1 u.[1]:18 Hence, we can get a rough value of an atom's mass in atomic mass units by working out the sum of the number of protons and the number of neutrons in the nucleus, which is called the mass number. The atomic mass of an atom is usually within 0.1 u of the mass number.
The number of protons an atom has determines what element it is. However, most elements in nature consist of atoms with different numbers of neutrons.[3] An atom of an element with a certain number of neutrons is called an isotope.[1]:44 For example, the element chlorine has two common isotopes: chlorine-35 and chlorine-37. Both isotopes of chlorine have 17 protons, but chlorine-37 has 20 neutrons, 2 more neutrons than chlorine-35, which has 18.[4] Each isotope has its own atomic mass, called its isotopic mass. In the case of chlorine, chlorine-35 has a mass of around 35 u, and chlorine-37 around 37 u.
As mentioned above, note that the atomic mass of an atom is not the same as its mass number. The mass number (symbol: A) of an atom is the sum of the number of protons and the number of neutrons in the nucleus.[1]:20 Mass numbers are always whole numbers with no units. Also, relative isotopic mass is not the same as isotopic mass, and relative atomic mass (also called atomic weight) is not the same as atomic mass.
A relative isotopic mass is the mass of an isotope relative to 1/12 of the mass of a carbon-12 atom. In other words, a relative isotopic mass tells you the number of times an isotope of an element is heavier than one-twelfth of an atom of carbon-12. The word relative in relative isotopic mass refers to this scaling relative to carbon-12. Relative isotopic mass is similar to isotopic mass and has exactly the same numerical value as isotopic mass, whenever isotopic mass is expressed in atomic mass units. However, unlike isotopic mass, relative isotopic mass values have no units.
Like relative isotopic mass, a relative atomic mass (symbol: Ar) is a ratio with no units. A relative atomic mass is the ratio of the average mass per atom of an element from a given sample to 1/12 the mass of a carbon-12 atom.[5] We find the relative atomic mass of a sample of an element by working out the abundance-weighted mean of the relative isotopic masses.[3] For example, to continue the chlorine example from above, if there is 75% of chlorine-35 and 25% of chlorine-37 in a sample of chlorine,[4]
{\displaystyle A_{r}={\frac {(35\times 75)+(37\times 25)}{100}}={\frac {(2625)+(925)}{100}}={\frac {3550}{100}}=35.5}
↑ 1.0 1.1 1.2 1.3 Moore, John T. (2010). Chemistry Essentials For Dummies. Wiley. ISBN 978-0-470-61836-3.
↑ "atomic mass unit". National Institute of Standards and Technology. Retrieved 2020-01-12.
↑ 3.0 3.1 Otter, Chris; Stephenson, Kay, eds. (2008). Salters Advanced Chemistry: Chemical Ideas (Third ed.). Heinemann. p. 17. ISBN 978-0-435631-49-9.
↑ 4.0 4.1 Salters Advanced Chemistry: Revise Chemistry For Salters AS (Second ed.). Heinemann. 2008. p. 3. ISBN 978-0-435631-54-3.
↑ Daintith, John, ed. (2008). A Dictionary of Chemistry (Sixth ed.). Oxford University Press. p. 457. ISBN 978-0-19-920463-2.
Atomic mass -Citizendium
Retrieved from "https://simple.wikipedia.org/w/index.php?title=Atomic_mass&oldid=8071487"
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Correct state and state estimation error covariance using tracking filter and JPDA - MATLAB correctjpda - MathWorks 日本
Joint probabilistic data association coefficients, specified as an (N+1)-element vector. The ith (i = 1, …, N) element of jpdacoeffs is the joint probability that the ith measurement in zmeas is associated with the filter. The last element of jpdacoeffs corresponds to the probability that no measurement is associated with the filter. The sum of all elements of jpdacoeffs must equal 1.
\begin{array}{l}{x}_{k}{}^{+}={x}_{k}{}^{â}+{K}_{k}\left(yâh\left({x}_{k}{}^{â}\right)\right)\\ {P}_{k}{}^{+}={P}_{k}{}^{â}â{K}_{k}{S}_{k}{K}_{k}{}^{T}\end{array}
where xk− and xk+ are the a priori and a posteriori state estimates, respectively, Kk is the Kalman gain, y is the actual measurement, and h(xk−) is the predicted measurement. Pk− and Pk+ are the a priori and a posteriori state error covariance matrices, respectively. The innovation matrix Sk is defined as
{S}_{k}={H}_{k}{P}_{k}{}^{â}{H}_{k}{}^{T}
In the workflow of a JPDA tracker, the filter needs to process multiple probable measurements yi (i = 1, …, N) with varied probabilities of association βi (i = 0, 1, …, N). Note that β0 is the probability that no measurements is associated with the filter. The measurement update equations for a discrete extended Kalman filter used for a JPDA tracker are
\begin{array}{l}{x}_{k}{}^{+}={x}_{k}{}^{â}+{K}_{k}\underset{i=1}{\overset{N}{â}}{\mathrm{β}}_{i}\left({y}_{i}âh\left({x}_{k}{}^{â}\right)\right)\\ {P}_{k}{}^{+}={P}_{k}{}^{â}â\left(1â{\mathrm{β}}_{0}\right){K}_{k}{S}_{k}{K}_{k}{}^{T}+{P}_{k}\end{array}
{P}_{k}={K}_{k}\underset{i=1}{\overset{N}{â}}\left[{\mathrm{β}}_{i}\left({y}_{i}âh\left({x}_{k}{}^{â}\right)\right){\left({y}_{i}âh\left({x}_{k}{}^{â}\right)\right)}^{T}â\left(\mathrm{δ}y\right){\left(\mathrm{δ}y\right)}^{T}\right]{K}_{k}{}^{T}
\mathrm{δ}y=\underset{j=1}{\overset{N}{â}}{\mathrm{β}}_{j}\left({y}_{j}âh\left({x}_{k}{}^{â}\right)\right)
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EUDML | Proof of Atiyah's conjecture for two special types of configurations. EuDML | Proof of Atiyah's conjecture for two special types of configurations.
Proof of Atiyah's conjecture for two special types of configurations.
Đoković, Dragomir Ž.. "Proof of Atiyah's conjecture for two special types of configurations.." ELA. The Electronic Journal of Linear Algebra [electronic only] 9 (2002): 132-137. <http://eudml.org/doc/125350>.
@article{Đoković2002,
author = {Đoković, Dragomir Ž.},
keywords = {Atiyah's conjecture; Hopf map; configuration of points in three-dimensional Euclidean space; complex projective line; configuration of points in three-dimensional Euclidean space},
title = {Proof of Atiyah's conjecture for two special types of configurations.},
AU - Đoković, Dragomir Ž.
TI - Proof of Atiyah's conjecture for two special types of configurations.
KW - Atiyah's conjecture; Hopf map; configuration of points in three-dimensional Euclidean space; complex projective line; configuration of points in three-dimensional Euclidean space
Atiyah's conjecture, Hopf map, configuration of
N
points in three-dimensional Euclidean space, complex projective line, configuration of
N
points in three-dimensional Euclidean space
Articles by Đoković
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EUDML | On the locally -convex algebra and a differential-geometric interpretation of it. EuDML | On the locally -convex algebra and a differential-geometric interpretation of it.
On the locally
m
-convex algebra
{L}_{\text{Γ}}\left(E\right)
and a differential-geometric interpretation of it.
Tsertos, Yannis. "On the locally -convex algebra and a differential-geometric interpretation of it.." Portugaliae Mathematica 54.2 (1997): 127-137. <http://eudml.org/doc/48008>.
@article{Tsertos1997,
author = {Tsertos, Yannis},
keywords = {topological algebra; tangent space; -convex algebra; group of invertible elements; -convex algebra},
title = {On the locally -convex algebra and a differential-geometric interpretation of it.},
AU - Tsertos, Yannis
TI - On the locally -convex algebra and a differential-geometric interpretation of it.
KW - topological algebra; tangent space; -convex algebra; group of invertible elements; -convex algebra
topological algebra, tangent space,
m
-convex algebra, group of invertible elements,
m
General theory of topological algebras
Topological algebras of operators
Articles by Tsertos
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LMIs in Control/Matrix and LMI Properties and Tools/Iterative Convex Overbounding - Wikibooks, open books for an open world
LMIs in Control/Matrix and LMI Properties and Tools/Iterative Convex Overbounding
Iterative convex overbounding is a technique based on Young’s relation that is useful when solving an optimization problem with a BMI constraint.
2 The LMI:Iterative Convex Overbounding
Consider the matrices
{\displaystyle Q=Q^{T}\in \mathbb {R} ^{n\times n},B\in \mathbb {R} ^{n\times m},R\in \mathbb {R} ^{m\times p},D\in \mathbb {R} ^{p\times q},S\in \mathbb {R} ^{q\times r}andC\in \mathbb {R} ^{r\times n}}
, where S and R are design variables in the BMI given by
{\displaystyle {\begin{aligned}\qquad Q+BRDSC+C^{T}S^{T}D^{T}R^{T}B^{T}<0\qquad (1)\\\end{aligned}}}
The LMI:Iterative Convex OverboundingEdit
Suppose that S0 and R0 are known to satisfy (1). The BMI of (1) is implied by the LMI
{\displaystyle {\begin{bmatrix}Q+\phi (R,S)+\phi ^{\text{T}}(R,S)&B(R-R0)U&C^{\text{T}}(S-S0)^{\text{T}}V^{\text{T}}\\*&W^{\text{-1}}&0\\*&*&-W\\\end{bmatrix}}<0\qquad (2)}
where Φ(R,S) = B(RDS0+R0DS-R0DS0)C, W>0 is an arbitrary matrix, D=UV, and the matrices U and VT have full column rank. The LMI of (2) is equivalent to the BMI of (1) when R = R0 and S = S0, and is therefore non-conservative for values of R and S and are close to the previously known solutions R0 and S0.
Alternatively, the BMI of (1) is implied by the LMI
{\displaystyle {\begin{bmatrix}Q+\phi (R,S)+\phi ^{\text{T}}(R,S)&Z^{\text{T}}U^{\text{T}}(R-R0)^{\text{T}}B^{\text{T}}+V(S-S0)C\\*&Z\\\end{bmatrix}}<0\qquad (3)}
where Z > 0 is an arbitrary matrix, D = UV, and the matrices U and VT have full column rank. Again, the LMI of (4) is equivalent to the BMI of (2) when R = R0 and S = S0, and is therefore non-conservative for values of R and S and are close to the previously known solutions R0 and S0.
A benefit of convex overbounding compared to a linearization approach, is that in addition to ensuring conservatism or error is reduced in the neighborhood of R = R0 and S = S0, the LMIs of (2) and (3) imply (1). Iterative convex overbounding is particularly useful when used to solve an optimization problem with BMI constraints.
Young’s Relation
Retrieved from "https://en.wikibooks.org/w/index.php?title=LMIs_in_Control/Matrix_and_LMI_Properties_and_Tools/Iterative_Convex_Overbounding&oldid=4010577"
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EUDML | Properties of a covariance matrix with an application to -optimal design. EuDML | Properties of a covariance matrix with an application to -optimal design.
Properties of a covariance matrix with an application to
D
-optimal design.
Zhu, Zewen; Coster, Daniel C.; Beasley, Leroy B.
Zhu, Zewen, Coster, Daniel C., and Beasley, Leroy B.. "Properties of a covariance matrix with an application to -optimal design.." ELA. The Electronic Journal of Linear Algebra [electronic only] 10 (2003): 65-76. <http://eudml.org/doc/122720>.
author = {Zhu, Zewen, Coster, Daniel C., Beasley, Leroy B.},
keywords = {simple linear regression; circulant correlation},
title = {Properties of a covariance matrix with an application to -optimal design.},
AU - Coster, Daniel C.
AU - Beasley, Leroy B.
TI - Properties of a covariance matrix with an application to -optimal design.
KW - simple linear regression; circulant correlation
simple linear regression, circulant correlation
Articles by Zhu
Articles by Coster
Articles by Beasley
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EUDML | A new characterization of some alternating and symmetric groups. EuDML | A new characterization of some alternating and symmetric groups.
A new characterization of some alternating and symmetric groups.
Khosravi, Amir; Khosravi, Behrooz
Khosravi, Amir, and Khosravi, Behrooz. "A new characterization of some alternating and symmetric groups.." International Journal of Mathematics and Mathematical Sciences 2003.45 (2003): 2863-2872. <http://eudml.org/doc/50724>.
@article{Khosravi2003,
author = {Khosravi, Amir, Khosravi, Behrooz},
keywords = {order components; finite simple -groups; alternating groups; symmetric groups; centralizers; finite simple -groups},
title = {A new characterization of some alternating and symmetric groups.},
AU - Khosravi, Amir
AU - Khosravi, Behrooz
TI - A new characterization of some alternating and symmetric groups.
KW - order components; finite simple -groups; alternating groups; symmetric groups; centralizers; finite simple -groups
order components, finite simple
{C}_{pp}
-groups, alternating groups, symmetric groups, centralizers, finite simple
{C}_{pp}
Articles by Khosravi
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Multiple Soliton Solutions for a New Generalization of the Associated Camassa-Holm Equation by Exp-Function Method
Yao Long, Yinghui He, Shaolin Li, "Multiple Soliton Solutions for a New Generalization of the Associated Camassa-Holm Equation by Exp-Function Method", Mathematical Problems in Engineering, vol. 2014, Article ID 418793, 7 pages, 2014. https://doi.org/10.1155/2014/418793
Yao Long,1 Yinghui He ,1 and Shaolin Li1
The Exp-function method is generalized to construct N-soliton solutions of a new generalization of the associated Camassa-Holm equation. As a result, one-soliton, two-soliton, and three-soliton solutions are obtained, from which the uniform formulae of N-soliton solutions are derived. It is shown that the Exp-function method may provide us with a straightforward, effective, and alternative mathematical tool for generating N-soliton solutions of nonlinear evolution equations in mathematical physics.
The investigation of the traveling wave solutions to nonlinear evolution equations (NLEEs) plays an important role in mathematical physics. A lot of physical models have supported a wide variety of solitary wave solutions. In the recent years, much efforts have been spent on this task and many significant methods have been established such as inverse scattering transform [1], Bäcklund and Darboux transform [2], Hirota bilinear method [3], homogeneous balance method [4], Jacobi elliptic function method [5], tanh-function method [6], Exp-function method [7], simple equation method [8], F-expansion method [9, 10], improved F-expansion method [11], and extended F-expansion method [12]. Here, we study a new generalization of the associated Camassa-Holm equation.
The Camassa-Holm (CH) equation where is a nonzero real constant, was derived as a model for shallow water waves by Camassa and Holm in 1993 [13]. This equation is integrable with the following Lax pair: Considerable interest was paid to the CH equation in recent decades about its integrability and exact solutions [14–20]. Schiff and Fisher showed that the Camassa-Holm equation possessed the Bäcklund transformations and an infinite number of local conserved quantities by using the loop group approach [15, 16]. Parker gave explicit multisoliton solutions for the CH equation by taking the Hirota bilinear method and a coordinate transformation [18]. Its structure and dynamics were investigated in the different parameter regimes. According to [18], there is a reciprocal transformation, , such that Let us apply the reciprocal transformation to the Lax pair (2) and define the following potential function : then (1) is transformed into the following associated Camassa-Holm (ACH) equation: Hone showed in [21] how the ACH equation (5) is related to Schrodinger operators and the KdV equation and described how to construct solutions of the ACH equation from tau-functions of the KdV hierarchy, including rational, N-soliton, and elliptic solutions.
Recently, integrable negative order flows, mixed equations, and the relationship of different hierarchies attracted much attention, including continuous and discrete cases, such as the negative KdV, mixed KdV, and Volterra lattice equations. In [22], Luo et al. introduced a new generalization of the associated Camassa-Holm equation; namely, where and are two arbitrary constants.
Apparently, (6) is reduced to the ACH equation (5) when we take , . For , , (6) gives the KdV equation. So, (6) may be called the ACH-KdV equation. In [22], Luo et al. show that (6) is integrable in the sense of Lax pair and construct some exact solutions of (6) by Darboux transformation through Lax pair.
The Exp-function method [7] proposed by He and Wu in 2006 provides us with a straightforward and effective method for obtaining generalized solitary wave solutions and periodic solutions of NLEEs. The method was used by many researchers to study various NLEEs. More recently, Marinakis [23] did very interesting work to generalize the Exp-function method for constructing N-soliton solutions of NLEEs. Marinakis chose the famous Korteweg-de Vries (KdV) equation to illustrate the generalized work and successfully obtained the known 2-soliton and 3-soliton solutions in a simple and straightforward way.
In the present paper, we would like to generalize the Exp-function method for constructing N-soliton solutions of ACH-KdV equation (6).
The rest of this paper is organized as follows. In Section 2, we give the description of the Exp-function method for constructing N-soliton solutions of NLEEs. In Section 3, we apply the method to (6). In Section 4, some conclusions and discussions are given.
2. Basic Idea of the Exp-Function Method for N-Soliton Solutions of NLEEs
In this section, we recall the Exp-function method [23] for N-soliton solutions of NLEEs. For a given NLEE, say, in two variables and , the Exp-function method for one-soliton solution is based on the assumption where , , , , , and are unknown constants and the values of and can be determined by balancing the linear term of highest order in (7) with the highest order nonlinear term.
In order to seek N-soliton solutions for integer , we generalize (8) to the following form: given the value of , it becomes which can be used to construct two-soliton solution.
When , (9) changes into which can be used to obtain three-soliton solution.
Substituting (10) into (7) and using Mathematica, then equating to zero each coefficient of the same order power of the exponential functions yields a set of equations. Solving the set of equations, we can determine the 2-soliton solution and the following 3-soliton solution by means of (11), provided they exist. If possible, we may conclude with the uniform formula of N-soliton solutions for any .
3. Multisoliton Solutions of the ACH-KdV Equation
In this section we apply Exp-function method to the ACH-KdV equation (6). We first remove the integral term in (6) by introducing the potential then substituting (12) into (6) yields
Suppose that (13) admits the one-soliton solution of the form where , , are undetermined constants, and is an arbitrary constant. Obviously, (14) is included in the same form as (7). Substituting (14) into (13) and then equating to zero each coefficient of the same order power of ( ) yield a set of equations for , , , and as follows: Solving these equations by Maple, one has Substituting (16) into (14), we have Using potential (12), we can get the one-soliton solutions of (6) as follows: where and , , and are arbitrary constants. The one-soliton solution (18) is shown in Figure 1.
Figures of solution (18) and with , , , , , . (a) Spatial plots in the intervals and ; (b) plan plots with , in the interval .
Next, we suppose that (13) has the 2-soliton solution in the form where ; ; , , , , , , , and are constants to be determined; and and are arbitrary constants. Obviously, (19) has the same form as (10). Substituting (19) into (13) and using manipulations similar to those illustrated above, we obtain where
Substituting (20) into (19), we have Using potential (12), we can get the two-soliton solutions of (6) as follows: where ; ; and , , , , , and are arbitrary constants. The two-soliton solution (18) is shown in Figure 2.
Figures of solution (23) and with , , , , , , , , . (a) Spatial plots in the intervals and ; (b)–(d) plan plots and with , , , , respectively.
From Figure 2 we can find the wave chase phenomenon. The speed of the first wave ; the speed of the second wave . As time increases, the second wave exceeds the first one.
In what follows, we now suppose that the three-soliton solution of (13) can be expressed as follows: where, ; ; ; ( ), , , and are constants to be determined; and , and are arbitrary constants. Obviously, (24) has the same form as (11). Substituting (24) into (13) and using manipulations similar to those illustrated above, we obtain where Substituting (25) into (24) and using potential (12), we can obtain the three-soliton solution of (6). The expression is so complicated; therefore we do not give out it in detail. The three-soliton solution is shown in Figure 3, from which wave chase phenomenon also can be found.
Three-soliton waves and with , , , , , , , , , , , . (a) Spatial plots in the intervals and ; (b)–(d) plan plots and with , , , .
When , similar computational work becomes more and more complicated since the coefficients of the exponential functions become a highly nonlinear system as shown in [23]. Fortunately, we can find a uniform formula of the N-soliton solutions by analyzing the obtained solutions (17), (22), and (24). We rewrite solutions (17), (22), and (24) in an alternative form: where ; where ( ), ; and where ( ), ( ).
The uniform formula of the N-soliton solutions can be constructed as follows: where ( ) and ( ).
In this paper, one-soliton, two-soliton, and three-soliton solutions of the ACH-KdV equation have been successfully obtained, from which the uniform formula of N-soliton solutions is derived. This is due to the generalization of the Exp-function method. Figures 1–3 imply that these obtained solutions have rich local structures. It may be important to explain some physical phenomena. The method with the help of mathematical software for generating 1-soliton, 2-soliton, and 3-soliton solutions is more simple and straightforward than Hirota bilinear method, without employing the bilinear operator defined in the Hirota bilinear method. The paper shows that the Exp-function method may provide us with a straightforward and effective mathematical tool for generating N-soliton solutions or testing its existence and can be extended to other NLEEs in mathematical physics.
This research is supported by the Natural Science Foundation of China (nos. 11161020, 11361023) as well as the Young and Middle-Aged Academic Backbone of Honghe University (no. 2014GG0105).
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Redox Reactions Chemistry NEET Practice Questions, MCQs, Past Year Questions (PYQs), NCERT Questions, Question Bank, Class 11 and Class 12 Questions, and PDF solved with answers
5mL of NHCl, 20 mL of N/2 H2SO4 and 30mL of N/3 HNO3 are mixed together and volume made one litre. The normality of the resulting solution is:
Subtopic: Equivalent Weight |
When a metal is burnt, its mass is increased by 24 percent. The equivalent mass of the metal will be:
1g of pure calcium carbonate was found to require 50 mL of dilute HCl for complete reactions. The strength of the HCl solution is given by:
The equivalent mass of H3PO4 in the following reaction is,
H3PO4 + Ca(OH)2
\to
CaHPO4 + 2H2O
1.520g of the hydroxide of a metal on ignition gave 0.995 g of oxide. The equivalent mass of metal is:
0.5 g of fuming H2SO4 (oleum) is diluted with water. This solution is completely neutralised by 26.7 mL of 0.4 N NaOH. The percentage of free SO3 in the sample is:
One g of a mixture of Na2CO3 and NaHCO3 consumes y equivalent of HCl for complete neutralisation. One g of the mixture is strongly heated, then cooled and the residue treated with HCl. How many equivalent of HCl would be required for complete neutralisation?
1. 2y equivalent
2. y equivalent
3. 3y/4 equivalent
The chloride of a metal contains 71% chlorine by mass and the vapour density of it is 50. The atomic mass of the metal will be:
The equivalent mass of Zn(OH)2 in the following reaction is equal to its,
Zn(OH)2 + HNO3
\to
Zn(OH)(NO3) + H2O :
1. Formula mass/1
3. 2 x formula mass
What will be the normality of a solution obtained by mixing 0.45 N and 0.60 N NaOH in the ratio 2:1 by volume?
1. 0.4 N
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Document similarities with BM25 algorithm - MATLAB bm25Similarity - MathWorks í•œêµ
Similarity Between Documents
Similarity to Query
Document Similarities Using Bag-of-Words Model
Evaluate BM25+ Document Similarity
TFScaling
DocumentLengthScaling
DocumentLengthCorrection
Document similarities with BM25 algorithm
similarities = bm25Similarity(documents)
similarities = bm25Similarity(documents,queries)
similarities = bm25Similarity(bag)
similarities = bm25Similarity(bag,queries)
similarities = bm25Similarity(___,Name,Value)
Use bm25Similarity to calculate document similarities.
By default, this function calculates BM25 similarities. To calculate BM11, BM15, or BM25+ similarities, use the 'DocumentLengthScaling' and 'DocumentLengthCorrection' arguments.
similarities = bm25Similarity(documents) returns the pairwise BM25 similarities between the specified documents. The score in similarities(i,j) represents the similarity between documents(i) and documents(j).
similarities = bm25Similarity(documents,queries) returns similarities between documents and queries. The score in similarities(i,j) represents the similarity between documents(i) and queries(j).
similarities = bm25Similarity(bag) returns similarities between the documents encoded by the specified bag-of-words or bag-of-n-grams model. The score in similarities(i,j) represents the similarity between the ith and jth documents encoded by bag.
similarities = bm25Similarity(bag,queries) returns similarities between the documents encoded by the bag-of-words or bag-of-n-grams model bag and the documents specified by queries. The score in similarities(i,j) represents the similarity between the ith document encoded by bag and queries(j).
similarities = bm25Similarity(___,Name,Value) specifies additional options using one or more name-value pair arguments. For instance, to use the BM25+ algorithm, set the 'DocumentLengthCorrection' option to a nonzero value.
documents = tokenizedDocument(textData)
Calculate the similarities between them using the bm25Similarity function. The output is a sparse matrix.
similarities = bm25Similarity(documents);
Visualize the similarities of the documents in a heat map.
heatmap(similarities);
title("BM25 Similarities")
The first three documents have the highest pairwise similarities which indicates that these documents are most similar. The last document has comparatively low pairwise similarities with the other documents which indicates that this document is less like the other documents.
Create an array of input documents.
"the fast fox jumped over the lazy dog"
"the dog sat there and did nothing"
8 tokens: the fast fox jumped over the lazy dog
7 tokens: the dog sat there and did nothing
Create an array of query documents.
"a brown fox leaped over the lazy dog"
"another fox leaped over the dog"];
queries = tokenizedDocument(str)
queries =
8 tokens: a brown fox leaped over the lazy dog
6 tokens: another fox leaped over the dog
Calculate the similarities between input documents and query documents using the bm25Similarity function. The output is a sparse matrix. The score in similarities(i,j) represents the similarity between documents(i) and queries(j).
similarities = bm25Similarity(documents,queries);
xlabel("Query Document")
ylabel("Input Document")
In this case, the first input document is most like the first query document.
Counts: [154×3527 double]
Vocabulary: [1×3527 string]
Calculate similarities between the sonnets using the bm25Similarity function. The output is a sparse matrix.
similarities = bm25Similarity(bag);
Visualize the similarities between the first five documents in a heat map.
heatmap(similarities(1:5,1:5));
The BM25+ algorithm addresses a limitation of the BM25 algorithm: the component of the term-frequency normalization by document length is not properly lower bounded. As a result of this limitation, long documents which do not match the query term can often be scored unfairly by BM25 as having a similar relevance to shorter documents that do not contain the query term.
BM25+ addresses this limitation by using a document length correction factor (the value of the 'DocumentLengthScaling' name-value pair). This factor prevents the algorithm from over-penalizing long documents.
Create two arrays of tokenized documents.
textData1 = [
documents1 = tokenizedDocument(textData1)
To calculate the BM25+ document similarities, use the bm25Similarity function and set the 'DocumentLengthCorrection' option to a nonzero value. In this case, set the 'DocumentLengthCorrection' option to 1.
similarities = bm25Similarity(documents1,documents2,'DocumentLengthCorrection',1);
xlabel("Query")
title("BM25+ Similarities")
Here, when compared with the example Similarity Between Documents, the scores show more similarity between the input documents and the first query document.
queries — Set of query documents
tokenizedDocument array | bagOfWords object | bagOfNgrams object | string array of words | cell array of character vectors
Set of query documents, specified as one of the following:
A tokenizedDocument array
A bagOfWords or bagOfNgrams object
A 1-by-N string array representing a single document, where each element is a word
A 1-by-N cell array of character vectors representing a single document, where each element is a word
To compute term frequency and inverse document frequency statistics, the function encodes queries using a bag-of-words model. The model it uses depends on the syntax you call it with. If your syntax specifies the input argument documents, then it uses bagOfWords(documents). If your syntax specifies bag, then it uses bag.
Example: bm25Similarity(documents,'TFScaling',1.5) returns the pairwise similarities for the specified documents and sets the token frequency scaling factor to 1.5.
'textrank' (default) | 'classic-bm25' | 'normal' | 'unary' | 'smooth' | 'max' | 'probabilistic'
TFScaling — Term frequency scaling factor
Term frequency scaling factor, specified as the comma-separated pair consisting of 'TFScaling' and a nonnegative scalar.
This option corresponds to the value k in the BM25 algorithm. For more information, see BM25.
DocumentLengthScaling — Document length scaling factor
Document length scaling factor, specified as the comma-separated pair consisting of 'DocumentLengthScaling' and a scalar in the range [0,1].
This option corresponds to the value b in the BM25 algorithm. When b=1, the BM25 algorithm is equivalent to BM11. When b=0, the BM25 algorithm is equivalent to BM15. For more information, see BM11, BM15, or BM25.
DocumentLengthCorrection — Document length correction factor
Document length correction factor, specified as the comma-separated pair consisting of 'DocumentLengthCorrection' and a nonnegative scalar.
This option corresponds to the value
\mathrm{δ}
in the BM25+ algorithm. If the document length correction factor is nonzero, then the bm25Similarity function uses the BM25+ algorithm. Otherwise, the function uses the BM25 algorithm. For more information, see BM25+.
similarities — BM25 similarity scores
BM25 similarity scores, returned as a sparse matrix:
Given a single array of tokenized documents, similarities is a N-by-N nonsymmetric matrix, where similarities(i,j) represents the similarity between documents(i) and documents(j), and N is the number of input documents.
Given an array of tokenized documents and a set of query documents, similarities is an N1-by-N2 matrix, where similarities(i,j) represents the similarity between documents(i) and the jth query document, and N1 and N2 represents the number of documents in documents and queries, respectively.
Given a single bag-of-words or bag-of-n-grams model, similarities is a bag.NumDocuments-by-bag.NumDocuments nonsymmetric matrix, where similarities(i,j) represents the similarity between the ith and jth documents encoded by bag.
Given a bag-of-words or bag-of-n-grams models and a set of query documents, similarities is a bag.NumDocuments-by-N2 matrix, where similarities(i,j) represents the similarity between the ith document encoded by bag and the jth document in queries, and N2 corresponds to the number of documents in queries.
The BM25 algorithm aggregates and uses information from all the documents in the input data via the term frequency (TF) and inverse document frequency (IDF) based options. This behavior means that the same pair of documents can yield different BM25 similarity scores when the function is given different collections of documents.
The BM25 algorithm can output different scores when comparing documents to themselves. This behavior is due to the use of the IDF weights and the document length in the BM25 algorithm.
Given a document from a collection of documents
\mathcal{D}
, and a query document, the BM25 score is given by
\text{BM25}\left(\text{document},\text{query};\mathcal{D}\right)=\underset{\text{word}â\text{ query}}{â}\left(\text{IDF}\left(\text{word;}\mathcal{D}\right)\frac{\text{Count}\left(\text{word},\text{document}\right)\left(k+1\right)}{\text{Count}\left(\text{word},\text{document}\right)+k\left(1âb+b\frac{|\text{document}|}{\stackrel{¯}{n}}\right)}\right),
Count(word,document) denotes the frequency of word in document.
\stackrel{¯}{n}
denotes the average document length in
\mathcal{D}
k denotes the term frequency scaling factor (the value of the 'TFScaling' name-value pair argument). This factor dampens the influence of frequently appearing terms on the BM25 score.
b denotes the document length scaling factor (the value of the 'DocumentLengthScaling' name-value pair argument). This factor controls how the length of a document influences the BM25 score. When b=1, the BM25 algorithm is equivalent to BM11. When b=0, the BM25 algorithm is equivalent to BM15.
\text{IDF}\left(\text{word},\mathcal{D}\right)
is the inverse document frequency of the specified word given the collection of documents
\mathcal{D}
The BM25+ algorithm is the same as the BM25 algorithm with one extra parameter. Given a document from a collection of documents
\mathcal{D}
and a query document, the BM25+ score is given by
{\text{BM25}}^{+}\left(\text{document},\text{query};\mathcal{D}\right)=\underset{\text{word}â\text{ query}}{â}\left(\text{IDF}\left(\text{word;}\mathcal{D}\right)\left(\frac{\text{Count}\left(\text{word},\text{document}\right)\left(k+1\right)}{\text{Count}\left(\text{word},\text{document}\right)+k\left(1âb+b\frac{|\text{document}|}{\stackrel{¯}{n}}\right)}+\mathrm{δ}\right)\right),
where the extra parameter
\mathrm{δ}
denotes the document length correction factor (the value of the 'DocumentLengthScaling' name-value pair). This factor prevents the algorithm from over-penalizing long documents.
BM11 is a special case of BM25 when b=1.
\mathcal{D}
\text{BM11}\left(\text{document},\text{query};\mathcal{D}\right)=\underset{\text{word}â\text{ query}}{â}\left(\text{IDF}\left(\text{word;}\mathcal{D}\right)\frac{\text{Count}\left(\text{word},\text{document}\right)\left(k+1\right)}{\text{Count}\left(\text{word},\text{document}\right)+k\left(\frac{|\text{document}|}{\stackrel{¯}{n}}\right)}\right).
\mathcal{D}
\text{BM15}\left(\text{document},\text{query};\mathcal{D}\right)=\underset{\text{word}â\text{ query}}{â}\left(\text{IDF}\left(\text{word;}\mathcal{D}\right)\frac{\text{Count}\left(\text{word},\text{document}\right)\left(k+1\right)}{\text{Count}\left(\text{word},\text{document}\right)+k}\right).
[1] Robertson, Stephen, and Hugo Zaragoza. "The Probabilistic Relevance Framework: BM25 and Beyond." Foundations and Trends® in Information Retrieval 3, no. 4 (2009): 333-389.
[2] Barrios, Federico, Federico López, Luis Argerich, and Rosa Wachenchauzer. "Variations of the Similarity Function of TextRank for Automated Summarization." arXiv preprint arXiv:1602.03606 (2016).
tokenizedDocument | bleuEvaluationScore | rougeEvaluationScore | cosineSimilarity | textrankScores | lexrankScores | mmrScores | extractSummary
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Electric Charges And Fields, Popular Questions: ICSE Class 12-science PHYSICS, Physics Part I - Meritnation
Nidhi Bansal asked a question
An infinite line charge is at the axis of a cylinder of length 1m & radius 7cm.If the electric field at any point on curved surface is 250N/C. Find the net electric flux through the cylinder.
Q.7). A point charge 25
\mu C
is located in the XY plane at the point of position vector
\stackrel{\to }{{r}_{0}}=\left(\stackrel{^}{i}+\stackrel{^}{j}\right)m
. What is the magnitude of electric field at the point of position vector
\stackrel{\to }{{r}_{1}}=\left(4\stackrel{^}{i}+5\stackrel{^}{j}\right)m
900\frac{V}{m}
9\frac{kV}{m}
90\frac{V}{m}
Mohini Agarwal asked a question
A cone of height h and base radius r is located in a uniform electric field parallel to its base . what amount of electric flux enters the cone ?
\stackrel{⇀}{E}=a\sqrt{x}\stackrel{^}{i}
\stackrel{\to }{E}
\mathcal{l}
\mathcal{l}
\mathcal{l}
\mathcal{l}
Nandana Rose asked a question
A charge of 10 μc is brought from point A (0,4 cm,0) to C (3 cm,0,0) via point B (0,0,6 cm) in vacuum. Calculate the work done if the charge at origin is 20 μc. please answer fast
Brinda asked a question
two balls of charges q1 tand q2 initially have the exact same velocity. Both the balls are subjected to same uniform electric field for the same time. As a result, the velocity of the first ball is reduced to half of its initial value and its direction changes by 60°. The direction of velocity of second ball is found to change by 90°. (i) The electric field and the initial velocity of the charged particle are inclined at what angle (ii) The new velocity of the second charged particle has a magnitude `x` times the initial velocity. What is the value of x (iii) If the specific charge ( charge to mass ratio) of the first ball is k, what is the specific charge of the second ball
Rajbir Cheema asked a question
What is the relation between charges q1, q2 and q3. If the electric flux through surface S2 is 8-times
of that through surface S1?
a soap bubble of radius r is blown so that its diameter is doubled if T is S.T of water the energy required to do this at constant temperature is
(a) 8 pi r2 T
(b) 12 pi r2
(c) 24 pi r2 T
(d) 16 pi r2 T
Find the Electric flux passing through infinite strip of small width h at a distance d (d>>h) from point charge q.
A point charge is placed at centre of the spherical gaussian surface . How will the flux change if
1. The sphere is replaced by a cube of same or different volume.
2. The second charge is placed near and outside the original sphere
3. A second charge is placed inside the sphere
4. The original charge is replaced by an electric dipole.
why is charge on body cannot be less than electronis charge?
Shejal Routray asked a question
what is the line of symmetry of dipole field?
Karthik Vishwanath asked a question
What must be the charge on each of a pair of pith balls suspended in air from the same point by strings 10 cm long if they repel each other to a separation of 8 cm? Mass of each pith ball is 1 gram and accleration due to gravity is 9.8 m/s2
Aditya Nath Pandey asked a question
four particles form a square. the charges are q1 = q4 = Q & q2 = q3 = q
(a) What is Q/q if the net electostatic force on particles 1& 4 is zero? (b) is there any value of q that makes the net electrostatic force on each of the four particles zero? Explain.
The energy of the electron in the ground state of hydrogen is -13.6 eV. Calculate the energy of the photon that would be emitted if the electron were to make a transition corresponding to the emission of the first line of a) Lyman series b) Balmer series of the hydrogen spectrum.
the sum of two point charges is 6 micro coloumb.they attract each other with a force of 0.9N,when kept 40cm apart in vacuum.calculate the charges.please do not give links
a small electric dipole is placed in x-y plane at the origin with its dipole moment directed along positive x-axis. the direction of electric field at point (2, 2root2, 0) 1)along positive z-axis 2)along positive y-axis 3)along negative y-axis 4)along negative z-axis
Helly Joshi asked a question
Looking at the various electric lines of forces between charges Q1 and Q2 , find the value of Q1/Q2.
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Generalized sidelobe canceler beamformer - MATLAB - MathWorks Deutschland
phased.GSCBeamformer
LMSStepSize
Generalized Sidelobe Cancellation on Uniform Linear Array
Generalized Sidelobe Cancellation in Two Directions
Generalized Sidelobe Cancellation
Generalized sidelobe canceler beamformer
The phased.GSCBeamformer System object™ implements a generalized sidelobe cancellation (GSC) beamformer. A GSC beamformer splits the incoming signals into two channels. One channel goes through a conventional beamformer path and the second goes into a sidelobe canceling path. The algorithm first pre-steers the array to the beamforming direction and then adaptively chooses filter weights to minimize power at the output of the sidelobe canceling path. The algorithm uses least mean squares (LMS) to compute the adaptive weights. The final beamformed signal is the difference between the outputs of the two paths.
Create the phased.GSCBeamformer object and set its properties.
beamformer = phased.GSCBeamformer
beamformer = phased.GSCBeamformer(Name,Value)
beamformer = phased.GSCBeamformer creates a GSC beamformer System object, beamformer, with default property values.
beamformer = phased.GSCBeamformer(Name,Value) creates a GSC beamformer object, beamformer, with each specified property Name set to the specified Value. You can specify additional name-value pair arguments in any order as (Name1,Value1,...,NameN,ValueN). Enclose each property name in single quotes.
Example: beamformer = phased.GSCBeamformer('SensorArray',phased.ULA('NumElements',20),'SampleRate',300e3) sets the sensor array to a uniform linear array (ULA) with default ULA property values except for the number of elements. The beamformer has a sample rate of 300 kHz.
Length of the signal path FIR filters, specified as a positive integer. This property determines the adaptive filter size for the sidelobe canceling path. The FIR filter for the conventional beamforming path is a delta function of the same length.
LMSStepSize — Adaptive filter step size factor
The adaptive filter step size factor, specified as a positive real-valued scalar. This quantity, when divided by the total power in the sidelobe canceling path, sets the actual adaptive filter step size that is used in the LMS algorithm.
Y = beamformer(X) performs GSC beamforming on the input, X, and returns the beamformed output, Y.
Create a GSC beamformer for a 11-element acoustic array in air. A chirp signal is incident on the array at
-5{0}^{\circ }
{0}^{\circ }
in elevation. Compare the GSC beamformed signal to a Frost beamformed signal. The signal propagation speed is 340 m/s and the sample rate is 8 kHz.
Create the microphone and array System objects. The array element spacing is one-half wavelength. Set the signal frequency to the one-half the Nyquist frequency.
fc = fs/2;
transducer = phased.OmnidirectionalMicrophoneElement('FrequencyRange',[20 20000]);
array = phased.ULA('Element',transducer,'NumElements',11,'ElementSpacing',lam/2);
Simulate a chirp signal with a 500 Hz bandwidth.
t = 0:1/fs:.5;
signal = chirp(t,0,0.5,500);
Create an incident wave arriving at the array. Add gaussian noise to the wave.
collector = phased.WidebandCollector('Sensor',array,'PropagationSpeed',c, ...
'SampleRate',fs,'ModulatedInput',false,'NumSubbands',512);
incidentAngle = [-50;0];
signal = collector(signal.',incidentAngle);
noise = 0.5*randn(size(signal));
recsignal = signal + noise;
Perform Frost beamforming at the actual incident angle.
frostbeamformer = phased.FrostBeamformer('SensorArray',array,'PropagationSpeed', ...
c,'SampleRate',fs,'Direction',incidentAngle,'FilterLength',15);
yfrost = frostbeamformer(recsignal);
Perform GSC beamforming and plot the beamformer output against the Frost beamformer output. Also plot the nonbeamformed signal arriving at the middle element of the array.
gscbeamformer = phased.GSCBeamformer('SensorArray',array, ...
'PropagationSpeed',c,'SampleRate',fs,'Direction',incidentAngle, ...
'FilterLength',15);
ygsc = gscbeamformer(recsignal);
plot(t*1000,recsignal(:,6),t*1000,yfrost,t,ygsc)
Zoom in on a small portion of the output.
idx = 1000:1300;
plot(t(idx)*1000,recsignal(idx,6),t(idx)*1000,yfrost(idx),t(idx)*1000,ygsc(idx))
legend('Received signal','Frost beamformed signal','GSC beamformed signal')
Create a GSC beamformer for an 11-element acoustic array in air. A chirp signal is incident on the array at
-5{0}^{\circ }
{0}^{\circ }
in elevation. Compute the beamformed signal in the direction of the incident wave and in another direction. Compare the two beamformed outputs. The signal propagation speed is 340 m/s and the sample rate is 8 kHz. Create the microphone and array System objects. The array element spacing is one-half wavelength. Set the signal frequency to the one-half the Nyquist frequency.
Create an incident wavefield hitting the array.
Perform GSC beamforming and plot the beamformer outputs. Also plot the nonbeamformed signal arriving at the middle element of the array.
'PropagationSpeed',c,'SampleRate',fs,'DirectionSource','Input port', ...
'FilterLength',5);
ygsci = gscbeamformer(recsignal,incidentAngle);
ygsco = gscbeamformer(recsignal,[20;30]);
plot(t*1000,recsignal(:,6),t*1000,ygsci,t*1000,ygsco)
legend('Received signal at element','GSC beamformed signal (incident direction)', ...
'GSC beamformed signal (other direction)','Location','southeast')
plot(t(idx)*1000,recsignal(idx,6),t(idx)*1000,ygsci(idx),t(idx)*1000,ygsco(idx))
The generalized sidelobe canceler (GSC) is an efficient implementation of a linear constraint minimum variance (LCMV) beamformer. LCMV beamforming minimizes the output power of an array while preserving the power in one or more specified directions. This type of beamformer is called a constrained beamformer. You can compute exact weights for the constrained beamformer but the computation is costly when the number of elements is large. The computation requires the inversion of a large spatial covariance matrix. The GSC formulation converts the adaptive constrained optimization LCMV problem into an adaptive unconstrained problem, which simplifies the implementation.
In the GSC algorithm, incoming sensor data is split into two signal paths as shown in the block diagram. The upper path is a conventional beamformer. The lower path is an adaptive unconstrained beamformer whose purpose is to minimize the GSC output power. The GSC algorithm consists of these steps:
Presteer the element sensor data by time-shifting the incoming signals. Presteering time-aligns all sensor element signals. The time shifts depend on the arrival angle of the signal.
Pass the presteered signals through the upper path into a conventional beamformer with fixed weights, wconv.
Also pass the presteered signals through the lower path into the blocking matrix, B. The blocking matrix is orthogonal to the signal and removes the signal from the lower path.
Filter the lower path signals through a bank of FIR filters. The FilterLength property sets the length of the filters. The filter coefficients are the adaptive filter weights, wad.
Compute the difference between the upper and lower signal paths. This difference is the beamformed GSC output.
Feed the beamformed output back into the filter. The filter adapts its weights using a least mean-square (LMS) algorithm. The actual adaptive LMS step size is equal to the value of the LMSStepSize property divided by the total signal power.
[1] Griffiths, L. J., and Charles W. Jim. "An alternative approach to linearly constrained adaptive beamforming." IEEE Transactions on Antennas and Propagation, 30.1 (1982): 27-34.
[3] Johnson, D.H., and Dan E. Dudgeon, Array Signal Processing, Englewood Cliffs: Prentice-Hall, 1993.
phased.FrostBeamformer | phased.MVDRBeamformer | phased.PhaseShiftBeamformer | phased.SubbandPhaseShiftBeamformer | phased.TimeDelayBeamformer | phased.TimeDelayLCMVBeamformer
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23 CFR § 658.17 - Weight. | CFR | US Law | LII / Legal Information Institute
PART 658 - TRUCK SIZE AND WEIGHT, ROUTE DESIGNATIONS - LENGTH, WIDTH AND WEIGHT LIMITATIONS
23 CFR § 658.17 - Weight.
(a) The provisions of the section are applicable to the National System of Interstate and Defense Highways and reasonable access thereto.
(b) The maximum gross vehicle weight shall be 80,000 pounds except where lower gross vehicle weight is dictated by the bridge formula.
(c) The maximum gross weight upon any one axle, including any one axle of a group of axles, or a vehicle is 20,000 pounds.
(d) The maximum gross weight on tandem axles is 34,000 pounds.
(e) No vehicle or combination of vehicles shall be moved or operated on any Interstate highway when the gross weight on two or more consecutive axles exceeds the limitations prescribed by the following formula, referred to as the Bridge Gross Weight Formula:
W=500\left(\frac{\mathrm{LN}}{N-1}+12N+36\right)
except that two consecutive sets of tandem axles may carry a gross load of 34,000 pounds each if the overall distance between the first and last axle is 36 feet or more. In no case shall the total gross weight of a vehicle exceed 80,000 pounds.
(f) Except as provided herein, States may not enforce on the Interstate System vehicle weight limits of less than 20,000 pounds on a single axle, 34,000 pounds on a tandem axle, or the weights derived from the Bridge Formula, up to a maximum of 80,000 pounds, including all enforcement tolerances. States may not limit tire loads to less than 500 pounds per inch of tire or tread width, except that such limits may not be applied to tires on the steering axle. States may not limit steering axle weights to less than 20,000 pounds or the axle rating established by the manufacturer, whichever is lower.
(g) The weights in paragraphs (b), (c), (d), and (e) of this section shall be inclusive of all tolerances, enforcement or otherwise, with the exception of a scale allowance factor when using portable scales (wheel-load weighers). The current accuracy of such scales is generally within 2 or 3 percent of actual weight, but in no case shall an allowance in excess of 5 percent be applied. Penalty or fine schedules which impose no fine up to a specified threshold, i.e., 1,000 pounds, will be considered as tolerance provisions not authorized by 23 U.S.C. 127.
(h) States may issue special permits without regard to the axle, gross, or Federal Bridge Formula requirements for nondivisible vehicles or loads.
(i) The provisions of paragraphs (b), (c), and (d) of this section shall not apply to single-, or tandem-axle weights, or gross weights legally authorized under State law on July 1, 1956. The group of axles requirement established in this section shall not apply to vehicles legally grandfathered under State groups of axles tables or formulas on January 4, 1975. Grandfathered weight limits are vested on the date specified by Congress and remain available to a State even if it chooses to adopt a lower weight limit for a time.
(j) The provisions of paragraphs (c) through (e) of this section shall not apply to the operation on Interstate Route 68 in Allegany and Garrett Counties, Maryland, of any specialized vehicle equipped with a steering axle and a tridem axle and used for hauling coal, logs, and pulpwood if such vehicle is of a type of vehicle as was operating in such counties on U.S. Routes 40 or 48 for such purposes on August 1, 1991.
(k) Any over-the-road bus, or any vehicle which is regularly and exclusively used as an intrastate public agency transit passenger bus, is excluded from the axle weight limits in paragraphs (c) through (e) of this section until October 1, 2009. Any State that has enforced, in the period beginning October 6, 1992, and ending November 30, 2005, a single axle weight limitation of 20,000 pounds or greater but less than 24,000 pounds may not enforce a single axle weight limit on these vehicles of less than 24,000 lbs.
(m) The provisions of paragraphs (b) through (e) of this section shall not apply to the operation, on I-99 between Bedford and Bald Eagle, Pennsylvania, of any vehicle that could legally operate on this highway section before December 29, 1995.
(n) Any vehicle subject to this subpart that utilizes an auxiliary power or idle reduction technology unit in order to promote reduction of fuel use and emissions because of engine idling, may be allowed up to an additional 400 lbs. total in gross, axle, tandem, or bridge formula weight limits.
(i) By written certification, the weight of the APU; and
(ii) By demonstration or certification, that the idle reduction technology is fully functional at all times.
(2) Certification of the weight of the APU must be available to law enforcement officers if the vehicle is found in violation of applicable weight laws. The additional weight allowed cannot exceed 400 lbs. or the weight certified, whichever is less.
[49 FR 23315, June 5, 1984, as amended at 59 FR 30420, June 13, 1994; 60 FR 15214, Mar. 22, 1995; 62 FR 10181, Mar. 5, 1997; 63 FR 70653, Dec. 22, 1998; 72 FR 7748, Feb. 20, 2007]
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Deconvolución sísmica - SEG Wiki
Deconvolución sísmica
This page is a translated version of the page Seismic deconvolution and the translation is 50% complete.
Seismic deconvolution is a general term for deconvolution methods designed to remove effects that tend to mask the primary reflected events on a seismogram. Such masking effects can be produced by the earth itself — in the form of absorption, reverberation (multiple reflections), and ghosting — whereas other masking effects result from the seismic source and receiver. To deconvolve seismic data, first we must supply certain required parameters. Then we can design and apply deconvolution filters. Basic to the understanding of deconvolution in the processing of reflection seismic data is development of a model of the layered earth.
An oil well drilled in a sedimentary basin will reveal the layering of sediments. When a wave of unit energy strikes an interface between two layers, some of the energy is reflected, and the remainder of the energy is transmitted. The amount of reflected energy depends on the reflection coefficient of the interface. If we plot the reflection coefficients of all these interfaces as a function of two-way traveltime, we obtain the so-called reflectivity function. This reflectivity gives vital information about the geologic structure. In the ideal case of distinct, well-defined layers, this reflectivity function would consist of a pip at each interface. The size of such a pip would be equal to the value of the reflection coefficient at that interface. The magnitudes of most reflection coefficients encountered in petroleum exploration are small, generally much smaller than one. For the simplified case of no multiples and no transmission losses, a powerful model of the reflection seismogram can be obtained by attaching the source wavelet to each pip on the reflectivity function.
In Chapter 9, in our discussion of equations 28 and 29 of that chapter, we examined the convolutional model
{\displaystyle x=s*a*m*i*\varepsilon }
, where s is the seismic source (or signature), a is the intrinsic seismic absorption operator, m is the multiple-reflection response, i is the instrument response, and
{\displaystyle \varepsilon }
is the reflectivity. (Note that in this book, we use the symbol a in two ways: as the inverse of the minimum-delay wavelet b and as the absorption operator. The meaning is made clear by the context in which the symbol appears.) The signature and the instrument response are more subject to our control, so usually they can be measured or estimated. In such cases, these responses are removable with signature deconvolution methods, as described in Chapter 9.
On the other hand, the earth controls the nature of both intrinsic seismic absorption and the multiple-reflection response, so these are not handled as easily. In Chapter 14, we deal with seismic absorption, so we will neglect it here. As a result, we are left with a further simplified convolutional model given by the signature-free trace
{\displaystyle z=m*\varepsilon }
. Seismic deconvolution is based on this simple convolutional model. We can think of the reflectivity
{\displaystyle \varepsilon }
as the filter input, the wavelet m as the impulse response of the filter, and the signature-free trace z as the filter output. In this model, we assume that the reflectivity series
{\displaystyle \varepsilon }
is white and that the multiple-reflection response m is a minimum-phase wavelet. The purpose of deconvolution in the present model is to remove the multiples from the observed seismogram, thus yielding the ideal seismogram (i.e., the reflectivity function). To describe how deconvolution works in practice, we must justify our model. We will do so by considering the earth as a stack of sedimentary layers.
Many great oil fields discovered in the early days of seismic prospecting were in areas that produced textbook-type seismograms. Such seismograms showed beautiful primary reflections that accurately represented the sedimentary structure, because in those areas, the sedimentary layers were characterized by interfaces with small and uncorrelated reflection coefficients (i.e., interfaces without major reflectivity magnitudes). In other words, such favorable seismic areas contained no major interfaces that would give rise to strong multiple reflections in the depth range of interest. Because of the smallness and randomness of the reflectivity in these sedimentary columns, the multiples tended to interact destructively with each other, with the net effect that no strong multiples tended to appear.
On land, the situation became quite different in areas where major multiples originated from near-surface limestone layers. At sea, the situation was different as well — the water layer introduced strong multiple energy in the form of so-called multiple reverberations. At that time, seismograms containing such water-layer-induced reverberations were known as “ringing” or “singing” records.
When one or more strongly reflecting interfaces exist in the sedimentary column, multiples from these reflectors start to build up and tend to mask the primary events. For example, in marine exploration, the water layer represents a nonattenuating medium bounded by two strongly reflecting interfaces, so it represents an energy trap. A seismic pulse generated in this energy trap will be reflected successively between the two interfaces. Those water reverberations will obscure reflections arriving from deeper horizons below the water layer. On land, a deep limestone layer bounded by strongly reflecting interfaces also can produce multiple reflections, which then interfere with primary reflections.
Modelado de la cola y modelado de la cabeza Modelo convolucional por partes
Procesamiento de la ondícula Algunas consideraciones
Modelos utilizados para la deconvolución
Predicción de mínimos cuadrados y suavizamiento
El filtro de error predicción
Deconvolución de pico
Deconvolución predictiva
Modelado de la cola y modelado de la cabeza
Modelo convolucional por partes
Modelo convolucional variable en tiempo
Modelo de coeficientes de reflexión aleatorios
Implementación de la deconvolución
Apéndice J: Ejercicios
Retrieved from "https://wiki.seg.org/index.php?title=Seismic_deconvolution/es&oldid=173956"
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EUDML | C...-Bounding Sets and Compactness. EuDML | C...-Bounding Sets and Compactness.
C...-Bounding Sets and Compactness.
P. Biström; Jesus A. Jaramillo
Biström, P., and Jaramillo, Jesus A.. "C...-Bounding Sets and Compactness.." Mathematica Scandinavica 75.1 (1994): 82-86. <http://eudml.org/doc/167305>.
@article{Biström1994,
author = {Biström, P., Jaramillo, Jesus A.},
keywords = {-bounding sets; compactness},
title = {C...-Bounding Sets and Compactness.},
AU - Biström, P.
AU - Jaramillo, Jesus A.
TI - C...-Bounding Sets and Compactness.
KW - -bounding sets; compactness
Peter Biström, Jesús Jaramillo, Mikael Lindström, Algebras of real analytic functions: Homomorphisms and bounding sets
{C}^{\infty }
-bounding sets, compactness
Articles by P. Biström
Articles by Jesus A. Jaramillo
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Home : Support : Online Help : Mathematics : Differential Equations : Lie Symmetry Method : Commands for PDEs (and ODEs) : LieAlgebrasOfVectorFields : LHPDE : GetIDBasis
get initial data basis as an IDBasis object from a LHPDE object
set initial data basis as an IDBasis object in a LHPDE object
a IDBasis object.
The GetIDBasis method returns the initial data basis as an IDBasis object for a LHPDE object.
The SetIDBasis method sets the IDBasis object in a LHPDE object.
To set an initial data basis in a LHPDE object, constructing an IDBasis object is required. See LieAlgebrasOfVectorFields[IDBasis] for more detail.
For an IDBasis object to be eligible to be set in a LHPDE object, their parametric derivatives must be the same. See example below.
These methods are associated with the LHPDE and IDBasis objects. For more detail, see Overview of the LHPDE object and Overview of the IDBasis object.
\mathrm{with}\left(\mathrm{LieAlgebrasOfVectorFields}\right):
\mathrm{Typesetting}:-\mathrm{Settings}\left(\mathrm{userep}=\mathrm{true}\right):
\mathrm{Typesetting}:-\mathrm{Suppress}\left([\mathrm{\xi }\left(x,y\right),\mathrm{\eta }\left(x,y\right)]\right):
\mathrm{E2}≔\mathrm{LHPDE}\left([\mathrm{diff}\left(\mathrm{\xi }\left(x,y\right),y,y\right)=0,\mathrm{diff}\left(\mathrm{\eta }\left(x,y\right),x\right)=-\mathrm{diff}\left(\mathrm{\xi }\left(x,y\right),y\right),\mathrm{diff}\left(\mathrm{\eta }\left(x,y\right),y\right)=0,\mathrm{diff}\left(\mathrm{\xi }\left(x,y\right),x\right)=0],\mathrm{indep}=[x,y],\mathrm{dep}=[\mathrm{\xi },\mathrm{\eta }]\right)
\textcolor[rgb]{0,0,1}{\mathrm{E2}}\textcolor[rgb]{0,0,1}{≔}[{\textcolor[rgb]{0,0,1}{\mathrm{\xi }}}_{\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{\mathrm{\eta }}}_{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{\mathrm{\xi }}}_{\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{\mathrm{\eta }}}_{\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{\mathrm{\xi }}}_{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{indep}}\textcolor[rgb]{0,0,1}{=}[\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{dep}}\textcolor[rgb]{0,0,1}{=}[\textcolor[rgb]{0,0,1}{\mathrm{\xi }}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{\eta }}]
B≔\mathrm{IDBasis}\left(\mathrm{E2},[\mathrm{\xi }\left(x,y\right)-y\mathrm{diff}\left(\mathrm{\xi }\left(x,y\right),y\right),\mathrm{\eta }\left(x,y\right)-x\mathrm{diff}\left(\mathrm{\xi }\left(x,y\right),y\right),-\mathrm{diff}\left(\mathrm{\xi }\left(x,y\right),y\right)]\right)
\textcolor[rgb]{0,0,1}{B}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{\mathrm{\xi }}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{\mathrm{\xi }}}_{\textcolor[rgb]{0,0,1}{y}}\right)\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{\eta }}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{\mathrm{\xi }}}_{\textcolor[rgb]{0,0,1}{y}}\right)\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{\mathrm{\xi }}}_{\textcolor[rgb]{0,0,1}{y}}]
No initial data basis is set in E2 yet.
\mathrm{GetIDBasis}\left(\mathrm{E2}\right)
\textcolor[rgb]{0,0,1}{\mathrm{FAIL}}
Their parametric derivatives must be the same,
\mathrm{GetParametricDerivatives}\left(B\right)
[\textcolor[rgb]{0,0,1}{\mathrm{\xi }}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{\mathrm{\xi }}}_{\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{\eta }}]
\mathrm{ParametricDerivatives}\left(\mathrm{E2}\right)
[\textcolor[rgb]{0,0,1}{\mathrm{\xi }}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{\mathrm{\xi }}}_{\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{\eta }}]
\mathrm{SetIDBasis}\left(\mathrm{E2},B\right)
[\textcolor[rgb]{0,0,1}{\mathrm{\xi }}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{\mathrm{\xi }}}_{\textcolor[rgb]{0,0,1}{y}}\right)\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{\eta }}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{\mathrm{\xi }}}_{\textcolor[rgb]{0,0,1}{y}}\right)\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{\mathrm{\xi }}}_{\textcolor[rgb]{0,0,1}{y}}]
\mathrm{B1}≔\mathrm{GetIDBasis}\left(\mathrm{E2}\right)
\textcolor[rgb]{0,0,1}{\mathrm{B1}}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{\mathrm{\xi }}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{\mathrm{\xi }}}_{\textcolor[rgb]{0,0,1}{y}}\right)\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{\eta }}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{\mathrm{\xi }}}_{\textcolor[rgb]{0,0,1}{y}}\right)\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{\mathrm{\xi }}}_{\textcolor[rgb]{0,0,1}{y}}]
\mathrm{type}\left(\mathrm{B1},'\mathrm{IDBasis}'\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
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AWGN Channel - MATLAB & Simulink - MathWorks Italia
Relationship Between EsN0 and EbN0
Relationship Between EsN0 and SNR
An AWGN channel adds white Gaussian noise to the signal that passes through it. You can create an AWGN channel in a model using the comm.AWGNChannel System object™, the AWGN Channel block, or the awgn function.
The following examples use an AWGN Channel: QPSK Transmitter and Receiver and General QAM Modulation over AWGN Channel.
Typical quantities used to describe the relative power of noise in an AWGN channel include
Signal-to-noise ratio (SNR) per sample. SNR is the actual input parameter to the awgn function.
Ratio of bit energy to noise power spectral density (EbN0). This quantity is used by Bit Error Rate Analysis app and performance evaluation functions in this toolbox.
Ratio of symbol energy to noise power spectral density (EsN0)
The relationship between EsN0 and EbN0, both expressed in dB, is as follows:
{E}_{s}/{N}_{0}\text{ (dB)}={E}_{b}/{N}_{0}\text{ (dB)}+10{\mathrm{log}}_{10}\left(k\right)
where k is the number of information bits per symbol.
In a communications system, k might be influenced by the size of the modulation alphabet or the code rate of an error-control code. For example, in a system using a rate-1/2 code and 8-PSK modulation, the number of information bits per symbol (k) is the product of the code rate and the number of coded bits per modulated symbol. Specifically, (1/2) log2(8) = 3/2. In such a system, three information bits correspond to six coded bits, which in turn correspond to two 8-PSK symbols.
The relationship between EsN0 and SNR, both expressed in dB, is as follows:
\begin{array}{l}{E}_{s}/{N}_{0}\text{ (dB)}=10{\mathrm{log}}_{10}\left({T}_{sym}/{T}_{samp}\right)+SNR\text{}\text{ (dB) for complex input signals}\\ {E}_{s}/{N}_{0}\text{ (dB)}=10{\mathrm{log}}_{10}\left(0.5{T}_{sym}/{T}_{samp}\right)+SNR\text{}\text{ (dB) for real input signals}\end{array}
where Tsym is the symbol period of the signal and Tsamp is the sampling period of the signal.
For a complex baseband signal oversampled by a factor of 4, the EsN0 exceeds the corresponding SNR by 10 log10(4).
Derivation for Complex Input Signals. You can derive the relationship between EsN0 and SNR for complex input signals as follows:
\begin{array}{c}{E}_{s}/{N}_{0}\text{ (dB)}=10{\mathrm{log}}_{10}\left(\left(S\cdot {T}_{sym}\right)/\left(N/{B}_{n}\right)\right)\\ =10{\mathrm{log}}_{10}\left(\left({T}_{sym}{F}_{s}\right)\cdot \left(S/N\right)\right)\\ =10{\mathrm{log}}_{10}\left({T}_{sym}/{T}_{samp}\right)+SNR\text{}\text{ (dB)}\end{array}
Bn = Noise bandwidth, in Hertz = Fs = 1/Tsamp.
Fs = Sampling frequency, in Hertz
Behavior for Real and Complex Input Signals. These figures illustrate the difference between the real and complex cases by showing the noise power spectral densities of a real bandpass white noise process and its complex lowpass equivalent.
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Arcade flyer for Tetris The Grand Master
JP: August, 1998
Tetris The Grand Master[a] is an arcade game released exclusively in Japan, and the first game of the TGM series. Players place pieces and clear lines as the game goes on, increasing their grade along the way.
At later levels, the game will force them to keep up with the high speeds using special techniques; in addition to the Initial Rotation System introduced in the game. The game is known for being one of Arika's popular games to this day, and later installments would introduce modes with their own gameplay and objective.
2.1 Scoring formula
3.1 Secret grade
4.1 Item blocks
5 Hidden modes
6 Speed timings
TGM's gameplay is heavily inspired by it's arcade predecessor, Sega's Tetris released 10 years earlier. It uses a modified rotation system, color scheme, and relies heavily on mechanics such as lock delay. Another game which inspired TGM is TETRIS SEMIPRO-68k, a fan game which was the first to introduce 20G gravity.
The main goal in TGM is to score points, awarding the player a higher grade. The game ends when a player reaches level 999. If the player scored enough points, they will be awarded with the grade S9. To achieve the grade GM, the player must also meet some time requirements during play. If the player tops out before reaching level 999, the game ends, awarding the player the current grade and it's "mastering time", the time at which the grade was awarded during gameplay.
Level has a unique implementation in TGM. The level counter will increase by 1 for every piece that enters the playfield. It will also increase by 1 for each line cleared.
When the player is about to increment the hundreds digit (e.g., level 399), only line clears will increase the level. Level 998 is treated similarly, with a final line clear required to reach 999 and finish.
Main article: IRS
Abbreviation for Initial Rotation System.
Normally a piece will appear in the rotation showed in the piece preview. With IRS, holding either the left or right rotation button will cause the piece to appear rotated 90 degrees. This allows the player a higher degree of freedom when placing pieces at higher game speeds.
Main article: ghost piece
Abbreviation for Temporary Landing System This system is a semi-transparent representation of where the piece will land if allowed to drop into the playfield. It is displayed up to level 100.
The Grand Master shares the same scoring mechanisms of many other Tetris games:
The player receive more points for clearing more lines at once.
Lines are worth more with each passing level.
The player receive points for forcing a piece down. (Although only when this results in cleared lines, unlike in some other games.)
The player receive a combo bonus for clearing lines with consecutive pieces.
The player receive a bravo bonus for clearing the entire playfield.
{\displaystyle {\text{Combo}}={\text{Previous Combo Value}}+(2\times {\text{Lines}})-2}
{\displaystyle {\text{Score}}=(\left\lceil ({\text{Level}}+{\text{Lines}})/4\right\rceil +{\text{Soft}})\times {\text{Lines}}\times {\text{Combo}}\times {\text{Bravo}}}
Level is the current level the player is on (before the lines are cleared).
Lines is the number of lines the player just cleared.
{\displaystyle \left\lceil ({\text{Level}}+{\text{Lines}})/4\right\rceil }
is rounded up.
Soft is the cumulative number of frames during which Down was held during the piece's active time. Manually locking pieces already on the ground will increase Soft by 1.
Locking a piece without clearing lines resets Combo to 1. Otherwise, the game updates Combo as follows, before calculating Score:
E.g., a double-triple-single combo will have combo values 3, 7, and 7 respectively.
Bravo is equal to 4 if this piece has cleared the screen, and otherwise is 1.
In TGM grade is entirely determined by score. As the player passes certain milestones, the game will assign the player the next grade.
Grade Conditions
GM Grade Conditions
300 12,000 (Grade 1) =<04:15:00
500 40,000 (Grade S4) =<07:30:00
(slightly higher than S9) =<13:30:00
Main article: Secret Grade Techniques
Secret grade is a hidden grading system that recognizes a ">" pattern in the playfield by leaving holes. The first hole becomes Secret grade 9, and each subsequent hole increasing the grade until row 19, the GM grade. The player must top out to have the Secret grade awarded. A minimum grade of 5 is needed to see the message.
If the arcade operator has enabled Vs. mode, two players can start a Vs. game. One player must first start a game, and during any point of that game the second player may press start to challenge them, and begin a game.
When the player clears 2 or more lines at once, those lines will be sent to their opponent's playfield as garbage lines. If the match does not end within 5 minutes, the game ends in a draw.
Item blocks can be used to attack or defend from the player's opponent.
Clearing lines will fill up the item meter adjacent to a player's playfield. When the meter is full an item block is dealt to the player. They may use it by clearing a line which the piece is part of.
There are 18 different item blocks.
DEATH BLOCK Makes the opponent's next block big.
NEGA FIELD Flip the playfield's occupied cells to empty, and empty cells to occupied.
SPIN FIELD Rotate the opponent's field for a short time.
180° FIELD Rotate the playfield upside down and move the newly rotated cells down.
SHOT GUN! Randomly shoot a group of holes in the opponent's stack.
HARD BLOCK The opponent's next piece must be cleared twice to be removed from the playfield.
LASER BLOCK Remove a column of holes in the opponent's stack. The attacker many move the laser by pressing left and right.
ROLL ROLL The opponent's next 3 pieces are auto rotating by a fixed interval.
TRANS FORM The opponent's next 3 pieces will change tetromino every time the piece is rotated.
X-RAY A partially invisible effect is applied to the opponent's stack for a short period.
PRESS FIELD Shrinks the opponent's playfield for a short period.
↑ DEL FIELD Deletes the upper half of the player's playfield.
↓ DEL FIELD Deletes the lower half of the player's playfield.
→ MOV FIELD Push every cell in the player's playfield to the right.
← MOV FIELD Push every cell in the player's playfield to the left.
DEL EVEN Deletes every 2nd row in the player's playfield.
FREE FALL Forces all cells to move down in the player's playfield removing any holes.
EXCHG FIELD Swaps the player's and opponent's playfield.
All codes must be entered at the title screen. Codes may be combined (i.e. Big 20G mode + Reverse Monochrome mode.) Scores achieved using codes will not be stored on the high score table.
Key: L = Left, D = Down, U = Up, R = Right
Immediately enables maximum gravity (20G).
Input code: DDDDDDDDCBA
Tetrominoes are twice normal size, simulating play in a 5x10 playfield
Input code: LLLLDCBA
Play in reverse! Pieces will spawn at the bottom of the playfied and "fall" upwards
Input code: DUUDCBA
All tetrominoes are monochrome
Input code: RRRUCBA
Shows the "Temporary Landing System" aka "ghost piece" for the entirety of the game, not just 0-100
Input code: ABCCBAACB
Uki mode
Instead of line clear sound effect when clearing multiple lines, a child's voice will say "Uki" repeatedly, the japanese word for the sound a monkey makes. The child will say "Waaahhh!" instead if the player scores a Tetris.
Input code: ABABABABABABABABB
No item VS (Play VS mode without item blocks):
Hold both player Start buttons together before the match begins.
Gravity does not increase uniformly, unlike many other tetris games. It rises and falls, depending on the level as shown in the table below.
The unit for gravity is G (rows per frame), as a fraction with a constant denominator of 256. This means G = Internal Gravity/256. For example, at levels 90 through 99, the gravity is 64/256G, or 1/4G.
Unlike TGM2, the line clear delay, lock delay, ARE and DAS do not change throughout the game. The following are the timing values adjusted to what a player would observe, including inclusive DAS counting.
The player's DAS charge is unmodified during line clear delay, the first 4 frames of ARE, the last frame of ARE, and the frame on which a piece spawns.
The tables below provide verified lists of archives and files to make sure that Tetris the Grand Master can be emulated properly. There aren't any guides, at this point, for dumping these required files from a PCB; but these files can be searched out on the Internet.
tgmj.zip 7fcc8a72b4bc1f3340e1db4724b0628d
coh3002c.zip 4f97958fe67444637b5c6700fb62849c
cpzn2.zip 99304c05983b337217ab4b56400a14ac
qsound_hle.zip 758388893761ea3ff0d820f60344a38e
cp10.ic652 26574f621b61ea09eef08e8a7eba5a65
m534002c-59.ic353 1254b215f64aee1f83895e0213a9ac82
coh-3002c.353 1254b215f64aee1f83895e0213a9ac82
dl-1425.bin 108b113a596e800a02fece73f784eeb0
Required TGM Files
ate_02.2e 591fe13727496e4037e53cc834e24ccd
ate-01m.3a b832063bef8d81ff6a318f39da6f4c74
ate-05m.3h 99740df1cf1201a4efe5d5b2c4806425
ate-06m.4h 2188d3a9be4769a21e44d33cbfbd2024
atej_04.2h 14e13f937163b6db734d3220a017bfc0
cp11 439f9c44559f2f4761d433dd4b96866d
The current version of MAME is capable of running Tetris the Grand Master. MAME can either be run as a standalone program on Windows or the Macintosh, or as a core through other Emulator frontends such as RetroArch or OpenEmu (Experimental). It is recommended that all of the BIOS files from the above archives are added into the "tgmj.zip" archive before importing the game into an emulator. If the tgmj.zip archive is imported without the BIOS files inside, the emulator will report numerous file requirement errors, or it will simply not start.
Capcom Sony ZN-2, JP
PCB, 244×245×50 mm
↑ Japanese: テトリス・ザ・グランドマスター Hepburn: Tetorisu Za Gurando Masutā
Tetris: The Grand Master on The Cutting Room Floor
Retrieved from "https://tetris.wiki/index.php?title=Tetris_The_Grand_Master&oldid=23127"
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The Sard conjecture on Martinet surfaces
1 June 2018 The Sard conjecture on Martinet surfaces
André Belotto da Silva, Ludovic Rifford
Duke Math. J. 167(8): 1433-1471 (1 June 2018). DOI: 10.1215/00127094-2017-0058
Given a totally nonholonomic distribution of rank
2
3
-dimensional manifold, we investigate the size of the set of points that can be reached by singular horizontal paths starting from the same point. In this setting, by the Sard conjecture, that set should be a subset of the so-called Martinet surface of
2
-dimensional Hausdorff measure zero. We prove that the conjecture holds in the case where the Martinet surface is smooth. Moreover, we address the case of singular real-analytic Martinet surfaces, and we show that the result holds true under an assumption of nontransversality of the distribution on the singular set of the Martinet surface. Our methods rely on the control of the divergence of vector fields generating the trace of the distribution on the Martinet surface and some techniques of resolution of singularities.
André Belotto da Silva. Ludovic Rifford. "The Sard conjecture on Martinet surfaces." Duke Math. J. 167 (8) 1433 - 1471, 1 June 2018. https://doi.org/10.1215/00127094-2017-0058
Received: 20 August 2016; Revised: 6 October 2017; Published: 1 June 2018
Secondary: 32S45 , 34H05
Keywords: control theory , differential forms , Differential geometry , resolution of singularities , Sard conjecture , sub-Riemannian geometry
André Belotto da Silva, Ludovic Rifford "The Sard conjecture on Martinet surfaces," Duke Mathematical Journal, Duke Math. J. 167(8), 1433-1471, (1 June 2018)
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structures - Maple Help
Home : Support : Online Help : Mathematics : Discrete Mathematics : Combinatorics : Combinatorial Structures : structures
Predefined Structures in the combstruct Package
combstruct[function](struct(args), size=n)
(optional) non-negative integer specifying the size of the object or the string 'allsizes'
Some special combinatorial classes are predefined in the combstruct package. They are Combination or Subset, Permutation, Partition, and Composition.
Combination or Subset:
Combinations (or subsets) of elements
The argument is a set, list, or non-negative integer. In the case of a non-negative integer n, it is treated as the set
{1,2,...,n}
. The default size is
'\mathrm{allsizes}'
Use the string 'allsizes' when the object is selected from all possible sizes. If the size is not specified, each structure has default behavior.
Permutations of elements
The argument is a set, list, or non-negative integer. In the case of a non-negative integer n, it is treated as the list
[1,2,...,n]
. The default size is the number of elements in the set/list.
Partitions of positive integers into sums of positive integers without regards to order.
The argument is a positive integer. The default size is 'allsizes'.
Composition of a positive integer n into a positive integer k parts of size at least 1, where the order of summands is meaningful.
The argument is a positive integer. The default size is
'\mathrm{allsizes}'
Combination and Subset are different names for the same structure.
You can define your own structure that is understood by the functions in the combstruct package. To create the structure Foo, create the procedures `combstruct/count/Foo`, `combstruct/draw/Foo`, `combstruct/allstructs/Foo`, and `combstruct/iterstructs/Foo`.
Each of these functions must take size as their first argument, a non-negative integer, the string
'\mathrm{allsizes}'
or the string
'\mathrm{default}'
, with the arguments that your structure needs. Thus,
\mathrm{draw}\left(\mathrm{Foo}\left(x,y,z\right),\mathrm{size}=n\right)
invokes `combstruct/draw/Foo`(n, x, y, z).
\mathrm{with}\left(\mathrm{combstruct}\right):
Count all possible subsets of
{1,2,3}
\mathrm{count}\left(\mathrm{Subset}\left({1,2,3}\right)\right)
\textcolor[rgb]{0,0,1}{8}
Draw a combination of
[1,2,3,4,5]
that has 4 elements.
\mathrm{draw}\left(\mathrm{Combination}\left(5\right),\mathrm{size}=4\right)
{\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}}
Count the permutations of
[a,a,b]
, using all three elements.
\mathrm{count}\left(\mathrm{Permutation}\left([a,a,b]\right)\right)
\textcolor[rgb]{0,0,1}{3}
Create an iterator over all the permutations of
[a,a,b]
that consist of only two elements.
\mathrm{it}≔\mathrm{iterstructs}\left(\mathrm{Permutation}\left([a,a,b]\right),\mathrm{size}=2\right):
Draw any partition of 9.
\mathrm{draw}\left(\mathrm{Partition}\left(9\right)\right)
[\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}]
List all the compositions of 3 in 2 parts.
\mathrm{allstructs}\left(\mathrm{Composition}\left(3\right),\mathrm{size}=2\right)
[[\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}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}]]
combstruct[count]
combstruct[draw]
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Validate quality of credit scorecard model - MATLAB validatemodel - MathWorks í•œêµ
Stats=4×2 table
PointsInfo=38×3 table
'CAP' — Cumulative Accuracy Profile. Plots the fraction of borrowers up to score “s†versus the fraction of defaulters up to score “s†('PctObs' versus 'Sensitivity' columns of T optional output argument). For more details, see Cumulative Accuracy Profile (CAP).
'ROC' — Receiver Operating Characteristic. Plots the fraction of non-defaulters up to score “s†versus the fraction of defaulters up to score “s†('FalseAlarm' versus 'Sensitivity' columns of T optional output argument). For more details, see Receiver Operating Characteristic (ROC).
'KS' — Kolmogorov-Smirnov. Plots each score “s†versus the fraction of defaulters up to score “s,†and also versus the fraction of non-defaulters up to score “s†('Scores' versus both 'Sensitivity' and 'FalseAlarm' columns of the optional output argument T). For more details, see Kolmogorov-Smirnov statistic (KS).
'TrueBads' — Cumulative number of “bads†up to, and including, the corresponding score.
'FalseBads' — Cumulative number of “goods†up to, and including, the corresponding score.
'TrueGoods' — Cumulative number of “goods†above the corresponding score.
'FalseGoods' — Cumulative number of “bads†above the corresponding score.
'Sensitivity' — Fraction of defaulters (or the cumulative number of “bads†divided by total number of “badsâ€). This is the distribution of “bads†up to and including the corresponding score.
'FalseAlarm' — Fraction of non-defaulters (or the cumulative number of “goods†divided by total number of “goodsâ€). This is the distribution of “goods†up to and including the corresponding score.
The scores of given observations are sorted from riskiest to safest. For a given fraction M (0% to 100%) of the total borrowers, the height of the CAP curve is the fraction of defaulters whose scores are less than or equal to the maximum score of the fraction M, also known as “Sensitivity.â€
The area under the CAP curve, known as the AUCAP, is then compared to that of the perfect or “ideal†model, leading to the definition of a summary index known as the accuracy ratio (AR) or the Gini coefficient:
AR=\frac{{A}_{R}}{{A}_{P}}
where AR is the area between the CAP curve and the diagonal, and AP is the area between the perfect model and the diagonal. This represents a “random†model, where scores are assigned randomly and therefore the proportion of defaulters and non-defaulters is independent of the score. The perfect model is the model for which all defaulters are assigned the lowest scores, and therefore, perfectly discriminates between defaulters and nondefaulters. Thus, the closer to unity AR is, the better the scoring model.
To find the receiver operating characteristic (ROC) curve, the proportion of defaulters up to a given score “s,†or “Sensitivity,†is computed.
This proportion is known as the true positive rate (TPR). Additionally, the proportion of nondefaulters up to score “s,“ or “False Alarm Rate,†is also computed. This proportion is also known as the false positive rate (FPR). The ROC curve is the plot of the “Sensitivity†vs. the “False Alarm Rate.†Computing the ROC curve is similar to computing the equivalent of a confusion matrix at each score level.
AR=2\left(AUROC\right)â1
The KS plot shows the distribution of defaulters and the distribution of non-defaulters on the same plot. For the distribution of defaulters, each score “s†is plotted versus the proportion of defaulters up to “s," or “Sensitivity." For the distribution of non-defaulters, each score “s†is plotted versus the proportion of non-defaulters up to “s," or “False Alarm." The statistic of interest is called the KS statistic and is the maximum difference between these two distributions (“Sensitivity†minus “False Alarmâ€). The score at which this maximum is attained is also of interest.
[1] “Basel Committee on Banking Supervision: Studies on the Validation of Internal Rating Systems.†Working Paper No. 14, February 2005.
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On Variational Inclusion and Common Fixed Point Problems in q-Uniformly Smooth Banach Spaces
2012 On Variational Inclusion and Common Fixed Point Problems in q-Uniformly Smooth Banach Spaces
Yanlai Song, Huiying Hu, Luchuan Ceng
We introduce a general iterative algorithm for finding a common element of the common fixed-point set of an infinite family of
{\lambda }_{i}
-strict pseudocontractions and the solution set of a general system of variational inclusions for two inverse strongly accretive operators in a q-uniformly smooth Banach space. Then, we prove a strong convergence theorem for the iterative sequence generated by the proposed iterative algorithm under very mild conditions. The methods in the paper are novel and different from those in the early and recent literature. Our results can be viewed as the improvement, supplementation, development, and extension of the corresponding results in some references to a great extent.
Yanlai Song. Huiying Hu. Luchuan Ceng. "On Variational Inclusion and Common Fixed Point Problems in q-Uniformly Smooth Banach Spaces." J. Appl. Math. 2012 (SI06) 1 - 20, 2012. https://doi.org/10.1155/2012/865810
Yanlai Song, Huiying Hu, Luchuan Ceng "On Variational Inclusion and Common Fixed Point Problems in q-Uniformly Smooth Banach Spaces," Journal of Applied Mathematics, J. Appl. Math. 2012(SI06), 1-20, (2012)
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