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Random Dirichlet series arising from records
We study the distributions of the random Dirichlet series with parameters (s; β) defined by [equation presented] where (In) is a sequence of independent Bernoulli random variables, In taking value 1
with probability 1=n^β and value 0 otherwise. Random series of this type are motivated by the record indicator sequences which have been studied in extreme value theory in statistics. We show that
when s > 0 and 0 < β ≤ 1 with s + β > 1 the distribution of S has a density; otherwise it is purely atomic or not defined because of divergence. In particular, in the case when s > 0 and β = 1, we
prove that for every 0 < s < 1 the density is bounded and continuous, whereas for every s > 1 it is unbounded. In the case when s > 0 and 0 < β < 1 with s + β > 1, the density is smooth. To show the
absolute continuity, we obtain estimates of the Fourier transforms, employing van der Corput's method to deal with number-theoretic problems. We also give further regularity results of the densities,
and present an example of a non-atomic singular distribution which is induced by the series restricted to the primes.
Funders Funder number
Israel Science Foundation
Japan Society for the Promotion of Science 26800029
• Random Dirichlet series
• Records
• The van der Corput lemma
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The working of PCA
Until now, you’ve learnt the two building blocks of PCA: Basis and variance. In the following video, we will make use of both the terms to make you understand the objective that PCA aims to achieve.
The steps of PCA as summarised in the above video are as follows:
• Find n new features – Choose a different set of n basis vectors (non-standard). These basis vectors are essentially the directions of maximum variance and are called Principal Components
• Express the original dataset using these new features
• Transform the dataset from the original basis to this PCA basis.
• Perform dimensionality reduction – Choose only a certain k (where k < n) number of the PCs to represent the data. Remove those PCs which have fewer variance (explain less information) than
PCA’s role in the ML pipeline almost solely exists as a dimensionality reduction tool. Basically, you choose a fixed number of PCs that explained a certain threshold of variance that you have chosen
and then uses only that many columns to represent the original dataset. This modified dataset is then passed on to the ML pipeline for further prediction algorithms to take place. PCA helps us in
improving the model performance significantly and helps us in visualising higher-dimensional datasets as well.
Additional Reading
As mentioned in the video, you can take a look at the Algorithm of PCA optional session to understand in detail about how PCA finds the new basis vectors using the eigendecomposition of the
covariance matrix method.
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Excel Solver - Non-Smooth Optimization
The most difficult type of optimization problem to solve is a non-smooth problem (NSP). Such a problem may not only have multiple feasible regions and multiple locally optimal points within each
region – because some of the functions are non-smooth or even discontinuous, derivative information generally cannot be used to determine the direction in which the function is increasing (or
decreasing). In other words, the situation at one possible solution gives very little information about where to look for a better solution.
In all but the simplest problems, it is impractical to exhaustively enumerate all of the possible solutions and pick the best one, even on a fast computer. Hence, most methods rely on some sort of
controlled random search, or sampling of possible solutions – combined with deterministic (non-random) methods for exploring the search space.
Genetic and Evolutionary Algorithms
The Evolutionary Solving method uses genetic and evolutionary algorithms to seek “good” solutions for non-smooth optimization problems. An evolutionary algorithm for optimization is different from
“classical” optimization methods in several ways. First, it relies in part on random sampling. This makes it a nondeterministic method, which may yield different solutions on different runs. (To
obtain the same solution on each run, you can set a Random Seed option for the Evolutionary Solving method.)
Second, where most classical optimization methods maintain a single best solution found so far, an evolutionary algorithm maintains a population of candidate solutions. Only one (or a few, with
equivalent objectives) of these is “best,” but the other members of the population are “sample points” in other regions of the search space, where a better solution may later be found. The use of a
population of solutions helps the evolutionary algorithm avoid becoming “trapped” at a local optimum, when an even better optimum may be found outside the vicinity of the current solution.
Third – inspired by the role of mutation of an organism’s DNA in natural evolution – an evolutionary algorithm periodically makes random changes or mutations in one or more members of the current
population, yielding a new candidate solution (which may be better or worse than existing population members). There are many possible ways to perform a “mutation,” and the Evolutionary Solver
actually employs five different mutation strategies. The result of a mutation may be an infeasible solution, and the Evolutionary Solver attempts to “repair” such a solution to make it feasible;
this is sometimes, but not always, successful.
Fourth – inspired by the role of sexual reproduction in the evolution of living things – an evolutionary algorithm attempts to combine elements of existing solutions in order to create a new
solution, with some of the features of each “parent.” The elements (e.g. decision variable values) of existing solutions are combined in a crossover operation, inspired by the crossover of DNA
strands that occurs in reproduction of biological organisms. As with mutation, there are many possible ways to perform a “crossover” operation – some much better than others – and the Evolutionary
Solver actually employs multiple variations of four different crossover strategies.
Fifth – inspired by the role of natural selection in evolution – an evolutionary algorithm performs a selection process in which the “most fit” members of the population survive, and the “least fit”
members are eliminated. In a constrained optimization problem, the notion of “fitness” depends partly on whether a solution is feasible (i.e. whether it satisfies all of the constraints), and partly
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Why entropy is logarithmic
Further Reading
In Entropy -- implications of the 2nd law of thermodynamicse, we defined the entropy, $S$ of a particular macrostate of a system as equal to $k_B \ln W$, where $W$ is the number of possible
arrangements of the system (microstates) corresponding to that macrostate. But why? Why not just say that entropy is the number of arrangements? Let's think through why it has to be defined this way.
We want to define entropy to be an extensive property. This means that if I have two systems A and B, the total entropy should be the entropy of A plus the entropy of B. This is like mass (2 kg + 2
kg = 4 kg), and not an intensive property like temperature. (If you combine two systems that are each at 300 K, you have a system at 300 K, not at 600 K!)
What happens to the number of possible arrangements when you combine two systems? If system A can be in 3 different arrangements and system B can be in 5 different arrangements, then there are $3 \
times 5 = 15$ possible combinations. They multiply! This '80s music video explains why.
So we can't just define entropy as the number of possible arrangements, because we need the entropy to add, not multiply, when we combine two systems.
How do you turn multiplication into addition? Just take the logarithm: $3 \times 5 = 15$, but $\ln 3 + \ln 5 = \ln 15$.
So that's why entropy is defined as a constant times ln $W$. $W \times$ (the number of arrangements) is a dimensionless number, so $\ln W$ is too.
The constant out in front could be any constant, but we use Boltzmann's constant, $k_B = 1.38 \times 10^{-23} \mathrm{ J/K}$. When we get to Gibbs free energy, we'll see that this constant has the
right units, we see that it's very convenient for entropy to be in units of energy/temperature.
Workout: Why entropy is logarithmic
Article 599
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How do you solve x / 30 - 1/(5x) = 1/6? | HIX Tutor
How do you solve #x / 30 - 1/(5x) = 1/6#?
Answer 1
$x = - 1 \mathmr{and} x = 6$
Write the equivalent equation, obtained by multiplying all terms by #30x#:
#x^2-6=5x and x!=0#
#x=-1 and x=6#
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Answer 2
To solve the equation x / 30 - 1/(5x) = 1/6, we can start by finding a common denominator for the fractions. The common denominator is 30x. Multiplying each term by 30x, we get:
x * (30x / 30) - (1/(5x)) * (30x / 30) = (1/6) * (30x / 30)
This simplifies to:
x^2 - 6 = 5x
Rearranging the equation:
x^2 - 5x - 6 = 0
Factoring the quadratic equation:
(x - 6)(x + 1) = 0
Setting each factor equal to zero:
x - 6 = 0 or x + 1 = 0
Solving for x:
x = 6 or x = -1
Therefore, the solutions to the equation are x = 6 and x = -1.
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Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
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5.2: Polynomials
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We begin with the definition of a term.
Definition: Term
A term is either a single number (called a constant term) or the product of a number and one or more variables.
For example, each of the following is a term.
\[ -5 \quad-3 x^{2} \quad 12 y^{2} z^{3} \quad 13 a^{2} b c^{3} \nonumber \]
Note how the first term is a single number, while the remaining terms are products of a number and one or more variables. For example, \(−3x^2\) is the product of \(−3\), \(x\), and \(x\).
Definition: Coefficient
When a term is a product of a number and one or more variables, the number is called the coefficient of the term. In the case of a term that is a single number, the number itself is called the
Thus, for example, the coefficients of the terms \[-5 \quad-3 x^{2} \quad 12 y^{2} z^{3} \quad 13 a^{2} b c^{3} \nonumber \]are \(−5\), \(−3\), \(12\), and \(13\), respectively.
Definition: Degree
The degree of a term is the sum of the exponents on each variable of the term. A constant term (single number with no variables) has degree zero.
Thus, for example, the degrees of the terms \[-5 \quad-3 x^{2} \quad 12 y^{2} z^{3} \quad 13 a^{2} b c^{3} \nonumber \]are \(0\), \(2\), \(5\), and \(6\), respectively. In the last example, note that
\(13a^2bc^3\) is equivalent to \(13a^2b^1c^3\), so adding exponents, we get:
\[\begin{aligned} \text { Degree of } 13 a^{2} b c^{3} &=\text { Degree of } 13 a^{2} b^{1} c^{3} \\ &=2+1+3 \\ &=6 \end{aligned} \nonumber \]
Definition: Monomial
The words monomial and term are equivalent.
Thus, \[-5 \quad-3 x^{2} \quad 12 y^{2} z^{3} \quad 13 a^{2} b c^{3} \nonumber \]are monomials.
Definition: Binomial
A binomial is a mathematical expression containing exactly two terms, separated by plus or minus signs.
For example, each of the mathematical expressions \[2 x+3 y \quad-3 a^{2}-3 b^{2} \quad x y+7 \quad-3 x^{2} y+5 x y^{2} \nonumber \]is a binomial. Each expression has exactly two terms.
Definition: Trinomial
A trinomial is a mathematical expression containing exactly three terms, separated by plus or minus signs.
For example, each of the mathematical expressions \[2 x^{2}+3 x+7 \quad a^{2}+2 a b+b^{2} \quad x^{4}-2 x^{2} y^{2}+3 y^{4} \nonumber \]is a trinomial. Each expression has exactly three terms.
A bicycle has two wheels, a binomial has two terms. A tricycle has three wheels, a trinomial has three terms. But once we get past three terms, the assignment of special names ceases and we use the
generic word polynomial, which means “many terms.”
Definition: Polynomial
A polynomial is a many-termed mathematical expression, with terms separated by plus or minus signs. The coefficients of a polynomial are the coefficients of its terms.
Each of the previous expressions, \[12 y^{2} z^{3} \quad-3 a^{2}-3 b^{2} \quad x^{4}-2 x^{2} y^{2}+3 y^{4} \nonumber \]though assigned the particular names monomial, binomial, and trinomial,
respectively, are also “many-termed” expressions and can also be called polynomials. However, because the word polynomial means “many terms,” we can also use the word polynomial to describe
mathematical expressions with more than three terms, such as: \[x^{4}-4 x^{3} y+6 x^{2} y^{2}-4 x y^{3}+y^{4} \nonumber \]The coefficients of \(x^{4}-4 x^{3} y+6 x^{2} y^{2}-4 x y^{3}+y^{4}\) are \(1
\), \(−4\), \(6\), \(−4\), and \(1\).
Ascending and Descending Powers
When asked to simplify a polynomial expression, we should combine any like terms we find, and when possible, arrange the answer in ascending or descending powers.
Example \(\PageIndex{1}\)
Simplify the following polynomial expression, arranging your answer in the descending powers of \(x\). Once you’ve completed that task, make a second arrangement, ordering your terms in ascending
powers of \(x\). \[2 x^{3}+7 x-3 x^{2}+11 x+8 x^{2}+11+15 x \nonumber \]
In order to arrange our answer in descending powers of \(x\), we want to place the term with the highest power of \(x\) first and the term with the lowest power of \(x\) last. We use the commutative
and associative properties to change the order and regroup, then we combine like terms.
\[\begin{aligned} 2 x^{3}+7 x &-3 x^{2}+11 x+8 x^{2}+11+15 x \\ &=2 x^{3}+\left(-3 x^{2}+8 x^{2}\right)+(7 x+11 x+15 x)+11 \\ &=2 x^{3}+5 x^{2}+33 x+11 \end{aligned} \nonumber \]
Note how the powers of \(x\) start at \(3\), then go down in order.
To arrange our final answer in ascending powers of \(x\), we put the lowest power of \(x\) first, then the highest power of \(x\) last, regrouping and combining like terms.
\[\begin{aligned} 2 x^{3}+7 x &-3 x^{2}+11 x+8 x^{2}+11+15 x \\ &=11+(7 x+11 x+15 x)+\left(-3 x^{2}+8 x^{2}\right)+2 x^{3} \\ &=11+33 x+5 x^{2}+2 x^{3} \end{aligned} \nonumber \]
Note how we start with the constant term, then the powers of \(x\) increase in order.
Exercise \(\PageIndex{1}\)
Simplify the following polynomial, and arrange your answer in ascending powers of \[3 x^{2}-5 x^{3}+8 x+9 x^{2}-7 x+2 x^{3} \nonumber \]
\(x+12 x^{2}-3 x^{3}\)
When we have a polynomial in a single variable, such as the polynomial in Example \(\PageIndex{1}\), arranging the terms in ascending or descending order is fairly straightforward. However, a
polynomial in two or more variables is a bit more difficult, and sometimes impossible, to arrange in a decent order.
Example \(\PageIndex{2}\)
Simplify the following polynomial expression, then arrange your answer in descending powers of \(x\).\[x^{3}+2 x y^{2}-6 x^{2} y+y^{3}-3 x y^{2}+4 x^{2} y \nonumber \]
We’ll again use the commutative and associative properties to change the order and regroup, putting the terms with the highest powers of \(x\) first, then follow with terms containing lower powers of
\(x\) in order.
\[\begin{aligned} x^{3}+2 x y^{2} &-6 x^{2} y+y^{3}-3 x y^{2}+4 x^{2} y \\ &=x^{3}+\left(-6 x^{2} y+4 x^{2} y\right)+\left(2 x y^{2}-3 x y^{2}\right)+y^{3} \\ &=x^{3}-2 x^{2} y-x y^{2}+y^{3} \end
{aligned} \nonumber \]
Note that this is a very natural order, the powers of \(x\) decrease while simultaneously the powers of \(y\) increase.
Exercise \(\PageIndex{2}\)
Simplify the following polynomial, and arrange your answer in descending powers of \(x\): \[-4 x^{2} y^{2}+3 x y^{3}+6 x^{3} y-x y^{3}+2 x^{2} y^{2} \nonumber \]
\(6 x^{3} y-2 x^{2} y^{2}+2 x y^{3}\)
Not all examples will have nice ordering presented in Example \(\PageIndex{2}\), with the powers of one variable descending while the powers of the other variable simultaneously ascends. Sometimes we
have to make some very subjective choices on the ordering of terms.
Example \(\PageIndex{3}\)
Simplify the following polynomial expression, then arrange your answer in some sort of reasonable order. \[a^{3} b^{3}+2 a^{2} b-3 a^{2} b^{3}+4 a^{3} b^{3}+5 a^{4}+3 a^{2} b+b^{5} \nonumber \]
Let’s try to arrange the terms so that the powers of a descend. Again, we use the commutative and associative properties to change the order and regroup.
\[\begin{aligned} a^{3} b^{3}+2 a^{2} b &-3 a^{2} b^{3}+4 a^{3} b^{3}+5 a^{4}+3 a^{2} b+b^{5} \\ &=5 a^{4}+\left(a^{3} b^{3}+4 a^{3} b^{3}\right)+\left(2 a^{2} b+3 a^{2} b\right)-3 a^{2} b^{3}+b^{5}
\\ &=5 a^{4}+5 a^{3} b^{3}+5 a^{2} b-3 a^{2} b^{3}+b^{5} \end{aligned} \nonumber \]
Note that in our final arrangement, the powers of \(a\) descend, but the powers of \(b\) bounce up and down, but at least we have the powers of \(a\) descending. That should help us spot if we’ve
missed a term while simplifying the given problem.
Exercise \(\PageIndex{3}\)
Simplify the following polynomial, and arrange your answer in ascending powers of \(b\): \[5 a^{3} b^{2}+4 a b^{3}-2 a^{2} b+3 a^{3} b^{2}-a b^{3} \nonumber \]
\(-2 a^{2} b+8 a^{3} b^{2}+3 a b^{3}\)
The Degree of a Polynomial
To find the degree of a polynomial, locate the term of the polynomial having the highest degree.
The degree of a polynomial
The degree of a polynomial is the degree of the term having the highest degree.
Finding the degree of a polynomial of a single variable is pretty easy.
Example \(\PageIndex{4}\)
What is the degree of the polynomial \(x^{3}-4 x^{2}+5-6 x+2 x^{7}\)?
First, let’s arrange the polynomial in descending powers of x.
\[2 x^{7}+x^{3}-4 x^{2}-6 x+5 \nonumber \]
Arranging the polynomial in descending powers of \(x\) makes it easier to see that the term of the polynomial with the highest degree is \(2x^7\). Therefore, the degree of the polynomial is \(7\).
Exercise \(\PageIndex{4}\)
What is the degree of the polynomial \(2 x^{3}+8 x^{2}+3 x^{4}+2 x+10\)?
Finding the degree of a polynomial of more than one variable is a little bit trickier.
Example \(\PageIndex{5}\)
What is the degree of the polynomial \(x^{4}-2 x^{3} y^{7}+y^{5}\)?
Note that the polynomial is already arranged in descending powers of \(x\), an arrangement that is probably as good as we are going to get. In the following table, we list the degree of each term.
Remember, the degree of any term is found by summing the exponents on its variables.
\[\begin{array}{cc}{\text { Term }} & {\text { Degree }} \\ \hline x^{4} & {4} \\ {-2 x^{3} y^{7}} & {10} \\ {y^{5}} & {5} \\ \hline\end{array} \nonumber \]
Hence, the term with the highest degree is \(-2 x^{3} y^{7}\), making \(10\) the degree of the polynomial.
Exercise \(\PageIndex{5}\)
What is the degree of the polynomial \(x^{2} y^{4}-6 x^{2} y^{2}+5 x^{2} y^{5}-2 x y\)?
Polynomial Functions
First we define what we mean by a polynomial function.
Polynomial function
A polynomial function is a function defined by a rule that assigns to each domain object a range object defined by a polynomial expression.
Advanced courses, such as multivariate calculus, frequently use polynomial functions of more than one variable such as \(f(x, y)=x^{2}+y^{2}\). However, in this course, our focus will be on
polynomial functions of a single variable, such as \(p(x)=3-4 x-9 x^{2}\) and \(q(x)=x^{3}-9 x^{2}+11\).
Example \(\PageIndex{6}\)
Given the polynomial function \(p(x)=x^{3}-8 x-11\), evaluate \(p(−3)\).
To evaluate \(p(−3)\), first restate the function definition, then replace each occurrence of the variable \(x\) with open parentheses.
\[\begin{aligned} p(x) &= x^{3}-8 x-11 \quad \color {Red} \text { Original function definition. } \\ p(\;\;) &= (\;\;)^{3}-8(\;\;)-11 \quad \color {Red} \text { Replace each occurrence of } x \text {
with open parentheses. } \end{aligned} \nonumber \]
Next, substitute \(−3\) for \(x\) in the open parentheses prepared in the last step.
\[\begin{aligned} p(-3) &= (-3)^{3}-8(-3)-11 \quad \color {Red} \text { Substitute }-3 \text { for } x \text { in the open parentheses positions.} \\ p(-3) &= -27-8(-3)-11 \quad \color {Red} \text {
Exponent first: }(-3)^{3}=-27 \\ p(-3) &= -27+24-11 \quad \color {Red} \text { Multiply: }-8(-3)=24 \\ p(-3) &= -14 \quad \color {Red} \text { Add. } \end{aligned} \nonumber \]
Hence, \(p(−3) = −14\). You can easily check this result on your calculator (see Figure \(\PageIndex{1}\)).
Figure \(\PageIndex{1}\): Calculator check.
Exercise \(\PageIndex{6}\)
Given the polynomial function \(p(x)=-3 x^{2}+7 x+4\), evaluate \(p(2)\).
The Graph of a Polynomial Function
One of the most important polynomial functions in all of mathematics and science is the polynomial having degree two.
Quadratic polynomial
The second degree polynomial having the form \[p(x)=a x^{2}+b x+c \nonumber \] is called a quadratic polynomial. The graph of this polynomial is called a parabola.
The parabola is approximately U-shaped. Some open upwards, some open downwards, depending on the sign of the leading term.
In Figure \(\PageIndex{2}\), the leading term of the parabola \(p(x)=2 x^{2}-8 x+6\) has positive two as its coefficient, so it opens upward.
Figure \(\PageIndex{2}\): The graph of \(p(x)=2 x^{2}-8 x+6\) opens up.
In Figure \(\PageIndex{3}\), the leading term of the parabola \(p(x)=-2 x^{2}-8 x-6\) has negative two as its coefficient, so it opens downward.
Figure \(\PageIndex{3}\): The graph of \(p(x)=-2 x^{2}-8 x-6\) opens down.
The sign of the leading term of \(p(x)=ax^2 +bx+c\) determines whether the parabola opens up or down.
• If \(a>0\), the parabola opens upward.
• If \(a<0\), the parabola opens downward.
The turning point of a parabola has a special name.
The vertex of a parabola
The graph of the second degree polynomial \(p(x)=ax^2+bx+c\) has a single turning point, called the vertex of the parabola.
Example \(\PageIndex{7}\)
Use your graphing calculator to sketch the graph of the quadratic polynomial \(p(x)=−3x^2 + 12x + 25\).
The degree of the polynomial \(p(x)=−3x^2 + 12x + 25\) is two, so it is a quadratic polynomial and its graph is a parabola. Moreover, its leading term has negative three as its coefficient, so we know
that the parabola opens downward. Enter \(y = −3x^2 + 12x + 25\) as \(Y 1=-3 * X \wedge 2+12 * X+25\) in the Y= menu (see the first image in Figure \(\PageIndex{4}\)), then select 6:ZStandard from the
ZOOM menu to produce the third image in Figure \(\PageIndex{4}\).
Figure \(\PageIndex{4}\): Sketching the graph of \(p(x)=−3x^2 + 12x + 25\).
Note that the graph in Figure \(\PageIndex{4}\) appears to have the U-shape of a parabola that opens downwards. Its vertex (turning point) is not visible, but one would surmise that it lies off the
top of the screen. We need to adjust the WINDOW parameters so that the vertex of the parabola is visible in the viewing screen. After some experimentation, we settle on the parameters shown in the
first image in Figure \(\PageIndex{5}\), then push the GRAPH button to produce the second image in Figure \(\PageIndex{5}\).
Figure \(\PageIndex{5}\): Adjust the WINDOW parameters so that the vertex is visible in the viewing screen.
In reporting your result on your homework, follow the Calculator Submission Guidelines from Chapter 3, Section2.
1. Draw axes with a ruler.
2. Label the horizontal axis \(x\) and the vertical axis \(y\).
3. Indicate the WINDOW parameters \(\mathrm{Xmin}, \mathrm{Xmax}, \mathrm{Ymin}\), and \(\mathrm{Ymax}\)\) at the end of each axis.
4. Freehand the curve and label it with its equation.
Exercise \(\PageIndex{7}\)
Use your graphing calculator to sketch the graph of the quadratic polynomial \(p(x)=2x^2 −5x−4\).
When the degree of the polynomial is larger than two, the number of turning points of the graph might increase. This makes for some very interesting curves. In more advanced courses, such as
intermediate and college algebra, you will be introduced to a variety of techniques that will help you determine appropriate viewing windows for the graphs of these higher degree polynomials.
However, in this introductory section, we will assist you by suggesting a good viewing window for each polynomial, one that will allow you to see all of the turning points of the graph of the
Example \(\PageIndex{8}\)
Use your graphing calculator to sketch the graph of the polynomial function \(p(x)=x^{4}-37 x^{2}+24 x+180\). Set your window parameters as follows: \(\mathbf{X} \min =-10, \mathbf{X} \max =10, \
mathbf{X} \operatorname{scl}=1, \mathbf{Y} \min =-1000, \mathbf{Y} \max = 1000,\) and \(\mathbf{Y} \operatorname{scl} =100\).
Enter the polynomial function in \(\mathbf{Y} \mathbf{1}\) of the Y= menu, then enter the suggested window parameters in the WINDOW menu (see Figure \(\PageIndex{6}\)).
Figure \(\PageIndex{6}\): Enter the polynomial and adjust the WINDOW parameters.
Push the GRAPH button on the top row of your calculator to produce the graph of the polynomial function shown in Figure \(\PageIndex{7}\).
Figure \(\PageIndex{7}\): The graph of \(p(x)=x^4 −37x^2 + 24x + 180\).
Sweet-looking curve!
Exercise \(\PageIndex{8}\)
Use your graphing calculator to sketch the graph of the quadratic polynomial p(x)=x3 −14x2 + 20x+ 60. Set your window parameters as follows: \(\mathbf{X} \min =-10, \mathbf{X} \max =20, \mathbf{X} \
operatorname{scl}=1, \mathbf{Y} \min =-200, \mathbf{Y} \max = 200,\) and \(\mathbf{Y} \operatorname{scl} =20\). | {"url":"https://math.libretexts.org/Bookshelves/Algebra/Elementary_Algebra_(Arnold)/05%3A_Polynomial_Functions/5.02%3A_Polynomials","timestamp":"2024-11-14T01:39:35Z","content_type":"text/html","content_length":"156242","record_id":"<urn:uuid:cf594e30-0357-47dc-bc33-92d4fd542edd>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00488.warc.gz"} |
One of the first decisions in a new project is which unit testing framework to use. Traditionally I’ve used CppUnit, so I pulled down the current release and started working.
This left me unhappy as the first test produced this compile-time error:
/usr/local/gcc-20140104/include/cppunit/TestAssert.h:109:6: note: template argument deduction/substitution failed:
cpput_eval.cc:13:5: note: deduced conflicting types for parameter ‘const T’ (‘int’ and ‘std::basic_string::size_type {aka long unsigned int}’)
CPPUNIT_ASSERT_EQUAL(4, str.size());
For a couple days I worked around this by casting the integer literal to a type that satisfied the calls, but eventually I got fed up.
So I looked for alternatives. I found fault with the first two choices, but joy with the third. Herein are some examples with discussion of what they reveal about the choices. The files are available
as a github gist.
The Test Criteria
Three specific assertions were found to cause trouble with various solutions, so the examples used below show all of them:
• Comparing a std::string size() with an integer literal;
• Pointer-equality testing for char * values;
• Comparing a floating point result to a specific absolute accuracy
In addition, these criteria are relevant:
• Verbosity: how much boilerplate do you have to add that isn’t really part of your test?
• Installation overhead: is it easy to build the library for specific compiler flags or is the assumption that you build it once and share it? This matters when playing with advanced language
feature flags such as -std=c++1y, which can affect linking test cases together.
• Assertion levels: when a test fails can you control whether the test keeps going or aborts (e.g., when following assertions would be invalid if the first fails).
• Assertion comparisons: can you express specific relations (not equal, greater than) or is it mostly a true/false capability?
Originally on SourceForge, this project has developed new life at freedesktop.org.
CppUnit comes with a standard configure/make/make install build process which installs the headers and the support library into the proper directories within a toolchain prefix. You need to provide a
main routine to invoke the test driver.
CppUnit provides only one level of assertion: the test case aborts when it fails. It also has limited ability to express specific requirements (for example, there is CPPUNIT_ASSERT_EQUAL(x,y) but no
Here’s what the tests looks like with CppUnit:
#include <cppunit/extensions/HelperMacros.h>
#include <string>
#include <cmath>
class testStringStuff : public CppUnit::TestFixture
void testBasic ()
const char * const cstr{"no\0no\0"};
const std::string str("text");
CPPUNIT_ASSERT_EQUAL(std::size_t{4}, str.size());
CPPUNIT_ASSERT(cstr != (cstr+3));
class testFloatStuff : public CppUnit::TestFixture
void testBasic ()
CPPUNIT_ASSERT_DOUBLES_EQUAL(11.045, std::sqrt(122.0), 0.001);
There’s a lot of overhead, what with the need to define and register the suites, though it didn’t really bother me until I saw what other frameworks require. And I did have to do that irritating
explicit cast to get the size comparison to compile.
The output is terse and all tests pass:
testFloatStuff::testBasic : OK
testStringStuff::testBasic : OK
OK (2)
Boost is a federated collection of highly-coupled but independently maintained C++ libraries covering a wide range of capabilities. It includes Boost.Test, the unit test framework used by boost
developers themselves.
Boost.Test can be used as a header-only solution, but I happened to install it in library form. This gave me a default main routine for invocation, though I did have to have a separate object file
with preprocessor defines which incorporated it into the executable.
Boost.Test also supports three levels of assertion. WARN is a diagnostic only; CHECK marks the test as failing but continues; and REQUIRE marks the test as failing and stops the test. There are also
a wide variety of conditions (EQUAL, NE, GT, …), each of which is supported for each level.
Here’s what the tests look like with Boost.Test:
#include <boost/test/unit_test.hpp>
#include <string>
#include <cmath>
const std::string str("text");
float fa[2];
const char * const cstr{"no\0no\0"};
BOOST_CHECK_EQUAL(4, str.size());
BOOST_CHECK_NE(fa, fa+1);
BOOST_CHECK_NE(cstr, cstr+3);
BOOST_CHECK_CLOSE(11.045, std::sqrt(122), 0.001);
This is much more terse than CppUnit, and seems promising. Here’s what happens when it runs:
Running 2 test cases...
butf_eval.cc(10): error in "StringStuffBasic": check cstr != cstr+3 failed [no == no]
butf_eval.cc(15): error in "FloatStuffBasic": difference{0.0032685%} between 11.045{11.045} and std::sqrt(122){11.045361017187261} exceeds 0.001%
*** 2 failures detected in test suite "Master Test Suite"
Um. Whoops?
Boost.Test silently treats the char* pointers as though they were strings, and does a string comparison instead of a pointer comparison. Which is not what I asked for, and not what BOOST_CHECK_NE
(x,y) will do with other pointer types.
Boost.Test also does not provide a mechanism for absolute difference in floating point comparison. Instead, it provides two relative solutions: BOOST_CHECK_CLOSE(v1,v2,pct) checks that v1 and v2 are
no more than pct percent different (e.g. 10 would be 10% different), while BOOST_CHECK_CLOSE_FRACTION(v1,v2,frac) does the same thing but using fractions of a unit (e.g. 0.1 would be 10% different).
Now, you can argue that there’s value in a relative error calculation. But to have two of them, and not have an absolute error check—that doesn’t work for me.
Boost.Test also has a few other issues. The released version has not been updated for four years, but the development version used internal to the Boost project has many changes, which are expected
to be released at some point in the future. From comments on the boost developers mailing list the documentation is generally agreed to be difficult to use, and has produced a rewritten version
(which, honestly, is what I had to use to try it out).
All in all, I don’t feel comfortable depending on Boost.Test.
Google Test
Google Test is another cross-platform unit test framework, which supports a companion mocking framework to support unit testing of capabilities that are not stand-alone.
The code comes with configure/make/install support, but also provides a single-file interface allowing it to be built easily within the project being tested with the same compiler and options as the
code being tested. You do need a separate main routine, but it’s a two-liner to initialize the tests and run them all.
Google Test supports two levels of assertion: failure of an ASSERT aborts the test, while failure of EXPECT fails the test but continues to check additional conditions. It also provides a wide
variety of conditions.
Here’s what the tests look like with Google Test:
#include <gtest/gtest.h>
#include <string>
#include <cmath>
TEST(StringStuff, Basic)
const std::string str("text");
const char * const cstr{"no\0no\0"};
ASSERT_EQ(4, str.size());
ASSERT_NE(cstr, cstr+3);
TEST(FloatStuff, Basic)
ASSERT_NEAR(11.045, std::sqrt(122.0), 0.001);
Even more terse than Boost.Test, because it doesn’t use something like GTEST_TEST or GTEST_ASSERT_EQ. To avoid conflict with user code I normally expect framework tools to provide their interfaces
within a namespace (literally for C++, or by using a standard identifier prefix where that wouldn’t work). Both CppUnit and Boost.Test do this for their macros, but for unit test code that doesn’t
get incorporated into an application I think it’s ok that this isn’t done.
And here’s what you get when running it:
[==========] Running 2 tests from 2 test cases.
[----------] Global test environment set-up.
[----------] 1 test from StringStuff
[ RUN ] StringStuff.Basic
[ OK ] StringStuff.Basic (0 ms)
[----------] 1 test from StringStuff (0 ms total)
[----------] 1 test from FloatStuff
[ RUN ] FloatStuff.Basic
[ OK ] FloatStuff.Basic (0 ms)
[----------] 1 test from FloatStuff (0 ms total)
[----------] Global test environment tear-down
[==========] 2 tests from 2 test cases ran. (0 ms total)
[ PASSED ] 2 tests.
A little more verbose than I’m accustomed to from CppUnit, but it’s tolerable. The most important bit is the last line tells you the overall success, so you only need to scroll up if something didn’t
Summarizing the individual tests for each criterion, with a bold answer being preferable from my subjective perspective:
Feature CppUnit Boost.Test Google Test
Handles size_t/int compares no yes yes
Handles char* compares yes no yes
Handles absolute float delta yes no yes
Verbosity high low low
Installation toolchain header-only or toolchain project
Assertion Levels one three two
Assertion Conditions few every many
So I’m now happily using Google Test as the unit test framework for new C++ projects.
In fact, I’ve also started to use Google Mock, which turns out to be even more cool and eliminates the biggest limitation on unit testing: what to do if the routine being tested normally needs a
heavy-weight and uncontrollable supporting infrastructure to satisfy its API needs. But I can’t really add anything beyond what you’ll can find on their wiki, so will leave it at that.
C++11 and integer rotate
About two months ago when I was starting to catch up on modern C++, I ran across John Regehr’s discussion of portable C rotate. From the initial code:
uint32_t rotl32a (uint32_t x, uint32_t n)
return (x<<n) | (x>>(32-n));
he evolves the solution to:
uint32_t rotl32c (uint32_t x, uint32_t n)
assert (n<32);
return (x<<n) | (x>>(-n&31));
which generates optimal code on x86 and avoids all undefined behavior. See the original post for full details.
In C++ I’d like to generalize this to any type that supports shift operations. To do this requires understanding exactly where the original version risked undefined behavior, and where the final
version does once it’s been generalized beyond
So here are the gotchas, with reference to the ISO/IEC 14882:2011(E) section and paragraph that discusses them.
• Integral promotion (4.5) is performed on both shift operands (5.8#1)
• Shift operations greater than or equal to the number of bits in the promoted left operand produce undefined behavior (section 5.8#1). Hence the assert in the final version, and the trickery of
, about which more later.
• Shifts on signed types with negative values are undefined (5.8#2,3). Left shifts on signed types with non-negative values are undefined if the shifted value exceeds the maximum representable
value in the unsigned version of the result type (colloquially, if a 1 bit is shifted out of the sign bit).
• Integral promotion is performed on the operand to unary minus, and the result of the operation is different depending on whether the operand is unsigned (5.3.2#1).
• Integral numbers might use a representation other than 2’s complement (3.9.1#7).
After all this is taken into account, one ends up with the following (see complete code in a test harness at this gist):
template <typename T>
rotl (T v, unsigned int b)
static_assert(std::is_integral<T>::value, "rotate of non-integral type");
static_assert(! std::is_signed<T>::value, "rotate of signed type");
constexpr unsigned int num_bits {std::numeric_limits<T>::digits};
static_assert(0 == (num_bits & (num_bits - 1)), "rotate value bit length not power of two");
constexpr unsigned int count_mask {num_bits - 1};
const unsigned int mb {b & count_mask};
using promoted_type = typename std::common_type<int, T>::type;
using unsigned_promoted_type = typename std::make_unsigned<promoted_type>::type;
return ((unsigned_promoted_type{v} << mb)
| (unsigned_promoted_type{v} >> (-mb & count_mask)));
Some commentary:
• Line 5 is a compile-time verification that the type is not a user-defined type, for which some of the other assumptions might not be valid.
• Line 6 protects against rotation of signed values, which are known to risk undefined behavior.
• Line 7 uses a standard-defined trait to find the number of bits in the representation of T.
• Line 8 makes sure we’re not dealing with some weird type where an upcoming mask operation won’t produce the right answer (e.g., the MSPGCC uint20_t type).
• Lines 9 and 10 use a bit mask to reduce the shift value to something for which it’s known the operation is defined; i.e. this function provides defined rotate behavior beyond what is mandated by
C++ for shift.
• Lines 11 and 12 deal with the possibility that the result of integral promotion of the (verified unsigned) type T might produce a signed type for which shift operations could produce undefined
• Lines 13 and 14 implement the rotate now that all the preconditions have been validated.
And, of course, the template when instantiated for uint32_t produces the same optimal code as the original.
In meta-commentary, the addition of static_assert in C++11 is an awesome enhancement, which can be combined with std::enable_if for some neat template metaprogramming techniques that still produce
comprehensible user diagnostics. The traits that provide implementation information on standard types are also a great enhancement for portable code. And the new using type alias capability makes
things more readable than the equivalent typedef approach.
BTW: Somebody might suggest that the second argument be unsigned char b, since it’s reasonable to assume the shift count will be less than 256 for any integral type (though not necessarily for
user-defined types). One reason not to do this is the classic argument that int is the native word size and there’s unlikely to be any benefit in using a smaller type. A second is more subtle and
• Per 4.5#1, a prvalue of type unsigned char can promote to a prvalue of type int if representation preconditions are satisfied.
• Per 5.3.1#8 the negation of an unsigned quantity is computed by subtracting its value from 2^n where n is the number of bits in the promoted operand. The implication is that the negation of a
signed quantity is computed by subtracting its value from zero.
• While the representation of -1 in (for example) 16-bit 2’s complement is 0xFFFF, its representation in 16-bit 1’s complement is 0xFFFE and its representation in 16-bit sign-magnitude is 0x8001.
What this means is -mb&count_mask will not give you the right answer in a non-2’s-complement implementation if mb isn’t at least the same rank (4.13) as int. It also means that -mb does not produce
the same value as 0-mb for all built-in integral types and processing environments.
Interesting stuff, IMO. | {"url":"https://www.pabigot.com/2014/01/","timestamp":"2024-11-14T17:40:45Z","content_type":"text/html","content_length":"47022","record_id":"<urn:uuid:8ce732e5-450a-4f8c-8e63-1e3f8bd0aad5>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00355.warc.gz"} |
The maths explained series: Quantitative risk analysis — Cydea
The maths explained series: Quantitative risk analysis
Tuesday, 22 August, 2023
In Qualitative and quantitative risk analysis, we talked about the difference between qualitative and quantitative risk analysis, and we made the case for the use of quantitative risk analysis in
cybersecurity. In this post we will look to demystify the maths commonly used by other industries to calculate their risk, and show how the same methods can be used to provide a more meaningful view
of your cybersecurity risk.
As mentioned in that earlier post, actuaries, portfolio managers, and other industries working with complex risk profiles use probability distributions, a form of quantitative risk analysis which
provides a view of the probability of different impact values within a range of outcomes.
These distributions use the same information, from the same experts, that would be used in a standard risk matrix, but also utilise tried and tested statistical techniques to provide a more accurate
view of the profile of each risk. That is, it is a model which acknowledges that there is a higher probability that a certain risk will result in a loss of £1,000 or more, than there is of the same
risk resulting in a loss of £100,000 or more.
The model uses two main concepts:
1. A subject matter expert could, with a little guidance and training, provide a reasonable estimate for the financial impact of a cybersecurity event (this is at the very worse, no less accurate
than the current system of them identifying which impact range an event fits into) - within a 90% confidence interval.
2. A cybersecurity event will never result in a negative loss, and so, as is commonly done for stock prices, can be modelled using a lognormal probability distribution.
Ok, what does any of that mean?
Let’s explain some terminology you may encounter when diving in.
What is a confidence interval?
A 90% confidence interval is just another way of saying that the expert is confident that 9 times out of 10 the impact of an event will fit between these 2 values. For more information on this, check
out Jamie’s article on Why is estimating an important skill?
What is a probability distribution?
A probability distribution assigns a probability to each possible outcome, and for measurable outcomes (like cost) this is often shown as a graph. For business risks, it is usually a graph showing
the likelihood of each monetary loss, for values in a set range, should a risk be realised.
To create probability distributions for cybersecurity risks, we utilise three main statistical concepts:
• Normal probability distributions
• Lognormal probability distributions
• Stochastic simulations (in particular Monte Carlo simulations)
Normal probability distributions
As the name suggests, these are the probability distributions which are normally seen, especially in nature. Human height, the birth weight of a puppy and the diameter of a tree are some of the many
events whose values are normally distributed, and therefore the probability that a specific person, puppy or tree will be a specific height, weight or diameter can be modelled using a normal
probability distribution.
These probability distributions are easily recognisable by the bell shape that they have when drawn as a graph. Imagine, for example, that we were to take all of the adult Springer Spaniels in the UK
and weigh them. Adult Springer Spaniels usually weigh between 18 and 23 kilograms. If we were to plot all of their recorded weights, we would get a distribution that looks similar to the graph below.
More dogs would weigh between 20 and 21 kilograms than between 21 and 22, or 19 and 20 kilograms. As you continue to move away from the centre of the graph, there would be fewer and fewer dogs who
were measured to be that weight. So the probability of an adult Springer Spaniel being 18kg is lower than the probability of them being 20kg.
The symmetrical nature of this distribution type means that if we have just two values of equal probability, we are able to calculate both the mean (the most probable value), and the standard
deviation (a standardised measure of how spread out the possible outcomes are). These two values are all that is needed to accurately replicate any normal distribution.
Lognormal probability distributions
A lognormal probability distribution is not symmetrical, instead it has a long tail to the right which allows for the possibility of extreme events, and cannot have a negative value. Consider if we
were to record the amount of prize money paid to gambling winners in the UK. The majority of prizes paid are relatively low (matching 2 or 3 numbers on the lottery or winning small amounts on scratch
cards), however there are a very small number who win prizes of millions of pounds.
This model is also well suited for cybersecurity risk as the realisation of a cyber risk is never going to result in a negative loss, and there is always a small possibility of an extreme loss
A particularly useful feature of a lognormal distribution is that if you take the natural logarithm of the predicted values, and plot the distribution of these, you will get a normal distribution.
The relationship between a lognormal and normal distribution is bidirectional, which means we can use the characteristics of a normal distribution to calculate the location parameter (peak) and the
scale parameter (spread of the distribution), by calculating the mean and standard deviation of the associated normal distribution.
So as with the normal distribution, only two values are needed to accurately reproduce the probability distribution. That 90% confidence interval, defined by the same SME who would decide which
impact bucket a risk belongs in for qualitative risk analysis, is all the information needed to produce a probability distribution which will give a more nuanced view of risk.
Stochastic simulations
Stochastic simulations are simulated events, which incorporate randomness. This kind of modelling is used for many purposes including financial analysis, weather forecasting and biochemistry. One of
the most well known stochastic models is the Monte Carlo simulation, a computerised simulation which utilises a random variable to simulate the probability of real-world events, while running the
same scenario multiple times. Given the same initial conditions and parameters, different outcomes are produced on each run. It has been proven many times that this model, given a suitably high
number of runs, is consistent with real world and expected outcomes.
How can these provide a view of cyber risk?
For cybersecurity risk modelling, we utilise a two-layered Monte Carlo simulation. First there is the probability that the event will occur at all, then there is the calculation of the event’s
impact, based on the probability distribution described earlier. See the graph below for a risk that has a 10% chance of occurring.
This means that the probabilities calculated for the graph only apply 10% of the time (so multiply them by 0.1) and the rest of the time the monetary loss (as a result of this specific risk) is zero.
It is relatively easy in Excel or Google sheets to then run a Monte Carlo simulation using these probabilities. The formulas used follow these steps:
• Pick a number between 0 and 1
• If this number is greater than the probability that the risk occurs (in this case, greater than 0.1), then the simulation marks that as the risk not occurring and records a loss of zero.
• If the probability is less than, or equal to the probability of the risk occurring, then we tell excel to calculate another random number between 0 and 1.
• Using that random number as the probability, and the location and scale parameters calculated earlier, excel returns the monetary value for that event occurring.
• This is repeated multiple times (thousands of times) and a graph of the results (counting how many are above each value) is produced to form a complete probability distribution for the risk.
This graph looks a little more like this:
The use cases and further usefulness of this method was described in the earlier blog post: Qualitative and quantitative risk analysis.
Join Ray as he talks about this topic at his webinar on 7th September 2023, 12:30-1:30pm on Vimeo Live.
Photo by Scott Graham on Unsplash | {"url":"https://cydea.com/blog/maths-explained-quantitative-risk-analysis/","timestamp":"2024-11-02T14:06:38Z","content_type":"text/html","content_length":"32411","record_id":"<urn:uuid:fd40d4c0-b1d1-4fa4-baa7-358c774c760b>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00551.warc.gz"} |
Law of Cosines: Solving for an Angle
Solve for an Angle with the Cosine Law #1
Solve for an Angle with the Cosine Law #2
Solve for an Angle with the Cosine Law #3
Solve for an Angle with the Cosine Law: Word Problem #1
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The Stacks project
Definition 15.28.1. Let $R$ be a ring. Let $\varphi : E \to R$ be an $R$-module map. The Koszul complex $K_\bullet (\varphi )$ associated to $\varphi $ is the commutative differential graded algebra
defined as follows:
1. the underlying graded algebra is the exterior algebra $K_\bullet (\varphi ) = \wedge (E)$,
2. the differential $d : K_\bullet (\varphi ) \to K_\bullet (\varphi )$ is the unique derivation such that $d(e) = \varphi (e)$ for all $e \in E = K_1(\varphi )$.
Comments (2)
Comment #8389 by Peng Du on
In line above tag/0628, better to denote the map by p: C(f)⟶A[1] (rather than [-1]). This might depend on your convention, but using [1] is consistent with all texts, both homotopically (e.g.
suspension in homotopy theory become [1] in triangulated categories) and homologically/cohomologically. The moral is, [1] always shifts a chain/cochain 1 step to the left (provided you draw all
arrows rightward), i.e. to the opposite of the direction of a chain/cochain.
Comment #8999 by Stacks project on
OK, I am not going to change the convention of shifting for chain complexes. However, for cochain complexes what you say is the convention we use. Almost all of the complexes in the Stacks
project are cochain complexes.
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such that no two of them are cons of r points out of n point Do... | Filo
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such that no two of them are cons of points out of point Do yourself-3 : 1. Find the number of ways of selecting 5 members from a committee of 5 men \& 2 women such that all women are always
included. 2. Out of first 20 natural numbers, 3 numbers are selected such that there is exartly one even number. How many different selections can be made ? 3. How many four letter words can be made
from the letters of the word 'PROBLEM'. How many of these start as well as end with a vowel ? 4. The number of ways in which 5 different books can be distributed among 10 people if each person can
get at most one book is :
a. 252
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such that no two of them are cons of points out of point Do yourself-3 : 1. Find the number of ways of selecting 5 members from a committee of 5 men \& 2 women such that all women are always
Question included. 2. Out of first 20 natural numbers, 3 numbers are selected such that there is exartly one even number. How many different selections can be made ? 3. How many four letter words can
Text be made from the letters of the word 'PROBLEM'. How many of these start as well as end with a vowel ? 4. The number of ways in which 5 different books can be distributed among 10 people if
each person can get at most one book is :
Updated Jan 14, 2023
Topic Coordinate Geometry
Subject Mathematics
Class Class 12
Answer Video solution: 1
Upvotes 59
Video 8 min | {"url":"https://askfilo.com/user-question-answers-mathematics/such-that-no-two-of-them-are-cons-of-points-out-of-point-do-33383038393439","timestamp":"2024-11-07T04:01:45Z","content_type":"text/html","content_length":"244791","record_id":"<urn:uuid:3922e7c9-57c4-401b-a2f6-09f7e91fe7c0>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00476.warc.gz"} |
What is the measure of DG?In circle D, mGEC is 230 degreesWhat is the measure of GDC?what is the measure of AC?Segment CO is congruent to segment HZwhich congruence statement is true?A. OZ is congruent to COB. CH is congruent to COZC. CH is congruent to HZOD. CO is congruent to HZin circle K, what is the value of x?A. x=30B. x=25C. x=20D. x=15
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How Many Feet In A 1 2 Mile - Life Answers HQ
Are you wondering how many feet are there in a 1/2 mile? Whether you’re a runner, a hiker, or simply curious, knowing the answer to this question can come in handy in many situations.
If you’re short on time, here’s a quick answer to your question: There are 2,640 feet in a 1/2 mile. However, if you want to learn more about this topic and understand the history behind the
measurement units, keep reading!
In this article, we’ll cover everything you need to know about feet, miles, and how they relate to each other. Here’s a brief overview of what we’ll include:
Why Do We Use Feet and Miles as Units of Measurement?
Measurement is a fundamental concept in science and human life. It helps us to quantify and compare physical quantities and objects. Over time, humans have developed various units of measurement that
differ from culture to culture and time to time. In the United States, the two most commonly used units of measurement for distance are feet and miles.
A Brief History of Measurement Units
The earliest known units of measurement were based on body parts such as the foot, hand, and cubit. The ancient Egyptians used cubits, which were the length of a forearm, to measure distances. The
Romans introduced a system of units based on the pace, which was the distance covered by a single step. Over time, measurement units became more standardized, and the metric system was introduced in
France in 1795.
Despite the introduction of the metric system, the United States has continued to use the customary system, which includes feet and miles. One reason for this is that the United States has a strong
connection to its British roots, and the British used the customary system until the 1960s. Additionally, the United States has a large infrastructure based on the customary system, and changing to
the metric system would be costly.
Advantages and Disadvantages of Using Feet and Miles
One advantage of using feet and miles is that they are familiar to most Americans. People have an intuitive understanding of how long a mile is and how tall six feet is. Additionally, the customary
system allows for easy conversions between units. For example, there are 5,280 feet in a mile, and 3 feet make up a yard.
However, the customary system also has some disadvantages. One drawback is that it is not as precise as the metric system. For example, a mile is approximately 1.609 kilometers, which is a more
precise measurement. Additionally, because the United States is one of the few countries that uses the customary system, it can cause confusion when communicating with people from other countries.
How to Convert Between Feet and Miles
Converting between feet and miles is a straightforward process. There are 5,280 feet in a mile, so to convert feet to miles, you divide the number of feet by 5280. For example, if you have 10,560
feet, you would divide that by 5280 to get 2 miles. To convert miles to feet, you multiply the number of miles by 5280. For example, if you have 3 miles, you would multiply that by 5280 to get 15,840
It’s important to note that there are other units of measurement for distance, such as kilometers and meters. If you need to convert between these units and feet or miles, you can use conversion
charts or online calculators to make the process easier.
Sources: National Institute of Standards and Technology
How Many Feet are in a Mile?
The Imperial System of Measurement, also known as the British Imperial System, is a system of units that was used in the United Kingdom and its colonies. Today, the Imperial System is still used in
some countries, including the United States, but it has been largely replaced by the metric system.
In the Imperial System, there are 5,280 feet in a mile. This may seem like a large number, but it’s important to remember that the Imperial System is based on the length of the human body. In fact,
the foot was originally defined as the length of the average human foot!
Converting Miles to Feet: The Formula
To convert miles to feet, you can use the following formula:
feet = miles x 5,280
For example, if you want to know how many feet are in 2 miles, you would use the formula like this:
feet = 2 x 5,280
feet = 10,560
So there are 10,560 feet in 2 miles.
Examples of Mile to Feet Conversions
Here are some examples of mile to feet conversions:
• 1/2 mile = 2,640 feet
• 1 mile = 5,280 feet
• 2 miles = 10,560 feet
• 5 miles = 26,400 feet
• 10 miles = 52,800 feet
Remember, to convert miles to feet, simply multiply the number of miles by 5,280. You can use this formula to convert any distance in miles to feet.
For more information on the Imperial System of Measurement, visit bipm.org.
How Many Feet are in a 1/2 Mile?
If you’re wondering how many feet are in a 1/2 mile, you’re not alone. Many people need to convert between different units of measurement for a variety of reasons. Fortunately, converting 1/2 mile to
feet is a straightforward process that you can easily do on your own.
Converting 1/2 Mile to Feet
One mile is equal to 5,280 feet. Therefore, half a mile is equal to 2,640 feet. This means that there are 2,640 feet in a 1/2 mile. To convert from miles to feet, you simply need to multiply the
number of miles by 5,280. Conversely, to convert from feet to miles, you would divide the number of feet by 5,280.
Examples of 1/2 Mile to Feet Conversions
If you’re still having trouble visualizing how many feet are in a 1/2 mile, consider the following examples:
• If you walk for half a mile, you will have taken approximately 1,320 steps.
• A football field is 100 yards long, or 300 feet. Therefore, two football fields placed end to end would be approximately 600 feet, or just under a quarter of a mile. To walk a half mile would be
like walking past four football fields.
Understanding how to convert between different units of measurement can be useful in many different situations. Whether you’re trying to plan a walking route, measure the distance between two
locations, or complete a math problem, knowing how many feet are in a 1/2 mile can be a valuable piece of information.
For more information on converting between units of measurement, check out Math is Fun or Khan Academy.
Other Units of Length and Distance
Aside from feet and miles, there are other units of length and distance that are used in different contexts. Here are some of them:
• Metric System: Meters and Kilometers – The metric system is the standard system of measurement in most countries and is used in scientific and technical fields. Meters (m) and kilometers (km) are
the most commonly used units of length and distance in the metric system. One kilometer is equivalent to 1,000 meters.
• Nautical Miles – Nautical miles (nm) are often used in maritime and aviation industries to measure distance. One nautical mile is equivalent to 1.15 statute miles or 1.85 kilometers.
Here’s a table that compares different units of length and distance:
Unit Equivalent to…
1 inch 2.54 centimeters
1 foot 12 inches or 0.3048 meters
1 yard 3 feet or 0.9144 meters
1 mile (statute) 5,280 feet or 1.609 kilometers
1 kilometer 0.6214 miles or 3,281 feet
1 nautical mile 1.15 statute miles or 1.852 kilometers
It’s important to note that different units of length and distance are used in different contexts, and it’s essential to use the appropriate unit for the task at hand. Knowing how to convert between
units can also be useful in some situations.
In conclusion, knowing how many feet are in a 1/2 mile can be useful in many situations, from planning your running route to estimating the length of a hiking trail.
In this article, we’ve covered the history of measurement units, the advantages and disadvantages of using feet and miles, how to convert between these units, and other units of length and distance.
We hope this guide has been informative and helpful. If you have any questions or comments, feel free to leave them below! | {"url":"https://www.lifeanswershq.com/how-many-feet-in-a-1-2-mile/","timestamp":"2024-11-08T22:16:54Z","content_type":"text/html","content_length":"109652","record_id":"<urn:uuid:d3cb9a73-19ed-42a6-9f69-db61237829e8>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00122.warc.gz"} |
Subsample to Sample Composers#
class IdentityComposer[source]#
Copy subsamples to samples verbatim.
class SplatComposer[source]#
A composer that overwrites current samples with subproblem samples.
See Examples.
Sample Processors#
class GreedyPathMerge[source]#
Dialectic-search merge operation [KS]. Generates a path from one input state, representing the thesis, to another input state, representing the antithesis, using a greedy method of single bit
flips selected by decreasing energy.
Returns the best sample on the path, which represents the synthesis.
Note: only the lowest-energy sample, is considered from either input state.
See Examples.
Kadioglu S., Sellmann M. (2009) Dialectic Search. In: Gent I.P. (eds) Principles and Practice of Constraint Programming - CP 2009. CP 2009. Lecture Notes in Computer Science, vol 5732. Springer,
Berlin, Heidelberg
class IsoenergeticClusterMove(seed=None, **runopts)[source]#
Isoenergetic cluster move (ICM), also know as Houdayer move.
ICM creates two new samples from a pair of samples by identifying, among connected variables, clusters with exactly complementary values and swapping one such randomly chosen cluster between the
two samples. The total energy of the two samples remains unchanged, yet such moves on variables reasonably grouped together can enable better exploration of the solution space.
seed (int, optional, default=None/current time) – Pseudo-random number generator seed.
Two states with at least one sample each. First state should also contain a relevant problem.
Two states from input with updated first sample in each.
Primitive Sample Operations#
This example runs one iteration of a SplatComposer composer, overwriting an initial solution to a 6-variable binary quadratic model of all zeros with a solution to a 3-variable subproblem that was
manually set to all ones.
import dimod
from hybrid.composers import SplatComposer
from hybrid.core import State, SampleSet
from hybrid.utils import min_sample
bqm = dimod.BinaryQuadraticModel({t: 0 for t in range(6)},
{(t, (t+1) % 6): 1 for t in range(6)},
0, 'BINARY')
composer = SplatComposer()
state0 = State.from_sample(min_sample(bqm), bqm)
state1 = state0.updated(subsamples=SampleSet.from_samples({3: 1, 4: 1, 5: 1}, 'BINARY', 0.0))
composed_state = composer.run(state1).result()
>>> print(composed_state.samples)
Response(rec.array([([0, 0, 0, 1, 1, 1], 1, 2)],
dtype=[('sample', 'i1', (6,)), ('num_occurrences', '<i8'), ('energy', '<i8')]), [0, 1, 2, 3, 4, 5], {}, 'BINARY')
This example runs one iteration of a GreedyPathMerge composer on a thesis and antithesis State to find a ground state of a square graph. By inverting the state of variable \(d\) and \(c\) in
samples_d and then variable \(a\) of the lowest energy sample of samples_a (second sample), the composer finds a path between these two samples that contains the ground state shown on the right of
the top figure.
import dimod
bqm = dimod.BinaryQuadraticModel({}, {'ab': 1.0, 'bc': 1.0, 'cd': 1.0, 'da': 1}, 0, 'SPIN')
samples_d = {'a': 1, 'b': 1, 'c': -1, 'd': -1}
samples_a = [{'a': -1, 'b': -1, 'c': 1, 'd': 1}, {'a': -1, 'b': 1, 'c': 1, 'd': 1}]
states = [hybrid.State.from_samples(samples_d, bqm),
hybrid.State.from_samples(samples_a, bqm)]
synthesis = GreedyPathMerge().next(states)
>>> print(synthesis.samples)
a b c d energy num_occ.
0 +1 +1 +1 +1 -4.0 1
[ 1 rows, 4 variables ] | {"url":"https://docs.ocean.dwavesys.com/en/latest/docs_hybrid/reference/composers.html","timestamp":"2024-11-15T04:32:37Z","content_type":"text/html","content_length":"56037","record_id":"<urn:uuid:712f4098-669f-47da-8caa-12320d7e83ba>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00796.warc.gz"} |
Flordia College Academy
Tampa, Florida
General Information:
• Math Challenge is free and no-sign up is necessary.
• All students are invited to participate.
• There will be 15 challenges throughout the school year.
• Refer to the Schedule Calendar for dates on when the challenges and their solutions will be available on this webpage.
How to Participate
1. A printed copy of the Math Challenge will be sent home in the communication folder. A copy can be printed from the site if an extra copy is needed.
2. Students will have about two weeks to complete the current Math Challenge and papers must be submitted on/before the due date. No late submissions will be accepted.
3. Math Challenge papers can be turned into the student’s homeroom teacher.
4. Students submitting the Math Challenge paper will be entered into a drawing to be done on the Friday after papers are due for prizes. Name and grade level must be on the paper in order to be
entered into the drawing.
5. Students who complete 12 out of the 15 Math Challenges will be eligible to participate in a schoolwide celebration in May!
6. Students/parents must complete the required number of problems on each Math Challenge for his/her grade level. (See paper for how many problems each grade level must complete.)
7. Parents and family members may and are encouraged to assist students with the problems in the Math Challenge. If help with strategies is needed you may visit mathinaction.org for tips.
8. Look for the link to the Math Challenge on the Falcon Flyer. Happy solving! | {"url":"https://www.mathinaction.org/fl-college-academy.html","timestamp":"2024-11-05T09:24:41Z","content_type":"text/html","content_length":"31500","record_id":"<urn:uuid:ed90ed01-1b70-495a-a842-a2b01174908f>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00831.warc.gz"} |
Quantum Computing is the study of a non-classical model of computation. Whereas traditional models of computing such as the Turing machine or Lambda calculus rely on “classical” representations
of computational memory, a quantum computation could transform the memory into a quantum superposition of possible classical states.^
A quantum computer is a device that could perform such complex computation.^
Quantum and classical computers both try to solve problems, but the way they manipulate data to get answers is fundamentally different. What makes quantum computers unique is the introduction of two
principles of quantum mechanics crucial for its operation – superposition and entanglement.
Superposition is the counterintuitive ability of a quantum object, like an electron, to simultaneously exist in multiple “states.” With an electron, one of these states may be the lowest energy level
in an atom while another may be the first excited level. If an electron is prepared in a superposition of these two states it has some probability of being in the lower state and some probability of
being in the upper. A measurement will destroy this superposition, and only then can it be said that it is in the lower or upper state.
Understanding superposition makes it possible to understand the basic component of information in quantum computing, the qubit. In classical computing, bits are transistors that can be off or on,
corresponding to the states 0 and 1. In qubits such as electrons, 0 and 1 simply correspond to states like the lower and upper energy levels discussed above. Qubits are distinguished from classical
bits, which must always be in the 0 or 1 state, by their ability to be in superpositions with varying probabilities that can be manipulated by quantum operations during computations.
Entanglement is a phenomenon in which quantum entities are created and/or manipulated such that none of them can be described without referencing the others. Individual identities are lost. This
concept is exceedingly difficult to conceptualize when one considers how entanglement can persist over long distances. A measurement on one member of an entangled pair will immediately determine
measurements on its partner, making it appear as if information can travel faster than the speed of light. This apparent action at a distance was so disturbing that even Einstein dubbed it “spooky”.
Quantum computers obtain their speedup by trying every possible answer to a problem in parallel. In reality a quantum computer leverages entanglement between qubits and the probabilities associated
with superpositions to carry out a series of operations (a quantum algorithm) such that certain probabilities are enhanced (i.e., those of the right answers) and others depressed, even to zero (i.e.,
those of the wrong answers). When a measurement is made at the end of a computation, the probability of measuring the correct answer should be maximized. The way quantum computers leverage
probabilities and entanglement is what makes them so different from classical computers.
WHY DO WE WANT IT?
The promise of developing a quantum computer sophisticated enough to execute Shor’s algorithm for large numbers has been a primary motivator for advancing the field of quantum computation. To develop
a broader view of quantum computers, however, it is important to understand that they will likely deliver tremendous speed-ups for only specific types of problems. Researchers are working to both
understand which problems are suited for quantum speed-ups and develop algorithms to demonstrate them. In general, it is believed that quantum computers will help immensely with problems related to
optimization, which play key roles in everything from defense to financial trading.
Multiple additional applications for qubit systems that are not related to computing or simulation also exist and are active areas of research, but they are beyond the scope of this overview. Two of
the most prominent areas are (1) quantum sensing and metrology, which leverage the extreme sensitivity of qubits to the environment to realize sensing beyond the classical shot noise limit, and (2)
quantum networks and communications, which may lead to revolutionary ways to share information.
Understanding Qubit
A quantum bit, or qubit, is the basic unit of information for a quantum computer, analogous to a bit in ordinary machines. But unlike a bit, which can have the value 0 or 1, a qubit can take on an
infinite number of values. Physicists call these the states of the qubit.
It turns out that specifying a qubit’s state is a lot like specifying your position on Earth using latitude and longitude. The yellow arrow in the figure below points to a spot at a particular
latitude and longitude. Notice how the arrow moves when you change the sliders.
Just as two numbers can pinpoint a spot on a globe, there are two numbers that determine a qubit state. In general, that state will be a combination of two special quantum states, which are called 0
and 1 to match the naming convention for ordinary bits. These special states are like the north and south pole on a globe.
Although there are an infinite number of possible qubit values, observing a qubit’s state—by making a quantum measurement—yields either 0 or 1. The result of a given measurement is probabilistic and
depends on the details of this combination. In particular, the chance of observing the qubit in one of the special states depends on its distance from either pole. (The opacity of the colored discs
on the right side of the animation below represents this chance, with more opaque colors indicating a higher probability for 0 or 1.) Qubits with values on the equator of the sphere are equally
likely to be 0 or 1 when measured, but they are all subtly different. As the arrow traces traces a path along the equator, different colors represent phase differences—numbers responsible for
interference effects when two qubits interact.
Different physical systems can store a qubit, such as the polarization of a photon, the spin of an electron or the amount of magnetic field that passes through the middle of a loop of superconductor.
Researchers at JQI freqently use atoms to store qubits. Creating a real qubit in a lab requires detailed knowledge of these platforms, but the abstraction of a qubit allows scientists to treat them
on equal an equal footing when studying their properties.
Finally, just as bits undergo digital logic gates like AND, OR and NOT, qubits, too, have gates. Visually, these are rotations of a qubit’s state to a new position on the surface of the sphere.
Quantum computations consist of preparing many qubits in a certain state, rotating them and making measurements.
BUILDING A QUANTUM COMPUTER?
Building quantum computers is incredibly difficult. Many candidate qubit systems exist on the scale of single atoms, and the physicists, engineers, and materials scientists who are trying to execute
quantum operations on these systems constantly deal with two competing requirements. First, qubits need to be protected from the environment because it can destroy the delicate quantum states needed
for computation. The longer a qubit survives in its desired state the longer its “coherence time.” From this perspective, isolation is prized. Second, however, for algorithm execution qubits need to
be entangled, shuffled around physical architectures, and controllable on demand. The better these operations can be carried out the higher their “fidelity.” Balancing the required isolation and
interaction is difficult, but after decades of research a few systems are emerging as top candidates for large-scale quantum information processing.
Superconducting systems, trapped atomic ions, and semiconductors are some of the leading platforms for building a quantum computer. Each has advantages and disadvantages related to coherence,
fidelity, and ultimate scalability to large systems. It is clear, however, that all of these platforms will need some type of error correction protocols to be robust enough to carry out meaningful
calculations, and how to design and implement these protocols is itself a large area of research. For an overview of quantum computing, with more detail regarding experimental implementations.
In this article, “quantum computing” has so far been used as a blanket term describing all computations that utilize quantum phenomena. There are actually multiple types of operational frameworks.
Logical, gate-based quantum computing is probably the best recognized. In it, qubits are prepared in initial states and then subject to a series of “gate operations,” like current or laser pulses
depending on qubit type. Through these gates the qubits are put in superpositions, entangled, and subjected to logic operations like the AND, OR, and NOT gates of traditional computation. The qubits
are then measured and a result obtained.
Another framework is measurement-based computation, in which highly entangled qubits serve as the starting point. Then, instead of performing manipulation operations on qubits, single qubit
measurements are performed, leaving the targeted single qubit in a definitive state. Based on the result, further measurements are carried out on other qubits and eventually an answer is reached.
A third framework is topological computation, in which qubits and operations are based on quasiparticles and their braiding operations. While nascent implementations of the components of topological
quantum computers have yet to be demonstrated, the approach is attractive because these systems are theoretically protected against noise, which destroys the coherence of other qubits.
Finally, there are the analog quantum computers or quantum simulators envisioned by Feynman. Quantum simulators can be thought of as special purpose quantum computers that can be programmed to model
quantum systems. With this ability they can target questions such as how high-temperature superconductors work, or how certain chemicals react, or how to design materials with certain properties.
Key Components of a Quantum Computer
In order to work with qubits for extended periods of time, they must be kept very cold. Any heat in the system can introduce error, which is why quantum computers are designed to create and operate
at temperatures near absolute zero.
Here’s a look at how a quantum computer’s dilution refrigerator, made from more than 2,000 components, exploits the mixing properties of two helium isotopes to create such an environment for the
qubits inside.
1. Qubit Signal Amplifier
One of two amplifying stages is cooled to a temperature of 4 Kelvin.
2. Input Microwave Lines
Attenuation is applied at each stage in the refrigerator in order to protect qubits from thermal noise during the process of sending control and readout signals to the processor.
3. Superconducting Coaxial Lines
In order to minimize energy loss, the coaxial lines that direct signals between the first and second amplifying stages are made out of superconductors.
4 Cryogenic Isolators
Cryogenic isolators enable qubits signals to go forward while preventing noise from compromising qubit quality.
5 Quantum Amplifiers
Quantum amplifiers inside of a magnetic shield capture and amplify processor readout signals while minimizing noise.
6 Cryoperm Shield
The quantum processor sits inside a shield that protects it from electromagnetic radiation in order to preserve its quality.
7 Mixing Chamber
The mixing chamber at the lowest part of the refrigerator provides the necessary cooling power to bring the processor and associated components down to a temperature of 15 mK — colder than outer
Future Trends in Quantum Computing
Quantum Computers are not destined to replace the processors in personal computers or smartphones anytime soon.Total enterprise Quantum Computing market revenue is expected to reach $9.1 billion
annually by 2030, up from $111.6 million in 2018.
For the most part, quantum computers will be best suited to addressing optimization problems, identifying patterns in data, and conducting complex simulations that would be too taxing for
traditional, or classical, computers. These issues will drive the global market for enterprise QC. But quantum computers have not yet demonstrated quantum supremacy or quantum advantage.
Significantly scaling the processing power, improving error correction abilities, and writing and refining quantum algorithms will be required before enterprises adopt QC en masse. Still, the QC
market is expected to grow strongly through 2030.
Real world applications of quantum computers will have a visible impact on the world and how companies and people engage with it.
Logistical and optimization problems:
These simple questions involve number crunching hundreds to thousands of variables at once, a feat that modern supercomputers just can't handle; so instead, they compute a small percentage of those
variables to help these companies manage their logistical needs in a less than optimal way. But with a quantum computer, it will slice through a mountain of variables without breaking a sweat.
Similar to the point above, the reason why the weather channel sometimes gets it wrong is because there are too many environmental variables for their supercomputers to process (that and sometimes
poor weather data collection). But with a quantum computer, weather scientists can not only forecast near-term weather patterns perfectly, but they can also create more accurate long-term climate
assessments to predict the effects of climate change.
Personalized Medicine:
Quantum computers will also allow Big Pharma to better predict how different molecules react with their drugs, thereby significantly speeding up pharmaceutical development and lowering prices.
Space exploration:
The space telescopes of today (and tomorrow) collect enormous amounts of astrological imagery data each day that tracks the movements of trillions of galaxies, stars, planets, and asteroids. Sadly,
this is far too much data for today's supercomputers to sift through to make meaningful discoveries on a regular basis. But with a mature quantum computer combined with machine-learning, all this
data can finally be processed efficiently, opening the door to the discovery of hundreds to thousands of new planets daily by the early-2030s.
Fundamental Sciences:
Similar to the points above, the raw computing power these quantum computers enable will allow scientists and engineers to devise new chemicals and materials, as well as better functioning engines
and of course, cooler Christmas toys.
This application is also a topic of excitement among researchers in the artificial intelligence (AI) field, as this improved natural learning capacity could accelerate progress in AI research by
decades. More on this in our Future of Artificial Intelligence series.
Data Encryption:
Banking, communication, national security services, the internet itself depends on reliable encryption to function. (Oh, and forget about the bitcoin as well, given its core dependence on
encryption.) If these quantum computers work as advertised, all of these industries will be at risk, at worst endangering the entire world economy until we build quantum encryption to keep pace.
Real-Time Language Translation:
In 20 years, language will no longer be a barrier to business and everyday interactions. For example, a person who only speaks English can more confidently enter into business relationships with
partners in foreign countries where English brands would have otherwise failed to penetrate, and when visiting said foreign countries, this person may even fall in love with a certain somebody who
only happens to speak Cantonese.
Sources / References:
Comments (1 Comment)
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Which of the following is an efferent neuron that is responsible for releasing a neurotransmiter that stimulates a muscle cell to contract?
Correct Answer : A
An efferent neuron that is responsible for releasing a neurotransmitter that stimulates a muscle cell to contract is a motor neuron ². Motor neurons carry signals from the brain to the peripheral
nervous system in order to initiate an action. The neurotransmitter acetylcholine (ACh) is released by motor neurons at the neuromuscular junction in skeletal muscle, causing the muscle to contract
The other options are incorrect because they do not accurately describe the type of neuron responsible for releasing a neurotransmitter that stimulates a muscle cell to contract. Interneurons are
found within the central nervous system and facilitate communication between sensory and motor neurons. Sensory neurons carry information from sensory receptors to the central nervous system.
Neuroglia are support cells for neurons and do not transmit nerve impulses.
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Related Questions
Correct Answer is C
The correct answer is c. Data are biased by the methodology. The researcher's data should be rejected because they are biased by the methodology used to gather them. By only using a small net, the
researcher excluded all large birds from the study. This means that the data do not accurately represent the average wing strength of all birds found in the American Northwest.
A. The data contradicting the control group is not a reason to reject the data in this case.
B. The data being different than expected is not a reason to reject the data in this case.
D. The data not being able to be displayed graphically is not a reason to reject the data in this case.
Correct Answer is C
The correct answer is c. Lacteal vessels. Lipids absorbed in the small intestine will first enter lacteal vessels, which are small lymphatic vessels located in the villi of the small intestine. These
vessels transport the absorbed lipids to the lymphatic system, where they eventually enter the bloodstream.
a. Veins and b. Arteries are blood vessels that transport blood throughout the body. Lipids absorbed in the small intestine do not directly enter these vessels.
d. Interstitial spaces are spaces between cells and tissues that contain interstitial fluid. Lipids absorbed in the small intestine do not directly enter these spaces.
Correct Answer is D
The tibia and fibula are located in the crural region of the body, which is the lower leg between the knee and ankle. The coxal region refers to the hip area, the antecubital region is the front of
the elbow, and the tarsal region is the ankle and foot.
Correct Answer is D
Most of the carbon dioxide from the blood moves into the alveoli by diffusion down a concentration gradient ¹. Carbon dioxide is always carried in the blood and is released into alveolar air during
expiration ¹. Respiratory gases move from higher concentration to lower concentration ¹. In alveolar air, when carbon dioxide is less than in blood, carbon dioxide is released ¹.
The other options are incorrect because they do not accurately describe the process by which most of the carbon dioxide from the blood moves into the alveoli. Passive transport using carrier
proteins, active transport using energy, and conversion to carbon monoxide is not the processes responsible for moving most of the carbon dioxide from the blood into the alveoli.
Correct Answer is A
The sequence of bases on the complementary strand of DNA would read5’AGCTAGCGT 3’ (Choice A). In DNA, the nitrogenous bases adenine (A) and thymine (T) pair together, and cytosine (C) and guanine (G)
pair together. The complementary strand is also antiparallel to the original strand, meaning that it runs in the opposite direction with the 5' end matching up with the 3' end of the original strand.
The other options do not accurately represent the complementary sequence of bases or the antiparallel orientation of the strands.
Correct Answer is D
Enzymes are a type of protein that catalyze chemical reactions in the body. Proteins are one of the four main classes of biological molecules, along with lipids, carbohydrates, and nucleic acids.
The other options are not classes of biological molecules that include enzymes.Lipids are a class of molecules that includes fats and oils, vitamins are organic compounds that are essential for
normal growth and nutrition, and carbohydrates are a class of molecules that includes sugars and starches.
Correct Answer is C
The correct answer is c. 100,000. The pH scale is a logarithmic scale, which means that each change of one pH unit represents a tenfold change in the hydrogen-ion concentration. A pH 4 solution has a
hydrogen-ion concentration that is 10^5 (or 100,000) times greater than that of a pH 9 solution.
a. 0.00001 is the hydrogen-ion concentration of a pH 9 solution as compared with a pH 4 solution.
b. 5 is the difference in pH units between a pH 4 solution and a pH 9 solution.
d. 50 is not the correct answer.
Correct Answer is C
The best reason for the prolonged preservation of the body is that it was frozen in the cold temperature of the Alps shortly after death and remained frozen until it was found. Freezing can preserve
a body by slowing down or stopping the decomposition process.
The other options are not as likely to have caused prolonged preservation.
Ultraviolet rays can damage molecules rather than preserve them. Toxins in food would not necessarily kill all bacteria that could cause decomposition. Blood loss from an arrow wound would not
necessarily clear all enzymes that could break down tissue.
Correct Answer is C
The correct answer is c. germ. If a mother's germ cell contains mutated DNA, this mutation can be passed to her offspring. Germ cells are the reproductive cells (eggs in females and sperm in males)
that carry genetic information from one generation to the next.
a. Somatic cells are all the other cells in the body that are not germ cells. Mutations in somatic cells are not passed on to offspring.
b.White blood cells are a type of somatic cell that plays a role in the immune system.
d. Stem cells are undifferentiated cells that have the ability to develop into different types of cells in the
Correct Answer is A
A chloride ion has a negative charge because it gained an electron. When an atom gains an electron, it becomes negatively charged because it now has more electrons than protons. In the case of a
chloride ion, the neutral chlorine atom gains an electron to become a negatively charged chloride ion.
The other options are incorrect because they do not result in a negative charge. Losing an electron would result in a positive charge. Losing or gaining a proton would change the identity of the atom
and is not related to the formation of a chloride ion.
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Percentage Calculator - Calculate Percent of a number
Percentage Calculator
Percentage calculator helps to calculate the exact percentage of numbers online.
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How Does Our Percent Calculator Work?
The percent calculator offered on SmallSEOTools works on smart algorithms that accurately find the percentage of the input values. The functionality of percent finder is based on the percentage
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How to Calculate It Manually?
The percentage calculation is based on a percentage formula. You can calculate percentage manually with the help of the following formula:
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The percentage difference calculator is designed to help users calculate the difference between two positive integers greater than zero.
How to Find the Percentage of Something?
You can find the percentage of something with the help of a percentage calculation formula. Here is an example that can help you understand it:
Anna’s monthly salary is $2500. She pays x% income tax on it. The amount deducted as income tax from her salary is $250. What percentage of Anna’s salary is deducted as income tax?
Percentage Calculatoin Formula:
P x V1 = V2
P = x
V1 = 2500
V2 = 250
x% x 2500 = 250
x% = 250/2500
x = 10%
Anna pays a 10% income tax on her salary every month.
What is the Percentage?
The percentage is the ratio of a number expressed as a fraction of 100. It is generally denoted with the percent symbol, “%.”
How to calculate percentage between two numbers?
You can calculate the percentage between two numbers with the help of a percentage calculator. This percent calculator online allows you to submit the two numbers and calculate percentage in seconds.
What Can I Calculate with a Percentage Converter?
A percentage converter allows you to make any calculation in percentages. Whether you want to calculate the ratio of a student’s marks in an examination or check what amount you’re paying in taxes,
you can easily get accurate results with the percentage finder.
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No! When identifying the percentage of a number, it cannot get greater than 100. However, when finding the difference of values at a certain point in terms of percentage, the percentage can be higher
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No! The percentage of a number cannot be negative, but you can get a negative percentage when finding the percentage difference between two values. Still, in such cases, you need to ignore the
negative sign and write the percentage in positive integers only.
Can a Percentage Be a Decimal?
No! A percentage cannot be a decimal because it is formed by multiplying the decimal value by 100.
Is it Possible to Calculate the Percentage Without a Percent Calculator?
Yes, you can calculate the percentage without an online percent calculator using the percentage calculation formula. However, if the values are complex, you may need help to calculate the percentage | {"url":"https://smallseotools.com/percentage-calculator/","timestamp":"2024-11-14T10:17:53Z","content_type":"text/html","content_length":"117991","record_id":"<urn:uuid:b4a2cffe-34f1-4c49-b8e6-cc3d6b51db69>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00665.warc.gz"} |
Physics - P2 Tpoic 2 - Controlling and using electric current
View mindmap
• P2 Topic 2 - Controlling and using electric current
□ Key terms
☆ current
○ the rate of flow of charge
☆ voltage
○ what pushes the current around the circuit
☆ resistance
○ something that slows the current down
☆ the balance
○ if you increase the voltage, then more current will flow
☆ potential differance
□ Energy transferred
☆ when electric charge goes through a change (gives up energy) of potential difference
☆ electrical power = potential difference x current
☆ energy transferred = current x potential difference x time
□ Energy conserved
☆ junctions
○ where the current either splits or rejoins in a parallel circuit
☆ current doesn't get used up or lost in a circuit
○ its conserved
○ the total current entering the junction is the same as the total leaving the junction
□ Resistance
☆ ammeter
○ measures current - must be placed in series
☆ voltmeter
○ measures voltage - must be placed in parallel around the component
☆ voltage-current graphs
○ fixed resistors
■ current is proportional to voltage
○ filament lamps
■ as temperature increases, resistance increases
○ diode
■ current will only flow in one direction
□ Devices and reisitance
☆ light dependent resistor
○ in bright light resistance falls
☆ thermistor
○ in hot weather, resistance drops
☆ resistors
○ get hot when an electric current passes through them
■ electrons collide with ions in the lattice
■ heat causes resistance to increase
○ Untitled
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Converting Fractions To Mixed Numbers Worksheet
Converting Fractions To Mixed Numbers Worksheet function as fundamental devices in the world of mathematics, giving an organized yet flexible system for learners to discover and grasp mathematical
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Converting Fractions To Mixed Numbers Worksheet
Converting Fractions To Mixed Numbers Worksheet -
You can convert an improper fraction into a mixed number just by dividing the top number numerator by the bottom number denominator The answer to the division is the whole number part of the mixed
number and the remainder of the division tells you what fraction is left over 14 5
Below are six versions of our grade 6 math worksheet on rewriting improper fractions as mixed numbers Improper fractions are fractions with a value greater than 1 mixed numbers combine a whole number
and a fraction These worksheets are pdf files Worksheet 1 Worksheet 2 Worksheet 3 Worksheet 4 Worksheet 5 Worksheet 6
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Converting To Mixed Numbers Worksheet By Teach Simple Lupon gov ph
Converting To Mixed Numbers Worksheet By Teach Simple Lupon gov ph
Convert between Improper Fractions and Mixed Numbers Worksheets Explore this pack of printable converting between improper fractions and mixed numbers worksheets and excel in the step by step
conversion of both an improper fraction to a mixed number and a mixed number to an improper fraction
Www k5learning Improper fractions to mixed numbers Grade 6 Fraction Worksheet Convert the fractions into mixed numbers 12 7 1 5 7 4 21 9 2 1
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From Theory to Real-World Application
Improper Fractions To Mixed Numbers Worksheet
Improper Fractions To Mixed Numbers Worksheet
Converting improper fractions into mixed numbers with denominator in place Liveworksheets transforms your traditional printable worksheets into self correcting interactive exercises that the students
can do online and send to the teacher
This page has worksheets for teaching basic fraction skills equivalent fractions simplifying fractions and ordering fractions There are also links to fraction and mixed number addition subtraction
multiplication and division Adding Fractions Mixed Numbers Add fractions with same and different denominators Also add mixed numbers
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Mixed Operations Fractions Worksheets
Convert Mixed Numbers Into Improper Fractions Denominators Not Exceeding Tenths Grade 4 Math
Check more of Converting Fractions To Mixed Numbers Worksheet below
Converting Fractions To Mixed Numbers Worksheet
10 Best Images Of Converting Mixed Numbers Worksheet Improper Fractions As Mixed Numbers
Changing Mixed Numbers To Improper Fractions Worksheet
Fractions With Mixed Numbers Worksheets
10 Best Images Of Converting Mixed Numbers Worksheet Improper Fractions As Mixed Numbers
Improper Fractions To Mixed Numbers Worksheets
Convert Fractions To Mixed Numbers K5 Learning
Below are six versions of our grade 6 math worksheet on rewriting improper fractions as mixed numbers Improper fractions are fractions with a value greater than 1 mixed numbers combine a whole number
and a fraction These worksheets are pdf files Worksheet 1 Worksheet 2 Worksheet 3 Worksheet 4 Worksheet 5 Worksheet 6
Improper Fraction Worksheets Math Salamanders
Improper Fraction Worksheets Welcome to our Improper Fraction Worksheets page Here you will find a wide range of free printable Fraction Worksheets which will help your child understand and practice
how to convert improper fractions to mixed numbers
Below are six versions of our grade 6 math worksheet on rewriting improper fractions as mixed numbers Improper fractions are fractions with a value greater than 1 mixed numbers combine a whole number
and a fraction These worksheets are pdf files Worksheet 1 Worksheet 2 Worksheet 3 Worksheet 4 Worksheet 5 Worksheet 6
Improper Fraction Worksheets Welcome to our Improper Fraction Worksheets page Here you will find a wide range of free printable Fraction Worksheets which will help your child understand and practice
how to convert improper fractions to mixed numbers
Fractions With Mixed Numbers Worksheets
10 Best Images Of Converting Mixed Numbers Worksheet Improper Fractions As Mixed Numbers
10 Best Images Of Converting Mixed Numbers Worksheet Improper Fractions As Mixed Numbers
Improper Fractions To Mixed Numbers Worksheets
Improper And Mixed Fractions Worksheet
Convert Between Improper Fractions And Mixed Numbers Worksheets
Convert Between Improper Fractions And Mixed Numbers Worksheets
The Worksheet For Adding Fraction Numbers To One Digit Number Is Shown In This Image | {"url":"https://alien-devices.com/en/converting-fractions-to-mixed-numbers-worksheet.html","timestamp":"2024-11-09T01:27:10Z","content_type":"text/html","content_length":"27089","record_id":"<urn:uuid:2f23b3a1-8369-41bb-8023-79d6647a2c9b>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00377.warc.gz"} |
--- _id: '19978' abstract: - lang: eng text: "We introduce the \\emph{Online Connected Dominating Set Leasing} problem\r\n(OCDSL) in which we are given an undirected connected graph $G = (V, E)$, a
set\r\n$\\mathcal{L}$ of lease types each characterized by a duration and cost, and a\r\nsequence of subsets of $V$ arriving over time. A node can be leased using lease\r\ntype $l$ for cost $c_l$ and
remains active for time $d_l$. The adversary gives\r\nin each step $t$ a subset of nodes that need to be dominated by a connected\r\nsubgraph consisting of nodes active at time $t$. The goal is to
minimize the\r\ntotal leasing costs. OCDSL contains the \\emph{Parking Permit\r\nProblem}~\\cite{PPP} as a special subcase and generalizes the classical offline\r\n\\emph{Connected Dominating Set}
problem~\\cite{Guha1998}. It has an $\\Omega(\\log\r\n^2 n + \\log |\\mathcal{L}|)$ randomized lower bound resulting from lower bounds\r\nfor the \\emph{Parking Permit Problem} and the \\emph{Online
Set Cover}\r\nproblem~\\cite{Alon:2003:OSC:780542.780558,Korman}, where $|\\mathcal{L}|$ is the\r\nnumber of available lease types and $n$ is the number of nodes in the input\r\ngraph. We give a
randomized $\\mathcal{O}(\\log ^2 n + \\log |\\mathcal{L}| \\log\r\nn)$-competitive algorithm for OCDSL. We also give a deterministic algorithm for\r\na variant of OCDSL in which the dominating
subgraph need not be connected, the\r\n\\emph{Online Dominating Set Leasing} problem. The latter is based on a simple\r\nprimal-dual approach and has an $\\mathcal{O}(|\\mathcal{L}| \\cdot\r\n\\
Delta)$-competitive ratio, where $\\Delta$ is the maximum degree of the input\r\ngraph." author: - first_name: Christine full_name: Markarian, Christine last_name: Markarian citation: ama: Markarian
C. Online Connected Dominating Set Leasing. arXiv:180502994. 2018. apa: Markarian, C. (2018). Online Connected Dominating Set Leasing. ArXiv:1805.02994. bibtex: '@article{Markarian_2018, title=
{Online Connected Dominating Set Leasing}, journal={arXiv:1805.02994}, author={Markarian, Christine}, year={2018} }' chicago: Markarian, Christine. “Online Connected Dominating Set Leasing.”
ArXiv:1805.02994, 2018. ieee: C. Markarian, “Online Connected Dominating Set Leasing,” arXiv:1805.02994. 2018. mla: Markarian, Christine. “Online Connected Dominating Set Leasing.” ArXiv:1805.02994,
2018. short: C. Markarian, ArXiv:1805.02994 (2018). date_created: 2020-10-12T12:42:54Z date_updated: 2022-01-06T06:54:17Z department: - _id: '63' language: - iso: eng publication: arXiv:1805.02994
status: public title: Online Connected Dominating Set Leasing type: preprint user_id: '15415' year: '2018' ... | {"url":"https://ris.uni-paderborn.de/record/19978.yaml","timestamp":"2024-11-14T17:00:15Z","content_type":"text/x-yaml","content_length":"3195","record_id":"<urn:uuid:4b22e44d-f979-4c78-8288-1be3f13f5d70>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00571.warc.gz"} |
Eastern University Master of Science in Data Science Half-way Review - Dustin K MacDonald
Eastern University Master of Science in Data Science Half-way Review
As a student in the Eastern University MS in Data Science program, I’ve been taking courses in Python, R, statistics and databases for the last 9 months. I’m half way through the program now, and
thought it would be helpful to potential students to provide more information on what each course in the program (that I’ve taken so far) is about.
This is a continuation of my REVIEW: Eastern University Master of Science in Data Science 2021. I expect to make another post once I’ve gotten further into the program.
Courses I’ve Completed:
• DTSC-520 Fundamentals of Data Science
• DTSC-550 Introduction to Statistical Modeling
• DTSC-660 Data Analytics in R
• DTSC-575 Principles of Python Programming
• DTSC-660 Data and Database Management with SQL
Upcoming courses:
• DTSC-670 Foundations of Machine Learning Models (starts August 30)
• DTSC-680 Applied Machine Learning
• DTSC-600 Information Visualization
• DTSC-690 Data Science Capstone: Ethical and Philosophical Issues in Data Science
• DTSC-691 Data Science Capstone: Applied Data Science
DTSC-520 Fundamentals of Data Science
This course is the first one in the program, so it covers a lot of material. It’s an introduction to Python and the Anaconda distribution, numpy, pandas, matplotlib, seaborn, and then the general
principles of data science.
This class is a whirlwind. There are optional coding assignments that you can complete but it is mostly theory based. You’ll be asked things on exams like what a specific slice of a string is, or how
many times a loop will repeat, etc., but you won’t have to submit detailed coding assignments.
This course has 4 exams currently, each one worth 25% of your grade.
DTSC-550 Introduction to Statistical Modeling
This course is, again, a theory course but this one focuses on statistics and R. It goes right from the basics of measures of central tendency into variance, covariance, standard deviation,
hypothesis testing (T-Tests, Z-Tests, and ANOVA.)
It also discusses parametric and nonparametric statistical testing and includes optional labs in R. I actually skipped the labs, which was a mistake (!) because it made 650, the course that came
after, quite a bit harder than it needed to be.
This course includes 5 exams, each worth 20% of your grade.
DTSC-650 Data Analytics in R
This course is an introduction to R. It continues the statistical education but focuses on applying all of the concepts you learned in 550 with the R programming language. While 550 briefly touches
on linear and logistic regression, 650 goes into depth with how to perform these in R and how to interpret the results.
650 also adds other components like using the AIC to assess model fit, how to interpret R-squared and how to use the Bonferroni correction to adjust p-values.
This course, when I took it, included 60% exams and 40% CodeGrade coding assignments.
CodeGrade assignments are repeatable coding assignments, where you’re given a dataset and then asked to answer questions on it.
For example, one of the datasets included a series of orders from a pizza place, and you might be asked, “Find the average number of orders delivered by Lisa on Fridays” or “Write a regression
predicting whether a customer got wine based on their total order price and day of the week.” These assignments were really enjoyable and helped solidify my knowledge of R, but they ended up taking
me a long time because I never used R until then.
There were 8 CodeGrade assignments making up 30% of the grade, and 4 exams making up 60% of the grade. My average CodeGrade assignment was 40 lines of code.
The last 10% of the grade was a large final project. This was also done in R, and it used a real dataset: https://www.kaggle.com/cdc/behavioral-risk-factor-surveillance-system.
We had to answer a variety of questions and also do our own analysis (exploratory data analysis and regressions, etc.) It was a lot of work but also a lot of fun. My assignment was about 400 lines of
R, and I cut down some of it by writing a function to calculate some specific summary statistics I wanted for the variables I had selected.
DTSC-575 Principles of Python Programming
This course is an introduction to Python course. It reviews what you did in 520 and adds on object-oriented programming. The first module is basically a review of the Python from 520, but it adds
some information on list comprehensions.
Module 2 goes over strings, string formatting (also from 520), conditionals, the walrus operator, loops (with the addition of the break and continue commands.)
Module 3 goes over how to create functions including giving them arguments and parameters, decorators and exceptions and how to use lambdas which are little self-contained one line programs.
Module 4 goes over object-oriented programming, how to create objects and classes and make them parent/childs of each other, which is called inheritance.
Module 5 is called “odds and ends” and it goes over how to do statistics in Python, including how to use the scipy package, and how to do different tests from 550 and 650 in Python including ANOVA,
t-test and linear regression.
This course was surprisingly short but packed a lot of material in. There are 24 small CodeGrade assignments. In contrast to 650 where there were 8 assignments averaging 40 lines of code each (320
lines in total), this course had 24 small assignments that were under 10 lines of code each (240 lines in total.)
I did need to look up the quadratic formula to answer one of these questions, but otherwise it was pretty straightforward.
DTSC-660 Data and Database Management with SQL
I completed 660 and 575 at the same time. In retrospect, that was a bad idea. This course turned out to have 20 hours of video, 5 exams and 4 assignments!
The first two modules focus on the basics of database design. This is a lot of theory and mostly involves just drilling the definitions and trying to understand how they all fit together.
The first assignment involves designing an entity-relationship (ER) diagram for a fictional business. Assignment 2 is designing a relational schema for a fictional business and answering some
questions about primary and foreign keys, among others.
Modules 3-6 were 1000% better than Modules 1 and 2. Starting in Module 3, the professor walks through PostgresSQL syntax and shows you how to achieve different tasks. This is a comprehensive course
(as the 20 hours of video indicate) – you will be well-versed in SQL when you are finished.
Assignment 3 is to write a short SQL query, worth 3%.
Assignment 4 is pretty big. It involves writing a number of SQL queries, some procedures, functions and triggers, all in PostgresSQL. Assignment 4 is worth 20% of the grade.
Finally, the last module, Module 6, is on Git and Github. I was really happy to see this module because I wanted to create a Github and start posting my contributions. I still need to do some more
project work (and get it looking nice – right now it’s just used as a repository for work-in-progress code.)
For DTSC-660, all the assignments total up to 44%. The 5 quizzes are worth 56%. There are 6 modules, so one module does not have a quiz, since it has Assignment 4 in it.
I have really enjoyed this program. I am super excited for DTSC-670, which is the Foundations of Machine Learning course. I’ve already gotten to Chapter 4 in the textbook and hope to get up to
Chapter 5 by the time the course starts (it actually goes up to Chapter 7 in the book.)
15 thoughts on “Eastern University Master of Science in Data Science Half-way Review”
1. Hi Dustin!
Thanks for your thorough review. I am wondering if you have any advice for a total newb as far as course load. I am working full time but have few commitments outside of my 40 hour work week
schedule. So far, which classes do you think can be reasonably combined and which ones would you recommend taking by themselves?
1. Hi Shea,
There is a suggested course outline given on the EU website. I’d recommend sticking with one course at a time if you can afford to, so that you can focus on each course. The suggested
guideline for 2 courses is:
Term 1: 520/550
Term 2: 650/660
Term 3: 575/670
Term 4: 600/680
Term 5: 690/691
If you’re doing one course at a time, you’d follow that same progression, you would just take 520 in Term 1 and 550 in term 2. The hardest courses are 660 (Intro to Database Design/SQL), 670
(Foundations of Machine Learning) and 680 (Applied Machine Learning.) 650 (Data Analytics with R) was probably my favorite course in the program even though I’d never used R before.
The “easiest” courses will differ person to person but I would say 575 and 600 are probably the most manageable. The suggested course outline pairs up one of those most difficult courses with
an easier course, you’ll notice.
To reiterate, I would start with one course at a time and only double up if you’re really sure you can handle the workload. Better to finish early than have to rush and not learn as deeply as
you would like.
Happy learning!
1. Thanks for the above schedule. That is the schedule I will attempt.
2. Hi Dustin!
Thank you for taking the time to provide such in depth info on this program. I was wondering how much time there is between the 7 week sessions. Thanks in advance!
1. Hi Antonia,
Most sessions have a week between them, this winter session actually has a month – and a few sessions have no gaps at all. You can see the full breakdown here: https://www.eastern.edu/about/
Hope this helps,
3. Hi Dustin,
Thanks for putting this together. You mention a textbook for DTSC-670, can you share the name of it?
1. Hi Joe,
The textbook for 670/680 is Hands-On Machine Learning with Scikit-Learn, Keras, and TensorFlow: Concepts, Tools, and Techniques to Build Intelligent Systems 2nd Edition by Aurélien Géron.
It’s pretty inexpensive to buy and a great reference.
4. Hello, I just signed up to start January 10th. Wondering if you have any additional reviews of any courses? This Blog definitely helped me decide that I wanted to apply. How far along are you
1. Hi Scooby,
This is the most recent review I’ve written. I just completed DTSC-680 the last term. On January 10 I’ll start DTSC-600 (Data Visualization) and DTSC-690 (the first of the two capstone
courses.) I’m really enjoying the program and I’ll definitely write more once I’m done the program in April.
1. Thanks for getting back with me and good luck!
5. Dustin my name is Joe Valovage I am getting my on line MBA in organizational management at Eastern. I am planning on taking the buisness antalytics concentration that is offered there. I had
questions for you about 550; 600 650 and 660. They are the required courses. If you can reach out to me that would be much appreciated. Thanks again for this wonderful overview you have provided.
1. Hi Joe,
I emailed you.
7. Hi Dustin,
I’m thinking about applying for my Masters in data science , i come from sociology major so I have zero experience or knowledge with data science . My question is, for someone like me, how many
classes would you recommend taking, if I still want to finishing in that 1 year mark ? And how long where you given to complete each assignments? And did you haveto to maintain a certain GPA,
before being put on academic probation?
1. Hi Joly,
In order to finish within 10 months you’ll need to take 2 classes at a time. I finished in 14 months by starting with one class and then doubling up later, dropping down to 1 class for the
harder ones. Each class is 7 weeks long, and you need a minimum GPA of 2.0 to avoid being put on academic probation. I would recommend taking one course at a time to start, and if you find
yourself comfortable you can go up to 2 courses. Some classes like 520, 550, 575 and 600 I found could be done more quickly. Others, like 660 and 680 I needed a lot more time on. Hope this
helps, good luck! | {"url":"https://dustinkmacdonald.com/eastern-university-master-of-science-in-data-science-half-way-review/","timestamp":"2024-11-10T23:53:07Z","content_type":"text/html","content_length":"96387","record_id":"<urn:uuid:26cdf2aa-4f8d-498f-bbdc-228ae00543a1>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00494.warc.gz"} |
Geometric similarity
The geometric similarity curriculum unit uses the context of designing images for mobile devices with different screen sizes to develop pupils' understanding of the variant and invariant properties
of geometrically similar polygons. i.e. what changes and what stays the same.
It introduces the following 'hard to teach' mathematical ideas:
• Identifying the variants and invariants in shapes that are mathematically similar, including identification of the scale factor of enlargements.
• Recognising the important one-to-one geometric correspondence of sides and vertices within mathematically similar polygons.
The software has been designed to offer:
• dynamic measurements and comparisons, driven by angle and scale factor sliders.
• support for structuring recording within tables.
• linking between geometrical manipulation and values in tables and the ratio checker.
The following resources are provided to support schools to involve more teachers in the department to teach the Cornerstone Maths geometric similarity unit with confidence.
□ A presentation to support a PD session to introduce teachers to the key mathematical ideas addressed by the unit and to gain hands-on experience with the software (PowerPoint)
The complete set of PD resources (zip file)
Landmark activities are those in which the use of the technology prompts pupils (and teachers) to have an 'aha' moment about the mathematics.
In the geometric similarity, the landmark activity challenges pupils' early definitions of geometric similarity that rely on properties of the side lengths. However, pupils will often need some
careful support to use the software productively - and have a personal 'aha' moment.
Watch the following video clips to see how different teachers have used the software to support this dialogue:
Example 1
Example 2
Some examples of pupils' written work
The nature of the Cornerstone Maths activities results in some rich opportunities for accurate assessment of pupils' mathematical understanding.
The following Investigations (and particular questions) from the Geometric similarity unit are especially effective:
□ Investigation 2, Question 4-5 "What is the relationship between an original and a mathematically similar shape?"
□ Investigation 3, Question 11-12 "Describe what a scale factor is. Describe how to use it..."
□ Investigation 4, Question 7 "Devise a set of instructions so that anyone can create mathematically similar enlargements."
□ Investigation 5, Question 2 "What is the relationship between corresponding angles in mathematically similar shapes?"
Some students' responses to these questions for a departmental discussion about assessment (
or ). | {"url":"https://www.ucl.ac.uk/ioe/departments-and-centres/centres/ucl-knowledge-lab/current-research/cornerstone-maths/curriculum-units-and-pd-resources/geometric-similarity","timestamp":"2024-11-13T21:35:53Z","content_type":"text/html","content_length":"57979","record_id":"<urn:uuid:c422e00e-7769-4bc2-88c9-c460c6eb7a6b>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00221.warc.gz"} |
TWO-AGGREGATOR TOPOLOGY OPTIMIZATION USING MULTIPLE PATHS IN DATA CENTER NETWORKS - Nexgen Technology
by nexgentech | Oct 25, 2017 | ieee project
In this paper we focus on the problem of dataaggregation using two aggregators in a data center network,where the source racks are allowed to split their data and send tothe aggregators using
multiple paths. We show that the problemof finding a topology that minimizes aggregation time is NP-hardfor k = 2, 3, 4, where k is the maximum degree of each ToR switch(number of uplinks in a
top-of-rack switch) in the data center. Wealso show that the problem becomes solvable in polynomial timefor k = 5 and 6 and conjecture the same for k > 6. Experimentalresults show that, for k = 6,
our topology optimization algorithmreduces the aggregation time by as much as 83.32% and reducestotal network traffic by as much as 99.5% relative to the torusheuristic, proposed in [1], which
readily proves the significantimprovement in performance achieved by the proposed algorithm.
This paper focuses on the two aggregator variant of theproblem addressed by us in, we haveexamined the single-aggregator network topology optimizationwith splitting (SANTOS) problem. Consequently,
much of therelated-work section of has been reproduced here andwe have added in a summary of the new results obtained byus in .Multi-path data aggregation has been studied by manyresearchers in the
past to seek better performance. Previouswork includes research by Rao et al.Xue et al. aothers. Rao et al. have worked on providing transmissiontime guarantees in sending a message of finite
length andobtaining a threshold on the maximum time difference betweentwo out of order packets of a sequential message, transmittedat constant rate, from a source to a destination in a
computernetwork. Xue [9] presents a polynomial time algorithmfor computing an optimal multi-path end-to-end routing totransmit a given message while the previously published pathbasedalgorithm for
this problem is sub-optimal. However ourproblem TANTOS is markedly different from these, becauseTANTOS is defined on a data-center network, where thetopology is constrained by the maximum degree of
each ToRswitch (k).
In this paper, we have explored the Two Aggregator NetworkTopology Optimization with Splitting (TANTOS) problem.We have proved that TANTOS is NP-hard for k = 2 usingreduction from the standard 2-way
Partition problem, wherek is the maximum degree of a ToR switch in the data centernetwork. We have formulated a new problem called the 3-way Partition problem and showed it to be NP-hard
usingreduction from the 2-way Partition problem. We have employedreduction from this newly formulated 3-way Partitionproblem to prove that TANTOS is NP-hard for k = 3. We haveproved that TANTOS is
NP-hard for k = 4 using reductionfrom the standard 2-way Partition problem. For k = 5 andk = 6, we have proposed polynomial time algorithms to solveTANTOS optimally by exploring all possible
instances of theproblem. Based on our observations in k = 5 and 6, we haveconjectured that TANTOS is polynomially solvable for k > 6.Through extensive experiments, we illustrated the
improvedperformance by our optimal algorithm for k = 6 compared toa 3D extension of the 2D torus heuristic proposed by Wang etal. in [1]. Our algorithm reduced the data aggregation time andtotal
network traffic by up to 83:32% and 99:5%, respectivelyrelative to Wang’s heuristic.
[1] G. Wang , T.S. Eugene Ng , A. Shaikh, ”Programming your network atrun-time for big data applications”, Proceedings of the first workshopon Hot topics in software defined networks (HotSDN), 2012.
[2] S. Das, S. Sahni, ”Network Topology Optimization for Data Aggregation”,IEEE/ACM International Symposium on Cluster, Cloud, andGrid Computing (CCGrid), 2014, pp 493-501.[3] S. Das, S. Sahni,
”Network topology optimization for data aggregationwith splitting”, IEEE International Symposium on Signal Processingand Information Technology (ISSPIT), 2014, pp 398-403.
[4] S. Das, S. Sahni, ”Network topology optimisation for data aggregationusing multiple paths”, International Journal on Metaheuristics, 2015, pp115-140.
[5] S. Das, S. Sahni, ”Two-aggregator network topology optimization withsplitting”, IEEE Symposium on Computers and Communication (ISCC),2015, pp 683-688.
[6] R. L. Graham, ”Bounds on Multiprocessing Timing Anomalies”, SIAMJOURNAL ON APPLIED MATHEMATICS, 1969, Volume 17, Number2, pp 416-429.
[7] E. G. Coffman Jr., R. Sethi, ”A generalized bound on LPT sequencing”,Proceedings of ACM SIGMETRICS conference on Computer performancemodeling measurement and evaluation, 1976.
[8] N. S. V. Rao, S. G. Batsell, ”QoS Routing via multiple paths usingbandwidth reservation”, Proceedings of the IEEE INFOCOM, 1998, pp11-18.[9] G. Xue, ”Optimal multi-path end-to-end data
transmission in networks”,Proceedings of ISCC, 2000, pp. 581-586.[10] A. Hammadi, L. Mhamdi, ”A survey on architectures and energyefficiency in data center networks”, Computer Communications,
2014,vol. 40, no. 0, pp. 1 21.
[11] D. Kliazovich, P. Bouvry, Y. Audzevich, S. Khan, ”Greencloud: apacket-level simulator of energy-aware cloud computing data centers”,Global Telecommunications Conference (GLOBECOM), 2010, pp.
[12] Y. Chen, R. Griffith, J. Liu, R.H. Katz, A.D. Joseph, ”Understanding tcpincast throughput collapse in datacenter networks”, Proceedings of the1st ACM Workshop on Research on Enterprise
Networking, WREN,2009, pp. 73-82. | {"url":"https://nexgenproject.com/two-aggregator-topology-optimization-using-multiple-paths-data-center-networks/","timestamp":"2024-11-10T15:56:09Z","content_type":"text/html","content_length":"91421","record_id":"<urn:uuid:e9ed5ecc-cc80-4e9b-909f-fe9e6020d8d9>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00301.warc.gz"} |
Introduction to PoolDilutionR
The equations in pdr_predict() come from the following equations in von Fischer and Hedin 2002.
Equation 5 describes the change in pool size over time in terms of \(P\), gross production, and \(k\), the first order rate constant of consumption.
\[m_t = \frac{P}{k} - (\frac{P}{k} - m_0) * e^{(-kt)}\]
\(m_t\) is the total pool size at time t.
\(m_0\) is the total pool size at time zero.
Equation 9 tracks the change in the heavy isotopologue over time in terms of consumption and its fractionation constant.
\[ n_t = n_0 * e^{(-k\alpha t)}\]
\(n_t\) is the pool size of the heavy isotopologue at time t.
\(n_0\) is the pool size of the heavy isotopologue at time zero.
\[ \alpha = \frac{k_{(\,^{13}C)\,}}{k_{(\,^{12}C)\,}} \]
\(k_{(\,^{13}C)\,}\) is the first-order rate constant for consumption of \(^{13}CH_4\)
\(k_{(\,^{12}C)\,}\) is the first-order rate constant for consumption of \(^{12}CH_4\).
In PoolDilutionR, we expand this equation to allow for the production of heavy molecules from authocthonous sources.
Equation 9, adjusted
\[n_t = \frac{p_{frac}}{k_{frac}} - ( \frac{p_{frac}}{k_{frac}} - 0) * e^{-k_{frac}t}\]
Where \(k_{frac}\) is equivalent to \(k*\alpha\) above, for whatever heavy isotope is applicable.
The isotopic composition of the pool over time is described in Equation 10.
\[ AP_t = \frac {n_t}{m_t} + AP_p \]
Which we can now simplify to:
Equation 10, adjusted \[AP_t = (\frac{n_t}{m_t}) * 100\]
Due to production of heavy isotopologue now being accounted for in our adjusted Equation 9.
Combined this looks like:
\[ AP_t = \left( \frac{\frac{p_{frac}}{k_{frac}} - ( \frac{p_{frac}}{k_{frac}} - 0) * e^{-k_{frac}t}} {\frac{P}{k} - (\frac{P}{k} - m_0) * e^{(-kt)}} \right) * 100 \]
Cost Function
The pdr_cost() provides feedback to pdr_predict() on the quality of each iteration of fitted rates and/or fractionation constants. The sum of errors are weighted by the standard deviation of the
observations, as well as a scaling factor, (\(N\)).
Equation 14:
\[E = \left(\sum_{t=1}^j\frac {AP_{obs}(t) - AP_{pred}(t)}{SD_{obs-AP}}\right) * N_{ap} + \left(\sum_{t=1}^j\frac {m_{obs}(t) - m_{pred}(t)}{SD_{obs-m}}\right) * N_{m}\]
\(SD_{obs-AP}\) is the standard deviation among all observations of atom percent for a single sample
\(SD_{obs-m}\) is the standard deviation among all observations of total pool size for a single sample
\[N_x = \frac{SD_{x_{observed}}}{SD_{x_{precision}}}\]
\(x\) is either atom percent (\(ap\)) or total pool size (\(m\))
\(SD_{x_{precision}}\) is the instrument precision for that variable as standard deviation (ie., standard precision)
Users have the ability to replace the default cost function (pdr_cost()) with their own, if desired.
Fractionation describes the tendency of heavier isotopes and isotopologues to resist chemical transformation, whether these be phase changes or chemical reactions, whether spontaneous or
enzyme-mediated. There are two major types of fractionation. Equilibrium fractionation occurs when phase changes favor the heavier isotopologue staying a lower energy state (Druhan, Winnick, and
Thullner 2019, Urey 1947). A typical example would be the relative enrichment of the ocean in heavy water \(H_2^{18}O\) due to preferential evaporation of light water \(H_2^{16}O\). The second major
type of fractionation is kinetic, and is classically associated enzyme selectivity. This is what drives the distinction in \(^{13}C\) signatures between C3 and C4 plants (O’Leary 1981).
Because our knowledge of earth system processes and enzyme diversity is rapidly expanding, additional fractionation constants will be added to pdr_fraction as part of future package versions. | {"url":"http://cran.fhcrc.org/web/packages/PoolDilutionR/vignettes/Intro_PoolDilutionR.html","timestamp":"2024-11-09T01:05:45Z","content_type":"text/html","content_length":"122569","record_id":"<urn:uuid:ba738f4e-e21b-4162-ba92-35204770dfdb>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00324.warc.gz"} |
Deep Reinforcement Learning
Write about "Deep Reinforcement Learning" in the context of "Deep Learning" for a "Artificial Intelligence" book. Start with an introduction tailored to "Deep Reinforcement Learning". Explain "Deep
Reinforcement Learning" and its related sub-topics in detail. Write code examples whenever applicable. Please write it in Markdown.
Deep Reinforcement Learning (DRL) is a subfield of machine learning that combines deep learning techniques with reinforcement learning algorithms. Reinforcement learning is an approach to learning
where an agent learns to make decisions in an environment by taking actions and receiving rewards or penalties.
In DRL, the deep learning component involves using artificial neural networks to approximate the state-value or action-value function, which is used by the reinforcement learning algorithm to decide
which actions to take. The neural network is trained using data generated by the reinforcement learning algorithm as it interacts with the environment.
DRL has been used to solve a wide range of problems, including game playing, robotic control, and autonomous navigation. For example, DeepMind's AlphaGo and AlphaGo Zero used DRL to defeat human
champions at the game of Go, and Google's DeepMind used DRL to develop an agent that could control a simulated robot to perform various tasks.
DRL is a powerful tool for solving complex problems in which the optimal solution is not known in advance, and it provides a framework for building agents that can learn to make decisions and improve
their performance over time through trial-and-error.
The goal of DRL is to train an artificial agent to make decisions and take actions in an environment in order to maximize a reward signal.
The mathematical foundation of DRL is based on the Markov Decision Process (MDP), which is a mathematical framework for modeling decision-making problems. In an MDP, the agent interacts with an
environment by taking actions and observing rewards. The environment is described by a state transition function, which defines the next state as a function of the current state and action, and a
reward function, which defines the reward for a given state and action.
The objective of the agent in DRL is to learn a policy, which is a mapping from states to actions. The policy is learned by estimating the expected return for each state, which is the expected
cumulative reward over time, starting from that state and following the policy. The expected return can be estimated using value-based methods, such as Q-learning or SARSA, or policy-based methods,
such as policy gradient methods.
In DRL, the policy is represented by a deep neural network, which takes the state as input and outputs the action. The neural network is trained using reinforcement learning algorithms, such as
Q-learning or policy gradient methods, to maximize the expected return. The training process involves repeatedly collecting experience by interacting with the environment and updating the neural
network parameters to improve the policy.
Mathematically, DRL can be described as an optimization problem, where the goal is to find the policy that maximizes the expected return. The optimization is typically performed using gradient- based
algorithms, such as stochastic gradient ascent, which updates the neural network parameters in the direction of the gradient of the expected return. The gradient can be estimated using Monte Carlo
methods, or by using the chain rule of differentiation in the case of policy gradient methods.
The main types of Deep Reinforcement Learning (DRL) include:
• Value-Based Methods: These methods estimate the expected future reward for each state-action pair and use this information to make decisions. Examples include Q-Learning and Deep Q-Networks
• Policy-Based Methods: These methods directly estimate the policy function, which maps states to actions, without estimating the value function. Examples include REINFORCE and Proximal Policy
Optimization (PPO).
• Actor-Critic Methods: These methods combine value-based and policy-based methods, using a critic to estimate the value function and an actor to directly estimate the policy. Examples include A3C
and DDPG.
• Model-Based Methods: These methods use a model of the environment to simulate future states and estimate the expected reward. Examples include Dyna-Q and Model-Based Reinforcement Learning.
Each type of DRL has its own strengths and weaknesses, and the choice of method depends on the specific problem being solved and the available computational resources. For example, value-based
methods can be used for problems with well-defined reward functions, while policy-based methods are more flexible and can handle problems with complex reward functions. Model-based methods can be
more computationally expensive, but they can provide a more complete understanding of the environment.
Value-Based Methods
Value-Based Methods of Deep Reinforcement Learning (DRL) estimate the expected future reward for each state-action pair, known as the value function. This information is then used to make decisions
about which actions to take. The main idea behind value-based methods is to use the value function to select the action that leads to the highest expected reward.
One of the most popular value-based methods is Q-Learning, which uses a table to store the estimated values for each state-action pair. The values are updated as the agent interacts with the
environment, using the Bellman equation to estimate the expected reward for each state-action pair.
Deep Q-Networks (DQN) is a variant of Q-Learning that uses a neural network to approximate the value function, instead of using a table. The neural network is trained using experience replay, where a
buffer stores a large number of experiences and the network is trained on a randomly selected batch of these experiences to reduce the correlation between successive updates.
Overall, value-based methods are effective in problems with well-defined reward functions, where the optimal policy can be determined by maximizing the expected reward. However, they can be limited
in problems with complex reward functions, as they do not directly estimate the policy function.
Policy-Based Methods
Policy-Based Methods of Deep Reinforcement Learning (DRL) directly estimate the policy function, which maps states to actions, without estimating the value function. The goal of these methods is to
directly optimize the policy, such that it maximizes the expected reward.
One popular policy-based method is REINFORCE, which uses Monte Carlo methods to estimate the gradient of the expected reward with respect to the policy parameters. The policy parameters are then
updated using gradient ascent to maximize the expected reward.
Another popular policy-based method is Proximal Policy Optimization (PPO), which combines ideas from value-based and policy-based methods. PPO uses a value function to provide a baseline for the
policy update, and it also uses a trust region constraint to ensure that the update to the policy is not too large. This makes PPO more stable and reliable than pure policy-based methods, such as
Overall, policy-based methods are flexible and can handle problems with complex reward functions, as they directly estimate the policy. However, they can be sensitive to the choice of hyperparameters
and the initialization of the policy parameters, and they may require more samples to converge compared to value-based methods.
Actor-Critic Methods
Actor-Critic Methods of Deep Reinforcement Learning (DRL) are a combination of value-based and policy-based methods. They consist of two components: an actor, which directly estimates the policy, and
a critic, which estimates the value function.
The actor and the critic work together to improve the policy. The actor takes actions in the environment and receives rewards, and the critic uses this information to estimate the value function. The
value function is then used to update the policy, by adjusting the policy parameters so that actions that lead to higher expected reward are more likely to be taken.
One popular actor-critic method is Advantage Actor-Critic (A2C), which uses the advantage function, which is the difference between the value function and the baseline, to update the policy. Another
popular method is Deep Deterministic Policy Gradients (DDPG), which is a variant of A2C that uses a deep neural network to approximate the policy and the value function.
Actor-critic methods are a good choice for problems where it is difficult to specify the reward function, as they directly estimate the policy and use the value function to provide a baseline for the
policy update. They are also computationally efficient, as they only require a single network to be trained, instead of two separate networks as in policy-based methods. However, they can still be
sensitive to the choice of hyperparameters, such as the learning rate, and they may require a large number of samples to converge.
Model-Based Methods
Model-Based Methods of Deep Reinforcement Learning (DRL) are a class of methods that incorporate a model of the environment into the reinforcement learning process. The model is used to simulate the
environment and to make predictions about the next state, reward, and action.
In model-based methods, the model is typically trained simultaneously with the policy, and the policy is updated based on the predictions made by the model. This allows the agent to learn about the
environment more efficiently, as it can explore the environment through the model, instead of having to interact with the real environment.
One popular model-based method is Model-Based Reinforcement Learning (MBRL), which uses a combination of model-based and value-based methods. MBRL trains a model of the environment and uses the model
to generate simulations, which are used to update the value function and the policy. Another popular model-based method is Dyna, which uses the model to plan ahead and make predictions about the
Model-based methods have the advantage of being more sample efficient, as they can use the model to generate simulations and avoid having to interact with the real environment as much. They can also
handle problems with partial observability, as they can use the model to fill in missing information. However, model-based methods can be computationally expensive, as they require training both a
model and a policy, and they can also suffer from model bias, if the model is inaccurate. | {"url":"https://cstopics.com/books/artificial-intelligence/06-deep-learning/06-deep-reinforcement-learning/","timestamp":"2024-11-04T00:44:02Z","content_type":"text/html","content_length":"112961","record_id":"<urn:uuid:1519082a-619e-4241-8af0-406640f90c52>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00522.warc.gz"} |
Live Online classes for kids from 1-10 | Upfunda Academy
What is Ratio and Proportion?
Ratio and proportion are mathematical concepts that help us make comparisons and solve problems in our daily lives. Whether we are dividing a pizza among friends or planning a trip, understanding
ratios and proportions can be very useful. In this article, we will explain these concepts in simple terms and provide you with some examples to help you understand them better.
Ratio is a way to compare two or more things that have something in common. For example, let's say we have 10 apples and 5 oranges. We can compare the number of apples to the number of oranges by
using a ratio. A ratio is a way of showing the relationship between two numbers.
For example,
The ratio of apples to oranges is 10:5 or 2:1. This means that for every 2 apples, there is 1 orange.
Another example,
Of a ratio is if we have a pizza that is cut into 8 slices and 3 of the slices are pepperoni, the ratio of pepperoni slices to total slices is 3:8.
Ratios are very useful in many situations, such as cooking, construction, and finance.
Proportion is an equation or statement that is used to depict that the two ratios or fractions are equivalent. Two equivalent ratios are always in proportion. Proportions refer to the equality of two
ratios and are denoted by the symbol (: :). They help us to solve for unknown quantities.
There are two types of proportions.
• Direct Proportion
• Inverse Proportion
Direct Proportion
Direct proportion is a concept in math that helps us understand how things change when we increase or decrease one of them. It means that as one thing increases, the other thing also increases, and
as one thing decreases, the other thing also decreases.
For example, let's say you are baking cookies and you need to use flour and sugar. If you increase the amount of flour, you will also need to increase the amount of sugar to keep the recipe balanced.
This is because the amount of sugar needed is directly proportional to the amount of flour used.
Inverse Proportion
Inverse proportion is a concept in mathematics that shows how two values are related to each other. Inverse proportion means that as one value increases, the other value decreases. For example,
imagine you are filling up a glass with water. If you increase the amount of water in the glass, the level of air in the glass decreases. This is because the amount of water and air are inversely
proportional to each other.
Ratio and Proportion Formula
The formula for ratio is expressed as a : b ⇒ a/b, where,
• a = the first term or antecedent.
• b = the second term or consequent.
Now, in order to express a proportion for the two ratios, a : b and c : d, we write it as
a:b::c:d ⟶ ab=cd�:�::�:�⟶��=��
• The two terms b and c are called mean terms.
• The two terms a and d are known as extreme terms.
• In a: b = c : d, the quantities a and b should be of the same kind with the same units, whereas, c and d may be separate of the same kind and of the same units.
• The proportion formula can be expressed as, a/b = c/d or a: b:: c : d.
• In proportion, the product of the means = the product of the extremes. Therefore, in the proportion formula a: b:: c : d, we get b × c = a × d.
Difference Between Ratio and Proportion
The difference between ratio and proportion can be seen in the following table.
Ratio Proportion
A comparison of two or more quantities that have the same unit of The equality of two ratios.
Can be written in different forms, such as 2:1 or 2/1. Two ratios are proportional if they have the same value.
Represents how many times one quantity is greater than another. When two ratios are proportional, it means that they have the same value.
Example: The ratio of boys to girls in a class is 2:3. Example: If the ratio of boys to girls in a class is 2:3, and there are 10 boys, then there must be 15 girls for the ratios to
be proportional.
Applications of Ratio and Proportion
Here are five possible applications of ratio and proportion that can be explained to 10-year-olds:
1. Cooking: When we cook, we use ratios and proportions to measure ingredients in the correct amounts. For example, a recipe may call for a ratio of 2 cups of flour to 1 cup of milk to make
2. Scaling drawings: Architects and designers use ratios and proportions to scale their drawings and create blueprints. For example, a blueprint may have a scale of 1 inch to 1 foot, which means
that 1 inch on the blueprint represents 1 foot in real life.
3. Finance: In finance, ratios and proportions are used to analyze financial statements and make investment decisions. For example, a company may use the ratio of its earnings to its revenue to
measure its profitability.
4. Construction: Engineers and construction workers use ratios and proportions to calculate the measurements of buildings and structures. For example, the ratio of the height of a building to its
width may be used to determine the slope of the roof.
5. Maps: Maps and globes use ratios and proportions to represent the size and distance of different locations. For example, a map may have a scale of 1 inch to 10 miles, which means that 1 inch on
the map represents 10 miles in real life.
Test your knowledge with Upfunda Quiz!
1. In an office, the working hours are 9.30 to 5.30 PM and in between 30 minutes are spent on lunch. Find the ratio of office hours to the time spent for lunch.
2. If X : y : : y : z then the correct statement is
3. First, Second and third terms of a proportion are 5, 120 and 40. Then the fourth term is Choose A) 960
4. A sum of money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets Rs. 1000 more than D, what is B's share?
Answer Key
1. C) 16:1
2. B) y2 = xz
3. A) 960
4. 2000 | {"url":"https://upfunda.academy/blog/7d94ac26-2b86-4aef-89ae-70008325d7f2","timestamp":"2024-11-09T00:07:44Z","content_type":"text/html","content_length":"39526","record_id":"<urn:uuid:52af6196-5b9a-4956-98e1-2c0658eb5eb8>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00792.warc.gz"} |
Critical discussion of the dividend discount model and capital asset pricing | 15 Writers
Level: Undergraduate 1st
Written by: Carl R
Executive Summary
Capital Asset Pricing Model (CAPM) and Dividend Discount Model (DDM) are among the most commonly used models in valuation. Their key appeal is simplicity, which is achieved by making rather strong
assumptions about the market and its participants. Although DDM is rooted in theoretically sound Discounted Cash Flow (DCF) models, its application is mainly limited to firms with large, stable
dividends. One if its key inputs, the discount rate, needs to be estimated using a cost of equity model such as CAPM. The CAPM makes numerous assumptions about the market that have been commonly
criticised. However, extensions of the model have been proposed that incorporate some of the anomalies in the observed pricing dynamics.
One of the key aspects of financial markets relevant to investors is determining whether a stock of overpriced or underpriced. An asset’s fundamental value is generally understood as the present
value of future cash flows, with numerous approaches having been proposed on how to estimate this value. The present essay focuses on two major models that have been commonly employed in stock
valuation, namely the Capital Asset Pricing Model (CAPM) and the Dividend Discount Model (DDM). While these models have received a lot of critique for making unrealistic assumptions, they remain
appealing due to their simplicity and intuitiveness.
Capital Asset Pricing Model (CAPM)
The main rationale of the CAPM is that risky investments should be more rewarding than risk-free assets (Fama and French, 2004). The model assumes that the expected returns on a risky asset should
exceed the returns on a risk-free asset by an amount that is proportional to equity premium (Fama and French, 1996). The latter represents the reward for investing in stocks over risk-free assets,
and is measured as the expected return on the market portfolio less the return on the risk-free asset (French, 2017). The coefficient of proportionality is called the asset beta, and it represents
the correlation of asset returns with market returns (Fama and French, 2004). Assuming that the market portfolio is efficient in terms of mean-variance optimisation, the CAPM implies that asset
returns linearly depend on the equity premium (Cochrane, 2017).
The CAPM makes several assumptions regarding investors and markets. Most importantly, the model assumes that investors act rationally, have homogeneous expectations, and are mean-variance optimisers
(Mayers, 1973; Galagedera, 2007). Another assumption is that the assets are traded publicly (French, 2017). Furthermore, it is assumed that investors are able to borrow and lend at the risk-free
rate, implying that there is no difference in the optimal portfolio between lenders and borrowers (Roll, 1977). In addition, the baseline CAPM makes simplifying assumptions such as there being no
taxes or transaction costs (Fama and French, 1996). More generally, the CAPM can be linked to the framework of the Efficient Market Hypothesis (EMH) which assumes that all available information is
incorporated in market prices (O’Sullivan, 2018).
Strengths and weaknesses
The CAPM is sufficiently simple to understand and easy to apply (Rossi, 2016). In practice, estimating the model only requires making decisions about the data to be used, such as the choice of the
estimation window, the frequency of the data, or the benchmark index (O’Sullivan, 2018). Furthermore, the model represents volatility of an asset relative to the market as a single value, beta, which
can be especially convenient when comparing different assets (Cochrane, 2017). Considering the ubiquity of discounted cash flow (DCF) models in valuation (Pinto et al., 2019), one of the key
applications of CAPM is estimating cost of equity as the discount factor for present value calculations (d'Amico and De Blasis, 2020).
However, CAPM has been heavily criticised for making strong assumptions about the market (Fama and French, 2004). For example, unrestricted risk-free borrowing and lending is not realistic and may
imply unlimited liability. As a result, collateral requirements would increase and limit the ability to reinvest profits (Fama and French, 1996). The assumptions of zero taxes and no transaction
costs are also problematic. Investors are likely to differ in terms of after-tax returns and as such would have different optimal portfolios of risky assets (Roll, 1977). Another unrealistic
assumption is that investors only care about mean and variance of returns for a single-period portfolio (Elbannan, 2015). This leads to a major weakness of CAPM, which is failing to account for
important dimensions of risk that are not captured by return variability (French, 2017).
In practice, market return is proxied by return on a market index such as S&P500. However, a major critique of the model is that the market portfolio is not observable which makes CAPM not testable
(Roll, 1977). Empirical tests appear to confirm that CAPM struggles to accurately describe stock return dynamics including the so-called market anomalies (Kroll et al., 1988; Reinganum, 1981; Fama
and French, 2004; Rossi, 2016). This has been attributed to the static nature of the CAPM as well as deviations of real markets from the model of rational, utility-maximising investors (Fama and
French, 2004; Brunnermeier et al., 2021). Notably, behavioural effects and biases such as overconfidence or herding may greatly distort asset pricing (Tversky and Kahneman, 1992; Hirshleifer, 2015).
Extensions and alternatives
Some of the CAPM’s limitations have been addressed in its extensions. In particular, extended models have been considered that explicitly address specific model assumptions such as assumptions on
borrowing restrictions (Roll, 1977), assets being publicly traded (Mayers, 1973), and investors having a single-period investment horizon (Merton and Samuelson, 1992; Barberis et al., 2015). For
example, intertemporal CAPM (ICAPM) is an extension of the base CAPM where investors are allowed to consider how their future wealth will be affected by current investment decisions (Fama and French,
2004; Khan, 2008). ICAPM implies that investors maximise the expected utility of lifetime consumption, and that current prices can be influenced by the uncertainty in future investment opportunities
(Elbannan, 2015).
Multi-factor models are probably among the most commonly used extensions of CAPM. Such models augment CAPM by introducing additional explanatory terms besides market risk. The most prominent example
is the Fama-French three-factor model which adds two new factors, namely size and value (Fama and French, 1996). Size refers to the differences in returns on portfolios of small and large stocks,
while value represents the difference in returns on stocks with high and low book-to-market ratio. It has been argued that the model successfully captures some of the dimensions of systematic risk
that are ignored by the CAPM (Fama and French, 2015). The three-factor model has been further extended to include other factors such as momentum, profitability, liquidity, and investment (Liu, 2006;
French, 2017; Blitz et al., 2018). Multi-factor models are often formulated within the Arbitrage Pricing Theory (Ross, 1978) where some of the unrealistic assumptions of the CAPM are relaxed.
Dividend Discount Model (DDM)
The Dividend Discount Model (DDM) describes prices as a function of several characteristics of future dividends, namely size, certainty, and timing (Barker, 1999; Lazzati and Menichini, 2015). DDM
posits that the share price is equal to the present value of the expected stream of dividend payments (Bask, 2020). The model is rooted in the more general perspective of viewing an asset’s intrinsic
long-term value as the present value of future cash flows (d'Amico and De Blasis, 2020). The key inputs that are required for applying DDM are future dividends and the measure of risk (Damodaran,
2012). Risk is represented by a discount rate which is usually taken to be the cost of equity and estimated using other techniques (Foerster and Sapp, 2005). Information about future dividends is
captured by two parameters, namely the dividend amount and the dividend growth rate (Irons, 2014).
In essence, DDM is rooted in discounted cash flow (DCF) methods as it views firm value as the dividend amount discounted at an appropriate rate (Drake and Fabozzi, 2008). However, it requires
significantly fewer inputs than a general DCF valuation while still allowing for some flexibility. In particular, the Modigliani-Miller hypothesis on dividends implies that it does not matter to
investors whether a firm pays out dividends (Brennan, 1971; Handley, 2008). Assuming the hypothesis is true, it is possible to apply DDM when the stock does not pay any dividends. Specifically, one
can replace the stock’s dividend with earnings per share, although this requires making assumptions on the earnings growth rate (Damodaran, 2012).
Strengths and weaknesses
The main appeal of DDM is its simplicity, with the model being easy to implement and intuitive to understand (Payne and Finch, 1999; Drake and Fabozzi, 2008). Fundamentally, DDM is based on the same
principles as DCF valuation, and as such shares the advantages of DCF methods such as the ability to capture the time value of money (Irons, 2014). DDM may produce accurate valuations as long as the
stocks pay out large and stable dividends (McLemore et al., 2015; Mugoša and Popović, 2015; Gacus and Hinlo, 2018), although it may still be less accurate than a full DCF valuation (Ivanovski et al.,
At the same time, the simplicity of DDM is also its main drawback. The model depends on several inputs, namely the discount factor and the dividend growth rate, and as such it is sensitive to
assumptions on these inputs (Payne and Finch, 1999; Ivanovski et al., 2015). The discount factor is often taken to be the cost of equity estimated using another model such as CAPM. It follows that
DDM inherits the weaknesses of the underlying cost of equity model (Drake and Fabozzi, 2008). The model’s structure and assumptions make it less applicable to firms that do not have a stable dividend
policy (d'Amico and De Blasis, 2020). Furthermore, the model may be less relevant for firms that are not paying out large dividends as a result of using share buybacks to reduce taxes. This leads to
lower dividend cash flow and consequently an underestimated firm value as implied by DDM (Damodaran, 2012). Another issue with DDM is its assumption of constant growth of dividends. This assumption
is not realistic and may lead to the overestimation of firm value. DDM may be less relevant for certain markets depending on their cyclicality and other industry-level factors. Notably, DDM appears
to be more widely used in the financial industry compared to other markets (Imam et al., 2008). Accuracy of DDM might also depend on the structure of risk in the market, and how this risk is
incorporated into the model (Bao and Feng, 2018).
Extensions and alternatives
While discounted cash flow (DCF) models are acknowledged as a theoretically sound valuation method, DDM is generally viewed as being too simplistic to fully realise the strengths of DCF models (Imam
et al., 2008). Nevertheless, numerous extensions and alternatives to DDM have been considered. Although the assumption of constant dividend growth is not realistic, it can be relaxed by considering a
multi-stage model (Drake and Fabozzi, 2008). A related extension of DDM is a Markov chain model where the dividend growth rate is represented by a Markov process (d'Amico and De Blasis, 2020). The
dividend growth increases or decreases with a certain probability in each period, which allows the model to provide a more flexible and realistic description of future dividend cash flow.
DDM ignores the value of certain assets and it may underestimate firm value if the dividend cash flow is reduced. Notably, this is the case for firms that practice share buybacks. Nevertheless, DDM
can be adjusted to directly incorporate the values of such assets in the dividends flow (Damodaran, 2012). Stochastic DDM can be used to allow dividends to be random and evolve according to some
evolution process, shifting away from the basic DDM setting of deterministic growth and discount rates (Agosto and Moretto, 2015). DDM has also been extended to a dynamic model with explicitly
endogenous choice of investment (Lazzati and Menichini, 2015). DDM can be coupled with APT which would allow for more accurate estimation of long-germ risk premium (Jawadi and Prat, 2017). Overall,
the practical implications of DDM seem to be limited as many companies pay no or little dividends, which makes full DCF models such as discounted free cash flow more useful (Imam et al., 2008; Pinto
et al., 2019).
CAPM and DDM provide simple and intuitive modelling frameworks at the cost of making strong assumptions about the market. DDM is valuation method rooted in DCF methods and it can be viewed as a
one-period DCF model. Its key drawback is that application to firms with unstable or small dividends may be problematic. The main input to DDM is cost of equity which can be estimated by employing a
model such as CAPM. The latter makes numerous assumptions about the market and its participants, but a variety of extensions have been proposed that more accurately describe pricing dynamics.
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Irons, R. (2014). Enhancing the dividend discount model to account for accelerated share price growth. Journal of Accounting and Finance, 14(4), pp.153-159.
Ivanovski, Z., Ivanovska, N., and Narasanov, Z. (2015). Application of dividend discount model valuation at Macedonian Stock Exchange. UTMS Journal of Economics, 6(1), pp.147-154.
Jawadi, F., and Prat, G. (2017). Equity prices and fundamentals: a DDM–APT mixed approach. Review of Quantitative Finance and Accounting, 49, pp.661-695.
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, 47(3-4), pp.225-252.
Payne, T. H., and Finch, J. H. (1999). Effective teaching and use of the constant growth dividend discount model. Financial Services Review, 8(4), pp.283-291.
Pinto, J. E., Robinson, T. R., and Stowe, J. D. (2019). Equity valuation: A survey of professional practice. Review of Financial Economics, 37(2), pp.219-233.
Reinganum, M. R. (1981). Misspecification of capital asset pricing: Empirical anomalies based on earnings' yields and market values. Journal of financial Economics, 9(1), pp.19-46.
Roll, R. (1977). A critique of the asset pricing theory's tests Part I: On past and potential testability of the theory. Journal of Financial Economics, 4(2), pp.129-176.
Ross, S. A. (1978). The current status of the capital asset pricing model (CAPM). The Journal of Finance, 33(3), pp.885-901.
Rossi, M. (2016). The capital asset pricing model: a critical literature review. Global Business and Economics Review, 18(5), pp.604-617.
Tversky, A. and Kahneman, D. (1992). Advances in prospect theory: Cumulative representation of uncertainty. Journal of Risk and Uncertainty, 5(4), pp.297-323. | {"url":"https://15writers.com/sample-essays/critical-discussion-of-the-dividend-discount-model-and-capital-asset-pricing/","timestamp":"2024-11-03T00:25:18Z","content_type":"text/html","content_length":"144360","record_id":"<urn:uuid:9e83b94e-a63e-46cd-94cc-1aea1f1291e0>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00740.warc.gz"} |
Dynamics of Hot Bose-Einstein Condensat
SciPost Submission Page
Dynamics of Hot Bose-Einstein Condensates: stochastic Ehrenfest relations for number and energy damping
by Rob G. McDonald, Peter S. Barnett, Fradom Atayee, Ashton S. Bradley
This Submission thread is now published as
Submission summary
Authors (as registered SciPost users): Ashton Bradley
Submission information
Preprint Link: https://arxiv.org/abs/1908.05809v3 (pdf)
Date accepted: 2020-01-13
Date submitted: 2019-12-17 01:00
Submitted by: Bradley, Ashton
Submitted to: SciPost Physics
Ontological classification
Academic field: Physics
• Atomic, Molecular and Optical Physics - Theory
• Mathematical Physics
Specialties: • Quantum Physics
• Statistical and Soft Matter Physics
Approach: Theoretical
Describing partially-condensed Bose gases poses a long-standing theoretical challenge. We present exact stochastic Ehrenfest relations for the stochastic projected Gross-Pitaevskii equation,
including both number and energy damping mechanisms, and all projector terms that arise from the energy cutoff separating system from reservoir. We test the theory by applying it to the centre of
mass fluctuations of a harmonically trapped prolate system, finding close agreement between c-field simulations and analytical results. The formalism lays the foundation to analytically explore
experimentally accessible hot Bose-Einstein condensates.
Author comments upon resubmission
We first address objection
_3) the Kohn theorem is violated and this shows the theory is not appropriate for real experiments_
There are a number of systems under current and future study for which the theory is applicable. For our purpose it forms a very useful test for the formalism. We have revised the manuscript to
clarify these points.
At the start of section 4 we have made substantial revisions of the text to point out several systems of interest where Kohn’s theorem is not satisfied by the system, and hence the SPGPE reservoir
theory provides a useful approximation. We have also emphasized that the SPGPE has been used to give a first-principles treatment of high temperature non-equilibrium experiments. In particular,
spontaneous vortex formation was described using one fitted parameter (ref [3] of revised manuscript), and the SPGPE gave a quantitative description of the experiment in ref [17] of the revised
manuscript, with no fitted parameters:
"While the time-independent reservoir approximation (TIRA) is not strictly applicable for scalar BEC in the purely harmonic trap, there are a number of physical systems where it is applicable: Kohn’s
theorem does not apply to a scalar Bose gas held in a harmonic trap if the trap becomes non harmonic at high energy. The theorem is also inapplicable for a harmonically trapped system if the
reservoir consists of second atomic species confined by a different trapping potential, as may occur during sympathetic cooling. Furthermore, any system that is not harmonically trapped will not obey
Kohn’s theorem, and is thus potentially amenable to the TIRA. Example systems in non-harmonic traps for which the theory is applicable include vortex decay in hard-wall confinement [46], soliton
decay in a 1D toroidal trap (where the present approach was first used) [19], and persistent current formation in a 3D toroidal trap, where SPGPE simulations [17] compare well with experiment [44].
In this work our approach is simply to test the formalism on a simple model system within the TIRA by integrating out the spatial degrees of freedom to find effective stochastic equations of motion
for the centre of mass. We stress that the TIRA approached used here is physically valid for (at least) two scenarios if immediate interest: non-harmonic trapping at high energies for a scalar BEC,
and sympathetic cooling involving two BEC components in different harmonic traps [14]."
These points of context are also mentioned briefly in footnote 3 (page 14) of the revised manuscript.
We have also substantially revised our conclusions to further clarify the physical applicability of the theory of the harmonically trapped system to real physical systems. In particular:
“We tested our stochastic Ehrenfest equations in two ways. Considering the centre of mass motion of a finite-temperature quasi-1D condensate near equilibrium, we tracked the size of the largest
projector corrections and saw they are indeed small. We also compared the steady-state correlations of position and momentum to analytic solutions derived by neglecting the projector corrections,
finding excellent agreement. Our chosen test system has the weakness that the centre of mass motion in a purely harmonic trap is not strictly amenable to the reservoir theory due to a violation of
Kohn's theorem. However, our treatment is physically relevant for non-harmonic trapping, multicomponent systems, and other systems of interest that physically violate Kohn's theorem, provided a
low-energy fraction is harmonically trapped. Indeed, since the thermal equilibrium properties involve small excursions from equilibrium, the confining potential is only required to be _locally_
harmonic near the trap minimum and any number of non-harmonic effects may intrude at larger distances. The centre of mass motion thus provides an excellent formal and numerical test of the SERs,
being one of the simplest states of motion to handle analytically. \par We have shown that SERs can be used to obtain analytic equations that agree with numerical solutions of the full SPGPE and
offer some physical insight into the open system dynamics. Future work will explore systems involving analytically tractable excitations such as vortex decay in hard-wall confinement [46], soliton
[39] and phase-slip dynamics [40] in toroidal confinement, sympathetic cooling [41,42], spinor BECs [23], and quantum turbulence in non-harmonic confinement [44-47].”
We hope that these changes are appropriate to address the remaining concerns regarding physical applicability.
_2) the equilibrium is not properly discusses and if it is the classical (field) equilibrium that I think is the case here, then it is actually not appropriate for the normal state of the gas, which
the authors are discussing_
We find this comment unclear. We do not discuss the normal state of the gas directly, as our theory is a stochastic field theory of the low-energy partially degenerate fraction of the gas. There is a
classical field equilibrium (within the truncated Wigner classical field approximation), that contains the condensate and a low energy normal fraction, but the latter must be extracted numerically
(typically via Penrose-Onsager).
A high-energy normal fraction of the gas appears explicitly via the reservoir, due to the choice of the energy cutoff - it must be chosen to separate the coherent region from the incoherent region of
phase space.
The normal fraction of the gas is thus included in the properties of the reservoir (gaussian statistics, chemical potential, temperature, and reservoir interaction rates), formulated using the
single-particle Wigner function for the high energy part of the field.
In short, we are not sure we understand the question, but we would be happy to provide an answer to a clarified question.
With Best Regards,
Ashton Bradley (on behalf of the authors).
List of changes
- Introduced acronym Stochastic Ehrenfest relation (SER)
- Minor change of wording in second to last paragraph of introduction:
“As a test we apply the Ehrenfest relations to the centre of mass fluctuations of a harmonically trapped system tightly confined along two spatial dimensions. We find that the analtyic solution of
the SER for the centre of mass is in close agreement with SPGPE simulations. ”
- Replaced center —> centre throughout
- Reordered the wording in the “v) Thermal equilibrium” paragraph at the end of section 3 on page 13, to improve clarity
- Introduced acronym TIRA (Time Independent Reservoir Approximation)
- Start of section 4, revised text:
"While the time-independent reservoir approximation (TIRA) is not strictly applicable for scalar BEC in the purely harmonic trap, there are a number of physical systems where it is applicable: Kohn’s
theorem does not apply to a scalar Bose gas held in a harmonic trap if the trap becomes non harmonic at high energy. The theorem is also inapplicable for a harmonically trapped system if the
reservoir consists of second atomic species confined by a different trapping potential, as may occur during sympathetic cooling. Furthermore, any system that is not harmonically trapped will not obey
Kohn’s theorem, and is thus potentially amenable to the TIRA. Example systems in non-harmonic traps for which the theory is applicable include vortex decay in hard-wall confinement [46], soliton
decay in a 1D toroidal trap (where the present approach was first used) [19], and persistent current formation in a 3D toroidal trap, where SPGPE simulations [17] compare well with experiment [44].
In this work our approach is simply to test the formalism on a simple model system within the TIRA by integrating out the spatial degrees of freedom to find effective stochastic equations of motion
for the centre of mass. We stress that the TIRA approached used here is physically valid for (at least) two scenarios if immediate interest: non-harmonic trapping at high energies for a scalar BEC,
and sympathetic cooling involving two BEC components in different harmonic traps [14]."
These points of context are also mentioned briefly in footnote 3 (page 14) of the revised manuscript.
- Conclusions revised:
“We tested our stochastic Ehrenfest equations in two ways. Considering the centre of mass motion of a finite-temperature quasi-1D condensate near equilibrium, we tracked the size of the largest
projector corrections and saw they are indeed small. We also compared the steady-state correlations of position and momentum to analytic solutions derived by neglecting the projector corrections,
finding excellent agreement. Our chosen test system has the weakness that the centre of mass motion in a purely harmonic trap is not strictly amenable to the reservoir theory due to a violation of
Kohn's theorem. However, our treatment is physically relevant for non-harmonic trapping, multicomponent systems, and other systems of interest that physically violate Kohn's theorem, provided a
low-energy fraction is harmonically trapped. Indeed, since the thermal equilibrium properties involve small excursions from equilibrium, the confining potential is only required to be _locally_
harmonic near the trap minimum and any number of non-harmonic effects may intrude at larger distances. The centre of mass motion thus provides an excellent formal and numerical test of the SERs,
being one of the simplest states of motion to handle analytically. \par We have shown that SERs can be used to obtain analytic equations that agree with numerical solutions of the full SPGPE and
offer some physical insight into the open system dynamics. Future work will explore systems involving analytically tractable excitations such as vortex decay in hard-wall confinement [46], soliton
[39] and phase-slip dynamics [40] in toroidal confinement, sympathetic cooling [41,42], spinor BECs [23], and quantum turbulence in non-harmonic confinement [44-47].”
Published as SciPost Phys. 8, 029 (2020) | {"url":"https://www.scipost.org/submissions/1908.05809v3/","timestamp":"2024-11-10T18:47:20Z","content_type":"text/html","content_length":"40528","record_id":"<urn:uuid:eb78fd86-c1a0-4007-a549-3e05dde9abb1>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00413.warc.gz"} |
a high school randomly selected 75 of the 200 seniors
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NCERT Solutions | Class 9 Maths Chapter 13 Surface Areas And Volumes | NCERTBOOKSPDF.COM
NCERT Solutions | Class 9 Maths Chapter 13 Surface Areas and Volumes
NCERT Class 9 Maths Solutions PDF: In this post, we have discussed the solution of the Maths class 9 book which is followed in all CBSE schools. Solutions are given below with proper Explanation and
utmost care has been taken to ensure that the solutions are correct. Answers provided will not only help in completing all the assignments but also help students in clearing their concepts. Students
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Exercise 13.3 Class 9 Maths Chapter 13 Surface Areas and Volumes
1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
3. Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find
(i) radius of the base and (ii) total surface area of the cone.
4. A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is ₹ 70.
5. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and
wastage in cutting is approximately 20 cm (Use π = 3.14).
6. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of ₹ 210 per 100 m2.
7. A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
8. A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each
of the cones is to be painted and the cost of painting is ₹ 12 per m2, what will be the cost of painting all these cones? (Use π = 3.14 and take √1.04 = 1.02)
1 thought on “NCERT Solutions | Class 9 Maths Chapter 13 Surface Areas and Volumes” | {"url":"https://ncertbookspdf.com/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-2/","timestamp":"2024-11-08T02:47:56Z","content_type":"text/html","content_length":"82547","record_id":"<urn:uuid:cb49bd8d-eba2-46b4-8a17-c5b8858a9943>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00146.warc.gz"} |
Excel Formula Python: Count Players with More Deposits than Last Month
In this tutorial, we will learn how to write an Excel formula in Python to count the number of players who have more deposits than last month. This formula utilizes the COUNTIF function in Google
Sheets to compare the values in the current month's deposits column with the values in the last month's deposits column. By applying the criteria ">"&B2, where B2 refers to the cell containing the
last month's deposit value, the COUNTIF function counts the instances where the current month's deposits are greater than the last month's deposits.
To implement this formula in Python, we can use the xlwings library, which allows us to interact with Excel files using Python. We will first open the Excel file and select the appropriate worksheet.
Then, we can use the xlwings Range object to access the columns containing the deposit values. We will iterate through the cells in the current month's deposits column and compare each value with the
corresponding value in the last month's deposits column. If the current month's deposit value is greater than the last month's deposit value, we will increment a counter variable. Finally, we will
return the value of the counter variable, which represents the number of players with more deposits than last month.
Let's consider an example to understand this formula better. Suppose we have a dataset with two columns: column B represents the last month's deposits, and column C represents the current month's
deposits. If we have the following values in columns B and C:
B C
The formula =COUNTIF(C:C, ">"&B2) would return 3, indicating that there are 3 players who have more deposits this month compared to last month.
In conclusion, by using the Excel formula =COUNTIF(C:C, ">"&B2) in Python, we can easily count the number of players who have more deposits than last month. This formula provides a convenient way to
analyze and track changes in deposit amounts over time.
Brief explanation
This formula uses the COUNTIF function in Google Sheets to count the number of players who have more deposits than last month. It compares the values in column C (current month deposits) with the
values in column B (last month deposits) and counts the instances where the current month deposits are greater than the last month deposits.
Step-by-step explanation
1. The COUNTIF function is applied to column C, which contains the current month deposits.
2. The criteria for COUNTIF is set as ">"&B2, where B2 refers to the cell containing the last month's deposit value.
3. The COUNTIF function then counts the number of cells in column C that meet the condition of having a deposit value greater than the last month's deposit value.
For example, if we have the following data in columns B and C:
| B | C |
| 100 | 120 |
| 150 | 90 |
| 200 | 210 |
| 180 | 160 |
| 120 | 130 |
The formula =COUNTIF(C:C, ">"&B2) would return 3, indicating that there are 3 players who have more deposits this month compared to last month. | {"url":"https://codepal.ai/excel-formula-generator/query/977kTznN/excel-formula-python-number-players-more-deposits-last-month","timestamp":"2024-11-03T20:09:26Z","content_type":"text/html","content_length":"97045","record_id":"<urn:uuid:c1747daf-5a22-4da0-bdb1-28407e30c87e>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00120.warc.gz"} |
C2.1 Unsymmetric Bending
• It is imperative that you follow the sign convention (given below) for the formula to work.
• z-axis must always be 90^o CCW from the y-axis.
• Moment direction must follow the right-hand rule.
• y and z are the distances of the point of interest from the centroid, along the y-axis and z-axis respectively.
• I[y] and I[z] are the moments of inertia about the y-axis and z-axis respectively.
• M[y] and M[z] are positive according to the right-hand rule, with the thumb pointing to the +ve y-axis and +ve z-axis respectively.
Due to two-directions of moments being applied, the neutral-axis (line of zero stress) tilts from the horizontal axis:
The orientation of the neutral-axis from the horizontal is set as ‘α’, and to get the formula for α we set σ[b] = 0 (since the bending stress is zero at the neutral-axis).
Sometimes the moment is given in terms of its magnitude and direction. In that case:
Let’s look at an example now.
• It is imperative that you follow the sign convention (given below) for the formula to work.
• z-axis must always be 90^o CCW from the y-axis.
• Moment direction must follow the right-hand rule.
• y and z are the distances of the point of interest from the centroid, along the y-axis and z-axis respectively.
• I[y] and I[z] are the moments of inertia about the y-axis and z-axis respectively.
• M[y] and M[z] are positive according to the right-hand rule, with the thumb pointing to the +ve y-axis and +ve z-axis respectively.
Due to two-directions of moments being applied, the neutral-axis (line of zero stress) tilts from the horizontal axis:
The orientation of the neutral-axis from the horizontal is set as ‘α’, and to get the formula for α we set σ[b] = 0 (since the bending stress is zero at the neutral-axis).
Sometimes the moment is given in terms of its magnitude and direction. In that case:
Let’s look at an example now. | {"url":"http://www.engineeringcorecourses.com/solidmechanics2/C2-bending/C2.1-unsymmetric-bending/theory/","timestamp":"2024-11-05T19:34:19Z","content_type":"text/html","content_length":"22145","record_id":"<urn:uuid:47115fc0-62a8-4121-a6fb-fa29f2f5adae>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00043.warc.gz"} |
Generating random numbers with a different probabilities
Answered: Steven Lord on 1 Dec 2021
If I want to generate random numbers between 0 and 150, however i want the probability of having numbers between 0 and 50 to be higher, how can i do it?
2 Comments
36 views (last 30 days)
Generating random numbers with a different probabilities
I see essentially this same question so often. The problem is, it is not enough to say that you want higher probabilities for some values. That is too vague a question to have an answer, because
there are infinitely many distributions that will satisfy your vaguely stated goal. Worse, you have not even stated if you need integer results, or if real numbers are desired.
Answers (3)
Hi there. These codes below are completely original and made by me for you. Each time you need to enter a probability value to the system. When the probability is closer to 1, the system gives more
digits from 0-50 range. I hope it works for you. Good luck.
clc; clear; close all;
choice = menu('Choose the case','probability=1','probability=0.7','probability=0.5','probability=0.3');
if choice==1
r = randi([0 50],1,125);
k = randi([50 150],1,25);
elseif choice==2
r = randi([0 50],1,100);
k = randi([50 150],1,50);
r = randi([0 50],1,75);
k = randi([50 150],1,75);
l=[r k];
2 Comments
Yusuf Suer Erdem on 1 Dec 2021
I never experienced that.
Assuming there are just two levels of probability, and that the numbers are real, not just integers, you could try:
p50 = 0.75; % probability of number less than 50
N = 10^5; % number of random numbers required
u = rand(N,1); % uniform random numbers
r(u<=p50) = u(u<=p50)*50; % random numbers less than 50
r(u>p50) = u(u>p50)*100 + 50; % random numbers between 50 and 150
1 Comment
If you know the probabilities you want each number to have you could use discretize. For instance if I want to generate numbers between 1 and 10 with the odd numbers being twice as likely:
P = repmat([2 1], 1, 5)
cumulativeP = [0 cumsum(P)./sum(P)]
r = rand(1, 1e5); % Random numbers in range (0, 1)
d = discretize(r, cumulativeP); % Bin the random numbers in r using the bins in cumulativeP
h = histogram(d, (1:11)-0.5, 'Normalization', 'probability'); % Show the results
The bars for 1, 3, 5, 7, and 9 are about twice as tall as the bins for 2, 4, 6, 8, and 10 as expected.
shouldBeCloseToP = h.Values./h.Values(end)
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Graphing Polynomial Functions
Learning Outcomes
• Draw the graph of a polynomial function using end behavior, turning points, intercepts, and the Intermediate Value Theorem.
We can use what we have learned about multiplicities, end behavior, and turning points to sketch graphs of polynomial functions. Let us put this all together and look at the steps required to graph
polynomial functions.
How To: Given a polynomial function, sketch the graph
1. Find the intercepts.
2. Check for symmetry. If the function is an even function, its graph is symmetric with respect to the y-axis, that is, f(–x) = f(x).
If a function is an odd function, its graph is symmetric with respect to the origin, that is, f(–x) = –f(x).
3. Use the multiplicities of the zeros to determine the behavior of the polynomial at the x-intercepts.
4. Determine the end behavior by examining the leading term.
5. Use the end behavior and the behavior at the intercepts to sketch the graph.
6. Ensure that the number of turning points does not exceed one less than the degree of the polynomial.
7. Optionally, use technology to check the graph.
Example: Sketching the Graph of a Polynomial Function
Sketch a possible graph for [latex]f\left(x\right)=-2{\left(x+3\right)}^{2}\left(x - 5\right)[/latex].
Show Solution
Try It
Sketch a possible graph for [latex]f\left(x\right)=\frac{1}{4}x{\left(x - 1\right)}^{4}{\left(x+3\right)}^{3}[/latex].
Check yourself with an online graphing tool when you are done.
Show Solution
Try it
Use an online graphing tool to find an odd degree function with one zero at (-3,0) whose multiplicity is 3 and another zero at (2,0) with multiplicity 2. The end behavior of the graph is:
as [latex]x\rightarrow-\infty, f(x) \rightarrow\infty[/latex] and as [latex]x\rightarrow \infty, f(x)\rightarrow -\infty[/latex]
The Intermediate Value Theorem
In some situations, we may know two points on a graph but not the zeros. If those two points are on opposite sides of the x-axis, we can confirm that there is a zero between them. Consider a
polynomial function f whose graph is smooth and continuous. The Intermediate Value Theorem states that for two numbers a and b in the domain of f, if a < b and [latex]f\left(a\right)\ne f\left(b\
right)[/latex], then the function f takes on every value between [latex]f\left(a\right)[/latex] and [latex]f\left(b\right)[/latex].
We can apply this theorem to a special case that is useful for graphing polynomial functions. If a point on the graph of a continuous function f at [latex]x=a[/latex] lies above the x-axis and
another point at [latex]x=b[/latex] lies below the x-axis, there must exist a third point between [latex]x=a[/latex] and [latex]x=b[/latex] where the graph crosses the x-axis. Call this point [latex]
\left(c,\text{ }f\left(c\right)\right)[/latex]. This means that we are assured there is a value c where [latex]f\left(c\right)=0[/latex].
In other words, the Intermediate Value Theorem tells us that when a polynomial function changes from a negative value to a positive value, the function must cross the x-axis. The figure below shows
that there is a zero between a and b.
A General Note: Intermediate Value Theorem
Let f be a polynomial function. The Intermediate Value Theorem states that if [latex]f\left(a\right)[/latex] and [latex]f\left(b\right)[/latex] have opposite signs, then there exists at least one
value c between a and b for which [latex]f\left(c\right)=0[/latex].
Example: Using the Intermediate Value Theorem
Show that the function [latex]f\left(x\right)={x}^{3}-5{x}^{2}+3x+6[/latex] has at least two real zeros between [latex]x=1[/latex] and [latex]x=4[/latex].
Show Solution
Try It
Show that the function [latex]f\left(x\right)=7{x}^{5}-9{x}^{4}-{x}^{2}[/latex] has at least one real zero between [latex]x=1[/latex] and [latex]x=2[/latex].
Show Solution
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Hooke's Law - BrainDuniya
What is Hooke’s Law?
By experiments on behavior of a metal wire under an axial pulling force, an English physicist “Robert Hooke” (1635-1703 A.D.), formulated a law which is popularly known as Hooke’s law. Hooke’s law
states that –
The elongation produced in a wire under axial pull is directly proportional to the applied pull.
Hooke’s Law and Young’s theory
Later, another scientist, “Thomas Young” stated that –
Strain produced in the wire is directly proportional to the stress developed in the wire.
Therefore, Young explained about the limitations of Hooke’s law. He modified the Hooke’s law to the more general form as follows –
1. Materials show elastic property only up to a certain limit of developed stress. Beyond this limit permanent deformation set up in the material body.
2. The maximum stress within which a material body regains its original size and shape after the removal of deforming force, is called elastic limit.
3. Within elastic limit, elongation per unit length i.e. strain produced in a wire is directly proportional to the applied pull per unit cross sectional area i.e. stress.
Therefore, within elastic limit –
\text {Stress} \ \propto \ \text {Strain}
So, \quad \left ( \frac {\text {Stress}}{\text {Strain}} \right ) = \text {Constant}
This constant of proportionality is called modulus of elasticity or coefficient of elasticity of the material.
Load – Extension Curve
120201 HOOKE’S LAW AND LOAD-EXTENSION CURVE
Robert Hooke use a device called “Extensometer” to do his experiment for measuring the extension of a wire under a tensile load.
Consider about the elastic curve of an ductile material as shown in figure. In this experiment, a wire made of an elastic material is hanged in the device and loaded under the axial tensile load. The
load is gradually increases and the extensions in the wire are recorded for each loading.
The values of applied tensile load are plotted along X axis and the resulted elongation is plotted along Y axis. The obtained curve is called “Load – Extension Curve”.
Hooke noted that, when the applied load is within certain limit, the obtained curve is a straight line passing through the origin. So, the extension in the length of wire is proportional to the load
This represents the correctness of Hooke’s law.
If a body gets deformed under the action of an external force, then at each section of the body an internal restoring force of reaction is set up which tends to restore the body into its original
The internal restoring force set up in the deformed body per unit area of cross-section is called stress.
The restoring force is equal but opposite to the external deforming force. Therefore, stress is expressed as –
\text {Stress} = \frac {\text {Applied force}}{\text {Area}}
The SI unit of stress is ( \text {N-m}^{-2} ) and the CGS unit is ( \text {dyne-cm}^{-2} ) .
Types of Stress
Different types of stresses developed in a deformed body are –
Longitudinal Stress
It is defined as the restoring force set up per unit cross-sectional area of a body when the length of the body changes in the direction of the deforming force.
If ( F_{axial} ) is the applied external force on a rigid body having area of cross section ( A )
Then, longitudinal stress is given by –
f = \left ( \frac {F_{axial}}{A} \right )
Longitudinal stress is of two types –
1. Tensile stress – It is the restoring force set up per unit cross-sectional area of a body when its length increases under a deforming force.
Tangential or Shearing Stress
When a deforming force acts tangentially to the surface of a body, it produces a change in the shape of the body.
If ( F_{tangential} ) is the applied external force on a rigid body tangentially over a surface area ( A )
Then, shearing stress is given by –
f = \left ( \frac {F_{tangential}}{A} \right )
The ratio of the change produced in any dimension of a rigid body to the original dimension is called strain.
Therefore, strain is given by –
Strain = \left ( \frac {\text {Change in dimension}}{\text {Original dimension}} \right )
Since, strain is a ratio of two similar quantities, so it is dimensionless and has no units.
Types of Strain
Different types of strain developed in a body are –
Longitudinal Strain
It is defined as the ratio of increase or decrease in length and original length.
Therefore, \quad \text {Longitudinal strain} = \left ( \frac {\text {Change in length}}{\text {Original length}} \right ) = \left ( \frac {\Delta l}{l} \right )
Volumetric Strain
It is defined as the ratio of change in volume and original volume.
Therefore, \quad \text {Volumetric strain} = \left ( \frac {\text {Change in volume}}{\text {Original volume}} \right ) = \left ( \frac {\Delta V}{V} \right )
Shear Strain
It is defined as the tangent of the angle θ (in radian), through which a face of body which was originally perpendicular to the fixed surface gets deformed.
Therefore, \quad \text {Shear strain} = \tan \theta = \left ( \frac {\text {Relative distance between two parallel planes}}{\text {Distance between parallel planes}} \right )
Stress – Strain curve & Hooke’s Law
Consider about the following figure. The figure shows a typical strain – strain curve relating to Hooke’s law of a ductile metal wire which is loaded by a gradually increasing axial load. This curve
is also known as elastic curve.
It has six distinct regions as shown in figure.
120202 HOOKE’S LAW AND YOUNG’S THEORY
(1) REGION OA –
The initial part OA of the graph is a straight line indicating that stress is proportional to strain. Upto the point A , Hooke’s law is obeyed. The point A is called the proportional limit. In this
region the wire is perfectly elastic.
(2) REGION AB –
After the point A , the stress is not proportional to the strain. However, if the external load is removed at any point between O and B , the curve is retraced along BAO and the wire attains its
original length.
The portion OB of the graph is called elastic region and the point B is called elastic limit or yield point. The stress corresponding to the yield point is called yield strength ( S_y )
Up to the point B , the elastic forces of the material are conservative forces i.e. when the load is removed and the material returns to its original size, then the work done in producing the
deformation is completely recovered.
(3) REGION BC –
Beyond the point B , the strain increases more rapidly than stress. If the load is removed at any point C , the wire does not return to its original length but it traces dashed line CE . Even on
reducing the stress to zero, a residual strain equal to OE is left in the wire. Thus the material is said to have acquired a permanent set.
The fact that the stress-strain curve is not retraced on reversing the strain is called elastic hysteresis.
(4) REGION CD –
If the load is increased beyond the point C , there is a large increase in the strain i.e. the length of the wire increases tremendously. In this region a neck is develop at some point D of the wire
and the wire ultimately breaks at this point. The point D is called the fracture point.
(5) REGION OB –
In the region from O to B , the material regains its original shape and size upon removal of external applied load. Hence this region is called elastic region of the curve.
(6) REGION BD –
In the region between B and D , the material gets permanent deformation. This region is called plastic region and the material is said to undergo plastic flow or plastic deformation.
The stress corresponding to the breaking point D is called ultimate strength or tensile strength of the material.
Types of materials based on elastic curve
On the basis of strain – strain curve, materials are classified into following types –
Ductile material
Materials which have large plastic range i.e. region between B and D of strain – strain curve, are called ductile materials.
As shown in the stress-strain curve, the fracture point is widely separated from the elastic limit point B . Such materials undergo an irreversible increase in length before breaking takes place at
point D . So, these materials can be drawn into thin wires.
Example –
Copper, Silver, Iron, Steel, Aluminium, etc.
The property by which any material can be drawn into thin wires by the application of tensile load is called ductility.
Brittle material
120203 STRESS-STRAIN CURVE FOR A BRITTLE MATERIAL
The materials which have very small plastic range i.e. short region between B to D of strain – strain curve are called brittle materials. Such materials break as soon as the stress is increased
beyond the elastic limit. Their breaking point D lies just close to their elastic limit point B as shown in figure.
Example –
Cast iron, Glass, Ceramics, etc.
The property by which any material breaks or fail to withstand as soon as the the stress increases above the elastic limit is called brittleness.
Malleable material
120204 LOAD-COMPRESSION CURVE FOR A MALLEABLE METAL
Stress – strain behavior of solid materials are different in case of tensile load and compressive load. Some materials show good plastic behavior under the action of compressive load.
When a material is compressed, a stage is reached beyond which it cannot recover its original shape after the deforming force is removed. This is called the elastic limit point A' for compression as
shown in figure. The solid then behaves like a plastic body. The yield point B' obtained under compression is called crushing point.
Between the points A' and B' , metals are said to be malleable i.e., they can be hammered or rolled into thin sheets.
Example –
Gold, Silver, Lead etc.
The property by which any material can be hammered and flattened into thin sheets under the action of compressive load is called malleability.
120205 STRESS-STRAIN CURVE FOR ELASTOMERS
The materials which can be elastically stretched to a large values of strain are called elastomers. These materials has a large range of elastic region but no well defined plastic region.
Elastomers do not obey Hooke’s law. Their Young’s modulus is also very small.
Example –
• Rubber can be stretched to several times of its original length but still it can regain its original length when the applied force is removed. It just breaks when pulled beyond a certain limit.
• In our body, the elastic tissue of aorta is an elastomer.
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Calorie per Kilogram
Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like latent heat finds its
use in a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps
in the conversion of different units of measurement like cal/kg to mJ/g through multiplicative conversion factors. When you are converting latent heat, you need a Calorie per Kilogram to Millijoule
per Gram converter that is elaborate and still easy to use. Converting cal/kg to Millijoule per Gram is easy, for you only have to select the units first and the value you want to convert. If you
encounter any issues to convert Calorie per Kilogram to mJ/g, this tool is the answer that gives you the exact conversion of units. You can also get the formula used in cal/kg to mJ/g conversion
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How To Calculate Pay Per Hour - Certified Calculator
How To Calculate Pay Per Hour
Determining your hourly rate is crucial for understanding your earnings and negotiating compensation. This calculator simplifies the process by allowing you to input your total pay and the hours
worked, providing you with an accurate hourly rate.
The formula for calculating the hourly rate is derived from the relationship between total pay, hours worked, and the hourly rate:
Hourly Rate=Total PayHours WorkedHourly Rate=Hours WorkedTotal Pay
How to Use
1. Enter your total pay in dollars.
2. Input the number of hours you’ve worked.
3. Click the “Calculate” button to find your hourly rate.
Suppose your total pay is $500, and you’ve worked 25 hours. Using the calculator, your hourly rate would be $20 per hour (\frac{$500}{25 \text{ hours}}).
1. Q: Why is it essential to know my hourly rate? A: Knowing your hourly rate helps you understand your earnings per hour, aiding in budgeting, financial planning, and negotiating fair compensation.
2. Q: Can I use this calculator for salaried positions? A: This calculator is designed for hourly positions. For salaried positions, you can calculate the hourly rate by dividing the annual salary
by the number of work hours in a year.
3. Q: What if I have different rates for different tasks? A: This calculator provides an average hourly rate. If you have varying rates, you may need to calculate the hourly rate for each task
Use this calculator to quickly determine your hourly rate based on your total pay and hours worked. Regularly assessing your hourly rate empowers you to make informed decisions about your work and
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Scheduling Model in LLVM - Part II
In the previous post, we covered the basics of scheduling model in LLVM. Specifically, per-operand tokens that connect an instruction with models that spell out processor-specific scheduling
properties like instruction latency, and the concept of processor resources with different sizes of buffer.
While I was planning to write how scheduling models are used in this post – namely, covering things like instruction scheduler and MCA – the draft was overwhelmed by the sheer amount of content
needed to cover just the substrate. In addition, I found that I missed some more advanced yet commonly used constructions in the previous post. So if you’ll excuse me, I’d like to procrastinate
writing about MachineScheduler and MCA, leaving it for future Min to worry, and dive into three important scheduling model constructions in this post: number of ProcResource units, ProcResGroup, and
super resource.
These three horsemen together enable scheduling models to express hierarchy structure – a concept that we have only scratched the surface previously. Modern microarchitectures often employ
complicated processor resource distributions and groupings, like having multiple execution pipes with asymmetric capabilities. It is of paramount importance to express those structures with the
things we’re about to cover. Without further ado, let’s start with the number of ProcResource units!
Number of units in a ProcResource
So far we’ve mentioned things like ProcResource<1> or ProcResource<2> several times without explaining the numbers in the template argument list. That specific argument stands for the number of units
in a processor resource.
This property is directly related to the throughput of this resource, namely, how many uops can it process in a given time. To give you a more concrete example, let’s see how LLVM’s scheduling model
calculates the reciprocal throughput – a synonym of inverse throughput – of an instruction.
1std::optional<double> Throughput;
2const MCSchedModel &SM = STI.getSchedModel();
3const MCWriteProcResEntry *I = STI.getWriteProcResBegin(&SCDesc);
4const MCWriteProcResEntry *E = STI.getWriteProcResEnd(&SCDesc);
5for (; I != E; ++I) {
6 if (!I->ReleaseAtCycle)
7 continue;
8 unsigned NumUnits = SM.getProcResource(I->ProcResourceIdx)->NumUnits;
9 double Temp = NumUnits * 1.0 / I->ReleaseAtCycle;
10 Throughput = Throughput ? std::min(*Throughput, Temp) : Temp;
12if (Throughput)
13 return 1.0 / *Throughput;
The code above is excerpted from MCSchedModel::getReciprocalThroughput: it scans through every write resources in this instruction (represented by its scheduling class, SCDesc) via each resource’s
index ProcResourceIdx. The throughput contributed by each resource used by this instruction is calculated by dividing the number of units (NumUnits) by ReleaseAtCycle, which is the number of cycles
reserved on this resource. We eventually take the largest inverse throughput (i.e. smallest throughput) among all the resources as the overall throughput of this instruction.
A single ProcResource with number of unit larger than one is equivalent to multiple identical ProcResource instances. For example, let’s say we have the following scheudling model:
1def IEX : ProcResource<3>;
3def : WriteRes<WriteIMul, [IEX]>;
In this model, we assign IEX (integer execution pipes) to WriteIMul – a SchedWrite token that represents integer multiplication instructions. This is equivalent to having three individual integer
pipes – IEX0, IEX1, and IEX2, where any of them can do multiplications:
1def IEX0 : ProcResource<1>;
2def IEX1 : ProcResource<1>;
3def IEX2 : ProcResource<1>;
Having (effectively) three available pipes also means that we can dispatch three multiplications in parallel! Take the following RISC-V assembly snippet as an example, assuming we’re dispatching them
into this model with an issue width of 6. Since there is no Read-After-Write (RAW) dependencies among the instructions, we can dispatch them in parallel.
mul a1, a1, a2
mul t4, t4, t5
mul r7, r7, r0
What we’re interested in here, is how each of them consumes processor resources. We can visualize this process with the following resource consumption table:
instruction IEX0 IEX1 IEX2
mul a1, a1, a2 Consumed Available Available
mul t4, t4, t5 Consumed Available Consumed
mul t0, t0, t1 Consumed Consumed Consumed
The instructions are dispatched from top to bottom. For each of the instruction, we randomly^1 look for an empty pipe to dispatch it into.
Alternatiely, we can rewrite this table into a more compact format:
instruction Consumed IEX units
mul a1, a1, a2 1 / 3
mul t4, t4, t5 2 / 3
mul t0, t0, t1 3 / 3
In this table, we focus on the number of consumed units in def IEX : ProcResource<3>, where 2 / 3 means “two out of three total units are consumed”. This table will come into handy later when we’re
discussing more advanced scheduling model concepts.
But for now, let’s step back for a second: if dispatching to ProcResource<3> is equivalent to doing the same thing against three individual ProcResource<1> where we can dispatch an instruction to any
of them…
Haven’t we seen something similar in the previous post already?
That’s right! It’s ProcResGroup. This is what we have after rewriting the same model with ProcResGroup:
1def IEX0 : ProcResource<1>;
2def IEX1 : ProcResource<1>;
3def IEX2 : ProcResource<1>;
5def IEX : ProcResGroup<[IEX0, IEX1, IEX2]>;
7def : WriteRes<WriteIMul, [IEX]>;
Both models express the fact that multiplication instructions can run on any of the three integer pipes.
But then it prompts a question: if they’re so similar, why do we have two different syntax in the first place?
The key, as it turns out, is the fact that we were dealing with three identical pipes in the previous example. In reality, we might not always have execution units with the same capabilities. For
example, here is a more realistic design:
In this design, only two out of three pipes are capable of doing multiplications; divisions and cryptographies, on the other hand, can only run on one of the pipes.
The rationale behind this design is that complex operations like division or cryptography usually take up a larger chip area and draw more power, while being less commonly used. So it’s pretty common
to have a heterogeneous layout where certain operations are only available in a subset of execution units.
With only a single def IEX : ProcResource<3>, it’ll be more difficult to express the resources used by each kind of instructions because currently there is no way to say something like “WriteIDiv
uses the second unit of IEX”:
1def IEX : ProcResource<3>
3// Simple arithmetics, like ADD
4def : WriteRes<WriteIALU, [IEX]>;
5// Multiplication
6def : WriteRes<WriteIMul, [/*IEX[0] and IEX[2]??*/]>;
7// Division
8def : WriteRes<WriteIDiv, [/*IEX[1]??*/]>;
9// Cryptography
10def : WriteRes<WriteCrypto, [/*IEX[2]??*/]>;
On the contrary, it’s much more straight forward to express it with the ProcResGroup we had introduced in the previous post:
1def IEX0 : ProcResource<1>;
2def IEX1 : ProcResource<1>;
3def IEX2 : ProcResource<1>;
5def IntegerArith : ProcResGroup<[IEX0, IEX1, IEX2]>;
6def IntegerMul : ProcResGroup<[IEX0, IEX2]>;
8// Simple arithmetics, like ADD
9def : WriteRes<WriteIALU, [IntegerArith]>;
10// Multiplication
11def : WriteRes<WriteIMul, [IntegerMul]>;
12// Division
13def : WriteRes<WriteIDiv, [IEX1]>;
14// Cryptography
15def : WriteRes<WriteCrypto, [IEX2]>;
As a quick recap: by consuming ProcResGroup<[IEX0, IEX2]>, a multiplication instruction might run on either IEX0 or IEX2 during runtime.
It is worth pointing out that with this model, we have to deal with resource consumptions that go across different ProcResource and ProcResGroup. For instance, when we dispatch a cryptography
instruction, the instruction not only consumes IEX2 but also effectively decreases the number of available units in IntegerArith and IntegerMul – which is what multiplication consumes – because IEX2
presents in both ProcResGroup.
In order to account for overlapping ProcResource and ProcResGroup, for each ProcResource or ProcResGroup used by an instruction, LLVM actually inserts an implicit processor resource usage for every
ProcResGroup it overlaps. Using the snippet above as an example, this is what it looks like after such “expansion”:
1// Simple arithmetics, like ADD
2def : WriteRes<WriteIALU, [IntegerArith]>;
3// Multiplication
4def : WriteRes<WriteIMul, [IntegerMul, IntegerArith]>;
5// Division
6def : WriteRes<WriteIDiv, [IEX1, IntegerArith]>;
7// Cryptography
8def : WriteRes<WriteCrypto, [IEX2, IntegerArith, IntegerMul]>;
A cryptography now consumes not only IEX2, but also one IntegerArith unit and one IntegerMul unit upon dispatch. So if we dispatch the following RISC-V instruction sequence^2:
mul s0, s0, a2
sha256sum0 a0, a1
mul a3, a3, t0
Here is what happens at cycle 0:
instruction IntegerArith IntegerMul IEX2
mul s0, s0, a2 1 / 3 1 / 2 0 / 1
sha256sum0 a0, a1 2 / 3 2 / 2 1 / 1
mul a3, a3, t0 3 / 3 FAIL TO CONSUME 1 / 1
The first multiplication instruction consumes both IntegerArith and IntegerMul. Because IntegerArith has overlapping resources with IntegerMul – IEX0 and IEX2, to be precise.
Similarly, when it comes to the sha256sum0 instruction, it increases the number of consumed resources on not just IEX2 but IntegerArith and IntegerMul as well. Lastly, for the last multiplication
instruction, its attempt to acquire IntegerMul will fail because we no longer have spare capacity in that resource, which causes the instruction to stall during the dispatch stage, namely, a dispatch
ProcResGroup gives you the ability to reference a subset of execution units, which is suitable for modeling units with heterogeneous capabilities. And as it turns out, there is actually a second way
to reference subsets of execution units – super resource.
Super resource
Super resource allows us to construct a hierarchy between two ProcResource instances (NOT ProcResGroup). In this relationship, the child ProcResource represents a subset of units from the parent
To give you a better idea, let’s see a real-world example from the Load / Store Unit (LSU) in AMD Zen3.
Image source: Chip and Cheese. Captured from the original image .
The diagram above shows the LSU part of Zen3’s microarchitecture. There are three arrows between load & store queues and L1 Data Cache, along with an equal number of AGUs (Address Generation Unit)
positioned above the queues.
You might notice that among all three arrows, which are load and store pipes, between the queues and L1 Data Cache, only two of them goes down (indicating stores) while there are three going up
(indicating loads). This reveals that all three available pipes are capable of loading data, while only two of them (don’t care which two of them though) can store data. Importantly, each pipe can
either load or store data at any given time, but not both simultaneously.
This structure is described by the following code in Zen3’s scheduling model:
1def Zn3LSU : ProcResource<3>;
3let Super = Zn3LSU in
4def Zn3Load : ProcResource<3> {
5 ...
8let Super = Zn3LSU in
9def Zn3Store : ProcResource<2> {
10 ...
Zn3Load and Zn3Store are processor resources representing the load and store pipes, respectively. Both of them designate Zn3LSU – which represents the entire LSU – as their super resource via the
Super field.
By designating Zn3LSU as their super resource, both Zn3Load and Zn3Store are essentially representing a subset of all three pipes from Zn3LSU – 2 pipes for Zn3Store and 3 for Zn3Load, coinciding with
what we saw from Zen3’s microarchitecture diagram earlier. Put it differently, a unit from Zn3LSU can either be allocated as a load or a store pipe, while no more than two store pipes are allowed to
exist at any given time.
LLVM implements super resource in a really similar way to how it implements ProcResGroup – by expanding ProcResource that has super resources. Let me explain this using the snippet below which shows
some Zn3Load and Zn3Store usages.
1def Zn3LSU : ProcResource<3>;
3let Super = Zn3LSU in
4def Zn3Load : ProcResource<3>;
6let Super = Zn3LSU in
7def Zn3Store : ProcResource<2>;
9// Loads, stores, and moves, not folded with other operations.
10defm : Zn3WriteResInt<WriteLoad, [Zn3AGU012, Zn3Load], ...>;
11defm : Zn3WriteResInt<WriteStore, [Zn3AGU012, Zn3Store], ...>;
In this snippet, WriteLoad – the SchedWrite for some of the X86 load instructions – uses Zn3AGU012 and Zn3Load while WriteStore – the SchedRead for some of the X86 store instrutions – has a similar
resource usage of Zn3AGU012 and Zn3Store.
LLVM effectively expands the Zn3Load and Zn3Store usages in these two SchedWrite entries into:
1defm : Zn3WriteResInt<WriteLoad, [Zn3AGU012, Zn3Load, Zn3LSU], ...>;
2defm : Zn3WriteResInt<WriteStore, [Zn3AGU012, Zn3Store, Zn3LSU], ...>;
That’s right! Similar to how ProcResGroup implicitly inserts resource usages of overlapping ProcResGroup, LLVM also implicitly inserts resource usages of super resource, Zn3LSU, into the list.
With the following sequence of X86 load and store instructions:
movq %r9, (%rbx) # store
movq 4(%r8), %rax # load
movq %r10, (%rcx) # store
They’ll have the following resource consumptions upon dispatch (Zn3AGU012 is omitted from this table for simplicity):
instruction Zn3Load Zn3Store Zn3LSU
movq %r9, (%rbx) 0 / 3 1 / 2 1 / 3
movq 4(%r8), %rax 1 / 3 1 / 2 2 / 3
movq %r10, (%rcx) 1 / 3 2 / 2 3 / 3
Whenever a store (e.g. movq %r9, (%rbx)) is being dispatched, it increases the counters of both Zn3Store and Zn3LSU. Similarly, a load instruction increases both Zn3Load and Zn3LSU counters.
Let’s use the following consecutive store instructions to show how we throttle the numebr of store pipes to 2:
movq %r9, (%rbx) # store
movq %rax, (%r8) # store
movq %r10, (%rcx) # store
This snippet produces the following resource consumption table:
instruction Zn3Load Zn3Store Zn3LSU
movq %r9, (%rbx) 0 / 3 1 / 2 1 / 3
movq %rax, (%r8) 0 / 3 2 / 2 2 / 3
movq %r10, (%rcx) 0 / 3 FAIL TO CONSUME 3 / 3
The last instruction fails to consume Zn3Store, because it only has a total of 2 units. In other words, the last instruction in this case is throttled by Zn3Store, despite the fact that there are
enought number of LSU pipes.
And that, is how Zen3 uses super reousrce to set a cap on the number of store pipes in its scheduling model.
ProcResGroup v.s. Super resource
So far, we have learned how to use ProcResGroup and super resource. Naturally we want to ask: what are their actual differences and, more importantly, when should I use them?
Conceptually, both super resource and ProcResGroup provide a way to reference a subset of a larger collection of hardware units. Super resource creates a “slice” of an existing ProcResource;
ProcResGroup approaches this from an opposite direction: it combines multiple smaller ProcResource into a larger set, so that we can either reference to the larger set or the original individual
The main difference between them comes up when execution pipes have certain kinds of partially overlapping capabilities, like this:
First, let’s (try to) describe this model with super resource in the following way:
1// Note: this is the WRONG approach
2def IEX : ProcResource<3>;
4let Super = IEX in {
5 def IntegerArith : ProcResource<2>;
6 def IntegerMul : ProcResource<2>;
7 def IntegerDiv : ProcResource<1>;
10// Simple arithmetics, like ADD
11def : WriteRes<WriteIALU, [IntegerArith]>;
12// Multiplication
13def : WriteRes<WriteIMul, [IntegerMul]>;
14// Division
15def : WriteRes<WriteIDiv, [IntegerDiv]>;
In hindsight, this model looks correct: two out of three pipes can be allocated to MUL or ALU, while only a single pipe can be used for divisions. But things start to get off the track when we run
the following RISC-V snippet throught this model.
mul a1, a1, a2
mul t4, t4, t5
div s0, s0, t0
First, let’s expand those WriteRes entries:
1// Simple arithmetics, like ADD
2def : WriteRes<WriteIALU, [IntegerArith, IEX]>;
3// Multiplication
4def : WriteRes<WriteIMul, [IntegerMul, IEX]>;
5// Division
6def : WriteRes<WriteIDiv, [IntegerDiv, IEX]>;
IntegerArith, IntegerMul, and IntegerDiv all have IEX as its super resource, which is implicitly inserted into the list of resource usages in all three entries.
With this expansion, we can pan out the (plausibly correct) resource consumption table:
instruction IntegerArith IntegerMul IntegerDiv IEX
mul a1, a1, a2 0 / 2 1 / 2 0 / 1 1 / 3
mul t4, t4, t5 0 / 2 2 / 2 0 / 1 2 / 3
div s0, s0, t0 0 / 2 2 / 2 1 / 1 3 / 3
Again, each instruction gets the resources they demanded and everything looks correct – until you realize that if both IEX1 and IEX2, the multiplication-capable pipes, have already been consumed, how
can the last instruction be dispatched to IEX1, the only pipe that is capable of doing division?
Now, you might try to fix this by assinging a different super resource to IntegerDiv, let’s say IntegerMul:
1// ????
2let Super = IntegerMul in
3def IntegerDiv : ProcResource<1>;
But then we will run into the same problem if we have the following RISC-V snippet, in which add instructions use IntegerALU:
add a1, a1, a2
add t4, t4, t5
div s0, s0, t0
Because the first two add instructions will already consume both IEX0 and IEX1 before division tries to grab IEX1 that is no longer available.
The root cause for the problem we have here is that we cannot declare both IntegerMul and IntegerALU as the super resource of IntegerDiv. Super resource is only effective if you can organize the
ProcResource hierarchy into a tree.
Take the processor model we used at the beginning of this post as the example, we can easily organize their processor resources into a tree as shown in the figure below.
On the other hand, the processor model we saw in this section can only be expressed with a DAG:
Of course, we can easily describe this model with DAG structure using ProcResGroup:
1def IEX0 : ProcResource<1>;
2def IEX1 : ProcResource<1>;
3def IEX2 : ProcResource<1>;
5def IntegerArith : ProcResGroup<[IEX0, IEX1]>;
6def IntegerMul : ProcResGroup<[IEX1, IEX2]>;
8// Simple arithmetics, like ADD
9def : WriteRes<WriteIALU, [IntegerArith]>;
10// Multiplication
11def : WriteRes<WriteIMul, [IntegerMul]>;
12// Division
13def : WriteRes<WriteIDiv, [IEX1]>;
After expansion, we effectively have the following WriteRes entries:
1// Simple arithmetics, like ADD
2def : WriteRes<WriteIALU, [IntegerArith]>;
3// Multiplication
4def : WriteRes<WriteIMul, [IntegerMul]>;
5// Division
6def : WriteRes<WriteIDiv, [IEX1, IntegerArith, IntegerMul]>;
Now WriteIDiv consumes not just IEX1 but also IntegerArith and IntegerMul – the predecessors of the division resource in the DAG we just saw.
If we run this model over one of the earlier snippets:
instruction IntegerArith IntegerMul IEX1
mul a1, a1, a2 0 / 2 1 / 2 0 / 1
mul t4, t4, t5 0 / 2 2 / 2 0 / 1
div s0, s0, t0 1 / 2 FAIL TO CONSUME 1 / 1
The division instruction is unable to be dispatched, because it failed to consume the IntegerMul resource – and this behavior is something we expect.
I hope you’re now convinced that ProcResGroup is more flexible and more generic than super resource, because it can express models with either tree or non-tree structures. This comes unsurprised as
ProcResGroup was actually invented later than super resource.
That said, super resource might come handy when we only care about the number of proceesor units and referencing the exact pipes is less important. For example, in an extreme situation where there
are a total of 12 execution pipes in a model, instead of spelling out all processor resources like this^3:
1def IEX0 : ProcResource<1>;
2def IEX1 : ProcResource<1>;
4def IEX11 : ProcResource<1>;
6// I make up these groupings, the point is that it's
7// quite cumbersome to referece every IEX pipes they use.
8def IntegerArith : ProcResGroup<[IEX0, IEX1, ...]>;
9def IntegerMul : ProcResGroup<[IEX6, IEX8, ...]>;
It’s certainly easier and more concise to write:
1def IEX : ProcResource<12>;
3let Super = IEX in {
4 def IntegerArith : ProcResource<12>;
5 def IntegerMul : ProcResource<6>;
To conclude, in this post we discussed several options to express processor resources with hierarchy structures. Notably, ProcResGroup and super resource. The takeaway is that ProcResGroup is
generally more flexible and versitile than the other options, but can be quite verbose in some cases, in which super resource or even jsut plain ProcResource with multiple units is more desirable.
1. The actual dispatching algorithm in real processors is much more complicated, but let’s just assume it looks for available pipes without any specific order. ↩︎
2. Again, using a processor with issue width of 6 ↩︎
3. Even we can simplify it with foreach and some other TableGen magics, I’m sure it’s still more verbose than using super resource. ↩︎
#llvm #compiler-instruction-scheduling | {"url":"https://myhsu.xyz/llvm-sched-model-1.5/","timestamp":"2024-11-03T10:53:01Z","content_type":"text/html","content_length":"79944","record_id":"<urn:uuid:5a76b7c2-0f56-4f5d-acec-a986b0f38766>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00785.warc.gz"} |
Transformer Models 101: Getting Began — Part 1
The complex math behind transformer models, in easy words
Image by Kerttu from Pixabay
It is not any secret that transformer architecture was a breakthrough in the sphere of Natural Language Processing (NLP). It overcame the limitation of seq-to-seq models like RNNs, etc for being
incapable of capturing long-term dependencies in text. The transformer architecture turned out to be the inspiration stone of revolutionary architectures like BERT, GPT, and T5 and their variants. As
many say, NLP is within the midst of a golden era and it wouldn’t be fallacious to say that the transformer model is where it began.
Need for Transformer Architecture
As said, necessity is the mother of invention. The normal seq-to-seq models were no good when it got here to working with long texts. Meaning the model tends to forget the learnings from the sooner
parts of the input sequence because it moves to process the latter a part of the input sequence. This loss of knowledge is undesirable.
Although gated architectures like LSTMs and GRUs showed some improvement in performance for handling long-term dependencies by discarding information that was useless along the option to remember
necessary information, it still wasn’t enough. The world needed something more powerful and in 2015, “attention mechanisms” were introduced by Bahdanau et al. They were used together with RNN/LSTM to
mimic human behaviour to concentrate on selective things while ignoring the remaining. Bahdanau suggested assigning relative importance to every word in a sentence in order that model focuses on
necessary words and ignores the remaining. It emerged to be a large improvement over encoder-decoder models for neural machine translation tasks and shortly enough, the applying of the eye mechanism
was rolled out in other tasks as well.
The Era of Transformer Models
The transformer models are entirely based on an attention mechanism which can also be referred to as “self-attention”. This architecture was introduced to the world within the paper “Attention is All
You Need” in 2017. It consisted of an Encoder-Decoder Architecture.
Fig. Transformer Model Architecture on high-level (Source: Writer)
On a high level,
• The encoder is liable for accepting the input sentence and converting it right into a hidden representation with all useless information discarded.
• The decoder accepts this hidden representation and tries to generate the goal sentence.
In this text, we’ll delve into an in depth breakdown of the Encoder component of the Transformer model. In the subsequent article, we will take a look at the Decoder component intimately. Let’s
The encoder block of the transformer consists of a stack of N encoders that work sequentially. The output of 1 encoder is the input for the subsequent encoder and so forth. The output of the last
encoder is the ultimate representation of the input sentence that’s fed to the decoder block.
Fig. Enoder block with stacked encoders (Source: Writer)
Each encoder block might be further split into two components as shown within the figure below.
Fig. Components of Encoder Layer (Source: Writer)
Allow us to look into each of those components one after the other intimately to grasp how the encoder block is working. The primary component within the encoder block is multi-head attention but
before we hop into the main points, allow us to first understand an underlying concept: self-attention.
Self-Attention Mechanism
The primary query that may pop up in everyone’s mind: Are attention and self-attention different concepts? Yes, they’re. (Duh!)
Traditionally, the attention mechanisms got here into existence for the duty of neural machine translation as discussed within the previous section. So essentially the eye mechanism was applied to
map the source and goal sentence. Because the seq-to-seq models perform the interpretation task token by token, the eye mechanism helps us to discover which token(s) from the source sentence to focus
more on while generating token x for the goal sentence. For this, it makes use of hidden state representations from encoders and decoders to calculate the eye scores and generate context vectors
based on these scores as input for the decoder. Should you want to learn more concerning the Attention Mechanism, please hop on to this text (Brilliantly explained!).
Coming back to self-attention, the essential idea is to calculate the eye scores while mapping the source sentence to itself. If you’ve got a sentence like,
“The boy didn’t cross the road because it was too wide.”
It is straightforward for us humans to grasp that word “it” refers to “road” within the above sentence but how will we make our language model understand this relationship as well? That is where
self-attention comes into the image!
On a high level, every word within the sentence is compared against every other word within the sentence to quantify the relationships and understand the context. For representational purposes,
you’ll be able to confer with the figure below.
Allow us to see intimately how this self-attention is calculated (in real).
• Generate embeddings for the input sentence
Find embeddings of all of the words and convert them into an input matrix. These embeddings might be generated via easy tokenisation and one-hot encoding or may very well be generated by embedding
algorithms like BERT, etc. The dimension of the input matrix might be equal to the sentence length x embedding dimension. Allow us to call this input matrix X for future reference.
• Transform input matrix into Q, K & V
For calculating self-attention, we’d like to rework X (input matrix) into three latest matrices:
– Query (Q)
– Key (K)
– Value (V)
To calculate these three matrices, we’ll randomly initialise three weight matrices namely Wq, Wk, & Wv. The input matrix X might be multiplied with these weight matrices Wq, Wk, & Wv to acquire
values for Q, K & V respectively. The optimal values for weight matrices might be learned through the process to acquire more accurate values for Q, K & V.
• Calculate the dot product of Q and K-transpose
From the figure above, we are able to imply that qi, ki, and vi represent the values of Q, K, and V for the i-th word within the sentence.
Fig. Example of dot product of Q and K-transpose (Source: Writer)
The primary row of the output matrix will inform you how word1 represented by q1 is said to the remaining of the words within the sentence using dot-product. The upper the worth of the dot-product,
the more related the words are. For intuition of why this dot product was calculated, you’ll be able to understand Q (query) and K (key) matrices by way of information retrieval. So here,
– Q or Query = Term you’re looking for
– K or Key = a set of keywords in your search engine against which Q is compared and matched.
As within the previous step, we’re calculating the dot-product of two matrices i.e. performing a multiplication operation, there are possibilities that the worth might explode. To make certain this
doesn’t occur and gradients are stabilised, we divide the dot product of Q and K-transpose by the square root of the embedding dimension (dk).
• Normalise the values using softmax
Normalisation using the softmax function will end in values between 0 and 1. The cells with high-scaled dot-product might be heightened moreover whereas low values might be reduced making the
excellence between matched word pairs clearer. The resultant output matrix might be considered a rating matrix S.
• Calculate the eye matrix Z
The values matrix or V is multiplied by the rating matrix S obtained from the previous step to calculate the eye matrix Z.
But wait, why multiply?
Suppose, Si = [0.9, 0.07, 0.03] is the rating matrix value for i-th word from a sentence. This vector is multiplied with the V matrix to calculate Zi (attention matrix for i-th word).
Zi = [0.9 * V1 + 0.07 * V2 + 0.03 * V3]
Can we are saying that for understanding the context of i-th word, we must always only concentrate on word1 (i.e. V1) as 90% of the worth of attention rating is coming from V1? We could clearly
define the necessary words where more attention must be paid to grasp the context of i-th word.
Hence, we are able to conclude that the upper the contribution of a word within the Zi representation, the more critical and related the words are to 1 one other.
Now that we all know tips on how to calculate the self-attention matrix, allow us to understand the concept of the multi-head attention mechanism.
Multi-head attention Mechanism
What is going to occur in case your rating matrix is biased toward a particular word representation? It’s going to mislead your model and the outcomes is not going to be as accurate as we expect.
Allow us to see an example to grasp this higher.
S1: “All is well”
Z(well) = 0.6 * V(all) + 0.0 * v(is) + 0.4 * V(well)
S2: “The dog ate the food since it was hungry”
Z(it) = 0.0 * V(the) + 1.0 * V(dog) + 0.0 * V(ate) + …… + 0.0 * V(hungry)
In S1 case, while calculating Z(well), more importance is given to V(all). It’s even greater than V(well) itself. There isn’t a guarantee how accurate this might be.
Within the S2 case, while calculating Z(it), all of the importance is given to V(dog) whereas the scores for the remaining of the words are 0.0 including V(it) as well. This looks acceptable because
the “it” word is ambiguous. It is sensible to relate it more to a different word than the word itself. That was the entire purpose of this exercise of calculating self-attention. To handle the
context of ambiguous words within the input sentences.
In other words, we are able to say that if the present word is ambiguous then it’s okay to present more importance to another word while calculating self-attention but in other cases, it might
probably be misleading for the model. So, what will we do now?
What if we calculate multiple attention matrices as an alternative of calculating one attention matrix and derive the ultimate attention matrix from these?
That’s precisely what multi-head attention is all about! We calculate multiple versions of attention matrices z1, z2, z3, ….., zm and concatenate them to derive the ultimate attention matrix. That
way we might be more confident about our attention matrix.
Moving on to the subsequent necessary concept,
Positional Encoding
In seq-to-seq models, the input sentence is fed word by word to the network which allows the model to trace the positions of words relative to other words.
But in transformer models, we follow a distinct approach. As an alternative of giving inputs word by word, they’re fed parallel-y which helps in reducing the training time and learning long-term
dependency. But with this approach, the word order is lost. Nonetheless, to grasp the meaning of a sentence appropriately, word order is incredibly necessary. To beat this problem, a latest matrix
called “positional encoding” (P) is introduced.
This matrix P is shipped together with input matrix X to incorporate the data related to the word order. For obvious reasons, the size of X and P matrices are the identical.
To calculate positional encoding, the formula given below is used.
Fig. Formula to calculate positional encoding (Source: Writer)
Within the above formula,
• pos = position of the word within the sentence
• d = dimension of the word/token embedding
• i = represents each dimension within the embedding
In calculations, d is fixed but pos and that i vary. If d=512, then i ∈ [0, 255] as we take 2i.
This video covers positional encoding in-depth if you happen to want to know more about it.
Visual Guide to Transformer Neural Networks — (Part 1) Position Embeddings
I’m using some visuals from the above video to clarify this idea in my words.
Fig. Positional Encoding Vector Representation (Source: Writer)
The above figure shows an example of a positional encoding vector together with different variable values.
Fig. Positional Encoding Vectors with constant i and d (Source: Writer)
Fig. Positional Encoding Vectors with constant i and d (Source: Writer)
The above figure shows how the values of PE(pos, 2i) will vary if i is constant and only pos varies. As we all know the sinusoidal wave is a periodic function that tends to repeat itself after a hard
and fast interval. We are able to see that the encoding vectors for pos = 0 and pos = 6 are equivalent. This isn’t desirable as we’d want different positional encoding vectors for various values of
This might be achieved by various the frequency of the sinusoidal wave.
Fig. Positional Encoding Vectors with various pos and that i (Source: Writer)
As the worth of i varies, the frequency of sinusoidal waves also varies resulting in several waves and hence, resulting in several values for each positional encoding vector. This is strictly what we
wanted to attain.
The positional encoding matrix (P) is added to the input matrix (X) and fed to the encoder.
Fig. Adding positional encoding to the input embedding (Source: Writer)
The following component of the encoder is the feedforward network.
Feedforward Network
This sublayer within the encoder block is the classic neural network with two dense layers and ReLU activations. It accepts input from the multi-head attention layer, performs some non-linear
transformations on the identical and at last generates contextualised vectors. The fully-connected layer is liable for considering each attention head and learning relevant information from them. For
the reason that attention vectors are independent of one another, they might be passed to the transformer network in a parallelised way.
The last and final component of the Encoder block is Add & Norm component.
Add & Norm component
This can be a residual layer followed by layer normalisation. The residual layer ensures that no necessary information related to the input of sub-layers is lost within the processing. While the
normalisation layer promotes faster model training and prevents the values from changing heavily.
Fig. Encoder components with Add & Norm layers included (Source: Writer)
Inside the encoder, there are two add & norm layers:
• connects the input of the multi-head attention sub-layer to its output
• connects the input of the feedforward network sub-layer to its output
With this, we conclude the inner working of the Encoders. To summarize the article, allow us to quickly go over the steps that the encoder uses:
• Generate embeddings or tokenized representations of the input sentence. This might be our input matrix X.
• Generate the positional embeddings to preserve the data related to the word order of the input sentence and add it to the input matrix X.
• Randomly initialize three matrices: Wq, Wk, & Wv i.e. weights of query, key & value. These weights might be updated through the training of the transformer model.
• Multiply the input matrix X with each of Wq, Wk, & Wv to generate Q (query), K (key) and V (value) matrices.
• Calculate the dot product of Q and K-transpose, scale the product by dividing it with the square root of dk or embedding dimension and at last normalize it using the softmax function.
• Calculate the eye matrix Z by multiplying the V or value matrix with the output of the softmax function.
• Pass this attention matrix to the feedforward network to perform non-linear transformations and generate contextualized embeddings.
In the subsequent article, we’ll understand how the Decoder component of the Transformer model works.
This is able to be all for this text. I hope you found it useful. Should you did, please don’t forget to clap and share it together with your friends.
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14C Calibration the Bayesian way
With what follows I want to demonstrate that the basic functionality of Oxcal is reflected in this tiny bit of code:
likelihood <- function(proposal){dnorm(calf(proposal),bp,std)}
collector <- bp <- 3600; std <- 30; last_prob <- likelihood(bp)
for (i in 1:10000) {
proposal <- rnorm(1, tail(collector,1), 3*std)
proposal_prob <- likelihood(proposal)
if ((proposal_prob/last_prob) > runif(1))
last_prob <- proposal_prob
} else {
proposal <- tail(collector,1)
collector <- c(collector, proposal)
The ¹⁴C cycle
¹⁴C is produced in the atmosphere by solar radiation, is than oxidised to CO[2] and than introduced into the foodchain. From than onward, the ¹⁴C decays.
The nonlinear nature of the ¹⁴C curve
The problem is, that the amount of ¹⁴C in the atmosphere was different during different times. This causes that there is no simple linear relationship between ¹⁴C years you get from the lab to the
real calendric cal bp or bc date. The expression of this is the wiggly nature of the calibration curve:
This becomes more obvious if we zoom in:
changes in solar activity
The main reason for that might be the fluctuation of the solar activity, that causes different amount of solar radiation and with that also different ¹⁴C values. One fluctuation is a well known 11
year cycle, but there are other, not so well behaving fluctuations, too.
calibration | using material of known age
To cope with that, we use samples of known age, eg. tree rings. Radiometric dating those gives us a estimation for the amount of ¹⁴C at different times. Since the resulting calibration curve is not
linear, there is no linear calculation possible to come up with the real age of things.
Real world calibration | The problem with the nonlinear nature
In an ideal world, you could use a ruler for translating a radiocarbon age to a calender age. But due to the wiggles of the curve, there are parts where a radiocarbon age could result in two or even
more possible calendric ages.
video of chunk calib_anim
Real world calibration | The problem nature of a ¹⁴C date
What makes things worse, is the fact, that you will never get a single radiocarbon age, but always a probability distribution with a ‘most likely’ value (the uncalibrated ¹⁴C age) with a standard
deviation (eg. +/- 25). On the pro side is that this distribution behaves statistically well, since it follows a normal distribution. On the con side: this enlarges the possible range where a
calibration gives multiple values, and it makes the calculation non-trivial.
There are several approaches for calibration, one I have shown in an earlier post, this time we will deal with the Bayesian way.
The bayesian approach to that problem
There are several buzzwords that show up when people talk about Bayesian calibration. Those are eg.
• Monte Carlo
• Markov Chain
• Metropolis–Hastings algorithm
The sound impressive, don’t they. But we will handle them one by one.
Monte Carlo
Monte Carlo methods are just a name for 'doing something including a random aspect', meaning eg. 'roll a dice'.
Markov Chain
Markov Chain means 'Something takes place in respect to the current state'. In this example, we start from a sunny day. There is a 90% chance that the next day will be sunny again, and a 10% chance
that it will be raining. If the next day will be one of those 10% where it is rainy, there is a 50% chance that it stays rainy, and a 50% chance that it will be sunny again. So, next day is rainy,
again 50/50 chances. If next day is sunny, again 90/10 chances. And so on.
Together it is MCMC
MCMC stands for ‘Monte Carlo Markov Chain’. This is a procedure that can be used for estimating parameter that can not be measured directly, based on an external evaluation.
To demonstrate that and to prepare for the real ¹⁴C calibration, we will estimate the value of Pi.
Excurse Value of Pi using MCMC
We have a square of 1 m and an circle with radius 1 m.
We throw darts. We count, how many darts are in the circle, and how many are in the surrounding square.
The darts are the random part (Monte Carlo). The amount of darts inside the circle is relative to the area, as it is the amount of darts within the square.
area of circle to area of square is:
\(\frac{\text{area of circle}}{\text{area of square}} = \frac{ r^{2} \cdot \pi }{ (2 \cdot r)^{2} } = \frac{ \pi }{ 4 } = \frac{\text{darts in circle}}{\text{darts in square}} = P\)
\(\pi = P \cdot 4\)
To calculate the value of Pi, we have to count the darts that are inside the circle and those inside the square, and divide their ratio by 4.
Now some R code follows:
We make a circle with radius of 0.5 and center it in a square (center = 0.5;0.5). We make a loop, that is repeated 10000 times. We randomly draw from a uniform distribution x and y for the point (we
throw the dart). We check if the dart is inside the circle. If so, we memorize it. Afterwards, we divide the amounts of dart inside the circle by the number of trials, divided by 4.
radius <- 0.5
center_x <- center_y <- 0.5
collector <- vector()
for (i in 1:10000) {
x <- runif(1)
y <- runif(1)
inside <- (x-center_x)^2 + (y - center_y)^2 < radius^2
collector <- c(collector,inside)
The result:
is close to Pi:
Experiment Live
Here you can see the experiment running live:
video of chunk pi_2
Anatomy of the algorithm
The algorithm we have used can be divided into several parts:
# Setup
radius <- 0.5
center_x <- center_y <- 0.5
collector <- vector()
# Loop
for (i in 1:10000) {
# Proposal (function)
x <- runif(1)
y <- runif(1)
# Likelihood (function)
inside <- (x-center_x)^2 + (y - center_y)^2 < radius^2
collector <- c(collector,inside)
• A setup, where we define start values
• A loop, within which things are repeated for some time
• A proposal function, that proposes candidates for our ‘desired’ result (darts in the circle).
• A likelihood function that determinates how likely it is that we archived a ‘desired’ result. In this case it is either in or out, so likelihood is either 0 or 1.
• At last, a collector, that collects our results.
Pi vs ¹⁴C Date
In this experiment, there are already several aspects that are similar to the ¹⁴C calibration that we intend to produce, but also some differences that we have to cope with:
Pi ¹⁴C Date
unknown value of Pi unknown distribution of calibrated date
choosing a random point choosing a random date
evaluation against the circle (inside/outside) evaluation against the uncalibrated date
deterministic evaluation criterion (inside/outside) probabalistic criterion (prob. distribution of uncalibrated date)
With some little effort we will be able to proceed from here to Bayesian calibration.
Toward the Bayesian calibration
Lets start with things that are similar:
Setup looks rather similar. We only need different things, like uncalibrated date and its standard deviation.
# Pi
radius <- 0.5
center_x <- center_y <- 0.5
collector <- vector()
# Calibration
bp <- 3600
std <- 30
collector <- vector()
collector[1] <- bp
Also the loop is rather similar:
# Pi
for (i in 1:10000) {
# Actions that are repeated
# 10000 times
# Calibration
for (i in 1:10000) {
# Actions that are repeated
# 10000 times
Proposal function
In case of Pi, we propose random x and y coordinates. For the calibration, we propose a calendric date. But we do not propose any random date, but one that depends on our previous proposal. This is a
Markov Chain, remember? We make a new proposal near to our previous one, and we make it more likely that it is nearer than that it is farther away. For that we use a normal distribution (rnorm).
# Pi
x <- runif(1)
y <- runif(1)
# Calibration
proposal <- rnorm(
video of chunk proposal_function_live
Putting together what we have now:
bp <- 3600
std <- 30
collector <- vector()
collector[1] <- bp
for (i in 1:10000) {
proposal <- rnorm(1, tail(collector,1), 3*std)
# What is missing is the likelihood function!
collector <- c(collector, proposal)
Differences to the Pi example
From now on, things are a bit different due to the differences of our examples:
Pi ¹⁴C Date
unknown value of Pi unknown distribution of calibrated date
choosing a random point choosing a random date
evaluation against the circle (inside/outside) evaluation against the uncalibrated date
deterministic evaluation criterion (inside/outside) probabalistic criterion (prob. distribution of uncalibrated date)
We cannot directly evaluate, if the date fits our calibrated date, because we do not know it yet. But we know the uncalibrated date, and will use that instead. Also, we do not have a inside/outside
of the circle criterion (0/1), but one that has a certain probability, so values between 0 and 1 are possible. We will cope with that.
Differences to the Pi example
Problem One
We evaluate against the uncalibrated date:
-> We have to backcalibrate the proposal
Problem Two
We have a probabilistic criterion, we only get a probability back if the proposal fits our uncalibrated date
-> We have to introduce some probabilistic (random) decision
Solution One: Backward calibration
In contrast to forward calibration, backward calibration is straight forward. We can use the ‘ruler approach’, since every calibrated date only reflects one uncalibrated one. The wiggles only go up
and down, luckily not left and right!
video of chunk backcalib_anim
intcal13 <- read.csv(
skip = 11, header=F)[,1:2]
colnames(intcal13) <- c("CAL BP", "14C age")
calf <- function(this_bp) {
approx(intcal13$`CAL BP`,
intcal13$`14C age`,
Likelihood function
For the likelyhood function, we need the shape of the uncalibrated date. With that we can now check how good a proposed date would fit our uncalibrated one. We can determine, how likely a proposal
would be given our uncalibrated date.
likelihood <- function(proposal){
pred = calf(proposal)
With that, we have a likelihood function!
Solution Two: Evaluation with a random component
• We take the last proposal, resp. its likelihood
• We compute the likelihood of the new proposal
• If the new proposal fits better to the uncalibrated date, we keep it
• If not: We still keep it with a probability equal to how worse the new proposal is compared to the prior one
-> Metropolis-Hastings algorithm
Solution Two: Evaluation with a random component
We start with the bp value of the uncalibrated date as our first proposal. Than we make a second proposal, in relation to the first one.
first_proposal <- bp
second_proposal <- rnorm(1, first_proposal, 3*std)
Our second proposal is:
With our likelihood function, we check how likely this second proposal is:
ratio <- likelihood(second_proposal)/likelihood(first_proposal)
## [1] 3.353998e-10
The second proposal is 3.3539984 * 10^-10 times likely compared to the first.
Solution Two: Evaluation with a random component
If the ratio is more than 1 (second proposal is better), we keep it.
If it is less than one (second proposal is worse), we keep it with a probability equal to the ratio. Meaning, we roll a dice…
alpha <- runif(1)
## [1] 0.2552167
Check if our random value is smaller than our ratio.
alpha < ratio
## [1] FALSE
That is false. So our second proposal will not be recorded, and will not be part of the Markov chain.
Everything together
The final code
likelihood <- function(proposal){dnorm(calf(proposal),bp,std)}
collector <- bp <- 3600; std <- 30; last_prob <- likelihood(bp)
for (i in 1:10000) {
proposal <- rnorm(1, tail(collector,1), 3*std)
proposal_prob <- likelihood(proposal)
if ((proposal_prob/last_prob) > runif(1))
last_prob <- proposal_prob
} else {
proposal <- tail(collector,1)
collector <- c(collector, proposal)
With every loop, our collector records those proposals that were accepted, either because they were better, or because the were choosen in the relegation round… due to the random factor.
The result
If we make a histogramm from the collected proposals, we get something that is essentially already the calibrated date:
hist(collector,probability=TRUE, breaks = 100)
Compare the result
We can compare the resulting histogram with a calibration from Oxcal: looks similar!
See it run
video of chunk algorithm_live
Adding stratigraphical informations
The Scenario
We have two samples, S_1 (3600, 30) and S_2 (3550, 40). We know from the stratigraphy, that S_1 must be younger than S_2.
We can easily introduce this information into our calibration with a little extra code.
The new code
First step: Calibrate multiple dates at once. For this, we add another dimension to our variables (bp, std, collector, last_prob), to reflect the fact that we now deal with multiple data:
bp <- c(3600, 3550); std <- c(30,40)
collector <- list(date1=bp[1], date2=bp[2])
last_prob <- c(likelihood(bp[1]), likelihood(bp[2]))
for (i in 1:20000) {
curr_date <- (i-1)%%2+1 # alter between 1 and 2
proposal <- rnorm(1, tail(collector[[curr_date]],1), 3*std[curr_date])
proposal_prob <- likelihood(proposal)[curr_date]
if ((proposal_prob/last_prob[curr_date]) > runif(1))
last_prob[curr_date] <- proposal_prob
} else {
proposal <- tail(collector[[curr_date]],1)
collector[[curr_date]] <- c(collector[[curr_date]], proposal)
The result
If we now let the code run, we can calibrate two dates at once:
The new code
Second step: Alter the Metropolis-Hastings -> Gibbs Sampling We add another condition:
If calibrating for the first date (S_1), every proposal must be younger than the last proposal for the second date (S_2), and vice versa,
If calibrating for the second date (S_2), every proposal must be older than the last proposal for the second date (S_1)
bp <- c(3600, 3550); std <- c(30,40)
collector <- list(date1=bp[1], date2=bp[2])
last_prob <- c(likelihood(bp[1]), likelihood(bp[2]))
for (i in 1:20000) {
curr_date <- (i-1)%%2+1 # alter between 1 and 2
proposal <- rnorm(1, tail(collector[[curr_date]],1), 3*std[curr_date])
proposal_prob <- likelihood(proposal)[curr_date]
# This is the new bit!
if (curr_date == 1 & proposal > tail(collector[[2]],1)) proposal_prob = 0
if (curr_date == 2 & proposal < tail(collector[[1]],1)) proposal_prob = 0
if ((proposal_prob/last_prob[curr_date]) > runif(1))
last_prob[curr_date] <- proposal_prob
} else {
proposal <- tail(collector[[curr_date]],1)
collector[[curr_date]] <- c(collector[[curr_date]], proposal)
The result
Now we can see the modeled date according to our stratigraphic information:
Both dates can again be compared against the Oxcal output:
Compare to Oxcal unmodeled
Compare to Oxcal modeled
Voila, Oxcal reverse engineered! | {"url":"http://resillience2020.archaeological.science/jekyll/update/blog/2017/01/23/bayesian_calibration.html","timestamp":"2024-11-04T17:28:05Z","content_type":"text/html","content_length":"67379","record_id":"<urn:uuid:d7c4ba49-42fb-4caa-9137-16e242fb2579>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00089.warc.gz"} |
Question ID - 155407 | SaraNextGen Top Answer
$ \quad$ A solid circular shaft, made of ductile material with yield stress $\sigma_{Y}=280 \mathrm{MPa}$, is subjected to a torque of $10 \mathrm{kNm}$. Using the Tresca failure theory, the smallest
radius of the shaft to avoid failure is _________$\mathrm{cm}$ (round off to two decimal places). | {"url":"https://www.saranextgen.com/homeworkhelp/doubts.php?id=155407","timestamp":"2024-11-13T18:39:04Z","content_type":"text/html","content_length":"14551","record_id":"<urn:uuid:284c624f-9b02-4faf-8879-7329133370a5>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00373.warc.gz"} |
Computer Science Assignment Template
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%% Template for AIMS Rwanda
Assignments %%% %%% %%% Author: AIMS Rwanda tutors %%% ### %%% %%% Email: tutors2017-18@aims.ac.rw %%% ### %%% %%% Copyright: This template was designed to be used for %%% ####### %%% %%% the
assignments at AIMS Rwanda during the academic year %%% ### %%% %%% 2017-2018. %%% ######### %%% %%% You are free to alter any part of this document for %%% ### ### %%% %%% yourself and for
distribution. %%% ### ### %%% %%% %%% %%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%% Ensure that you do not write the questions before each of the solutions because it is not necessary. %%%%%% \
documentclass[12pt,a4paper]{article} %%%%%%%%%%%%%%%%%%%%%%%%% packages %%%%%%%%%%%%%%%%%%%%%%%% \usepackage{graphicx} \usepackage{tabulary} \usepackage{amsmath} \usepackage{fancyhdr} \usepackage
{amssymb} \usepackage{amsthm} \usepackage{placeins} \usepackage{amsfonts} \usepackage{graphicx} \usepackage[all]{xy} \usepackage{tikz} \usepackage{verbatim} \usepackage[left=2cm,right=2cm,top=
3cm,bottom=2.5cm]{geometry} \usepackage{hyperref} \usepackage{caption} \usepackage{subcaption} \usepackage{multirow} \usepackage{psfrag} %%%%%%%%%%%%%%%%%%%%% students data %%%%%%%%%%%%%%%%%%%%%%%% \
newcommand{\student}{\textbf{Azamuke Denish}} \newcommand{\course}{\textbf{Demo Template for Msc. CS}} \newcommand{\assignment}{\textbf{1}} %%%%%%%%%%%%%%%%%%% using theorem style
%%%%%%%%%%%%%%%%%%%% \newtheorem{thm}{Theorem} \newtheorem{lem}[thm]{Lemma} \newtheorem{defn}[thm]{Definition} \newtheorem{exa}[thm]{Example} \newtheorem{rem}[thm]{Remark} \newtheorem{coro}[thm]
{Corollary} \newtheorem{quest}{Question}[section] %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \usepackage{lipsum}%% a garbage package you don't need except to create examples. \usepackage{fancyhdr} \
pagestyle{fancy} %\lhead{Azamuke Denish} \rhead{ \thepage} %\cfoot{\textbf{AIMS Rwanda Academic Year 2020 - 2021}} \renewcommand{\headrulewidth}{0.4pt} \renewcommand{\footrulewidth}{0.4pt}
%%%%%%%%%%%%%% Shortcut for usual set of numbers %%%%%%%%%%% \newcommand{\N}{\mathbb{N}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb
{C}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%555 \begin{document} %%%%%%%%%%%%%%%%%%%%%%% title page %%%%%%%%%%%%%%%%%%%%%%%%%% \thispagestyle{empty} \begin{center} \includegraphics
[scale = 0.6]{cs4.png} %\textbf{AFRICAN INSTITUTE FOR MATHEMATICAL SCIENCES \\[0.5cm] %(AIMS RWANDA, KIGALI)} \vspace{0.5cm} \end{center} %%%%%%%%%%%%%%%%%%%%% assignment information %%%%%%%%%%%%%%%%
\noindent \rule{17cm}{0.2cm}\\[0.3cm] Name: \student \hfill Assignment Number: \assignment\\[0.1cm] Course: \course \hfill Date: \today\\ \rule{17cm}{0.05cm} \vspace{1.0cm}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section*{Problem 1} Given that \textbf{Airqo}, \textit{Africa's leading air quality monitoring, research and analytics network company} has used its
public RSA key $(n, e)$ for years. After a security check they had to change it to $(n, e')$, with the same $n$ but with a different number $e'$ which is relatively prime to $e$. A customer had
previously sent his message $\overline{a}$ which was encoded with the old key. After he got the news of the security check he encodes this same message $\overline{a}$ with the new public key. \\How
can an attacker get $\overline{a}$ from the knowledge of the old and new encrypted message $\overline{c_1}$ and $\overline{c_2}$ respectively using only the public keys? You are required to evaluate
this for the example where $n = 247, e = 11, e' = 17, c_1 = 24, c_2 = 93.$\\\\ \newpage \begin{thebibliography}{99} \bibitem{am} Bitter, R., Mohiuddin, T., \& Nawrocki, M. (2017). LabVIEW: {\em
Advanced programming techniques.\em} CRC press. \end{thebibliography} \end{document} | {"url":"https://ko.overleaf.com/latex/templates/computer-science-assignment-template/tykxrfjzykjc","timestamp":"2024-11-08T18:26:42Z","content_type":"text/html","content_length":"40608","record_id":"<urn:uuid:856839c1-49af-437e-a0c8-7de6e3cdbe1a>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00526.warc.gz"} |
Critical Value in Statistics: Definition, Importance, and detailed example - Learning Desk
In statistics, a critical value is a threshold or boundary point used in hypothesis testing and constructing confidence intervals. It is a specific value derived from the sampling distribution of a
test statistic, such as the t-distribution or the normal distribution, and is used to determine whether to reject or fail to reject the null hypothesis.
In this article, we will discuss the definition of critical value, the Formula, and the importance of critical value. Also, with the help of examples,the topic will be explained easily. After
completely understanding this article, anyone can defend this article.
Critical Value in Statistics
The critical value in statistics is those value when we compared to a test statistic in hypothesis testing to find out the null hypothesis are rejected are not. The null hypothesis is a statement
that assumes there is no significant difference or relationship between variables, while the alternative hypothesis contradicts the null hypothesis by asserting that there is a significant difference
or relationship.
Hypothesis testing involves collecting sample data and comparing it to a critical value to make an inference about the population.
Critical value critical value in sctatistics formula depends on the test static distribution so, there are different kind of formula which is used to find the critical value. One of the most common
critical formulas is to be used which is an interval of confidence or level of significance that can be used to find the critical value.
Importance of Critical Value
Critical values play a central role in hypothesis testing. They help determine whether the observed data provide sufficient evidence to favor the alternative hypothesis inrejecting the null
By comparing the computed test statistic to the critical value, statisticians can make informed decisions about the validity of their hypotheses and draw meaningful conclusions from their analyses.
Critical values are used to construct confidence intervals, which provide a range of plausible values for a population parameter. The critical value determines the width of the interval and reflects
the desired level of confidence.
Confidence intervals are valuable in statistical inference as they help quantify the uncertainty associated with sample estimates and provide a measure of the precision of the estimated parameter.
Also Read : How to Make a Productive Study Plan for CBSE Class 10 New Academic Session 2023-2024?
The critical value is chosen based on the desired significance level, and it establishes the threshold for determining statistical significance. A lower significance level corresponds to a more
stringent test, requiring stronger evidence to reject the null hypothesis.
Critical values provide a clear and objective criterion for decision-making in statistical analysis. By comparing the test statistic to the critical value, statisticians can determine whether to
reject or fail to reject the null hypothesis.
This helps researchers and decision-makers make informed choices based on the evidence available from the data.
Critical value in statistics are often used to standardize test statistics, transforming them into a common scale that facilitates comparison and inference across different datasets and
This standardization enables researchers to draw conclusions based on a set of predefined critical values rather than relying on specific parameter values or sample characteristics.
• Reproducibility and Consistency:
Critical values provide a standardized and consistent approach to statistical analysis. By adhering to predefined critical values and significance levels, researchers can ensure that their findings
are replicable and comparable across studies. This consistency promotes transparency and allows for the objective evaluation of results by other researchers.
How to find critical value in statistics?
Example 1:
Let’s say we have a sample of 80 observations, and we want to determine whether the sample mean differs from the population mean by a significant amount at a 90% level of confidence.
Step 1:Set up hypotheses:
H[0]: μ = μ[0] (population mean)
H[a]: μ ≠ μ[0] (sample mean is different from population mean)
Step 2:Significance level:
Significance level = 1 – 0.90 = 0.10
because the confidence level is 90%.
The two-tailed significance level is 0.10 divided by 2 for each tail of the distribution, resulting in α/2 = 0.05 for each tail.
Step 3: Critical value:
To find the critical value, we need to determine the z-value corresponding to the significance level and the two tails.
Since the sample size is relatively large (n = 80) and assuming the population standard deviation is unknown, we can use the standard normal distribution (z-distribution).
Using statistical software or a standard normal distribution table,we find that the critical z-value for a significance level of 0.05 (two-tailed) is approximately ±1.645.
Step 4:Make a decision:
In this step, we compare the calculated test statistic (z-value) with the critical value to make a decision.
If the calculated test statistic falls outside the range of -1.645 to +1.645, we reject the null hypothesis and conclude that the sample mean is significantly different from the population mean at a
90% confidence level.
Example 2:
Suppose a one-tailed t-test is being conducted on data with a sample size of 12 at
α = 0.025. Critical value?
To find the critical value for a one-tailed t-test here, with a sample size of 12 and α = 0.025, we need to consult the statistical software or t-distribution table.
Step 1:
sample size (n)= 12
α (signific interval) = 0.025
Degree of freedom= 12 – 1 = 11
Step 2:
To solve this, we use t distribution table of statistical software
Using a one-tailed table (0.025, 11) = 2.718.
Therefore, the critical value for the given one-tailed t-test with a sample size of 12 and α = 0.025 is approximately 2.718.
In this article, we have discussed the definition of critical value, the Formula, and the importance of critical value. Also, with the help of examples, the topic will be explained easily. After
studying this article anyone can defend this topic easily.
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vtkWeightedTransformFilter Class Reference
#include <vtkWeightedTransformFilter.h>
Detailed Description
transform based on per-point or per-cell weighting functions.
vtkWeightedTransformFilter is a filter that can be used to "skin" structures and to create new and complex shapes. Unlike a traditional transform filter (which has one transform for a data set) or an
assembly (which has one transform per part or group of parts), a weighted transform produces the weighted sum of transforms on a per-point or per-cell basis.
Each point or cell in the filter's input has an attached DataArray that contains tuples of weighting functions, one per point or cell. The filter also has a set of fixed transforms. When the filter
executes, each input point/cell is transformed by each of the transforms. These results are weighted by the point/cell's weighting factors to produce final output data.
Linear transforms are performance-optimized. Using arbitrary transforms will work, but performance may suffer.
As an example of the utility of weighted transforms, here's how this filter can be used for "skinning." Skinning is the process of putting a mesh cover over an underlying structure, like skin over
bone. Joints are difficult to skin because deformation is hard to do. Visualize skin over an elbow joint. Part of the skin moves with one bone, part of the skin moves with the other bone, and the
skin in the middle moves a little with each.
Weighted filtering can be used for a simple and efficient kind of skinning. Begin with a cylindrical mesh. Create a FloatArray with two components per tuple, and one tuple for each point in the mesh.
Assign transform weights that linear interpolate the distance along the cylinder (one component is the distance along the cylinder, the other is one minus that distance). Set the filter up to use two
transforms, the two used to transform the two bones. Now, when the transforms change, the mesh will deform so as to, hopefully, continue to cover the bones.
vtkWeightedTransformFilter is also useful for creating "strange and complex" shapes using pinching, bending, and blending.
Weighted combination of normals and vectors are probably not appropriate in many cases. Surface normals are treated somewhat specially, but in many cases you may need to regenerate the surface
Cell data can only be transformed if all transforms are linear.
See also:
vtkAbstractTransform vtkLinearTransform vtkTransformPolyDataFilter vtkActor
Definition at line 80 of file vtkWeightedTransformFilter.h.
Member Typedef Documentation
Constructor & Destructor Documentation
vtkWeightedTransformFilter::vtkWeightedTransformFilter ( ) [protected]
vtkWeightedTransformFilter::~vtkWeightedTransformFilter ( ) [protected]
Member Function Documentation
static vtkWeightedTransformFilter* vtkWeightedTransformFilter::New ( ) [static]
Create an object with Debug turned off, modified time initialized to zero, and reference counting on.
Reimplemented from vtkPointSetAlgorithm.
virtual const char* vtkWeightedTransformFilter::GetClassName ( ) [virtual]
static int vtkWeightedTransformFilter::IsTypeOf ( const char * name ) [static]
Return 1 if this class type is the same type of (or a subclass of) the named class. Returns 0 otherwise. This method works in combination with vtkTypeMacro found in vtkSetGet.h.
Reimplemented from vtkPointSetAlgorithm.
virtual int vtkWeightedTransformFilter::IsA ( const char * name ) [virtual]
Return 1 if this class is the same type of (or a subclass of) the named class. Returns 0 otherwise. This method works in combination with vtkTypeMacro found in vtkSetGet.h.
Reimplemented from vtkPointSetAlgorithm.
static vtkWeightedTransformFilter* vtkWeightedTransformFilter::SafeDownCast ( vtkObject * o ) [static]
void vtkWeightedTransformFilter::PrintSelf ( ostream & os,
vtkIndent indent
) [virtual]
Methods invoked by print to print information about the object including superclasses. Typically not called by the user (use Print() instead) but used in the hierarchical print process to combine the
output of several classes.
Reimplemented from vtkPointSetAlgorithm.
unsigned long vtkWeightedTransformFilter::GetMTime ( ) [virtual]
Return the MTime also considering the filter's transforms.
Reimplemented from vtkObject.
virtual void vtkWeightedTransformFilter::SetWeightArray ( const char * ) [virtual]
WeightArray is the string name of the DataArray in the input's FieldData that holds the weighting coefficients for each point. The filter will first look for the array in the input's PointData
FieldData. If the array isn't there, the filter looks in the input's FieldData. The WeightArray can have tuples of any length, but must have a tuple for every point in the input data set. This array
transforms points, normals, and vectors.
virtual char* vtkWeightedTransformFilter::GetWeightArray ( ) [virtual]
WeightArray is the string name of the DataArray in the input's FieldData that holds the weighting coefficients for each point. The filter will first look for the array in the input's PointData
FieldData. If the array isn't there, the filter looks in the input's FieldData. The WeightArray can have tuples of any length, but must have a tuple for every point in the input data set. This array
transforms points, normals, and vectors.
virtual void vtkWeightedTransformFilter::SetTransformIndexArray ( const char * ) [virtual]
TransformIndexArray is the string name of the DataArray in the input's FieldData that holds the indices for the transforms for each point. These indices are used to select which transforms each
weight of the DataArray refers. If the TransformIndexArray is not specified, the weights of each point are assumed to map directly to a transform. This DataArray must be of type UnsignedShort, which
effectively limits the number of transforms to 65536 if a transform index array is used. The filter will first look for the array in the input's PointData FieldData. If the array isn't there, the
filter looks in the input's FieldData. The TransformIndexArray can have tuples of any length, but must have a tuple for every point in the input data set. This array transforms points, normals, and
virtual char* vtkWeightedTransformFilter::GetTransformIndexArray ( ) [virtual]
TransformIndexArray is the string name of the DataArray in the input's FieldData that holds the indices for the transforms for each point. These indices are used to select which transforms each
weight of the DataArray refers. If the TransformIndexArray is not specified, the weights of each point are assumed to map directly to a transform. This DataArray must be of type UnsignedShort, which
effectively limits the number of transforms to 65536 if a transform index array is used. The filter will first look for the array in the input's PointData FieldData. If the array isn't there, the
filter looks in the input's FieldData. The TransformIndexArray can have tuples of any length, but must have a tuple for every point in the input data set. This array transforms points, normals, and
virtual void vtkWeightedTransformFilter::SetCellDataWeightArray ( const char * ) [virtual]
The CellDataWeightArray is analogous to the WeightArray, except for CellData. The array is searched for first in the CellData FieldData, then in the input's FieldData. The data array must have a
tuple for each cell. This array is used to transform only normals and vectors.
virtual char* vtkWeightedTransformFilter::GetCellDataWeightArray ( ) [virtual]
The CellDataWeightArray is analogous to the WeightArray, except for CellData. The array is searched for first in the CellData FieldData, then in the input's FieldData. The data array must have a
tuple for each cell. This array is used to transform only normals and vectors.
virtual void vtkWeightedTransformFilter::SetCellDataTransformIndexArray ( const char * ) [virtual]
The CellDataTransformIndexArray is like a TransformIndexArray, except for cell data. The array must have type UnsignedShort.
virtual char* vtkWeightedTransformFilter::GetCellDataTransformIndexArray ( ) [virtual]
The CellDataTransformIndexArray is like a TransformIndexArray, except for cell data. The array must have type UnsignedShort.
virtual void vtkWeightedTransformFilter::SetTransform ( vtkAbstractTransform * transform,
int num
) [virtual]
Set or Get one of the filter's transforms. The transform number must be less than the number of transforms allocated for the object. Setting a transform slot to NULL is equivalent to assigning an
overriding weight of zero to that filter slot.
virtual vtkAbstractTransform* vtkWeightedTransformFilter::GetTransform ( int num ) [virtual]
Set or Get one of the filter's transforms. The transform number must be less than the number of transforms allocated for the object. Setting a transform slot to NULL is equivalent to assigning an
overriding weight of zero to that filter slot.
virtual void vtkWeightedTransformFilter::SetNumberOfTransforms ( int num ) [virtual]
Set the number of transforms for the filter. References to non-existent filter numbers in the data array is equivalent to a weight of zero (i.e., no contribution of that filter or weight). The
maximum number of transforms is limited to 65536 if transform index arrays are used.
virtual int vtkWeightedTransformFilter::GetNumberOfTransforms ( ) [virtual]
Set the number of transforms for the filter. References to non-existent filter numbers in the data array is equivalent to a weight of zero (i.e., no contribution of that filter or weight). The
maximum number of transforms is limited to 65536 if transform index arrays are used.
virtual void vtkWeightedTransformFilter::AddInputValuesOn ( ) [virtual]
If AddInputValues is true, the output values of this filter will be offset from the input values. The effect is exactly equivalent to having an identity transform of weight 1 added into each output
virtual void vtkWeightedTransformFilter::AddInputValuesOff ( ) [virtual]
If AddInputValues is true, the output values of this filter will be offset from the input values. The effect is exactly equivalent to having an identity transform of weight 1 added into each output
virtual void vtkWeightedTransformFilter::SetAddInputValues ( int ) [virtual]
If AddInputValues is true, the output values of this filter will be offset from the input values. The effect is exactly equivalent to having an identity transform of weight 1 added into each output
virtual int vtkWeightedTransformFilter::GetAddInputValues ( ) [virtual]
If AddInputValues is true, the output values of this filter will be offset from the input values. The effect is exactly equivalent to having an identity transform of weight 1 added into each output
int vtkWeightedTransformFilter::RequestData ( vtkInformation * ,
vtkInformationVector ** ,
vtkInformationVector *
) [protected, virtual]
This is called by the superclass. This is the method you should override.
Reimplemented from vtkPointSetAlgorithm.
Member Data Documentation
The documentation for this class was generated from the following file:
Generated on Mon Sep 27 19:01:28 2010 for VTK by | {"url":"https://vtk.org/doc/release/5.6/html/a02164.html","timestamp":"2024-11-03T16:20:59Z","content_type":"text/html","content_length":"47580","record_id":"<urn:uuid:9590ecde-96de-4194-bd0a-b7b53905c4cc>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00762.warc.gz"} |
6. Translation
• In pure translation we assume that the center of mass is undergoing some motion (inertia) and possibly accelerating. Later on we will be able to consider simultaneous translation and rotation.
• In a dynamic system the translation of a rigid body is affected by an imbalance in the forces acting on it.
Problem 6.1 Given an initial (t=0) state of x=5, v=2, a=3, find the system state 5 seconds later assuming constant acceleration.
6.1 Mathematical Properties
• Effects that can apply forces on a rigid body include,
• We should also consider the energy and power transferred in translation. Keep in mind that energy is conserved. In real systems it doesn’t appear or disappear by itself.
6.1.1 Gravity And Other Fields
• As we know well, gravity causes many objects to move.
• We can model gravity as a simple force acting on the center of mass.
• The magnitude of the gravitational force is proportional to the mass of the object, and the direction of the force depends upon the location of the nearest large body. For us gravity is typically
down, so we could write the gravitational force as,
• Note that the units balance out to provide forces in Newtons, or as accelerations.
• When we deal with other fields we can develop a force equation proportional to the appropriate effect. These might include,
• NOTE: If the field is not uniformly applied to all of the mass, the resultant force will not go through the center of mass, and the result will be less force applied to pure translation, and the
unbalanced forces will add to angular acceleration.
6.1.2 Mass and Inertia
• If we sum the forces acting on a body, and set them equal to the inertial forces, we get a powerful set of equations capable of dealing with most cases of translation. This is called d’Alembert’s
• Consider the simple example below,
• We can also draw the inertia force on the free body diagram, acting on the center of mass. (Note: the inertial force is always opposite to the motion, therefore make it opposite to the direction of
positive motion). In d’Alembert’s equation this is done by putting it after the equals sign.
Problem 6.2 Given the parameters, find the acceleration.
6.1.3 Friction
• Previously you have dealt with static (dry coulomb) friction. This applies when the body is not moving.
• When the body begins to move the nature of the friction changes.
• The simplest model of dynamic friction assumes a constant friction force. After the maximum static friction force is exceeded, the object begins to move, and the friction force drops to a lower
value. Note that as the applied force ’F’ increases (hence velocity too) the friction force eventually diminishes. Therefore this model is good only at lower speeds.
• Friction dissipates energy from a system through heating, sound and vibration. The reduction of energy tends to make the system more stable.
Problem 6.3 Find the acceleration of the block for both angles indicated.
6.1.4 Springs
• Springs are extremely common in most products.
• They usually take advantage of the Modulus of Elasticity (Young’s modulus) to produce a force proportional to deformation.
• For springs that undergo a simple elongation or compression, we can use a simple relationship: Hooke’s law.
• Hooke’s law is valid for springs as long as thy are deforming elastically. Once they deform plastically they either fail, or the spring constant changes (usually becomes larger) and the undeformed
length changes.
• Springs can be thought of as energy storage elements, much like capacitors and inductors.
• We typically assume that springs are massless. This allows us to ignore the transmission delay of forces along the spring (due to inertia). And, when we consider typical applications of springs
this is almost always valid.
• Before the springs were shown with one end fixed. We will also have to deal with springs that have both ends moving. In this case we could rewrite Hooke’s Law based on the redefined directions.
• ASIDE: a spring has a natural or undeformed length. When at this length it is neither in tension or compression
• Energy in a spring is the square of the deformation
Problem 6.4 Given the spring coefficients and desired deflections find F1 and F2 separately. (don’t try to solve both at the same time.)
6.1.5 Damping
• Friction is one technique for reducing energy in a system, and damping is another.
• Damping forces are actually proportional to the velocity rate: the faster you try to move them, the more they push back. The symbol below is for a ’dashpot’.
• These components are very popular in cars (and many other products) as shock absorbers.
• Typically these devices are made using a cylinder with a viscous fluid. As the cylinder is move the fluid is forced to move between two chambers (ahead and behind the piston). To move between the
chambers it must pass through a small orifice. The faster it moves though the more it fights back.
• To make the presence of viscous damping between two surfaces more clear, we can add a crosshatching between touching surfaces.
• Another effect that we often deal with is aerodynamic drag. The drag force increases as the square of velocity. The magnitude of the drag force coefficient ’D’ is approximated theoretically, or
measured experimentally.
• The drag force coefficient is a function of,
type of flow (laminar vs. turbulent)
Problem 6.5 If we are pushing the cylinder below, what is the force for the two velocities?
6.1.6 Cables And Pulleys
• Cables are useful when transmitting tension forces.
• The centerline of the cable becomes the centerline for the force.
• If the force becomes compressive, the cable goes limp, and will not transmit force.
• A cable can be define a number of ways,
• Pulleys allow us to change the direction of a force acting through a cable.
• Typically we assume that a pulley is massless and frictionless (although we can assume both using material we will cover in rotation). If this is the case then the tension in the cable on both
sides of the pulley is equal.
• We can also assume the pulley is fixed, and the cable must slide over the surface. This creates friction, and hence a resistive force.
Given the force and friction coefficient, find the force F required to lift/drop the mass slowly.
6.1.7 Contact Points And Joints
• Points of contact determine how separate rigid bodies interact.
• These points of contact will transmit action/reaction forces and moments between rigid bodies.
• When drawing FBDs for a system that has multiple rigid bodies, we must be careful to put the equal magnitude, opposite direction forces on joined rigid bodies.
• When looking at joints between rigid bodies, we should consider their degrees of freedom. Each degree of freedom will typically have a force or moment component equal to zero. If this is the case
we can remove it from the free body diagrams.
• In 2D planar problems we can transmit up to two force components and one moment. Typical joints include,
• In 3D spatial problems, we can transmit up to three force components, and three moment components. Typical joints include,
6.2 System Examples
• In these systems there are some basic steps to solving the problems,
1. Assign letters/numbers to designate components (if not already done): this will allow you to refer to components in your calculations.
2. Define a position and directions for any moving masses. This will include the selection of reference points.
3. Draw free body diagrams for each component, and add forces (inertia is optional).
4. Write equations for each component by summing forces.
5. Combine the equations by eliminating unwanted variables.
6. Develop a final equation that relates input (forcing functions) to outputs (results).
Problem 6.6 Develop the equation relating the input force to the motion of the cart for the problem below. Assume that in the initial position the spring is compressed a distance ‘d’.
Problem 6.7 Develop the equation relating the input force to the motion (in terms of ‘x’) of the left side cart for the problem below.
Problem 6.8 Develop differential equations for the system shown.
• Consider the examples below,
• In the previous example not the similarity between springs/resistors in series/parallel.
6.3 Problems
Problem 6.9 1. Find the effective damping coefficients for the pairs below, | {"url":"https://engineeronadisk.com/V3/engineeronadisk-98.html","timestamp":"2024-11-13T19:33:08Z","content_type":"text/html","content_length":"20326","record_id":"<urn:uuid:bab8e436-b246-4ae0-b81d-7442633af9d2>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00583.warc.gz"} |
Radius of a Graph | Lexique de mathématique
Radius of a Graph
In this graph, because any vertex can be connected to any other by a chain of two edges, they are all centres of the graph and their distance is 2. Therefore, the radius of this graph is 2. | {"url":"https://lexique.netmath.ca/en/radius-of-a-graph/","timestamp":"2024-11-05T03:47:21Z","content_type":"text/html","content_length":"63347","record_id":"<urn:uuid:8795c854-154d-4fb2-81f4-a87e75240900>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00617.warc.gz"} |
Resolve force into components along different paths
• Thread starter Color_of_Cyan
• Start date
Even memorize the 5-12-13 triangle - it comes up sometimes, and it is just a 3-4-5 triangle multiplied by 2. I have it memorized as "5-12-13-15" but that is only because I was born with an extra
finger. :-)In summary, you use the law of sines to find the components of F1 along the u and V axes. For part B, Fu = 386 lb and Fv = 283 lb. For part C, Fu = 286 lb and Fv = 186 lb. These values can
be found by drawing a vector head-to-tails summation diagram and using the geometry of the resulting triangle. The dot
Homework Statement
A. Determine magnitude of resultant force (of F1 and F2 acting on the point)
B. Resolve F1 into components along the u and V axes and determine the magnitudes of these components.
C. Resolve F1 into components along the u and V axes and determine the magnitudes of these components.
Homework Equations
law of sines,
Pythagorean theorem
The Attempt at a Solution
I just need help with parts B and C.
A goes :
Fx = 150(cos60) + 200(cos45)= 216 lb
Fy = 150(sin60) + 200(sin45) = -11.5 lb
(216^2 + 11.5^2)^1/2 means
R = 216 lb at 3 deg under the U axis because tan^-1(-11.5/216) = -3 For part B the only guide to find how to solve these before is use law of sines, but
trying to solve for resolving the components of F1 along a and B:
200 / (sin45) = F(u) / 1
F(u) = 282 lb
but it is supposedly wrong.
Any hints?
Last edited by a moderator:
It is useful to draw the vector triangles to help you know what to do.
Trying to remember equations won't help. But you do know how to do trigonometry - and you know how to handle triangles that are not right-angle triangles? And you know about dot and cross products?
B. resolve along u and V axis
This means either:
1. ##\vec{F}_1=a\hat{u}+b\hat{V}## (where the hats indicate a unit vector)
2. the amount of ##\vec{F}_1## along direction ##\hat{u}## and ##\hat{V}##.
You'll have to figure out which depending on how your course is taught.
The second one comes from the dot product.
The first one is trickier - you you need to draw the vector head-to-tails summation diagram where one vector is along the V direction and the other vector is along the u direction (lynchpin: the
magnitude can be negative - you can use the fact that the u-direction is horizontal to help your sketch). You can use the geometry of the resulting triangle to help you.
Last edited:
I drew you this image to help you with part B. If you can get this you'll know how to do part C
Now you use law of sines to find F_u and F_v. let's see if that gives you an idea on how to go along with this. I hope you can see how i got the angles.
That's the one! That's really close to doing the homework for the OP though.
Hopefully OP can see how that was drawn.
If not, it may be easier to do F2 first.
Thanks, I had the triangle wrong. I did not really know that it was supposed to be parallel to the other axis. Yes, Simon, it was asking for the first one.
Part B: 200 lb/(sin 30) = Fu/(sin 105) ---> Fu = 386 lb
Part C: 200 / (sin 30) = Fv / (sin 45) ---> Fv = 283 lbThe second one might be helpful to know later on too. Can I ask how the second case comes from the dot product? I'm guessing you would just be
looking for the magnitude and it would probably go something like: (magnitude in direction) = (orginal force) / cos (angle between original force and direction) ?
Can I ask how the second case comes from the dot product?
From the geometric interpretation of the dot product...
i.e. for when you consider forces that do not act through the center of mass...
The amount of ##\vec{u}## in direction ##\vec{v}## is ##\vec{u}\cdot\vec{v}/|\vec{v}| = |\vec{u}|\cos\theta## where ##\theta## is the angle between them.
Note: the angles you were given, 30, 60, 45, are "nice" angles:
##\sin(30)=\cos(60)=\frac{1}{2}## ; ##\cos(30)=\sin(60)=\frac{\sqrt{3}}{2}## ... 2-1-√3 triangle
##\sin(45)=\cos(45)=\frac{1}{\sqrt{2}}## ... 1-1-√2 triangle
... it is worth memorizing these, and the angles in a 3-4-5 triangle.
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FAQ: Resolve force into components along different paths
1. What is meant by resolving a force into components along different paths?
Resolving a force into components along different paths means breaking down a single force into its vertical and horizontal components along different axes or directions. This is done to better
understand and analyze the effects of a force on an object in a specific direction.
2. Why is it important to resolve a force into components?
Resolving a force into components allows for a more accurate analysis of the forces acting on an object. It also simplifies calculations and makes it easier to understand the overall effects of the
force on an object.
3. How is a force resolved into components along different paths?
A force can be resolved into components along different paths using trigonometric principles. The vertical component can be found by multiplying the magnitude of the force by the sine of the angle
between the force and the vertical axis. The horizontal component can be found by multiplying the magnitude of the force by the cosine of the angle between the force and the horizontal axis.
4. What are the benefits of resolving a force into components along different paths?
Resolving a force into components allows for a better understanding of the direction and magnitude of the force acting on an object. This information is useful in engineering and physics, as it can
help determine the stability and motion of objects.
5. Can a force be resolved into components along any path?
No, a force can only be resolved into components along mutually perpendicular axes. This means that the force must be broken down into components along axes that are at a 90-degree angle to each | {"url":"https://www.physicsforums.com/threads/resolve-force-into-components-along-different-paths.668235/","timestamp":"2024-11-09T11:21:43Z","content_type":"text/html","content_length":"100204","record_id":"<urn:uuid:df32dfeb-54f4-4d6a-ada9-7fafd2bbc48d>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00858.warc.gz"} |
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on mobile-bed streams with spatially extensive floodplain vegetation. The mathematical framework for the sediment transport operator is described in Simpson and Castelltort (2006) and Davy and Lague
(2009). This operator calculates an explicit sediment mass balance within the water column at every cell in order to handle the local disequilibria between entrainment and deposition that arise due
to strong spatial variability in shear stress in complex flows. The vegetation drag operator uses the mathematical approach of Nepf (1999) and Kean and Smith (2006), treating vegetation as arrays of
objects (cylinders) that the flow must go around. Compared to methods that simulate the increased roughness of vegetation with a modified Manning's n, this method better accounts for the effects of
drag on the body of the flow and the quantifiable differences between vegetation types and densities (as stem diameter and stem spacing). This operator can simulate uniform vegetation as well as
spatially-varied vegetation across the domain. The vegetation drag module also accounts for the effects of vegetation on turbulent and mechanical diffusivity, following the equations in Nepf (1997,
ANUGA is a hydrodynamic modelling tool that allows users to model realistic flow problems in complex 2D geometries. Examples include dam breaks or the effects of natural hazards such as riverine
flooding, storm surges and tsunami. The user must specify a study area represented by a mesh of triangular cells, the topography and bathymetry, frictional resistance, initial values for water level
(called stage within ANUGA), boundary conditions and forces such as rainfall, stream flows, windstress or pressure gradients if applicable. ANUGA tracks the evolution of water depth and horizontal
momentum within each cell over time by solving the shallow water wave governing equation using a finite-volume method. ANUGA also incorporates a mesh generator that allows the user to set up the
geometry of the problem interactively as well as tools for interpolation and surface fitting, and a number of auxiliary tools for visualising and interrogating the model output. Most ANUGA components
are written in the object-oriented programming language Python and most users will interact with ANUGA by writing small Python scripts based on the ANUGA library functions. Computationally intensive
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response of bifurcations to external changes.
Biogenic mixing of marine sediments
Blocklab treats landscape evolution in landscapes where surface rock may be released as large blocks of rock. The motion, degradation, and effects of large blocks do not play nicely with standard
continuum sediment transport theory. BlockLab is intended to incorporate the effects of these large grains in a realistic way.
CAESAR is a cellular landscape evolution model, with an emphasis on fluvial processes, including flow routing, multi grainsize sediment transport. It models morphological change in river catchments.
CASCADE combines elements of two exploratory morphodynamic models of barrier evolution -- barrier3d (Reeves et al., 2021) and the BarrierR Inlet Environment (brie) model (Nienhuis & Lorenzo-Trueba,
2019) -- into a single model framework. Barrier3d, a spatially-explicit cellular exploratory model, is the core of CASCADE. It is used within the CASCADE framework to simulate the effects of
individual storm events and SLR on shoreface evolution; dune dynamics, including dune growth, erosion, and migration; and overwash deposition by individual storms. BRIE is used to simulate
large-scale coastline evolution arising from alongshore sediment transport processes; this is accomplished by connecting individual Barrier3d models through diffusive alongshore sediment transport.
Human dynamics are incorporated in cascade in two separate modules. The first module simulates strategies for preventing roadway pavement damage during overwashing events, including rebuilding
roadways at sufficiently low elevations to allow for burial by overwash, constructing large dunes, and relocating the road into the barrier interior. The second module incorporates management
strategies for maintaining a coastal community, including beach nourishment, dune construction, and overwash removal.
CHILD computes the time evolution of a topographic surface z(x,y,t) by fluvial and hillslope erosion and sediment transport.
CICE is a computationally efficient model for simulating the growth, melting, and movement of polar sea ice. Designed as one component of coupled atmosphere-ocean-land-ice global climate models,
today’s CICE model is the outcome of more than two decades of community collaboration in building a sea ice model suitable for multiple uses including process studies, operational forecasting, and
climate simulation.
CLUMondo is based on the land systems approach. Land systems are socio-ecological systems that reflect land use in a spatial unit in terms of land cover composition, spatial configuration, and the
management activities employed. The precise definition of land systems depends on the scale of analysis, the purpose of modelling, and the case study region. In contrast to land cover classifications
the role of land use intensity and livestock systems are explicitly addressed. Each land system can be characterized in terms of the fractional land covers.<br>Land systems are characterized based on
the amount of forest in the landscape mosaic and the management type ranging from swidden cultivation to permanent cultivation and plantations.
Caesar Lisflood is a geomorphological / Landscape evolution model that combines the Lisflood-FP 2d hydrodynamic flow model (Bates et al, 2010) with the CAESAR geomorphic model to simulate erosion and
deposition in river catchments and reaches over time scales from hours to 1000's of years. Featuring: Landscape evolution model simulating erosion and deposition across river reaches and catchments A
hydrodynamic 2D flow model (based on the Lisflood FP code) that conserves mass and partial momentum. (model can be run as flow model alone) designed to operate on multiple core processors (parallel
processing of core functions) Operates over a wide range to spatial and time scales (1km2 to 1000km2, <1year to 1000+ years) Easy to use GUI
Calculate the hypsometric integral for each pixel at the catchment. Each pixel is considered a local outlet and the hypsometric integral is calculated according to the characteristics of its
contributing area.
Calculate wave-generated bottom orbital velocities from measured surface wave parameters. Also permits calculation of surface wave spectra from wind conditions, from which bottom orbital velocities
can be determined.
Calculates non-equilibrium suspended load transport rates of various size-density fractions in the bed
Calculates shear velocity associated with grain roughness
Calculates the bedload transport rates and weights per unit area for each size-density. NB. Bedload transport of different size-densities is proportioned according to the volumes in the bed.
Calculates the constant terminal settling velocity of each size-density fraction's median size from Dietrich's equation.
Calculates the critical Shields Theta for the median size of a distribution and then calculates the critical shear stress of the ith, jth fraction using a hiding function
Calculates the critical shear stress for entrainment of the median size of each size-density fraction of a bed using Yalin and Karahan formulation, assuming no hiding
Calculates the gaussian or log-gaussian distribution of instantaneous shear stresses on the bed, given a mean and coefficient of variation.
Calculates the logrithmic velocity distribution called from TRCALC
Calculates the total sediment transport rate in an open channel assuming a median bed grain size
Calculation of Density Stratification Effects Associated with Suspended Sediment in Open Channels. This program calculates the effect of sediment self-stratification on the streamwise velocity and
suspended sediment concentration profiles in open-channel flow. Two options are given. Either the near-bed reference concentration Cr can be specified by the user, or the user can specify a shear
velocity due to skin friction u*s and compute Cr from the Garcia-Parker sediment entrainment relation.
Calculation of Sediment Deposition in a Fan-Shaped Basin, undergoing Piston-Style Subsidence
Calculator for 1D Subaerial Fluvial Fan-Delta with Channel of Constant Width. This model assumes a narrowly channelized 1D fan-delta prograding into standing water. The model uses a single grain size
D, a generic total bed material load relation and a constant bed resistance coefficient. The channel is assumed to have a constant width. Water and sediment discharge are specified per unit width.
The fan builds outward by forming a prograding delta front with an assigned foreset slope. The code employs a full backwater calculation.
Calculator for 1D Subaerial Fluvial Fan-Delta with Channel of Constant Width. This model assumes a narrowly channelized 1D fan-delta prograding into standing water. The model uses a single grain size
D, a generic total bed material load relation and a constant bed resistance coefficient. The channel is assumed to have a constant width. Water and sediment discharge are specified per unit width.
The fan builds outward by forming a prograding delta front with an assigned foreset slope. The code employs the normal flow approximation rather than a full backwater calculation.
CarboCAT uses a cellular automata to model horizontal and vertical distributions of carbonate lithofacies
ChesROMS is a community ocean modeling system for the Chesapeake Bay region being developed by scientists in NOAA, University of Maryland, CRC (Chesapeake Research Consortium) and MD DNR (Maryland
Department of Natural Resources) supported by the NOAA MERHAB program. The model is built based on the Rutgers Regional Ocean Modeling System (ROMS, http://www.myroms.org/) with significant
adaptations for the Chesapeake Bay. The model is developed to provide a community modeling system for nowcast and forecast of 3D hydrodynamic circulation, temperature and salinity, sediment
transport, biogeochemical and ecosystem states with applications to ecosystem and human health in the bay. Model validation is based on bay wide satellite remote sensing, real-time in situ
measurements and historical data provided by Chesapeake Bay Program. http://ches.communitymodeling.org/models/ChesROMS/index.php
Cliffs features: Shallow-Water approximation; Use of Cartesian or spherical (lon/lat) coordinates; 1D and 2D configurations; Structured co-located grid with (optionally) varying spacing; Run-up on
land; Initial conditions or boundary forcing; Grid nesting with one-way coupling; Parallelized with OpenMP; NetCDF format of input/output data. Cliffs utilizes VTCS-2 finite-difference scheme and
dimensional splitting as in (Titov and Synolakis, 1998), and reflection and inundation computations as in (Tolkova, 2014). References: Titov, V.V., and C.E. Synolakis. Numerical modeling of tidal
wave runup. J. Waterw. Port Coast. Ocean Eng., 124(4), 157–171 (1998) Tolkova E. Land-Water Boundary Treatment for a Tsunami Model With Dimensional Splitting. Pure and Applied Geophysics, 171(9),
2289-2314 (2014)
Coastal barrier model that simulates storm overwash and tidal inlets and estimates coastal barrier transgression resulting from sea-level rise.
Code for estimating long-term exhumation histories and spatial patterns of short-term erosion from the detrital thermochronometric data.
Code functionality and purpose may be found in the following references: References # Zhang L., Parker, G., Stark, C.P., Inoue, T., Viparelli, V., Fu, X.D., and Izumi, N. 2015, "Macro-roughness model
of bedrock–alluvial river morphodynamics", Earth Surface Dynamics, 3, 113–138. # Zhang, L., Stark, C.P., Schumer, R., Kwang, J., Li, T.J., Fu, X.D., Wang, G.Q., and Parker, G. 2017, "The
advective-diffusive morphodynamics of mixed bedrock-alluvial rivers subjected to spatiotemporally varying sediment supply" (submitted to JGR)
Computes transient (semi-implicit numerical) and steady-state (analytical and numerical) solutions for the long-profile evolution of transport-limited gravel-bed rivers. Such rivers are assumed to
have an equilibrium width (following Parker, 1978), experience flow resistance that is proportional to grain size, evolve primarily in response to a single dominant "channel-forming" or
"geomorphically-effective" discharge (see Blom et al., 2017, for a recent study and justification of this assumption and how it can be applied), and transport gravel following the Meyer-Peter and
Müller (1948) equation. This combination of variables results in a stream-power-like relationship for bed-material sediment discharge, which is then inserted into a valley-resolving Exner equation to
compute long-profile evolution.
CruAKtemp is a python 2.7 package that is a data component which serves to provide onthly temperature data over the 20th century for permafrost modeling. The original dataset at higher resolution can
be found here: http://ckan.snap.uaf.edu/dataset/historical-monthly-and-derived-temperature-products-771m-cru-ts The geographical extent of this CRUAKtemp dataset has been reduced to greatly reduce
the number of ocean or Canadian pixels. Also, the spatial resolution has been reduced by a factor of 13 in each direction, resulting in an effective pixel resolution of about 10km. The data are
monthly average temperatures for each month from January 1901 through December 2009. | {"url":"https://csdms.colorado.edu/csdms_wiki/index.php?title=Property:Extended_model_description&limit=50&offset=40&from=&until=&filter=","timestamp":"2024-11-03T13:43:13Z","content_type":"text/html","content_length":"121720","record_id":"<urn:uuid:f3f4504d-dd05-4751-b2ed-fb1c6669a4c4>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00501.warc.gz"} |
Per (Pelle) Lindström
Per Lindstöm in 2001
Per (Pelle) Lindström
First professor of Logic and the University of Gothenburg
This is a much condensed version of the obituary by Väänänen and Westerståhl in Theoria 2010 (76) pages 100-107.
Per Lindström, or Pelle Lindström as he insisted on being called, was born on April 9, 1936, and spent most of his academic life at the Department of Philosophy, University of Gothenburg, in Sweden,
where he was employed first as a lecturer (‘docent’) and, from 1991 until his retirement in 2001, as a Professor of Logic.
Lindström is most famous for his work in model theory. In 1964 he made his first major contribution, the so-called Lindström’s test for model completeness. In 1966 he proved the undefinability of
well-order in L[ω[1]ω] (obtained independently and in more generality by Lopez-Escobar). The same year he also introduced the concept of a Lindström quantifier, which has now become standard in model
theory, theoretical computer science, and formal semantics.
It was his 1969 paper ‘On extensions of elementary logic’ (in Theoria), where he presented his famous characterizations of first-order logic—Lindström’s Theorem—in terms of properties such as
compactness, completeness, and Löwenheim-Skolem properties, that was first recognized as a major contribution to logic. It laid the foundation of what has become known as abstract model theory. The
proof was based on Ehrenfeucht-Fraïssé games, a concept he came up with independently, and on a new proof of interpolation. Several other characterizations of first-order logic followed in later
Beginning at the end of the 1970’s, Lindström turned his attention to the study of formal arithmetic and interpretability. He started a truly systematic investigation of this topic, which had been
somewhat dormant since Feferman’s pioneering contributions in the late 1950’s. In doing so he invented novel technically advanced tools, for example, the so-called Lindström fixed point construction,
a far-reaching application of Gödel’s diagonalization lemma to define arithmetical formulas with specific properties. Pelle Lindström had an exceptionally clear and concise style in writing
mathematical logic. His 1997 book, Aspects of Incompleteness, remains a perfect example: it provides a systematic introduction to his work in arithmetic and interpretability. The book is short but
rich in material.
Throughout his life, Pelle Lindström also took an active interest in philosophy. He participated in the debate following Roger Penrose’s new version of the argument that Gödel’s Incompleteness
Theorems show that the human mind is not mechanical. He presented his own philosophy of mathematics, which he called ‘quasi-realism’, in a paper in The Monist in 2000. It is based on the idea that
the ‘visualizable’ parts of mathematics are beyond doubt (and that classical logic holds for them). He counted as visualizable not only the ω-sequence of natural numbers but also arbitrary sets of
numbers, the latter visualizable as branches in the infinite binary tree, whereas nothing similar can be said for sets of sets of numbers, for example.
Pelle Lindström passed away in Gothenburg, Sweden, on August 21, 2009, after a short period of illness. | {"url":"https://logic-gu.se/lindstrom-lectures/per-lindstrom/","timestamp":"2024-11-05T09:23:58Z","content_type":"text/html","content_length":"11322","record_id":"<urn:uuid:aaa20602-3e0c-4ca7-a329-b89efd29a67e>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00488.warc.gz"} |
The Cicada Principle, revisited with CSS variables
Many of today’s web crafters were not writing CSS at the time Alex Walker’s landmark article The Cicada Principle and Why it Matters to Web Designers was published in 2011. Last I heard of it was
in 2016, when it was used in conjunction with blend modes to pseudo-randomize backgrounds even further.
So what is the Cicada Principle and how does it relate to web design in a nutshell? It boils down to: when using repeating elements (tiled backgrounds, different effects on multiple elements etc),
using prime numbers for the size of the repeating unit maximizes the appearance of organic randomness. Note that this only works when the parameters you set are independent.
When I recently redesigned my blog, I ended up using a variation of the Cicada principle to pseudo-randomize the angles of code snippets. I didn’t think much of it until I saw this tweet:
This made me think: hey, maybe I should actually write a blog post about the technique. After all, the technique itself is useful for way more than angles on code snippets.
The main idea is simple: You write your main rule using CSS variables, and then use :nth-of-*() rules to set these variables to something different every N items. If you use enough variables, and
choose your Ns for them to be prime numbers, you reach a good appearance of pseudo-randomness with relatively small Ns.
In the case of code samples, I only have two different top cuts (going up or going down) and two different bottom cuts (same), which produce 2*2 = 4 different shapes. Since I only had four shapes, I
wanted to maximize the pseudo-randomness of their order. A first attempt looks like this:
pre {
clip-path: polygon(var(--clip-top), var(--clip-bottom));
--clip-top: 0 0, 100% 2em;
--clip-bottom: 100% calc(100% - 1.5em), 0 100%;
pre:nth-of-type(odd) {
--clip-top: 0 2em, 100% 0;
pre:nth-of-type(3n + 1) {
--clip-bottom: 100% 100%, 0 calc(100% - 1.5em);
This way, the exact sequence of shapes repeats every 2 * 3 = 6 code snippets. Also, the alternative --clip-bottom doesn’t really get the same visibility as the others, being present only 33.333% of
the time. However, if we just add one more selector:
pre {
clip-path: polygon(var(--clip-top), var(--clip-bottom));
--clip-top: 0 0, 100% 2em;
--clip-bottom: 100% calc(100% - 1.5em), 0 100%;
pre:nth-of-type(odd) {
--clip-top: 0 2em, 100% 0;
pre:nth-of-type(3n + 1),
pre:nth-of-type(5n + 1) {
--clip-bottom: 100% 100%, 0 calc(100% - 1.5em);
Now the exact same sequence of shapes repeats every 2 * 3 * 5 = 30 code snippets, probably way more than I will have in any article. And it’s more fair to the alternate --clip-bottom, which now
gets 1/3 + 1/5 - 1/15 = 46.67%, which is almost as much as the alternate --clip-top gets!
You can explore this effect in this codepen:
Or, to better explore how different CSS creates different pseudo-randomness, you can use this content-less version with three variations:
Of course, the illusion of randomness is much better with more shapes, e.g. if we introduce a third type of edge we get 3 * 3 = 9 possible shapes:
I also used primes 7 and 11, so that the sequence repeats every 77 items. In general, the larger primes you use, the better the illusion of randomness, but you need to include more selectors, which
can get tedious.
So this got me thinking: What else would this technique be cool on? Especially if we include more values as well, we can pseudo-randomize the result itself better, and not just the order of only 4
different results.
So I did a few experiments.
Pseudo-randomized color swatches, with variables for hue, saturation, and lightness.
Which one looks more random? Why do you think that is?
Admittedly, this one can be done with just longhands, but since I realized this after I had already made it, I figured eh, I may as well include it 🤷🏽‍♀️
It is also really cool when combined with pseudo-random colors (just hue this time):
Lots of things here:
• Using translate and transform together to animate them separately without resorting to CSS.registerPropery()
• Pseudo-randomized horizontal offset, animation-delay, font-size
• Technically we don’t need CSS variables to pseudo-randomize font-size, we can just set the property itself. However, variables enable us to pseudo-randomize it via a multiplier, in order to
decouple the base font size from the pseudo-randomness, so we can edit them independently. And then we can use the same multiplier in animation-duration to make smaller snowflakes fall slower!
In general, the larger the primes you use, the better the illusion of randomness. With smaller primes, you will get more variation, but less appearance of randomness.
There are two main ways to use primes to create the illusion of randomness with :nth-child() selectors:
The first way is to set each trait on :nth-child(pn + b) where p is a prime that increases with each value and b is constant for each trait, like so:
:nth-child(3n + 1) { property1: value11; }
:nth-child(5n + 1) { property1: value12; }
:nth-child(7n + 1) { property1: value13; }
:nth-child(11n + 1) { property1: value14; }
:nth-child(3n + 2) { property2: value21; }
:nth-child(5n + 2) { property2: value22; }
:nth-child(7n + 2) { property2: value23; }
:nth-child(11n + 2) { property2: value24; }
The benefit of this approach is that you can have as few or as many values as you like. The drawback is that because primes are sparse, and become sparser as we go, you will have a lot of “holes”
where your base value is applied.
The second way (which is more on par with the original Cicada principle) is to set each trait on :nth-child(pn + b) where p is constant per trait, and b increases with each value:
:nth-child(5n + 1) { property1: value11; }
:nth-child(5n + 2) { property1: value12; }
:nth-child(5n + 3) { property1: value13; }
:nth-child(5n + 4) { property1: value14; }
:nth-child(7n + 1) { property2: value21; }
:nth-child(7n + 2) { property2: value22; }
:nth-child(7n + 3) { property2: value23; }
:nth-child(7n + 4) { property2: value24; }
This creates a better overall impression of randomness (especially if you order the values in a pseudo-random way too) without “holes”, but is more tedious, as you need as many values as the
prime you’re using.
What other cool examples can you think of? | {"url":"https://verou.me/blog/2020/07/the-cicada-principle-revisited-with-css-variables/","timestamp":"2024-11-13T06:22:28Z","content_type":"text/html","content_length":"18970","record_id":"<urn:uuid:02002a18-0f34-4cc4-bf1d-01b91e7afb8d>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00734.warc.gz"} |
Analysis and
Analysis and Partial Differential Equations
Analysis is the study of sequences and functions and their main mathematical properties such as convergence, continuity, and integrability. These concepts underpin the theory of calculus with
numerous applications in engineering and the natural sciences. An important field within analysis is the study of partial differential equations (PDE), which are equations for functions of several
variables and their respective derivatives. The research group in MSCS is particularly interested in PDE arising from fluid dynamics or quantum mechanics, as well as in harmonic analysis and
integrable systems. Connections to probability theory and computational mathematics are also explored.
Feb 10 2025
Monday, 4:00 pm–4:50 pm
636 SEO
Feb 24 2025
Monday, 4:00 pm–4:50 pm
636 SEO
Mar 31 2025
Monday, 4:00 pm–4:50 pm
636 SEO | {"url":"https://mscs.uic.edu/research-groups/analysis-partial-differential-equations/","timestamp":"2024-11-05T15:53:21Z","content_type":"text/html","content_length":"227199","record_id":"<urn:uuid:1ef06194-bd83-4977-af3a-a79983a33ba1>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00806.warc.gz"} |
Square Foot to Square Meter Converter (ft² to m²)
Square Foot to Square Meter Converter
“Maximize Precision: Convert Square Feet to Square Meters Instantly.”
How to Convert Square Feet to Square Meters?
To convert area from Square Feet (ft²) to Square Meters (m²), use the formula:
Square Meters (m²) = Square Feet (ft²) / 10.7639
This formula is based on the conversion factor where 1 square foot equals approximately 0.092903 square meters.
Steps to Convert Square Feet to Square Meters:
To convert an area from Square Feet to Square Meters:
1. Divide the Square Feet value by 10.7639.
Example Conversion:
To convert 500 Square Feet (ft²) to Square Meters (m²):
500 ft² / 10.7639 = 46.4515 square meters
So, 500 Square Feet is approximately equal to 46.4515 Square Meters.
Square Foot to Square Meter Conversion Formula
To convert square feet to square meters, you can use the following formula:
Square meters = Square feet × 0.0929
Example of Square Foot to Square Meter Conversion
Example 1: 100 Square Foot to Square Meter Conversion
For example, let’s convert 100 square feet to square meters:
Square meters = 100 × 0.0929 Square meters = 9.29 square meters
Square Foot to Square Meter Conversion Table
Here’s a conversion table for square feet to square meters for the first 20 entries:
Square Feet Square Meters
1 0.0929
2 0.1858
3 0.2787
4 0.3716
5 0.4645
6 0.5574
7 0.6503
8 0.7432
9 0.8361
10 0.929
11 1.0219
12 1.1148
13 1.2077
14 1.3006
15 1.3935
16 1.4864
17 1.5793
18 1.6722
19 1.7651
20 1.858
Square Foot to Square Meter Converter FAQs
How do I convert square feet to square meters?
To convert square feet to square meters, multiply the number of square feet by 0.092903. This is because 1 square meter equals approximately 0.092903 square feet.
Why is the conversion factor for square feet to square meters 0.092903?
The conversion factor 0.092903 is derived from the relationship between feet and meters. Since 1 foot is approximately 0.3048 meters, squaring this value (0.3048 × 0.3048) gives 0.092903 square
meters per square foot.
Can I use an online calculator to convert square feet to square meters?
Yes, there are many online calculators available where you can input the value in square feet, and it will automatically convert it to square meters. These tools are convenient for quick conversions.
Is the conversion from square feet to square meters consistent across all contexts?
Yes, the conversion factor of 0.092903 square meters per square foot remains the same in all contexts.
How many square meters are in 500 square feet?
To convert 500 square feet to square meters, multiply by 0.092903. The result is approximately 46.45 square meters.
Related Posts
Related Tags
Square meters calculator, How to convert square feet to square meters in Excel, Square meter to square feet converter app, 1 Square feet to cm, Meter to square meter, 1 square feet to inch, 1 square
feet means, Feet to meters
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Residual Income Formula | Calculator (Examples With Excel Template)
Updated July 26, 2023
Residual Income Formula (Table of Contents)
What is Residual Income Formula?
In Corporate finance, the term “residual income” refers to the amount of operating income generated in excess of the minimum required return or the desired income.
As such, residual income can be seen as a performance assessment tool for the company to see how efficiently it able to utilize its business assets. In fact, the residual income is the performance
indicator for the companies just like return on investment for portfolio managers. The formula for residual income can be derived by deducting the product of the minimum required rate of return and
average operating assets from the operating income. Mathematically, Residual Income is represented as,
Residual Income = Operating Income – Minimum Required Rate of Return * Average Operating Assets
Examples of Residual Income Formula (With Excel Template)
Let’s take an example to understand the calculation of Residual Income in a better manner.
Residual Income Formula – Example #1
Let us take the example of an investment center that had an operating income of $1,000,000 during the year by using operating assets worth $5,000,000. Calculate the residual income of the investment
center if the minimum required rate of return is 18%.
Residual Income is calculated using the formula given below
Residual Income = Operating Income – Minimum Required Rate of Return * Average Operating Assets
• Residual Income = $1,000,000 – 18% * $5,000,000
• Residual Income = $100,000
Therefore, the residual income of the investment center stood at $100,000.
Residual Income Formula – Example #2
Let us take the example of a company with operating income during the current year of $80,000. The company has an operating asset base of $500,000, while the cost of capital is 12% as per the latest
annual report. Calculate the residual income of the company during the year.
Residual Income of the company is calculated using the formula given below
Residual Income = Operating Income – Minimum Required Rate of Return * Average Operating Assets
• Residual Income = $80,000 – 12% * $500,000
• Residual Income = $20,000
Therefore, the residual income of the company during the year is $20,000.
Residual Income Formula – Example #3
Let us take the example of a company which has recently acquired a new unit as a diversification of its existing operation. The value of operating assets of the unit is $200,000 at the beginning of
the year and $250,000 at the end of the year. During the year, the unit generated operating income of $50,000. As per the corporate strategy, the minimum required rate of return from the unit is 15%.
Calculate whether the unit is able to generate any residual income during the year.
Average Operating Assets is calculated as
• Average Operating Assets = ($200,000 + $250,000) / 2
• Average Operating Assets = $225,000
Residual Income is calculated using the formula given below
Residual Income = Operating Income – Minimum Required Rate of Return * Average Operating Assets
• Residual Income = $50,000 – 15% * $225,000
• Residual Income = $16,250
Therefore, the company is able to generate a residual income of $16,250 during the year.
The formula for residual income can be calculated by using the following steps:
Step 1: Firstly, determine the minimum required rate of return expected by the investor based on their investment strategy, risk appetite, investment horizon, and current market return. In fact, most
cases companies use the cost of capital as the minimum required rate of return.
Step 2: Next, determine the operating assets or the total capital employed by the company in the operations. In most cases, the average of the value of the operating assets at the beginning of the
year and at the end of the year is used.
Step 3: Next, calculate the minimum required income based on the minimum required rate of return (step 1) and the average operating assets (step 2) as shown below.
Minimum Required Income = Minimum Required Rate of Return * Average Operating Assets
Step 4: Next, determine the operating income of the company which is an income statement item.
Step 5: Finally, the formula for residual income can be derived by deducting the minimum required income (step 3) from the operating income (step 4) as shown below.
Residual Income = Operating Income – Minimum Required Income
Residual Income = Operating Income – Minimum Required Rate of Return * Average Operating Assets
Relevance and Uses of Residual Income Formula
It is important to understand the concept of residual income because it is usually used in the performance assessment of capital investment, department or business unit. A positive residual incomes
implies that the unit has been able to generate more return than the minimum required rate, which is desirable. As such, the higher the residual income, the better it is considered by the company.
However, there can be instances when a project or business unit has failed the test for return on investment due to the low rate of return but has cleared the test for residual income on the back of
nominal positive dollar value, which can be very tricky and requires management call. Another major disadvantage of residual income technique is that it favors bigger investments against smaller ones
because it assesses on the basis of the absolute dollar amount.
Residual Income Formula Calculator
You can use the following Residual Income Calculator
Operating Income
Minimum Required Rate of Return
Average Operating Assets
Residual Income
Residual Income = Operating Income - Minimum Required Rate of Return * Average Operating Assets
0 - 0 * 0 = 0
Recommended Articles
This is a guide to the Residual Income Formula. Here we discuss How to Calculate Residual Income along with practical examples. We also provide a Residual Income Calculator with a downloadable excel
template. You may also look at the following articles to learn more – | {"url":"https://www.educba.com/residual-income-formula/","timestamp":"2024-11-04T04:55:21Z","content_type":"text/html","content_length":"341347","record_id":"<urn:uuid:47fbd06d-a25b-4669-995f-0a541ae557ac>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00334.warc.gz"} |
How do you find the integral of int cotx dx? | HIX Tutor
How do you find the integral of #int cotx dx#?
Answer 1
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Answer 2
To find the integral of (\int \cot(x) , dx), you can use the following steps:
1. Rewrite (\cot(x)) as (\frac{\cos(x)}{\sin(x)}).
2. Perform a substitution: Let (u = \sin(x)), then (du = \cos(x) , dx).
3. Rewrite the integral in terms of (u): (\int \frac{1}{u} , du).
4. Integrate (\frac{1}{u}) with respect to (u): (\ln|u| + C).
5. Substitute back (u = \sin(x)): (\ln|\sin(x)| + C).
So, the integral of (\int \cot(x) , dx) is (\ln|\sin(x)| + C), where (C) is the constant of integration.
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Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
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process electricty from coal
The Energy Information Administration lists the heat rate for different types of power plants, and the average operating efficiencies of thermal power plants in the in 2020 were: Natural gas: 44%
efficient, meaning 56% of the energy in the gas was lost, with 44% of the energy turned into electricity. Coal: 32% efficient.
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Natural gas is burned to produce electricity following the same general process used in a coal power plant (figure (PageIndex{n})). Oil is occasionally used to generate electricity as well.
Figure (PageIndex{n}): This combustion chamber burns either natural gas or oil. Fuel flows through a natural gas line or from oil storage into the ...
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The steam spins turbines to generate electricity. In Australia in 2017 coal was used to produce about 60% of the nation's electricity requirements. Using brown coal for power generation is
problematic because of the high water content. It crumbles easily on exposure to the air which reduces its value as a fuel and requires specialised storage.
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Coal produces about 211 pounds (96 kilograms) of heattrapping carbon dioxide per million BTUs of energy produced, compared to natural gas which produces about 117 pounds (53 kilograms) and ...
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Dwindling coal inventories and higher global coal demand than available supply led to less coalfired electricity generation during most of 2021. Generation declined because operators of electric
generation plants wanted to ensure they had sufficient supply to meet demand in the 202122 winter heating season. In late 2021 and into 2022, coal ...
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The process details the steps in the production of electricity. Looking from an overall perspective, it is readily apparent that energy production involves the combination of coal and oxygen
undergoing various chemical processes including heating that results in gases that then power two different types of turbines to produce electricity | Band: 5
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Fossil fuels are a common source of energy for electricity generation in Canada. In 2016, Canada got about % of its electricity from coal, % from natural gas and % from oil and diesel. Most of
the electricity in Alberta, Saskatchewan, Nova Scotia, and Nunavut comes from fossil fuels. Other provinces also use fossil fuels to generate ...
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Or follow us on Google News! coalrelated CO 2 emissions decreased by 7%, or 68 million metric tons (MMmt), in 2022 relative to 2021. This decrease was largely due to an 8% decline in coal ...
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Coalfired power plants are the largest single source of CO2 emissions in the US power sector, and accounted for 24 percent of all energyrelated carbon dioxide emissions in 2016 . ... However,
capturing and compressing CO2 is a very energyintensive process, causing large reductions in the amount of net energy the plant is able to produce. ...
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The process heat energy demand was 199 petajoules or around 35% of total energy used in New Zealand in 2016. Around half of the process heat demand was met by burning coal or natural gas, the
remaining demand was largely met by electricity, bioenergy5, using geothermal energy directly, and liquid fossil fuels ( diesel).6 Key facts7 ...
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Alternatively, electricity from natural gas may be derived by piping natural gas underground to power plants. Similar to the process with coal, the power plants burn natural gas to boil water to
produce steam. The steam spins the blades of a turbine that are connected to a generator. The generator then spins magnets to generate electricity.
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Coal is an abundant natural resource that can be used as a source of energy, as a chemical source from which numerous synthetic compounds (, dyes, oils, waxes, pharmaceuticals, and pesticides)
can be derived, and in the production of coke for metallurgical is a major source of energy in the production of electrical power using steam generation.
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Coal still leads the charge when it comes to electricity, representing % of global power generation in 2022, followed by natural gas at %, and hydroelectric at %. Source: Energy Institute. Over
threequarters of the world's total coalgenerated electricity is consumed in just three countries. China is the top user of coal, making ...
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The latter three ranks are commonly referred to as "black coal" while lignite is commonly called "brown coal". Coal is Australia's largest energy resource. At the end of 2019, Australia's
recoverable Economic Demonstrated Resources were 75,428 million tonnes (Mt) of black coal and 73,865 Mt of brown coal.
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Steam coal, also known as thermal coal, is used in power stations to generate electricity. First coal is milled to a fine powder, which increases the surface area and allows it to burn more
quickly. In pulverised coal combustion (PCC) systems, the powdered coal is blown into the combustion chamber of a boiler where it is burnt at high ...
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Coal generated less than 2% of Britain's electricity in 2020, despite being the largest single energy source seven years the country's electricity gets cleaner every year, there ...
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Petroleum. Hydrocarbon gas liquids. Natural gas. Coal. Nuclear energy. These energy sources are called nonrenewable because their supplies are limited to the amounts that we can mine or extract
from the earth. Coal, natural gas, and petroleum formed over thousands of years from the buried remains of ancient sea plants and animals that lived ...
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Coal is the biggest single source of energy for electricity production and its share is growing. The efficiency of converting coal into electricity matters: more efficient power plants use less
fuel and emit less climatedamaging carbon dioxide. This book explores how efficiency is measured and reported at coalfired power plants.
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Coal is a combustible black or brownishblack sedimentary rock with a high amount of carbon and hydrocarbons. Coal is classified as a nonrenewable energy source because it takes millions of years
to form. Coal contains the energy stored by plants that lived hundreds of millions of years ago in swampy forests. Layers of dirt and rock covered the ...
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"There's a major issue around captive coal power stations in Indonesia, that runs the risk of derailing or slowing that JETP process," said Leo Roberts, an analyst at climate think tank E3G.
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Step 1: Mining. The first step in the process of generating electricity from coal is to mine it. Coal is typically found in underground mines or in openpit mines. The coal is extracted using
large machinery, such as draglines and shovels. Once the coal is mined, it is transported to a power plant via truck, train, or conveyor belt.
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IELTS WRITING TASK 1. You should spend about 20 minutes on this task. The diagram shows how electricity is produced from coal. Summarise the information by selecting and reporting the main
features, and make comparisons where relevant. Write at least 150 words. SAMPLE ANSWER: The diagram illustrates the process of generating electricity from coal.
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Coal fired power plants follow the Rankine cycle in order to complete this process. Since they require plenty of water to be circulated in this cycle, coal power plants need to be located near a
body of water. The process of coal fired plants can be seen below in Figure 3. Figure 3. The process of a coal fired power plant to convert coal into ...
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The method used for generating electricity is the same as with a coal power plant or a bioenergy power plant. It is only the fuel that differentiates the power plant. ... Although the Barsebäck
Plant has been decommissioned, we would be happy to tell you about the process and other aspects related to the operation of a nuclear power plant.
WhatsApp: +86 18838072829 | {"url":"https://panirecord.fr/process_electricty_from_coal/7628.html","timestamp":"2024-11-02T02:17:45Z","content_type":"application/xhtml+xml","content_length":"23749","record_id":"<urn:uuid:549b4ddc-e3cd-4b9e-8615-0c4b93359df4>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00882.warc.gz"} |
Exploring Polymorphism
Let's explore some different ways in which polymorphism presents itself. Consider the following example of the union design pattern:
/** * An interface that represents an operation on two doubles */ public interface IBinaryOp { double apply( double x, double y); // all interface methods are public and abstract by default }
/** * An IBinaryOp that adds two numbers */ public class AddOp implements IBinaryOp { public double apply( double x, double y) { return x+y; } }
/** * An IBinaryOp that multiplies two numbers */ public class MultOp implements IBinaryOp { public double apply( double x, double y) { return x*y; }
public String getDescription() { return "MultOp is a multiplying function."; } }
Exercise 2.1
Is the following legal code? IBinaryOp bop = new IBinaryOp();
Exercise 2.2
Is the following legal code? IBinaryOp bop = new AddOp();
Exercise 2.3
Given the above declaration and assignment of bop, is the following assignment then possible? bop = new MultOp();
Exercise 2.4
Suppose we have bop = new AddOp(); , what is the result of bop.apply(5,3) ?
Exercise 2.5
Suppose we now say bop = new MultOp(), what is the result of bop.apply(5,3) now?
Exercise 2.6
Suppose we have some variable, called myOp of type IBinaryOp what is the result of myOp.apply(5,3)?
Exercise 2.7
Suppose we have bop = new MultOp(), is it legal to call bop.getDescription() ?
Exercise 2.8
Is the following legal code? AddOp aop = new AddOp()
Exercise 2.9
Given the declaration in the previous exercise, is the following legal? Aop = new MultiOp()
Exercise 2.10
Suppose we have definitions of aop and bop from above. Is the following legal? That is, can we compile and run the following statement without error? bop = aop;
Exercise 2.11
Is the converse legal as well? That is, using the above definitions, can we compile and run the following statement? aop = bop; | {"url":"https://www.opentextbooks.org.hk/zh-hant/ditatopic/8186","timestamp":"2024-11-12T00:27:35Z","content_type":"text/html","content_length":"168018","record_id":"<urn:uuid:54ef7a51-ecde-4485-93ec-6d905e3d6412>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00122.warc.gz"} |
Negotiating different disciplinary discourses: biology students’ ritualized and exploratory participation in mathematical modeling activities
Non-mathematics specialists’ competence and confidence in mathematics in their disciplines have been highlighted as in need of improvement. We report from a collaborative, developmental research
project which explores the conjecture that greater integration of mathematics and biology in biology study programs, for example through engaging students with Mathematical Modeling (MM) activities,
is one way to achieve this improvement. We examine the evolution of 12 first-semester biology students’ mathematical discourse as they engage with such activities in four sessions which ran
concurrently with their mandatory mathematics course and were taught by a mathematician with extensive experience with MM. The sessions involved brief introductions to different aspects of MM,
followed by small-group work on tasks set in biological contexts. Our analyses use the theory of commognition to investigate the tensions between ritualized and exploratory participation in the
students’ MM activity. We focus particularly on a quintessential routine in MM, assumption building: we trace attempts which start from ritualized engagement in the shape of “guesswork” and evolve
into more productively exploratory formulations. We also identify signs of persistent commognitive conflict in the students’ activity, both intra-mathematical (concerning what is meant by a “math
task”) and extra-mathematical (concerning what constitutes a plausible solution to the tasks in a biological sense). Our analyses show evidence of the fluid interplay between ritualized and
exploratory engagement in the students’ discursive activity and contribute towards what we see as a much needed distancing from operationalization of the commognitive constructs of ritual and
exploration as an unhelpfully dichotomous binary.
• Assumption building
• Commognition
• Discourse
• Mathematics in biology
• Rituals and explorations
• University mathematics education
• Person: Research Group Member, Academic, Teaching & Research | {"url":"https://research-portal.uea.ac.uk/en/publications/negotiating-different-disciplinary-discourses-biology-students-ri","timestamp":"2024-11-09T06:04:13Z","content_type":"text/html","content_length":"57299","record_id":"<urn:uuid:64b2568f-e146-4e63-9b8b-90f4745ab581>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00601.warc.gz"} |
Zeta Functions for Two-Dimensional Shifts of Finite Typesearch
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Zeta Functions for Two-Dimensional Shifts of Finite Type
eBook ISBN: 978-0-8218-9457-6
Product Code: MEMO/221/1037.E
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Zeta Functions for Two-Dimensional Shifts of Finite Type
eBook ISBN: 978-0-8218-9457-6
Product Code: MEMO/221/1037.E
List Price: $60.00
MAA Member Price: $54.00
AMS Member Price: $36.00
• Memoirs of the American Mathematical Society
Volume: 221; 2013; 60 pp
MSC: Primary 37; Secondary 82; 11
This work is concerned with zeta functions of two-dimensional shifts of finite type. A two-dimensional zeta function \(\zeta^{0}(s)\), which generalizes the Artin-Mazur zeta function, was given
by Lind for \(\mathbb{Z}^{2}\)-action \(\phi\). In this paper, the \(n\)th-order zeta function \(\zeta_{n}\) of \(\phi\) on \(\mathbb{Z}_{n\times \infty}\), \(n\geq 1\), is studied first. The
trace operator \(\mathbf{T}_{n}\), which is the transition matrix for \(x\)-periodic patterns with period \(n\) and height \(2\), is rotationally symmetric. The rotational symmetry of \(\mathbf
{T}_{n}\) induces the reduced trace operator \(\tau_{n}\) and \(\zeta_{n}=\left(\det\left(I-s^{n}\tau_{n}\right)\right)^{-1}\).
The zeta function \(\zeta=\prod_{n=1}^{\infty} \left(\det\left(I-s^{n}\tau_{n}\right)\right)^{-1}\) in the \(x\)-direction is now a reciprocal of an infinite product of polynomials. The zeta
function can be presented in the \(y\)-direction and in the coordinates of any unimodular transformation in \(GL_{2}(\mathbb{Z})\). Therefore, there exists a family of zeta functions that are
meromorphic extensions of the same analytic function \(\zeta^{0}(s)\). The natural boundary of zeta functions is studied. The Taylor series for these zeta functions at the origin are equal with
integer coefficients, yielding a family of identities, which are of interest in number theory. The method applies to thermodynamic zeta functions for the Ising model with finite range
□ Chapters
□ 1. Introduction
□ 2. Periodic patterns
□ 3. Rationality of $\zeta _{n}$
□ 4. More symbols on larger lattice
□ 5. Zeta functions presented in skew coordinates
□ 6. Analyticity and meromorphic extensions of zeta functions
□ 7. Equations on $\mathbb {Z}^{2}$ with numbers in a finite field
□ 8. Square lattice Ising model with finite range interaction
• Permission – for use of book, eBook, or Journal content
• Book Details
• Table of Contents
• Requests
Volume: 221; 2013; 60 pp
MSC: Primary 37; Secondary 82; 11
This work is concerned with zeta functions of two-dimensional shifts of finite type. A two-dimensional zeta function \(\zeta^{0}(s)\), which generalizes the Artin-Mazur zeta function, was given by
Lind for \(\mathbb{Z}^{2}\)-action \(\phi\). In this paper, the \(n\)th-order zeta function \(\zeta_{n}\) of \(\phi\) on \(\mathbb{Z}_{n\times \infty}\), \(n\geq 1\), is studied first. The trace
operator \(\mathbf{T}_{n}\), which is the transition matrix for \(x\)-periodic patterns with period \(n\) and height \(2\), is rotationally symmetric. The rotational symmetry of \(\mathbf{T}_{n}\)
induces the reduced trace operator \(\tau_{n}\) and \(\zeta_{n}=\left(\det\left(I-s^{n}\tau_{n}\right)\right)^{-1}\).
The zeta function \(\zeta=\prod_{n=1}^{\infty} \left(\det\left(I-s^{n}\tau_{n}\right)\right)^{-1}\) in the \(x\)-direction is now a reciprocal of an infinite product of polynomials. The zeta function
can be presented in the \(y\)-direction and in the coordinates of any unimodular transformation in \(GL_{2}(\mathbb{Z})\). Therefore, there exists a family of zeta functions that are meromorphic
extensions of the same analytic function \(\zeta^{0}(s)\). The natural boundary of zeta functions is studied. The Taylor series for these zeta functions at the origin are equal with integer
coefficients, yielding a family of identities, which are of interest in number theory. The method applies to thermodynamic zeta functions for the Ising model with finite range interactions.
• Chapters
• 1. Introduction
• 2. Periodic patterns
• 3. Rationality of $\zeta _{n}$
• 4. More symbols on larger lattice
• 5. Zeta functions presented in skew coordinates
• 6. Analyticity and meromorphic extensions of zeta functions
• 7. Equations on $\mathbb {Z}^{2}$ with numbers in a finite field
• 8. Square lattice Ising model with finite range interaction
Permission – for use of book, eBook, or Journal content
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Precalculus - Online Tutor, Practice Problems & Exam Prep
Welcome back, everyone. So, up to this point, we've spent a lot of time talking about vectors and what they look like. Now recall that visually, vectors are these arrows drawn in space, and we've
talked about how these arrows can be stretched or shrunk, or combined with other arrows to create resultant vectors. Well, what we're going to be talking about in this video is position vectors and
component form. And this might sound a bit random and confusing but don't sweat it because it turns out all we're really going to be learning about in this video is a simple way to write numbers that
represent our vector. And I think you're going to find mathematically it's actually very intuitive. So without further ado, let's get right into things, because this is an important concept to
Now let's say we have this vector, which we'll call vector \( \mathbf{v} \). As you can see, this is an arrow drawn in space. Now what we can do is represent this as a position vector, and a position
vector is simply a vector where the initial point is drawn at the origin. So, if we want to draw vector \( \mathbf{v} \) as a position vector, I just need to relocate this so the initial point is at
the origin. And that right there is the position vector, and that's all there is to it. It's just moving your vector to the origin of the graph.
Now the question becomes, how can we write this vector with some kind of numbers? Well, we can see that we have some numbers here on the graph, and we can use these numbers to figure out what our
vector is. Because what we do is we represent these vectors using component form, where we have an x-component and a y-component. And all these components do is tell you the length of the vector in
the x and y directions. So you can see in the x-direction, we need to go 3 units to the right, so we'd have 3. And then in the y-direction, we need to go 2 units up, so we'd have 2. So this vector is
(3,2), and that's all there is to it. As you can see, it's really straightforward.
Now it turns out that there are also ways that you can represent these vectors if you don't have a position vector. So say that you were given some initial point of the vector, like a point right
there, and then a terminal point over here. You could figure out what the vector is in component form by using this equation down here. And to really put this equation to use and understand it rather
than just looking at it, let's actually try an example where we have to do this. So, in this example, we are told if a vector has initial (0,2,3) and a terminal (0,3,5), without drawing the vector,
write the vector in component form. So we're not allowed to just graph this immediately and figure out what it looks like. What we need to do is use this equation to figure out what our vector is.
But this equation is actually pretty simple to use. So all I need to do is recognize that our vector \( \mathbf{v} \) is going to be the difference in the x components, the difference in the y
So we can see that we have the final x minus the initial x, and then we're going to have the final y minus the initial y. Now I can see what these values are based on the points above. So if I go
ahead and go to this first point, you see this first point is (0,2,3). I can see that the second point is (0,3,5). So I'm going to have for the x points is the difference between 3-2. So we're going
to have 3 minus 2. And the reason that I put 3 first is that notice it's the final x minus the initial x. It's going to be 0.2 minus 0.1. So we have 3 minus 2, and then we're going to subtract the
y-values. So the final y is 5, and the initial y is 3. This is what our vector is going to be. So we're going to have 3 minus 2 which is 1, and we're going to have 5 minus 3 which is 2. And that
right there is the solution. That is vector \( \mathbf{v} \).
So this is how you can solve problems when you can't initially draw them or don't initially have some kind of graph of the vector. Now if we want to know what this vector looks like, we actually can
use this graph just for reference. Well, I can see that our vector is (1,2). And if we draw this as a position vector, starting at the origin of our graph, we're going to go 1 to the right, and we're
going to go 2 up. And that right there would be our vector \( \mathbf{v} \). So as you can see, whatever you're dealing with these types of vectors, you're going to have the x-component, which is how
far we travel in the x-direction, and the y-component, which is how far we travel in the y-direction. And that's always going to be the case when using component form. So that is how you can
represent vectors using numbers, and how you can draw position vectors which are at the origin of your graph. So hope you found this video helpful. Thanks for watching. | {"url":"https://www.pearson.com/channels/precalculus/learn/patrick/14-vectors/vectors-in-component-form?chapterId=24afea94","timestamp":"2024-11-10T21:13:37Z","content_type":"text/html","content_length":"496759","record_id":"<urn:uuid:bc2788c3-3957-4dd5-9ed3-6931af028c07>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00046.warc.gz"} |
Algebra 1 - Writing Linear Equations
This post explains and gives practice opportunities related to TEKS A.2C:
write linear equations in two variables given a table of values, a graph, and a verbal description;
Students learn two new formulas for linear functions, the point-slope form and the standard form. Using what they learned in 8th grade math about how to find the slope and how to write a linear
function in slope-intercept form, students now explore how to write linear functions given a set of coordinates in tabular form.
Sometimes students are given a story problem and have to use that to generate an equation for a linear function in either slope-intercept or standard form.
STAAR Practice
Between 2016 and 2024 (including redesign practice), this readiness standard has been tested 18 times on the STAAR test. Videos explaining the problems can be found below. If you'd rather take a quiz
over these questions, click here. The videos below are linked to the questions in the quiz as answer explanations after the quiz is submitted.
To view all the posts in this Algebra 1 TEKS review series, click here. | {"url":"https://www.fiveminutemath.net/post/algebra-1-writing-linear-equations-1","timestamp":"2024-11-01T20:23:44Z","content_type":"text/html","content_length":"1050594","record_id":"<urn:uuid:30976cfc-ff9c-44ba-9d08-50e87db5770e>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00553.warc.gz"} |
September 2016 LSAT Question 12 Explanation
If Jefferson is assigned to area 2, then which one of the following must be true?
Please explain
Please explain why A is the correct answer choice?
@sprozes, thanks for your question.
Based on Rule 1, we know that M is in area 3. Based on this question, we know that we need to put J in area 2. This gives us:
2: J
3: M
Now, let's look at O, since it has significant restrictions. Rule 2 tells us that O cannot go in area 1, so it can either go in area 2 or in area 3. Let's look at what happens if we put O in area 2:
2: JO
3: M
Because of Rule 4, this means that we need to also put K in area 2. Since the game tells us that we can put no more than three rangers in a single area, area 2 is full. Now, we need to put P in area
3 because Rule 2 says P cannot go in area 1 and area 2 is full. We also need to put L in area three because Rule 3 says L needs to go with either K or M, and the area that K is in is full. This gives
2: JOK
3: MPL
However, this is an invalid scenario because there are no rangers in area 1 and the game dictates that there must be at least one ranger in each area. Therefore, O cannot go in area 2.
So, O must be in area 3. This changes our setup to:
2: J
3: MO
Let's consider which variables can go in area 1. J, M, and O can't, as they're already placed elsewhere. P can't because of Rule 2. This leaves us with only L or K.
When L is placed in area 1:
1: L
2: J
3: MO
Rule 3 dictates that L is either with M or K, and since L is in area 1 and M is in area 3, we must place K in area 1 to satisfy this rule. We are left with P, which can be placed in either area 2 or
area 3. Our final setup is:
1: LK
2: J (possibly P)
3: MO (possibly P)
When K is placed in area 1:
1: K
2: J
3: MO
Rule 3 says that L must be with either K or M, so we have the option of putting L in area 1 or in area 3. Similarly, we have the option of putting P in either area 2 or area 3 (though we cannot put
both P and L in area 3, as it would be past full). This gives us:
1: K (possibly L)
2: J (possibly P)
3: MO (possibly L) (possibly P)
Now, let's evaluate each answer choice. Remember, we are asked which MUST always be true, not whIch could be true.
(A) is correct because we see that K is in area 1 in each of our final two scenarios.
(B) is incorrect because L could be in area 3 in our second scenario.
(C) is incorrect because O is always in area 3.
(D) is incorrect because P could be in area 3 in both of our scenarios.
(E) is incorrect because P could be in area 2 in both of our scenarios.
Does that make sense? Please reach out with any other questions and best of luck with your studies! | {"url":"https://testmaxprep.com/lsat/community/100005356-please-explain","timestamp":"2024-11-15T04:46:48Z","content_type":"text/html","content_length":"67591","record_id":"<urn:uuid:9bbcb327-056c-4846-8943-804257414afd>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00331.warc.gz"} |
An empirical application of a stochastic volatility model with GH skew Student's t-distribution to the volatility of Latin-American stock returns
Using daily stocks returns data of a set of Latin-American countries (Argentina, Brazil, Chile, Mexico and Peru) for the sample period 1996:01–2013:12, we estimate a stochastic volatility model
incorporating both leverage effects and skewed heavy-tailed disturbances through of the GH Skew Student's t-distribution based on Bayesian estimation method proposed by Nakajima and Omori (2012). Two
alternative models are estimated, one using an alternative Skew Student's t-distribution and the other using a symmetric Student's t-distribution. The results suggest the presence of leverage effects
in all markets except for Peru where the evidence is unclear. In addition, there is evidence of asymmetries and heavy tails in the Argentina and S&P500 markets while in the other countries there is
no robust evidence of such characteristics. Using the Bayes factor, the results indicate that the SVGHSkewt model dominates the other two models for the cases of Peru, Argentina, Brazil and S&P500
whereas the simple SVt model is preferred for the markets of Mexico and Chile. Similar findings are obtained after performing a robustness analysis regarding the priors of the parameters associated
with the skewness and the tails of the distribution. | {"url":"https://cris.pucp.edu.pe/en/publications/an-empirical-application-of-a-stochastic-volatility-model-with-gh-2","timestamp":"2024-11-07T20:45:05Z","content_type":"text/html","content_length":"38499","record_id":"<urn:uuid:9cd363de-8974-4e9d-89c9-027f2020058b>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00054.warc.gz"} |
No.1316 Original Retro & PG problems
Gani Ganapathi & Nicolas Dupont JF – R2017-18
(India / France)
Definition: (click to show/hide)
No.1316 Gani Ganapathi &
Nicolas Dupont
India / France
original – 08.08.2018 Solution: (click to show/hide)
white Bf1c1 Ke1 Qd1 Ph2g2f2e2d2c2b2a2 Sg1b1 Rh1a1 black Bf8c8 Ke8 Qd8 Ph7g7f7e7d7c7b7a7 Sg8b8 Rh8a8
PG 11.5 (14+15)
Immun Chess
33 Comments
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August 9, 2018 09:50
Is it not same as Strict Circe?
August 9, 2018 09:51
Sorry strict five is different
August 9, 2018 16:23
Jacobi found an alternative in 11 moves: 1.b3 Sc6 2.Ba3 Sd4 3.Bd6 exd6 4.f4 d5 5.f5 Bd6 6.f6 Se7 7.fxe7 f5 8.Sc3 Kf7 9.e8=S Rxe8 10.Sb1 Re6 11.c3 Rg6.
August 9, 2018 19:21
Reply to Paul Rãican
PG 11.5 cannot be cooked in 11 moves because the side to move will be different. Shorter solution (cook) can only be 10.5 or 9.5 etc. I am surprised that Paul Raican, a retro expert makes such a
A similar claim of cook was made for our WCCT entry and it was rightly rejected by the judging countries.
August 10, 2018 01:34
A nice idea. Promotion must be to queen, because a knight cannot lose a tempo.
August 10, 2018 06:01
Although the main idea of this problem is to lose a tempo for a PG 11.5, any shorter solution, whether it is in 11 or less moves is valid, unless the authors stipulate that the problem is an exact
proof game in 11.5 moves. All proof games are assumed to be shortest proof games (SPG), unless otherwise specified. The stipulation of a PG as an exact proof game is rather unpleasant and composers
tend to avoid it. If possible, a better solution is to add an additional half move (like Paul Raican’s suggestion of 12…Sc2+, or preferably a non-checking move) to make the problem sound as a
shortest PG and preserve the tempo play, as well.
August 10, 2018 07:33
Thierry Le Gleuher told me that as the retros editor of Phénix he would consider a PG 11.5 cooked if there is a solution in 11 moves or less, unless the stipulation says “exact”. This is in agreement
with Paul and Kostas. As for the current problem, adding the half-move 12… Kf6 works (C+ Jacobi, separate tests in 12.0, 11.5, 11.0, …). The extra move doesn’t have to be check — the goal is only to
force 12 white moves.
August 10, 2018 11:46
IMO, SPG 11.5 asks for the shortest proofgame with black to move, and since there’s no proofgame in less moves this is indeed the shortest profgame.
See e.g. P1004019 for a composition with stipulation ‘Shortest proof game? (a) white to move (b) black to move’.
August 10, 2018 15:07
Joost, the stipulation “Shortest Proof Game” does not need to specify the side to move, or the number of moves that are needed to reach the diagram. In the case of 1316, an SPG would be in 11 moves,
not in 11.5. If you specify a problem as an SPG and further add which side’s turn it is to move as in P1004019, then you are looking only for the SPG ending with a white/black move, not for the
(absolute) shortest PG.
In order to avoid all controversy, and especially the stipulation: “exact PG 11.5” or “SPG with Black to move”, composers prefer to add a “tail” move (like 12…Sc2+ or 12…Kf6, as proposed in this
case). I remember a PG by Reto Aschwanden (I cannot find it now) that had a shorter solution (cook) and this remedy of adding an extra “tail” move was not possible, so the composer had to specify
that his problem was an exact PG, instead of an SPG. If my memory serves me well, Reto’s problem was selected in the award of that informal tourney, even as an exact PG.
If we try to compare a PG with a helpmate, then the set play in a helpmate would be equivalent to a PG in 11 moves (going backwards) for a solution in 11.5 moves. Even in helpmates, where the
stipulation specifies the side to move, we still stipulate the set play. This would make sense in a PG if there was a unique solution in 11 moves (not the case here) and this solution was in contrast
with the solution in 11.5 moves. I can think of my own P1288899 as such an example.
In any case, No.1316 is not an SPG (it would be cooked as an SPG), so the stipulation has to change (see the first paragraph of this comment). I would definitely prefer the added 12th move for Black,
if this was my own problem, but it is not. Let’s wait and see what the authors will choose to do.
August 10, 2018 20:44
Reply to Kostas Prentos
Kostas, I think the Reto composition is P1080421 , but I don’t have my old Probleemblad issues anymore, so I can’t verify.
August 11, 2018 01:02
Reply to Joost
Thank you, Joost. This is the Reto PG. It is cooked in 21.5 moves, but not in 22 moves. I don’t have any old Probleemblad issues with me, but according to WinChloe, the award was published in a
special issue (December 2009).
August 10, 2018 16:32
Is is true that, in the modern approach, PG is often used to denote in fact SPG, and exact PG to denote in fact non-shortest PG.
Nevertheless another approach seems to be used both in the PDB and in WinChloe. For example P1288899 is labelled a) BP in 20.0 b) BP in 20.5, and a) PJ in 20.0 b) PJ in 20.5 respectively, and not b)
exact BP/PJ in 20.5. So it seems here that PG in X moves means PG in exactly X moves.
I think it is the best option (rendez-vous at the restaurant in 3 hours means in exactly 3 hours, not in 1, 2 or 3 hours), even if I can understand the logic of the others. So to avoid any confusion,
I agree to add 12… Kf6 in 1316.
August 10, 2018 18:00
Reply to dupont
PDB uses the word “genau” extensively. The search stip=”genau” and k=”unique proof game” returns 27 problems, including twins that differ by 0.5 (P1091679, P1339295).
August 10, 2018 19:16
Reply to François Labelle
In both examples (P1091679, P1339295) the word “genau” could have been omitted, imho. It is clear that in both problems the twin b is not an SPG.
Proof games with tempo play often have the same difficulties as 1316. See for example the discussion on whether the added “tail” move 16…Ke7 was necessary or not,
in this fine proof game. It is easy to forget this detail, especially since the PG solving engines don’t search for solutions in less than the stipulated moves.
August 10, 2018 17:40
Good choice, Nicolas. Although there are exceptions to this rule, it is a well established convention that a PG is a shortest PG. Yet, I don’t agree that P1288899 is a good example to support the
opposite opinion. The twin a is an SPG, while the twin b obviously is not (as both twins have the same final position). I believe it is all quite clear, without the need to stipulate that twin b is
an exact PG.
I have always operated under the assumption that a PG in n moves is in fact an SPG. To this end, stipulating the number of moves is a courtesy to the reader/solver by the composer and could be
omitted without changing anything in the solution. Like with classical retros, the least explaining that has to be done by the composer in the form of the stipulation, the better.
Finally, stipulating that a PG in n moves is not an SPG, but an exact PG is not always bad. See for example the popular P0000811 that owes its reputation to the fact that it is a PG in exactly 4
moves and not in 3 or 3.5.
August 10, 2018 17:56
I don’t understand the logic, if there is one at all.
11.5 is AS EXACT a number as it could ever be.
For a PG it means 11 pairs of halfmoves + 0.5 of a pair, so White made the last move.
So far, everything is clear and there’s no room for any arbitrarily imposed bureaucratic conventions.
PG 11.0 is not a solution, since the stipulated position is not matched.
PG 10.5 hypothetically might reach the required position and if such PG is real, that could require an agreement and some convention as a default.
(Even this wouldn’t be an issue for me. As a default, PG 11.5 asks for any play reaching the diagram in 11.5 moves, where 11.5 is actually 11.5, no less, no more. The stipulation tells about what has
ALREADY HAPPENED – the exact position and number of moves. “Forward stipulations” tell what could happen, and the default logic of stipulation could be different.)
August 10, 2018 19:47
Reply to Nikola Predrag
Nikola, if you do not specify that the solution is in exactly 11.5 moves, then any shorter solution is a cook. The magic word here is “exactly”. A PG is supposed to be a shortest PG, the fastest way
to reach the given position. To avoid confusion, you can specify that the problem is in exactly 11.5 moves, but this is really not the best available choice for the PG composer, only the last resort.
August 10, 2018 20:30
Ok, François and Kostas – the PDB is using “exact PG” for a non-shortest game (in the case there are no twins). But it is not true for WinChloe! For example the famous Orban game is labelled “PJ in
4.0” and not “exact PJ in 4.0”.
So it seems that at least WinChloe is sharing Nikola’s argument – 4.0 means 4.0, no different length could be considered a cook as it doesn’t fill the stipulation.
As I said before, this option also sounds the best to my mind, even if I accept (and Gani too) to add a tail move in 1316 in order to avoid any conflict, but without being convinced by its full
August 10, 2018 20:49
Kostas, you may claim that 11.5 means 173.666 or whatever you wish, and you may bureaucratically impose it to the community. But, is that what you really want?
Direct simple hierarchy of logic is the best default.
11.5 is the EXACT number and the default meaning is exact.
WHY on earth would anyone distort such a clear logic?
You might argue that 10.5 moves leading to THE SAME POSITION makes a short solution. We could accept that as a default for possibly practical reasons, but first you present them convincingly!
However, PG 11.0 can NEVER reach THE EXACT position as the stipulated one. So, what are we talking about?
PG means ANY PG, and if the number of moves is given, both the position and number are known.
For ‘shortest PG’, just stipulate SPG, WITHOUT a number.
But no, upside-down logic claims that indicating ‘shortest’ is not necessary (as if that was obvious???) and that 11.5 is not exactly 11.5 unless explicitly indicated!??
Anyway, you say:
-“A PG is supposed to be a shortest PG, the fastest way to reach the given position”-
Do we really have to read again the rules of the game to comprehend what is a position?
August 11, 2018 02:24
Reply to Nikola Predrag
Nikola, this is not a convention I invented, nor am I trying to impose it to anyone. You are free to follow whatever logic suits you best.
August 11, 2018 03:33
Kostas, it’s by no means about you as an individual.
And it’s not about me as an individual.
Logic suits nobody, there is THE logic or no logic.
It’s terrifying how easily we surrender to ‘out-of-mind’ conventions!
Position after 11.5 moves becomes a position after 11.0 moves.
I am very sorry that Nicolas has surrendered. And one by one, the others will surrender without a single convincing argument.
WHO has invented that ‘convention’ and why do you (or anyone) accept it?
August 11, 2018 10:41
Nikola, I was not going to engage in an exercise in futility, but your last question intrigued me a bit. I don’t know the answer, but this link may be of some help. I don’t own the book “Shortest
Proof Games” by Gerd Wilts and Andrey Frolkin. My guess is the authors may offer a historic account.
As to the question “why I accept it”, the answer is “because it makes perfect sense to me”. There is no way in hell I would knowingly allow a cook such as this in one of my own compositions. Others
can do whatever pleases them.
I don’t believe there is anything else I can add without repeating my previous comments.
August 11, 2018 13:30
The example no 8 from the aforementioned book “Shortest Proof Games” by Michel Caillaud & Jacques Rotenberg, 2HM Europe Echecs 1991 (PDB – P0001681) has the same stipulation mentioned by Joost
“Shortest Proof Game?” with two twins: a) White to move and b) Black to move.
That means between 1991 and 2001 there was probably no general objection against using the stipulation “Shortest Proof Games” without specifying the number of moves in such tempo motivated cases.
For the second question: in any stipulation, not specifically for proof games, any way to reach to the conclusion in less than the specified number of moves is considered to be a shorter solution.
Therefore, the accepted convention for such cases is to add the word ‘exact’ in the stipulation.
This convention was not invented by the retro composers – it was just imposed in the retro genre like other existing conventions from chess composition. The logic for this convention is perhaps to
force the solver find the author’s intention and not claim he found a shorter solution to reach the aim.
August 11, 2018 23:00
Different genres have different principles and meaning of the features. And the outcome of a logic procedure may (expectedly) be different.
There’s one logic but the input is essentially different for different genres.
There’s obviously no curiosity about convincing arguments.
August 12, 2018 07:46
This subject was discussed on the MatPlus forum back in 2012:
August 13, 2018 15:12
Reply to Geoff Foster
True. It had been discussed then also. What was the conclusion? 🙂
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Mathematics for Elementary Teachers
Problem 6
Harriet is part of a group of five children who share four pies. Jeff is part of a group of seven children who share four pies. Jean is part of a group of seven children who share six pies.
1. Who gets more pie, Harriet or Jeff? Justify your answer!
2. Who gets more pie, Jeff or Jean? Justify your answer!
3. Who gets more pie, Harriet or Jean? Justify your answer!
Problem 7
Yesterday was Zoe’s birthday, and she had a big rectangular cake. Today, leftover cake is shown here.
Draw a picture of the original (whole) cake and explain your work.
Problem 8
Use benchmarks and intuitive methods to arrange the fractions below in ascending order. Explain how you decided. (The point of this problem is to think more and compute less!):
Problem 9
Which of these fractions has the larger value? Justify your choice.
Problem 10
Solve each division problem. Look for a shortcut, and explain your work.
Problem 11
Yoko says
because she cancels the sixes:
But note:
So is Yoko right? Does her cancelation rule always work? If it does not always work, can you find any other example where it works? Can you find every example where it works?
Problem 12
Jimmy says that a fraction does not change in value if you add the same amount to the numerator and the denominator. Is he right? If you were Jimmy’s teacher, how would you respond?
Problem 13
1. Shelly says that if
2. Rob says that if
Problem 14
Jill, her brother, and another partner own a pizza restaurant. If Jill owns
Problem 15
John spent a quarter of his life as a boy growing up, one-sixth of his life in college, and one-half of his life as a teacher. He spent his last six years in retirement. How old was he when he died?
Problem 16
Nana was planning to make a red, white, and blue quilt. One-third was to be red and two-fifths was to be white. If the area of the quilt was to be 30 square feet, how many square feet would be blue?^
Ku’u Hae Aloha (My Beloved Flag), Hawaiian cotton quilt from Waimea, before 1918, Honolulu Academy of Arts.
Problem 17
Rafael ate one-fourth of a pizza and Rocco ate one-third of it. What fraction of the pizza did they eat?
Problem 18
Problem 18 (Tangrams). Tangrams^[2] are a seven-piece puzzle, and the seven pieces can be assembled into a big square.
1. If the large square shown above is one whole, assign a fraction value to each of the seven tangram pieces. Justify your answers.
2. The tangram puzzle contains a small square. If the small square (the single tangram piece) is one whole, assign a fraction value to each of the seven tangram pieces. Justify your answers.
3. The tangram set contains two large triangles. If a large triangle (the single tangram piece) is one whole, assign a fraction value to each of the seven tangram pieces. Justify your answers.
4. The tangram set contains one medium triangle. If the medium triangle (the single tangram piece) is one whole, assign a fraction value to each of the seven tangram pieces. Justify your answers.
5. The tangram set contains two small triangles. If a small triangle (the single tangram piece) is one whole, assign a fraction value to each of the seven tangram pieces. Justify your answers
Problem 19
Mikiko said her family made two square pizzas at home. One of the pizzas was 8 inches on each side, and the other was 12 inches on each side. Mikiko ate
of a pizza. Do you agree with Mikiko’s calculation? Did she eat
Problem 20
Look at the triangle of numbers. There are lots of patterns here! Find as many as you can. In particular, try to answer these questions:
1. What pattern describes the first number in each row?
2. How is each fraction related to the two fractions below it?
3. Can you write down the next two rows of the triangle?
Problem 21
Marie made a sheet cake at home, but she saved some to bring to work and share with her co-workers the next day. Answer these questions about Marie’s cake. (Draw a picture!)
1. Suppose Marie saved
2. What if Marie saved
3. What if she saved
Problem 22
An elementary school held a “Family Math Night” event, and 405 students showed up. Two-thirds of the students who showed up won a door prize. How many students won prizes?
Problem 23
For each picture shown:
• What multiplication problem is represented?
• What is the product?
Problem 24
For each problem, use only the digits 0, 1, 2,. . . , 9 at most once each in place of the variables. Find the value closest to 1. Note that
Problem 25
A town plans to build a community garden that will cover
Problem 27
The family-sized box of laundry detergent contains 35 cups of detergent. Your family’s machine requires
Problem 28
Jessica bikes to campus every day. When she is one-third of the way between her home and campus, she passes a grocery store. When she is halfway to school, she passes a Subway sandwich shop. This
morning, Jessica passed the grocery store at 8:30am, and she passed Subway at 8:35am. What time did she get to campus?
Problem 29
If you place a full container of flour on a balance scale and place on the other side a
Problem 31
Lily was flying to San Francisco from Honolulu. Halfway there, she fell asleep. When she woke up, the distance remaining was half the distance traveled while she slept. For what fraction of the trip
was Lily asleep?
1. Image used under Creative Commons CC0 1.0 Universal Public Domain Dedication. ↵
2. Tangram image from Wikimedia Commons, public domain. ↵ | {"url":"https://pressbooks.oer.hawaii.edu/mathforelementaryteachers/chapter/problem-bank-4/","timestamp":"2024-11-02T21:18:45Z","content_type":"text/html","content_length":"107697","record_id":"<urn:uuid:2ec68385-c015-41e5-b0dd-70e59217b70b>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00040.warc.gz"} |
Perform localization of molecular orbitals obtained from a previous calculation.
For example, we can localize the occupied molecular orbitals (using IBOs of Knizia) after a DFT calculation:
dft1 := dft(
structure( molecule = water )
ao = '6-31G*'
xc = PBE
load = 'dft1'
orbital_type = 'occupied'
method = 'ibo'
This command can appear in the global context.
Specify the exponent for the Boys localization functional.
□ The type is int
□ The default is 2
□ The value must be one of:
☆ 2 - Use an exponent of 2 for the Boys localization functional.
☆ 4 - Use an exponent of 4 for the Boys localization functional.
Threshold for determining null space in incomplete Cholesky decomposition.
□ The type is real
□ The default is 1.0e-9
□ The value must be greater than 0.0
Specify the version of IAOs to be constructed.
□ The type is string
□ The default is standard
□ The value must be one of:
☆ standard - Use the standard definition of IAOs. See this paper for details.
☆ simplified - Use the simplified definition of IAOs. See this paper for details.
☆ economical - Use the economical IAOs. This is based on the simplified IAOs, but without the orthonormalization of the depolarized MOs.
Specify the exponent for the IBO localization functional.
□ The type is int
□ The default is 4
□ The value must be one of:
☆ 2 - Use an exponent of 2 for the IBO localization functional.
☆ 4 - Use an exponent of 4 for the IBO localization functional.
Name of result set from which to load the molecular structure, basis set, and orbitals.
This option is mandatory.
□ The type is string
□ The default is previous
Maximum number of iterations.
□ The type is int
□ The default is 50
Specify the method of localization.
□ The type is string
□ The default is ibo
□ The value must be one of:
☆ pipek - Use Pipek-Mezey localization method.
☆ boys - Use Boys localization method.
☆ ibo - Use intrinsic bond orbital (IBO) localization method of Knizia.
☆ cholesky - Use partial Cholesky decomposition of the density matrix to generate local orbitals.
Specify the MINAO basis for making the IAOs. The MINAO basis that should be used depends to the basis used in loaded result set. For DFT and HF calculations use the cc-pVTZ-MINAO basis. For xTB
calculations use the GFN-xTB-MINAO basis.
□ The type is string-lowered
□ The default is cc-pVTZ-MINAO
Specify the name of a set of results.
This option is deprecated.
□ The type is string
□ There is no default value.
Threshold for detecting the occupied space for fractional occupations (e.g. results from a fermi smearing). In the closed shell integer occupation case, the occupation numbers will be passed as
is. If fractional occupation numbers are present, a sanity check will be performed to ensure all occupation numbers are close to 0 or 2 within occ_threshold and are in descending order.
Otherwise, an error will be thrown if values between occ_threshold and 2-occ_threshold are detected.
□ The type is real
□ The default is 1.0e-8
Specify set of orbitals on which localization is performed.
□ The type is string
□ The default is occupied
□ The value must be one of:
☆ occupied - All occupied molecular orbitals.
☆ virtual - All virtual molecular orbital.
☆ core - Core occupied molecular orbitals.
☆ valence occupied - Valence occupied molecular orbitals.
☆ valence virtual - Valence virtual molecular orbitals.
☆ non-valence virtual - Non-valence virtual molecular orbitals.
☆ doubly occupied - Doubly occupied molecular orbitals (for ROHF orbitals only).
☆ valence doubly occupied - Valence doubly occupied molecular orbitals (for ROHF orbitals only).
☆ valence singly occupied - Valence singly occupied molecular orbitals (for ROHF orbitals only).
Specify a list of orbital types to be localized separately. Each entry of the list should be an orbital type defined in orbital_type, and should not have overlapping orbitals with the other
orbital groups in the list. For example: orbital_types = ['valence occupied', 'valence virtual'] specifies that both valence occupied and valence virtual orbitals will be localized separately.
However, orbital_types = ['valence occupied', 'occupied'] is not a valid input because occupied orbitals include valence occupied ones.
If valence virtual is requested without also requesting non-valence virtual then the non-valence virtual orbitals will still be generated, but not localized. Instead they will be canonicalized
when the Fock matrix is available (i.e. the Fock matrix will be diagonalized in the non-valence virtual subspace).
□ The type is [string]
□ There is no default value.
Ordering of the localized orbitals.
□ The type is string
□ The default is fock
□ The value must be one of:
☆ none - Do not re-order localized orbitals. The orbitals are not guaranteed to be ordered by any specific criterion.
☆ fock - The orbitals are ordered as increasing diagonal elements of the Fock matrix in the localized orbital basis. This requires the Fock matrix in the original AO basis to be available
when reading previous result set; otherwise, will fall back to none ordering.
Specify the exponent for the Pipek-Mezey localization functional.
□ The type is int
□ The default is 2
□ The value must be one of:
☆ 2 - Use an exponent of 2 for the Pipek-Mezey localization functional.
☆ 4 - Use an exponent of 4 for the Pipek-Mezey localization functional.
Print level.
□ The type is int
□ There is no default value.
□ The value must be one of:
☆ -2 - No output
☆ -1 - Minimum output
☆ 0 - Output that doesn't scale with system size
☆ 1 - Output that scales linearly with system size
☆ 2 - (Debugging) output that scales quadratically with system size
☆ 3 - (Debugging) output that scales cubically with system size
Convergence threshold for localization, using orbital-stability conditions defined by Pipek-Mezey.
□ The type is real
□ The default is 1.0e-9
Whether to use the IAOs constructed from the occupied orbitals to localize all types of orbitals, or to use a specific set of IAOs made from each type of orbitals.
□ The type is bool
□ The default is false | {"url":"https://software.entos.ai/qcore/documentation/utility_commands/localize/","timestamp":"2024-11-08T08:55:29Z","content_type":"text/html","content_length":"43852","record_id":"<urn:uuid:090c4b32-6669-4a33-8b41-de228a16273f>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00667.warc.gz"} |
mp_arc 17-3
17-3 Wolfgang Orthuber
Geometrical appearance of circumference as statistical consequence (271K, pdf) Jan 5, 17
Abstract , Paper (src), View paper (auto. generated pdf), Index of related papers
Abstract. Because identical fermions (elementary particles) have (except spacetime coordinates) exactly the same features everywhere, these are (per proper time) a multiple mapping of the same.
This mapping also leads to the geometrical appearance (of spacetime) and it provides a set of possibilities which can be selected (like "phase space"). Selection of possibilities means
information. New selection of possibilities means decision resp. creation of information. This paper should motivate to a more consequent information theoretical approach (not only in quantum
mechanics but) also towards spacetime geometry. It is a short supplement to previously published material, where it was shown that proper time is proportional to the sum of return probabilities
of a Bernoulli Random Walk. The probabilities at every point in such a walk result from "OR" operation of incoming paths. The probability of a "AND" operation at a certain point can be
interpreted as meeting probability of two simultaneous and independent Bernoulli Random Walks. If no direction is preferred (p=1/2), after n steps this meeting probability (of two simultaneous
symmetric Bernoulli Random Walks resp. BRWs) in the common starting point goes for large n to 1/(2pi n), which is the inverse of the circumference of a circle with radius n. So if a BRW pair
denotes two commonly starting simultaneous independent BRWs (each with p=1/2), after n steps (in case of large n) in the average 1 of (2pi n) BRW pairs meet again in its original starting point.
Likewise due to the limited speed of light our knowledge of surrounding is the more delayed, the greater the distance n is. Therefore there are the more (geometric) possibilities of return ((2pi
n) possibilities for multiples of the same fermion on a circle with radius n), the greater the distance (the radius) n is. This shows a basic example for a connection between statistical results
and geometrical appearance.
Files: 17-3.src( 17-3.comments , 17-3.keywords , 17_01_05_wgeoapp1.pdf.mm ) | {"url":"http://kleine.mat.uniroma3.it/mp_arc-bin/mpa?yn=17-3","timestamp":"2024-11-03T02:55:07Z","content_type":"text/html","content_length":"3299","record_id":"<urn:uuid:861a45cf-cb2c-431e-b2c9-5be29f9dc50d>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00595.warc.gz"} |
Activation Functions
Activation function
Activation functions in artificial neural networks define the output of a node given an input or set of inputs. Standard integrated circuits can be seen as digital networks of these activation
functions, which are either "ON" or "OFF". Nonlinear activation functions allow networks to solve complex problems with fewer nodes.
2 courses cover this concept
This is a deep-dive into the details of deep learning architectures for visual recognition tasks. The course provides students with the ability to implement, train their own neural networks and
understand state-of-the-art computer vision research. It requires Python proficiency and familiarity with calculus, linear algebra, probability, and statistics.
Brown University's Deep Learning course acquaints students with the transformative capabilities of deep neural networks in computer vision, NLP, and reinforcement learning. Using the TensorFlow
framework, topics like CNNs, RNNs, deepfakes, and reinforcement learning are addressed, with an emphasis on ethical applications and potential societal impacts. | {"url":"https://cogak.com/concept/766","timestamp":"2024-11-09T22:11:13Z","content_type":"text/html","content_length":"119031","record_id":"<urn:uuid:38c0d58d-0c30-463f-a60c-8f2390461e8d>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00517.warc.gz"} |
1. Introduction2. The Magnitude of the Self-Generated Magnetic Field3. The Structure of the Geomagnetic Field4. Changes of Polarity of the Geomagnetic Field5. Secular Variation and the Westward Drift6. ConclusionsCite this paperAppendixReferences
JHEPGCJournal of High Energy Physics, Gravitation and Cosmology2380-4327Scientific Research Publishing10.4236/jhepgc.2016.21004JHEPGC-62524ArticlesPhysics&Mathematics The Geomagnetic Field
ngelFierros Palacios[1]^*Instituto de Investigaciones Eléctricas, División de Energías Alternas, Mexico City, México* E-mail:afierros@iie.org.mx25122015020133404 June 2015accepted 28 December 31
December 2015© Copyright 2014 by authors and Scientific Research Publishing Inc. 2014This work is licensed under the Creative Commons Attribution International License (CC BY). http://
In this paper a solution to the problem of the self-generated magnetic field of the Earth is pro-posed. The solution is based on the existence of a steady-state current distribution localized in some
region inside the convective zone of the planet, constituted by the fluid Outer Core. The magnitude of the self-generated magnetic field is obtained and it is shown to be a dipolar field.
The Geomagnetic Field
In order to properly pose the problem on the origin and structure of the Geomagnetic Field, and according to the arguments that follow, it is proposed that the origin may be located in the Outer Core
and not in the Inner Core, as commonly believed. According to the specialized literature [1] , the Earth is composed by the Crust, the Mantle and the Central Core. The Crust is the thin outer region
of the Earth. The Mantle is a region that goes from the Crust to the Central Core. It is totally solid, although throughout geologic times it can behave as a plastic material.
On the other hand, the Central Core is usually divided into two regions, called the Outer Core and the Inner Core. There is enough evidence to prove that the Outer Core is a fluid, while the Inner
Core is solid [2] .
This latter region resembles a sphere composed almost totally by high temperature molteniron, possibly constant along all its volume. This region can be considered as a huge mass of isothermal molten
Due to the fact that this hot mass is under great pressure, the value of its density is very big, almost near that of the solid iron [2] . Thus it is very hard that under such conditions the
necessary temperature differences for the production of convective currents can be reached in this place.
On the other hand, the values of pressure, density and temperature in the Outer Core increase continuously from those at the surface of the planet to those at the Inner Core. Does, in the outer Core
the proper conditions exist to generate the thermal gradients necessary to produce convective streams.
Consequently, it is being proposed here to consider the Outer Core as the convective zone of the planet, and that it is in this place where the origin of the Earth’s self-generated magnetic field can
be located.
As in the case of gaseous star, it is assumed that the Geomagentic Field is generated by a steady-state current distribution localized in some region inside the convective zone, which is produced by
a process of maximum ionisation. Moreover, due to the fact that the Outer Core is dragged along by the planet’s rotation, this region revolves around the terrestrial axis with a differential
rotational velocity, due to the fact that it is composed by a very viscous fluid.
Let us consider the case of a huge mass of very viscous compressible and conducting fluid, isolated in space and with inwardly increasing values of pressure, density and temperature; that revolves
around its own axis with rotational velocity v(x, t) and is under the influence of a magnetic field H(x, t). Here, x is the distance measured from any internal point to the center of the object, and
t is the time.
Additionally, this huge concentration of matter is found distributed in a configuration that has spherical symetry.
The following equation is used to determine the dynamic state of this huge mass of fluid
This last relation is the momentum balance equation of magnetohydrodynamics (MHD) [3] . In (1) f is the body force per unit mass, r (x, t) the mass density, and
; (2)
are the components of the generalized stress tensor [3] , d[ij] the components of the Kroenecker delta, p (x, t) the pressure and in the third term of the right hand side of equation (2), one has the
components of Maxwell’s magnetic stress tensor [4] .
On the other hand
are the components of the viscosity stress tensor, with h and z the coefficients of viscosity [3] [7] .
Let us consider that the above described object is the Earth’s Central Core. In this case, the rotational velocity of this central region, is not an explicit function of the variable x; due to the
fact that v = v (l, t) with l the latitude.
Beside, one can suppose that H(x,t) is the self-generated magnetic field by the Earth. Since that object revolves around it’s own axis in such a way that the regime is steady, ¶v/¶t = 0, and then,
the term on the left hand side of (1) is zero; that is to say,
where the definitions of hydrodynamic derivative [3] [7] and a well known formula in vector analysis [3] [10] were used.
Consequently, from (1) one gets the following result
; (4)
because in that case, the bodyforce per unit mass is g, the acceleration of terrestrial gravity. From the last equation, and due to the fact that ¶/¶x^j = ¶/¶x^id[ij], the following result is
; (5)
due to that. The last relation is the equation that governs the state of equilibrium of the Earth’s Central Core, which in this case is magneto mechanical. Equation (5) can be integrated considering
that r = r (R, t), with R the Central Core radius, to obtain the following result
From the specialized literature [2] one obtains that the pressure is given by the next expression
; (7)
where C[V] is the specific heat at constant volume, T the temperature, g the thermodynamic Grüneisen parameter [5] [6] and p[o] is an initial ambient pressure [2] [5] [6] such that it is possible to
assume that the hydrostatic equation is satisfied [7] ; i.e.,
According to (7) and (8), one obtains from (6) that
; (9)
Consequently, it can be said that the magnitude of the geomagnetic field in the inner regions of the Earth varies like the square root of the product of the mass density and the temperature; both
calculated at those regions.
It is considered that within the theoretical framework of MHD, H = B [3] [4] , and then the Earth’s self-generated magnetic field satisfies the basic laws of magnetostatics, which in their
differential form are the following relationships [8]
; (12)
where j is the steady-state current distribution localized in some region of the convective zone, and c is the velocity of light in empty space.
According to (11), B(x) must be the curl of some vector field A(x), called the vector potential [8] ; that is,
For a steady-state current distribution localized in a relative small region of space, the vector potential is given by the following relation [8]
; (14)
where is a distance measured relative to a suitable origin within the localized steady-state current distribution [9] , and x is the coordinate of a point at a great distance from the localized
steady-state current distribution.
Expanding the denominator of (14) in powers of until the lowest order of approximation, the following expression is obtained for a given component of A(x) [8]
; (15)
The fact that j is a localized divergence less steady-state current distribution, allows simplification and transformation of (15). In fact, it can be shown that [8] [9]
Thus, the first term in (15), corresponding to the monopole term in the electrostatic expansion is therefore absent. The integral in the second term in (15) can then be written as follows [8] [9]
It is customary to define the magnetic moment density, or magnetization as [8] [9]
and its integral as the magnetic moment m; that is
Then, the vector potential from the second term in (15) is the magnetic dipole vector potential
This is the lowest non-vanishing term in the expansion of A for a localized steady-state current distribution. The corresponding magnetic induction B can be calculated directly by evaluating the curl
of the last equation [8] [9] ; that is,
; (19)
where n is a unit vector in the direction x. The magnetic induction B has exactly the form of the field of a dipole. Far away from the localized steady-state current distribution, the magnetic
induction B is that of a magnetic dipole of dipole momentum given by (17).
From geological formation like frozen lava flows or sediment allayers, it is frequently found to have repeated alterations of magnetic polarity; apparently due to a process of ionic reordering in the
Outer Core of the Earth [2] . That mechanism appears to be responsible of reversal of the Geomagnetic Field; the last of which occurred 700,000 years ago [2] .
In order to give a possible explanation of such mechanism, let us consider the average field produced by a system of charges each one of them with charge q, in steady motion, at large distances from
the point where the field is calculated. It can be demonstrated that the magnetic moment of that system is given by the following expression [3] .
; (20)
Where c is the velocity of light in the empty space, r the vector radius of the point where the field is calculated, and v = dr/dt is the average velocity of each charge. On the other hand, and
according to (19)
; (21)
where r is the distance from the Outer Core to the Earth surface, and we considered that n = 1. From the previous two equations, we have that
The polarity of the magnetic induction is close related with the sign of the system of charges. Thus, in one geological cycle there could be surges of ionized particles mainly of negative charge, and
in the following, mainly of positive charges. Then the polarity of the Geomagnetic Field to be as expected, according to the kind of charged particles in each cycle. This is enough to give a
heuristic explanation of the process of changes of polarity of the magnetic induction of the Earth.
The Geomagnetic field can be represented by a magnetic dipole situated in the Earth’s Outer Core, considered as the convective zone of the Planet; and having a dipole moment with its axis inclined
about 11˚ to the Earth’s geographic axes. It has been known for over 400 years that undergoes a secular variation due to a steady progressive change in magnetic declination, or angle between magnetic
north and geographic north [2] . That movement is due to a steady west ward drift of the Geomagnetic Field, which is closely related to the eccentric dipole position [2] .
Let’s consider the Outer Core as a charged body spinning with differential velocity, due to the fact that it is composed by a very viscous fluid, in the self-generated magnetic field. Further, it has
a system of particles each one of them with charge q, and mass m, in steady motion with velocities, due to a convective process. That system of particles constitutes the steady state current
distribution localized, which is the mechanism responsible of the generation of the Geomagnetic Field.
In order to propose a solution of that problem, let us consider the system of particles and its interaction with the self-generated magnetic field. The time rate of change of the total angular
momentum of the system, is equal to the total impressed torque [15] ; so that
It will be assumed that all the particles have the same q/m ratio; in such a way that
; (24)
where Q is the total charge, and m the total mass.
Thus, the total angular momentum of the system of particles is
with p = mv the linear momentum of that system.
As the field is uniform, there will be no net force on the system, but there will be a net torque, approximately given by [15]
; (26)
where m is the magnetic moment of the system; so that
In Theoretical Physics, there is a unique relationship between the angular momentum of the system and its magnetic moment, given by the following expression [15] [16]
in such a way that the equation of motion is
This is the equation of motion for a constant vector which is rotating about the direction of B with an angular velocity
The motion is similar to what in Nuclear Physics, is called the Larmor precession, and the angular velocity is known as the Larmor frequency. The uniform precession of a system of charges in a
magnetic field holds true provided the center of mass is at rest [15] . For any system of charged particles, therefore, the total angular momentum, and with it the magnetic moment, rotates with the
angular velocity (30) around the direction of the field, while its absolute magnitude and the angle which it makes with this direction remain fixed. In other words, both vectors will undergo a Larmor
precession, with the only requirement that all charges have the same q/m ratio. On the other hand, the Outer Core revolves around the geographic axis with a differential rotational velocity. The
interaction between both independent movements has as a consequence the secular variation, due to a westward drift, of the Geomagnetic Field.
In the present paper, a fundamental hypothesis is made which assumes that the origin of the self-generated Geomagnetic Field may be located in the Outer Core, considered as the convective zone of the
Earth. This geomagnetic field is produced by some special mechanism, like the one producing the self-generated magnetic field in all gaseous stars [9] . In fact, according to the density and
temperature conditions, some region should exist in the convective zone that has a maximum of ionization.
The electrically charged particles are moved by the convective streams across that region, making their contribution to the steady-state current distribution localized, and move away being
continuously replaced by other particles [9] .
Since this current distribution is produced by the high ionization in the region, and the process depends on the density and temperature conditions in that region, the magnitude of the self-generated
dipolar Geomagnetic Field is a function of those variables [9] , as it can be easily seen from Equation (9).
Finally, the changes of polarity of the Geomagnetic Field apparently are due to a process of ionic reordering in the Outer Core, in such a way that in one geological cycle the magnetic induction has
the (N-S) polatiry, and in the following the (S-N) polarity.
The Goemagnetic Field undergoes a secular variation due to a westward drift, which can be related with a combination of the Larmor precession of the total angular momentum, and also the magnetic
moment of the system of charged particles, around the self-generated magnetic field, and the differential rotational movement of the Outer Core around the terrestrial axes, which is observed over
large areas of the Earth.
Angel FierrosPalacios, (2016) The Geomagnetic Field. Journal of High Energy Physics, Gravitation and Cosmology,02,33-40. doi: 10.4236/jhepgc.2016.21004
Let us consider any point at the surface of the Outer Core. The strength of the dipole field can be obtained from equation (19) and using the following data [2]
Consequently, the corresponding magnetic induction is
Now, the value of C[V] con be calculated from (9), to take advantage of the last result and the next data
to obtain that
In order to estimate the strength of the magnetic induction at the Earth’s Equator, the magnitude of B[c] in the Crust can be calculated; where the corresponding data has the following values, and
using again equation (9), i.e.
Then, in the Crust and near the Earth’s surface one obtains that
The strength of magnetic induction on the Earth’s surface and at the Equator is equal to 0.307 gauss [2] . As it can be easily seen, the theoretical calculation and the direct measurement are
practically equal.
Finally, it is important to mention what follows: Concerning the elaboration of an alternative theoretical scheme on the origin and structure of the Geomagnetic field, many researchers have engaged
themselves, to the self- exited dynamo models [11] . Unfortunately, the results obtained by them are far from satisfactory even now [12] . The model was initially proposed in 1919 by J. Larmor [13]
with the purpose of giving an explanation to the phenomenon of Sunspots. That suggestion was quickly rejected for being inadequate and inconsistent to the astronomical observations about the
phenomenon [14] .
However, the model was used, throughout 40 or more years, in order to try giving an explanation of the origin and structure of the magnetic field self-generated by gaseous stars [11] .
Also, by means of this model, the idea is to explain the origin and structure of the magnetic field self-gener- ated by the Earth. In this case it has not been possible to give any satisfactory
explanation concerning the basic characteristics of the geomagnetic field either [2] .
Bullen, E.K. (1979) El interior de la Tierra. El redescubrimiento de la Tierra. Consejo Nacional de Ciencia y Tecnología, México.Stacey, F.D. (1977) Physics of the Earth. Second Edition. John Wiley &
Sons. New York. Santa Barbara. Londo Sydney. Toronto.Fierros Palacios, A. (2006) The Hamilton-Type Principle in Fluid Dynamics. Fundamentals and Applications to Mag- netohydrodynamics,
Thermodynamics, and Astrophysics. Springer-Verlag, Wien.Landau, L.D. and Lifshitz, E.M. (1960) Electrodynamics of Continuous Media. Addison-Wesley Publishing Co., London.Landau, L.D. and Lifshitz,
E.M. (1958) Statistical Physics. Pergamon Press LTD: London-Paris and Adisson-Wesley Publishing Co.Callen, H.B. (1960) Thermodynamics. Jonh Wiley & Sons, Inc. NewYork. London Sydney.Landau, L.D. and
Lifshitz, E.M. (1959) Fluid Mechanics. Addison-Wesley Publishing Co., London.Jackson, J.D. (1962) Classical Electrodynamics. John Wiley & Sons, Inc., New York, London. http://dx.doi.org/10.1063/
1.3057859Fierros Palacios, A. (2002) The Magnetic Field in the Stability of the Stars. Submitted.Spigel, M.R. (1959) Vector Analysis and Introduction to Tensor Analysis. Shaum Publishing Co., New
York.Parker, E.N. (1955) Hidromagnetic Dynamo Models. Astrophysical Journal, 122, 293-314. http://dx.doi.org/10.1086/146087Cowling, T.G. (1981) The Present Status of Dynamo Theory. Annual Review of
Astronomy and Astrophysics, 19, 115- 135. http://dx.doi.org/10.1146/annurev.aa.19.090181.000555Larmor, J. (1919) Brit Assoc. Reports, 159.Cowling, T.G. (1934) Monthly Notices. Royal Astronomical
Society, 94, 39-48. http://dx.doi.org/10.1093/mnras/94.1.39Goldstein, H. (1959) Classical Mechanics. Addison-Wesley Publishing, Inc. U.S.A., London.Landau, L.D. and Lifshitz, E.M. (1962) The
Classical Theory of Fields. Pergamon Press, Addison-Wesley Publishing, Inc., London, Paris. | {"url":"https://www.scirp.org/xml/62524.xml","timestamp":"2024-11-02T00:12:39Z","content_type":"application/xml","content_length":"33728","record_id":"<urn:uuid:78659b6a-06ad-4142-82a3-09eea8256c9e>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00583.warc.gz"} |
erand48(3): generate uniformly distributed | Linux Man Page
drand48, erand48, lrand48, nrand48, mrand48, jrand48, srand48, seed48, lcong48 — generate uniformly distributed pseudo-random numbers
#include <stdlib.h>
double drand48(void);
double erand48(unsigned short xsubi[3]);
long int lrand48(void);
long int nrand48(unsigned short xsubi[3]);
long int mrand48(void);
long int jrand48(unsigned short xsubi[3]);
void srand48(long int seedval);
unsigned short *seed48(unsigned short seed16v[3]);
void lcong48(unsigned short param[7]);
Feature Test Macro Requirements for glibc (see feature_test_macros(7)):
All functions shown above: _XOPEN_SOURCE
|| /* Glibc since 2.19: */ _DEFAULT_SOURCE
|| /* Glibc versions <= 2.19: */ _SVID_SOURCE
These functions generate pseudo-random numbers using the linear congruential algorithm and 48-bit integer arithmetic.
The drand48() and erand48() functions return nonnegative double-precision floating-point values uniformly distributed over the interval [0.0, 1.0).
The lrand48() and nrand48() functions return nonnegative long integers uniformly distributed over the interval [0, 2^31).
The mrand48() and jrand48() functions return signed long integers uniformly distributed over the interval [-2^31, 2^31).
The srand48(), seed48() and lcong48() functions are initialization functions, one of which should be called before using drand48(), lrand48() or mrand48(). The functions erand48(), nrand48() and
jrand48() do not require an initialization function to be called first.
All the functions work by generating a sequence of 48-bit integers, Xi, according to the linear congruential formula:
Xn+1 = (aXn + c) mod m, where n >= 0
The parameter m = 2^48, hence 48-bit integer arithmetic is performed. Unless lcong48() is called, a and c are given by:
a = 0x5DEECE66D
c = 0xB
The value returned by any of the functions drand48(), erand48(), lrand48(), nrand48(), mrand48() or jrand48() is computed by first generating the next 48-bit Xi in the sequence. Then the appropriate
number of bits, according to the type of data item to be returned, is copied from the high-order bits of Xi and transformed into the returned value.
The functions drand48(), lrand48() and mrand48() store the last 48-bit Xi generated in an internal buffer. The functions erand48(), nrand48() and jrand48() require the calling program to provide
storage for the successive Xi values in the array argument xsubi. The functions are initialized by placing the initial value of Xi into the array before calling the function for the first time.
The initializer function srand48() sets the high order 32-bits of Xi to the argument seedval. The low order 16-bits are set to the arbitrary value 0x330E.
The initializer function seed48() sets the value of Xi to the 48-bit value specified in the array argument seed16v. The previous value of Xi is copied into an internal buffer and a pointer to this
buffer is returned by seed48().
The initialization function lcong48() allows the user to specify initial values for Xi, a and c. Array argument elements param[0-2] specify Xi, param[3-5] specify a, and param[6] specifies c. After
lcong48() has been called, a subsequent call to either srand48() or seed48() will restore the standard values of a and c.
For an explanation of the terms used in this section, see attributes(7).
Interface Attribute Value
drand48(), erand48(), lrand48(), nrand48(), mrand48(), jrand48(), srand48(), seed48(), lcong48() Thread safety MT-Unsafe race:drand48
The above functions record global state information for the random number generator, so they are not thread-safe.
Conforming to
POSIX.1-2001, POSIX.1-2008, SVr4.
This page is part of release 5.04 of the Linux man-pages project. A description of the project, information about reporting bugs, and the latest version of this page, can be found at https://
Referenced By
drand48_r(3), rand(3), random(3), random_r(3), stress-ng(1), zshmodules(1).
The man pages erand48(3), jrand48(3), lcong48(3), lrand48(3), mrand48(3), nrand48(3), seed48(3) and srand48(3) are aliases of drand48(3).
2017-09-15 Linux Programmer's Manual | {"url":"https://dashdash.io/3/erand48","timestamp":"2024-11-10T09:29:06Z","content_type":"text/html","content_length":"21247","record_id":"<urn:uuid:bb523a76-4a81-49d7-9935-955df7fc911b>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00807.warc.gz"} |
[algorithm] What exactly does big ? notation represent? - SyntaxFix
First let's understand what big O, big Theta and big Omega are. They are all sets of functions.
Big O is giving upper asymptotic bound, while big Omega is giving a lower bound. Big Theta gives both.
Everything that is ?(f(n)) is also O(f(n)), but not the other way around.
T(n) is said to be in ?(f(n)) if it is both in O(f(n)) and in Omega(f(n)).
In sets terminology, ?(f(n)) is the intersection of O(f(n)) and Omega(f(n))
For example, merge sort worst case is both O(n*log(n)) and Omega(n*log(n)) - and thus is also ?(n*log(n)), but it is also O(n^2), since n^2 is asymptotically "bigger" than it. However, it is not ?(n^
2), Since the algorithm is not Omega(n^2).
A bit deeper mathematic explanation
O(n) is asymptotic upper bound. If T(n) is O(f(n)), it means that from a certain n0, there is a constant C such that T(n) <= C * f(n). On the other hand, big-Omega says there is a constant C2 such
that T(n) >= C2 * f(n))).
Do not confuse!
Not to be confused with worst, best and average cases analysis: all three (Omega, O, Theta) notation are not related to the best, worst and average cases analysis of algorithms. Each one of these can
be applied to each analysis.
We usually use it to analyze complexity of algorithms (like the merge sort example above). When we say "Algorithm A is O(f(n))", what we really mean is "The algorithms complexity under the worst^1
case analysis is O(f(n))" - meaning - it scales "similar" (or formally, not worse than) the function f(n).
Why we care for the asymptotic bound of an algorithm?
Well, there are many reasons for it, but I believe the most important of them are:
1. It is much harder to determine the exact complexity function, thus we "compromise" on the big-O/big-Theta notations, which are informative enough theoretically.
2. The exact number of ops is also platform dependent. For example, if we have a vector (list) of 16 numbers. How much ops will it take? The answer is: it depends. Some CPUs allow vector additions,
while other don't, so the answer varies between different implementations and different machines, which is an undesired property. The big-O notation however is much more constant between machines
and implementations.
To demonstrate this issue, have a look at the following graphs:
It is clear that f(n) = 2*n is "worse" than f(n) = n. But the difference is not quite as drastic as it is from the other function. We can see that f(n)=logn quickly getting much lower than the other
functions, and f(n) = n^2 is quickly getting much higher than the others.
So - because of the reasons above, we "ignore" the constant factors (2* in the graphs example), and take only the big-O notation.
In the above example, f(n)=n, f(n)=2*n will both be in O(n) and in Omega(n) - and thus will also be in Theta(n).
On the other hand - f(n)=logn will be in O(n) (it is "better" than f(n)=n), but will NOT be in Omega(n) - and thus will also NOT be in Theta(n).
Symetrically, f(n)=n^2 will be in Omega(n), but NOT in O(n), and thus - is also NOT Theta(n).
^1Usually, though not always. when the analysis class (worst, average and best) is missing, we really mean the worst case. | {"url":"https://syntaxfix.com/question/26769/what-exactly-does-big-notation-represent","timestamp":"2024-11-15T03:39:04Z","content_type":"text/html","content_length":"43485","record_id":"<urn:uuid:4c6cc030-5f83-4e22-b44e-7fd16d0b4a80>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00584.warc.gz"} |
Quantifying comparison of large detrital geochronology data sets
The increase in detrital geochronological data presents challenges to existing approaches to data visualization and comparison, and highlights the need for quantitative techniques able to evaluate
and compare multiple large data sets. We test five metrics commonly used as quantitative descriptors of sample similarity in detrital geochronology: The Kolmogorov-Smirnov (K-S) and Kuiper tests, as
well as Cross-correlation, Likeness, and Similarity coefficients of probability density plots (PDPs), kernel density estimates (KDEs), and locally adaptive, variable-bandwidth KDEs (LA-KDEs). We
assess these metrics by applying them to 20 large synthetic data sets and one large empirical data set, and evaluate their utility in terms of sample similarity based on the following three criteria.
(1) Similarity of samples from the same population should systematically increase with increasing sample size. (2) Metrics should maximize sensitivity by using the full range of possible
coefficients. (3) Metrics should minimize artifacts resulting from sample-specific complexity. K-S and Kuiper test p-values passed only one criterion, indicating that they are poorly suited as
quantitative descriptors of sample similarity. Likeness and Similarity coefficients of PDPs, as well as K-S and Kuiper test D and V values, performed better by passing two of the criteria.
Cross-correlation of PDPs passed all three criteria. All coefficients calculated from KDEs and LA-KDEs failed at least two of the criteria. As hypothesis tests of derivation from a common source,
individual K-S and Kuiper p-values too frequently reject the null hypothesis that samples come from a common source when they are identical. However, mean p-values calculated by repeated subsampling
and comparison (minimum of 4 trials) consistently yield a binary discrimination of identical versus different source populations. Cross-correlation and Likeness of PDPs and Cross-correlation of KDEs
yield the widest divergence in coefficients and thus a consistent discrimination between identical and different source populations, with Cross-correlation of PDPs requiring the smallest sample size.
In light of this, we recommend acquisition of large detrital geochronology data sets for quantitative comparison. We also recommend repeated subsampling of detrital geochronology data sets and
calculation of the mean and standard deviation of the comparison metric in order to capture the variability inherent in sampling a multimodal population. These statistical tools are implemented using
DZstats, a MATLAB-based code that can be accessed via an executable file graphical user interface. It implements all of the statistical tests discussed in this paper, and exports the results both as
spreadsheets and as graphic files.
ASJC Scopus subject areas
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Current Search: Khadka, Bal K.
Solving approximate SVP in an Ideal Lattice using a cluster.
Khadka, Bal K., Magliveras, Spyros S., Graduate College
The shortest vector problem SVP is de ned as follows: For a given basis B of an integral lattice L fi nd a vector v in L whose length is minimal. Here we present the result of our experiments
based on a hill climbing algorithm using a computer cluster and a number of parallel executions of a standard basis reduction technique, such as LLL, to successfully reduce an initial basis of L.
We begin by reducing ideal lattices of relatively small dimension and progressively reduce ideal lattices of...
Show moreThe shortest vector problem SVP is de ned as follows: For a given basis B of an integral lattice L fi nd a vector v in L whose length is minimal. Here we present the result of our
experiments based on a hill climbing algorithm using a computer cluster and a number of parallel executions of a standard basis reduction technique, such as LLL, to successfully reduce an initial
basis of L. We begin by reducing ideal lattices of relatively small dimension and progressively reduce ideal lattices of higher dimension, beating several earlier published solutions to the
approximate SVP problem.
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Date Issued
Document (PDF)
New LS[3][2,3,2^8] Geometric Large Sets.
Hurley, Michael Robert, Khadka, Bal K., Magliveras, Spyros S., Graduate College
Let V be an n-dimensional vector space over the field of q elements. By a geometric t-[qn,k,λ] design we mean a collection D of k-dimensional subspaces if V, called blocks, such that every
tdimensional subspace T of V appears in exactly λ blocks in D. In a recent paper Braun, Kohnert, Ӧstergård, and Wassermann constructed the first ever known large set LS[N][2,k,qn], namely an LS
[3][2,3,28] under a cyclic group G of order 255. In this work we construct an additional 8 large sets with the same...
Show moreLet V be an n-dimensional vector space over the field of q elements. By a geometric t-[qn,k,λ] design we mean a collection D of k-dimensional subspaces if V, called blocks, such that
every tdimensional subspace T of V appears in exactly λ blocks in D. In a recent paper Braun, Kohnert, Ӧstergård, and Wassermann constructed the first ever known large set LS[N][2,k,qn], namely
an LS[3][2,3,28] under a cyclic group G of order 255. In this work we construct an additional 8 large sets with the same parameters, using the L3 algorithm for lattice basis-reduction.
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Date Issued
Document (PDF)
Techniques in Lattice Basis Reduction.
Khadka, Bal K., Magliveras, Spyros S., Florida Atlantic University, Charles E. Schmidt College of Science, Department of Mathematical Sciences
The mathematical theory of nding a basis of shortest possible vectors in a given lattice L is known as reduction theory and goes back to the work of Lagrange, Gauss, Hermite, Korkin, Zolotarev,
and Minkowski. Modern reduction theory is voluminous and includes the work of A. Lenstra, H. Lenstra and L. Lovasz who created the well known LLL algorithm, and many other researchers such as L.
Babai and C. P. Schnorr who created signi cant new variants of basis reduction algorithms. The shortest...
Show moreThe mathematical theory of nding a basis of shortest possible vectors in a given lattice L is known as reduction theory and goes back to the work of Lagrange, Gauss, Hermite, Korkin,
Zolotarev, and Minkowski. Modern reduction theory is voluminous and includes the work of A. Lenstra, H. Lenstra and L. Lovasz who created the well known LLL algorithm, and many other researchers
such as L. Babai and C. P. Schnorr who created signi cant new variants of basis reduction algorithms. The shortest vector (SVP) and closest vector (CVP) problems, presently considered
intractable, are algorithmic tasks that lie at the core of many number theoretic problems, integer programming, nding irreducible factors of polynomials, minimal polynomials of algebraic numbers,
and simultaneous diophantine approximation. Lattice basis reduction also has deep and extensive connections with modern cryptography, and cryptanalysis particularly in the post-quantum era. In
this dissertation we study and compare current systems LLL and BKZ, and point out their strengths and drawbacks. In addition, we propose and investigate the e cacy of new optimization techniques,
to be used along with LLL, such as hill climbing, random walks in groups, our lattice di usion-sub lattice fusion, and multistage hybrid LDSF-HC technique. The rst two methods rely on the
sensitivity of LLL to permutations of the input basis B, and optimization ideas over the symmetric group Sm viewed as a metric space. The third technique relies on partitioning the lattice into
sublattices, performing basis reduction in the partition sublattice blocks, fusing the sublattices, and repeating. We also point out places where parallel computation can reduce runtimes
achieving almost linear speedup. The multistage hybrid technique relies on the lattice di usion and sublattice fusion and hill climbing algorithms. Unlike traditional methods, our approach brings
in better results in terms of basis reduction towards nding shortest vectors and minimal weight bases. Using these techniques we have published the competitive lattice vectors of ideal lattice
challenge on the lattice hall of fame. Toward the end of the dissertation we also discuss applications to the multidimensional knapsack problem that resulted in the discovery of new large sets of
geometric designs still considered very rare. The research introduces innovative techniques in lattice basis reduction theory and provides some space for future researchers to contemplate
lattices from a new viewpoint.
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Date Issued
Subject Headings
Cryptography., Combinatorial analysis., Group theory.
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VA to kVA Calculator Online
Home » Simplify your calculations with ease. » Electrical »
VA to kVA Calculator Online
The VA to kVA calculator simplifies the conversion of electrical power from Volt-Amperes (VA) to Kilovolt-Amperes (kVA). It’s a handy tool used in electrical engineering to transform the apparent
power, denoted in VA, into kilovolt-amperes, a unit often utilized in sizing electrical systems.
Formula of VA to kVA Calculator
The conversion formula is straightforward: kVA = VA / 1000
This simple calculation allows quick transformation between these power units, aiding in various electrical applications, from sizing electrical equipment to understanding power consumption.
General Terms Table:
Here’s a helpful table of common electrical power terms that users often search for, aiding in easy reference without the need for manual calculations:
Power Unit Abbreviation Equivalent
VA Volt-Amperes Same as VA
kVA Kilovolt-Amperes 1 kVA = 1000 VA
kW Kilowatts Actual Power Consumed
HP Horsepower 1 HP ≈ 0.746 kW
This table serves as a quick reference guide for individuals dealing with electrical systems or calculations.
Example of VA to kVA Calculator
Let's consider an example to illustrate the conversion. If you have a device with a power rating of 5000 VA, using the formula kVA = VA / 1000, the calculation would be: kVA = 5000 / 1000 = 5 kVA
This example demonstrates how to convert VA to kVA using the calculator's simple formula.
Most Common FAQs:
Q: Why is VA to kVA conversion necessary?
A: The conversion to kVA is vital in understanding the apparent power of an electrical system. It helps in proper equipment sizing, ensuring efficient and safe power usage.
Q: What's the difference between kVA and kW?
A: While kVA represents apparent power (VA rating), kW denotes actual power consumed. It’s essential to consider both in electrical systems to assess efficiency accurately.
Q: Is VA to kVA conversion relevant for household appliances?
A: Understanding power units is beneficial for assessing the electrical needs of various devices, ensuring proper circuit sizing and preventing overload issues.
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Differential Equations and Linear Algebra, 2.4b: Second Order Equations With Damping
From the series: Differential Equations and Linear Algebra
Gilbert Strang, Massachusetts Institute of Technology (MIT)
A damped forced equation has a particular solution y = G cos(ωt – α). The damping ratio provides insight into the null solutions.
Published: 27 Jan 2016
I'm coming back to the number one example, but not the easiest example, of a second order equation with an oscillating forcing term, cosine omega t. We have to know the answer to this problem. And
it's a little messy, but the method is not messy. The method is straightforward.
So let me begin by looking for the rectangular form. I call this the rectangular form. It separates the cosine with its amplitude and the sine with its amplitude into two separate pieces. So if I'm
looking for that solution, and m and n are the numbers I want to find, how do I proceed?
It's a case of undetermined coefficients, M and N. And the way to determine them is substitute this into the equation and match the cosine term and find M and N. And the way we find M and N, we need
two equations for two quantities, M and N.
And imagine this substituted in there. I'll get some cosines. So the cosines on one side will match the cosine on the other side. And also from the derivative, I'll get some sines and they should
match 0 because I have no sine omega t on the right hand side.
So I have two equations, matching the sines, matching the cosines. And I solve those. Two equations, two unknowns. And I just write the answer down.
M involves a C minus omega squared. M is coming from the cosines. And we get cosines from that term and that term. Divided by some number, D, that I'll write down.
And N is just B omega divided by that same D. And now I'll write down D. That's C minus A omega squared squared plus B omega squared. This is what comes out from the two equations for M and N.
I just solve those equations. This D here is the two by two determinant if we think about the linear algebra behind two equations. And that's what it is. And so the answer now is in terms of A, C, B,
and D, which is a mixture of all of A, B, and C. That's the solution.
Only I always want to show you a different form of the solution. And in this case, a better form. Because the most important physical quantity is the magnitude. How large does y get? What is the
amplitude of this?
This is a sinusoid. And we remember that every sinusoid can be written in a polar form. Says that y of t is some amplitude of G, the gain, times a cosine of omega t with a shift, with a lag, with an
angle alpha. So I have two numbers now.
That's the gain. And this is the phase shift alpha. And that's an attractive form because it has only one term. The two numbers, G and alpha, get put into a single term where we can see the magnitude
of the oscillation.
And what does that come out to be? I won't go through all the steps. I'll just write down what G turns out to be. G turns out to be-- it comes from there-- and it's 1 over the square root of D. Well,
G is the square root of M squared plus N squared.
The square root of M squared plus N squared. And if I put M squared and N squared, then I have D over D squared. I get that answer. That's the gain.
Let me write that word, gain, again. Because you got it there. Here it is again. And as always, the tangent of alpha is the N over the M, which is just B omega over C minus A omega squared. I like
that polar form.
And I feel I should just do an example. I didn't do any of the algebra in this video. But you know where the algebra came from. It came from substituting the form we expect for the solution. And of
course, that form that we expect is the form we get provided omega, the driving frequency, is different from omega N.
Well, no. I guess we're all right even if omega is omega N, because we have a damping term. So that's the answer.
So an example. Why not an example? y double prime plus y prime plus 2y equals cosine of t. That's a simple example. I took omega to be 1, you see. And there is omega. And then A is 1, B is 1, C is 2.
We can evaluate everything. In fact, I think M and N are 1/2. D, by the way, will be 1 squared plus 1 squared. That's 2 square root. Sorry. D will be 2. 1 squared plus 1 squared.
So what do I know? Do I know the rectangular form? Yes. Rectangular form is 1/2. 1/2 for both the cosine and the sine. 1/2 of cosine t plus sine t. That's the rectangular form.
Two simple things, but I have to add them. And in my mind, I don't necessarily see how the cosine adds to the sine. But the sinusoidal identity, the polar form, gives it to me. So what is it in polar
So G, the gain, is going to be 1 over the square root of 2. At the highest point, the cosine and the sine are the same. They're both 1 over the square root of 2. I have two of them. So I get 1 over
the square root of 2 cosine of t minus pi over 4 is the angle, the phase lag.
When I add the cosine and the sine, I get a sinusoid that's sitting over pi over 4, 45 degrees. So those are the two forms. So in a nice example, we certainly got a nice answer. We certainly did.
So that is the-- worked out, more or less worked out, in principle, worked out-- is the solution to what I think of as the most important application when the forcing term is a cosine. So it gives
oscillating motion. It gives a phase shift. And it gives these formulas.
The only thing I would add is that I need to comment on better notation. So I have used in these formulas A, B, and C. But those have meaning as mass, damping constant, spring constant. M, B, and K.
And it's combinations of those that come in. So let me just take this moment to say better notation. Or maybe I should say engineering notation instead of A, B, C, which are mass, damping, spring
Well, that's already better to use letters that have a meaning. But the small but very important point is that two combinations of A, B, C, M, B, K are especially good. One is the natural frequency
that we've seen already, square root of C over A.
Square root of K over M. So that gives us one important combination of A and C. And the other one is the damping ratio. And it's called zeta. And that damping ratio is B over the square root of 4ac.
Ha! You'll say, where does that come from? Or I can use these letters, B over the square root of 4mk. That damping ratio is, so to speak, it's the right dimensionless quantity. The dimensions of this
ratio are just numbers.
Those two quantities have the same dimension. And we can see that because in the quadratic formula comes-- you remember that in a quadratic formula comes the square root of b squared minus 4ac? Now
if you see a formula that has b squared minus 4ac in it, you know that these must have the same units.
Otherwise, subtraction would be a crime. So they have the same ratio and the same units and therefore the ratio is dimensionless. Let me write that word. Dimensionless.
So conclusion. I could rewrite the answer in terms of these quantities omega n and zeta. I won't do that here. That can wait for another time.
But just to say since we've found a solution to the most important application with cosine omega t there, since we found the solution, appropriate to comment that we could write the answer in terms
of omega n, the natural frequency, and z, zeta, the damping ratio. Thank you.
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Mental Math and Number Talk Resources (Frames, Dot Cards, Grids, Number Charts) - The Teachers' Cafe
Check out our free worksheet options and tips on making your students successful!
Why Mental Maths Isn’t Going Away
With calculators, computers and smartphones, some wonder about the benefit of mental maths. After all, this skill enables students to quickly and accurately make calculations in their head. But can’t
technology do that faster for you anyway? So what is the benefit of mental calculations in today’s world?
Keeps Your Focus on the Bigger Maths Problem
Lacking mental calculation skills can be a big roadblock to success in higher maths, and STEM lessons further on. As students advance in mathematics, simple calculations are still present. And,
students who must use a calculator for these calculations can lose sight of the bigger problem they're trying to solve. Imagine that you visit another country. Wanting to speak the local language,
but not fluent yet, you bring along a translation tool. Unfortunately, while you’re searching for the right verb conjugation, you may forget what you meant to say next! Forgetting even a single word
could change the whole meaning of your sentence. Forgetting a step in a higher maths problem will lead to incorrect answers too. With mental calculation skills, students can focus their mental energy
on the higher level maths they're trying to learn.
Improve Estimation Skills
Doing calculations quickly in their heads can boost students' estimation abilities. This allows students to double-check the answers given by their calculators. For example, if students are
multiplying 189 by 21, they might estimate that the answer will be around 4000 (200 x 20). If their calculator reports an answer of 2,079, they’ll realize that they must have hit a wrong button.
Then, they can redo the calculation and get the right answer.
Strengthen Mathematical Flexibility and Creativity
The problems our students will face in their careers will undoubtedly require creativity and flexibility. Number sense skills can help our students become mental calculation champions! And, it will
spark creativity and flexible thinking at the same time! Number sense is our ability to recognize the patterns in how numbers can work together. Using these pattern-based shortcuts, students can
minimize the mental load of doing calculations. Not only is it fun, but it teaches students that there are many ways to get to the right answer - an important part of STEM learning!
Make Your Students Addition and Subtraction Superstars
Good news - we've got even more free math worksheets for you here! (they work great for your homeschool lessons too)
Practice makes perfect! And, there are some calculations that are just best to memorize. For example, students should memorize 1) how to add zero, 2) how to count on by 1, 10 or 100, and 3) how to
double single-digit numbers. Dot plates can be useful to practice these mental calculations. Simply explain to students which operation you want to practice with them. Then, randomly show them a dot
card, and they give the answer. For example, if you’re practicing doubles, and you show them a 6-dot card, they should respond with 12.
Another great tool is the 10-frame. This framework helps students identify what numbers add up to 10. For example, you can share a ten frame with three dots in it. Students can then count the empty
spaces to see that 7 more dots would complete the frame and give you 10 total. With this knowledge, students can use the commutative property of addition to quickly identify and add by tens. For
example, adding 8 + 9 + 2 + 5 + 1 + 5 is the same as (8 + 2) + (9 + 1) + (5 + 5).
Finally, our students are really going to shine when they use decomposition in their mental calculations. For example, students can swap tricky numbers like 4, 6, or 9 with “friendly” numbers that
are easier to work with, like 5’s and 1’s, 10’s, 100’s. When these nearby friendly numbers are decomposed, the calculation becomes easy. For example, if students need to add 99 to 276, they may
recognize that 99 is the same thing as the friendly number 100, minus 1. So, they can add 100 to 276, and then subtract 1 from that answer.
Make Your Students Multiplication and Division Dynamos
Build off your students' multiplication and division skills with this engaging worksheet game - that teaches students to budget, the FUN way!
Don't put away those dot cards just yet! There is also a level of calculation with multiplication and division that requires memorization. For example, multiplying 0 through 10, 100, 1000, etc.
should be memorized. Flash cards and dot cards are invaluable classroom and homeschool resources that can help students to practice these skills until they are second nature.
With these facts memorized, students can then learn strategies to maximize their effectiveness! The commutative property helps students multiply a series of numbers quickly. For example, 4 x 13 x 5
can be solved by multiplying the 4 and 5 together first to get 20. Then students can reduce the problem to 20 x 13, a much simpler proposition.
Decomposition is another strategy that can simplify the calculation for students. For example, 20 = 2 x 10, so we could also write 20 x 13 as 2 x 10 x 13. Again, using the commutative property, we
can do 2 x 13 first to get 26, and then we get 26 x 10 for a total of 260.
Knowing that multiplying by one doesn't change an answer is another powerful trick. For example, 1/2 x 2 = 1. So, students can multiply a problem by 1/2 and 2 without changing the answer. For
example, we can multiply 5 x 240 by 1/2 and 2. Then, we can use the commutative property to get (5 x 2) x (240 x 1/2). This simplifies to 10 x 120, giving us 1200.
Make it a Game and Learning Takes Off
Flashcards and boring worksheets simply won’t cut it! Repetitive practice is necessary to master these skills, but it doesn't have to be dull. You just need some go-to games to keep students engaged
in the classroom, and during their homeschool lesson:
"Which Doesn’t Belong" Number Talk
Use a simple 4-square frame like in the “One of These Things... Activity” worksheet linked below. Fill each square with a different number or equation. Then, challenge students to identify one of the
four options that they feel doesn’t belong in the set and explain why. There is no wrong answer to this exercise, which makes it a great opportunity for a number talk! As a class, let students share
their answers and thinking. This type of discussion is perfect for developing everyone’s number sense.
"How Many Ways" Number Talk
In this talk, challenge students to see how many ways they can come up with as a class for solving a given problem. Share one problem at a time. Then, students take turns sharing different strategies
they used to solve the problem. For example, with the problem 38 + 47, students might solve by doing a) 30 + 40 + 8 + 7 or b) 35 + 3 + 47 or c) 40 - 2 + 47. Consider some friendly competition too!
Big groups can compete to see who comes up with the highest number of different strategies. Each new method shared is a new tool in their maths toolbelt!
Discover the Rule
In this game, provide a series of numbers or equations to students. Then, students work in small groups to determine the rule. For example, students could be given the list: 21, 9, 12, 15. They then
determine the numbers are all multiples of 3. Or students could be given 12 + 18, 7 + 13, 5 + 25, 16 + 14, 1 + 29, etc. and have to determine that the ones places always add to ten. This can spark
some great discussion!
Stand Up, Sit Down
Also a quick way to do a formative check for understanding, this game gets students moving! Share a problem, and have students stand up or sit down based on what they think the answer is. In a
similar variation, if you had four answer options, students could move to one of the four corners of the room. Consider inviting a student from each answer choice to share their reason for choosing
that answer. Then, give all students the option to change their answer choice. See if students can come to a consensus as a class on the correct answer.
Up to 101 Game
Turn a worksheet into a board game! All you need is a 1 to 100 chart, place markers and one 10-sided die (or two 6-sided dice). Each student's marker starts at 0, and students take turns rolling the
dice. With each turn, students can choose to add, subtract, multiply or divide their place number on the chart by the number on the dice. The goal is to get to 101, but they cannot go over that
number. For example, if a student starts at zero and rolls a 7, they can add 7 to 0 and go to the 7 spot. On their next turn if they roll an 8, they could go to 15 by adding or 56 by multiplying. If
a student is on 99 and rolls a 9, they must subtract or divide, since they can't go over 101. The first student to land exactly on 101 wins!
Enjoy Free Resources To Make Your Students Shine!
Don’t let mental maths stress out your students, or you! And while we’re at it, let’s make it fun to learn!
Help yourself to the free worksheet options and resources below.
You'll be maximizing student learning, and they'll be having fun in no time!
Ready to Supercharge Your Classroom?
These printable activities gamify your lesson to thrill students into learning.
Students will solve immersive puzzles, overcome critical-thinking challenges, and pull together as a team.
Each worksheet is designed to maximise fun, develop key skills, and do all the hard prep-work for you.
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Geometric Distribution - Definition, Formula, Mean, Examples
Probability theory is a important branch of math which deals with the study of random occurrence. One of the essential theories in probability theory is the geometric distribution. The geometric
distribution is a distinct probability distribution that models the amount of experiments required to get the initial success in a sequence of Bernoulli trials. In this article, we will define the
geometric distribution, extract its formula, discuss its mean, and offer examples.
Explanation of Geometric Distribution
The geometric distribution is a discrete probability distribution that describes the number of tests required to accomplish the initial success in a sequence of Bernoulli trials. A Bernoulli trial is
an experiment that has two possible outcomes, usually referred to as success and failure. Such as tossing a coin is a Bernoulli trial since it can either come up heads (success) or tails (failure).
The geometric distribution is used when the experiments are independent, which means that the outcome of one trial does not affect the outcome of the upcoming trial. Furthermore, the chances of
success remains unchanged across all the trials. We could denote the probability of success as p, where 0 < p < 1. The probability of failure is then 1-p.
Formula for Geometric Distribution
The probability mass function (PMF) of the geometric distribution is specified by the formula:
P(X = k) = (1 - p)^(k-1) * p
Where X is the random variable that portrays the amount of trials needed to get the initial success, k is the number of tests required to achieve the first success, p is the probability of success in
an individual Bernoulli trial, and 1-p is the probability of failure.
Mean of Geometric Distribution:
The mean of the geometric distribution is defined as the expected value of the number of trials required to obtain the first success. The mean is given by the formula:
μ = 1/p
Where μ is the mean and p is the probability of success in a single Bernoulli trial.
The mean is the expected number of trials required to get the initial success. For example, if the probability of success is 0.5, then we anticipate to attain the first success after two trials on
Examples of Geometric Distribution
Here are few primary examples of geometric distribution
Example 1: Tossing a fair coin till the first head shows up.
Imagine we flip a fair coin until the initial head shows up. The probability of success (getting a head) is 0.5, and the probability of failure (obtaining a tail) is also 0.5. Let X be the random
variable which depicts the number of coin flips needed to obtain the initial head. The PMF of X is provided as:
P(X = k) = (1 - 0.5)^(k-1) * 0.5 = 0.5^(k-1) * 0.5
For k = 1, the probability of obtaining the first head on the first flip is:
P(X = 1) = 0.5^(1-1) * 0.5 = 0.5
For k = 2, the probability of achieving the first head on the second flip is:
P(X = 2) = 0.5^(2-1) * 0.5 = 0.25
For k = 3, the probability of obtaining the initial head on the third flip is:
P(X = 3) = 0.5^(3-1) * 0.5 = 0.125
And so on.
Example 2: Rolling an honest die till the initial six appears.
Let’s assume we roll an honest die up until the first six turns up. The probability of success (achieving a six) is 1/6, and the probability of failure (getting all other number) is 5/6. Let X be the
random variable which portrays the number of die rolls needed to get the initial six. The PMF of X is stated as:
P(X = k) = (1 - 1/6)^(k-1) * 1/6 = (5/6)^(k-1) * 1/6
For k = 1, the probability of obtaining the initial six on the first roll is:
P(X = 1) = (5/6)^(1-1) * 1/6 = 1/6
For k = 2, the probability of obtaining the initial six on the second roll is:
P(X = 2) = (5/6)^(2-1) * 1/6 = (5/6) * 1/6
For k = 3, the probability of getting the initial six on the third roll is:
P(X = 3) = (5/6)^(3-1) * 1/6 = (5/6)^2 * 1/6
And so forth.
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The geometric distribution is an essential theory in probability theory. It is applied to model a wide range of real-world phenomena, for example the count of tests required to obtain the first
success in several scenarios.
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offer customized and productive tutoring services to support you be successful. Contact us today to plan a tutoring session and take your math skills to the next level. | {"url":"https://www.longbeachinhometutors.com/blog/geometric-distribution-definition-formula-mean-examples","timestamp":"2024-11-12T16:10:46Z","content_type":"text/html","content_length":"74394","record_id":"<urn:uuid:6f2c82dd-cdc1-4d49-838b-de37efd715eb>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00179.warc.gz"} |
Bubble Sort in Python - AskPython (2024)
Let’s study one of the most intuitive and easiest to learn sorting algorithms, and implement Bubble Sort in Python. We’ll start by understanding sorting itself, and then we’ll get to sorting via
bubble sort, and finally, we’ll see how to implement it in Python.
Importance of Sorting Algorithms
What is sorting? And why is it so important? These are the questions that we will try to answer in this section.
From the books in a library and the words in a dictionary to the entries of a database and the instructions in a processor, we’ve experienced sorting numerous times.
“In computer science, sorting is the act of arranging things in an ordered sequence.” – Wikipedia
This means when we sort things, we need to know the criteria upon which we will arrange the sequence given to us. For the purposes of this tutorial, we shall assume the criteria is the value of a
number, and we shall sort a given sequence of numbers.
In computer science, the most important purpose of sorting is to produce efficient algorithms. Binary Search is an exceptionally fast searching algorithm that will not be possible in an unsorted
collection of objects.
Almost all set operations work very fast on sorted data.
Apart from making efficient algorithms, sorting is used when the very requirement of a program is to sort something, like a program that works with a deck of cards. Consequently, sorting algorithms
are one of the most fundamental concepts a programmer must know.
Understanding Bubble Sort Algorithm
Think of how in a glass of soda, the bubbles inside rise up. The bubbles represent the greatest/smallest element in a given sequence, and the bubble’s rising movements represent how the greatest/
smallest element moves to the end/beginning of the sequence.
This is how Bubble Sort works, and why it has the name.
To put it simply, we go through the sequence multiple times, and every time, we swap several pairs of elements in a manner that the greatest/smallest element in the sequence ends up at one of the
ends of the sequence.
For the sake of this tutorial, we shall consider the given array, and we shall sort it in increasing order of the value of the numbers.
Now, the algorithm of Bubble Sort works like this for sorting in increasing order:
1. Consider two variables i and j. i represents the number of elements we have sorted or the number of times we have gone through the list because every time we go through the list we sort one item
for certain.
j represents a position in the list, so if we say that j is 3, then we are talking about the third number in the list, which is 11.
2. Consider n as the number of elements in the list.
3. Let i be equal to 0. Because we have not gone through the list and no elements are sorted.
4. Let j be equal to 1. So we are starting with the number in the first position.
5. If the number at position j is greater than the number at position j+1, then we need to swap the numbers at positions j and j+1. This is because the list is in increasing order, so the number
that comes before cannot be greater than the number that comes after.
6. Increase j by 1. So now, we can look at the next pair of numbers.
7. If j is not n-i, go to step 5, otherwise, we stop the loop and go to the next step. In this loop, every time a swap occurs, the greater element moves toward the end of the list. This is the
behavior of Bubble Sort, the greatest elements bubble towards the end of the list. If i represents the number of elements already sorted, then the last i elements of the list are in their correct
position (because they bubbled their way through during the i number of times we went through the loop), so we don’t need to check the last i elements as it will only waste time, and hence the
loop ends when j is equal to n-i.
8. Increase i by 1. If we ended the loop when j reached the end, we have gone through the list one more time and one more element is sorted.
9. If i is not n-1, then go to step 4, otherwise, we stop the loop with i and go to the next step. As you might have noticed, there are two loops, the inner one with j is responsible for sorting one
more element, and we have a total of n elements to sort, which is handled by the outer loop that runs on i. If i becomes n-1, it means n-1 elements are sorted, which automatically means that the
last element is also in its correct position, and that means the entire sequence is sorted, and so we stop.
10. The sequence is sorted.
Now, you may want to try this on the given sequence, and that is what we’ll do now.
Bubble Sort Example
Given sequence: 12, 16, 11, 10, 14, 13
Number of elements (n): 6
Let’s start-
• Step 1: Variables i and j representing sorted elements and position.
• Step 2: n is 6. n = 6
• Step 3: Set i as 0. i = 0
• Step 4: Set j as 1. j = 1
• Step 5: Comparing positions j and j+1, the element at position 1 (12) is not greater than the one at 2 (16).
• Step 6: Increment j. j = 2
• Step 7: j (2) is not n-i (6), so we go to step 5.
• Step 5: Position 2 (16) is greater than position 3 (11), so we swap.
• Sequence: 12, 11, 16, 10, 14, 13
• Step 6: Increment j. j = 3
• Step 7: 3 is not 6, so we go to step 5.
• Step 5: 16 is greater than 10, so we swap. Sequence: 12, 11, 10, 16, 14, 13
• Step 6: Increment j. j = 4
• Step 7: 4 is not 6, so we go to step 5.
• Step 5: 16 is greater than 14, so we swap. Sequence: 12, 11, 10, 14, 16, 13
• Step 6: Increment j. j = 5
• Step 7: 5 is not 6, so we go to step 5.
• Step 5: 16 is greater than 13, so we swap. Sequence: 12, 11, 10, 14, 13, 16
• Step 6: Increment j. j = 6
• Step 7: j (6) is equal to n-i (6), so we move on to step 8. Notice that the greatest element (16) is at the end, and we have sorted one element for certain.
• Step 8: Increase i. i = 1
• Step 9: i (1) is not n-1 (5), so we repeat it all over from step 4, and the loop continues, the resulting changes in the sequence will look like this:
11, 12, 10, 14, 13, 16
11, 10, 12, 14, 13, 16
11, 10, 12, 14, 13, 16
11, 10, 12, 13, 14, 16
10, 11, 12, 13, 14, 16
10, 11, 12, 13, 14, 16
10, 11, 12, 13, 14, 16
10, 11, 12, 13, 14, 16
10, 11, 12, 13, 14, 16
10, 11, 12, 13, 14, 16
10, 11, 12, 13, 14, 16
After this, i becomes 5, which is n-1, so the loop ends and the algorithm tells us that the list is sorted. It also seems that the list may end up getting sorted before the algorithm finishes, which
just means that the given sequence was somewhat sorted before it was given to the algorithm.
Implementing Bubble Sort in Python
Now that we have the algorithm ready, we can start to implement each step in Python. There are some things to note:
The sequence will be represented by a list, and lists have indexes instead of positions, and indexes go from 0 to size-1 instead of 1 to size, so that will need to be adjusted, and here’s how the
algorithm will look:
def bubble_sort(sequence): n = len(sequence) for i in range(n-1): for j in range(n-i-1): if(sequence[j] > sequence[j+1]): sequence[j], sequence[j+1] = sequence[j+1], sequence[j]
Let us use an example and sort it using this algorithm:
Note that this algorithm sorts the list in place, but it is very simple to change the algorithm so that it returns a sorted list instead.
In this tutorial, we studied what sorting is and where it is used, then we learned how Bubble Sort works, we came up with an algorithm and implemented Bubble sort in Python.
Bubble Sort is one of many sorting algorithms and it is far from the best one but it is very easy to implement. The reason it is not used too often is that it has a complexity of O(n^2), which means
if the number of elements in the list is doubled, the time it takes to sort them using this algorithm will increase by four times.
So for a very large amount of data, this algorithm becomes inefficient. Nevertheless, knowing Bubble Sort as a programmer is important and I hope you learned something. | {"url":"https://flashgamz.info/article/bubble-sort-in-python-askpython","timestamp":"2024-11-14T00:39:28Z","content_type":"text/html","content_length":"72789","record_id":"<urn:uuid:24a72cff-5fbb-41bd-bc26-88082157fff3>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00719.warc.gz"} |
Energy, Christiaan Huygens, and the Wonderful Cycloid—Theory versus Experiment
Department of Physics and Project Unit, Sapir Academic College, Sderot 79165, Israel
Hemdat Hadarom Academic College of Education, Netivot 80200, Israel
Department of Physics, Ben-Gurion University of the Negev, Beer Sheva Campus 84990, Israel
The Department of Science and Technology Education, Ben-Gurion University of Negev, Beer Sheva Campus 84990, Israel
Department of Solar Energy and Environmental Physics, Jacob Blaustein Institute for Desert Research, Ben-Gurion University of the Negev, Sede-Boker Campus 84990, Israel
Author to whom correspondence should be addressed.
These authors contributed equally to this work.
Submission received: 4 February 2018 / Revised: 14 March 2018 / Accepted: 8 April 2018 / Published: 16 April 2018
The cycloid is one of the most intriguing objects in the classical physics world, at once solving the brachistochrone and isochronous curve problems. Historically, the cycloid shape has been employed
to great success in many physical contexts. We discuss one such case, presenting the longitude problem as a pathway into an in-depth discussion of the analytical solution of a point mass motion along
a cycloid. The classical solution is presented, and the modifications needed for a rolling ball along a cycloid rail are made. A comparison is then made between the two cases, and we show that the
difference in most physical cases between the point mass and the rolling ball is at most ~7%. Next, an experiment is presented in which the isochronous nature of the cycloid path is tested, to
different degrees of success. The results are discussed and several possible origins of the discrepancy between the theory and the experimental results are identified. We conclude with a discussion
of skidding and slipless rolling.
1. Introduction
In 1662 the Royal British Academy announced a large monetary prize for building a precise naval clock that would enable solving the problem of finding longitude in the high seas. One of the problems
that made navigating in the oceans difficult was finding longitude. While finding latitude through simple astronomical observations was relatively easy, finding longitude was a most difficult task.
The British navy that ruled the oceans paid a high price for it. On 22 October 1707, three British navy ships making their way from Gibraltar to England crashed into the rocks off one of the islands,
32 km from England’s southwest edge. In this accident, 2000 British sailors were killed [
]. This loss emphasized the need to solve the longitude problem. In 1714, the British Parliament announced the Longitude Act, which offered a prize of 20,000 pounds (a huge amount of money at the
time) to anyone who could solve the longitude problem [
]. Already in the second half of the 17th century, it was clear that the solution to the problem lay in building a precise clock. Because one hour equals 15 degrees, it was understood that the
measurement needed to be exact to the number of seconds in an arc. Measuring the difference between the two clocks—one that shows the time in a home port and one that shows the local time on a
ship—enabled finding longitude in a precise way [
]. The said clock also needed to overcome the changes in weather and ship movements. For example, a regular pendulum clock was influenced by the changes in temperature that caused changes in the
length of the string and therefore also caused changes in the pendulum’s cycle times [
The Dutch scientist Christiaan Huygens (1629–1695) responded to the challenge and decided to upgrade Galileo’s pendulum clock. His idea was to build an isochronous curve, meaning, a curved lane, upon
which the motion time of the ball would not be dependent on the starting point. This pendulum would enable the building of clocks that were more precise than regular pendulum clocks, in which a mass
moves along a pathway shaped as a circular arc. Huygens worked on this problem and on developing clocks for a period of almost 40 years, between 1656 and 1693. He succeeded in demonstrating that the
desired curve was a cycloid (
Figure 1
)—one of the most famous curves in mathematics that also solves the brachistochrone curve problem—finding the quickest pathway between two points. On 7 January 1657, he wrote, “These days I have
discovered a method of building clocks, by means of which it will be possible to measure time so precisely that it will be possible to measure longitude even in the ocean” [
]. As proof that the cycloid is an isochronous pathway, he published a book entitled
Horologium oscillatorium
in 1673. Huygens proved this argument in an ingenious way (see
Appendix A
) based on basic mechanical reasoning, without using infinitesimal mathematics [
In this manuscript, we explain Huygens’ theory by comparing the experiment with the measurement of a ball’s dependency on motion time from its starting height during its motion along different rails.
This article is organized as follows: In the remainder of
Section 1
, we analytically discuss the cycloid and the usual approximations made, and we study their impact using computer simulations.
Section 2
presents the methods and the experimental system, and
Section 3
reviews the results. We conclude the paper in
Section 4
, and provide some appendices for the sake of thoroughness.
1.1. The Cycloid: Coordinates, Equations of Motion, And Approximations
In this section, we lay the theoretical groundwork for this article. We first derive the cycloid coordinates (Equation (4)). We then develop the Lagrangian and equations of motion for a point mass on
a cycloid path, or, equivalently, a cycloid pendulum (Equation (6)); switch to canonical coordinates; and find the associated $ω$ (Equation (11)). Finally, we add a rotational degree of freedom,
which accounts for the slipless roll dynamics, and, in turn, for the energy stored in the rotation of the ball (Equations (14) and (15)). This is the framework necessary for understanding the
isochronous movement and for making sense of the experiments described in the next sections.
1.2. Cycloid Coordinates
The cycloid form is derived by tracking a point on the circumference of a spinning wheel moving along a plane. Tracking a point on the circumference relative to the wheel’s center, where the wheel is
spinning clockwise, is given by
${ x = a c o s ( θ ) y = − a s i n ( θ ) .$
Adding the motion of the wheel on the plane, we have the following connection between
$x c m$
and the rotation angle:
Thus, we are left with the following cycloid coordinates:
${ x = a ( θ + c o s ( θ ) ) y = a ( 1 − s i n ( θ ) ) .$
The introduction of a phase shifts these coordinates to a more canonical form:
${ x = a ( θ + s i n ( θ ) ) y = a ( 1 − c o s ( θ ) )$
which reach phase and
$x 0$
calibration identical to the ones in
Figure 1
1.3. Equations of Motion for a Cycloid
Perhaps the best way to develop the equations of motion for the cycloid motion is through Lagrangian mechanics. This has the added value, as we shall show, of identifying generalized coordinates that
simplify the problem greatly. We emphasize that in this following treatment we neglect the rotational degree of freedom. The kinetic term is simply given by
$K = m v 2 2 = m 2 ( x ˙ 2 + y ˙ 2 ) = m a 2 θ ˙ 2 [ 1 + cos ( θ ) ] .$
The potential energy
is given by
$V = m g y = m g a [ 1 − cos ( θ ) ] .$
Thus, the full Lagrangian is given by
$𝓛 = m a 2 θ ˙ 2 [ 1 + cos ( θ ) ] − m g a [ 1 − cos ( θ ) ] .$
The above Lagrangian yields the following equation of motion, and angular momentum:
${ θ ¨ = − g a [ sin ( θ ) 1 + cos ( θ ) ] P θ = L = m a 2 θ ˙ [ 1 + cos ( θ ) ] .$
Note that the angular velocity is not a conserved quantity. This is a good reason to find the conserved quantities in this system.
This Lagrangian can be written as
$𝓛 = 2 m a 2 θ ˙ 2 cos 2 θ 2 − 2 m g a sin 2 θ 2 .$
We now move to the following generalized coordinate
$s = 4 a s i n θ 2 s ˙ = 2 a θ ˙ c o s θ 2$
This allows us to write the Lagrangian in the form
$𝓛 = m s ˙ 2 2 − m g s 2 8 a$
which is nothing but a harmonic oscillator Lagrangian in
, with the associated angular velocity
Note that the angular velocity for s is conserved and that $T = 4 π = a / g$.
1.4. "Get the Ball Rolling"—Correcting for Angular Kinetic Energy
Thus far, the equation of motion was simple and analytically solvable. Now, we add the rotation of the ball itself, with its radius
$r .$
The moment of inertia for a solid ball is given by (c.f. [
The connection between
, the cycloid angle, and the rotation of the ball around its central axis is given by
$r d ϕ = a 2 + 2 cos ( θ ) d θ ⇒ ω = a θ ˙ 2 + 2 cos ( θ ) r .$
So, the added term to the Lagrangian is given by
$K ω = I ω 2 2 = m r 2 ω 2 5 = 2 m a 2 [ 1 + cos ( θ ) ] θ ˙ 2 5 .$
The corrected Lagrangian is given by
$ℒ = 7 m a 2 θ ˙ 2 [ 1 + cos ( θ ) ] 5 − m g a [ 1 − cos ( θ ) ] .$
The equation of motion is then given by
$θ ¨ = sin ( θ ) 14 ( 1 + cos ( θ ) ) [ 7 θ ˙ 2 − 5 g a ]$
which, interestingly enough, does not seem to be dependent on the radius of the ball itself, nor on its mass.
1.5. The Influence of Slipping While Rolling in Motion on Cycloid Pathways
As mentioned above, the effect of rolling on the motion in a cycloid path is not dependent on the radius of the rolling ball. This can be seen in Equation (15), where the term for angular
acceleration does not depend on $r$. This is a consequence of the connection between the pathway angle $θ$ and the rolling angle of the ball $ϕ$ as shown in Equation (13). An interesting question,
then, is to what extent slipping while rolling might change the movement and the period.
In order to answer that, we find the equation of motion for rolling, while applying instead of Equation (13) a slightly different connection:
$ω = ( 1 − α ) a θ ˙ 2 + 2 cos ( θ ) r$
is the relative part of the movement that is slipping; when there is no slipping,
$α = 0$
, and we should revert to a rolling without slipping. This is obviously a crude model which bundles the slipping into some fractional quantity, disregarding the position in which the slipping had
occurred, etc.
The equations of motion are then given by
$d P θ d t = − sin ( θ ) [ 5 + 2 ( 1 − α ) 2 5 m a 2 θ ˙ 2 + m g a ]$
$d P θ d t = 2 ( 5 + 2 ( 1 − α ) 2 ) 5 m a 2 [ ( 1 + cos ( θ ) θ ¨ − sin ( θ ) θ ˙ 2 ) ]$
Now we add by hand the friction term to get
$θ ¨ = sin ( θ ) 1 + cos ( θ ) [ θ ˙ 2 2 − ( 5 g a 2 ( 5 + 2 ( 1 − α ) 2 ) ) ( 1 + α μ k cot ( α ) θ ˙ | θ ˙ | ) ] .$
One can easily see that when
$α = 0 ,$
we indeed revert back to the slipless roll case (
Figure 2
2. Methods
The Experimental System
The system was built in a laboratory at the Davidson Institute and included three main rails: a cycloid that was created using a circle with a radius of 16 cm; an inclined plane with a sloping degree
$30 ∘$
; and a pathway that was composed of a flexible rail, which could be used to change its shape (see
Figure 3
Figure 4
). A steel ball can move along each one of the rails. The steel balls are held by permanent magnets that are in a small mobile structure that can be moved along the length of the rail and change the
height from which the ball can start moving. The release of the ball is performed by an electromagnet. The electromagnet creates a magnetic field that is reversed in its direction relative to the
magnetic field of the permanent magnet that holds the ball. The ball is released from rest by pressing a small switch on the side of the rail. The switch is additionally connected to a PC for timing
purposes. The balls on the cycloid’s rail and on the inclined plane can be released simultaneously. The motion of the balls along the length of the rails can be well approximated by the sliding of a
point mass, which is a conclusion we reached with the help of an experiment on the inclined plane as well as precise analysis of the motion (see above). In the next paragraph, several experiments are
presented which can be performed with the system. Every experiment has a suitable method by which the ball’s time in motion can be measured [
3. Results and Discussion
In this experiment, the motion time of a small metal ball whose mass is m = 32.76 g and whose radius is 1 cm was measured as a function of the initial height h on two different rails: the cycloid and
the inclined plane (see
Figure 3
). In each of the rails, we changed the starting height several times and measured the total motion time from the moment of its release all the way to its bumping into the metal board that was placed
at the bottom of the rail. For accuracy’s sake, we averaged over three measurements for each height. (There was a sound sensor on the metal board that was at the bottom of the rail, with which the
motion time was measured—see a detailed description in the frame in
Figure 4
). The experiment’s results are presented in the graphs in
Figure 2
Figure 5
Figure 6
Figure 7
The results show that while the duration of movement in the descent of the inclined plane is dependent on the starting height, the cycloid is indeed an isochronous (of equal times) curve, meaning
that the duration of the ball’s motion is not dependent on the starting point along the length of the cycloid path, and it stays almost completely constant.
It is shown that with initial heights that are less than 30 cm, the ball that moves down the descent of the inclined plane arrives first, while when
> 30 cm, the ball that moves in the cycloid rail “wins” [
The explanation for this is that when h > 30 cm, the starting points of the two rails become closer to each other, meaning that the problem becomes identical to the brachistochrone problem (the
fastest path), and the cycloid pathway is the fastest out of all the possible pathways that connect the starting point with the finishing point [
]. This could also be analyzed by progressively introducing “kinks” in an inclined path, such that in the continuum limit we get a smooth curve that is the cycloid [
]. A comparison between the duration time achieved in the experiment with the cycloid rail and the theoretical duration time is called for. The formula for frictionless, rotationless motion time
along a cycloid is used (as is shown above):
is the radius of the circle that created the cycloid. In the experiment, we measured the motion time from a certain height all the way down to the bottom of the cycloid, meaning that we measured the
duration of one quarter of a motion [
]. The result was 0.444 s, in comparison with the theoretical duration of one quarter of a motion T/4 = 0.402 s (substituting g = 980 cm/s
and r = 16 cm into Equation (16)). This means that a relative deviation of 10.45% was recorded.
In our opinion, the duration time measured is bigger than the calculated time per Equation (20) mostly because the ball is not in a slipless slide regime. Part of its starting potential energy turns
into circular kinetic energy while another part goes to heat due to slip friction [
]. This was not considered in Equation (16) or the theoretical analysis above. It is possible to compare this result with the measurement by using the optical gateway of several movements (Experiment
C described later on). There, the time interval recorded for a quarter of a motion was 0.451 s. The relative deviation in this case has increased to 12.2%. We estimate that this difference stems
mainly from the fact that the measurement in the third experiment was performed without repetitions, as compared with the three repetitions of the measurements that were performed with the sound
sensor [
]. When we compare the theory to the experiment in ball sliding on an incline without sliding, we can write the equation of motion of the ball sliding and the sum of momentums across the center of
mass of the balls. The equations are
$Σ F x = m g sin β − f s = m a Σ F y = N − m g cos β = 0 Σ τ C M = f s R = m α = m a R$
, here, denoting the ball’s radius, and
the acceleration. By substituting the moment of inertia of a ball around a central axis,
$I = 2 / 5 m R 2$
in Equation (2), and assuming that the ball is on the threshold of movement, the static friction force can be written as
Thus, we obtain a condition for rolling without slippage to be obtained when
Figure 7
shows the moving time of a ball as a function of the initial height. By neglecting air friction and rolling friction (i.e., a simple skid in a sloping plane), it can be shown that the final velocity
of the ball starting to slide from rest is
$2 g h$
. The use of motion formulas is accelerated and related as
$x = h / sin β$
, and when
is the distance of the slippage along the slope and
is the slope angle, we get
$t s l i d i n g = 2 x v = 2 h sin β 2 g h = 1 sin β 2 h g .$
Of course, we expect this theoretical time to be less than the measured time because, in reality, there is energy deposited in the rotational kinetic energy as well as energy losses due to work
wasted against skid, rolling friction, and friction with the air (
Figure 6
The Energy Consideration
Assuming the motion of the ball is a nonslip roll, we can calculate the real rolling time by using energy conservation and the fact that the mass center acceleration of the ball is constant.
According to the law of conservation of energy, we can write
$m g h = m v 2 2 + I ω 2 2 + μ r m g cos β ⋅ x$
in which the second expression on the right-hand side expresses the kinetic energy of the ball due to rotation around its axis and is the angular speed of the ball, and
is the moment of inertia of the ball around its axis and equal to
$2 / 5 m R 2$
is the radius of the ball. For friction without rolling we can use the connection between the angular velocity and the linear velocity of the center of mass
$v = R ω$
. The last expression on the right-hand side expresses the work that is wasted as a result of the skid friction. Rolling friction results from tiny deformations of the surface resulting from the body
rolling on it [
]. The rotational friction coefficient
$μ r$
is usually small compared with the coefficient of kinetic friction between the surfaces (this is the great advantage inherent in the invention of the wheel). It is worth noting that the roll is
caused by the static friction between the ball and the surface, but because in the nonslip rotation the point of contact between the surface and the ball is at rest relative to the surface, there is
no loss of energy as a result of the static friction [
]. However, the rolling friction still works. Of course, this also means that skid and slipless roll are mutually exclusive. Because the skid nature is sparse (meaning most of the time there is no
appreciable skid), and the difference between point mass sliding and ball rotation is slight, we choose to encode the energy loss due to friction while skidding in a friction term.
For rolling without sliding, the velocity of the center mass is
$v − R ω$
where ω is the angular velocity. Assuming the ball starts from rest and letting
$μ r$
be the coefficient of friction rolling, we obtain an updated expression for the ball’s motion time as a function of height
$T s l i d i n g = 2 sin α 0.7 h g ( 1 − μ r cot β ) .$
We measured it indirectly by using the energy conservation law and the smoothness of the ball along a symmetrical inclined plane that we created using the flexible track (the slope angle of the
flexible sloping plane was 34.420). The reduction in the initial potential energy should be equal to the work of the rolling friction force:
$Δ E p = m g Δ h = μ r m g cos θ$
where the height difference
$Δ h$
is between the starting point of the roll and the end point [
]. Several measurements were performed and yielded
= 0.044, which is a fairly reasonable value for the rolling friction coefficient.
According to energy conservation, the initial potential energy is equal to the sum of the kinetic energy, the circular kinetic energy, and the work invested in the friction of the roll (Equation
(23)); therefore, we now separately express each of the energy components so that they can be calculated from
, which we can write as
$v = a t = g ( sin β − μ s cos β ) ≡ g γ t$
$μ s$
is the static friction between the ball and rail and
$γ = g ( sin α − μ s cos α )$
. Therefore, we obtain for the kinetic energy
$E k = m v 2 2 = m g 2 γ 2 t 2 2 .$
The static coefficient of friction between the ball and the rail can be estimated from mechanical considerations (Equation (26)), and by Equation (25) we can get
$μ s ≥ 0.185$
for our system. Because the motion is rolling without slipping, for the purpose of calculating the energy balance, the value is selected as
$μ s = 0.185$
. The circular kinetic energy can be expressed using angular velocity and the use of angular acceleration formulas:
$ω ≈ α t = μ s m g cos β ⋅ R I .$
Thus, in general,
$ω 2 = 25 4 μ s 2 g 2 cos 2 β R 2 t 2 .$
We get
$I ω 2 2 = 5 4 m μ s 2 g 2 cos 2 β t 2 .$
The work of rolling force friction can be expressed as follows:
$W f r = μ r m g cos β ⋅ x = μ r m g h cot β .$
Using the expressions in Equations (28), (31) and (32), the overall mechanical energy can be calculated as a function of height and compared with the initial potential energy (by neglecting the air
resistance) [
]. In
Figure 7
, it can be seen that the fit to Equation (24) is very good for all initial heights.
4. Conclusions
The isochronous qualities of the cycloid were studied, both analytically and experimentally. We saw that there is a discrepancy between theory and experiment, on the order of 10%. We attribute this
disparity to some skidding along the path, specifically in the initial stages of movement, which results in some energy loss to heat. Additional factors are the rolling deformations and the point
mass sliding vs rolling ball approximation that might contribute a 1% disparity between theory and experiment. The experimental setups that were introduced are relatively inexpensive and easy to
build. This might lead us to consider incorporating the cycloid into the physics curriculum, even at the high school level. The system allows investigation of the motion of small balls along curved
paths and helps in demonstrating the amazing properties of a cycloid which is both brachistochrone and tautochrone. In principle, students can measure the ball velocity using a light gate and the
elapsed time using Audacity software and plot the ball velocity versus time along the cycloid path. This will allow them to calculate the acceleration and will help them to understand that the final
velocity depends only on the initial height and that, due to energy conservation, the acceleration and times along different ramps can be different.
Author Contributions
Yuval Ben Abu, Conceptualization, Data curation, Formal analysis, Funding acquisition, Investigation, Methodology, Project administration, Resources, Software, Supervision, Validation, Visualization,
Writing—original draft, Writing—review & editing. Ira Wolfson, Methodology, Data curation, Formal analysis, Haim Eshach, professional advice. Hezi Yizhaq, Conceptualization, Supervision and
Writing—review & editing.
Conflicts of Interest
The authors declare no conflict of interest.
Appendix A. Huygens’ Proof That a Cycloid Is an Isochronic Curve 2
The calculation of the cycle time of the cycloid plot is usually done using integral calculus and the geometric properties of the cycloid, and Huygens succeeded in solving the problem without the use
of integrals but by brilliant intuition. This is a good opportunity to trace the thinking of one of the great scientists of the 20th century. First, Huygens uses a geometric feature of the cycloid
that is shown in
Figure A1
The tangent
$A t F$
to the cycloid, which is also the direction of the velocity of the point moving on the track, cuts the circle at the highest point;
$A t B t$
is the circle diameter, and the point
$B t$
is the lowest point. Therefore, the peripheral angle that rests on it is straight, and the
$A t F B t$
triangle is a straight triangle. In this triangle,
$A t B t = 2 r sin α$
, and
is the radius of the circle, making
$y = E t B t = 2 r sin 2 α$
; this produces very important characteristics of the cycloid:
Figure A1. The tangent to a cycloid at a point $A t$ passes through the point $F$ that is the highest point of the circle and creates an angle $α$ with the diameter $F B t$. The proof that the
tangent passes at the highest point is that the direction of the tangent $A t F$ is the same as the velocity of the point on the cycloid, and that velocity is equal to two equal velocities; one is
the right velocity parallel to the x axis and the other is the tangential velocity resulting from the circular motion. Using geometric considerations, it can be shown that $F B t$ is indeed the
diameter of the circle and, therefore, $A t B t$ is a normal for the cycloid. This feature can also be demonstrated by writing the tangent equation to the cycloid and finding its points of
intersection with the circle.
Suppose that a point mass moves on the cycloid in
Figure A1
Figure A2
and that at time
$t = 0$
it is at a point
$C 0$
at distance
above the plane. The goal is to find the movement time
of the mass from the starting point
to the point at the bottom of the cycloid. Assuming that there is no loss of energy and that the movement is only smooth motion, then the cycle time of the movement (up to
$C 2 τ$
and back) will be
$T = 4 τ$
. We are interested in finding out the dependence of
. Suppose that at time
the mass is at a point
$C t$
at distance
above the plane. From the energy conservation law, we can express the speed of the mass at this point:
Now look at the projection of the position of the mass on the vertical
$C 0 B ’$
. At time
, this projection is at point
$C t ’$
, and at time
, it is at point
$B ’$
after a vertical distance
. The vertical component
of the velocity of the mass at point
$C t$
creates an angle
with the velocity and, therefore,
$w = v cos α$
. By Equation (30),
$cos α = 2 r − y 2 r$
, while
$y = 2 r − h$
, and this is the
$C t$
from the upper right. Therefore, we obtain
$cos α = h / 2 r$
, and for
This is where Huygens’ brilliance comes into play when he notices that the vertical velocity component of a circular motion is at a constant velocity in a circle with diameter
similar to w (
Figure A2
). To prove this, we will mark the
$C t ’ ’$
point on the circle opposite to the point
$C t ’$
and the length of the segment
$C t ’ C t ’ ’$
(according to the Pythagorean theorem in the triangle
$O C t ’ C t ’ ’$
). The movement of the mass on the cycloid is projected to move on the semicircle of its diameter
, because in both cases the vertical distance is the same, so the vertical component velocity must be the same (because the times are equal). Thus, they are similar triangles (both triangles are
right-angled and they have two vertical sides, respectively). Note that the length of the vertical line coming from a point
$C t ’$
and from a triangulation similarity we can draw
is the speed along the circular motion. From Equations (32) and (33), we conclude that this speed is
This speed is constant and equal to
$H 2 ω$
when the angular velocity
$ω = g r$
. In the motion of the point mass from
$C 0$
, a half-circle is moved from point
$C 0$
to point
$B ’$
at a time equal to the half-time surrounding
$τ = 2 π ω$
, independent of
. The cycle time along the cycloid will therefore be
In conclusion, Huygens’ proof is based on the idea that the velocity vector can be displayed along the cycloid as a sum of two velocities: the vertical axis is a circular motion at angular velocity
independent of the starting height and the horizontal axis.
Figure A2. Huygens’ proof is based on the idea that the velocity vector along the cycloid can be presented as a sum of two speeds: the vertical axis is a circular motion at angular velocity
independent of the starting height along the semicircle $C 0 C t ’ ’ B ’$ and the horizontal axis is a variable speed movement that increases with time.
Figure 1. A cycloid is the curve that is created by the pathway of the point that is on the perimeter of the wheel without sliding. In this sketch, the parametric equations of a cycloid are
presented, where a is the radius of the wheel and θ is the angle that is defined in the illustration. When the wheel completes a full circle, the angle changes from 0 to 2π.
Figure 2. Rolling motion of a ball on a cycloid path, with slippage. $α$ is the relative slipping factor, which represents how much of the movement is slipping and how much is slipless rolling. $μ k$
was set to 0.4, which is approximately the kinetic friction coefficient for steel on steel.
Figure 3. The experimental system: the cycloid, B; the inclined plane, C; and the flexible railing. In the inset, one can see the electromagnets that were used in a controlled release of the balls on
railings A and B.
Figure 4.
A schematic description of the experimental system for measuring the duration of the ball’s motion on one of the rails [
Figure 5. Duration times of the small ball’s movement on the cycloid (black crosses) and on the inclined plane (red squares) as a function of the starting height as it was measured in the experiment.
Figure 6. The motion time of the ball along the sloped plain versus the initial height. The blue dots indicate the measured time, the black line indicates the theoretical time of the plot without
rolling, and the red line describes the sliding time according to Equation (19); the correspondence between the theory and the experiment is very good.
Figure 7. The total mechanical energy (in units of erg) as a function of the initial height $h$ (in cm). The blue line describes the potential energy $m g h$, and the light-blue line indicates the
total energy calculated using Equations (28), (31) and (32). The remaining lines indicate kinetic energy, rotational energy, and the work of rolling friction force as a function of the initial
height. The correspondence between the calculation and the theory is very good and indicates that the motion of the ball down the sloping plane is a rolling without smoothing.
© 2018 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http:/
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Ben-Abu, Y.; Wolfson, I.; Eshach, H.; Yizhaq, H. Energy, Christiaan Huygens, and the Wonderful Cycloid—Theory versus Experiment. Symmetry 2018, 10, 111. https://doi.org/10.3390/sym10040111
AMA Style
Ben-Abu Y, Wolfson I, Eshach H, Yizhaq H. Energy, Christiaan Huygens, and the Wonderful Cycloid—Theory versus Experiment. Symmetry. 2018; 10(4):111. https://doi.org/10.3390/sym10040111
Chicago/Turabian Style
Ben-Abu, Yuval, Ira Wolfson, Haim Eshach, and Hezi Yizhaq. 2018. "Energy, Christiaan Huygens, and the Wonderful Cycloid—Theory versus Experiment" Symmetry 10, no. 4: 111. https://doi.org/10.3390/
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Optimizing campaigns with casino mathematics - Content Garden
The world of online advertising is a world of randomness and probabilities. Readers click on items in specific contexts with a certain probability, they bounce from reading an article after a certain
time with a certain probability and so on.
In this sense, there are definitely some similarities between our work at Content Garden and gambling in a casino: In roulette you may bet on ‘red’ or ‘black’, whereas for online advertising, you
might choose variant A or B of your content. The exact performance of a teaser, an article, or a campaign is not known beforehand – but data can be collected, and mathematical models can be used to
make predictions, quantify our uncertainty, and use our knowledge to maximize performance.
The Exploration-Exploitation Dilemma
Imagine you got a voucher from a casino to play the one-armed bandit slot machines 1000 times for free. The casino has three different machines as illustrated in Figure 1. If you pull the arm of one
of the slot machines, with probability p you win 1€, otherwise you get nothing. The casino tells you that the payout probabilities p[red],p[green],p[blue] are different for the different machines,
but neither it tells you the exact values nor if any machine is working better than any other one. Your task is therefore to explore, i. e., to try out the different machines to gain knowledge about
their performance, but at the same time you want to exploit what you already know and play the machine you think performs best as often as possible to maximize your profit. This is the classic form
of a Multi-Armed Bandit problem (that’s the official name mathematicians use to refer to this problem) exemplifying what is called the Exploration-Exploitation Dilemma.
The problem of finding the optimal strategy to play the various slot machines is so straightforward to understand but so hard to solve that Allied scientists suggested dropping it over Germany during
World War II in order to distract German scientists from their military work with ”Discussion of Dr Gittins’ paper.” The actual, optimal solution (via Gittins indices) is complex and shows
exponential scaling with respect to the number of pulls, which makes it inapplicable in most scenarios. However, there are well-established heuristic strategies with much better scaling behavior,
(almost) equivalent performance and manageable implementation effort.
The connection to playing out advertising content in the best possible way suggests itself immediately: The slot machines correspond to different content and their payout rate is equivalent to the
considered (unknown) KPI, e. g., the click-through rate.
A traditional approach in online advertising: A/B testing
Viewing the problem through the eyes of a marketer, you may think that the problem is a textbook example for A/B testing. You are absolutely right! You could start by trying the red, green, and blue
machines equally until you reach a specific level of significance for your hypothesis that, e. g., the red machine is the best option. Then, you would turn off the green and blue machines and only
continue with the red one until you run out of pulls as shown in the left part of Figure 2. However, classical A/B testing hast a few weaknesses, which I’d like to briefly summarize below:
• The number of runs, i.e., the number of lever pulls or ad impressions for your content may be too low to reach a reliable level of significance for the decision to turn off one or multiple tested
versions completely. You may not reach the 95% level of significance in your free 1000 runs on the slot machines, leaving the casino with 333 pulls on each machine.
• The level of significance needs to be specified manually. Is the common value of 95% optimal or should you rather use 90 or 99%?
• Runs/impressions are wasted unoptimized in the beginning, as every variant is played with equal fraction.
• You ultimately decide on a version when you have reached your level of significance. If you have chosen 95%, there is still a probability of 5% (1 out of 20!) that you are only observing your
hypothesis by chance and that a different version would actually have performed better – but you have decided to turn it off forever.
• What if you have some knowledge a priori? For example, before you start playing the slot machines, your friend may call you, telling you that last week, she observed that the green machine
performed best for her. Similarly, you may want to optimize three different teasers, one of which is very similar to a teaser you used in another campaign over a year ago. In classic A/B testing,
it is hardly possible to make use of this a priori knowledge.
State of the art algorithm: Thompson Sampling
Actually, there is an algorithm that can unwind all of the shortcomings mentioned above. It is called Thompson Sampling, named after William R. Thompson, who first described it in his famous 1933
paper “On the likelihood that one unknown probability exceeds another in view of the evidence of two samples”.
The basic concept of the algorithm is probability matching: Imagine we collected our first batch of data, i. e., some successes & failures of a handful of trials on the three slot machines. If
mathematics tells us that, based on the observed data, the probability of being the best machine is 50% for the red, 30% for the green and 20% for the blue one, the Thompson Sampling strategy is to
play the red machine with exactly 50%, the green one with 30% and the blue one with 20% probability in the next run. It thus matches the probabilities of being the optimal variant (= the best machine
or content) with the fraction of times this variant should be played – thereby automatically balancing exploration and exploitation in a very stable and performant manner.
The crucial step, obviously, is to establish the correct probability distributions over the success rates of the different machines. This can be achieved using Bayes’ Theorem – a concept that is
definitely worth another blog article. The resulting distributions may look like the curves depicted in Figure 3. The more runs have been carried out with a slot machine, the more we know about its
success rate – the sharper the respective probability distribution will be. In Figure 3 it looks like the red machine is most likely to perform the best. However, it still seems possible that the
blue one actually performs better. We can’t be entirely sure because we haven’t collected a lot of data for it yet. The most reasonable strategy is thus to play the red machine most of the time, but
also sacrifice some pulls on the blue and even the green machine to gain further knowledge.
From the point on where these probability distributions are established, the Thompson Sampling algorithm is pretty simple: In each run, draw random numbers from the established probability
distributions (visualized by the dots along the x-axis in Figure 3), and play the machine that belongs to the largest random number. Then, update the probability distribution with the data observed
(success or failure) and repeat the process. Over time, the runs will be distributed among the different machines as visualized on the right part of Figure 2.
It is even possible to use a priori knowledge by starting from probability distributions that are centered around what is believed to be the success rate before any data have been observed! So it can
be combined with any kind of machine learning model that is able to predict the success rate out of previously seen data from similar slot machines/ad content.
Sounds nice, but what’s the benefit of all this?
At Content Garden, the Thompson Sampling strategy is used to test and optimize teasers against each other with respect to their click-through rates. The algorithm is so stable and reliable that no
manual control is necessary from day 1. All the shortcomings of classical A/B testing are overcome, and even small campaigns can be optimized to a certain extent, although obviously optimization is
more effective for larger campaigns.
Moreover, it enables us to produce more ‘explorative’ content, i. e., to try out new approaches to different topics without running the risk of messing up a campaign because Thompson Sampling will
sort out underperforming content quickly. In combination with a click-through rate prediction model, it is even possible to include data from identical teaser image & text data that ran before on
different placements or in different campaigns to the optimization process. Depending on the campaign, a 20-30% increase in click-through rates is possible compared to manual optimization strategies.
And that has nothing to do with luck, but with pure mathematics – applied to successful native advertising. | {"url":"https://content-garden.com/optimizing-campaigns-with-casino-mathematics","timestamp":"2024-11-14T04:01:32Z","content_type":"text/html","content_length":"250097","record_id":"<urn:uuid:a8a5c1d2-ae58-44a3-9425-5fdd937f2ae7>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00231.warc.gz"} |
Measuring Tactical Alpha Part II: Examples and Analysis
When we left off in Part 1, we promised to examine how select Global Tactical Asset Allocation products stack up against the Global Market Portfolio from the perspective of several performance
measures – particularly Sharpe ratio, alpha and information ratio. Without further adieu:
Figure 1. Performance comparison of Global Tactical Asset Allocation products vs. ETF Proxy Global Market Portfolio, Jun 1, 2011 – Nov 28, 2014
Figure 2. Performance comparison of global risk parity products vs. ETF Proxy Global Market Portfolio, Jun 1, 2011 – Nov 28, 2014
Analysis: GestaltU, Data from Yahoo Finance and Bloomberg
A few notes about these tables. First, where stats are labeled (Incep), they are calculated from June 2011, or the product’s inception if it launched subsequent to that date, through the end of
November 2014. Second, CAGR numbers are annualized, except where a fund has been operating for less than 1 year. All risk-adjusted performance numbers are annualized from daily data, regardless of
the length of track record (daily ratios are multiplied by sqrt(252)). Betas, alphas and t-scores are all since inception, and all relative metrics (IR, alpha, beta, t-scores) are relative to the
Global Market Portfolio and based on daily observations.
So what story do these tables tell? Well, first off the Global Market Portfolio hasn’t been a tough bogey to beat in terms of raw returns over the past three years or so, with less than 6% annualized
returns. For comparison, the S&P (SPY ETF) has returned over 16% annualized over the period, and a US balanced fund (Vanguard US Balanced ETF) has gained 11% per year. Bear in mind US markets
represent over 30% of the global index, so international diversification has been quite a performance drag.
I know many of you with US-centric portfolios are patting yourself on the back. Ain’t self attribution bias grand? Make no mistake, you are US-centric because of home market bias, not superior
forecasting abilities, but I will be the first to admit that it’s better to be lucky than smart. I can state with some confidence that US-centric investors are unlikely to experience the same
relative success over the next three years. If that’s the case, what are you going to do about it?
In terms of returns relative to the GMP, GTAA funds are a mixed bag. The fund with the highest returns appears to be SMIDX, the SMI Dynamic Allocation fund, but this is somewhat of a red herring
because the fund has less than 1/2 the operating history of most other funds. On a risk adjusted basis, JP Morgan’s Efficiente (EFFE) mandate has delivered the highest risk adjusted performance, in
terms of Sharpe, Sortino, and Omega over the entire observation period. More importantly, given its low beta and high alpha scores, EFFE has generated its returns with very little reliance on
performance from the underlying indexes. This is a critical point, as funds with a high correlation to the GMP are vulnerable to a negative shift in performance when global markets turn at the end of
this cycle.
Investor legend Rob Arnott’s GTAA behemoth, PAAIX, managed under the PIMCO banner, deserves an honourable mention. It also surpassed the GMP’s Sharpe ratio over the past few years, and delivered the
second lowest alpha and beta of any fund, despite lower absolute returns.
We included the Good Harbor Tactical Core US fund in our analysis, despite the fact that it is US focused, because it highlights the risk of trying to market time strictly between the stocks and
bonds of one market. This is the difference between market timing and GTAA: you make just one bet.We deal with this concept in more detail in our new paper (see below). In our testing, we’ve observed
that market timing between stocks and bonds or stocks and cash is a much more difficult challenge than spreading bets across multiple asset classes, and Good Harbor’s unfortunate recent performance
lends credence to our own findings.
Given higher average structural allocations to bonds in risk parity funds, products in this class have clearly benefitted from the global race to the bottom in long rates, as average Sharpe ratios
are meaningfully higher than average GTAA Sharpe ratios. I strongly suspect this will reverse when the rate cycle finally turns (which admittedly could be quite a while). Setting aside QSPIX for a
moment as a special case, note that Invesco’s Balanced Risk portfolio sports the highest Sharpe, Sortino, and Omego ratios over the past 3+ years, as well as the lowest beta and highest alpha. This
is a large fund, with $10 billion in AUM according to Morningstar, yet it continues to deliver stellar returns year after year. Not for nothing, it has also generated the highest annualized returns
over this recent period.
We mentioned QSPIX is a special case, and it is. This fund, managed by AQR’s esteemed Andrea Frazzini and Ronen Israel, is based on a concept described in a 2012 paper by Antti Ilmamen, Ronen Israel,
and Tobias Moskowitz, entitled “Investing with Style: The Case for Style Investing” (currently behind AQR paywall). Antti Ilmamen is one of the greatest investment thinkers alive today, and his books
are required reading for every aspiring asset allocator. The authors present compelling evidence of the magnitude, persistence, and structurally low correlations, of the four primary sources of style
premia: value, momentum, carry and ‘defensive’. Across all asset classes covered, the authors demonstrate that style premia correlations averaged -0.22, and ranged between -0.6 and +0.21 from 1990 –
2012. Long-term Sharpe ratios for style premia composites across all asset class buckets range from 0.9 for value to 1. 37 for carry over the same period. In simulation, when normalized to a 10%
volatility, a combination style premia composites across all asset classes delivered a Sharpe of 2.52 before fees and expenses.
Of course, the authors are aware of the many frictions and pitfalls involved in implementing the strategy, so they included an analysis of the net historical performance after accounting for trading
costs (Sharpe declines to 1.9); discounting for model overfitting (Sharpe declines to 0.98), and; risk-management and fees (Sharpe ratio declines to 0.85). This seems to be to be quite a conservative
target (see Figure 5.)
Obviously, given the low expected average correlation with traditional 60/40 portfolios, and the high expected Sharpe ratio, QSPIX should substantially improve overall portfolio Sharpe, even with
small allocations. For example, a 10% allocation to QSPIX carved out of a 60/40 portfolio might raise overall Sharpe from 0.3 to 0.44, according to the authors.
Overall, I’d say the short snapshot of performance we’ve seen over the past year since inception would not cause me to reject the possibility that QSPIX will deliver against expectations. However,
the fund may be mildly vulnerable to liquidity shocks, as it has a gross leverage ratio of 8x (!!), so it should not play the role of a tail hedge in portfolios. In my opinion, the best structural
tail hedge is a good CTA fund.
So what can we conclude from our analysis? This article wasn’t meant to recommend, or point fingers, at any particular strategy, but rather to highlight how we might think about the performance of
global allocation funds, and what observed performance features might make them attractive. Above all, before committing any capital to these products, we would focus our scrutiny on the process
underlying the strategy. What factors do the managers believe are driving returns? What evidence do they have that their methodology is effective? We would want to see much longer trading histories,
analyze performance in multiple trading regimes, and understand how the strategy might interact with other holdings in portfolios. Where a long-term live history isn’t available (or even if one is
available), we would be keen to see simulations of historical performance using the same process, and understand all the ‘moving parts’ that might affect the character of the strategy.
That said, if we only have live returns to go on, we would focus on performance relative to the only true passive global benchmark, the GMP, rather than making comparisons with specific regional
indexes. Specifically, we would seek to harvest as much true alpha as possible relative to the GMP, as strategies with high alphas are less reliant on strong global market performance to deliver
returns. After all, aren’t we after diversification? Next we would look at overall risk metrics, especially volatility, but with one eye on drawdowns and beta. Only then would we start to care about
absolute returns and Sharpe ratios.
One other metric, Omega ratio, stands out as meaningful, since unlike all of the other performance metrics above, it makes no assumptions about the distribution of returns. The utility of Sharpe,
Sortino, alpha, and beta all depend on the assumption of normally distributed returns, but Omega accounts for the fact that returns often stray far from normality, especially over shorter horizons.
The formula for Omega ratio looks fancy, but it’s actually easy to calculate. First, since the Omega ratio reflects the relative probability of achieving returns above a minimum required return
(MRR), we must first choose an MRR. We chose to use the risk free rate, which is currently zero, and which makes our calculations really easy. But here is the general formula in Excel-friendly
Note that the returns variable refers to the vector of returns, so this is a matrix formula. In order for Excel to calculate it, you must hold down both the CTRL key and the ENTER key at the same
In any event, you will note that on this measure, and relative to a 0% risk free rate over the period studied, GTAA funds compare favourably relative to the GMP, almost across the board. This
suggests that, after accounting for higher moments of the return distributions, an investor would have a higher probability of achieving positive returns using GTAA than the GMP. An interesting
observation indeed.
Overall, there are a few worthy examples of successful GTAA mandates and several risk parity products worth considering for active global diversification. I should also mention that Meb Faber’s
Cambria has recently launched a very interesting new GTAA ETF, GMOM, based on newer additions to Meb’s ubiquitous paper, “A Quantitative Approach to Tactical Asset Allocation“. Well worth a look.
Lastly, we are excited to get our own GTAA track record audited so that we can add our own numbers to this list as we launch our new firm, ReSolve Asset Management, in the new year.
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Prabir Chandra Bhattacharyya,
DOI NO:
Dimension of Numbers,Dynamics of Numbers,,Quadratic Equation,Rectangular Bhattacharra’s Coordinates,significance of roots of a Quadratic Equation,
In this paper, the author has opened a new horizon in the theory of quadratic equations. The author proved that the value of x which satisfies the quadratic equation cannot be the only criteria to
designate as the root or roots of an equation. The author has developed a new mathematical concept of the dimension of a number. By introducing the concept of the dimension of number the author
structured the general form of a quadratic equation into two forms: 1) Pure quadratic equation and 2) Pseudo quadratic equation. First of all the author defined the pure and pseudo quadratic
equations. In the case of pure quadratic equation ax^2+bx+c=0 , the root of the equation will be a two-dimensional number having one root only while in the case of pseudo quadratic equation ax^2+bx+c
=0, the root of the equation will be a one-dimensional number having two roots only. The author proved that all pseudo quadratic equation is factorizable but all factorizable quadratic equation is
not a pseudo quadratic equation. The author begs to differ from the conventional theorem: "A quadratic equation has two and only two roots." By introducing the concept that any quadratic surd is a
two-dimensional number, the author developed a new theorem: “In a quadratic equation with rational coefficients, irrational roots cannot occur in conjugate pairs” and proved it. Any form of quadratic
equation ax^2+bx+c=0, can be solved by the application of the ‘Theory of Dynamics of Numbers’ even if the discriminant b^2-4ac<0 in real number only without introducing the concept of an imaginary
number. Therefore, the question of imaginary roots does not arise in the method of solution of any quadratic equation
I. Bhattacharyya, Prabir Chandra. : “AN INTRODUCTION TO THEORY OF DYNAMICS OF NUMBERS: A NEW CONCEPT”. J. Mech. Cont. & Math. Sci., Vol.-17, No.-1, January (2022). pp 37-53
II. Bhattacharyya, Prabir Chandra. : “A NOVEL CONCEPT IN THEORY OF QUADRATIC EQUATION”. J. Mech. Cont. & Math. Sci., Vol.-17, No.-3, March (2022) pp 41-63
III. Bhattacharyya, Prabir Chandra. : “AN INTRODUCTION TO RECTANGULAR BHATTACHARYYA’S CO-ORDINATES: A NEW CONCEPT”. J. Mech. Cont. & Math. Sci., Vol.-16, No.-11, November (2021). pp 76-86.
IV. Boyer, C. B. & Merzbach, U. C. (2011). A history of mathematics. New York: John Wiley & Sons.
V. Cajori, F., (1919). A History of Mathematics 2nd ed., New York: The Macmillan Company.
VI. Dutta, B.B. ( 1929). The Bhakshali Mathematics, Calcutta, West Bengal: Bulletin of the Calcutta Mathematical Society.
VII. Datta, B. B., & Singh, A. N. (1938). History of Hindu Mathematics, A source book. Mumbai, Maharashtra: Asia Publishing House.
VIII. Gandz, S. (1937). The origin and development of the quadratic equations in Babylonian, Greek, and Early Arabic algebra. History of Science Society, 3, 405-557.
IX. Gandz, S. (1940). Studies in Babylonian mathematics III: Isoperimetric problems and the origin of the quadratic equations. Isis, 3(1), 103-115.
X. Hardy G. H. and Wright E. M. “An Introduction to the Theory of Numbers”. Sixth Edition. P. 52.
XI. Katz, V. J. (1997), Algebra and its teaching: An historical survey. Journal of Mathematical Behavior, 16(l), 25-36.
XII. Katz, V., J. (1998). A history of mathematics (2nd edition). Harlow, England: Addison Wesley Longman Inc.
XIII. Katz Victor, (2007). The Mathematics of Egypt, Mesopotamia, China, India and Islam: A source book 1st ed., New Jersey, USA: Princeton University Press.
XIV. Kennedy, P. A., Warshauer, M. L. & Curtin, E. (1991). Factoring by grouping: Making the connection. Mathematics and Computer Education, 25(2), 118-123.
XV. Ling, W. & Needham, J., (1955). Horner’s method in Chinese Mathematics: Its root in the root extraction procedures of the Han Dynasty, England: T’oung Pao.
XVI. Nataraj, M. S., & Thomas, M. O. J. (2006). Expansion of binomials and factorisation of quadratic expressions: Exploring a vedic method. Australian Senior Mathematics Journal, 20(2), 8-17.
XVII. Rosen, Frederic (Ed. and Trans). (1831). The algebra of Mohumed Ben Muss. London: Oriental Translation Fund; reprinted Hildesheim: Olms, 1986, and Fuat Sezgin, Ed., Islamic Mathematics and
Astronomy, Vol. 1. Frankfurt am Main: Institute for the History of Arabic-Islamic Science 1997.
XVIII. Smith, D. (1951). History of mathematics, Vol. 1. New York: Dover. Smith, D. (1953). History of mathematics, Vol. 2. New York: Dover. Stols, H. G. (2004).
XIX. Smith, D. (1953). History of mathematics, Vol. 2. New York: Dover.
XX. Thapar, R., (2000). Cultural pasts: Essays in early Indian History, New Delhi: Oxford University Press.
XXI. Yong, L. L. (1970). The geometrical basis of the ancient Chinese square-root method. The History of Science Society, 61(1), 92-102.
XXII. http://en. wikipedia.org/wiki/Shridhara
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Books written by Ken Ross
• Both volumes of Abstract Harmonic Analysis (1963, 1968, co-authored with Edwin Hewitt) have recently been reprinted by Springer-Verlag in relatively inexpensive paper-back. The ISBN numbers for
these paper-backs must be used in ordering them. They are ISBN 0-387-94190-8 for volume 1 and ISBN 0-387-58318-1 for volume 2. For an interesting story about a result that could have been in
section 9, if we had known it, see subgroups of compactly generated groups. See also representations of group algebras.
• The second edition of my Springer-Verlag book Elementary Analysis: The Theory of Calculus is available as of April 2013. The first edition was published in 1980.
• The fifth edition of Discrete Mathematics, written in collaboration with Charles R. B. Wright, was published in 2003. The publisher is Prentice-Hall and the ISBN number is 0-13-065247-4. All
praises should be directed to me: rossmath@pacinfo.com; send all complaints to Wright wright@uoregon.edu.
• I have a book A MATHEMATICIAN AT THE BALLPARK: Odds and Probabilities for Baseball Fans, published in the summer of 2004 by Pi Press. A paperback edition came out February 27, 2007 and is
published by Plume, a division of Penguin. It includes a new appendix on fantasy baseball written by my friend Dan Schlewitz. In August 2008 it was remaindered.
Ken Ross, Mathematics Department, University of Oregon, Eugene Or 97403 USA
email: kenross.math@gmail.com or ross1@uoregon.edu
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UP Board Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions - UP Board Solutions
UP Board Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions
In this chapter, we provide UP Board Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions for Hindi medium students, Which will very helpful for every student in their exams.
Students can download the latest UP Board Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions pdf, free UP Board Solutions Class 12 Maths Chapter 2 Inverse Trigonometric Functions
book pdf download. Now you will get step by step solution to each question. Up board solutions कक्षा १२ गणित पीडीऍफ़
UP Board Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions
All Chapter UP Board Solutions For Class 12 Maths Hindi Medium
All Subject UP Board Solutions For Class 12 Hindi Medium
I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give
complete solutions so you got good marks in your exam.
यदि यह UP Board solutions से आपको सहायता मिली है, तो आप अपने दोस्तों को upboardsolutionsfor.com वेबसाइट साझा कर सकते हैं।
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This number is a prime.
The largest prime factor of 123456. Note that 643(64*3) = 123456. [De Montagu]
643 millimetres is the record monthly rainfall in Sydney (from June 1950).
643 = (3^8 + 8^3)/(3 + 8). [Trotter]
There is a 6-4-3 double play in baseball. [Patterson]
According to "The New Book of Prime Number Records," the product of the first 643 primes plus one is prime. This integer consists of 2038 digits and is known in MATHEMATICA as Euclid [643]. Also
note that 643 is a prime. [Schiffman]
As Alois P. Heinz discovered, 643 is the smallest nonprime among integers a(n) greater than 1 so that multiplying the n-th highly composite number by a(n) will give a highly composite number. a(643)
= 6 is not prime. [Post]
The smallest zero in "Cald's Sequence" occurs at the prime number 643. [Brockhaus]
If A=2, B=3, C=5, D=7, ... , Z=101, then 'A TWIN PRIME NUMBER' is a twin prime number. [Homewood]
(6^6 + 4^4 + 3^3) is divisible by 643. Are there any other multi-digit primes with this property? [Gaydos]
The 643-meter-high Tokyo Skytree is one of the most iconic towers in Japan and the world. [Yuki]
(There are 6 curios for this number that have not yet been approved by an editor.)
Printed from the PrimePages <t5k.org> © G. L. Honaker and Chris K. Caldwell | {"url":"https://t5k.org/curios/page.php?number_id=1464","timestamp":"2024-11-05T09:30:49Z","content_type":"text/html","content_length":"11456","record_id":"<urn:uuid:d2174264-c389-425d-afcb-88fac3b79887>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00812.warc.gz"} |
Applications of Interval Computations:
Applications of Interval Computations:
a book
• Mathematical Modeling and Industrial Mathematics
• Numerical Analysis
• Control and Optimization
• Expert Systems
Applications of Interval Computations
edited by
R. Baker Kearfott
University of Southwestern Louisiana, Lafayette, USA
Vladik Kreinovich
University of Texas at El Paso, USA
Applied Optimization
Volume 3
Applications of Interval Computations contains primarily survey articles of actual industrial applications of numerical analysis with automatic result verification and of interval representation of
Underlying topics include:
• branch and bound algorithms for global optimization,
• constraint propagation,
• solution sets of linear systems,
• hardware and software systems for interval computations, and
• fuzzy logic.
Actual applications described in the book include:
• economic input-output models,
• quality control in manufacturing design,
• a computer-assisted proof in quantum mechanics,
• medical expert systems,
• and others.
A realistic view of interval computations is taken: the articles indicate when and how overestimation and other challenges can be overcome.
An introductory chapter explains the content of the papers in terminology accessible to mathematically literate graduate students. The style of the individual, refereed contributions has been made
uniform and understandable, and there is an extensive book-wide index.
Audience:Valuable to students and researchers interested in automatic result verification.
Contents and Contributors:
1. Applications of Interval Computations: An Introduction;
R.B. Kearfott, V. Kreinovich.
2. A Review of Techniques in the Verified Solution of Constrained Global Optimization Problems;
R.B. Kearfott.
3. The Shape of the Symmetric Solution Set;
G. Alefeld, et al.
4. Linear Interval Equations: Computing Enclosures with Bounded Relative Overestimation is NP-Hard;
J. Rohn.
5. Quality Improvement Via Optimization of Tolerance Intervals During the Design Stage;
S. Hadjihassan, et al.
6. Applications of Interval Computations to Regional Economic Input-Output Models;
M.E. Jerrell.
7. Interval Arithmetic in Quantum Mechanics;
C.L. Fefferman, L.A. Seco.
8. Interval Computations on the Spreadsheet;
E. Hyvonen, S. De Pascale.
9. Solving Optimization Problems with Help of the UniCalc Solver;
A.L. Semenov.
10. Automatically Verified Arithmetic on Probability Distributions and Intervals;
D. Berleant.
11. Nested Intervals and Sets: Concepts, Relations to Fuzzy Sets, and Applications;
H.T. Nguyen, V. Kreinovich.
12. Fuzzy Interval Inference Utilizing the Checklist Paradigm and BK-Relational Products;
L.J. Kohout, W. Bandler.
13. Computing Uncertainty in Interval Based Sets;
L.M. Rocha, et al.
14. Software and Hardware Techniques for Accurate, Self-Validating Arithmetic;
M.J. Schulte, E.E. Swartzlander, Jr.
15. Stimulating Hardware and Software Support for Interval Arithmetic;
G.W. Walster.
Kluwer Academic Publishers, Dordrecht
Date of publishing: January 1996
444 pp.
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Multiplication Of Matrices Worksheet
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Revision #1 to TR23-146 | 27th November 2023 13:49
On Testing Isomorphism to a Fixed Graph in the Bounded-Degree Graph Model
We consider the problem of testing isomorphism to a fixed graph in the bounded-degree graph model. Our main result is that, for almost all $d$-regular $n$-vertex graphs $H$,
testing isomorphism to $H$ can be done using $\tildeO({\sqrt n})$ queries.
This result is shown to be optimal (up to a polylog factor) by a matching lower bound, which also holds for almost all graphs $H$.
The performance of our tester depends on natural graph parameters of the fixed ($n$-vertex) graph $H$ such as its diameter and the minimum radius of ``distinguishing neighborhoods'' (i.e., the
minimum $r=r(n)$ such that the ``$r$-neighborhoods'' of the $n$ different vertices are pairwise non-isomorphic).
Changes to previous version:
We discovered that Lemma 2.1 was previously proved by Mossel and Sun, and revised the paper accoringly; that is, omitted our own proof sketch of Lemma 2.1.
TR23-146 | 27th September 2023 15:45
On Testing Isomorphism to a Fixed Graph in the Bounded-Degree Graph Model
We consider the problem of testing isomorphism to a fixed graph in the bounded-degree graph model. Our main result is that, for almost all $d$-regular $n$-vertex graphs $H$,
testing isomorphism to $H$ can be done using $\tildeO({\sqrt n})$ queries.
This result is shown to be optimal (up to a polylog factor) by a matching lower bound, which also holds for almost all graphs $H$.
The performance of our tester depends on natural graph parameters of the fixed ($n$-vertex) graph $H$ such as its diameter and the minimum radius of ``distinguishing neighborhoods'' (i.e., the
minimum $r=r(n)$ such that the ``$r$-neighborhoods'' of the $n$ different vertices are pairwise non-isomorphic). | {"url":"https://eccc.weizmann.ac.il/report/2023/146/","timestamp":"2024-11-12T00:54:20Z","content_type":"application/xhtml+xml","content_length":"23016","record_id":"<urn:uuid:e95eb304-f73e-48e5-a009-b4226b3d6b58>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00092.warc.gz"} |
Bankers' Rounding
A number of people have pointed out to me over the years that VBScript's Round function is a bit weird. It seems like it should be pretty straightforward -- you pick the integer closest to the number
you've got, end of story. But what about, say, 1.5? There are two closest integers. Do you go up or down?
The Round function goes to the nearest integer, and if there are two nearest integers then it goes to the even one. 1.5 rounds to 2, 0.5 rounds to 0.
Why's that? Why not just arbitrarily say that we always round down in this situation? Why round down sometimes and up some other times? There actually is a good reason!
This algorithm is called the Bankers' Rounding algorithm because, unsurprisingly, it's used by bankers. Suppose a data source provides data which is often in exactly split quantities -- half dollars,
half cents, half shares, whatever -- but they wish to provide rounded-off quantities. Suppose further that a data consumer is going to derive summary statistics from the rounded data -- an average,
Ideally when you are taking an average you want to take an average of the raw data with as much precision as you can get. But in the real world we often have to take averages of data which has lost
some precision. In such a situation the Banker's Rounding algorithm produces better results because it does not bias half-quantities consistently down or consistently up. It assumes that on average,
an equal number of half-quantities will be rounded up as down, and the errors will cancel out.
If you don't believe me, try it. Generate a random list of numbers that end in 0.5, round them off, and average them. You'll find that Bankers' Rounding gives you closer results to the real average
than "always round down" averaging.
The Round, CInt and CLng functions in VBScript all use the Banker's Rounding algorithm.
There are two other VBScript functions which turn floats into integers. The Int function gives you the first integer less than or equal to its input, and the Fix function gives you the first integer
closer to zero or equal to its input. These functions do not round to the nearest integer at all, they simply truncate the fractional part.
UPDATE: What about FormatNumber? See this post.
• Anonymous
September 26, 2003
It's interesting to note another side effect of Bankers' Rounding. Notice any interest paid out is odd, while loans are even? It's to get that extra half after rounding.
• Anonymous
September 26, 2003
Those sneaky petes!
• Anonymous
September 26, 2003
After blogging about this myself (specifically for VB/VBA, not VBScript):http://ewbi.blogs.com/develops/2003/09/round_and_round.htmlI was surprised to see how many folks don't know this, or like
me always forget it. While I've gotten no comments or trackbacks, I have had hundreds of hits from Google searches.The interesting thing I was trying to point out in my post was that the Format
function does not use banker's rounding. For me, this turned out to be a good thing because it allowed me to catch a problem with my own replacement Round function, which I use when banker's
isn't what I want.Thanks for the explanation.
• Anonymous
September 26, 2003
Quick update to my prior post pointing out that VBScript's FormatNumber, like VBA's Format and FormatNumber, does not apparently use banker's rounding:http://ewbi.blogs.com/develops/2003/09/
quick_addition_.htmlThanks again.
• Anonymous
February 19, 2004
i tried format number (to 0 decimal points), and it still rounded up.
I am dividing numbers w/ decimals and need the final result to be rounded down to a whole number, never rounded up.
What am I missing?
• Anonymous
February 20, 2004
Apparently you're missing the last paragraph of this post -- that paragraph where I point out that Int rounds down and Fix rounds positive numbers down, negative numbers up.
Also, the next day's article is about FormatNumber.
• Anonymous
March 10, 2004
It's not just for averages.
Suppose you want to round and then sum the following amounts:
I've run into a calculation where I'm dividing a payment between principal and intrest, and without the bankers round the sum of the two would not add up right, making the occasional payment off
by a penny.
• Anonymous
March 10, 2004
> It's not just for averages.
Clearly -- Bankers' rounding produces better averages because it produces better sums.
The question of mean error accrued per operation is particularly important when dealing with subtractions.
If rounding tends to bias subtractions one way or another, and you perform MANY subtractions on subrahends that are close in size, then the net bias grows as the number of operations grows. That
can be very, very bad. (And if its money, pretty soon you've got a salami attack on your hands!)
• Anonymous
November 11, 2005
Doesn't this still introduce a small amount of bias since you round up all the odd values(1,3,5,7,9) and round down the evens(2,4,6,8) and there is one more odd than even? It is significantly
better than always rounding up but is there a procedure that can prevent all rounding bias?
• Anonymous
November 12, 2005
Kathleen, you've forgotten to list zero as an even number. Then there are as many evens as odds and your posited bias disappears.
• Anonymous
April 21, 2006
PingBack from http://www.datapoohbah.com/tech/?p=353
• Anonymous
April 28, 2006
Hmmmm. Seems to me that the Round function is inconsistent here. The round function works differently in Excel VBA than it does in Word VBA. Maybe someone up there in geniusland should have
created a separate BRound (Banker's Round) function for Bankers and kept the Round function working consistently across all Office products. Seems to me that VBA programmers would like the same
function with the same function name to work the same way across ALL products. Isn't consistency important anymore?
• Anonymous
April 28, 2006
The comment has been removed
• Anonymous
April 28, 2006
I stand corrected. The round function is consistent (banker's round) across VBA for Excel and Word. What I meant to say was that the round function in Excel, =ROUND(0.5,0) = 1 is inconsistent
with it's VBA counterpart (round(0.5,0) = 0). This is inconsistent, no?
May I suggest that the Round function VBA help file be enhanced to include the fact that Banker's round is being used and not an arithmetic round. I think this would have saved me and maybe
others considerable time.
• Anonymous
December 20, 2006
What is your evidence for the assertion that bankers use "Banker's rounding"? I have never met a banker that has ever heard of this method of rounding. IRS rules specify the 5-up method, as does
Euro conversions, ... The method is valid, and IMHO is best practice, hence it would be useful if it were available as a cell formatting option (without physically changing the underlying values)
in Excel. It has been the ASTM standard standard since the early 1940's; but bankers???
• Anonymous
January 22, 2007
However it works it would be nice if the documentation actually said what method was adopted rather than leaving us to find out the hard way.....
• Anonymous
June 15, 2007
Actually, I think its flawed for certain numbers. I'm embroiled in a row elsewhere, and I'm researching how to do the 'tricky' numbers. Try this dim i i = 50/111.111 wscript.echo i wscript.echo
round(i,2) 'should be 0.46 wscript.echo round(i,1) 'should be 0.4
• Anonymous
June 15, 2007
Oops, sorry typo! second one should be .5, as the input is 0.45000045000045
• Anonymous
June 16, 2007
The actual output for the last two is 0.45 and 0.5, and both are correct. I am very confused why you think this is a bug. Why do you think that it would round to 0.46? Obviously the input is
much, much, much closer to 0.45 than 0.46. It would be a pretty perverse implementation of Round which rounded to the one farther away!
• Anonymous
November 13, 2007
but, round function: 0.0451 -> 0.05 0.0450 -> 0.04 ??
• Anonymous
November 14, 2007
Again, I do not understand why this is a question. 0.0451 could be rounded to 0.04 or 0.05. 0.05 is closer. We round to the closest one. It would be stupid to round to the one farther away.
0.0450 could be rounded to 0.04 or 0.05. Neither is closer than the other. We round to the "even" one.
• Anonymous
April 09, 2008
Many thanks for this gem... I will go with rounding up as i am providing comparisons. Well worth knowing though...
• Anonymous
September 09, 2008
Thank goodness that the .Net Framework team thought to put in an overload for Math.Round that includes the MidpointRounding.AwayFromZero so we could bypass this ridiculous concept. Math.Round
should represent mathematics principles, not accounting or banking principles. I agree with the commenter above that suggested a BRound function or perhaps a BankersMath.Round or
Accounting.Round. In any and all cases, Math.Round should have followed mathematic standards. In any case, thanks for the description of what is happening and why my Round calls weren't working.
• Anonymous
April 17, 2009
Kathleen Barnes said: "Doesn't this still introduce a small amount of bias since you round up all the odd values(1,3,5,7,9) and round down the evens(2,4,6,8) and there is one more odd than even?"
yes is can still add bias but not for that reason. while Eric is right that the number of odd and even numbers is equal, what really matters is the distribution of odd and even numbers in your
sample when doing repeated rounding (or at least what distribution you might expect when choosing how to round). clearly if you are rounding and summing a series of only odd numbers you will see
a bias. there are many more types of rounding with different properties and trade offs as well of course e.g. stochastic rounding @Dale, it may be called bankers rounding but this is sound maths
used in many scientific and engineering uses e.g. DSP not just accounting (essentially its probably the right thing to do unless you know your samples are not uniformly distributed). conversely
rounding down is one of the easiest things to do but is seldom correct.
• Anonymous
October 21, 2010
Why not make an standardization of this situation? | {"url":"https://learn.microsoft.com/en-us/archive/blogs/ericlippert/bankers-rounding","timestamp":"2024-11-05T08:51:11Z","content_type":"text/html","content_length":"49491","record_id":"<urn:uuid:73298a3c-397c-44c3-8a00-bf1ae994c17b>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00311.warc.gz"} |
Mathematics – Science4All
Tag Archives: Mathematics
Notations do not matter to the essence of mathematics. But poor notations can be misleading. Notations based on exponents, radicals and logarithms definitely are. They are very distinct, even though
they are supposed to describe very similar relations between numbers. The triangle of power is a recently proposed alternative. In short, I am convinced!
Category Theory, Isomorphism, Functor (More Hiking in Modern Math World 7/7)
The Greatest Challenge of Mathematics | White Group Maths
This is a guest post I wrote on White Group Mathematics on December 30, 2012.
The Harmonious Mathematics of Music
It was when hearing the sounds of hammers that Pythagoras realized the ubiquity of numbers in mathematical harmony. He would go on laying down the mathematical foundations of music, based on octaves,
perfect fifths and major thirds. This mathematics of music would then become the favourite playground of all musicians, from Beethoven to Gangnam Style.
The Limitless Vertigo of Cantor’s Infinite
No one believed him. Not even fellow mathematicians. They thought he was wrong. They thought he was crazy. Even he ended up doubting himself and went crazy. And yet, he had mathematically proved it
all. Georg Cantor had figured out how to manipulate the infinite. Even more remarkable, he showed that there were actually several infinities; and some are bigger than others!
You've probably learned early on that there are three primary colours. But why three? And why these three? Surprisingly, the answer lies in the beautiful mathematics of linear algebra and (high)
dimension spaces!
This article retraces the endless pursuit of the infinite that is at the basis of mathematical analysis. From the first approximations of pi to the shape of our limitless universe, from the essential
usefulness of differential equations to the troubles with infinite sums, we present the great ideas of mathematical geniuses all along History.
The power of algebra lies in abstraction, and abstraction is basically forgetting. By retracing the History of algebra from its roots to more recent advancements, this article unveils the numerous
breakthrough in our understanding of the world, by abusing of the power of forgetting.
The Cubic Ball of the 2014 FIFA World Cup
I know this sounds crazy. Even stupid. But Adidas did design a cubic ball, called brazuca, for the 2014 World Cup. And, yet, this cubic ball is rounder than any previous ball in football History. How
is it possible? This article explains it.
The Addictive Mathematics of the 2048 Tile Game
2048 is the Internet sensation of the year. This very addictive game has been downloaded hundred of millions of times. Interestingly, this game raises plenty of intriguing mathematical questions.
This article unveils some of them!
Univalent Foundations of Mathematics
In an effort to make mathematics more computable, a consortium of today's greatest mathematicians have laid out new foundations. Amazingly, they all lie upon one single axiom, called univalence. The
goal of this axiom is to make formal mathematics more similar to informal mathematics. With univalence, our Arabic numbers aren't just like natural numbers; They are natural numbers. Univalence also
has unforeseen and mesmerizing consequences.
Homotopy Type Theory and Higher Inductive Types
In this article, we explore the possibilities allowed by higher inductive types. They enable a much more intuitive formalization of integers and new mind-blowing definitions of the (homotopical)
circle and sphere.
Type Theory: A Modern Computable Paradigm for Math
In 2013, three dozens of today's brightest minds have just laid out new foundation of mathematics after a year of collective effort. This new paradigm better fits both informal and
computationally-checkable mathematics. There is little doubt that it will fundamentally change our perspective on rigorous knowledge, and it could be that, in a few decades, the book they published
turns out to be the bedrock of all mathematics, and, by extension, all human knowledge! Have a primer of this upcoming revolution, with this article on type theory, the theory that the book builds
The Tortuous Geometry of the Flat Torus
Take a square sheet of paper. Can you glue opposite sides without ever folding the paper? This is a conundrum that many of the greatest modern mathematicians, like Gauss, Riemann, and Mandelbrot,
couldn't figure out. While John Nash did answer yes, he couldn't say how. After 160 years of research, Vincent Borrelli and his collaborators have finally provided a revolutionary and breathtaking
example of a bending of a square sheet of paper! And it is spectacularly beautiful!
The Most Beautiful Equation of Math: Euler’s Identity
In 1988, Euler's identity was elected most beautiful theorem of mathematics. It has been widely taught worldwide. But have you ever stopped to really sense the meaning of this incredible formula?
This article does.
The New Big Fish Called Mean-Field Game Theory
In recent years, at the interface of game theory, control theory and statistical mechanics, a new baby of applied mathematics was given birth. Now named mean-field game theory, this new model
represents a new active field of research with a huge range of applications! This is mathematics in the making!
The Revolutionary Galois Theory
In 1832, Évariste Galois died. He was 20. The night before his death, he wrote a legendary letter to his friend, in which he claims to have found a mathematical treasure! Sadly, this treasure had
long been buried in total indifference! It took nearly a century to rediscover it! Since then, Galois' legacy has become some of the finest pure mathematics, which represents a hugely active field of
research today with crucial applications to cryptography. Galois' work is now known as Galois theory. In essence, it unveils the hidden symmetries of numbers!
Linear Algebra and Higher Dimensions
Linear algebra is a one of the most useful pieces of mathematics and the gateway to higher dimensions. Using Barney Stinson's crazy-hot scale, we introduce its key concepts.
Numbers and Constructibility
Last summer, I got to discover Morellet's artwork on inclined grids. Amazingly, this artwork is a display of the irrationality of $\sqrt{2}$! It's also a strong argument for the existence of this
number. In this article, after discussing that, I take readers further by discussing what numbers can be constructed geometrically, algebraically, analytically or set theoretically using the power of
Logarithms and Age Counting
Amusingly, the age difference between a 45-year-old man and a 25-year-old woman doesn't seem as big as the age difference between them 20 years earlier, when the woman was a little 5-year-old girl.
This remark was the insight the late science popularizer Albert Jacquart liked to give to his readers to explain logarithms. This article pays tribute to the great scientist by introducing age
difference as he liked to tell it. | {"url":"https://www.science4all.org/tag/mathematics-2/","timestamp":"2024-11-03T23:16:17Z","content_type":"text/html","content_length":"47250","record_id":"<urn:uuid:c3444be2-7a53-4af6-9162-866f00598207>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00312.warc.gz"} |
Configuring the churn rate formula
On August 15, 2022 we retired the Churn Rate Formula setting. Accounts opened prior to this date that had selected the Shopify formula can continue to configure this setting. All other accounts,
including new accounts opened on or after August 15, 2022 use the standard formula.
ChartMogul offers two churn rate formulas depending on your business model: B2B or B2C.
Our Standard Formula generally works best for B2B subscription businesses. This is because the formula calculates how a given metric (e.g., number of customers, subscriptions, or MRR) has increased
(or decreased) between the start and end of a period.
For B2C subscription businesses, we recommend the Shopify Formula. This formula calculates how a given metric (e.g., number of customers, subscriptions, MRR) has increased (or decreased) between the
start and end of a period by calculating and summing together the increase (or decrease) for each day in the period.
The Shopify Formula is better suited for businesses that add a high volume of new customers each month — usually with little expansion or contraction MRR — and mainly offer monthly subscriptions
customers can cancel at any time.
Learn about each of ChartMogul’s churn rate charts, including how your churn rate formula affects the calculation of each rate:
Resources and further reading:
Configuring the Churn Rate Formula
Access your Churn Rate Formula setting by clicking Settings & Data > Data Settings > Subscription Analytics. From there, select the formula that best suits your needs.
Learn more about Data Settings in ChartMogul. | {"url":"https://help.chartmogul.com/hc/en-us/articles/207340419-Configuring-the-churn-rate-formula","timestamp":"2024-11-07T18:33:48Z","content_type":"text/html","content_length":"30588","record_id":"<urn:uuid:72b49f87-d320-4960-a217-3452a4f2358e>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00225.warc.gz"} |
Linear regression in Python
Regression algorithms predict continuous values from predictor variables. Predicting the price of a house based on its characteristics is a good example of regression analysis.In this article, I
will implement univariate (one-variable) linear regression in python.
Linear regression is an algorithm that will find a straight line that comes as close as possible to a set of points. Dots represent training data.
Our dots in orange are the input data. They are represented by the couple (x_{i}, y_{i}). The values x_{i} are the predictor variables, and y_{i} is the observed value (the price of a house for
example). We seek to find a straight line F(x) = \alpha*x + \beta such that, whatever x_{i}, we want F(x_{i}) \approx y_{i}.
In other words, we want a line that is as close as possible to all the points of our training data.
The problem we are trying to solve and its dataset are those of a course I took on Andrew NG’s Machine Learning on Coursera. At the time I had to implement the solution in MATLAB. I can assure you it
was not my cup of tea. 😉
Suppose you are the CEO of a food truck franchise. You are considering different cities to open a new point of sale. The chain already has trucks in different cities and you have data for city
profits and populations.You want to use this data to help you choose the city to open a new point of sale there.
This problem is of the supervised learning type which can be modeled by a linear regression algorithm. It is of the supervised type because for each city having a certain number of population
(predictive variable X), we have the gain made in the latter (the variable we are trying to predict: Y).
Training data is in CSV format. The data is separated by commas. The first column represents the population of a town and the second column shows the profit of a walking truck in that town. A
negative value indicates a loss.\endThe number of records of our input data is 97.
To solve this problem, we will predict the profit (the variable Y) according to the size of the population (the predictive variable X)
First of all, it will be necessary to read and load the data contained in the CSV file. Python offers via its Pandas library classes and functions to read various file formats including CSV.
The read_csv() function returns a DataFrame. It is an array of two dimensions containing, respectively, the size of the population and the profits made. To be able to use the regression libraries of
Python, it will be necessary to separate the two columns in two Python variables.
#selection of the first column of our dataset (the size of the population) X = df.iloc[0:len(df),0] #selection of second columns of our dataset (the profit made) Y = df.iloc[0:len(df),1]
The X and Y variables are now simple arrays containing 97 elements.
The len() function returns the size of an arrayThe iloc function allows you to retrieve data by its positioniloc[0:len(df),0] will retrieve all data from line 0 to line 97 (which is len(df)) located
at column index 0
Before modeling a machine learning problem, it is often helpful to understand the data. To achieve this, we can visualize them in graphs to understand their dispersion, deduce the correlations
between the predictive variables, etc.
Sometimes it is not possible to visualize the data because there are too many predictor variables. This is not the case here, we only have two variables: population and profits.
We can use a scatter plot type graph to visualize the data:
It is clear that there is a linear correlation between the variables. And that the more the size of the population increases, the more the profit does the same.
The Python code for making this point cloud is as follows:
import matplotlib.pyplot as plt axes = plt.axes() axes.grid() # draw a grid for better readability of the graph plt.scatter(X,Y) # X and Y are the variables we extracted in the previous paragraph
Matplotlib is the python library allowing to make graphs of several types:HistogramsPoint Clouds,Draw function curvesPie plot diagramsetc.
Now that we better understand our data, we will tackle the heart of the problem: Find a predictive function F(X) that will take a population size as input, and produce an estimate of the expected
gain as output. The idea of the game is that the prediction is close to the observed value F(X) \approx Y.
Note: For the sake of simplicity, I have chosen not to split my data from the CSV file into Training Set and Test Set. This good practice, to be applied in your ML problems, helps to avoid
over-learning. In this article, our data will be used both to train our regression algorithm and also as a test set.
To use linear regression with one variable (univariate), we will use the scipy.stats module. The latter has the linregress function, which allows you to do linear regression.
from scipy import stats #linregress() returns several return variables. We will be interested # especially on slope and intercept slope, intercept, r_value, p_value, std_err = stats.linregress(X, Y)
After the linregress() function has returned the parameters of our model: slope and intercept, we can make predictions. Indeed, the prediction function will be of the form:
We can write this function F(x) in python as follows:
Thanks to this function, we can make a prediction on our 97 populations which will give us a straight line.
#the variable fitLine will be an array of predicted values from the array of variables X fitLine = predict(X) plt.plot(X, fitLine, c='r')
Indeed, we can clearly see that the red line approaches all the points of the data set as closely as possible. Pretty isn’t it? 🙂
If we take the 22nd line of our CSV file by chance, we have the population size which is: 20.27 * 10,000 people and the gain made was: 21.767 * $10,000
We obtain an estimated gain close to the true observed gain (with a certain degree of error) | {"url":"https://akyalab.com/linear-regression-in-python/","timestamp":"2024-11-13T18:46:05Z","content_type":"text/html","content_length":"103970","record_id":"<urn:uuid:9b149558-466f-4d9a-a274-1ee7b344e2bd>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00104.warc.gz"} |
430+ Limits and Derivatives Questions with Solution JEE Math
Features and benefits:
1. You will get all 430+ Limits and derivatives problems and their solution in pdf format
2. All limits and derivatives mcq jee are solved with correct answer.
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9. Solving Limits and Derivatives questions can prepare you for JEE Main and JEE Advance Exam.
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6. Personalized to individual learning style.
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8. Highlighting and underlining for emphasis.
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11. You will know Problem solving skills and Tricks.
12. MCQ pdf is also usefule for improving concepts of Limits and derivatives, Limits and derivatives class 11,
13. Dowload Limits and derivatives class 11 pdf, Limits and derivatives class 11 Notes.
430+ Limits and Derivatives Questions with Solution JEE Math :
Features and benefits of 430+ Limits and Derivatives Questions with Solution pdf
1. You will get all 430+ Limits and derivatives problems and their solution in pdf format
2. All limits and derivatives mcq jee are solved with correct answer.
3. Solving Limits and derivatives MCQ can reinforce your concepts and problem-solving skills
4. This will improve your speed and accuracy.
5. You will learn short tricks to solve tough problems.
6. Solving 430+ questions can help to identify areas of weakness and gaps in concept knowledge
7. Solving more questions can expose individuals to a wider range of question types and formats.
8. Regular practice can help you achieve your learning goals and objectives
9. Solving Limits and Derivatives questions can prepare you for JEE Main and JEE Advance Exam.
Learn More..
1. Solving Limits and Derivatives mcq questions can enhance critical thinking and analytical skills.
2. Organized structure for easy understanding of 430+ Limits and Derivatives Problems with Solution.
3. Accurate representation of content according to JEE Main and Advance syllabus.
4. Key points and ideas are summarized.
5. Clear and concise language for easy reading.
6. Personalized to individual learning style.
7. Use of diagrams and graphs.
8. Highlighting and underlining for emphasis.
9. Clarify complex concepts in notes.
10. Exclusively designed for JEE Main and Advance Aspirants.
11. You will know Problem solving skills and Tricks.
12. MCQ pdf is also usefule for improving concepts of Limits and derivatives, Limits and derivatives class 11,
13. Dowload Limits and derivatives class 11 pdf, Limits and derivatives class 11 Notes.
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Calculate GSC CTR Stats By Position Using Python for SEO | importSEM
Estimated Read Time: 5 minute(s)
Common Topics: data, ctr, position, python, int
Last week SEO Clarity came out with a new SERP CTR study. The numbers were lower than I expected even as an average for all queries. It got me thinking. What is MY average CTR by position? Turns out,
it’s much higher. This is likely due to good SEO by optimizing the title, meta, and rich snippets. These calculations can be easily done in Google Sheets, but I wanted to try it in Python and make it
an app. Thus this will be a very short tutorial.
In this Python SEO tutorial, we’ll take a standard 1000 record Google Search Console 12-month performance export (Clicks, Impressions, CTR, Position) and sum/mean those metrics by position. Note, if
you are a very large site, the 1000 records will only be a snapshot, you’ll want to get much more data via the GSC API before running this. Let’s do it! As always, be careful copying the code as
indents are not always preserved.
Not interested in the tutorial? Head straight for the app here!
Requirements and Assumptions
• Python 3 is installed locally or Google Colab, and basic Python syntax is understood.
• Google Search Console performance data “Queries.csv” export file.
Import and Install Modules
• pandas: for storing the information in a table form
• NumPy: for easy dataframe type converting
First, we very simply import the 2 modules listed above that the script requires.
import pandas as pd
import numpy as np
Next, we read the Google Search Console performance CSV into a pandas dataframe. Note, when you export the data from GSC, it will download a zip file. Unzip and you’ll want to use Queries.csv from
that zip.
df = pd.read_csv("Queries.csv")
Now we create a counter variable x for a loop coming up, a variable y for you to adjust how many positions you want to process, and create our empty dataframe to store the calculations per SERP
x = 1
y = 9
d = {'Position': [], 'Sum Clicks': [], 'Sum Impressions':[], 'Avg CTR':[],'Min CTR':[],'Max CTR':[],'Max CTR KW':[]}
df2 = pd.DataFrame(data=d)
Next, we loop by the number of positions you set y to starting with the value of x which is 1 since there technically is no position 0. For each position range x to x.9, we filter the initial
dataframe. Sort the dataframe by CTR descending, remove the percent sign so we can process and retype the CTR column from string to float.
while x < y:
df1 = df[(df['Position'] >=x) & (df['Position'] < x+1)]
df1 = df1.sort_values('CTR',ascending=False)
df1['CTR'] = df1['CTR'].str.replace('%','')
df1['CTR'] = df1['CTR'].astype(np.float16)
There may be times when you don’t have a keyword in a numerical position, say 11, so we need to use try/except to detect that and keep moving on. We then simply use the pandas functions, mean(), min
(), max(), and sum() to make those calculations along with grabbing the first record’s keyword which will be the highest CTR.
ctr = int(round((df1['Clicks'].sum()/df1['Impressions'].sum())*100))
ctr_min = int(df1['CTR'].min())
ctr_max = int(df1['CTR'].max())
ctr_max_kw = df1.iloc[0]['Top queries']
clicks = int(df1['Clicks'].sum())
impressions = int(df1['Impressions'].sum())
Those calculations will result in a default of several decimal places. To remove those, we’ll set the columns to int.
df2['Avg CTR'] = df2['Avg CTR'].astype(int)
df2['Min CTR'] = df2['Min CTR'].astype(int)
df2['Max CTR'] = df2['Max CTR'].astype(int)
df2['Position'] = df2['Position'].astype(int)
df2['Sum Clicks'] = df2['Sum Clicks'].astype(int)
df2['Sum Impressions'] = df2['Sum Impressions'].astype(int)
Now we handle positions we don’t have in the data and just pass on through, increasing the counter to get the next position ready and move back up to the start of the loop.
x += 1
Once all the positions are processed via the while loop above, we create a dictionary list with all the data for a single position
data = {'Position': int((x)),'Sum Clicks':clicks,'Sum Impressions':impressions,'Avg CTR':ctr,'Min CTR':ctr_min,'Max CTR':ctr_max,'Max CTR KW':ctr_max_kw}
df2 = df2.append(data, ignore_index=True)
For the presentation, we want to add the percent sign, but you can’t if it’s an int value so we convert it to str, and add the percent sign via a simple lambda function. Finally, we display the
df2 = df2.astype(str)
df2['Avg CTR'] = df2['Avg CTR'].apply(lambda x: x + "%")
df2['Min CTR'] = df2['Min CTR'].apply(lambda x: x + "%")
df2['Max CTR'] = df2['Max CTR'].apply(lambda x: x + "%")
Sample Output
Now you have the framework to look at your Google Search Console data by position. Lots of future potential to further feature this script out with more calculations and perhaps some data blending
from different data sources. Now go supercharge your SEO! Enjoy!
Don’t forget to try the streamlit app here!
Now get out there and try it out! Follow me on Twitter and let me know your Python SEO applications and ideas!
How can Python be used to calculate Google Search Console (GSC) Click-Through Rate (CTR) stats by position for SEO analysis?
Python scripts can be crafted to fetch GSC data, calculate CTR stats based on position, and provide insights into the performance of keywords at different positions.
Which Python libraries are commonly used for calculating GSC CTR stats by position?
Commonly used Python libraries for this task include pandas for data manipulation, matplotlib for visualization, and seaborn for statistical data visualization.
What specific steps are involved in using Python to calculate GSC CTR stats by position for SEO?
The process includes fetching GSC data, extracting relevant information, grouping data by position, calculating CTR for each position, and using Python functions for analysis and visualization.
Are there any considerations or limitations when using Python for this calculation?
Consider the accuracy of position data, potential variations in CTR calculations, and the need for a clear understanding of the correlation between position and CTR. Regular updates to the analysis
may be necessary.
Where can I find examples and documentation for calculating GSC CTR stats by position using Python?
Explore online tutorials, documentation for relevant Python libraries, and resources specific to SEO data analysis for practical examples and detailed guides on calculating GSC CTR stats with Python.
Latest posts by Greg Bernhardt
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Nikola Tesla and numbers 3, 6, 9: The secret key to free energy? - Sens de la Vie
Nikola Tesla and numbers 3, 6, 9: The secret key to free energy?
by pierre
written by pierre
Nikola Tesla brought us a lot, but he also left with secrets and left us some leads, like indications on the numbers 3, 6, 9.
Tesla is not only the inventor of the alternating current we all use today, his discoveries have gone far beyond that.
In fact, he created revolutionary inventions such as wireless radio communications, turbine engines, helicopters (even though Da Vinci had originally had the idea), fluorescent and neon tubes,
torpedoes and X-rays, among others.
In addition to his countless inventions and futuristic creations, Nikola Tesla was also known for his eccentricities.
Such as the use of hotel rooms, the number of which could be divided by 3, the cleaning of plates with 18 napkins, or the rotation of 3 rounds around a block before entering a building.
No one knows the reason behind Nikola Tesla’s mysterious behaviour.
Interestingly, Tesla recounted many times that she had experienced intense flashes of light, which were followed by moments of creativity and intense clarity.
Nikola Tesla was able to imagine and see an invention in his mind during these “moments of clarity” almost in holographic detail.
He said that he could even turn these visions in all directions, disassemble them piece by piece, and so he knew exactly how he was going to build his Inventions based on his experiences, much like
Iron Man’s hero with his holographic images.
In addition to many other strange things, Nikola Tesla had calculated the nodal points around the planet, and they were probably related to the numbers three, six and nine.
Nikola Tesla and numbers 3, 6 and 9
Tesla said that these figures were extremely important.
He was obsessed with numbers 3, 6 and 9.
He understood a fundamental fact, unknown to many, which is the universal language of mathematics.
A science discovered by man, not invented by him.
Tesla took into account the numerical models that occur in the universe, such as star formation, embryonic cell development, and many others that some call “God’s plan”.
There is a fundamental system to which nature seems to respond: “The powers of the binary system”, where the model starts from one and continues to double the number.
Thus, cells and embryos are developed, for example, according to the following scheme: 1, 2, 4, 8, 16, 32, 64, 128, 256, etc.
Marko Rodin discovered inside the Vortex Math (the science of torus anatomy) a repetitive pattern: 1, 2, 4, 8, 7, 5, 1, 2, 2, 4, 8, 7, 5, 1, 2, 4, and so on to infinity.
Here, numbers 3, 6 and 9 do not exist and, according to Rodin, this is due to the fact that these numbers represent a vector from the third to the fourth dimension, called a “flow field”.
This field is a higher dimensional energy, which has an influence on the energy circuit of the other six numbers.
Going even further, Randy Powell, a student of Marko Rodin, says it is the secret key to free energy, which Tesla sought until the last days of her life.
However, if we look beyond Tesla itself, we notice that whatever the culture, number three has always been extremely important.
Video: Tesla and the numbers 3, 6, 9
See also:
The belief in one God in ancient Egypt, teaching of Hermes
Source: www.ancient-code.com
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Ratios & Revenue (OG Quant Review #115)
Your Answer
A pharmaceutical company received $3 million in royalties on the first $20 million in sales of the generic equivalent of one of its products and then $9 million in royalties on the next $108 million
in sales. By approximately what percent did the ratio of royalties to sales decrease from the first $20 million in sales to the next $108 million in sales?
(A) 8%
(B) 15%
(C) 45%
(D) 52%
(E) 56%
Answer: C
In the explanation, it says "the percent decrease in the royalties to sales ratios is 100 times the quotient of the difference in the ratios divided by the ratio of royalties to sales for the first
$20 million in sales". What?? How did they get to this? Can anyone explain it a bit better?
Thank you in advance! | {"url":"https://www.beatthegmat.com/ratios-revenue-og-quant-review-115-t275786.html?sid=e80eb58ff5daeddb77d246848833ccac","timestamp":"2024-11-03T19:00:40Z","content_type":"text/html","content_length":"762723","record_id":"<urn:uuid:3b412636-9936-4a1e-92d2-6129448975a2>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00529.warc.gz"} |
Sin3x Formula in terms of sinx [with Proof] - iMath
Sin3x formula in terms of sinx is as follows: 3sinx – 4sin^3x. The formula can be expressed as sine 3x = 3 sine x – 4 sine cube x. In this post, we will learn how to prove sin3x formula along with
some examples. This is very useful in the area of Trigonometry to solve various trigonometric equations, identities, etc.
Sin3x Formula
Sin3x is the sine trigonometric function of a triple angle 3x. The formula of sin 3x in terms of sinx is given below.
sin3x=3sinx – 4sin^3x.
sin3x in Terms of sin x
To derive the sin3x formula, we will use the following trigonometric formulas:
1. sin(a+b) = sin acos b +cos a sin b
2. sin 2x=2sin x cos x
3. cos 2x=1-2sin^2x
4. sin^2x+cos^2x=1
Below are the steps to be followed in order to establish the formula of sin 3x.
Step 1: At first, we will write 3x as 2x+x, and then we will apply the above formula 1. By doing so, we get that
sin 3x=sin(2x+x)
= sin 2x cos x +cos 2x sin x
Step 2: Now, we will apply the formulas of sin2x and cos2x which are given in (2) and (3) above. Thus, we have
sin 3x=(2sin x cos x) cos x +(1-2sin^2x)sin x
= 2sin x cos^2 x+sin x-2sin^3x
= 2sin x (1-sin^2x)+sin x-2sin^3x as we know that cos^2x=1-sin^2x follows from the above formula (4).
= 2sin x -2sin^3x +sin x -2sin^3x
= 3sin x -4sin^3x .
Hence, the formula of sin 3x is 3sin x -4sin^3x.
Also Read:
Question: Find the value sin270 degree.
In the above formula of sin3x, that is, sin3x=3sinx-4sin^3x, we put x=90 degree. Thus, we obtain that
sin 270°=3sin 90° – 4sin^390°
= 3×1 – 4×1^3 as we know that sin90°=1
= 3-4
= -1
Thus, the value of sin270 degree is -1.
Q1: What is the sin3x formula in terms of sinx?
Answer: The sin3x formula in terms of sinx is given by sin3x = 3sinx -4sin^3x. | {"url":"https://www.imathist.com/sin3x-formula-in-terms-of-sinx/","timestamp":"2024-11-09T17:21:22Z","content_type":"text/html","content_length":"179318","record_id":"<urn:uuid:ebb38feb-3b72-441b-b379-af37ab4d2655>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00100.warc.gz"} |
1) Heat flow through a wall is one dimensional when the temperature of the wall varies in one direction only.
a) The small thickness of the wall causes the temperature gradient in that direction to be large. Further, if the air temperatures in and outside the house remain constant, and then heat transfer
through the wall of a house can be modeled as steady and one-dimensional.
b) The temperature of the wall in this case will depend on one direction only (say the x-direction) and can be expressed as T(x).
c) But
d) Fourier’s Law of heat conduction
e) Consider a plane wall of thickness L and average thermal conductivity k. The two surfaces of the wall are maintained at constant temperatures of T[1] and T[2] | {"url":"https://mechanicalengineering.softecksblog.in/5762/","timestamp":"2024-11-05T15:51:23Z","content_type":"text/html","content_length":"125623","record_id":"<urn:uuid:68ae7e88-486c-40bd-b40a-2e70a2b3569f>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00761.warc.gz"} |
Calculating number of ion channels in lipid bilayer
• Thread starter af86
• Start date
In summary, using the above information, the number of channels present in a lipid bilayer is 1.27E-17.
Homework Statement
Calculate number of channels present in a lipid bilayer
q = 1.6^-19; D (diffusion constant) = 1.2E-9; C (Concentration); 4.36; k = 1.38E-23; T=298; d = 2.5E-9
Therefore G[1] = 1.27E-17
Channel area = 7.07E-18
measured conductance (from injecting Na+ concentration of 100mM) = 149.24uS
Homework Equations
G1 (conductance) = q^2DC/kTd
Gm = g/[tex]\pi[/tex]r^2
The Attempt at a Solution
Conductance of a single channel = 1.27E-12 x 7.07E-18
Number of channels = Gm/G1 ??
Calculating the number of channels is what is getting to me. I have the area of the channel, and I have a measured conductance from injecting a concentration
The conductances sum as you mention. I would think that N = Gm/G1 just as you suppose. Did you not get the right answer?
Yep. Tried that and my answer came out a few order of magnitudes out...double and triple checked all my work and still..could be that the answer given is incorrect.
Not sure and without access to my books/notes, may be unable to help here: in your modelling, do you assume that a leak conductance is present?
Thanks denverdoc.
I figured it out (I think). But yes, leak conductance was taken into account; I think the only way to go is to divide Gm/G1.
Your welcome for what is is was worth--amazing how much one forgets--once upon a time,many moons ago, I was a post doc in an electrophysiology lab.
So I thought I figured this out - but I haven't!
I have some values, as a test, and I can't seem to get it right.
Can anybody please help!? I know it uses the equations above, but something is going wrong in the last step when I'm trying to calculate the number of channels and conductance...I know what the
answers should be; number of channels is 3E6
bilayer area = 1.1E-6m^2
bilayer thickness = 2.5E-9m
channel radius = 1.5E-9m
channel thickness = 2.5E-9
diffusion constant = 1.2E-9
dielectric constant = 2.3
change in energy (i.e [tex]\Delta[/tex]U or sometimes [tex]\Delta[/tex]E)
can anybody help please?
FAQ: Calculating number of ion channels in lipid bilayer
1. How do you calculate the number of ion channels in a lipid bilayer?
To calculate the number of ion channels in a lipid bilayer, you will need to know the surface area of the bilayer and the average area occupied by one ion channel. This can be determined using
various techniques such as patch-clamp electrophysiology or imaging methods. Once you have these values, simply divide the surface area by the area of one ion channel to get the total number of ion
channels in the bilayer.
2. What factors affect the number of ion channels in a lipid bilayer?
The number of ion channels in a lipid bilayer can be influenced by various factors such as the type and concentration of ions present, the composition and fluidity of the bilayer, and the presence of
other molecules such as proteins or lipids that may interact with the ion channels. These factors can alter the number, activity, and localization of ion channels in the bilayer.
3. Can the number of ion channels in a lipid bilayer change over time?
Yes, the number of ion channels in a lipid bilayer can change over time. This can be due to various physiological processes such as channel trafficking, insertion or removal of channels from the
membrane, or changes in the lipid composition of the bilayer. Additionally, external factors such as temperature, pH, and signaling molecules can also affect the number of ion channels in the
4. Are there any mathematical models for predicting the number of ion channels in a lipid bilayer?
Yes, there are various mathematical models and simulations that can be used to predict the number of ion channels in a lipid bilayer. These models take into account factors such as channel density,
channel activity, and bilayer properties to estimate the number of channels present. However, experimental validation is necessary to confirm the accuracy of these predictions.
5. Why is it important to know the number of ion channels in a lipid bilayer?
Understanding the number of ion channels in a lipid bilayer is crucial for studying their function and regulation. It can also provide valuable insights into the overall membrane properties and its
role in cellular processes such as signal transduction and ion homeostasis. Additionally, knowing the number of ion channels can aid in the development of new drugs and therapeutic strategies
targeting these channels. | {"url":"https://www.physicsforums.com/threads/calculating-number-of-ion-channels-in-lipid-bilayer.364397/","timestamp":"2024-11-03T18:24:01Z","content_type":"text/html","content_length":"89143","record_id":"<urn:uuid:062ea070-e53d-402c-aec3-56684227d5ad>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00508.warc.gz"} |
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