text
stringlengths 256
16.4k
|
|---|
The proton to electron mass ratio: $$\mu={m_p\over m_e}= {\alpha^2\over \pi r_pR_H}=1836.15267\;\;\leftarrow\;significant!!!$$
& the Planck mass to electron mass ratio:$${m_{\ell}\over m_e}={\alpha^2\over{4\pi\ell R_H}}=2.3893048e+22$$
using $$m_pr_p=4\ell m_{\ell}=r_em_e\;where\;r_e={\alpha^2\over\pi R_H}$$
$\alpha=$ fine-structure constant
$r_p=$ proton radius
$R_H=$ Rydberg constant
$r_e={\alpha^2\over\pi R_H}=$ will define this term later (effective radius??? atomic radius?)
$\ell=$ Planck $\ell$ength
$\pi=3.1415926535897932384626433832795028841971693993751058209749445923078164\dots$
This is significant proof that Nassim Haramein's proton radius solution IS correct as it works in harmony with mainstream solutions to solve for natural dimensionless constants that have never before been analytically determined (except for the Rydberg constant, $R_H$).
(see previous post on proton-electron mass ratio and Planck mass to electron mass ratio for Google calculator links for ratio calculations)
The Surfer, OM-IV
|
Background: The arena is fixed particle number nonrelativistic quantum mechanics. The state space is $$ \mathbf{H}(1)=\mathcal H\otimes\mathcal S, $$ where $\mathcal H$ is an "orbital" state space ($L^2(\mathbb R^3)$ if one'd like), and $\mathcal S$ is the spin space, which for a particle of spin $s$ is identifiable with $\mathbb C^{2s+1}$.
Because in this case it is (in my opinion) more tractable notationally, I will identify $\mathcal H$ with $L^2(\mathbb R^3)$ (position-basis wavefunctions), and a basis for $\mathcal S$ is $\left|\sigma\right\rangle$, where $\sigma=-s,-s+1/2,...,+s$. We can identify $\left|\sigma\right\rangle$ with the $\sigma$th standard basis vector of $\mathbb C^{2s+1}$.
A general 1-particle state is then $$ \Psi(\mathbf x)=\sum_\sigma\psi_\sigma(\mathbf x)\left|\sigma\right\rangle, $$ which is just a $2s+1$-component wavefunction.
The $k$-particle fermionic state space is $$ \mathbf H_A(k)=\bigwedge^k(\mathcal H\otimes\mathcal S), $$ and the $k$-particle bosonic state space is $$ \mathbf H_S(k)=\bigvee^k(\mathcal H\otimes\mathcal S). $$
In case $k=2$, the construction of a general state is easy. The "unsymmetric" state space is $\mathcal H\otimes\mathcal S\otimes\mathcal H\otimes\mathcal S\simeq \mathcal H\otimes\mathcal H\otimes\mathcal S\otimes\mathcal S$, and in the product spin space $\mathcal S^2\equiv\mathcal S\otimes\mathcal S$, and we may easily construct a basis, it is $$\Sigma_{\sigma_1\sigma_2}\equiv \left|\sigma_1\sigma_2\right\rangle\equiv\left|\sigma_1\right\rangle\otimes \left|\sigma_2\right\rangle, $$ which we may identify with $(2s+1)\times(2s+1)$ square matrices, particularily, $\Sigma_{\sigma_1\sigma_2}$ is the matrix whose $(\sigma_1,\sigma_2)$th element is 1, and the rest is zero.
A general "unsymmetric" state is $$ \Psi(\mathbf x_1,\mathbf x_2)=\sum_{\sigma_1,\sigma_2}\psi_{\sigma_1\sigma_2}(\mathbf x_1,\mathbf x_2)\Sigma_{\sigma_1\sigma_2}. $$
To find symmetric or skew-symmetric states, one may decompose $$ \Sigma_{\sigma_1\sigma_2}=S_{\sigma_1\sigma_2}+A_{\sigma_1\sigma_2}, $$ where $S_{\sigma_1\sigma_2}=\Sigma_{(\sigma_1\sigma_2)}$ and $A_{\sigma_1\sigma_2}=\Sigma_{[\sigma_1,\sigma_2]}$. Here I am ignoring the normalizations of the bases, as that is conceptually not relevant. There are $2s^2+3s+1$ independent $S_{\sigma_1\sigma_2}$ states, and $2s^2+s$ independent $A_{\sigma_1\sigma_2}$ states and together they span $\mathcal S^2$, so we can express a general
symmetric state as $$ \Psi_S(\mathbf x_1,\mathbf x_2)=\sum_{\sigma_1\le\sigma_2}\psi^S_{\sigma_1\sigma_2}(\mathbf x_1,\mathbf x_2)S_{\sigma_1\sigma_2}+\sum_{\sigma_1<\sigma_2}\psi^A_{\sigma_1\sigma_2}(\mathbf x_1,\mathbf x_2)A_{\sigma_1,\sigma_2}, $$ where the $\psi^S_{\sigma_1\sigma_2}(\mathbf x_1,\mathbf x_2)$ wavefunctions are symmetric with respect to the exchange of the positions $\mathbf x_1,\mathbf x_2$ and the wavefunctions $\psi^A_{\sigma_1\sigma_2}(\mathbf x_1,\mathbf x_2)$ are skew.
A general
skew-symmetric state is expressed then by the same combination but with the $\psi^S_{\sigma_1\sigma_2}(\mathbf x_1,\mathbf x_2)$ being skew in the position vectors and the functions $\psi^A_{\sigma_1\sigma_2}(\mathbf x_1,\mathbf x_2)$ being symmetric in the position vectors. Question:
Assume I want to do the same thing but with $k$ particles instead of 2. The situations seems much more complicated, mainly because the spin bases $$ \Sigma_{\sigma_1...\sigma_k} $$ now cannot be split into
totally symmetric and skew-symmetric parts while still spanning the set of all "unsymmetric" spin states.
So, if I want to construct all bosonic and all fermionic $k$-particle states, what is the analogue of what I have done with 2-particle states?
I have a feeling this might actually be quite untractable generally (we want to have the decomposition of $\bigwedge V\otimes W$ into direct sums of exterior (or symmetric) products, which by this answer here: https://math.stackexchange.com/questions/211561/exterior-power-of-a-tensor-product/553405 seems untractable), however I assume this is a very important situation for applied quantum mechanics, so in this case, how to construct symmetry-respecting states?
|
Difference between revisions of "Inertia"
(→Derivation)
Line 42: Line 42:
where <math>J = mr^{2}</math> is called the '''[https://en.wikipedia.org/wiki/Moment_of_inertia moment of inertia]''' (kg.m<sup>2</sup>).
where <math>J = mr^{2}</math> is called the '''[https://en.wikipedia.org/wiki/Moment_of_inertia moment of inertia]''' (kg.m<sup>2</sup>).
−
Note that in physics, the moment of inertia <math>J</math> is denoted as <math>I</math>. In electrical engineering, the convention is for the letter "i" to always be reserved for current, and is often replaced by the letter "j", e.g. the complex number operator i in mathematics is j in electrical engineering.
+
Note that in physics, the moment of inertia <math>J</math> is denoted as <math>I</math>. In electrical engineering, the convention is for the letter "i" to always be reserved for current, and is often replaced by the letter "j", e.g. the complex number operator i in mathematics is j in electrical engineering.
== Normalised Inertia Constants ==
== Normalised Inertia Constants ==
− + + + + + + +
== Generator Inertia ==
== Generator Inertia ==
Revision as of 05:31, 27 August 2018
In power systems engineering, "inertia" is a concept that typically refers to rotational inertia or rotational kinetic energy. For synchronous systems that run at some nominal frequency (i.e. 50Hz or 60Hz), inertia is the energy that is stored in the rotating masses of equipment electro-mechanically coupled to the system, e.g. generator rotors, fly wheels, turbine shafts.
Derivation
Below is a basic derivation of power system rotational inertia from first principles, starting from the basics of circle geometry and ending at the definition of moment of inertia (and it's relationship to kinetic energy).
The length of a circle arc is given by:
[math] L = \theta r [/math]
where [math]L[/math] is the length of the arc (m)
[math]\theta[/math] is the angle of the arc (radians) [math]r[/math] is the radius of the circle (m)
A cylindrical body rotating about the axis of its centre of mass therefore has a rotational velocity of:
[math] v = \frac{\theta r}{t} [/math]
where [math]v[/math] is the rotational velocity (m/s)
[math]t[/math] is the time it takes for the mass to rotate L metres (s)
Alternatively, rotational velocity can be expressed as:
[math] v = \omega r [/math]
where [math]\omega = \frac{\theta}{t} = \frac{2 \pi \times n}{60}[/math] is the angular velocity (rad/s)
[math]n[/math] is the speed in revolutions per minute (rpm)
The kinetic energy of a circular rotating mass can be derived from the classical Newtonian expression for the kinetic energy of rigid bodies:
[math] KE = \frac{1}{2} mv^{2} = \frac{1}{2} m(\omega r)^{2}[/math]
where [math]KE[/math] is the rotational kinetic energy (Joules or kg.m
2/s 2 or MW.s, all of which are equivalent) [math]m[/math] is the mass of the rotating body (kg)
Alternatively, rotational kinetic energy can be expressed as:
[math] KE = \frac{1}{2} J\omega^{2} [/math]
where [math]J = mr^{2}[/math] is called the
moment of inertia (kg.m 2).
Note that in physics, the moment of inertia [math]J[/math] is normally denoted as [math]I[/math]. In electrical engineering, the convention is for the letter "i" to always be reserved for current, and is therefore often replaced by the letter "j", e.g. the complex number operator i in mathematics is j in electrical engineering.
Normalised Inertia Constants
The moment of inertia can be expressed as a normalised quantity called the
inertia constant H, calculated as the ratio of the rotational kinetic energy of the machine at nominal speed to its rated power (VA): [math]H = \frac{1}{2} \frac{J \omega_0^{2}}{S_{b}}[/math]
where [math]\omega = 2 \pi \frac{n}{60}[/math] is the nominal mechanical angular frequency (rad/s)
[math]n[/math] is the nominal speed of the machine (revolutions per minute) [math]S_{b}[/math] is the rated power of the machine (VA) Generator Inertia
The moment of inertia for a generator is dependent on its mass and apparent radius, which in turn is largely driven by its prime mover type.
|
What’s in a name? that which we call a rose
By any other name would smell as sweet…
—William Shakespeare
In our daily lives, we are accustomed to giving multiple names to a single object. My wife calls me Xander, while my brothers and sisters call me Alex, and my students call me Mr. H. The titles, honorifics, and diminutives that people use to address me do not alter my fundamental self; they do not imply that multiple people inhabit my body, nor that there are multiple mes. Somehow, we are able to tolerate the ambiguity of calling the same person by multiple names without too much confusion.
Moreover, we seem to recognize that changing an object’s name does not change the underlying object. We seem to understand that people and objects have a kind of Platonic ideal associated with them, no matter what we call them. Romeo is Juliet’s lover, no matter his family name; and a rose has a sweet smelling blossom, no matter how we label it.
Even in mathematics, many labels are often used to describe a single object. Take, for example, the number one. We can right a Hindu-Arabic numeral to represent this number: 1. The Romans gave it a different name: I. If we had a very precise scale, we might use another symbol: 1.000. Or we could get the quantity by combining four quarters: \(\frac{4}{4}\). All of these symbols are different, by they equate to the same abstract mathematical concept: one.
So far, I have said nothing controversial. The various symbols and names for the number one that I listed above should be familiar to most people, and we should all be able to agree that they are equivalent. Now consider the following symbol: \(0.99\overline{9}\).
[1]
When confronted with this string of numerals, most mathematicians would instantly recognize it as yet another name for one. But, for some reason, many laypeople refuse to accept the equality \(0.99\overline{9}=1\).
As is customary when making a claim in mathematics, I will prove that these two symbols are, in fact, equal. In fact, I will provide not one, but three moderately rigorous proofs of this identity. Now, these proofs are going to be a bit mathy. There is some notation that you may have to mull over, and if you are not a terribly mathy person, then they may not be all that convincing. If this is the case, please simply try to keep in mind that many symbols can refer to the same thing. In short, a one by any other name…
Proof 1
Consider the quantity \(\frac{1}{3}\). When written as a decimal, we get \(\frac{1}{3} = 0.33\overline{3}\). Now consider multiplying both sides of this equation by 3. That is, consider that \(3\times\frac{1}{3} = 3\times 0.33\overline{3}\). Evaluating the left side of the equation, we have \(3\times\frac{1}{3} = \frac{3}{3} = 1\). Now consider the right side of the equation. Recall that when we multiply by 3, we are taking our original quantity, and adding it to itself three times. Using the notation that we learned in elementary school, it would look like this:
Clearly, in every decimal place, we would have \(3+3+3 = 9\). Hence \(3\times 0.33\overline{3} = 0.99\overline{9}\). Since we still have equality between the right and left sides of the equation, this gives us \(1 = 0.99\overline{9}\).
Proof 2
Let \(x = 0.99\overline{9}\). If we multiply both sides of this identity by 10, we will end up with the same string of digits on the right, but the decimal will move one place to the right. Thus \(10x = 9.99\overline{9}\). From each side of the equation, we will subtract \(x\): \(10x – x = 9.99\overline{9} – 0.99\overline{9}\). This gives us \(9x = 9\). Solving for \(x\), we have \(x = 1\). Substituting this value into the original identity gives us the desired result, which is \(1 = 0.99\overline{9}\).
Proof 3
This final proof is the most notationally intensive, but is also (in some ways) the most illuminating. It uses some properties of limits in order to get at the same identity. Note that \(0.9 = 1-\frac{1}{10}\). Similarly, \(0.99 = 1-\frac{1}{100}\), and so on. This means that we can rewrite \(0.99\overline{9}\) as
\(0.99\overline{9} = 1-\frac{1}{100\ldots} = 1-\frac{1}{10^\infty}.\)
While I think that the meaning of the last expression is clear, let me try to explain it. The idea is that the denominator of the fraction is a one followed by an infinite number of zeros. That being said, no mathematician would ever actually write something like $10^\infty$. Infinity (i.e. \(\infty\)) is not actually a number—it is really more of a concept. We cannot actually raise a number to the power of infinity. Normally, a mathematician would use limit notation to convey the idea. A mathematician would write
\(0.99\overline{9}=\lim_{n\to\infty}\left(1-\frac{1}{10^n}\right)\).
What this means is that, in some sense, we are going to let \(n\) get bigger and bigger and bigger. Eventually, \(n\) is going to be bigger than any number that we could possibly describe. Once it gets to be that big—that is, once we have let \(n\) grow to infinity—we are going to examine what happens.
First off, one of the nice things about limits is that we can break up sums and differences without changing the result. So, we end up with
\(\lim_{n\to\infty}\left(1-\frac{1}{10^n}\right) = \left(\lim_{n\to\infty} 1\right) – \left(\lim_{n\to\infty}\frac{1}{10^n}\right)\).
In the first term of the expression, there are no \(n\)s. This makes life easy. When \(n=1\), the first term will be 1. When \(n=100\), the first term will be 1. When \(n=10^{100}\), the first term will still be 1. No matter how big \(n\) gets to be, the first term will always be 1. So we can simplify the expression a little bit:
\(\left(\lim_{n\to\infty} 1\right) – \left(\lim_{n\to\infty}\frac{1}{10^n}\right) = 1-\left(\lim_{n\to\infty}\frac{1}{10^n}\right)\).
Now all we have to do is deal with the second term in the expression. Basically, as \(n\) gets larger and larger, the fraction with \(10^n\) in the denominator will get smaller and smaller. If \(n\) gets so large as to be infinite, the denominator will be infinite, and the fraction will simplify to zero. That is, \(\lim_{n\to\infty}\frac{1}{10^n} = 0\). This gives us the final identity
\(1-\left(\lim_{n\to\infty}\frac{1}{10^n}\right) = 1-0 = 1\).
Therefore \(0.99\overline{9} = 1\).
Notes: [1] The bar over the trailing 9 indicates that we keep writing 9s forever. Thus the symbol is a zero, followed by a decimal point, followed by an infinite number of 9s. The symbol is read “zero point nine nine nine bar.” ↩
|
While this reaction is highly regioselective, it is not 100% regioselective. Consider the nearly identical reactions:
$$\begin{align}\ce{CCl4 + H2C=CH-(CH2)5CH3 ->[(PhCO2)2][90\text{-}105ºC] Cl3C-CH2-CHCl-(CH2)5CH3} \hspace{.63cm} \text{75%}^{[1]}\\ \ce{CCl4 + H2C=CH-COH-(CH3)2 ->[(PhCO2)2][80ºC] Cl3C-CH2-CHCl-CH(OH)-(CH2)5CH3} \hspace{.63cm} \text{70%}^{[2]}\\\end{align}$$
Granted these reactions use radical initiators instead of light, but that wouldn't make the difference between 100% and 70-75% regioselectivity. The reality is that at the start of the reaction, $\ce{.CCl3}$ and $\ce{.Cl}$
are both competing to attack the alkene, and $\ce{.Cl}$ does in fact attack faster. Looking at the activation energies for the abstraction of hydrogen atoms by different radicals, we gain insight into their relative stability:
$$ \small\begin{array}{lcc}\hline\text{Radical} & \ce{CH3-H} & \ce{CH3CH2-H} & \ce{(CH3)2CH-H} & \ce{(CH3)C-H}^{[3]} \\\hline\ce{.F} & \text{1-1.5} & \text{<1} & & \text{<1} \\\ce{.Cl} & \text{3.4} & \text{1.1} \\\ce{.Br} & \text{17.5} & \text{13.0} & \text{9.5} & \text{6.9} \\\ce{.CH3} & \text{14.0} & \text{11.6} & \text{9.6} & \text{8.1} \\\ce{.CF3} & \text{10.9} & \text{8.0} & \text{6.5} & \text{4.9} \\\ce{.CCl3} & \text{17.9} & \text{14.2} & \text{10.6} & \text{7.7*} \\\hline_{^* \text{units are } \mathrm{kcal\ mol^{-1}}}\end{array}$$
As this reaction continues, however, chlorine atoms are abstracted from $\ce{CCl4}$, creating the product along with $\ce{.CCl3}$ radicals. Abstraction of $\ce{.CCl3}$ is highly disfavored due to steric effects. Thus, the radical chain reaction is as follows:
$$\ce{CCl4 ->[$h\nu$] .CCl3 + Cl.}$$
$$\ce{.CCl3 + CH2=CH-R -> Cl3C-CH2-\stackrel{\displaystyle .}{\ce{C}}H-R}$$
$$\ce{Cl3C-CH2-\stackrel{\displaystyle .}{\ce{C}}H-R + CCl4 -> Cl3C-CH2-CHCl-R + .CCl3}$$
The less than ideal yields of the radical addition of $\ce{CCl4}$ can be attributed to the strength of the $\ce{C-Cl}$ bond (higher activation energies correspond to slower rates of reactions). Because $\ce{CCl4}$ is in competition with the polymerization of the $\ce{.CCl3}$-alkene adduct, $\ce{BrCCl3}$ and $\ce{CBr4}$ are generally preferred for their higher selectivity, owed to the lower strength of the $\ce{C-Br}$ bond. One such example is shown below.
$$\ce{BrCCl3 + H2C=C(C2H5)2 ->[$h \nu$] Cl3C-CH2-CBr(C2H5)2} \hspace{1cm} {\text{91%}^{[4]}}$$
$^{[1]}$M. S. Kharasch, E. W. Jensen, and W. H. Urry, J. Am. Chem. Soc., 69, 1100 (1947). $^{[2]}$P. D. Klemmensen, H. Kolind-Andersen, H. B. Madsen, and A. Svendsen, J. Org. Chem., 44, 416 (1979). $^{[3]}$F. A. Carey, R. J. Sundberg, Advanced Organic Chemistry Part A: Structure and Mechanisms, Springer Science, New York, 2007, pg. 1034. $^{[4]}$M. S. Kharasch and M. Sage, J. Org. Chem., 14, 537 (1949).
|
I am new to differential geometry and I am trying to understand Gaussian curvature. The definitions found at Wikipedia and Wolfram sites are too mathematical. Is there any intuitive way to understand Gaussian curvature?
For a
intuitive understanding, imagine a flat sheet of paper (or just grab one in your hand). It has zero Gaussian curvature. If you take that sheet and bend it or roll it up into a tube or twist it into a cone, its Gaussian curvature stays zero.
Indeed, since paper isn't particularly elastic, pretty much anything you can do to the sheet that still lets you flatten it back into a flat sheet without wrinkles or tears will preserve its Gaussian curvature.
Now take that sheet and wrap it over a sphere. You'll notice that you have to wrinkle the sheet, especially around the edges, to make it conform to the sphere's surface. That's because a sphere has positive Gaussian curvature, and so the circumference of a circle drawn on a sphere is less than $\pi$ times its diameter. The wrinkles on the paper are where you have to fold it to get rid of that excess circumference.
Similarly, if you tried to wrap the sheet of paper over a saddle-shaped surface, you'd find that you would have to tear it (or crumple it in the middle) to make it lie on the surface. That's because, on a surface with negative Gaussian curvature, the circumference of a circle is longer than $\pi$ times its diameter, and so, to make a flat sheet lie along such a surface, you either have to tear it to increase the circumference, or wrinkle it in the middle to reduce the radius.
Indeed, in nature, plants can produce curved or wrinkled leaves simply by altering the rate at which the edges of the leaf grow as compared to the center, which alters the Gaussian curvature of the resulting surface, as in this picture of ornamental kale:
I know you're looking for an intuitive explanation, but I've always believed that intuition ought to come from concrete mathematical facts, if possible. Otherwise, you have no way of knowing whether or not the intuition someone feeds you actually matches the formal mathematics. (On that note, +1 for Joseph O'Rourke's answer.)
The Gaussian curvature, $K$, is given by $$K = \kappa_1 \kappa_2,$$ where $\kappa_1$ and $\kappa_2$ are the principal curvatures. Just from this definition, we know a few things:
For $K$ to be a large positive number, then $\kappa_1$ and $\kappa_2$ should both be large and have the same sign (i.e. both positive or both negative).
For $K$ to be zero, either $\kappa_1 = 0$ or $\kappa_2 = 0$.
For $K$ to be a large negative number, then $\kappa_1$ and $\kappa_2$ should both be large but have
oppositesigns.
Now recall that $\kappa_1(p)$ and $\kappa_2(p)$ are the maximum and minimum normal curvatures of all curves passing through $p.$ So:
$K(p) > 0$ means that the curves through $p$ of extremal normal curvature "curve the same way" (such as the red curve and the green curve). So, points in the purple region have $K > 0$. In some sense, the surface is shaped like an elliptic paraboloid there (like a bowl).
$K(p) = 0$ means that one of the curves through $p$ of extremal curvature has zero normal curvature (such as the yellow curve). So, points along the yellow curve have $K = 0$. In some sense, the surface is shaped like a parabolic cylinder there (like a bent piece of paper).
$K(p) < 0$ means that the curves through $p$ of extremal curvature "curve in opposite ways" (such as the blue curve and green curve). For example, points in the gray region have $K < 0$. In some sense, the surface is shaped like a hyperbolic paraboloid there (a saddle).
In fact, using Dupin's Indicatrix (which is really just a 2nd-order Taylor expansion) we can make rigorous the notion of being "locally like" an elliptic paraboloid, a cylinder, or a hyperbolic paraboloid.
One way to view the Gaussian curvature $K$ is as an area deficit, a comparison between the area $\pi r^2$ of a flat disk of radius $r$, to the area of a geodesic disk on the surface with intrinsic radius $r$. Let $A(r)$ be this latter area centered on a point of the surface. Then $$K = \lim_{r \rightarrow 0} \frac {12 (\pi r^2 - A(r) )}{\pi r^4}$$ You can see in the numerator the term $\pi r^2 - A(r)$ is exactly the area deficit. So at a point where $A(r)$ is smaller than the flat area, $K$ is positive.
This formulation was discovered by Diquet. There is a similar formulation based on a circumference deficit, due to Bertrand and Puiseux.
The Gaussian curvature is the ratio of the solid angle subtended by the normal projection of a small patch divided by the area of that patch.
The fact that this ratio is based totally on the definition of distance within the surface (independent of the embedding of the surface; that is, bending and twisting, etc.) is Gauss' Theorema Egregium.
|
$\dfrac{d}{dx}{\, (\cos{x})} \,=\, -\sin{x}$
The differentiation or derivative of cos function with respect to a variable is equal to negative sine. This formula is read as the derivative of $\cos{x}$ with respect to $x$ is equal to negative $\sin{x}$.
If $x$ is used to represent a variable, then the cosine function is written as $\cos{x}$ in mathematics. The derivative of the cos function with respect to $x$ is written in mathematical form as follows.
$\dfrac{d}{dx}{\, (\cos{x})}$
Mathematically, the differentiation of the $\cos{x}$ function with respect to $x$ is also written as $\dfrac{d{\,(\cos{x})}}{dx}$ and also written as ${(\cos{x})}’$ in simple form.
The derivative of the cos function can be written in terms of any variable.
$(1) \,\,\,$ $\dfrac{d}{dr}{\, (\cos{r})} \,=\, -\sin{r}$
$(2) \,\,\,$ $\dfrac{d}{dz}{\, (\cos{z})} \,=\, -\sin{z}$
Learn how to derive the derivative of the cosine function from first principle in differential calculus.
Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.
|
I'm trying to calculate the properties of the combustion process using propane and nitrous oxide. When I tried to nail down the combustion temperature, the result looks just off to me. I went through several times with the process but since the reaction should produce less enthalpy change than using pure oxygen gas and my results is above the standard combustion temperature with pure oxygen, it is definitely wrong somewhere.
The process, how I got my solution is the following:
$$C_3H_8 + 10 N_2O => 3 CO_2 + 4 H_2O + 10 N_2 + \Delta_rH$$
$$\Delta_rH = \Delta_fH_p - \Delta_fH_r$$
Where the r subscript means reactants and p is for products.
$$\Delta_fH(C_3H_8) = -104.7 kJ/mol$$ $$\Delta_fH(CO_2) = -393.5 kJ/mol$$ $$\Delta_fH(H_2O) = -241.8 kJ/mol$$ $$\Delta_fH(N_2O) = +82.05 kJ/mol$$ $$\Delta_fH(N_2) = 0 kJ/mol$$
Which amounts to the following conclusion:
$$\Delta_rH = (3 * -393.5 kJ + 4 * -241.8 kJ) - (1 * -104.7 kJ + 10 * 82.05 kJ) = -2863.5 kJ$$ $$\Delta_rH = -2863.5 kJ / 0.484 kg = -5916.3223 kJ/kg $$
This value is for 10 mols of Nitrous Oxide and 1 mol Propane. Since both molecules have the molar mass around 44 g / mol, it's essentially the stoichometric ratio of 1 to the 10.
So far, so good (I think). The result seems reasonable to me, as I went through the same process using only O
2, which ended up considerable more as expected (it only takes 5 part of O 2 so the reaction produces more enthalpy with the same mass, plus O 2 molar weight is 32.)
In the next step I used the following formula:
$$\Delta_rH = \sum_{i=1}^m n_i \int_{T_0}^{T_r}C_pdT $$
I assumed here that the:$$\int_{T_0}^{T_r}C_pdT$$part can be simplified with the assumption that the C
p won't change with the temperature. I'm well aware that this isn't the exact case, and I've found that for example water vapour has a particular big swing over the temperature in terms of heat capacity on a constant pressure. However, I tried to work with the values I could find around the internet, and in the case of water vapour I worked with the temperature that looked like a quite good mean value. So I continued like this:
$$\int_{T_0}^{T_r}C_pdT = \Delta T * \sum_{i=1}^m n_iC_p$$ $$T * \sum_{i=1}^m n_iC_p = \Delta_rH$$ $$T = \frac{\Delta_rH} {\sum_{i=1}^m n_iC_p}$$ $$T = 2863500 J / (3 * 37.135 J/K + 4 * 53.1 J/K + 10 * 29.12 J/K) = 4656.06 K$$
This value is obviously not good, given that the pure oxygen reaction combustion temperature with propane is 3093 K Ref. Now, is there some deeply flawed assumption in my process? Can it be that the simplification of that integral expression led me down on a wrong calculation in the first place? If so, how can I access to the exact functions of these heat capacities for the gases I'm using?
|
Quadratic programming (QP) involves minimizing or maximizing an objective function subject to bounds, linear equality, and inequality constraints. Example problems include portfolio optimization in finance, power generation optimization for electrical utilities, and design optimization in engineering.
Quadratic programming is the mathematical problem of finding a vector \(x\) that minimizes a quadratic function:
\[\min_{x} \left\{\frac{1}{2}x^{\mathsf{T}}Hx + f^{\mathsf{T}}x\right\}\]
Subject to the constraints:
\[\begin{eqnarray}Ax \leq b & \quad & \text{(inequality constraint)} \\A_{eq}x = b_{eq} & \quad & \text{(equality constraint)} \\lb \leq x \leq ub & \quad & \text{(bound constraint)}\end{eqnarray}\]
The following algorithms are commonly used to solve quadratic programming problems:
Interior-point-convex:solves convex problems with any combination of constraints Trust-region-reflective:solves bound constrained or linear equality constrained problems
For more information about quadratic programming, see Optimization Toolbox™.
|
Shinichi Mochizuki of Kyoto divided the steps needed to prove the 1985 conjecture by Oesterlé and Masser into four papers listed at the bottom of the Nature article above.
Up to a few exceptions to be proved separately, a strengthening of Fermat's Last Theorem Four days ago, Nature described a potentially exciting development in mathematics, namely number theory: The newly revealed proof works with mathematical structures such as the Hodge theaters (a theater with Patricia Hodge is above, I hope it's close enough) and with canonical splittings of the log-theta lattice (yes, the word "splitting" is appropriate above, too).
What is the conjecture about and why it's important, perhaps more important than Fermat's Last Theorem itself?
First, before I tell you what it is about, let me say that, as shown by Goldfeld in 1996, it is "almost stronger" than Fermat's Last Theorem (FLT) i.e. it "almost implies" FLT. What does "almost" mean? It means that it only implies a weakened FLT in which the exponent has to be larger than a certain large finite number.
I am not sure whether all the exponents for which the \(abc\) theorem doesn't imply FLT have been proved before Wiles or whether the required Goldfeld bound is much higher. Please tell me if you know the answer: what's the minimum Goldfeld exponent?
Off-topic: When Ben Bernanke was a kid... Via AA
Recall that Wiles proved Fermat's Last Theorem in 1996 and his complicated proof is based on elliptic curves. That's also true for Mochizuki's new (hopefully correct) proof. However, Mochizuki also uses Teichmüller theory, Hodge-Arejekekiv theory, log-volume computations and log-theta lattices, and various sophisticated algebraic structures generalizing simple sets, permutations, topologies and matrices. To give an example, one of these object is the "Hodge theater" which sounds pretty complicated and cultural. ;-)
I am not gonna verify the proof although I hope that some readers will try to do it. But let me just tell everybody what the FLT theorem and the \(abc\) theorem are.
Fermat's Last Theorem says that if positive integers \(a,b,c,n\) obey\[
a^n+b^n = c^n,
\] then it must be that \(n\leq 2\). Indeed, you can find solutions with powers \(1,2\) such as \(2+3=5\) and \(3^2+4^2=5^2\) but you will fail for all higher powers. Famous mathematicians have been trying to prove the theorem for centuries but at most, they were able to prove it for individual exponents \(n\), not the no-go theorem for all values of \(n\).
The \(abc\) conjecture says the following.
For any (arbitrarily small) \(\epsilon\gt 0\), there exists a (large enough but fixed) constant \(C_\epsilon\) such that each triplet of relatively prime (i.e. having no common divisors) integers \(a,b,c\) that satisfies \[
a+b=c
\] the following inequality still holds:\[
\Large \max (\abs a, \abs b, \abs c) \leq C_\epsilon \prod_{p|(abc)} p^{1+\epsilon}.
\] That's it. I used larger fonts because it's a key inequality of this blog entry.
In other words, we're trying to compare the maximum of the three numbers \(\abs a,\abs b,\abs c\) with the "square-free part" of their product (product in which we eliminate all copies of primes that appear more than once in the decomposition). The comparison is such that if the "square-free part" is increased by exponentiating it to a power \(1+\epsilon\), slightly greater than one but arbitrarily close, the "square-free part" will be typically be smaller than \(abc\) and even \(a,b,c\) themselves but the ratio how much it is smaller than either \(a\) or \(b\) or \(c\) will never exceed a bound, \(C_\epsilon\), that may be chosen to depend on \(\epsilon\) but it isn't allowed to depend on \(a,b,c\).
See e.g. Wolfram Mathworld for a longer introduction.
Now, you may have been motivated to jump to the hardcore maths and verify all the implications and proofs that have been mentioned above or construct your own.
|
1. Definition & Example
A
permutationof size$n$ is a bijection of $\{1,\ldots,n\}$. $\mathfrak{S}_n$ denotes the set of all such permutations.
There are two standard ways to denote $\pi \in \mathfrak{S}_n$.
The
one-line notationis given by $\pi = [\pi(1),\ldots,\pi(n)]$. E.g., $\pi = [5,4,2,3,1]$ says that
$$\pi(1)=5,\pi(2)=4,\pi(3)=2,\pi(4)=3,\pi(5)=1.$$
The
cycle notationis given by $\pi = (a^{(1)}_1, \ldots, a^{(1)}_{k_1}) \cdots (a^{(m)}_1, \ldots, a^{(m)}_{k_m})$ which means that every integer in a cycle is mapped to the next integer in this cycle, $\pi(a^{(j)}_i) = a^{(j)}_{i+1}$. E.g., $$\pi = (1,5)(2,4,3) = [5,4,2,3,1].$$ Often, fixed points are suppressed in the cycle notation.
The six permutations of size 3
There are $n! = 1\cdot 2 \cdot 3 \cdots n$ permutations of size $n$, see A000142.
2. Additional information
2.1. The Rothe diagram of a permutation
A way to visualize a permutation is by drawing its
Rothe diagramintroduced by H.A. Rothe in 1800. It is given by the collection of boxes $$R(\sigma) = \{ (\sigma_j,i) : i < j, \sigma_i > \sigma_j \}.$$ Alternatively, $R(\sigma)$ is by marking all boxes $b_i = (i,\sigma_i)$ and cross out all boxes below and to the right of $b_i$.
The Rothe diagram of $\sigma = [4,3,1,5,2]$
The
essential setof the Rothe diagram is the collection of boxes $(i,j)$ for which $(i+1,j)$ and $(i,j+1)$ are both not in the Rothe diagram. The Rothe diagram can also be used to understand the Bruhat orderon permutations, see [Man01]. It is also closely related to the Lehmer code for a permutation, described below.
2.2. Euler-Mahonian statistics
L. Carlitz proved in [Car54] that the generating function for the bistatistic $(\operatorname{des},\operatorname{maj})$ satisfies a nice recurrence: let $$B_n(q,t) = \sum_{\sigma \in \mathbf{S}_n} t^{\operatorname{des}(\sigma)} q^{\operatorname{maj}(\sigma)} = \sum_{k=0}^n B_{n,k}(q) t^k.$$ Then the coefficient $B_{n,k}(q)$ satisfy the recurrence $$B_{n,k}(q) = [k+1]_q B_{n-1,k}(q) + q^k[n-k]_q B_{n-1,k-1}(q), \qquad B_{0,k}(q) = \delta_{0,k}.$$
A permutation statistic is called
Eulerianif it is equidistributed with the number of descents (St000021), Mahonianif it is equidistributed with the major index (St000004), and Euler-Mahonianif it is equidistributed with the bistatistic $(\operatorname{des},\operatorname{maj})$, see e.g. [FZ94].
Two additional examples of Euler-Mahonian statistics are
2.3. The Lehmer code and the major code of a permutation
The
Lehmer codeencodes the inversions (St000018) of a permutation. It is given by $L(\sigma) = l_1 \ldots l_n$ with $l_i = \# \{ j > i : \sigma_j < \sigma_i \}$. In particular, $l_i$ is the number of boxes in the $i$-th column of the Rothe diagram. The Lehmer code of $\sigma = [4,3,1,5,2]$ is for example given by $32010$. The Lehmer code $L : \mathfrak{S}_n\ \tilde\longrightarrow\ S_n$ is a bijection between permutations of size $n$ and sequences $l_1\ldots l_n \in \mathbf{N}^n$ with $l_i \leq i$.
The
major code$M(\sigma)$ of a permutation $\sigma \in \mathfrak{S}_n$ is, similarly to the Lehmer code, a way to encode a permutation as a sequence $m_1 m_2 \ldots m_n$ with $m_i \geq i$. To define $m_i$, let $\operatorname{del}_i(\sigma)$ be the normalized permutation obtained by removing all $\sigma_j < i$ from the one-line notation of $\sigma$. The $i$-th index is then given by
$$m_i = \operatorname{maj}(\operatorname{del}_i(\sigma)) - \operatorname{maj}(\operatorname{del}_{i-1}(\sigma)).$$ For example, the permutation $[9,3,5,7,2,1,4,6,8]$ has major code $[5, 0, 1, 0, 1, 2, 0, 1, 0]$ since $$\operatorname{maj}([8,2,4,6,1,3,5,7]) = 5, \quad \operatorname{maj}([7,1,3,5,2,4,6]) = 5, \quad \operatorname{maj}([6,2,4,1,3,5]) = 4,$$ $$\operatorname{maj}([5,1,3,2,4]) = 4, \quad \operatorname{maj}([4,2,1,3]) = 3, \quad \operatorname{maj}([3,1,2]) = 1, \quad \operatorname{maj}([2,1]) = 1.$$ Observe that the sum of the major code of $\sigma$ equals the major index of $\sigma$.
2.4. Special classes of permutations
2.4.1. Pattern-avoiding permutations
A permutation $\sigma$
avoidsa permutation $\tau$ if no subword of $\sigma$ in one-line notation appears in the same relative order as $\tau$. E.g. $[4,2,1,3]$ is not $[3,1,2]$-avoiding since the subword $[4,1,3]$ has the same relative order as $[3,1,2]$. See also the wiki page on permutation pattern and the references therein. Each subword of $\sigma$ having the same relative order as $\tau$ is called occurrenceof $\tau$.
$$ \#\big\{ \sigma \in \mathfrak{S}_n : \sigma \text{ avoids } \tau \big\} = \frac{1}{n+1}\binom{2n}{n}, $$
where $\tau$ is any permutation in $\mathfrak{S}_3$.
A permutation $\sigma$
avoidsa pattern $ab\!\!-\!\!c$ for $\{a,b,c\} = \{1,2,3\}$ if it avoids the permutation $[a,b,c]$ with the additional property that the positions of $a$ and $b$ in $\sigma$ must be consecutive. In other words, a pattern $ab\!\!-\!\!c$ in $\sigma$ is a subword $\sigma_i,\sigma_i+1,\sigma_j$ with $\sigma_i > \sigma_j > \sigma_{i+1}$. In particular, usual three-pattern-avoidance is of the form $a\!\!-\!\!b\!\!-\!\!c$.
Mesh patterns, introduced by [BrCl11], constitute a further generalization. A mesh pattern is a pair $(\tau, R)$, where $\tau\in\mathfrak{S}_k$ and $R$ is a subset of $\{0,\dots,k\}\times\{0,\dots,k\}$. An occurrence of $(\tau, R)$ in a permutation $\pi$ is an occurrence $(t_1,\dots,t_k)$ of $\tau$ in $\pi$, satisfying the following additional requirement: for all $(i,j)\in R$, the permutation $\sigma$ has no values between $t_j$ and $t_{j+1}$ in the positions between the positions of $t_i$ and $t_{i+1}$,
where we set $t_0=0$ and $t_k=n+1$.
Thus, classical patterns are mesh patterns with empty $R$ and vincular patterns are mesh patterns with a shaded column.
2.5. Classes of permutations in Schubert calculus dominantif the following equivalent conditions hold:
its Lehmer code is a partition, this is if $l_1 \geq l_2 \geq \ldots \geq l_n$,
its Rothe diagram is the shape of a partition. In particular, $\sigma$ is dominant if and only if $\sigma^{-1}$ is dominant.
it is $[1,3,2]$ avoiding.
The Schubert polynomial $S_\sigma(x,y)$ for a dominant permutation $\sigma$ is given by $S_\sigma(x,y) = \prod_{(i,j) \in \lambda(\sigma)} (x_i - y_j),$ where $\lambda(\sigma)$ is the shape of the Rothe diagram of $\sigma$ [Man01, Proposition 2.6.7].
Grassmannianif the following equivalent conditions hold:
$l_1 \leq l_2 \leq \ldots \leq l_r$ and $l_i = 0$ for for some $r$ and all $i>r$,
$\sigma$ has a unique descent (St000021).
The Schubert polynomial $S_\sigma(x)$ for a Grassmannian permutation $\sigma$ is given by $S_\sigma(x) = s_{\lambda(\sigma)}(x_1,\ldots,x_n)$, where $s_\lambda$ is the Schur function for the shape $\lambda$ [Man01, Proposition 2.6.8].
bigrassmannianif $\sigma$ and $\sigma^{-1}$ are both Grassmannian.
Bigrassmannian permutations can be used to describe the Bruhat order on permutations: $\sigma \leq \tau$ in Bruhat order if and only if every bigrassmannian permutation dominated by $\sigma$ is also dominated by $\tau$ [LS96].
vexillaryif the following equivalent conditions hold: its Rothe diagram is, up to a permutation of its rows and columns, the diagram of a partition,
there is no sequence $i < j < k < l$ with $\sigma_j < \sigma_i < \sigma_l < \sigma_k$,
its essential set lies on a piecewise linear curve oriented $SW-NE$.
The Schubert polynomials for vexillary permutations are exactly the
flagged Schur functions, see [Man01, Section 2.6.5] for a detailed description. The term vexillarycomes from the Latin vexillum for a flag-like object used in the Classical Era of the Roman Empire [Man01, p. 65] and is due to A. Lascoux and M.-P. Schützenberger.
3. Properties
The
cycle typeof a permutation $\pi \in \mathfrak{S}_n$ is given by the integer partition of $n$ formed by the lengths of the cycles of $\pi$. E.g., $\pi = (1,4,3,7)(2)(5,6)$ has cycle type $[4,2,1]$. Two permutations $\pi$ and $\tau$ in $\mathfrak{S}_n$ are conjugated(i.e., $\pi = \sigma \tau \sigma^{-1}$ for some $\sigma \in \mathfrak{S}_n$) if and only if their cycle types coincide.
4. Remarks
The symmetric group admits a beautiful representation theory.
5. References [BrCl11] P. Brändén and A. Claesson, Electron. J. Combin, 2011.
6. Sage examples
7. Technical information for database usage Permutations are graded by size. The database contains all permutations of size at most 7.
|
This example shows how to add text to a chart, control the text position and size, and create multiline text.
Add text next to a particular data point using the
text function. In this case, add text to the point . The first two input arguments to the
text function specify the position. The third argument specifies the text.
By default, text supports a subset of TeX markup. Use the TeX markup
\pi for the Greek letter . Display an arrow pointing to the left by including the TeX markup
\leftarrow. For a full list of markup, see Greek Letters and Special Characters in Chart Text.
x = linspace(0,10,50);y = sin(x);plot(x,y)txt = '\leftarrow sin(\pi) = 0';text(pi,sin(pi),txt)
By default, the specified data point is to the left of the text. Align the data point to the right of the text by specifying the
HorizontalAlignment property as
'right'. Use an arrow pointing to the right instead of to the left.
x = linspace(0,10,50); y = sin(x); plot(x,y) txt = 'sin(\pi) = 0 \rightarrow'; text(pi,sin(pi),txt,'HorizontalAlignment','right')
Specify the font size for text by setting the
FontSize property as a name-value pair argument to the
text function. You can use a similar approach to change the font size when using the
title,
xlabel,
ylabel, or
legend functions.
x = linspace(0,10,50); y = sin(x); plot(x,y) txt = '\leftarrow sin(\pi) = 0'; text(pi,sin(pi),txt,'FontSize',14)
The text function creates a Text object.
Text objects have properties that you can use to customize the appearance of the text, such as the
HorizontalAlignment or
FontSize.
You can set properties in two ways:
Use name-value pairs in the
text command, such as
'FontSize',14.
Use the
Text object. You can return the
Text object as an output argument from the
text function and assign it to a variable, such as
t. Then, use dot notation to set properties, such as
t.FontSize = 14.
For this example, change the font size using dot notation instead of a name-value pair.
x = linspace(0,10,50);y = sin(x);plot(x,y)txt = '\leftarrow sin(\pi) = 0';t = text(pi,sin(pi),txt)
t = Text (\leftarrow sin(\pi) = 0) with properties: String: '\leftarrow sin(\pi) = 0' FontSize: 10 FontWeight: 'normal' FontName: 'Helvetica' Color: [0 0 0] HorizontalAlignment: 'left' Position: [3.1416 1.2246e-16 0] Units: 'data' Show all properties t.FontSize = 14;
Display text across multiple lines using a cell array of character vectors. Each element of the cell array is one line of text. For this example, display a title with two lines. You can use a similar approach to display multiline text with the
title,
xlabel,
ylabel, or
legend functions.
x = linspace(0,10,50); y = sin(x); plot(x,y) txt = {'Plotted Data:','y = sin(x)'}; text(4,0.5,txt)
Include a variable value in text by using the
num2str function to convert the number to text. For this example, calculate the average
y value and include the value in the title. You can use a similar approach to include variable values with the
title,
xlabel,
ylabel, or
legend functions.
x = linspace(0,10,50); y = sin(x); plot(x,y) avg = mean(y); txt = ['Average height: ' num2str(avg) ' units']; text(4,0.5,txt)
Add text anywhere within the figure using the
annotation function instead of the
text function. The first input argument specifies the type of annotation. The second input argument specifies the position of the annotation in units normalized to the figure. Remove the text box border by setting the
EdgeColor property to
'none'. For more information on text box annotations, see the
annotation function.
x = linspace(0,10,50); y = sin(x); plot(x,y) annotation('textbox',[.9 .5 .1 .2],'String','Text outside the axes','EdgeColor','none')
|
Just skimmed over the links:
Highview:
Using ambient heat to warm it, the process
recovers around 50 per cent of the electricity that is fed in, says Highview's chief executive Gareth Brett. The efficiency rises to around 70 per cent if you harness waste heat from a nearby industrial or power plant to heat the cryogen to a higher than ambient temperature, which increases the turbine's force, he says.
Batteries under development .. have efficiencies of around 80 to 90 per cent, but cost around \$4000 per kilowatt of generating capacity. Cryogenic storage would cost just \$1000 per kilowatt because it requires fewer expensive materials, claims Brett.
Here's some evidence that people who know about gas and liquid air are interested in the technology: strategic partnership between Messer and Highview
Actually Highview has a presentation on their web-site that contains a lot of information. Note that they propose to run a 4-stage process against heat (ca. 100 °C), not ambient temp.
Here are some values for energy storage efficiency comparison found after a very quick search:
Batteries (Wiki) German Wikipedia has lower efficiency values Energy conversion table (Wikipedia)
Pump-stored hydroelectricity: 70/75 - 80% (Wiki + de.Wiki)
(Note the difference between technology that is actually in use, so efficiencies that are achieved in practical use are known and prototypes/scientific studies about achievable efficiency.)
Dearman claims 70% efficiency for the engine. Unfortunately they don't say % of what (remember the low-temp heater manufacturers claiming > 100% efficiency because they used achievable efficiency with high temp exhaust gas as their denominator?). Neither do they say something about the power at which they have this efficiency (ideal thermodynamic efficiency is at P $\rightarrow$ 0 ($t_{cycle} \rightarrow \infty$)) Neither can we assume that they start their calculation taking into account the efficiency of producing liquid nitrogen. Nor storage loss.
=> so far, Highview's stationary plant concept looks more sound.
Let's have a look what thermodynamics says:
ideal efficiency
boiling T of liquid air: 77 K ambient T: assume 290 K => $\eta_{ideal} = 1 - \frac{77}{290}$ = 73% waste heat: assume 370 K => $\eta_{ideal} = 1 - \frac{77}{370}$ = 79%
(in other words, Dearman claims close-to-ideal efficiency)
energy density of liquid $\ce{N2}$: 620 kJ/l = 712 kJ/kg (from presentation at Highview)
The Linde process is said to need approx. 500 kWh / t = 1750 kJ/kg of air (that would be ≈40% efficiency; no real source, just something someone answered in a German-language chemistry forum) Here's short report that gives ≈20% efficiency for producing liquid air in Egypt. They also give storage loss (ca. 0.17 \%/d)
Thus, the 50% (achieved by Highview) to 70/80% (short term goal Highview against waste heat, quote for Dearman's engine) mean that
either the liquefaction was not part of the efficiency calculation, or they have a super efficient air liquefaction that is about 2 - 4 times as efficient as the two numbers I found. (Anyone around here who knows? Linde, Messer, AL?)
Update: just had a closer look at the Highview presentation. Their 50 - 80% are cycle efficiency based on "Available enthalpy for cold cycle ~300kJ/kg", so less than half of the energy density they give. Calculating with 1750 kJ/kg to produce the kg liquid air, we're at 17% (close to the "Egyptian" 20% efficiency for producing liquid air), of which they recover 0.5 - 0.8.
Conclusion: electric -> liquid air -> electric cycle with an overall efficiency in the order of magnitude of 10%
|
Prove that $\operatorname{trace}(ABC) = \operatorname{trace}(BCA) = \operatorname{trace}(CAB)$ if $A,B,C$ matrices have the same size.
closed as off-topic by user26857, mrp, John B, Daniel W. Farlow, Nick Peterson Mar 3 '17 at 23:25
This question appears to be off-topic. The users who voted to close gave this specific reason:
" This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, mrp, John B, Daniel W. Farlow, Nick Peterson
Is it already known that $\operatorname{Tr}(XY) = \operatorname{Tr}(YX)$ when $X$ and $Y$ are square matrices of the same size?
If it is, then simply set $X= AB$ and $Y = C$. It will give you $\operatorname{Tr}(ABC) = \operatorname{Tr}(CAB)$. You can get $\operatorname{Tr}(ABC) = \operatorname{Tr}(BCA)$ in a similar fashion.
Hint
$$tr(ABC)=\sum_i (ABC)_{ii}=(ABC=A(BC))=\sum_i\sum_j A_{ij}(BC)_{ji}= \sum_i\sum_j\sum_k A_{ij}B_{jk}C_{ki};$$
now you can exchange the order of the matrices to arrive at the thesis as each of the $A_{ij}$, $B_{jk}$ and $C_{ki}$ are scalars (considering matrices over $\mathbb R$, for example). We arrive at
$$tr(ABC)=\sum_i\sum_j\sum_k A_{ij}B_{jk}C_{ki}=\sum_i\sum_j\sum_kB_{jk}C_{ki}A_{ij}=(BCA=(BC)A)=tr(BCA), $$
and so on.
Hint: use the definition of trace.
$$\text{Tr}(ABC)=\sum_i\sum_j\sum_k A_{ij}B_{jk}C_{ki}.$$
I will help you by proving that $\text{tr}(AB) = \text{tr}(BA)$ for a $m \times n$ matrix $A$ and a $n \times m$ matrix $B$ first. Let $C = AB$ and $D = BA$. Then using the definition of matrix multiplication we find that $c_{ij} = \sum_{k=1}^{n} a_{ik}b_{kj}$ and $d_{ij} = \sum_{m=1}^{n} b_{im}a_{mj}$. Then,
\begin{align*} \text{tr}(C) & = \sum_{m = 1}^n c_{mm} \\ & = \sum_{m=1}^n\sum_{k=1}^na_{mk}b_{km} \\ & = \sum_{k=1}^n\sum_{m=1}^nb_{km}a_{mk} \\ & = \sum_{k=1}^nd_{kk} \\ & = \text{tr}(D). \end{align*} Now, for $ABC$ take $AB = A_1$ and $C = A_2$ and apply the idenity proven above.
|
Difference between revisions of "Inertia"
(→Normalised Inertia Constants)
(→Normalised Inertia Constants)
Line 50: Line 50:
: <math>H = \frac{1}{2} \frac{J \omega_0^{2}}{S_{b}}</math>
: <math>H = \frac{1}{2} \frac{J \omega_0^{2}}{S_{b}}</math>
−
where <math>\omega_{0} = 2 \pi \times \frac{n}{60}</math> is the nominal mechanical angular frequency (rad/s)
+
where
+
<math>\omega_{0} = 2 \pi \times \frac{n}{60}</math> is the nominal mechanical angular frequency (rad/s)
:: <math>n</math> is the nominal speed of the machine (revolutions per minute)
:: <math>n</math> is the nominal speed of the machine (revolutions per minute)
:: <math>S_{b}</math> is the rated power of the machine (VA)
:: <math>S_{b}</math> is the rated power of the machine (VA)
Revision as of 05:54, 27 August 2018
In power systems engineering, "inertia" is a concept that typically refers to rotational inertia or rotational kinetic energy. For synchronous systems that run at some nominal frequency (i.e. 50Hz or 60Hz), inertia is the energy that is stored in the rotating masses of equipment electro-mechanically coupled to the system, e.g. generator rotors, fly wheels, turbine shafts.
Derivation
Below is a basic derivation of power system rotational inertia from first principles, starting from the basics of circle geometry and ending at the definition of moment of inertia (and it's relationship to kinetic energy).
The length of a circle arc is given by:
[math] L = \theta r [/math]
where [math]L[/math] is the length of the arc (m)
[math]\theta[/math] is the angle of the arc (radians) [math]r[/math] is the radius of the circle (m)
A cylindrical body rotating about the axis of its centre of mass therefore has a rotational velocity of:
[math] v = \frac{\theta r}{t} [/math]
where [math]v[/math] is the rotational velocity (m/s)
[math]t[/math] is the time it takes for the mass to rotate L metres (s)
Alternatively, rotational velocity can be expressed as:
[math] v = \omega r [/math]
where [math]\omega = \frac{\theta}{t} = \frac{2 \pi \times n}{60}[/math] is the angular velocity (rad/s)
[math]n[/math] is the speed in revolutions per minute (rpm)
The kinetic energy of a circular rotating mass can be derived from the classical Newtonian expression for the kinetic energy of rigid bodies:
[math] KE = \frac{1}{2} mv^{2} = \frac{1}{2} m(\omega r)^{2}[/math]
where [math]KE[/math] is the rotational kinetic energy (Joules or kg.m
2/s 2 or MW.s, all of which are equivalent) [math]m[/math] is the mass of the rotating body (kg)
Alternatively, rotational kinetic energy can be expressed as:
[math] KE = \frac{1}{2} J\omega^{2} [/math]
where [math]J = mr^{2}[/math] is called the
moment of inertia (kg.m 2).
Note that in physics, the moment of inertia [math]J[/math] is normally denoted as [math]I[/math]. In electrical engineering, the convention is for the letter "i" to always be reserved for current, and is therefore often replaced by the letter "j", e.g. the complex number operator i in mathematics is j in electrical engineering.
Normalised Inertia Constants
The moment of inertia can be expressed as a normalised quantity called the
inertia constant H, calculated as the ratio of the rotational kinetic energy of the machine at nominal speed to its rated power (VA): [math]H = \frac{1}{2} \frac{J \omega_0^{2}}{S_{b}}[/math]
where [math]H[/math] is the inertia constant (s)
[math]\omega_{0} = 2 \pi \times \frac{n}{60}[/math] is the nominal mechanical angular frequency (rad/s) [math]n[/math] is the nominal speed of the machine (revolutions per minute) [math]S_{b}[/math] is the rated power of the machine (VA) Generator Inertia
The moment of inertia for a generator is dependent on its mass and apparent radius, which in turn is largely driven by its prime mover type.
Based on actual generator data, the normalised inertia constants for different types and sizes of generators are summarised in the table below:
Machine type Number of samples MVA Rating Inertia constant H Min Median Max Min Median Max Steam turbine 45 28.6 389 904 2.1 3.2 5.7 Gas turbine 47 22.5 99.5 588 1.9 5.0 8.9 Hydro turbine 22 13.3 46.8 312.5 2.4 3.7 6.8 Combustion engine 26 0.3 1.25 2.5 0.6 0.95 1.6
|
Huge cardinal
Huge cardinals (and their variants) were introduced by Kenneth Kunen in 1972 as a very large cardinal axiom. Kenneth Kunen first used them to prove that the consistency of the existence of a huge cardinal implies the consistency of $ZFC+$"there is a $\aleph_2$-saturated ideal over $\omega_1$". [1]
Contents Definitions
Their formulation is similar to that of the formulation of superstrong cardinals. A huge cardinal is to a supercompact cardinal as a superstrong cardinal is to a strong cardinal, more precisely. The definition is part of a generalized phenomenon known as the "double helix", in which for some large cardinal properties $n$-$P_0$ and $n$-$P_1$, $n$-$P_0$ has less consistency strength than $n$-$P_1$, which has less consistency strength than $(n+1)$-$P_0$, and so on. This phenomenon is seen only around the $n$-fold variants as of modern set theoretic concerns. [2]
Although they are very large, there is a first-order definition which is equivalent to $n$-hugeness, so the $\theta$-th $n$-huge cardinal is first-order definable whenever $\theta$ is first-order definable. This definition can be seen as a (very strong) strengthening of the first-order definition of measurability.
Elementary embedding definitions $\kappa$ is almost $n$-huge with target $\lambda$iff $\lambda=j^n(\kappa)$ and $M$ is closed under all of its sequences of length less than $\lambda$ (that is, $M^{<\lambda}\subset M$). $\kappa$ is $n$-huge with target $\lambda$iff $\lambda=j^n(\kappa)$ and $M$ is closed under all of its sequences of length $\lambda$ ($M^\lambda\subset M$). $\kappa$ is almost $n$-hugeiff it is almost $n$-huge with target $\lambda$ for some $\lambda$. $\kappa$ is $n$-hugeiff it is $n$-huge with target $\lambda$ for some $\lambda$. $\kappa$ is super almost $n$-hugeiff for every $\gamma$, there is some $\lambda>\gamma$ for which $\kappa$ is almost $n$-huge with target $\lambda$ (that is, the target can be made arbitrarily large). $\kappa$ is super $n$-hugeiff for every $\gamma$, there is some $\lambda>\gamma$ for which $\kappa$ is $n$-huge with target $\lambda$. $\kappa$ is almost huge, huge, super almost huge, and superhugeiff it is almost $1$-huge, $1$-huge, etc. respectively. Ultrafilter definition
The first-order definition of $n$-huge is somewhat similar to measurability. Specifically, $\kappa$ is measurable iff there is a nonprincipal $\kappa$-complete ultrafilter, $U$, over $\kappa$. A cardinal $\kappa$ is $n$-huge iff there is some cardinal $\lambda$, a nonprincipal $\kappa$-complete ultrafilter, $U$, over $\mathcal{P}(\lambda)$, and cardinals $\kappa=\lambda_0<\lambda_1<\lambda_2...<\lambda_{n-1}<\lambda_n=\lambda$ such that:
$$\forall i<n\forall x\subseteq\lambda(ot(x\cap\lambda_{i+1})=\lambda_i\rightarrow x\in U)$$
Where $ot(X)$ is the order-type of the poset $(X,\in)$. [1] This definition is, more intuitively, making $U$ very large, like most ultrafilter characterizations of large cardinals (supercompact, strongly compact, etc.).
Consistency strength and size
Hugeness exhibits a phenomenon associated with similarly defined large cardinals (the $n$-fold variants) known as the
double helix. This phenomenon is when for one $n$-fold variant, letting a cardinal be called $n$-$P_0$ iff it has the property, and another variant, $n$-$P_1$, $n$-$P_0$ is weaker than $n$-$P_1$, which is weaker than $(n+1)$-$P_0$. [2] In the consistency strength hierarchy, here is where these lay (top being weakest): measurable = $0$-superstrong = almost $0$-huge = super almost $0$-huge = $0$-huge = super $0$-huge $n$-superstrong $n$-fold supercompact $(n+1)$-fold strong, $n$-fold extendible $(n+1)$-fold Woodin, $n$-fold Vopěnka $(n+1)$-fold Shelah almost $n$-huge super almost $n$-huge $n$-huge super $n$-huge $(n+1)$-superstrong
All huge variants lay at the top of the double helix restricted to some natural number $n$, although each are bested by I3 cardinals (the critical points of the I3 elementary embeddings). In fact, every I3 is preceeded by a stationary set of $n$-huge cardinals, for all $n$. [1]
Similarly, every huge cardinal $\kappa$ is almost huge, and there is a normal measure over $\kappa$ which contains every almost huge cardinal $\lambda<\kappa$. Every superhuge cardinal $\kappa$ is extendible and there is a normal measure over $\kappa$ which contains every extendible cardinal $\lambda<\kappa$. Every $(n+1)$-huge cardinal $\kappa$ has a normal measure which contains every cardinal $\lambda$ such that $V_\kappa\models$"$\lambda$ is super $n$-huge". [1]
In terms of size, however, the least $n$-huge cardinal is smaller than the least supercompact cardinal. Assuming both exist, for any $\kappa$ which is supercompact and has an $n$-huge cardinal above it, there are $\kappa$ many $n$-huge cardinals less than $\kappa$. [1]
Every $n$-huge cardinal is $m$-huge for every $m\leq n$. Similarly with almost $n$-hugeness, super $n$-hugeness, and super almost $n$-hugeness. Every almost huge cardinal is Vopěnka (therefore the consistency of the existence of an almost-huge cardinal implies the consistency of Vopěnka's principle). [1]
References Kanamori, Akihiro. Second, Springer-Verlag, Berlin, 2009. (Large cardinals in set theory from their beginnings, Paperback reprint of the 2003 edition) www bibtex The higher infinite. Kentaro, Sato. Double helix in large large cardinals and iteration ofelementary embeddings., 2007. www bibtex
|
Optimal control of the coefficient for the regional fractional $p$-Laplace equation: Approximation and convergence
1.
Department of Mathematical Sciences, George Mason University, Fairfax, VA 22030, USA
2.
University of Puerto Rico, Rio Piedras Campus, Department of Mathematics, College of Natural Sciences, 17 University AVE. STE 1701, San Juan PR 00925-2537, USA
In this paper we study optimal control problems with the regional fractional $p$-Laplace equation, of order $s \in \left( {0,1} \right)$ and $p \in \left[ {2,\infty } \right)$, as constraints over a bounded open set with Lipschitz continuous boundary. The control, which fulfills the pointwise box constraints, is given by the coefficient of the regional fractional $p$-Laplace operator. We show existence and uniqueness of solutions to the state equations and existence of solutions to the optimal control problems. We prove that the regional fractional $p$-Laplacian approaches the standard $p$-Laplacian as $s$ approaches 1. In this sense, this fractional $p$-Laplacian can be considered degenerate like the standard $p$-Laplacian. To overcome this degeneracy, we introduce a regularization for the regional fractional $p$-Laplacian. We show existence and uniqueness of solutions to the regularized state equation and existence of solutions to the regularized optimal control problem. We also prove several auxiliary results for the regularized problem which are of independent interest. We conclude with the convergence of the regularized solutions.
Keywords:Regional fractional p-Laplace operator, non-constant coefficient, quasi-linear nonlocal elliptic boundary value problems, optimal control. Mathematics Subject Classification:35R11, 49J20, 49J45, 93C73. Citation:Harbir Antil, Mahamadi Warma. Optimal control of the coefficient for the regional fractional $p$-Laplace equation: Approximation and convergence. Mathematical Control & Related Fields, 2019, 9 (1) : 1-38. doi: 10.3934/mcrf.2019001
References:
[1]
D. Adams and L. Hedberg,
[2]
L. Ambrosio, N. Fusco and D. Pallara,
[3]
H. Antil and S. Bartels,
Spectral approximation of fractional PDEs in image processing and phase field modeling,
[4] [5]
V. Benci, P. D'Avenia, D. Fortunato and L. Pisani,
Solitons in several space dimensions:Derrick's problem and infinitely many solutions,
[6] [7] [8]
J. Bourgain, H. Brezis and P. Mironescu, Another look at sobolev spaces, in
[9]
J. Bourgain, H. Brezis and P. Mironescu,
Limiting embedding theorems for $W^{s,p}$ when $s\uparrow1$ and applications,
[10]
L. Brasco, E. Parini and M. Squassina,
Stability of variational eigenvalues for the fractional $p$-Laplacian,
[11] [12] [13]
L. Caffarelli, J.-M. Roquejoffre and Y. Sire,
Variational problems for free boundaries for the fractional Laplacian,
[14]
L. Caffarelli, S. Salsa and L. Silvestre,
Regularity estimates for the solution and the free boundary of the obstacle problem for the fractional Laplacian,
[15]
E. Casas, P. Kogut and G. Leugering,
Approximation of optimal control problems in the coefficient for the $p$-Laplace equation. I. Convergence result,
[16] [17]
A. Di Castro., T. Kuusi and G. Palatucci,
Local behavior of fractional $p$-minimizers,
[18] [19] [20] [21] [22]
P. Drábek and J. Milota,
[23]
A. Elmoataz, M. Toutain and D. Tenbrinck,
On the $p$-Laplacian and ∞-Laplacian on graphs with applications in image and data processing,
[24]
L. Evans, Partial differential equations and Monge-Kantorovich mass transfer, in
[25]
C.G. Gal and M. Warma,
On some degenerate non-local parabolic equation associated with the fractional $p$-Laplacian,
[26]
P. Grisvard,
[27]
A. Jonsson and H. Wallin, Function spaces on subsets of ${\bf R}^n$,
[28]
O. Kupenko and R. Manzo, Approximation of an optimal control problem in coefficient for variational inequality with anisotropic
[29] [30]
J.-L. Lions and E. Magenes,
[31] [32]
V. Maz'ya and S. Poborchi,
[33] [34] [35]
F. Murat and L. Tartar,
[36]
I. Pan and S. Das,
[37] [38]
R. Showalter,
[39]
L. Tartar, Problèmes de contrôle des coefficients dans des équations aux dérivées partielles, in
[40]
D. Valério and J. Sá da Costa,
[41] [42] [43]
M. Warma,
The fractional relative capacity and the fractional Laplacian with Neumann and Robin boundary conditions on open sets,
[44]
M. Warma, The fractional Neumann and Robin type boundary conditions for the regional fractional
[45]
M. Warma,
Local Lipschitz continuity of the inverse of the fractional
[46]
show all references
References:
[1]
D. Adams and L. Hedberg,
[2]
L. Ambrosio, N. Fusco and D. Pallara,
[3]
H. Antil and S. Bartels,
Spectral approximation of fractional PDEs in image processing and phase field modeling,
[4] [5]
V. Benci, P. D'Avenia, D. Fortunato and L. Pisani,
Solitons in several space dimensions:Derrick's problem and infinitely many solutions,
[6] [7] [8]
J. Bourgain, H. Brezis and P. Mironescu, Another look at sobolev spaces, in
[9]
J. Bourgain, H. Brezis and P. Mironescu,
Limiting embedding theorems for $W^{s,p}$ when $s\uparrow1$ and applications,
[10]
L. Brasco, E. Parini and M. Squassina,
Stability of variational eigenvalues for the fractional $p$-Laplacian,
[11] [12] [13]
L. Caffarelli, J.-M. Roquejoffre and Y. Sire,
Variational problems for free boundaries for the fractional Laplacian,
[14]
L. Caffarelli, S. Salsa and L. Silvestre,
Regularity estimates for the solution and the free boundary of the obstacle problem for the fractional Laplacian,
[15]
E. Casas, P. Kogut and G. Leugering,
Approximation of optimal control problems in the coefficient for the $p$-Laplace equation. I. Convergence result,
[16] [17]
A. Di Castro., T. Kuusi and G. Palatucci,
Local behavior of fractional $p$-minimizers,
[18] [19] [20] [21] [22]
P. Drábek and J. Milota,
[23]
A. Elmoataz, M. Toutain and D. Tenbrinck,
On the $p$-Laplacian and ∞-Laplacian on graphs with applications in image and data processing,
[24]
L. Evans, Partial differential equations and Monge-Kantorovich mass transfer, in
[25]
C.G. Gal and M. Warma,
On some degenerate non-local parabolic equation associated with the fractional $p$-Laplacian,
[26]
P. Grisvard,
[27]
A. Jonsson and H. Wallin, Function spaces on subsets of ${\bf R}^n$,
[28]
O. Kupenko and R. Manzo, Approximation of an optimal control problem in coefficient for variational inequality with anisotropic
[29] [30]
J.-L. Lions and E. Magenes,
[31] [32]
V. Maz'ya and S. Poborchi,
[33] [34] [35]
F. Murat and L. Tartar,
[36]
I. Pan and S. Das,
[37] [38]
R. Showalter,
[39]
L. Tartar, Problèmes de contrôle des coefficients dans des équations aux dérivées partielles, in
[40]
D. Valério and J. Sá da Costa,
[41] [42] [43]
M. Warma,
The fractional relative capacity and the fractional Laplacian with Neumann and Robin boundary conditions on open sets,
[44]
M. Warma, The fractional Neumann and Robin type boundary conditions for the regional fractional
[45]
M. Warma,
Local Lipschitz continuity of the inverse of the fractional
[46]
[1]
Olha P. Kupenko, Rosanna Manzo.
On optimal controls in coefficients for ill-posed non-Linear elliptic Dirichlet boundary value problems.
[2]
Peter I. Kogut, Olha P. Kupenko.
On optimal control problem for an ill-posed strongly nonlinear elliptic equation with $p$-Laplace operator and $L^1$-type of nonlinearity.
[3] [4]
John R. Graef, Lingju Kong, Qingkai Kong, Min Wang.
Positive solutions of nonlocal fractional boundary value problems.
[5] [6] [7]
Bo You, Yanren Hou, Fang Li, Jinping Jiang.
Pullback attractors for the non-autonomous quasi-linear complex Ginzburg-Landau equation with $p$-Laplacian.
[8] [9]
Lu Yang, Meihua Yang, Peter E. Kloeden.
Pullback attractors for non-autonomous quasi-linear parabolic equations with dynamical boundary conditions.
[10]
Arrigo Cellina.
The regularity of solutions to some variational problems, including the
[11] [12]
Sofia Giuffrè, Giovanna Idone.
On linear and nonlinear elliptic boundary value problems in the plane with discontinuous coefficients.
[13]
Peter I. Kogut.
On approximation of an optimal boundary control problem for linear elliptic equation with unbounded coefficients.
[14]
Vasily Denisov and Andrey Muravnik.
On asymptotic behavior of solutions of the Dirichlet problem in half-space for linear and quasi-linear elliptic equations.
[15]
Maria Rosaria Lancia, Alejandro Vélez-Santiago, Paola Vernole.
A quasi-linear nonlocal Venttsel' problem of Ambrosetti–Prodi type on fractal domains.
[16]
Kais Hamza, Fima C. Klebaner.
On nonexistence of non-constant volatility in the Black-Scholes
formula.
[17]
Boumediene Abdellaoui, Ahmed Attar, Abdelrazek Dieb, Ireneo Peral.
Attainability of the fractional hardy constant with nonlocal mixed boundary conditions: Applications.
[18]
Pasquale Candito, Giovanni Molica Bisci.
Multiple solutions for a Navier boundary value problem involving the $p$--biharmonic operator.
[19] [20]
2018 Impact Factor: 1.292
Tools Metrics Other articles
by authors
[Back to Top]
|
Search
Now showing items 1-5 of 5
Measurement of electrons from beauty hadron decays in pp collisions at root √s=7 TeV
(Elsevier, 2013-04-10)
The production cross section of electrons from semileptonic decays of beauty hadrons was measured at mid-rapidity (|y| < 0.8) in the transverse momentum range 1 < pT <8 GeV/c with the ALICE experiment at the CERN LHC in ...
Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
Elliptic flow of muons from heavy-flavour hadron decays at forward rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV
(Elsevier, 2016-02)
The elliptic flow, $v_{2}$, of muons from heavy-flavour hadron decays at forward rapidity ($2.5 < y < 4$) is measured in Pb--Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE detector at the LHC. The scalar ...
Centrality dependence of the pseudorapidity density distribution for charged particles in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2013-11)
We present the first wide-range measurement of the charged-particle pseudorapidity density distribution, for different centralities (the 0-5%, 5-10%, 10-20%, and 20-30% most central events) in Pb-Pb collisions at $\sqrt{s_{NN}}$ ...
Beauty production in pp collisions at √s=2.76 TeV measured via semi-electronic decays
(Elsevier, 2014-11)
The ALICE Collaboration at the LHC reports measurement of the inclusive production cross section of electrons from semi-leptonic decays of beauty hadrons with rapidity |y|<0.8 and transverse momentum 1<pT<10 GeV/c, in pp ...
|
Can a doable experiment prove that the objective reality doesn't exist?
Here's a rare example of the media hype that leads the reader to a basically correct conclusion about quantum mechanics.
As I have often argued, quantum mechanics fundamentally requires the description of the phenomena to be observer-dependent. An observer must know what he observes, what the result is, and the answers to these questions are in principle subjective. Consequently, the wave function or the density matrix, its collapse, and the precise predictions of the future measurements are subjective or observer-dependent, too. There is no way to objectively label phenomena as measurements or non-measurements and there is no viable way (and no way that would be compatible with relativity) to make the collapse of the wave function – describing the change of the observer's knowledge – as an objectively real collapse.
Wigner's friend experiment is the simplest thought experiment that shows the point. In that thought experiment (which is now claimed to become a real experiment), Eugene Wigner observes a lab in which his friend observes a quantum experiment. For the friend, the collapse occurs as soon as the friend sees something. But for Wigner himself who hasn't observed the particle inside, the system keeps on evolving as a superposition in which all options have nonzero amplitudes and are capable of reinterfering.
Today, the MIT Technology review promoted an Italian experimental paper:
A quantum experiment suggests there’s no such thing as objective reality (MIT)Massimiliano Proietti and 7 co-authors have performed an experiment with six photons. Experimental rejection of observer-independence in the quantum world (quant-ph arXiv February 2019)
It's not the "basic" Wigner's friend setup but an "extended" one. For that setup, one may derive a Bell-like inequality – assuming that the perceptions of both people are real – and this Bell-like inequality is violated by the quantum mechanical predictions and has been shown to be violated by 5 sigma in the Italian experiment. OK, I call it "Italian" because of some surnames but their affiliations are in Edinburgh, Grenoble, and Innsbruck – no Italy.
I still don't understand how it's possible – a "contradiction" between the people's perspective is only possible when the interference between the "not yet collapsed" options is actually measured which will never be possible in practice, I think. But they must make some "more classical" assumptions about the interpretation of the experiment and this is being experimentally violated. I am going to look at the details, after a day with the monster group.
At any rate, in principle, the conclusion is surely correct. In the Wigner's friend setup, the descriptions of the phenomena by the two people must be fundamentally distinct and impossible to unify into one "collective" let alone "objective" description. Different observers have different lists of observations with their generally different results, and therefore generally different predictions for the future and different microscopic descriptions what happened.
P.S.: The inequality they test and violate isn't really a Bell's inequality but a Clauser-Horne-Shimony-Holt inequality (equation 2) which is derived assuming an objective probability distribution. When the variables \(A_x,B_y\) take values in \(a,b\in \{-1,+1\}\), then the average values\[
\langle A_x B_y\rangle = \sum_{a,b} ab\cdot P(A_x\!=\!a,B_y\!=\!b)
\] obey the CHSH inequality (yes, you have heard about CHSH before)\[
\langle A_1 B_1\rangle + \langle A_1 B_0\rangle + \langle A_0 B_1\rangle - \langle A_0 B_0\rangle \leq 2.
\] OK, they seem to assume that all the photons' degrees of freedom are described by pure states according to both observers. This seems wrong to me. The observation of the photons by Wigner's friend creates an entanglement with the degrees of freedom in the friend's body, so the state of the photons themselves become mixed according to Wigner (I mean the external observer), not pure. So I think that when these things are done correctly, it's not possible to find this kind of a "contradiction" in practice.
|
It's hard to say just from the sheet music; not having an actual keyboard here. The first line seems difficult, I would guess that second and third are playable. But you would have to ask somebody more experienced.
Having a few experienced users here, do you think that limsup could be an useful tag? I think there are a few questions concerned with the properties of limsup and liminf. Usually they're tagged limit.
@Srivatsan it is unclear what is being asked... Is inner or outer measure of $E$ meant by $m\ast(E)$ (then the question whether it works for non-measurable $E$ has an obvious negative answer since $E$ is measurable if and only if $m^\ast(E) = m_\ast(E)$ assuming completeness, or the question doesn't make sense). If ordinary measure is meant by $m\ast(E)$ then the question doesn't make sense. Either way: the question is incomplete and not answerable in its current form.
A few questions where this tag would (in my opinion) make sense: http://math.stackexchange.com/questions/6168/definitions-for-limsup-and-liminf http://math.stackexchange.com/questions/8489/liminf-of-difference-of-two-sequences http://math.stackexchange.com/questions/60873/limit-supremum-limit-of-a-product http://math.stackexchange.com/questions/60229/limit-supremum-finite-limit-meaning http://math.stackexchange.com/questions/73508/an-exercise-on-liminf-and-limsup http://math.stackexchange.com/questions/85498/limit-of-sequence-of-sets-some-paradoxical-facts
I'm looking for the book "Symmetry Methods for Differential Equations: A Beginner's Guide" by Haydon. Is there some ebooks-site to which I hope my university has a subscription that has this book? ebooks.cambridge.org doesn't seem to have it.
Not sure about uniform continuity questions, but I think they should go under a different tag. I would expect most of "continuity" question be in general-topology and "uniform continuity" in real-analysis.
Here's a challenge for your Google skills... can you locate an online copy of: Walter Rudin, Lebesgue’s first theorem (in L. Nachbin (Ed.), Mathematical Analysis and Applications, Part B, in Advances in Mathematics Supplementary Studies, Vol. 7B, Academic Press, New York, 1981, pp. 741–747)?
No, it was an honest challenge which I myself failed to meet (hence my "what I'm really curious to see..." post). I agree. If it is scanned somewhere it definitely isn't OCR'ed or so new that Google hasn't stumbled over it, yet.
@MartinSleziak I don't think so :) I'm not very good at coming up with new tags. I just think there is little sense to prefer one of liminf/limsup over the other and every term encompassing both would most likely lead to us having to do the tagging ourselves since beginners won't be familiar with it.
Anyway, my opinion is this: I did what I considered the best way: I've created [tag:limsup] and mentioned liminf in tag-wiki. Feel free to create new tag and retag the two questions if you have better name. I do not plan on adding other questions to that tag until tommorrow.
@QED You do not have to accept anything. I am not saying it is a good question; but that doesn't mean it's not acceptable either. The site's policy/vision is to be open towards "math of all levels". It seems hypocritical to me to declare this if we downvote a question simply because it is elementary.
@Matt Basically, the a priori probability (the true probability) is different from the a posteriori probability after part (or whole) of the sample point is revealed. I think that is a legitimate answer.
@QED Well, the tag can be removed (if someone decides to do so). Main purpose of the edit was that you can retract you downvote. It's not a good reason for editing, but I think we've seen worse edits...
@QED Ah. Once, when it was snowing at Princeton, I was heading toward the main door to the math department, about 30 feet away, and I saw the secretary coming out of the door. Next thing I knew, I saw the secretary looking down at me asking if I was all right.
OK, so chat is now available... but; it has been suggested that for Mathematics we should have TeX support.The current TeX processing has some non-trivial client impact. Before I even attempt trying to hack this in, is this something that the community would want / use?(this would only apply ...
So in between doing phone surveys for CNN yesterday I had an interesting thought. For $p$ an odd prime, define the truncation map $$t_{p^r}:\mathbb{Z}_p\to\mathbb{Z}/p^r\mathbb{Z}:\sum_{l=0}^\infty a_lp^l\mapsto\sum_{l=0}^{r-1}a_lp^l.$$ Then primitive roots lift to $$W_p=\{w\in\mathbb{Z}_p:\langle t_{p^r}(w)\rangle=(\mathbb{Z}/p^r\mathbb{Z})^\times\}.$$ Does $\langle W_p\rangle\subset\mathbb{Z}_p$ have a name or any formal study?
> I agree with @Matt E, as almost always. But I think it is true that a standard (pun not originally intended) freshman calculus does not provide any mathematically useful information or insight about infinitesimals, so thinking about freshman calculus in terms of infinitesimals is likely to be unrewarding. – Pete L. Clark 4 mins ago
In mathematics, in the area of order theory, an antichain is a subset of a partially ordered set such that any two elements in the subset are incomparable. (Some authors use the term "antichain" to mean strong antichain, a subset such that there is no element of the poset smaller than 2 distinct elements of the antichain.)Let S be a partially ordered set. We say two elements a and b of a partially ordered set are comparable if a ≤ b or b ≤ a. If two elements are not comparable, we say they are incomparable; that is, x and y are incomparable if neither x ≤ y nor y ≤ x.A chain in S is a...
@MartinSleziak Yes, I almost expected the subnets-debate. I was always happy with the order-preserving+cofinal definition and never felt the need for the other one. I haven't thought about Alexei's question really.
When I look at the comments in Norbert's question it seems that the comments together give a sufficient answer to his first question already - and they came very quickly. Nobody said anything about his second question. Wouldn't it be better to divide it into two separate questions? What do you think t.b.?
@tb About Alexei's questions, I spent some time on it. My guess was that it doesn't hold but I wasn't able to find a counterexample. I hope to get back to that question. (But there is already too many questions which I would like get back to...)
@MartinSleziak I deleted part of my comment since I figured out that I never actually proved that in detail but I'm sure it should work. I needed a bit of summability in topological vector spaces but it's really no problem at all. It's just a special case of nets written differently (as series are a special case of sequences).
|
Equidistribution of zeros of Random orthogonal polynomials
Speaker
Dr. Koushik Ramachandran Oklahoma State University, USA
When Mar 08, 2018
from 04:00 PM to 05:00 PM
Where LH 006 Add event to calendar vCal
iCal
Abstract: We study the asymptotic distribution of zeros for the random polynomials \(Pn(z)\) = \(\sum^n_{k=0}\xi_kB_k(z),\) where \(\{\xi_k\}^\infty_{k=0}\) are non-trivial i.i.d. complex random variables. Polynomials \(\{B_k\}^\infty_{k=0}\) are deterministic, and are selected from a standard basis such as Bergman or Szeg˝o polynomials associated with a Jordan domain G bounded by an analytic curve. We show that the zero counting measures of \(P_n\) converge almost surely to the equilibrium measure on the boundary of G if and only if \(E[log^+|\xi_0|] < \infty\). This talk is based on joint work with Igor Pritsker.
|
The exact value of secant of $45$ degrees is derived in geometry and it can also be derived in trigonometry by a trigonometric identity. In geometry, the $\sec{(45^°)}$ value can be derived in theoretical and practical approaches. Now, let’s learn the ways of deriving the $\sec{\Big(\dfrac{\pi}{4}\Big)}$ value in mathematics one after one.
The secant of $45$ degrees value is derived theoretically in geometry by the direct relation between lengths of opposite and adjacent sides when angle of right triangle is $45^°$. In fact, the lengths of opposite and adjacent sides are equal if the angle of right angled triangle is $50^g$.
For example, $\Delta RPQ$ is a right triangle with $45^°$ angle. The length of both opposite and adjacent side is represented by $l$ and the length of hypotenuse is denoted by $r$. Now, express the lengths of all three sides mathematically by Pythagorean Theorem.
${PQ}^2 = {PR}^2 + {QR}^2$
$\implies r^2 = l^2 + l^2$
$\implies r^2 = 2l^2$
$\implies r = \sqrt{2}.l$
$\implies \dfrac{r}{l} = \sqrt{2}$
Actually, $r$ and $l$ are lengths of hypotenuse and adjacent side (or opposite side) respectively.
$\implies \dfrac{Length \, of \, Hypotenuse}{Length \, of \, Adjacent \, side} = \sqrt{2}$
This ratio of lengths of hypotenuse to adjacent side is calculated when angle of right triangle is $45^°$. Hence, it is represented by $\sec{(45^°)}$
$\therefore \,\,\, \sec{(45^°)} = \sqrt{2}$
$\sec{(45^°)} = \sqrt{2} = 1.4142135623\ldots$
Even though you don’t know the relation between opposite and adjacent side when angle of right triangle is $50^g$, you can evaluate the value of $\sec{\Big(\dfrac{\pi}{4}\Big)}$ approximately on your own by using geometrical tools. It is actually done by constructing a right triangle with $45$ degrees angle.
The above five steps construct a right triangle, denoted by $\Delta NLM$ geometrically. Now, use the $\Delta NLM$ to find the value of $\sec{\Big(\dfrac{\pi}{4}\Big)}$ on your own by using geometrical tools.
$\sec{(45^°)} = \dfrac{Length \, of \, Hypotenuse}{Length \, of \, Adjacent \, side}$
$\implies \sec{(45^°)} \,=\, \dfrac{LM}{LN}$
As you know that, the $\Delta NLM$ is constructed geometrically by taking the length of hypotenuse as $9 \, cm$ but the length of adjacent side is unknown. It is essential to know this for finding the value of $\sec{(50^g)}$.
The length of adjacent side can be measured by ruler and you will observe that it is nearly $6.35 \, cm$.
$\implies \sec{(45^°)} \,=\, \dfrac{LM}{LN} = \dfrac{9}{6.35}$
$\,\,\, \therefore \,\,\,\,\,\, \sec{(45^°)} \,=\, 1.4173228346\ldots$
The value of secant of $\dfrac{\pi}{4}$ can also be derived in trigonometry by the reciprocal identity of cosine function.
$\sec{(45^°)} = \dfrac{1}{\cos{(45^°)}}$
Replace the exact value of cos of $45$ degrees in fraction form.
$\implies \sec{(45^°)} = \dfrac{1}{\dfrac{1}{\sqrt{2}}}$
$\implies \sec{(45^°)} = 1 \times \dfrac{\sqrt{2}}{1}$
$\implies \sec{(45^°)} = 1 \times \sqrt{2}$
$\,\,\, \therefore \,\,\,\,\,\, \sec{(45^°)} = \sqrt{2}$
As per theoretical geometrical method, the value of $\sec{(45^°)}$ is $\sqrt{2}$ or $1.4142135623\ldots$ and the value $\sec{(45^°)}$ is $1.4173228346\ldots$ as per practical geometrical method. You observe that there is a slight difference between them when you compare them.
The value of $\sec{(45^°)}$, obtained from theoretical geometrical method is exact value but the value of $\sec{(45^°)}$, obtained from practical geometrical method is approximate value because it is calculated by measuring the approximate value of adjacent side by the ruler.
The approximate value of $\sec{\Big(\dfrac{\pi}{4}\Big)}$ is often considered as $1.4142$ in mathematics.
Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.
|
Trichotomy (mathematics)
In mathematics, the Law of
Trichotomy states that every real number is either positive, negative, or zero. [1] More generally, trichotomy is the property of an order relation < on a set X that for any x and y, exactly one of the following holds: x
In mathematical notation, this is
\forall x \in X \, \forall y \in X \, ( ( x < y \, \land \, \lnot (y < x) \, \land \, \lnot( x = y )\, ) \lor \, ( \lnot(x < y) \, \land \, y < x \, \land \, \lnot( x = y) \, ) \lor \, ( \lnot(x < y) \, \land \, \lnot( y < x) \, \land \, x = y \, \, ) ) \,. \forall x \in X \, \forall y \in X \, ( x < y \, \lor \, y < x \, \lor \, x = y ) \,.
In classical logic, this
axiom of trichotomy holds for ordinary comparison between real numbers and therefore also for comparisons between integers and between rational numbers. The law does not hold in general in intuitionistic logic.
In ZF set theory and Bernays set theory, the law of trichotomy holds between the cardinal numbers of well-orderable sets even without the axiom of choice. If the axiom of choice holds, then trichotomy holds between arbitrary cardinal numbers (because they are all well-orderable in that case).
[2]
More generally, a binary relation
R on X is trichotomous if for all x and y in X exactly one of xRy, yRx or x= y holds. If such a relation is also transitive it is a strict total order; this is a special case of a strict weak order. For example, in the case of three element set { a, b, c} the relation R given by aRb, aRc, bRc is a strict total order, while the relation R given by the cyclic aRb, bRc, cRa is a non-transitive trichotomous relation.
|
(The society is limited to people with PhDs after 1990, occasioning the title of this post, a reference to a song about a bar limited to people under 21, a reference you will not get unless your PhD was granted well before 1990.)
I can't blog all the great papers and discussions, so I'll pick one of particular interest, Itamar Drechsler, Alexi Savov, and Philipp Schnabl's "Model of Monetary Policy and Risk Premia"
This paper addresses a very important issue. The policy and commentary community keeps saying that the Federal Reserve has a big effect on risk premiums by its control of short-term rates. Low interest rates are said to spark a "reach for yield," and encourage investors, and too big to fail banks especially, to take on unwise risks. This story has become a central argument for hawkishness at the moment. The causal channel is just stated as fact. But one should not accept an argument just because one likes the policy result.
Nice story. Except there is about zero economic logic to it. The
level of nominal interest ratesand the risk premiumare two totally different phenomena. Borrowing at 5% and making a risky investment at 8%, or borrowing at 1% and making a risky investment at 4% is exactly the same risk-reward tradeoff.
In equations, consider the basic first order condition for investment, \[ 0 = E \left[ \left( \frac{C_{t+1}}{C_t} \right)^{-\gamma} (R_{t+1}-R^f_t) \right] \] \[ 1 = E \left[ \beta \left( \frac{C_{t+1}}{C_t} \right)^{-\gamma} \right] R_t^f \] Risk aversion \(\gamma\) controls the risk premium in the first equation, and impatience \(\beta\) controls the risk free rate in the second equation. The level of risk free rates has nothing to do with the risk premium.
Yes, higher risk aversion or consumption volatility would increase precautionary saving and lower interest rates in the second equation, holding \(\beta\) fixed. But that is the "wrong" sign -- lower interest rates are associated with higher, not lower, risk premiums.
Worse, that "wrong" sign is what we see in the data. Risk premiums are high in the early part of recessions, when interest rates are low. Risk premiums are low in booms, when interest rates are high. OK, I'm a bit defensive because "by force of habit" with John Campbell was all about producing that correlation. But that is the pattern in the data. I made a graph above of the Federal Funds rate (blue) and the spread between BAA bonds and treasuries (green, right scale). You can see the risk premium higher just when rates fall at the early stage of every recession, and premiums low at the peaks of the booms, when rates are at their peaks.
So, if one has this belief about Fed policy, there must be some other effect driving a big negative correlation between risk premiums and rates, yet the Fed can
causepremiums to go up or down a bit more by raising or lowering rates.
Every time I ask people -- policy types, central bankers, Fed staff, financial journalists -- about this widely held belief, I get basically psychological and institutional rather than economic answers. Fund managers, insurance companies, pension funds, endowments, have fixed nominal rate of return targets. People have nominal illusions and don't think 8% with 1% short rates is a lot better than 10% with 9% short rates. Maybe. But basing monetary policy on the notion that all investors are total morons seems dicey. For one thing, the minute the Fed starts to exploit rules of thumb, smart investors change the rules of thumb. Segmented markets and institutional constraints are written in sand, not stone, and persist only as long as they are not too costly.
OK, enter Drechsler, Savov, and Schnabl. They have a real, economic model of the phenomenon. That's great. We may disagree, but the only way to understand this issue is to write down a model, not to tell stories.
The model is long and hard, and I won't pretend I have it all right. I think I digest it down to one basic point. Banks had (past tense) to hold non-interest-bearing reserves against deposits. This is a source of nominal illusion. If banks have to hold some non-interest bearing cash for every investment they make, then the effective cost of funds is higher when the nominal rate is higher. We are, in effect, mismeasuring \(R^f\) in my equation.
This makes a lot of sense. Except... Before 2007 non-interest-bearing reserves were really tiny, $50 billion dollars out of $9 trillion of bank credit. Quantitatively, the induced nominal illusion is small. Also, while it's fun to write models in which all funds must channel through intermediaries, there are lots of ways that money goes directly from savers to borrowers, like mortgage-backed securities, without paying the reserve tax. Banks aren't allowed to hold equities, so this channel can't work at all for the idea that low rates fuel stock "bubbles."
And now, reserves will pay interest.
At the conference, Alexi disagreed with this interpretation. He showed the following graph:
Fed funds are typically higher than T bills, and the spread is higher when interest rates are higher. They interpret this quantity (p.3) as the "external finance spread." Fed funds represent a potential use of funds, and the shadow value of lending. Alexi cited another mechanism too: "sticky" deposits generate a relationsip (at least temporary) between interest rate levels and real bank funding costs. So by whatever mechanism, they say, you can see that cost of funds vary with the level of interest rates. In response to my sort of graph, yes, lots of other things push risk premiums around generating the negative correlation, but allowing the causal effect.
Read the paper for more. I have come to praise it not to criticize it. Real, solid, quantiative economic models are just what we need to have a serious discussion. This is a really important and unsolved question, which I will close by restating:
Does monetary policy, by controlling the level of short term rates, substantially affect risk premiums? If so, how?
Of course, maybe the answer is "it doesn't."
|
There are four important things to remember here.
The first is that you can factor numbers and here we have $84 = 12\cdot 7$
The second is that $\log_x(y\cdot z) = \log_x(y) + \log_x(z)$
The third is that $\log_x(y) = \dfrac{\log_z(y)}{\log_z(x)}$
Finally, remember that $\log_x(x)=1$
These are true for all positive real values of $x,y,z$ different than $1$.
So... we have $\log_{12}(5)=a$ and $\log_{12}(7)=b$
Using these, since these logarithms are the same base we can find $\log_5(7)$ as being $\frac{b}{a}$, but that isn't quite what we are interested in finding, but it is close.
Rather, we note that $\log_{12}(84)=\log_{12}(12\cdot 7) = \log_{12}(12)+\log_{12}(7)=1+b$
Now we can do our division to change the logarithm's base to $5$ and we get:
$$\log_5(84) = \dfrac{\log_{12}(84)}{\log_{12}(5)} = \dfrac{1+b}{a}$$
|
The Basel problem can be solved by simple integration!
Recall that the Basel problem is to determine the value of $\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \ldots$.
Consider $f(x) = \ldots + e^{-3x} + e^{-2x} + e^{-1x} + e^{0x} + e^{1x} + e^{2x} + e^{3x} + \ldots$. By double integrating $f$, we get $h(x) = \ldots + \frac{e^{-3x}}{3^2} + \frac{e^{-2x}}{2^2} + \frac{e^{-1x}}{1^2} + \frac{x^2}{2} + \frac{e^{1x}}{1^2} + \frac{e^{2x}}{2^2} + \frac{e^{3x}}{3^2} + \ldots + C_1x + C_2 $. If we can figure out what $h$ amounts to with $C_1$ and $C_2$ chosen to be zero, then we can readily solve the Basel problem, as the Basel sum then amounts to simply $h(0)/2$.
Let us go through this in detail. First, observe that $f(x)$, which is to say $\ldots + e^{-3x} + e^{-2x} + e^{-1x} + e^{0x} + e^{1x} + e^{2x} + e^{3x} + \ldots$, is unchanged by multiplication by $e^x$; thus, wherever $e^x \neq 1$, we must have $f(x) = 0$, at least in some sense.
(When $e^x$
does equal $1$, on the other hand, $f(x)$ is an infinite sum of $1$s; this can be made sense of, with $f$ turning out to be interpretable as a kind of integrable entity slightly more general than familiar finite-valued functions, but we won't need to worry about this for now.
For those who care pedantically about formal technicalities, I should also note that our $f(x) = 0$ result doesn't hold in the traditional formalization of series convergence, as $f(x)$ is not a convergent series (its terms do not approach $0$, for example), but this equation still holds in a good enough sense for our purposes; for example, $f$ is "Cesàro summable", a slight extension of traditional summation-by-convergence which agrees with traditional summation where it is well-defined, and which suffices to formally carry out our argument, should we care to be formal)
Next, consider integrating $f$ once, to obtain the particular antiderivative $g(x) = \ldots + \frac{e^{-3x}}{-3} + \frac{e^{-2x}}{-2} + \frac{e^{-1x}}{-1} + x + \frac{e^{1x}}{1} + \frac{e^{2x}}{2} + \frac{e^{3x}}{3} + \ldots$ [note the middle $x$ term, slightly different from all the rest].
On any range where $e^x \neq 1$, we must have that $g(x)$ is constant (as its derivative is $f(x) = 0$). In particular, we must have that $g(x)$ is constant as $x$ ranges between $0$ and $2 \pi i$, the range to which we will restrict all attention from hereon out. But what particular constant is $g(x)$ on this range? Well, the answer is the same as the average value of $g(x)$ on this range.
But the average value of $e^{nx}$ over this range for nonzero $n$ is zero (as that average value must be invariant under multiplication by any value of $e^{nx}$ inside the range, and some (indeed, most!) of those values will differ from $1$; essentially, $e^{nx}$ spins around a circle $n$ times, whose average value is the origin zero). So in computing the average value of $g(x)$ on our range, all terms vanish except for the average value of $x$, which will be $\pi i$ (halfway between the range's two endpoints).
Thus, $g(x) = \pi i$ for $x$ between $0$ and $2 \pi i$. (The same reasoning also tells us that $g(x) = - \pi i$ for $x$ between $0$ and $-2 \pi i$, for what it's worth, and more generally, that $g(x)$ steps up by $2 \pi i$ whenever $x$ passes a multiple of $2 \pi i$)
Integrating again, we get the particular antiderivative $h(x) = \ldots + \frac{e^{-3x}}{3^2} + \frac{e^{-2x}}{2^2} + \frac{e^{-1x}}{1^2} + \frac{x^2}{2} + \frac{e^{1x}}{1^2} + \frac{e^{2x}}{2^2} + \frac{e^{3x}}{3^2} + \ldots$. This must equal $\pi i x + C$ for $x$ between $0$ and $2 \pi i$.
Again, to determine the value of this constant $C$, we can consider the question of the average value of $h(x)$ throughout our range of interest, which again reduces to the average value of the one non-exponential term $\frac{x^2}{2}$. By the "power rule" of calculus, the average value of $\frac{x^2}{2}$ between $0$ and $2 \pi i$ is $\frac{(2 \pi i)^2/3}{2} = -\frac{2}{3} \pi^2$. This must therefore also equal the average value of $\pi i x + C$ on our range, which is $(\pi i)^2 + C = - \pi^2 + C$. Thus, we get $C = \pi^2 - \frac{2}{3} \pi^2 = \frac{1}{3} \pi^2$.
And therefore $h(0) = \frac{1}{3} \pi^2$. And since, as we noted before, the Basel sum comes out to $h(0)/2$, we find that the Basel sum is $\frac{1}{6} \pi^2$. Ta-da!
What's more, we can carry on integrating in the same way, determining a polynomial expression in $x$ for $\ldots + \frac{e^{-3x}}{3^n} + \frac{e^{-2x}}{2^n} + \frac{e^{-1x}}{1^n} + \frac{x^n}{n!} + \frac{e^{1x}}{1^n} + \frac{e^{2x}}{2^n} + \frac{e^{3x}}{3^n} + \ldots$ for each natural number $n$, and by evaluating this at $x = 0$, determining the value of $\ldots + \frac{1}{(-3)^n} + \frac{1}{(-2)^n} + \frac{1}{(-1)^n} + 0 + \frac{1}{1^n} + \frac{1}{2^n} + \frac{1}{3^n} + \ldots$. This will come out to zero for odd $n$ (the negatively indexed and positively indexed terms cancelling each other out as negated mirror images), but for even $n$, we will get interesting results (which we can divide by two, in just the same fashion as we did here, to restrict attention to just the positively indexed terms of this summation).
For example, in just the same way, with two more integrations, we find that $\frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \ldots = \frac{1}{90} \pi^4$. And with yet two more integrations, we find that $\frac{1}{1^6} + \frac{1}{2^6} + \frac{1}{3^6} + \ldots = \frac{1}{945} \pi^6$. And in general, for each (positive) even $n$, we find that $\frac{1}{1^n} + \frac{1}{2^n} + \frac{1}{3^n} + \ldots$ is some rational coefficient times $\pi^n$ (with a simple recursive computation for that rational coefficient). Ta-da again! Infinitely many "Ta-da!"-worthy results, all at once!
|
Author: Subhransu Maji
Background
In the improved B-CNN paper published at BMVC 2017 we showed that the matrix square root is an effective way to normalize covariance matrices used for classification tasks. While the square root and its gradient can be computed via a SVD decomposition, this is not efficienly implemented on existing libraries for GPUs. The BMVC paper presented some GPU friendly routines for computing the matrix square root and its gradient.
Here we discuss a two extensions that allows simpler and faster gradients through automatic differentiation and iterative methods for solving the Lyapunov equation.
Given a positive semi definite (PSD) matrix $A$ the square root is defined as a matrix $Z$ such that $ZZ=A$. This can be computed by first computing the SVD of the matrix $A=U\Sigma U^T$ after which the the square root can be obtained as $Z=U\Sigma^{1/2}U^T$. However, currently SVD lacks GPU support hence instead we proposed to use Newton Schulz iterations after scaling the matrix. Assume $||A|| < 1$. The square root can be obtained by initializing $Y_0=A$ and $Z_0=I$ and iterating: $$ Y_{k+1} = \frac{1}{2}Y_k(3I - Z_k Y_k), Z_{k+1} = \frac{1}{2}(3I - Z_k Y_k)Z_k. $$ Then $Y_k \rightarrow A^{1/2}$, and $Z_k \rightarrow A^{-1/2}$. We found that 5 iterations were sufficient to match the accuracy and 10 iterations sufficient to match the square root to numerical precision on a floating point GPU. This can be 5-20x faster than computing the SVD depending on the GPU and software, especially since these iterations can be implemented in batch mode in languages like pytorch.
Training requires gradient computation. In the paper we obtained this by solving a Lyapunov equation:$$ A^{1/2} \left( \frac{\partial L}{\partial A}\right) + \left( \frac{\partial L}{\partial A}\right) A^{1/2} = \frac{\partial L}{\partial Z}.$$Given the SVD of the matrix $A$ and the gradient ${\partial L}/{\partial Z}$ the solution to the Lyapunov equation can be obtained in closed form.The Lyapunov gradients are also more numerically stable than matrix backpropagation gradients with SVD (e.g., Ionescu et al.). The former depends on $\min(1/(\lambda_i + \lambda_j))$ while the latter depends on $\min(1/(\lambda_i - \lambda_j)).$ where $\lambda_i$ is the $i^{th}$ eigenvalue of $Z$.The Lyapunov equation has a closed form solution given the SVD hence during training SVD can be used as both the forwad and backward, while iterative methods can be used as forward at test time.This means that the training is slower.While training speed is less critical than testing speed it would be nice to also be able to train faster. Below are two approaches that implement the gradients efficiently. Gradients by autograd
The gradients can be automatically obtained if the Newton Schulz iterations are implemented in a language that supports automatic differentiation. For example here is a code snippet in PyTorch. The first few lines scale the matrix since the iterations are locally convergent.
def sqrt_newton_schulz_autograd(A, numIters, dtype): batchSize = A.data.shape[0] dim = A.data.shape[1] normA = A.mul(A).sum(dim=1).sum(dim=1).sqrt() Y = A.div(normA.view(batchSize, 1, 1).expand_as(A)); I = Variable(torch.eye(dim,dim).view(1, dim, dim). repeat(batchSize,1,1).type(dtype),requires_grad=False) Z = Variable(torch.eye(dim,dim).view(1, dim, dim). repeat(batchSize,1,1).type(dtype),requires_grad=False) for i in range(numIters): T = 0.5*(3.0*I - Z.bmm(Y)) Y = Y.bmm(T) Z = T.bmm(Z) sA = Y*torch.sqrt(normA).view(batchSize, 1, 1).expand_as(A) error = compute_error(A, sA) return sA, error This works quite well.For example on my GPU with a batch size of 32 and 512x512 random matrices the time taken to compute the gradient with 5 iterations is about 5-10% of the time taken for forward.Thus gradients are nearly free! Gradients by iterative Lyapunov solver
The drawback of the autograd approach is that a naive implementation stores all the intermediate results.Thus the memory overhead scales linearly with the number of iterations which is problematic for large matrices.In general one can tradeoff memory with computation during backpropagation.For example one can store the values of $T_k$, $Z_k$ and $Y_k$ at every $\sqrt{T}$ iterations and recompute the remaining from the previous checkpoint during backward propagation.This strategy requires $\sqrt{T}$ memory but is $2\times$ slower.Without any no additional memory overhead one can recompute the intermediate values on the fly from the input at step of the gradient but that would be $T^2 \times$ slower.
Can we obtain gradients with no memory overhead without taking a hit in speed?It turns out we can do this by using an iterative method for solving the Lyapunov eqauation implemented in the following code snippet: def lyap_newton_schulz(z, dldz, numIters, dtype): batchSize = z.shape[0] dim = z.shape[1] normz = z.mul(z).sum(dim=1).sum(dim=1).sqrt() a = z.div(normz.view(batchSize, 1, 1).expand_as(z)) I = torch.eye(dim,dim).view(1, dim, dim).repeat(batchSize,1,1).type(dtype) q = dldz.div(normz.view(batchSize, 1, 1).expand_as(z)) for i in range(numIters): q = 0.5*(q.bmm(3.0*I - a.bmm(a)) - a.transpose(1, 2).bmm(a.transpose(1,2).bmm(q) - q.bmm(a)) ) a = 0.5*a.bmm(3.0*I - a.bmm(a)) dlda = 0.5*q return dlda
In practice the iterative solver is as fast as the autograd but requires
no checkpointing. This is great if both the forward and backward are run to convergence.However, I've not tested what happens in real problems when the method is run only for a few iterations (e.g. only once!).The experiments in the BMVC paper showed that even a single iteration is useful.
Take a look at the source code in Matlab and Python here https://github.com/msubhransu/matrix-sqrt for a detailed comparison of the methods. The implementation is fairly straightforwad using PyTorch's autograd support.
References:
Ionescu, Catalin et al. "Matrix backpropagation for deep networks with structured layers." ICCV 15 Lin, Tsung-Yu, and Subhransu Maji. "Improved Bilinear Pooling with CNNs." BMVC 17 Higham, Nicholas J. "Functions of matrices: theory and computation. Society for Industrial and Applied Mathematics", 2008. NIPS 2017 autodiff workshop: https://autodiff-workshop.github.io
|
This is slightly contrived, but consider a situation where you have two balls, of mass $M$ and $m$, where $M=16\times100^N\times m$ for some integer $N$. The balls are placed against a wall as shown:
We push the heavy ball towards the lighter one and the wall. The balls are assumed to collide elastically with the wall and with each other. The smaller ball bounces off the larger ball, hits the wall and bounces back. At this point there are two possible solutions: the balls collide with each other infinitely many times until the larger ball reaches the wall (assume they have no size), or the collisions from the smaller ball eventually cause the larger ball to turn around and start heading in the other direction - away from the wall.
In fact, it is the second scenario which occurs: the larger ball eventually heads away from the wall. Denote by $p(N)$ the number of collisions between the two balls before the larger one changes direction, and gaze in astonishment at the values of $p(N)$ for various $N$:
\begin{align}p(0)&=3\\p(1)&=31\\p(2)&=314\\p(3)&=3141\\p(4)&=31415\\p(5)&=314159\\\end{align}
and so on. $p(N)$
is the first $N+1$ digits of $\pi$!
This can be made to work in other bases in the obvious way.
See 'Playing Pool with $\pi$' by Gregory Galperin.
|
@egreg It does this "I just need to make use of the standard hyphenation function of LaTeX, except "behind the scenes", without actually typesetting anything." (if not typesetting includes typesetting in a hidden box) it doesn't address the use case that he said he wanted that for
@JosephWright ah yes, unlike the hyphenation near box question, I guess that makes sense, basically can't just rely on lccode anymore. I suppose you don't want the hyphenation code in my last answer by default?
@JosephWright anway if we rip out all the auto-testing (since mac/windows/linux come out the same anyway) but leave in the .cfg possibility, there is no actual loss of functionality if someone is still using a vms tex or whatever
I want to change the tracking (space between the characters) for a sans serif font. I found that I can use the microtype package to change the tracking of the smallcaps font (\textsc{foo}), but I can't figure out how to make \textsc{} a sans serif font.
@DavidCarlisle -- if you write it as "4 May 2016" you don't need a comma (or, in the u.s., want a comma).
@egreg (even if you're not here at the moment) -- tomorrow is international archaeology day: twitter.com/ArchaeologyDay , so there must be someplace near you that you could visit to demonstrate your firsthand knowledge.
@barbarabeeton I prefer May 4, 2016, for some reason (don't know why actually)
@barbarabeeton but I have another question maybe better suited for you please: If a member of a conference scientific committee writes a preface for the special issue, can the signature say John Doe \\ for the scientific committee or is there a better wording?
@barbarabeeton overrightarrow answer will have to wait, need time to debug \ialign :-) (it's not the \smash wat did it) on the other hand if we mention \ialign enough it may interest @egreg enough to debug it for us.
@DavidCarlisle -- okay. are you sure the \smash isn't involved? i thought it might also be the reason that the arrow is too close to the "M". (\smash[t] might have been more appropriate.) i haven't yet had a chance to try it out at "normal" size; after all, \Huge is magnified from a larger base for the alphabet, but always from 10pt for symbols, and that's bound to have an effect, not necessarily positive. (and yes, that is the sort of thing that seems to fascinate @egreg.)
@barbarabeeton yes I edited the arrow macros not to have relbar (ie just omit the extender entirely and just have a single arrowhead but it still overprinted when in the \ialign construct but I'd already spent too long on it at work so stopped, may try to look this weekend (but it's uktug tomorrow)
if the expression is put into an \fbox, it is clear all around. even with the \smash. so something else is going on. put it into a text block, with \newline after the preceding text, and directly following before another text line. i think the intention is to treat the "M" as a large operator (like \sum or \prod, but the submitter wasn't very specific about the intent.)
@egreg -- okay. i'll double check that with plain tex. but that doesn't explain why there's also an overlap of the arrow with the "M", at least in the output i got. personally, i think that that arrow is horrendously too large in that context, which is why i'd like to know what is intended.
@barbarabeeton the overlap below is much smaller, see the righthand box with the arrow in egreg's image, it just extends below and catches the serifs on the M, but th eoverlap above is pretty bad really
@DavidCarlisle -- i think other possible/probable contexts for the \over*arrows have to be looked at also. this example is way outside the contexts i would expect. and any change should work without adverse effect in the "normal" contexts.
@DavidCarlisle -- maybe better take a look at the latin modern math arrowheads ...
@DavidCarlisle I see no real way out. The CM arrows extend above the x-height, but the advertised height is 1ex (actually a bit less). If you add the strut, you end up with too big a space when using other fonts.
MagSafe is a series of proprietary magnetically attached power connectors, originally introduced by Apple Inc. on January 10, 2006, in conjunction with the MacBook Pro at the Macworld Expo in San Francisco, California. The connector is held in place magnetically so that if it is tugged — for example, by someone tripping over the cord — it will pull out of the socket without damaging the connector or the computer power socket, and without pulling the computer off the surface on which it is located.The concept of MagSafe is copied from the magnetic power connectors that are part of many deep fryers...
has anyone converted from LaTeX -> Word before? I have seen questions on the site but I'm wondering what the result is like... and whether the document is still completely editable etc after the conversion? I mean, if the doc is written in LaTeX, then converted to Word, is the word editable?
I'm not familiar with word, so I'm not sure if there are things there that would just get goofed up or something.
@baxx never use word (have a copy just because but I don't use it;-) but have helped enough people with things over the years, these days I'd probably convert to html latexml or tex4ht then import the html into word and see what come out
You should be able to cut and paste mathematics from your web browser to Word (or any of the Micorsoft Office suite). Unfortunately at present you have to make a small edit but any text editor will do for that.Givenx=\frac{-b\pm\sqrt{b^2-4ac}}{2a}Make a small html file that looks like<!...
@baxx all the convertors that I mention can deal with document \newcommand to a certain extent. if it is just \newcommand\z{\mathbb{Z}} that is no problem in any of them, if it's half a million lines of tex commands implementing tikz then it gets trickier.
@baxx yes but they are extremes but the thing is you just never know, you may see a simple article class document that uses no hard looking packages then get half way through and find \makeatletter several hundred lines of trick tex macros copied from this site that are over-writing latex format internals.
|
Searching for just a few words should be enough to get started. If you need to make more complex queries, use the tips below to guide you.
Purchase individual online access for 1 year to this journal.
Impact Factor 2019: 0.808
The journal
Asymptotic Analysis fulfills a twofold function. It aims at publishing original mathematical results in the asymptotic theory of problems affected by the presence of small or large parameters on the one hand, and at giving specific indications of their possible applications to different fields of natural sciences on the other hand. Asymptotic Analysis thus provides mathematicians with a concentrated source of newly acquired information which they may need in the analysis of asymptotic problems.
Article Type: Research Article
Abstract: In this paper we investigate the asymptotic behavior of the solutions of the one-dimensional generalized porous-thermo-elasticity problem. First, we prove that when only thermal damping is present the decay of solutions is slow (generically). Second, we show that if the porous dissipation is also present the decay of the solutions is controlled by an exponential.
Keywords: generalized porous-thermo-elasticity, exponential stability, semigroup of contractions
Citation: Asymptotic Analysis, vol. 49, no. 3,4, pp. 173-187, 2006
Article Type: Research Article
Abstract: In this paper we study the asymptotic behavior of the viscoelastic system with nondissipative kernels. We show that the uniform decay of the energy depends on the decay of the kernel, the positivity of the kernel in t=0 and some smallness condition. That is, if the kernel g∈C2 (${\mathbb{R}}$ + ) with g(0)>0, decays exponentially to zero then the solution decays exponentially to zero. On the other hand, if the kernel decays polynomially as t−p then the corresponding solutions also decays polynomially to zero with the same rate of decay.
Keywords: materials with memory, asymptotic stability, indefinite dissipation
Citation: Asymptotic Analysis, vol. 49, no. 3,4, pp. 189-204, 2006
Authors: Penot, Jean-Paul
Article Type: Research Article
Abstract: We provide conditions ensuring the existence of a solution to a noncoercive minimization problem. These conditions are also necessary. They enlarge the applicability of previous results due to Baiocchi, Buttazzo, Gastaldi and Tomarelli and their extensions by Auslender. A particular attention is given to the case of quasiconvex objective functions.
Keywords: asymptotic cone, asymptotic function, coercivity, existence, minimization, recession
Citation: Asymptotic Analysis, vol. 49, no. 3,4, pp. 205-215, 2006
Article Type: Research Article
Abstract: We consider the following stationary Keller–Segel system from chemotaxis \[\Delta u-au+u^{p}=0,\quad u>0\ \hbox{in}\ {\varOmega},\qquad \frac{\curpartial u}{\curpartial \nu}=0\quad \hbox{on}\ \curpartial {\varOmega},\] where Ω⊂$\mathbb{R}$ 2 is a smooth and bounded domain. We show that given any two positive integers K,L, for p sufficiently large, there exists a solution concentrating in K interior points and L boundary points. The location of the blow-up points is related to the Green function. The solutions are obtained as critical points of some finite-dimensional reduced energy functional. No assumption on the symmetry, geometry nor topology of the domain is needed.
Citation: Asymptotic Analysis, vol. 49, no. 3,4, pp. 217-247, 2006
Authors: Jakobsen, Espen R.
Article Type: Research Article
Abstract: Recently, Krylov, Barles, and Jakobsen developed the theory for estimating errors of monotone approximation schemes for the Bellman equation (a convex Isaacs equation). In this paper we consider an extension of this theory to a class of non-convex multidimensional Isaacs equations. This is the first result of this kind for non-convex multidimensional fully non-linear problems. To get the error bound, a key intermediate step is to introduce a penalization approximation. We conclude by (i) providing new error bounds for penalization approximations extending earlier results by, e.g., Bensoussan and Lions, and (ii) obtaining error bounds for approximation schemes for the …penalization equation using very precise a priori bounds and a slight generalization of the recent theory of Krylov, Barles, and Jakobsen. Show more
Keywords: nonlinear degenerate elliptic equation, obstacle problem, variational inequality, penalization method, Hamilton–Jacobi–Bellman–Isaacs equation, viscosity solution, finite difference method, control scheme, convergence rate
Citation: Asymptotic Analysis, vol. 49, no. 3,4, pp. 249-273, 2006
Authors: Marchand, Fabien
Article Type: Research Article
Abstract: We show that under a smallness assumption on the L∞ norm of the initial data, there is a propagation of any Sobolev regularity greater than −1/2 for some weak solutions of the critical dissipative quasi-geostrophic equation. We also prove the existence of solutions in a space close to the homogeneous Sobolev space $\dot{H}$ 1 , theses solutions are given by a fixed point theorem.
Citation: Asymptotic Analysis, vol. 49, no. 3,4, pp. 275-293, 2006
Authors: Negulescu, Claudia
Article Type: Research Article
Abstract: In a previous work [N. Ben Abdallah and C. Negulescu, A one-dimensional transport model with small coherence lengths, Transp. Theory and Stat. Phys. 31(4–6) (2002), 559–578] a one-dimensional transport model accounting for both, quantum phenomena and smallness of coherence lengths, has been analyzed. In the limit of infinitely small coherence lengths, this model leads to the classical Vlasov equation. In this paper the two-dimensional situation is considered by adding to the one-dimensional transport direction a transversal confined one. The transport direction is splitted into several regions of a size comparable to the coherence length. A quantum model, based on the …2D Schrödinger equation, describes the electron transport within each cell, while on larger distances, statistics defined on the interfaces of the cells, determine the electron motion. Reflection and transmission coefficients, deduced from the wavefunctions, solutions of the Schrödinger equation, relate neighboring statistics with one another. In the limit of small coherence lengths, we obtain a collisionless subband model, which is classical in the transport direction and keeps quantum features in the confined one. Show more
Keywords: Schrödinger–Poisson equation, open boundary conditions, reflection–transmission-coefficients, quantum/classical subband model
Citation: Asymptotic Analysis, vol. 49, no. 3,4, pp. 295-329, 2006
Authors: Monneau, R.
Article Type: Research Article
Abstract: We study how three-dimensional linearized elasticity for thin plates can be approximated by a two-dimensional projection. The classical approach using formal asymptotic expansions in powers of the thickness in the Hellinger–Reissner formulation, only provides error estimates in the H1 norm for the displacements, assuming at least L2 regularity for the applied forces, plus additional regularity for some components. Here we make use of elliptic regularity theory. We prove a 3d–2d interior error estimate between the 3d displacement solution and its 2d projection. Moreover the constants involved in our estimate are independent on the particular geometry of the plate. …Our approach yields an H2 error estimate, assuming only L2 regularity for the applied forces, which is optimal from the point of view of elliptic regularity theory. We also obtain interior Wk,p and Ck,α error estimates. Show more
Keywords: linear elasticity, plate theory, error estimate, Kirchhoff–Love theory, two-dimensional projection
Citation: Asymptotic Analysis, vol. 49, no. 3,4, pp. 331-344, 2006
Inspirees International (China Office)
Ciyunsi Beili 207(CapitaLand), Bld 1, 7-901 100025, Beijing China Free service line: 400 661 8717 Fax: +86 10 8446 7947 china@iospress.cn
For editorial issues, like the status of your submitted paper or proposals, write to editorial@iospress.nl
如果您在出版方面需要帮助或有任何建, 件至: editorial@iospress.nl
|
This question already has an answer here:
With this given information I am able to calculate $\phi$ limits for $P$ portion over $T$ using the following method:
$$ \begin{split} \tan \beta = z/y,\\ z=R\cos\theta, \end{split}\quad \iff \quad y={R\cos\theta\over\tan\beta} $$
$$ \begin{split} R\sin\theta\sin\phi={R\cos\theta\over\tan\beta}, \\ \sin\phi={1\over\tan\beta\tan\theta}, \end{split}\quad \iff \quad \phi=\arcsin{1\over\tan\beta\tan\theta}.$$ So the $\phi$ can go until $\pi -\phi$.
Now consider my $P$ is not a hemisphere: rather it is a spherical cap, where the cap size can be defined by an angle $\gamma$.
What will be my $\phi$ in this case to calculate the $P$ portion over the line $T$?
|
I am quite confused about the objective function of the bayesian model averaging in the paper "Bayesian Averaging of Classifiers and the overfitting Problem".1
On the section 2, here is the first equation:
Let $n$ be the training set size, $\mathbf{x}$ examples in the training set, $\mathbf{c}$ the corresponding class labels and $h$ a model (or hypothesis) in the model space $H$. Then, by Bayes’ theorem, and assuming the examples are drawn independently, the posterior probability of $h$ given $(\mathbf{x}, \mathbf{c})$ is given by:
$Pr(h|\mathbf{x}, \mathbf{c})=\frac{Pr(h)}{Pr(\mathbf{x}, \mathbf{c})}\prod_{i=1}^{n}Pr(\mathbf{x_i},\mathbf{c_i}|h)$ (1)
where $Pr(h)$ is the prior probability of $h$, and the product of $Pr(\mathbf{x_i},\mathbf{c_i}|h)$ terms is the likelihood.
I could understand that the Eq(1) uses conditional independence.
In order to compute the likelihood it is necessary to compute the probability of a class label $\mathbf{c_i}$ given an unlabeled example $\mathbf{x_i}$ and a hypothesis $h$, since $Pr(\mathbf{x_i}, \mathbf{c_i}|h) = Pr(\mathbf{x_i}|h)Pr(\mathbf{c_i}|\mathbf{x_i}, h)$. This probability, $Pr(\mathbf{c_i}|\mathbf{x_i}, h)$, can be called the noise model, and is distinct from the classification model $h$, which simply produces a class prediction with no probabilities attached.
I can understand the above as well.
Then
Finally, an unseen example $x$ is assigned to the class that maximizes: $Pr(c|x,\mathbf{x},\mathbf{c}, H)=\sum_{h\in H}Pr(c|x,h)Pr(h|\mathbf{x},\mathbf{c})$ (4)
I have two questions:
I don't understand how to deduce the Eq(4); In those euqations of Bayesian Model Averaging, which are variables? I don't understand how to train it.
Thank you in advance.
1 Domingos, P., (2000) "Bayesian Averaging of Classifiers and the Overfitting Problem" Proceedings of the Seventeenth International Conference on Machine Learning, pp.223-230
|
Division Ring Norm is Continuous on Induced Metric Space
Jump to navigation Jump to search
Theorem
Let $\struct {R, \norm {\,\cdot\,}}$ be a normed division ring.
Proof
Let $x_0 \in R$.
Let $\epsilon \in \R_{\gt 0}$.
Let $x \in R: \norm {x - x_0} \lt \epsilon$.
Then:
\(\displaystyle \size {\norm {x} - \norm {x_0} }\) \(\le\) \(\displaystyle \norm {x - x_0}\) Reverse triangle inequality \(\displaystyle \) \(\lt\) \(\displaystyle \epsilon\)
$\blacksquare$
|
The set of integers $\Bbb Z$ under ordinary addition is cyclic. Both $1$ and $-1$ are generators. But I am a bit confused how can $1$ generate $0$ and how $-1$ generates $0$? What is the order of $1$ and $-1$ on this group of integers?
The subgroup generated by a set of elements of a group is the smallest subgroup that contains all the elements. A group by definition always includes the identity element.
The order of an element is the size of the subgroup generated by it. Both $1$ and $-1$ generate all of $(\mathbb Z,+)$ which has infinitely many elements, thus the order of them is infinite (as there are infinitely elements in the group).
Note that for any element $n$ of $\mathbb Z$, the generated subgroup consists of all multiples of $n$, which for non-zero $n$ consists of infinitely many elements as well, thus every non-zero integer has infinite order.
On the other hand, one can easily check that $\{0\}$ is a subgroup all be itself, called the trivial subgroup, and therefore this is the group generated by $0$ (note that again, it consists of all multiples of $0$, since $0n=0$ for all $n\in\mathbb Z$). Since the trivial subgroup only contains one element (namely $0$), the order of $0$ is $1$.
Indeed, it is easy to see that for
any group $G$, its identity element generates the trivial subgroup and therefore is of order $1$. Also, the group of any non-identity element $g$ contains at least two elements (namely the identity and $g$ itself), and therefore is of order $\ge 2$.
As a side note, even if we ignore that $0$ is by definition in the subgroup of $(\mathbb Z,+)$ generated by $1$, we can easily get it through the supported operations: Since $1$ is in the set, and the set is closed under negation, $-1$ is also in the set. And since the set is also closed under addition, $1+(-1)=0$ is also in the set. However while this works for groups, the same idea fails for other algebraic structures like monoids where you also have an identity, but generally no inverse; an example of this would be $(\mathbb Z_{>0},\times)$.
Therefore the right way to think of it is that the neutral element is already there by definition, as this works for all algebraic structures.
In $(\mathbb Z,+)$ there is no positive integer $m$ such that $m \times 1 =0$. We say that the order of $1$ (and of every other integer apart from $0$) is infinite. The order of $0$ is $1$.
gandalf61's answer is technically correct. But let me try to offer a more intuitive explanation.
You can think of the group $G =(\mathbb{Z}, +)$ in terms of functions which permutes integers, i.e, functions of the type $\mathbb{Z} \to \mathbb{Z}$. For example, think of $7_G$ as the function that adds $7$ to every integer, i.e, $7_G(x) = 7 + x$. And when you are composing two elements of this group, what you are really doing is function composition. For example, $7_G + 2_G = 9_G$ is justified since for any $x \in \mathbb{Z}$, $7_G(2_G(x)) = 7 + (2 + x) = 9 + x = 9_G(x)$. Similarly, negatives correspond to function inverses, and so on. (This perspective is called group actions)
Now, there is an elelement $0_G$, which represents the identity function of this type. This function fixes every element of $\mathbb{Z}$, in other words, it does nothing.
We could notice that if we had the function $1_G$ (and it's inverse), we could repeatedly compose them (sarting from $0_G$) to get any desired element of this group of functions. To get $0_G$ out of the generators is trivial, since you do not have to apply it any number of times at all.
This is the idea that if you apply a function $0$ number of times, you have not changed the argument at all and hence you get the identity function. (Put more complicatedly, the $\text{id}$ morphism is the identity in the monoid of morphisms.) This is the same spirit as you get $1$ if you multiply something $0$ number of times in a ring, i.e, $a^0 = 1$.
|
Ian Miller's answer is the nicest and most efficient solution to the problem.
Just for your curiosity, I shall give you another one using Taylor series since you will use them a lot during your studies.
First, changing variable $x=\frac \pi 2+y$ $$\frac{\sqrt[4]{ \sin (x)} - \sqrt[3]{ \sin (x)}}{\cos^2(x)}=\frac{\sqrt[4]{ \cos(y)} - \sqrt[3]{ \cos (y)}}{\sin^2(y)}$$ Now, using the generalized binomial expansion and Taylor series around $y=0$ $$cos^a(y)=1-\frac{a y^2}{2}+\left(\frac{a^2}{8}-\frac{a}{12}\right) y^4+O\left(y^6\right)$$ Using it, the numerator write $$\frac{y^2}{24}+\frac{y^4}{1152}+O\left(y^6\right)$$ and the denominator is $\approx y^2$. Then, the limit.
We could go further and use $$\sin(y)=y-\frac{y^3}{6}+O\left(y^4\right)$$ So the denominator is $$y^2-\frac{y^4}{3}+O\left(y^5\right)$$ which makes $$\frac{\sqrt[4]{ \cos(y)} - \sqrt[3]{ \cos (y)}}{\sin^2(y)}=\frac{\frac{y^2}{24}+\frac{y^4}{1152}+O\left(y^6\right) }{y^2-\frac{y^4}{3}+O\left(y^5\right) }$$ and the long division yields to $$\frac{\sqrt[4]{ \cos(y)} - \sqrt[3]{ \cos (y)}}{\sin^2(y)}=\frac{1}{24}+\frac{17 }{1152}y^2+O\left(y^3\right)$$ which shows the limit and also how it is approached.
|
If $f \in R(\alpha)$ on $[a,b]$ and if for every monotonic function $f : $
$$\int_a ^b f~ d \alpha = 0 $$
then, prove that $\alpha$ must be constant on $[a,b]$
Proof:
By integration by parts :
$\int_a ^b f~ d \alpha + \int_a ^b \alpha~ df = f(b) \alpha (b) -f(a) \alpha (a) $ . Substituting $\int_a ^b f~ d \alpha = 0 $ , we get :
$\int_a ^b \alpha~ df = f(b) \alpha (b) -f(a) \alpha (a) $
Given any point $c \in [a, b)$, we may choose a monotonic function $f$ defined as follows :
$f(x) = \begin{cases} 0 & x \leq c \\ 1 & x > c \end{cases} $
*So, we have $\int_a ^b \alpha ~df= \alpha(c) = \alpha(b)$, that is, $ \alpha$ is constant in $[a,b]$
I don't understand why choosing that function yields the result I marked with *
|
Erin Carmody successfully defended her dissertation under my supervision at the CUNY Graduate Center on April 24, 2015, and she earned her Ph.D. degree in May, 2015. Her dissertation follows the theme of
killing them softly, proving many theorems of the form: given $\kappa$ with large cardinal property $A$, there is a forcing extension in which $\kappa$ no longer has property $A$, but still has large cardinal property $B$, which is very slightly weaker than $A$. Thus, she aims to enact very precise reductions in large cardinal strength of a given cardinal or class of large cardinals. In addition, as a part of the project, she developed transfinite meta-ordinal extensions of the degrees of hyper-inaccessibility and hyper-Mahloness, giving notions such as $(\Omega^{\omega^2+5}+\Omega^3\cdot\omega_1^2+\Omega+2)$-inaccessible among others.
Erin has accepted a professorship at Nebreska Wesleyan University for.the 2015-16 academic year.
Erin is also an accomplished artist, who has had art shows of her work in New York, and she has pieces for sale. Much of her work has an abstract or mathematical aspect, while some pieces exhibit a more emotional or personal nature. My wife and I have two of Erin’s paintings in our collection:
|
$\{s_n\}$ is defined by $$s_1 = 0; s_{2m}=\frac{s_{2m-1}}{2}; s_{2m+1}= {1\over 2} + s_{2m}$$
The following is what I tried to do.
The sequence is $$\{0,0,\frac{1}{2},\frac{1}{4},\frac{3}{4},\frac{3}{8},\frac{7}{8},\frac{7}{16},\cdots \}$$
So the even terms $\{E_i\} = 1 - 2^{-i}$ and the odd terms $\{O_k\} = \frac{1}{2} - 2^{-k}$ and each of them has a limit of $1$ and $\frac{1}{2}$, respectively.
So, the upper limit is $1$ and the lower limit is $1\over 2$, am I right ?
Does this also mean that $\{s_n\}$ has no limits ?
Is my denotation $$\lim_{n \to \infty} \sup(s_n)=1 ,\lim_{n \to \infty} \inf(s_n)={1 \over 2} $$ correct ?
|
$\cot{(A+B)}$ $\,=\,$ $\dfrac{\cot{B}\cot{A}\,–\,1}{\cot{B}+\cot{A}}$
$\dfrac{\cot{B}\cot{A}\,–\,1}{\cot{B}+\cot{A}}$ $\,=\,$ $\cot{(A+B)}$
The cot of angle sum identity is called as cot of sum of two angles identity or cot of compound angle identity. It is usually used in two cases in mathematics.
The cot of angle sum identity is written in mathematical form in several ways in which $\cot{(A+B)}$, $\cot{(x+y)}$ and $\cot{(\alpha+\beta)}$ are popular in the world. You can even write cot of sum of two angles formula in terms of any two angles.
$(1) \,\,\,\,\,\,$ $\cot{(A+B)}$ $\,=\,$ $\dfrac{\cot{B}\cot{A}\,–\,1}{\cot{B}+\cot{A}}$
$(2) \,\,\,\,\,\,$ $\cot{(x+y)}$ $\,=\,$ $\dfrac{\cot{y}\cot{x}\,–\,1}{\cot{y}+\cot{x}}$
$(3) \,\,\,\,\,\,$ $\cot{(\alpha+\beta)}$ $\,=\,$ $\dfrac{\cot{\beta}\cot{\alpha}\,–\,1}{\cot{\beta}+\cot{\alpha}}$
You learned how to expand cot of sum of two angles function and also learned how to write its expansion in simplified form. It is your time to learn how to derive cot of angle sum identity in trigonometric mathematics.
Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.
|
Basically 2 strings, $a>b$, which go into the first box and do division to output $b,r$ such that $a = bq + r$ and $r<b$, then you have to check for $r=0$ which returns $b$ if we are done, otherwise inputs $r,q$ into the division box..
There was a guy at my university who was convinced he had proven the Collatz Conjecture even tho several lecturers had told him otherwise, and he sent his paper (written on Microsoft Word) to some journal citing the names of various lecturers at the university
Here is one part of the Peter-Weyl theorem: Let $\rho$ be a unitary representation of a compact $G$ on a complex Hilbert space $H$. Then $H$ splits into an orthogonal direct sum of irreducible finite-dimensional unitary representations of $G$.
What exactly does it mean for $\rho$ to split into finite dimensional unitary representations? Does it mean that $\rho = \oplus_{i \in I} \rho_i$, where $\rho_i$ is a finite dimensional unitary representation?
Sometimes my hint to my students used to be: "Hint: You're making this way too hard." Sometimes you overthink. Other times it's a truly challenging result and it takes a while to discover the right approach.
Once the $x$ is in there, you must put the $dx$ ... or else, nine chances out of ten, you'll mess up integrals by substitution. Indeed, if you read my blue book, you discover that it really only makes sense to integrate forms in the first place :P
Using the recursive definition of the determinant (cofactors), and letting $\operatorname{det}(A) = \sum_{j=1}^n \operatorname{cof}_{1j} A$, how do I prove that the determinant is independent of the choice of the line?
Let $M$ and $N$ be $\mathbb{Z}$-module and $H$ be a subset of $N$. Is it possible that $M \otimes_\mathbb{Z} H$ to be a submodule of $M\otimes_\mathbb{Z} N$ even if $H$ is not a subgroup of $N$ but $M\otimes_\mathbb{Z} H$ is additive subgroup of $M\otimes_\mathbb{Z} N$ and $rt \in M\otimes_\mathbb{Z} H$ for all $r\in\mathbb{Z}$ and $t \in M\otimes_\mathbb{Z} H$?
Well, assuming that the paper is all correct (or at least to a reasonable point). I guess what I'm asking would really be "how much does 'motivated by real world application' affect whether people would be interested in the contents of the paper?"
@Rithaniel $2 + 2 = 4$ is a true statement. Would you publish that in a paper? Maybe... On the surface it seems dumb, but if you can convince me the proof is actually hard... then maybe I would reconsider.
Although not the only route, can you tell me something contrary to what I expect?
It's a formula. There's no question of well-definedness.
I'm making the claim that there's a unique function with the 4 multilinear properties. If you prove that your formula satisfies those (with any row), then it follows that they all give the same answer.
It's old-fashioned, but I've used Ahlfors. I tried Stein/Stakarchi and disliked it a lot. I was going to use Gamelin's book, but I ended up with cancer and didn't teach the grad complex course that time.
Lang's book actually has some good things. I like things in Narasimhan's book, but it's pretty sophisticated.
You define the residue to be $1/(2\pi i)$ times the integral around any suitably small smooth curve around the singularity. Of course, then you can calculate $\text{res}_0\big(\sum a_nz^n\,dz\big) = a_{-1}$ and check this is independent of coordinate system.
@A.Hendry: It looks pretty sophisticated, so I don't know the answer(s) off-hand. The things on $u$ at endpoints look like the dual boundary conditions. I vaguely remember this from teaching the material 30+ years ago.
@Eric: If you go eastward, we'll never cook! :(
I'm also making a spinach soufflé tomorrow — I don't think I've done that in 30+ years. Crazy ridiculous.
@TedShifrin Thanks for the help! Dual boundary conditions, eh? I'll look that up. I'm mostly concerned about $u(a)=0$ in the term $u'/u$ appearing in $h'-\frac{u'}{u}h$ (and also for $w=-\frac{u'}{u}P$)
@TedShifrin It seems to me like $u$ can't be zero, or else $w$ would be infinite.
@TedShifrin I know the Jacobi accessory equation is a type of Sturm-Liouville problem, from which Fox demonstrates in his book that $u$ and $u'$ cannot simultaneously be zero, but that doesn't stop $w$ from blowing up when $u(a)=0$ in the denominator
|
Ok, suppose $M,N$ are Riemannian manifolds and $F:M\to N$ is a smooth map between them. In a book I have here they consider that $\dim M=m \geq \dim N=n$ and that $x\in M$ is a regular point such that the derivative $DF(x):T_x M\to T_{F(x)}N$ is surjective. With this conditions, note that $\ker DF(X)^\perp$ is isomorphic to $T_{F(x)}N$.
So far so good, the problem comes when they start to talk about the determinant of $DF(x)$. As far as I know, we talk about determinant in the context of matrices. So here started my trouble, for $DF(x)$ is a linear map between abstract spaces, there is no obvious way to make $DF(x)$ into a matrix.
After a search on Google, I got these two definitions, which I wanted to share here to be sure they are valid. For the first one, fortunately, I had some knowledge of differential forms, so I didn't got so confused when seeing it.
Definition 1: given any base $(v_1,\ldots,v_n)$ of $\ker DF(x)^\perp$, we define $$\det (DF(x)) = \frac{\omega(DF(x)v_1,\ldots,DF(x)v_n)}{\mu(v_1,\ldots,v_n)},$$where $\omega$ is the volume form on $N$ and $\mu$ is the volume form on $M$.
The second definition was told me by a friend and I couldn't find something useful on the Internet about it.
Definition 2: let $(\varphi, U)$ be a orthonormal coordinate system on $x$ and $(\psi, V)$ a orthonormal coordinate system on $F(x)$. Then $\det (DF(x)) = \det \left(D\Phi(\varphi^{-1}(x))\right)$, where $\Phi=\psi^{-1}\circ F\circ\varphi$.
This second definition also makes sense, assuming that the determinant doesn't depend on the charts chosen. In fact, this second definition is better, because it reduces the problem to the computation of a determinant of some matrix, which is something familiar. Just to be clear, $\varphi$ is a map from $U\subset \mathbb{R}^m$ to $M$ and $\psi$ is a map from $V\subset \mathbb{R}^n$ to $N$.
I have more than one question, some of them probably you will find easy to answer (assuming you are used to Riemannian Geometry).
1) This is a terminology question. Instead saying $(\varphi,U)$ is an orthonormal coordinate system, could I say it is an orthonormal chart ? I'm asking this because looks like it's common to use the term charts, except when you are gonna say it is orthogonal or orthonormal. In this case I found most of people prefer to use it with the term coordinate system. 2) From my reading, $(\varphi,U)$ is an orthogonal (orthonormal) coordinate system (or chart?) when $\left(\frac{\partial}{\partial x_1}|_x,\ldots,\frac{\partial}{\partial x_m}|_x\right)$ is an orthogonal (orthonormal) base for $T_xM$. Is my understanding correct? 3) The definitions 1 and 2 are really equivalent? I got a little suspicious about using "orthonormal" in definition 2. Maybe it could be "orthogonal", I don't know. PS: questions 1 and 2 are important but are technical details necessary to ask question 3. The last question is the most important of all.
|
Answer: The three denominations are $65$, $72$ and $97$.How did I detect this answer?I searched the list of primitive Pythagorean triples at [link to triple list], while using the Frobenius applet available at [applet link].Based on my search I know that the answer to the puzzle is unique; still, I would like to see a clean mathematical argument for this ...
None of the numbers less than 134 can be obtained with this set of stamps.All of the numbers in the range 134-143 can be obtained with a single stamp.None of the numbers between 143 and 2 * 134 can be obtained with a combination of these stamps.More generally, numbers that lie in the range between (n-1)143 and n(134) cannot be obtained.Because the stamps ...
The general picture here is as follows: if you have two positive integers $m,n$ with no common factor then every integer bigger than $mn-m-n$ can be written as $am+bn$ with $a,b$ non-negative integers, but $mn-m-n$ itself can't.Proof:First, suppose $mn-m-n = am+bn$. Then $(b+1)n$ is a multiple of $m$, and therefore (since $m,n$ have no common factor) so ...
This problem is an instance of the Frobenius coin problem. In fact, the coin values form an arithmetic sequence, whose Frobenius number has the special form:$$g(a, a+d, \ldots, a+sd) = \left(\left\lfloor\frac{a-2}{s}\right\rfloor+1\right)a + (d-1)(a-1)-1$$Here we have:$$\begin{align}g(999, 1000, 1001) &= \\ g(999, 999+1, 999+2\times 1) &= \...
I think it is-The following facts hold-There can be at most one 1000-coin. Otherwise we replace 2 1000-coins with (999+1001)There can be exactly one type of coin from 999 and 1001, otherwise we replace two of them with 2 1000-coins.There can be at most 500, 999-coins. Other wise we can replace 501, 999-coins with 499, 1001-coins and one 1000 coin....
with 5-cent stamps and 17-cent stamps you can find any number that ends with 4 and 9 with the equation below after 39 (considering you have at least one 5-cent and one 17-cent stamp);$5n+34\ where\ n>0$Similarly, you can find any number that ends with 2,7,1,6,3,8,0,5 after the numbers stated in the equations below:$5n+17\ where\ n\geqslant 0$ after ...
Consider the following equivalent representations:$$\begin{align}1000 + 1000 &= 999 + 1001 \\499 \times 1001 + 1000 &= 501 \times 999 \\500 \times 999 + 1000 &= 500\times 1001\end{align}$$The first shows that a unique representation can have no more than one $1000$, and cannot contain both $999$ and $1001$.The second shows that a ...
Some conditions for a representation $999d_1 + 1000d_2 + 1001d_3 = N$ not to be unique.If $d_2 \ge 2$, then $(d_1+1, d_2-2, d_3+1)$ is also a valid representation of $N$. For $N$ to have a unique representation, we must have $d_2 = 0$ or $1$.If both $d_1 \ge 1$ and $d_2 \ge 1$, then $(d_1-1, d_2+2, d_3-1)$ is a valid representation of $N$, so at least ...
The amounts can be written as 999$x$ + 1000$y$ + 1001$z$. If we want to add 1 to it:We can increase $y$ and decrease $x$ by 1.We can increase $z$ and decrease $y$ by 1.If neither can be done ($x$ and $y$ = 0), we can increase $x$ by 500 and decrease $z$ by 499 at the very least.So, 1001*498=498498 can't rise to 498499 and any amount starting from ...
I solved this by considering modulus.Once we can make a number x from the stamps, we can then make any x + 5y by adding y more 5-cent stamps. So, once we can make a number x for each modulus of 5, we can make all the remaining numbers.So, we can make a chart of n, where n is the number of 17-cent stamps, show its value in cents (n*17), and show which ...
I'm not sure if this even qualifies as a puzzle; it's more of a math problem. There's a simple formula given on Wikipedia for finding the Frobenius number for a given arithmetic sequence.We can now substitute the values $a=134$, $d=1$ and $s=9$. Using it, we get the answer 2009 as already given by @IvoBeckers
|
Difference between revisions of "Inertia"
(→Derivation)
(→Derivation)
Line 46: Line 46:
:* In physics, the moment of inertia <math>J</math> is normally denoted as <math>I</math>. In electrical engineering, the convention is for the letter "i" to always be reserved for current, and is therefore often replaced by the letter "j", e.g. the complex number operator i in mathematics is j in electrical engineering.
:* In physics, the moment of inertia <math>J</math> is normally denoted as <math>I</math>. In electrical engineering, the convention is for the letter "i" to always be reserved for current, and is therefore often replaced by the letter "j", e.g. the complex number operator i in mathematics is j in electrical engineering.
−
:* Moment of inertia
+
:* Moment of inertia also as <math>WR^{2}</math> or <math>WK^{2}</math>, where <math>WK^{2} = \frac{1}{2} WR^{2}</math>. WR<sup>2</sup> literally stands for weight x radius squared.
−
:* WR<sup>2</sup> is often used with imperial units of lb.ft<sup>2</sup>
+
:* WR<sup>2</sup> is often used with imperial units of lb.ft
+ +
<sup>2</sup>
== Normalised Inertia Constants ==
== Normalised Inertia Constants ==
Revision as of 06:29, 27 August 2018
In power systems engineering, "inertia" is a concept that typically refers to rotational inertia or rotational kinetic energy. For synchronous systems that run at some nominal frequency (i.e. 50Hz or 60Hz), inertia is the energy that is stored in the rotating masses of equipment electro-mechanically coupled to the system, e.g. generator rotors, fly wheels, turbine shafts.
Contents Derivation
Below is a basic derivation of power system rotational inertia from first principles, starting from the basics of circle geometry and ending at the definition of moment of inertia (and it's relationship to kinetic energy).
The length of a circle arc is given by:
[math] L = \theta r [/math]
where [math]L[/math] is the length of the arc (m)
[math]\theta[/math] is the angle of the arc (radians) [math]r[/math] is the radius of the circle (m)
A cylindrical body rotating about the axis of its centre of mass therefore has a rotational velocity of:
[math] v = \frac{\theta r}{t} [/math]
where [math]v[/math] is the rotational velocity (m/s)
[math]t[/math] is the time it takes for the mass to rotate L metres (s)
Alternatively, rotational velocity can be expressed as:
[math] v = \omega r [/math]
where [math]\omega = \frac{\theta}{t} = \frac{2 \pi \times n}{60}[/math] is the angular velocity (rad/s)
[math]n[/math] is the speed in revolutions per minute (rpm)
The kinetic energy of a circular rotating mass can be derived from the classical Newtonian expression for the kinetic energy of rigid bodies:
[math] KE = \frac{1}{2} mv^{2} = \frac{1}{2} m(\omega r)^{2}[/math]
where [math]KE[/math] is the rotational kinetic energy (Joules or kg.m
2/s 2 or MW.s, all of which are equivalent) [math]m[/math] is the mass of the rotating body (kg)
Alternatively, rotational kinetic energy can be expressed as:
[math] KE = \frac{1}{2} J\omega^{2} [/math]
where [math]J = mr^{2}[/math] is called the
moment of inertia (kg.m 2).
Notes about the moment of inertia:
In physics, the moment of inertia [math]J[/math] is normally denoted as [math]I[/math]. In electrical engineering, the convention is for the letter "i" to always be reserved for current, and is therefore often replaced by the letter "j", e.g. the complex number operator i in mathematics is j in electrical engineering. Moment of inertia is also referred to as [math]WR^{2}[/math] or [math]WK^{2}[/math], where [math]WK^{2} = \frac{1}{2} WR^{2}[/math]. WR 2literally stands for weight x radius squared. Moment of inertia is also referred to as [math]WR^{2}[/math] or [math]WK^{2}[/math], where [math]WK^{2} = \frac{1}{2} WR^{2}[/math]. WR WR 2is often used with imperial units of lb.ft 2or slug.ft 2. Conversions factors: 1 lb.ft 2= 0.04214 kg.m 2 1 slug.ft 2= 1.356 kg.m 2 1 lb.ft WR Normalised Inertia Constants
The moment of inertia can be expressed as a normalised quantity called the
inertia constant H, calculated as the ratio of the rotational kinetic energy of the machine at nominal speed to its rated power (VA): [math]H = \frac{1}{2} \frac{J \omega_0^{2}}{S_{b}}[/math]
where [math]H[/math] is the inertia constant (s)
[math]\omega_{0} = 2 \pi \times \frac{n}{60}[/math] is the nominal mechanical angular frequency (rad/s) [math]n[/math] is the nominal speed of the machine (revolutions per minute) [math]S_{b}[/math] is the rated power of the machine (VA) Generator Inertia
The moment of inertia for a generator is dependent on its mass and apparent radius, which in turn is largely driven by its prime mover type.
Based on actual generator data, the normalised inertia constants for different types and sizes of generators are summarised in the table below:
Machine type Number of samples MVA Rating Inertia constant H Min Median Max Min Median Max Steam turbine 45 28.6 389 904 2.1 3.2 5.7 Gas turbine 47 22.5 99.5 588 1.9 5.0 8.9 Hydro turbine 22 13.3 46.8 312.5 2.4 3.7 6.8 Combustion engine 26 0.3 1.25 2.5 0.6 0.95 1.6 Relationship between Inertia and Frequency
TBA
|
OpenCV 3.1.0
Open Source Computer Vision
In this tutorial you will learn how to:
Principal Component Analysis (PCA) is a statistical procedure that extracts the most important features of a dataset.
Consider that you have a set of 2D points as it is shown in the figure above. Each dimension corresponds to a feature you are interested in. Here some could argue that the points are set in a random order. However, if you have a better look you will see that there is a linear pattern (indicated by the blue line) which is hard to dismiss. A key point of PCA is the Dimensionality Reduction. Dimensionality Reduction is the process of reducing the number of the dimensions of the given dataset. For example, in the above case it is possible to approximate the set of points to a single line and therefore, reduce the dimensionality of the given points from 2D to 1D.
Moreover, you could also see that the points vary the most along the blue line, more than they vary along the Feature 1 or Feature 2 axes. This means that if you know the position of a point along the blue line you have more information about the point than if you only knew where it was on Feature 1 axis or Feature 2 axis.
Hence, PCA allows us to find the direction along which our data varies the most. In fact, the result of running PCA on the set of points in the diagram consist of 2 vectors called
eigenvectors which are the principal components of the data set.
The size of each eigenvector is encoded in the corresponding eigenvalue and indicates how much the data vary along the principal component. The beginning of the eigenvectors is the center of all points in the data set. Applying PCA to N-dimensional data set yields N N-dimensional eigenvectors, N eigenvalues and 1 N-dimensional center point. Enough theory, let’s see how we can put these ideas into code.
The goal is to transform a given data set
X of dimension p to an alternative data set Y of smaller dimension L. Equivalently, we are seeking to find the matrix Y, where Y is the Karhunen–Loève transform (KLT) of matrix X:
\[ \mathbf{Y} = \mathbb{K} \mathbb{L} \mathbb{T} \{\mathbf{X}\} \]
Organize the data set
Suppose you have data comprising a set of observations of
p variables, and you want to reduce the data so that each observation can be described with only L variables, L < p. Suppose further, that the data are arranged as a set of n data vectors \( x_1...x_n \) with each \( x_i \) representing a single grouped observation of the p variables. Calculate the empirical mean
Place the calculated mean values into an empirical mean vector
u of dimensions \( p\times 1 \).
\[ \mathbf{u[j]} = \frac{1}{n}\sum_{i=1}^{n}\mathbf{X[i,j]} \]
Calculate the deviations from the mean
Mean subtraction is an integral part of the solution towards finding a principal component basis that minimizes the mean square error of approximating the data. Hence, we proceed by centering the data as follows:
Store mean-subtracted data in the \( n\times p \) matrix
B.
\[ \mathbf{B} = \mathbf{X} - \mathbf{h}\mathbf{u^{T}} \]
where
h is an \( n\times 1 \) column vector of all 1s:
\[ h[i] = 1, i = 1, ..., n \]
Find the covariance matrix
Find the \( p\times p \) empirical covariance matrix
C from the outer product of matrix B with itself:
\[ \mathbf{C} = \frac{1}{n-1} \mathbf{B^{*}} \cdot \mathbf{B} \]
where * is the conjugate transpose operator. Note that if B consists entirely of real numbers, which is the case in many applications, the "conjugate transpose" is the same as the regular transpose.
Find the eigenvectors and eigenvalues of the covariance matrix
Compute the matrix
V of eigenvectors which diagonalizes the covariance matrix C:
\[ \mathbf{V^{-1}} \mathbf{C} \mathbf{V} = \mathbf{D} \]
where
D is the diagonal matrix of eigenvalues of C.
Matrix
D will take the form of an \( p \times p \) diagonal matrix:
\[ D[k,l] = \left\{\begin{matrix} \lambda_k, k = l \\ 0, k \neq l \end{matrix}\right. \]
here, \( \lambda_j \) is the
j-th eigenvalue of the covariance matrix C
This tutorial code's is shown lines below. You can also download it from here.
Read image and convert it to binary
Here we apply the necessary pre-processing procedures in order to be able to detect the objects of interest.
Extract objects of interest
Then find and filter contours by size and obtain the orientation of the remaining ones.
Extract orientation
Orientation is extracted by the call of getOrientation() function, which performs all the PCA procedure.
First the data need to be arranged in a matrix with size n x 2, where n is the number of data points we have. Then we can perform that PCA analysis. The calculated mean (i.e. center of mass) is stored in the
cntr variable and the eigenvectors and eigenvalues are stored in the corresponding std::vector’s. Visualize result
The final result is visualized through the drawAxis() function, where the principal components are drawn in lines, and each eigenvector is multiplied by its eigenvalue and translated to the mean position.
The code opens an image, finds the orientation of the detected objects of interest and then visualizes the result by drawing the contours of the detected objects of interest, the center point, and the x-axis, y-axis regarding the extracted orientation.
|
I'm using
chemfig (perhaps improperly) for relational algebra graphs. Here, for example, is a simple one:
\documentclass{article}\usepackage[italian]{babel}\usepackage[utf8]{inputenc}\usepackage[T1]{fontenc}\usepackage{chemfig}\usepackage{amsmath}\usepackage{amssymb}\usepackage{relsize}\newcommand{\select}{\sigma}\newcommand{\project}{\pi}\newcommand{\join}{\mathlarger{\mathlarger{\mathlarger{\mathlarger{\Join}}}}}\begin{document} \chemfig{R-[:30]\join(-[:90]\project_{CodR,NomeR})-[:330] \select_{Argomento='moto'}-[:270]A}\end{document}
It results in:
I'd like to make the atoms containing
\select and
\project "right-aligned", so that the bond would touch the sigma and the pi. I didn't find a solution in the chemfig documentation.
If what I want is not possible with
chemfig, are there viable alternatives?
EDIT: Maybe there is a way by redefining
\printatom? I found an example in the docs in which it shows how to change font, so maybe it's possible to make a custom box.
|
Suppose we have IID random variables $X_1,\dots,X_n$ with distribution $\mathrm{Ber}(\theta)$. We are going to observe a sample of the $X_i$'s in the following way: let $Y_1,\dots,Y_n$ be independent $\mathrm{Ber}(1/2)$ random variables, suppose that all the $X_i$'s and $Y_i$'s are independent, and define the sample size $N=\sum_{i=1}^n Y_i$. The $Y_i$'s indicate which of the $X_i$'s are in the sample, and we want to study the fraction of successes in the sample defined by $$ Z = \begin{cases} \frac{1}{N}\sum_{i=1}^n X_i Y_i & \text{if}\quad N > 0\, , \\ 0 & \text{if} \quad N = 0 \, . \end{cases} $$ For $\epsilon>0$, we want to find an upper bound for $\mathrm{Pr}\!\left(Z \geq \theta + \epsilon\right)$ that decays exponentially with $n$. Hoeffding's inequality doesn't apply immediately because of the dependencies among the variables.
We can draw a connection to Hoeffding's inequality in a fairly direct way.
Note that we have $$ \{ Z > \theta + \epsilon\} = \big\{\sum_i X_i Y_i > (\theta + \epsilon)\sum_i Y_i \big\} = \big\{ \sum_i (X_i - \theta - \epsilon) Y_i > 0 \} \>. $$
Set $Z_i = (X_i - \theta - \epsilon)Y_i + \epsilon/2$ so that the $Z_i$ are iid, $\mathbb E Z_i = 0$ and $$ \mathbb P( Z > \theta + \epsilon ) = \mathbb P\big(\sum_i Z_i > n \epsilon/2\big) \leq e^{-n \epsilon^2/2}\>, $$ by a straightforward application of Hoeffding's inequality (since the $Z_i \in [-\theta-\epsilon/2,1-\theta-\epsilon/2]$ and so take values in an interval of size one).
There is a rich and fascinating related literature that has built up over the last several years, in particular, on topics related to random matrix theory with various practical applications. If you are interested in this sort of thing, I highly recommend:
R. Vershynin,
Introduction to the non-asymptotic analysis of random matrices, Chapter 5 of Compressed Sensing, Theory and Applications. Edited by Y. Eldar and G. Kutyniok. Cambridge University Press, 2012.
I think the exposition is clear and provides a very nice way to get quickly acclimated to the literature.
Details to take care of the $N=0$ case.$$\begin{align} \{Z\geq\theta+\epsilon\} &= \left(\{Z\geq\theta+\epsilon\} \cap \{N=0\}\right) \cup \left(\{Z\geq\theta+\epsilon\} \cap \{N>0\}\right) \\&= \left(\{0\geq\theta+\epsilon\} \cap \{N=0\}\right) \cup \left(\{Z\geq\theta+\epsilon\} \cap \{N>0\}\right) \\&= \left(\emptyset \cap \{N=0\}\right) \cup \left(\{Z\geq\theta+\epsilon\} \cap \{N>0\}\right) \\&= \left\{\sum_{i=1}^n X_iY_i\geq(\theta+\epsilon)\sum_{i=1}^n Y_i\right\} \cap \{N>0\} \\&\subset \left\{\sum_{i=1}^n X_iY_i\geq(\theta+\epsilon)\sum_{i=1}^n Y_i\right\} \\&= \left\{\sum_{i=1}^n (X_i-\theta-\epsilon)Y_i\geq 0\right\} \\&= \left\{\sum_{i=1}^n \left((X_i-\theta-\epsilon)Y_i+\epsilon/2\right)\geq n\epsilon/2\right\} \, .\end{align}$$ For Alecos.$$\begin{align}\mathrm{E}\!\left[\sum_{i=1} ^n W_i\right]&=\mathrm{E}\!\left[I_{\{\sum_{i=1}^n Y_i=0\}}\sum_{i=1} ^n W_i\right] + \mathrm{E}\!\left[I_{\{\sum_{i=1}^n Y_i>0\}}\sum_{i=1} ^n W_i\right] \\&=\mathrm{E}\!\left[I_{\{\sum_{i=1}^n Y_i>0\}}\frac{\sum_{i=1} ^n Y_i}{\sum_{i=1}^n Y_i}\right]=\mathrm{E}\!\left[I_{\{\sum_{i=1}^n Y_i>0\}}\right]=1-1/2^n \, .\end{align}$$
This answer keeps mutating. The current version does not relate to the discussion I had with @cardinal in the comments (although it was through this discussion that I thankfully realized that the conditioning approach did not appear to lead anywhere).
For this attempt, I will use another part of Hoeffding's original 1963 paper, namely section 5 "Sums of Dependent Random Variables".
Set $$W_i \equiv \frac {Y_i}{\sum_{i=1}^nY_i}, \qquad \sum_{i=1}^nY_i \neq 0, \qquad \sum_{i=1}^nW_i=1, \qquad n\geq 2$$
while we set $W_i =0$ if $\sum_{i=1}^nY_i = 0$.
Then we have the variable
$$Z_n= \sum_{i=1}^nW_iX_i, \qquad E(Z_n) \equiv \mu_n$$
We are interested in the probability
$$\mathrm{Pr}(Z_n\geq \mu_n +\epsilon), \qquad \epsilon < 1-\mu_n$$
As for many other inequalities, Hoeffding starts his reasoning by noting that $$\mathrm{Pr}(Z_n\geq \mu_n +\epsilon) = E\left[\mathbf 1_{\{Z_n-\mu_n -\epsilon \geq 0\}}\right]$$ and that
$$\mathbf 1_{\{Z_n-\mu_n -\epsilon\geq 0\}} \leq \exp\Big\{h(Z_n-\mu_n -\epsilon)\Big\}, \qquad h>0$$
For the dependent-variables case, as Hoeffding we use the fact that $\sum_{i=1}^nW_i=1$ and invoke Jensen's inequality for the (convex) exponential function, to write
$$e^{hZ_n} = \exp\left\{h\left(\sum_{i=1}^nW_iX_i\right)\right\} \leq \sum_{i=1}^nW_ie^{hX_i}$$
and linking results to arrive at
$$\mathrm{Pr}(Z_n\geq \mu_n +\epsilon) \leq e^{-h(\mu_n+\epsilon)}E\left[\sum_{i=1}^nW_ie^{hX_i}\right]$$
Focusing on our case, since $W_i$ and $X_i$ are independent, expected values can be separated,
$$\mathrm{Pr}(Z_n\geq \mu_n +\epsilon) \leq e^{-h(\mu_n+\epsilon)}\sum_{i=1}^nE(W_i)E\left(e^{hX_i}\right)$$
In our case, the $X_i$ are i.i.d Bernoullis with parameter $\theta$, and $E[e^{hX_i}]$ is their common moment generating function in $h$, $E[e^{hX_i}] = 1-\theta +\theta e^h$. So
$$\mathrm{Pr}(Z_n\geq \mu_n +\epsilon) \leq e^{-h(\mu_n+\epsilon)}(1-\theta +\theta e^h)\sum_{i=1}^nE(W_i)$$
Minimizing the RHS with respect to $h$, we get
$$e^{h^*} = \frac {(1-\theta)(\mu_n+\epsilon)}{\theta(1-\mu_n-\epsilon)}$$
Plugging it into the inequality and manipulating we obtain
$$\mathrm{Pr}(Z_n\geq \mu_n +\epsilon) \leq \left(\frac {\theta}{\mu_n+\epsilon}\right)^{\mu_n+\epsilon}\cdot \left(\frac {1-\theta}{1-\mu_n-\epsilon}\right)^{1-\mu_n-\epsilon}\sum_{i=1}^nE(W_i)$$
while
$$\mathrm{Pr}(Z_n\geq \theta +\epsilon) \leq \left(\frac {\theta}{\theta+\epsilon}\right)^{\theta+\epsilon}\cdot \left(\frac {1-\theta}{1-\theta-\epsilon}\right)^{1-\theta-\epsilon}\sum_{i=1}^nE(W_i)$$
Hoeffding shows that
$$\left(\frac {\theta}{\theta+\epsilon}\right)^{\theta+\epsilon}\cdot \left(\frac {1-\theta}{1-\theta-\epsilon}\right)^{1-\theta-\epsilon} \leq e^{-2\epsilon^2}$$
Courtesy of the OP (thanks, I was getting a bit exhausted...) $$\sum_{i=1}^n E(W_i) =1-1/2^n$$
So, finally, the "dependent variables approach" gives us $$\mathrm{Pr}(Z_n\geq \theta +\epsilon) \leq (1-\frac 1{2^n})e^{-2\epsilon^2} \equiv B_D$$
Let's compare this to Cardinal's bound, that is based on an "independence" transformation, $B_I$. For our bound to be tighter, we need
$$B_D=(1-\frac 1{2^n})e^{-2\epsilon^2} \leq e^{-n\epsilon^2/2}=B_I$$
$$\Rightarrow \frac {2^n-1}{2^n} \leq \exp\left\{\left(\frac {4-n}{2}\right)\epsilon^2\right\}$$
So for $n\leq 4$ we have $B_D \leq B_I$. For $n \geq 5$, pretty quickly $B_I$ becomes tighter than $B_D$ but for very small $\epsilon$, while even this small "window" quickly converges to zero. For example, for $n=12$, if $\epsilon \geq 0.008$, then $B_I$ is tighter. So in all, Cardinal's bound is more useful.
COMMENT To avoid misleading impressions regarding Hoeffding's original paper, I have to mention that Hoeffding examines the case of a deterministic convex combination of dependent random variables. Specificaly, his $W_i$'s are numbers, not random variables, while each $X_i$ is a sum of independent random variables, while the dependency may exist between the $X_i$'s. He then considers various "U-statistics" that can be represented in this way.
|
A mathematical operation of dividing an algebraic term by its unlike term is called the division of unlike algebraic terms.
The division of any two unlike algebraic terms is written by displaying a division sign between them for calculating their quotient. The quotient of the unlike terms is actually calculated by eliminating the equivalent literal factors in literal coefficients and also calculating the numerical coefficients of them. The quotient of any two unlike algebraic terms is always an algebraic term.
$6xy$ and $3x^2$ are two unlike algebraic terms, and divide the term $6xy$ by $3x^2$.
Write the division of unlike algebraic terms in mathematical form.
$6xy \div 3x^2$ $\implies$ $\dfrac{6xy}{3x^2}$
Factorize the each algebraic term for eliminating the equivalent factors.
$\implies$ $\dfrac{6xy}{3x^2} \,=\, \dfrac{6 \times x \times y}{3 \times x^2}$ $\implies \dfrac{6xy}{3x^2} \,=\, \dfrac{6}{3} \times \dfrac{x}{x^2} \times y$
Find the quotient of the unlike terms by cancelling equivalent literal factors by the quotient rules of exponents, and then finding the quotient of the numerical coefficients.
$\implies$ $\dfrac{6xy}{3x^2} \,=\, \require{cancel} \dfrac{\cancel{6}}{\cancel{3}} \times \dfrac{\cancel{x}}{\cancel{x^2}} \times y$
$\implies$ $\dfrac{6xy}{3x^2} \,=\, 2 \times \dfrac{1}{x} \times y$ $\therefore \,\,\,\,\,\,$ $\dfrac{6xy}{3x^2} \,=\, \dfrac{2y}{x} \,\,$ (or) $\,\, 2x^{-1}y$
In this way, the division of any two unlike algebraic terms can be calculated in algebraic mathematics in three simple steps. The quotient of any two unlike terms is an algebraic temr.
Look at the following examples to learn how to divide an algebraic term by its unlike term.
$(1) \,\,\,\,\,\,$ $\dfrac{-b}{2a}$ $\,=\,$ $-\dfrac{1}{2}a^{-1}b$
$(2) \,\,\,\,\,\,$ $\dfrac{5cd^2}{cd}$ $\,=\,$ $\require{cancel} \dfrac{5 \times \cancel{c} \times \cancel{d^2}}{\cancel{c} \times \cancel{d}}$ $\,=\, 5d$
$(3) \,\,\,\,\,\,$ $\dfrac{14e}{7f^2}$ $\,=\,$ $\require{cancel} \dfrac{\cancel{14}}{\cancel{7}} \times \dfrac{e}{f^2}$ $\,=\,$ $2ef^{-2}$
$(4) \,\,\,\,\,\,$ $\dfrac{0.5gh}{5g}$ $\,=\,$ $\require{cancel} \dfrac{\cancel{0.5}}{\cancel{5}} \times \dfrac{\cancel{g}}{\cancel{g}} \times h$ $\,=\,$ $0.1h$
$(5) \,\,\,\,\,\,$ $\dfrac{ij^4}{k}$ $\,=\,$ $ij^4k^{-1}$
Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.
|
Moduli interpretations for noncongruence modular curves (2017, published in Mathematische Annalen)
Let $G$ be a finite 2-generated group. In this paper I study Teichmuller structures of level $G$ (or just $G$-structures) on elliptic curves $E$, which roughly correspond to a $G$-torsor on $E$, etale away from the origin. The corresponding moduli stacks $\mathcal{M}(G)$ are defined over $\mathbb{Q}$ (or more generally any $\mathbb{Z}[1/|G|]$-scheme), and over $\mathbb{C}$, $\mathcal{M}(G)_{\mathbb{C}}$ is a disjoint union of "modular curves" $\mathcal{H}/\Gamma$, where $\Gamma\le\text{SL}_2(\mathbb{Z})$ is finite index. If $G$ is abelian, then $\Gamma$ is a congruence subgroup, and one recovers the standard congruence modular curves. If $\Gamma$ is sufficiently nonabelian, then $\mathcal{M}(G)_{\mathbb{C}}$ is a union of noncongruence modular curves. By a result of Asada, for every finite index $\Gamma$, $\mathcal{H}/\Gamma$ can be realized as a component of such a moduli space, for a suitable $G$. As applications we outline a connection to the inverse Galois problem, and show that noncongruence modular forms for $\Gamma$ have bounded denominators at primes not dividing $|G|$.
Arithmetic monodromy actions on pro-metabelian fundamental groups of once-punctured elliptic curves (joint with Pierre Deligne, 2017, on arXiv)
If $G$ is a 2-generated abelian group, then it is easy to see that $G$ structures correspond to classical congruence level structures. When $G$ is nonabelian, it is often difficult to tell if the components of $\mathcal{M}(G)$ will be congruence or noncongruence. In this paper, we show that if $G$ is metabelian, then $G$-structures are again equivalent to classical congruence level structures. We do this by studying the outer automorphism group $\text{Out}(\widehat{M})$ of the rank 2 free profinite metabelian group $\widehat{M}$, which can be identified with the maximal pro-metabelian quotient of the fundamental group of a punctured elliptic curve over $\overline{\mathbb{Q}}$, and show that the natural image of $\pi_1(\mathcal{M}(1)_{\mathbb{Q}})$ in $\text{Out}(\pi_1(E^\circ_{\overline{\mathbb{Q}}})^{metabelian})\cong\text{Out}(\widehat{M})$ is isomorphic to its image in $\text{Out}(\pi_1(E^\circ_{\overline{\mathbb{Q}}})^{ab}) = \text{GL}_2(\widehat{\mathbb{Z}})$, where $\mathcal{M}(1)$ is the moduli stack of elliptic curves. The result is perhaps best expressed as follows: If $G$ is a finite 2-generated metabelian group of exponent $e$, then there is a sandwich: $$\mathcal{M}((\mathbb{Z}/e^2)^2)_{\mathbb{Q}}\rightarrow\mathcal{M}(G)_\mathbb{Q}\rightarrow\mathcal{M}(G^{ab})_\mathbb{Q}$$ where the second map is surjective, and the first map is surjective onto a connected component of $\mathcal{M}(G)_\mathbb{Q}$. Moreover, we show that all the components of $\mathcal{M}(G)_\mathbb{Q}$ are isomorphic, and so this sandwich gives a "complete" picture of $\mathcal{M}(G)_\mathbb{Q}$.
Katz modular forms (notes)
In these notes I give a detailed expository account of Katz's definition of modular forms and the $q$-expansion principle, in a way which makes sense for noncongruence subgroups of $SL(2,\mathbb{Z})$, or more precisely, for general stacks finite etale over the moduli stack of elliptic curves.
|
My inorganic lab had us do an XRD measurement, but I've never been explained how to interpret the data.
Question:
Calculate the unit cell dimensions $a$, $b$, and $c$, for $\ce{YBa2Cu3O7}$ from the indexed X-ray powder pattern provided in [the textbook]. Explain why the crystals are nearly tetragonal in terms of the atomic structure of the compound. The following formula is useful:
$$\sin^2{\theta} = \frac{\lambda^2}{4}\left(\frac{h^2}{a^2}+\frac{k^2}{b^2}+\frac{l^2}{c^2}\right)$$ The three numbers, $hkl$, called Miller indices, indicate the direction of the scattering plane. The Miller indices can be read from the diffractogram provided. For simple reflections of the type $h00$, $0k0$, and $00l$, the value of $h$, $k$, or $l$ corresponds to
nin Bragg's law. Bragg's law:
$$n\lambda = 2d\sin{\theta}$$
The spectrum:
(This is a scan from my textbook, while working on it I circled some items. Just ignore that.)
My Attempt:
$$\lambda = 0.154~\mathrm{nm}$$
Scouring the internet I found the formula for tetragonal atomic structure: the axes $a = b \neq c$, and the axis angles $\alpha = \beta = \gamma = 90^\circ$.
Solving for $c$: $$\begin{align} \sin^2{(22.9^\circ)} &= \frac{\lambda^2}{4} \left(\frac{h^2}{a^2}+\frac{k^2}{b^2}+\frac{l^2}{c^2}\right) \\ &=\frac{0.154^2}{4} \left(\frac{0^2}{a^2}+\frac{0^2}{b^2}+\frac{3^2}{c^2}\right) \\ 0.151 &= 0.0237\left(\frac{3^2}{c^2}\right) \\ c &= 1.188 \end{align}$$
For $a$ and $b$:
$$\begin{align} \sin^2{(22.9^\circ)} &= \frac{\lambda^2}{4} \left(\frac{h^2}{a^2} + \frac{k^2}{b^2} + \frac{l^2}{c^2}\right) \\ &= \frac{0.154^2}{4} \left(\frac{1^2}{a^2} + \frac{0^2}{b^2} + \frac{0^2}{c^2} \right) \\ 0.151 &= 0.0237 \left(\frac{1^2}{a^2}\right) \\ a &= 0.156 \end{align}$$
Is this the right approach? If I have the right answer, what should the units be?
|
The policy issue is this: we're in a recession. Interest rates are zero, and can't go lower. The Fed is desperately trying to goose the economy. Lots of people (most of the recent Jackson Hole Fed conclave) are advising "open-mouth operations," and "managing expectations," that the key to current prosperity is for the Fed to make statements about what it will do in the future; and these statements on their own, with no concrete action, will "increase demand" and lower today's unemployment. The Fed has been convinced, with more and more "forward guidance" as part of its strategy. For example, the latest FOMC statement made history by promising zero interest rates as long as unemployment stays above 6 and a half percent and inflation below two and a half.
Does any of this make any sense?
There are piles of complicated new-Keynesian models on this topic. Ivan's paper gets right to the core, giving a very simple model that explains their logic. The version on p. 20 is the simplest and clearest of all. We only really need one equation
log consumption growth = (substitution elasticity) x real interest rate
$$\frac{dc(t)}{dt} = \sigma^{-1} r(t) $$
(Hang in there, non-economists, you don't really even need the equations; everything is in words too. Also, this is an experiment with Latex macros in blogger. You should see a nice equation above. Let me know if it doesn't work in your browser.) When real interest rates are high, people choose to save more, so consumption is lower today and higher in the future. Thus, high consumption growth goes with high interest rates, and vice versa.
Now, forget about capital, so consumption equals output. Assume the Fed can set the real interest rate to whatever it wants. Also, think of everything as deviations from "potential output" or "trend." The level of consumption becomes the "output gap." Thus, by controlling interest rates, the Fed controls the
growth rateof consumption, output, and the output gap.
Now, here's the key. If the Fed raises the interest rate, and hence the
growth rateof consumption, the levelof consumption (now the output gap) falls. New-Keynesian models (correctly) anchor the level in the future; consumption (output) sooner or later must be equal to "potential." This is the key new-Keynesian mechanism by which higher interest rates lower output, and quite different from the static stories you may remember from old-Keynesian models. (I call this "IS" curve the "interntemporal substitution" curve.)
You might have thought that raising the growth rate of consumption would just be a good thing; implicitly anchoring the level of consumption at last year's value. That would be a mistake. The only way to raise the growth rate of consumption next year is for the level this year to drop. (In this economy without capital.) Then we grow faster by catching up.
Furthermore, if people expect high interest rates for many years, there will be many years of strong consumption growth, and today's level will be very depressed. In equations, we find the level of consumption (output gap) by integrating all its future growth rates (bottom of p. 19)
$$x(t) = \sigma^{-1} \int_t^\infty r(t) dt$$
In sum, the level of today's consumption, output, and output gap all depend on today's interest rate
andpeople's expectations of the path of future interest rates.
You can start to see how managing expectations of future interest rates might be an attractive idea. If we can't do anything about today's interest rates, lowering expected future rates will have the same effect on today's consumption.
In this paper's model, as most new-Keynesian thinking, our current problem is that the "natural rate" of interest, required to keep us at potential output, is sharply negative. (This is all exogenous.) With "only" 2% inflation and nominal interest rates stuck at zero, the Fed cannot deliver anything less than a negative 2% real rate of interest. If the "natural rate" is something like negative 5%, then we are stuck at a 3% "too high" real interest rate. Taking differences to "trend" or "potential," our situation is equivalent to a too-high real interest rate that the Fed can't do anything about.
The solid line in Figure 3 of the paper, reproduced above, shows this situation. The lines plot differences relative to potential output, or "output gaps," so being on the horizontal axis is the best outcome. We have a "negative natural rate" until time T, when the world returns to normal, the Fed regains control, and consumption and output revert to the trend line. You see that the too-high growth of consumption between now and T, and the too-low level of consumption (output gap) now.
Well, says Ivan, why not pull the whole line up? Suppose the Fed could promise now that, at time T, it will set interest rates
too low,from the point of view of that time? This will encourage people to consume at time T rather than in periods after that, i.e. it will engineer the negatively sloped dashed line from T onwards. If people knew there was going to be this boom at T, consumption today would be less depressed. (Again, the way the graph works is that you work from right to left, and the Fed controls the output and consumption growth ratesfrom right to left in order to produce today's level.)
More formally, Ivan assumes that the Fed wants to minimize the squared deviations from potential. A policy like the dashed one produces smaller sum-of-squared output deviations than the solid line we're facing now.
This pretty little graph illustrates lots of policy advice you're hearing from inside and outside the Fed. According to this view, the key thing the Fed can do to raise "demand" today is to promise that it will keep interest rates low longer than it normally would do -- in the period after T -- engineering a bout of output that is a little too high (in a bigger model, inflationary).
But the graph, and the paper, illustrate the central problem: At time T, the Fed will not want to keep rates low. "Too much" consumption means inflation, and in this model too much output (beyond "potential") is just as bad as too little. When time T hits, and the "natural rate" returns to normal, the optimal thing for the Fed to do
looking forwardis to set the interest rate equal to the natural rate, and follow the solid green line. In turn, people today know this, which is why their expectation for consumption at time T is at the solid line level, which is why consumption today is depressed.
Chicago Fed Chair Charlie Evans describes the needed policy as "Odyssean." As Odysseus realized in having himself tied to the mast, the ability to commit yourself today that you will do things tomorrow, things that you will not choose tomorrow if you will have the choice, can improve overall performance. This is a deep principle of good policy, and its violation describes many of our current problems. If the government refuses to commit today that it will not bail people out in the future, then people will take risky actions, and the bailout will recur.
And this is the central problem for this little parable in describing the Fed's actions. Let's read the Fed's Odyssean revelation (the FOMC statement) a little more carefully, with the model in mind:
"..the Committee expects that a highly accommodative stance of monetary policy will remain appropriate for a considerable time...;" "currently anticipates that this exceptionally low range for the federal funds rate will be appropriate at least as long as the unemployment rate remains above 6-1/2 percent..." "the Committee will also consider other information."There is no commitment at all in this. It's a description of how the Fed thinks it will feel in the future, but nowhere in here is what the new-Keynesian model demands: a commitment to do things in the future that the Fed will not want to do when the time comes.
The problem is deep. How can the Fed have the power to take the "discretionary" action today, in response to the current situation, but that action is somehow to commit itself to doing something in the future that it very much will not want to do when the time comes -- that it has so far explicitly and loudly promised that it will not do, and has built up 30 years of reputation against doing -- undershoot and cause inflation? Why can't the Fed 2015 convene another Jackson Hole conference, bring out Charlie Plosser and Jeff Lacker's friends to say "we're heading to a repetition of the 1970s, it's time to become new-Monetarists and promise that we'll be looking at monetary aggregates not unemployment rates?" It's so much hot air, and people know it.
True commitment requires legal, institutional, or constitutional restraints. It at least needs language like "the Committee commits to keep a highly accomodative stance of monetary policy long after it is appropriate" or better an institutional commitment "and to that end no further FOMC meetings will be scheduled until inflation hits 3%." But no institution gives up its discretion easily. (To be clear, I think this would all be a terrible idea; more below. I'm explaining the central commitment/discretion problem highlighted by Ivan's model.)
The paper, of course achieves the same thing in much greater generality, while also presenting this gorgeously transparent example. This example makes it crystal clear that the reason for "forward guidance" in new-Keynesian models is not to raise inflation and thus lower real interest rates -- it works here with constant inflation. "The reason for holding the interest rate at zero is not to promote inflation [and hence negative real rates] as is commonly assumed."..."The real reason for committing to zero interest rates is to create the expectations of a future consumption boom" (p. 4)
Cautioning implicitly against the Fed's "twist" policies aimed at lowering long-term interest rates "I show that optimal policy... actually raises long run interest rates.... This cautions against simple assessments of monetary policy centered around the lowering of yields at long maturities." In the full model, promising the inflationary boom raises long term rates. I.e. exactly the opposite of the Fed's "quantitative easing" operations.
To be clear, I don't buy any of this: that our current problems would all be solved if interest rates could be negative 5%; i.e. that the "natural rate of interest" is sharply negative; that the economy is being strangled by tight credit; and that committing to a repetition of the late 1970s would be a great way to escape our current troubles. I don't think you can analyze the situation in a model without capital, as saving translates to investment demand when there is capital. (Christiano, Eichenbaum and Rebelo's When is the Government Spending Multiplier Large? makes this point, next on my review list.) I don't think a model that says we are experiencing a low level and high growth rates makes much sense. I think the Fed is even more powerless than these models posit.
But this is the great point of clean, theoretical models. They are not black boxes that make predictions. They are collections of clear "if, then" statements. I see the core of the new-Keynesian forward-guidance argument clearly in Ivan's model. If I disagree with the "then," I have to find what I don't like about the "if." And he very clearly shows the difference between "commitment" in the new-Keynesian model vs. how it is translated to policy practice, for better or worse.
|
In my course I often see questions that ask me to calculate the limit of sequences such as: $$\lim\limits_{n \to \infty}{\sqrt [n]{a_n}} $$
How do I handle these questions?
A related question is to show that as ${a_n\to\infty}$
then$${\sqrt [n]{a_n}} > \left(1+\frac {1}{n}\right)$$
for almost every $n$.
I don't know the answer to the first question, so I'm having trouble with the second.
Thanks
|
I am doing IGCSE Maths, and am having a few problems with function notation. I understand the form
f(x).
What does the form
f: x ↦ y mean? Could you also give one or two examples?
And, if possible, state your source. Thank you.
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community
It means that $f$ is a function that takes the value $x$ to the value $y$. For instance, $$f: x\mapsto x^2$$ is an alternate way of writing $f(x) = x^2$.
$f:x \mapsto y$ means that $f$ is a function which takes in a value $x$ and gives out $y$.
But,
$f: \mathbb{N} \to \mathbb{N}$ means that $f$ is a function which takes a natural number as domain and results in a natural number as the result.
As it is evident from math.stackexchange notation — the symbol $\mapsto$ reads as "
maps to". This is backed up by Wikipedia article on functions:
... the notation $\mapsto$ ("maps to", an arrow with a bar at its tail) ...
There is another arrow-symbol, which also used for mapping $\rightarrow$, which might be a bit confusing. The difference between two (as it is mentioned in the linked answer, as well as in the answer by MathEnthusiast):
Example (borrowed from here):
$$f:R \rightarrow R$$ $$x \mapsto x^2$$
It means that: under $f$, any element $x \subset R$ gets mapped to the element $x∗x=x^2$ (which is also an element of $R$).
|
So I found out about the gamma function yesterday and I spent a bunch of time trying to evaluate certain values like $0.5!=\Gamma \left(1.5\right)$. I used multiple integration by parts, and in the end I always get $0=0$. How can someone compute gamma function values for all real numbers manually? I want to evaluate the integral by hand without using previous simplified equations for certain values. Thank you.
First off, the sad truth is that there are no known closed forms of the Gamma function for irrational values.
So, if you wanted to approximate the Gamma function for irrational values, presumably by hand, you might wish to implement the following limit formula, a consequence of the Bohr-Mollerup theorem.
$$\Gamma(s)=\lim_{n\to\infty}\frac {n^s(n!)}{s^{(n+1)}}$$
Where we use the rising factorial to denote $s^{(n+1)}=s(s+1)\dots(s+n)$.
As an example,
$$\Gamma(1/2)=\lim_{n\to\infty}\frac{\sqrt n(n!)}{(1/2)^{(n+1)}}=\lim_{n\to\infty}\frac{\sqrt n(n+1)4^{n+1}(n!)^2}{(2n+2)!}$$
Approximating with $n=9$, one finds that
$$\Gamma(1/2)\approx1.70264$$
Which is pretty close to $\sqrt\pi=1.77245$.
Anyways, to the fun stuff!
It's easy to see by integration by parts that
$$\Gamma(x+1)=x\Gamma(x)$$
And thus,
$$\Gamma(3/2)=\frac12\Gamma(1/2)$$
From there, I recommend browsing through these answers.
A perhaps more useful formula might be
$$\Gamma(s)\Gamma(1-s)=\frac\pi{\sin(\pi s)}$$
A variety of proofs from the integral definition may be found from One-line proof of the Euler's reflection formula. However, this formula provides no light on how to evaluate other values, such as
$$\Gamma(1/3)=\frac{\pi^{2/3}2^{2/3}}{3^{1/12}\operatorname{agm}(1,3^{1/4}2^{-1/2})^{1/2}}$$
$$\Gamma(1/4)=\frac{2^{1/2}\pi^{3/4}}{\operatorname{agm}(1,2^{-1/2})}$$
where $\operatorname{agm}$ is the arithmetic-geometric mean defined by
$$\operatorname{agm}(x,y)=\lim_{n\to\infty}a_n\\a_0=x,~g_0=y\\a_{n+1}=\frac{a_n+g_n}2,\quad g_n=\sqrt{a_ng_n}$$
This is the closest thing to a closed form one can provide for rational values that are not halves of integers.
Portions on how to find values at such values are given in this PDF.
To find the values of $\Gamma(1/2)$ and $\Gamma(1/4)$, note that
$$\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}=\int_0^1t^{x-1}(1-t)^{y-1}~\mathrm dt$$
Setting $x=y=\frac12$, we find that
$$\Gamma(1/2)^2=\int_0^1t^{-1/2}(1-t)^{-1/2}~\mathrm dt$$
Let $t=u^2$ to get
$$\Gamma(1/2)^2=2\int_0^1\frac{\mathrm du}{\sqrt{1-u^2}}$$
This may be solved either geometrically (consider the arc length formula applied to a circle) or with a trig substitution, and it ultimately yields
$$\Gamma(1/2)^2=\pi\\\Gamma(1/2)=\sqrt\pi$$
For $\Gamma(1/4)$, let $2x=y=\frac12$ to get
$$\frac{\Gamma(1/4)}{\Gamma(3/4)}\sqrt\pi=\int_0^1t^{-3/4}(1-t)^{-1/2}~\mathrm dt$$
Let $t=u^4$ to get
$$\frac{\Gamma(1/4)}{\Gamma(3/4)}\sqrt\pi=4\int_0^1\frac{\mathrm du}{\sqrt{1-u^4}}$$
This is a complete elliptic integral, which gives
$$\frac{\Gamma(1/4)}{\Gamma(3/4)}\sqrt\pi=4\mathrm K(i)=\frac{2\pi}{\operatorname{agm}(1,\sqrt2)}$$
Applying the reflection formula,
$$\Gamma(3/4)=\frac\pi{\sin(\pi/4)\Gamma(1/4)}=\frac{\pi\sqrt2}{\Gamma(1/4)}$$
$$\Gamma(1/4)^2=\frac{2^{3/2}\pi^{3/2}}{\operatorname{agm}(1,\sqrt2)}$$
$$\Gamma(1/4)=\frac{2^{3/4}\pi^{3/4}}{\operatorname{agm}(1,\sqrt2)^{1/2}}$$
By useing $\operatorname{agm}(ax,ay)=a\operatorname{agm}(x,y)=a\operatorname{agm}(y,x)$, the above may be rewritten as
$$\Gamma(1/4)=\frac{2^{1/2}\pi^{3/4}}{\operatorname{agm}(1,2^{-1/2})^{1/2}}$$
|
I have been going back through some Kleppner problems and have a doubt concerning problem 6.18. It states:
Find the period of a pendulum consisting of a disk of mass $M$ and radius $R$ fixed to the end of a rod of length $l$ and mass $m$. How does the period change if the disk is mounted to the rod by a frictionless bearing so that it is perfectly free to spin?
The first part (with the disk not free to spin) was reasonably straightforward. My only doubt was that I assumed the moments of inertia of the disk and rod could be added; this seems reasonable, but I don't quite know how to justify it rigorously (side-note: if anybody could give a hint on this, it would be highly appreciated). My result was:
$$T=2\pi\sqrt{\frac{MR^2/2+Ml^2+ml^2/3}{gl(M+m/2)}}$$
For the second part, the issue I had was mainly conceptual... So when the disk is free to spin, it's no longer part of the rigid body; so it won't contribute to the moment of inertia, right? The problem comes here: earlier, to calculate the torque on the rigid body about the pivot, I said that:
$$\tau=R_{CM}\times W$$
Where $R_{CM}$ is the center of mass of the rigid body. This formula comes from a summation over the torques on every small mass in the rigid body, so I figured that the torque on the rigid body, once the disk was no longer a part of it, would depend only on the center of mass of the rod (the disk would no longer affect the 'effective' center of mass that the torque acts on). This gives
$$T=2\pi\sqrt{\frac{2l}{3g}}$$
I checked my answer with this website (pages 5-7) afterwards, and the first part agreed but the second part was in disagreement; the problem was that in that site, $R_{CM}$ was still 'affected' by the spinning disk. Why is this so? (I explained above why I think that $R_{CM}$ should not contribute.)
|
X
Search Filters
Format
Subjects
Language
Publication Date
Click on a bar to filter by decade
Slide to change publication date range
1. Measurement of the ratio of the production cross sections times branching fractions of B c ± → J/ψπ ± and B± → J/ψK ± and ℬ B c ± → J / ψ π ± π ± π ∓ / ℬ B c ± → J / ψ π ± $$ \mathrm{\mathcal{B}}\left({\mathrm{B}}_{\mathrm{c}}^{\pm}\to \mathrm{J}/\psi {\pi}^{\pm }{\pi}^{\pm }{\pi}^{\mp}\right)/\mathrm{\mathcal{B}}\left({\mathrm{B}}_{\mathrm{c}}^{\pm}\to \mathrm{J}/\psi {\pi}^{\pm}\right) $$ in pp collisions at s = 7 $$ \sqrt{s}=7 $$ TeV
Journal of High Energy Physics, ISSN 1029-8479, 1/2015, Volume 2015, Issue 1, pp. 1 - 30
The ratio of the production cross sections times branching fractions σ B c ± ℬ B c ± → J / ψ π ± / σ B ± ℬ B ± → J / ψ K ± $$ \left(\sigma...
B physics | Branching fraction | Hadron-Hadron Scattering | Quantum Physics | Quantum Field Theories, String Theory | Classical and Quantum Gravitation, Relativity Theory | Physics | Elementary Particles, Quantum Field Theory
B physics | Branching fraction | Hadron-Hadron Scattering | Quantum Physics | Quantum Field Theories, String Theory | Classical and Quantum Gravitation, Relativity Theory | Physics | Elementary Particles, Quantum Field Theory
Journal Article
2. Precise determination of the mass of the Higgs boson and tests of compatibility of its couplings with the standard model predictions using proton collisions at 7 and 8 TeV
European Physical Journal C, ISSN 1434-6044, 05/2015, Volume 75, Issue 5, p. 1
Properties of the Higgs boson with mass near 125 GeV are measured in proton-proton collisions with the CMS experiment at the LHC. Comprehensive sets of...
TRANSVERSE-MOMENTUM | TOP-PAIR | RATIOS | NLO | RESUMMATION | ELECTROWEAK CORRECTIONS | BROKEN SYMMETRIES | HADRON COLLIDERS | LHC | QCD CORRECTIONS | PHYSICS, PARTICLES & FIELDS | Physics - High Energy Physics - Experiment | Physics | High Energy Physics - Experiment
TRANSVERSE-MOMENTUM | TOP-PAIR | RATIOS | NLO | RESUMMATION | ELECTROWEAK CORRECTIONS | BROKEN SYMMETRIES | HADRON COLLIDERS | LHC | QCD CORRECTIONS | PHYSICS, PARTICLES & FIELDS | Physics - High Energy Physics - Experiment | Physics | High Energy Physics - Experiment
Journal Article
3. Observation of Two Excited B-c(+) States and Measurement of the B-c(+) (2S) Mass in pp Collisions at root s=13 TeV
PHYSICAL REVIEW LETTERS, ISSN 0031-9007, 04/2019, Volume 122, Issue 13
Journal Article
The European Physical Journal C, ISSN 1434-6044, 3/2016, Volume 76, Issue 3, pp. 1 - 52
New sets of parameters (“tunes”) for the underlying-event (UE) modelling of the pythia8, pythia6 and herwig++ Monte Carlo event generators are constructed...
Nuclear Physics, Heavy Ions, Hadrons | Measurement Science and Instrumentation | Nuclear Energy | Quantum Field Theories, String Theory | Physics | Elementary Particles, Quantum Field Theory | Astronomy, Astrophysics and Cosmology | PHYSICS, PARTICLES & FIELDS | Nuclear energy | Distribution (Probability theory) | Collisions (Nuclear physics) | Analysis | Physics - High Energy Physics - Experiment | High Energy Physics - Experiment | PHYSICS OF ELEMENTARY PARTICLES AND FIELDS | Regular - Experimental Physics
Nuclear Physics, Heavy Ions, Hadrons | Measurement Science and Instrumentation | Nuclear Energy | Quantum Field Theories, String Theory | Physics | Elementary Particles, Quantum Field Theory | Astronomy, Astrophysics and Cosmology | PHYSICS, PARTICLES & FIELDS | Nuclear energy | Distribution (Probability theory) | Collisions (Nuclear physics) | Analysis | Physics - High Energy Physics - Experiment | High Energy Physics - Experiment | PHYSICS OF ELEMENTARY PARTICLES AND FIELDS | Regular - Experimental Physics
Journal Article
5. Measurement of the ratio of the production cross sections times branching fractions of B-c(+/-) -> J/psi pi(+/-) and B-+/- -> J/psi K-+/- and B(B-c(+/-) -> J/psi pi(+/-)pi(+/-)pi(-/+))/B(B-c(+/-) -> J/psi pi(+/-)) in pp collisions at root s=7 Tev
JOURNAL OF HIGH ENERGY PHYSICS, ISSN 1029-8479, 01/2015, Issue 1
Journal Article
Physics Letters B, ISSN 0370-2693, 09/2012, Volume 716, Issue 1, pp. 30 - 61
Results are presented from searches for the standard model Higgs boson in proton–proton collisions at and 8 TeV in the Compact Muon Solenoid experiment at the...
CMS | Higgs | Physics | PARTON DISTRIBUTIONS | SEARCH | PHYSICS, NUCLEAR | COLLIDERS | STANDARD MODEL | ELECTROWEAK CORRECTIONS | BROKEN SYMMETRIES | ASTRONOMY & ASTROPHYSICS | QCD CORRECTIONS | PP COLLISIONS | SPECTRUM | MODEL HIGGS-BOSON | PHYSICS, PARTICLES & FIELDS | Analysis | Collisions (Nuclear physics) | Searching | Elementary particles | Decay | Higgs bosons | Standard deviation | Solenoids | Standards | Bosons | Physics - High Energy Physics - Experiment | High Energy Physics - Experiment | PHYSICS OF ELEMENTARY PARTICLES AND FIELDS
CMS | Higgs | Physics | PARTON DISTRIBUTIONS | SEARCH | PHYSICS, NUCLEAR | COLLIDERS | STANDARD MODEL | ELECTROWEAK CORRECTIONS | BROKEN SYMMETRIES | ASTRONOMY & ASTROPHYSICS | QCD CORRECTIONS | PP COLLISIONS | SPECTRUM | MODEL HIGGS-BOSON | PHYSICS, PARTICLES & FIELDS | Analysis | Collisions (Nuclear physics) | Searching | Elementary particles | Decay | Higgs bosons | Standard deviation | Solenoids | Standards | Bosons | Physics - High Energy Physics - Experiment | High Energy Physics - Experiment | PHYSICS OF ELEMENTARY PARTICLES AND FIELDS
Journal Article
7. Azimuthal anisotropy of charged particles with transverse momentum up to 100 GeV/c in PbPb collisions at sNN=5.02 TeV
Physics Letters B, ISSN 0370-2693, 01/2018, Volume 776, Issue C, pp. 195 - 216
Journal Article
8. Azimuthal anisotropy of charged particles with transverse momentum up to 100GeV/c in PbPb collisions at root S-NN=5.02 TeV
PHYSICS LETTERS B, ISSN 0370-2693, 01/2018, Volume 776, pp. 195 - 216
The Fourier coefficients v(2) and v(3) characterizing the anisotropy of the azimuthal distribution of charged particles produced in PbPb collisions at root...
PERSPECTIVE | ELLIPTIC FLOW | High-pT | CMS | QUARK-GLUON PLASMA | ROOT-S(NN)=130 GEV | PHYSICS, NUCLEAR | Parton energy loss | SUPPRESSION | Flow | DEPENDENCE | DISTRIBUTIONS | PLUS AU COLLISIONS | ASTRONOMY & ASTROPHYSICS | QGP | Jet quenching | AU+AU COLLISIONS | SPECTRA | PHYSICS, PARTICLES & FIELDS
PERSPECTIVE | ELLIPTIC FLOW | High-pT | CMS | QUARK-GLUON PLASMA | ROOT-S(NN)=130 GEV | PHYSICS, NUCLEAR | Parton energy loss | SUPPRESSION | Flow | DEPENDENCE | DISTRIBUTIONS | PLUS AU COLLISIONS | ASTRONOMY & ASTROPHYSICS | QGP | Jet quenching | AU+AU COLLISIONS | SPECTRA | PHYSICS, PARTICLES & FIELDS
Journal Article
|
This is an exercise from my calculus class. The function is defined as $x\sin (1/y)+y\sin (1/x)$ if $x\neq0 $ and $y\neq0 $, and $0$ if $x=0 $ or $y=0$.
I'm pretty confident the limit exists and should be $0$, because: $$\lim_{(x,y)\to(0,0)}[x\sin (1/y)+y\sin (1/x)]=\lim_{(x,y)\to(0,0)}[x\sin (1/y)]+\lim_{(x,y)\to(0,0)}[y\sin (1/x)]$$
And: $x\leq x\sin(1/y)\leq x$, so $\lim_{(x,y)\to(0,0)}[x\sin (1/y)]=0$ right? (The same can be said for $\lim_{(x,y)\to(0,0)}[y\sin (1/x)]$)
However, I tried checking my answer, and according to Wolfram Alpha the limit doesn't exist. Is this because I'm wrong, or is it just because $x\sin (1/y)+y\sin (1/x)$ is undefined for $y\neq0 $ and $y\neq0 $
|
The
large cash bag is a rare item won from Treasure Hunter. When opened, it will give the player coins based on the player's total level. During the weekend of 1 February 2013, the gold received from the bag was doubled.
The coins received from a large cash bag is a random number between $ 90X $ and $ 110X $, where $ X $ depends on the player's total level
[1]. In the formula below, $ L $ is your total level.
$ X = \begin{cases} 750 + \left ( \frac{L}{27} - 1 \right ) \times 22 & \text{if }L < 324 \\ 1000 + \left ( \frac{L}{27} - 12 \right ) \times 75 & \text{if }324 \le L < 1404 \\ 4000 + \left ( \frac{L}{27} - 52 \right ) \times 125 & \text{if }1404 \le L < 2268\\ 8000 + \left ( \frac{L}{27} - 84 \right ) \times 133 & \text{if }2268 \le L \end{cases} $
This means that it is possible to receive the following amounts of coins at the threshold total levels:
Total level Coins 27 67,500 to 82,500 coins 324 90,000 to 110,000 coins 1404 360,000 to 440,000 coins 2268 720,000 to 880,000 coins 2715 918,170 to 1,122,207 coins Drop sources This list was created dynamically. For help, see the FAQ. To force an update of this list, click here. For an exhaustive list of all known sources for this item, see here.
Source Combat level Quantity Rarity Antique chest N/A 1 Rare Big event mystery box N/A 1 Rare Big event mystery box (Aiding the Exile) N/A 1 Rare Big event mystery box (Deathbeard's Demise) N/A 1 Rare Big event mystery box (Mental Health Awareness Week) N/A 1 Rare Big event mystery box (Rune Capers) N/A 1 Rare Big event mystery box (Zodiac Festival) N/A 1 Rare Event mystery box (Deathbeard's Demise) N/A 1 Rare Event mystery box (Going Like Clockwork) N/A 1 Rare Event mystery box (Mental Health Awareness Week) N/A 1 Rare Event mystery box (Rune Capers) N/A 1 Rare Event mystery box (Zodiac Festival) N/A 1 Rare Shiny four-leaf clover necklace N/A 1 Rare Shiny three-leaf clover necklace N/A 1 Rare Shiny two-leaf clover necklace N/A 1 Rare Sparkling four-leaf clover necklace N/A 1 Rare Sparkling three-leaf clover necklace N/A 1 Rare Sparkling two-leaf clover necklace N/A 1 Rare Twitch loot crate N/A 1 Rare Big event mystery box (Going Like Clockwork) N/A 1 Very rare Event mystery box N/A 1 Very rare Event mystery box (Aiding the Exile) N/A 1 Very rare References
Exclusive rewards (Limited) Large (Empty • Full)
|
Searching for just a few words should be enough to get started. If you need to make more complex queries, use the tips below to guide you.
Purchase individual online access for 1 year to this journal.
Impact Factor 2019: 0.808
The journal
Asymptotic Analysis fulfills a twofold function. It aims at publishing original mathematical results in the asymptotic theory of problems affected by the presence of small or large parameters on the one hand, and at giving specific indications of their possible applications to different fields of natural sciences on the other hand. Asymptotic Analysis thus provides mathematicians with a concentrated source of newly acquired information which they may need in the analysis of asymptotic problems.
Article Type: Research Article
Abstract: In this paper, we study the Navier–Stokes equations in a domain with small depth. With this aim, we introduce a small adimensional parameter ε related to the depth. First we make a change of variable to a domain independent of ε and then we use asymptotic analysis to study what happens when ε becomes small. In this way we obtain a model for ε small that, after coming back to the original domain and without making a priori assumptions about velocity or pressure behavior, gives us a shallow water model including a new diffusion term.
Keywords: asymptotic analysis, shallow waters with viscosity
Citation: Asymptotic Analysis, vol. 43, no. 4, pp. 267-285, 2005
Authors: Demengel, Françoise
Article Type: Research Article
Abstract: In this paper we study necessary and sufficient conditions on f and the first eigenfunctions for the 1-Laplacian, for equations of the form \[\cases{-\mathrm{div}(\sigma)-\lambda =fu^{q-1},\quad u\geqslant 0\ \hbox{in}\ {\varOmega},\cr \noalign{\vskip3pt}\sigma\cdot \nabla u=|\nabla u|,\quad |\sigma|_{L^{\infty}({\varOmega})}\leqslant 1,\ u\in W_{0}^{1,1}({\varOmega})\cr}\] when q≤N/(N−1) and λ≥λ1 , λ1 is the first eigenvalue for the 1-Laplacian.
Citation: Asymptotic Analysis, vol. 43, no. 4, pp. 287-322, 2005
Authors: Hillairet, M.
Article Type: Research Article
Abstract: We consider the Burgers–Hopf equation on \[$\mathbb {R}$ outside a finite set of (moving) points coupled with transmission conditions in these points prescribing the dynamics of these points. We address the problem of asymptotic collisions. We give new proof of result in J.L. Vàzquez and E. Zuazua preprint, May 2004, and complementary results.
Keywords: fluid-solid interactions, long-time behavior, lack of collision
Citation: Asymptotic Analysis, vol. 43, no. 4, pp. 323-338, 2005
Authors: Leung, Anthony W.
Article Type: Research Article
Abstract: This article considers the dynamics of a coupled system of incompressible Navier–Stokes equations with second-order wave equations. The system may be used to approximate the interaction of ionized plasma particles with an electromagnetic field. Under appropriate assumptions and provided that the viscosity is sufficiently large, we prove the existence of an invariant manifold. Moreover, the manifold is attractive as t→+∞ for all close neighboring solutions.
Keywords: coupled parabolic–hyperbolic system, invariant manifold, Navier–Stokes equations, asymptotic stability, electro- magnetic waves
Citation: Asymptotic Analysis, vol. 43, no. 4, pp. 339-357, 2005
Inspirees International (China Office)
Ciyunsi Beili 207(CapitaLand), Bld 1, 7-901 100025, Beijing China Free service line: 400 661 8717 Fax: +86 10 8446 7947 china@iospress.cn
For editorial issues, like the status of your submitted paper or proposals, write to editorial@iospress.nl
如果您在出版方面需要帮助或有任何建, 件至: editorial@iospress.nl
|
A new Higgs tadpole cancellation condition reformulating the hierarchy problem Strings 2013 [talks] is underway.The first hep-ph paper today probably got to that exclusive place because the authors were excited and wanted to grab the spot. Andre de Gouvea, Jennifer Kile, and Roberto Vega-Morales of Illinois chose the title \(H\rightarrow \gamma\gamma\) as a Triangle Anomaly: Possible Implications for the Hierarchy ProblemThey point out a curious feature of the diagrams calculating the Higgs boson decay to two photons (yes, it's the process that seemed to have a minor excess at the LHC but this excess went away): while the diagram is finite, one actually gets different results according to the choice of the regularization.
\({\Huge \Rightarrow}\)
In particular, the \(d=4\) direct calculation leads to a finite result but it actually violates the gauge invariance so it can't be right. It should be disturbing for you that wrong results may arise from quantum field theory calculations even if you don't encounter any divergence.
However, the right fix is known: work with a regularization, typically dimensional regularization, that automatically respects the gauge invariance. Using dim reg, in \(d=4-\epsilon\) dimensions, one automatically gets the right result. However, it still disagrees with the wrong result computed directly in \(d=4\).
While this episode doesn't mean that QFT is ill-defined or inconsistent and we actually know how to do things correctly, the finite-yet-wrong result in \(d=4\) surely sounds bizarre. The authors propose a new condition on quantum field theories: this strangeness shouldn't be there. In other words, the wrong, Ward-identity-violating terms in the \(d=4\) calculation should cancel. When they cancel, the \(d=4\) calculation will agree with the correct dim reg \(d=4-\epsilon\) calculation.
The paper suggests that this cancellation is a new general principle of physics that constrains the allowed spectra of particles and fields and that should be added next to the usual triangle diagram gauge anomaly cancellation conditions in the Standard Model and similar gauge theories.
Note that the triangle anomaly diagrams may be blamed on linear divergences in the integrals. Here, the new type of an "anomaly" that should be canceled is also related to the linearly divergent part of certain integrals because they behave differently under the shift of momenta. So even the computational origin of their new "anomaly" resembles the case of the chiral anomaly. In some sense, the "new" anomaly only differs from the well-known triangle anomaly by its replacement of one external gauge boson with the Higgs boson.
These diagrams that have to cancel are close to some Higgs tadpole diagrams – Feynman diagrams you would use to compute the shift of the Higgs vacuum expectation value (vev). The "tadpole cancellation conditions" are well-known to string theorists but they weren't really discussed in the context of ordinary 4D quantum field theories yet. I suppose that there should be a more natural way to phrase and justify the Higgs tadpole cancellation condition. The condition looks like eqn (36)\[
3ge^2M_W + \frac{e^2 g m_H^2}{2M_W} +\sum_{\rm scalars} 2\lambda_S v e_s^2-\!\!\sum_{\rm fermions}\!\! 2\lambda_f^2 v e_f^2 = 0
\] Supersymmetry seems to be the only known natural principle that cancels the new "anomaly". The authors have only checked it by some uninspiring brute force calculation in the MSSM as a function of several parameters. I guess that there's a simple proof that supersymmetry – unbroken or broken at an arbitrary scale – cancels the new "anomaly" condition.
It's probably true and they probably realize that the new condition is mostly equivalent to the usual unbearable lightness and naturalness of the Higgs' being. However, if you might phrase the condition for naturalness as a version of an anomaly cancellation condition, it would probably be (or at least look) much more inevitable than the usual arguments discussing the hierarchy problem.
|
The Gaussian Plane Waves method (GPW) solves the DFT Kohn-Sham equations efficiently. It uses gaussians as basisset, and planewaves as auxiliary basis. This is similar at the Resolution of Identity (RI) methods but with a different basisset.
In GPW the whole density is transferred to plane waves, and one has the density $n(r)=\sum_{i j} P_{i j} \phi_i (r) \phi_j(r)$ in the gaussian basis set and the density $\tilde n$ taken on a uniform grid.
Then $\tilde n$ is used to calculate the hartree potential $V_H$ (via an FFT based poisson solver) and the exchange and correlation potential $V_\text{xc}$. These potential are then transferred back to gaussian basis set by integrating them with $\phi_i(r)\phi_j(r)$. To make the collocation and integration perfectly consistent with each other one can set
FORCE_EVAL%DFT%QS%MAP_CONSISTENT this normally adds only a very small overhead to the calculation and is
$n$ and $\tilde n$ are not equal, and this introduces an error in the calculation. $\tilde n$ converges toward $n$ when the cutoff (that controls the grid spacing) goes to infinity (and gridspacing to 0). Which cutoff is sufficient to represent a density depends on how sharp is the gaussian basis set (or that of the potential, but it is always broader).
For historical reasons the density of the grid is given as the energy (in Ry) of the highest reciprocal vector that can be represented on the grid. This can be roughly given as $0.5(\pi/dr)^2$ where $dr$ is the gridspacing in Bohr. The characteristic length of a gaussian with exponent A is given by $1/\sqrt{a}$ (up to a factor 2 depending on the convention used). This means that the cutoff to represent in the same way a gaussian depends linearly on the exponent. Thus one can get a first guess for an acceptable guess can be take from the knowledge that for water with $\alpha_H=47.7$ a good cutoff for doing MD is 280 Ry.
It turns out that if one wants to put the whole density on the grid, the core electrons of even the simplest atoms cannot be represented, thus one has to remove to core electrons and use pseudopotentials for the atoms. In Cp2k we use the Godeker-Theta-Hutter pseudopotentials, these are harder than other pseudopotentials, but also more accurate.
$\tilde n$ is optimized for the electrostatic part, but is used also to calculate the exchange and correlation potential. Because of this, and because the GTH pseudopotential goes almost to 0 close to the core of the atom, the xc potential, especially for gradient corrected functionals, converges badly. Instead of using very high cutoffs one can perform a smoothing of the density, and calculate the derivatives on the grid with other methods than the G-space based derivatives.For MD of water using a cutoff of 280 Ry
XC_SMOOTH_RHO NN10 and
XC_DERIV SPLINE2_SMOOTH (in the
FORCE_EVAL%DFT%XC%XC_GRID section) give good results, please note that these options renormalize the total energy, and the amount of renormalization is dependent on the cutoff. Thus energies with different cutoffs cannot be easily compared, only interaction energies or forces can be calculated.
Methods that do not redefine the total energy are
XC_SMOOTH_RHO NONE and
XC_DERIV equal to either
PW, SPLINE3 or
SPLINE2. These are listed from the one that assumes more regularity (
PW the the one that assumes less regularity
SPLINE2. Normally
SPLINE2 is a good choice, but for high cutoffs (600 Ry for water)
SPLINE3 is better. The default (
PW) is not bad, but generally inferior to the others.
The QS part of Cp2k uses basis sets that are a linear combination gaussian functions. As the GPW method uses pseudopotentials one cannot use basis set of other programs, at least the core part should be optimized for cp2k. The polarization and augmentation functions can be taken from dunning type basis sets.
Cp2k basis normally are build with an increasing accuracy level starting from SZV (single Z valence, normally only for really crude results), DZVP (double Z valence and one polarization function, already suitable for MD, and give reasonable results), TZV2P (triple Z valence and two polarization functions, fairly good results), QZV3P (quadruple Z valence and three polarization functions, good results). Note that the discussion about the quality of the result is very indicative, and valid for condensed phase MD, for gas phase to reduce the BSSE an augmented basis (aug-) with diffuse basis functions is needed, if you calculate properties that depend on the polarization probably also (aug-) and polarization functions will be important. In any case you should perform a convergence check for your properties with respect to the basis set.
A good set of basis set are available in the BASIS_SET and GTH_BASIS_SETS files. Other can be created with a program, and you can ask on the list or try to optimize them with the program that is part of cp2k.
The basis set to use for an element is set in the
FORCE_EVAL%SUBSYS%KIND section with its name with the
BASIS_SET keyword.
The GTH pseudopotential that one has to use has a large library of elements available in the POTENTIAL file. The pseudopotential has to be optimized for the exchange correlation functional that one is using. Normally changing from one functional to the other is not too difficult with the optimization program that is part of cp2k. If your element-functional combination is missing ask on the forum.
The pseudopotential to use for an element is set in the
FORCE_EVAL%SUBSYS%KIND section with its name.
In DFT the choice exchange correlation functional is an important issue, because unfortunately results can depend on it. Here we don't want to discuss how to select it, normally it is a good idea to use the same one as what is used in your field, so that you can more easily compare your results with the literature. An important choice though is whether to use exact exchange or an hybrid functional or not. Hybrid functionals normally improve the description of barriers and radicals, but are much more costly. In general one should start with a GGA and meta GGA and switch to hybrid functionals to check the results or if needed.
In cp2k the xc functional is selected in the
FORCE_EVAL%DFT%TDDFPT%XC section. Common functional combinations can be selected directly as section parameters of the
XC_FUNCTIONAL for example with the
POTENTIAL keyword.</p>
&XC_FUNCTIONAL BLYP &END XC_FUNCTIONAL
but more complex combination have to be chosen by directly setting the underlying subsections.
Apart from cutoff and basis set with the GPW method there are two other important switches that control the convergence of the method. One is
FORCE_EVAL%DFT%QS%EPS_DEFAULT that controls the convergence of integral calculation, overlap cutoff, neighboring lists… and sets a slew of other
EPS_* keywords so that the KS matrix is calculated with the requested precision. The other is
EPS_SCF which controls the convergence of the scf procedure. In general one should have an error $e_f$ on the forces one should have $\sqrt{\text{eps}_\text{scf}} < e_f $ and $\sqrt{\text{eps}_\text{default}} < \text{eps}_\text{scf}$, and in general around 1.e-3 is the error that you need to have reasonable MD.
To perform the SCF and find the groundstate there are two main methods: the traditional diagonalization method accelerated by the DIIS procedure, and the orbital transform method.
FORCE_EVAL%DFT%SCF%SCF_GUESS RESTART is a parameter that controls how the initial density is built, normally one uses either
ATOMIC or
RESTART. During an MD this setting normally influences only the initial startup.
Traditional diagonalization is like most other DFT codes, it diagonalizes the KS matrix to get a new density from the n lowest eigenvectors.The density for the next scf cycle is built at the beginning just mixing the new and the old density with the
FORCE_EVAL%DFT%SCF%MIXING factor. If the new density is close enough to the old density
FORCE_EVAL%DFT%SCF%EPS_DIIS then the DIIS procedure is activated.
Diagonalization works well, but for difficult or big systems the OT method is better (and in general there is no reason not to use it as default). The OT method directly minimizes the the electronic energy with respect to the wavefunctions. It uses a clever parametrization of the wavefunctions so that the orthogonality constraint becomes a linear constraint.To activate OT adding the section
FORCE_EVAL%DFT%SCF%OT is enough.
The advantage of the OT method, are that being a direct method it always converges to something, that it needs less memory than the diagonalization (important for big systems), and each SCF iteration is faster (but it might need a little more iterations). Anyway normally it is faster than diagonalization.
The speed of convergence of the OT method the preconditioner choice (
FORCE_EVAL%DFT%SCF%PRECONDITIONER) is important.
FULL_KINETIC is a good choice for very large systems, but for smaller or difficult systems
FULL_ALL is normally better. The quality of the preconditioner is dependent on how close one is to the solution. Especially with expensive preconditioners like
FULL_ALL being closer might improve much the preconditioner and the convergence. For this reason it might be worthwhile to rebuild the preconditioner after some iterations of the SCF. This might be reached using setting
FORCE_EVAL%DFT%SCF%MAX_SCF to a small value, and activating the outer scf loop addint the
FORCE_EVAL%DFT%SCF%OUTER_SCF section, and setting its
EPS_SCF equal to the one of the scf loop, and chooing a outer
MAX_SCF big enough (the total number of iteration is the product of internal and external MAX_SCF.
Also the minimizer of the OT method can be changed with
FORCE_EVAL%DFT%SCF%OT%MINIMIZER, the default is
CG that uses a very robust conjugated gradient method. Less robust but often faster is the DIIS method.
|
Question is to solve for Galois Group of $x^4-2x^2-2$ over $\mathbb{Q}$.
I know the roots of this polynomial are $\sqrt{1+\sqrt{3}},-\sqrt{1+\sqrt{3}},\sqrt{1-\sqrt{3}},-\sqrt{1-\sqrt{3}}$.
But, $\sqrt{2}i=\sqrt{1+\sqrt{3}}.\sqrt{1-\sqrt{3}}$. So, I concluded that splitting field would be $\mathbb{Q}(\sqrt{1+\sqrt{3}},\sqrt{2}i)$ (for some reason i am not very sure about, i have not written splitting field to be $\mathbb{Q}(\sqrt{1-\sqrt{3}},\sqrt{2}i)$).
So, i have splitting field as $\mathbb{Q}(\sqrt{1+\sqrt{3}},\sqrt{2}i)$.
What i do usually after i know that extension is of the form $F(a,b)$ with $F(a)\cap F(b)=F$, I calculate Galois group of $F(a,b)/F(b)$ and Galois group of $F(a,b)/F(a)$ and write their product as in $G=Gal(F(a,b)/F)$
I do the same here for $Gal(\mathbb{Q}(\sqrt{1+\sqrt{3}},\sqrt{2}i)/\mathbb{Q}(\sqrt{1+\sqrt{3}}))$, this is of order 2, sending $\sqrt{2}i\rightarrow -\sqrt{2}i$ (fixing $\sqrt{1+\sqrt{3}}$) So, I have $Gal(\sqrt{1+\sqrt{3}},\sqrt{2}i)/\mathbb{Q}(\sqrt{1+\sqrt{3}}))\cong \mathbb{Z}_2$.
The problem is with $Gal(\mathbb{Q}(\sqrt{1+\sqrt{3}},\sqrt{2}i)/\mathbb{Q}(\sqrt{2}i))$, I Have to prove that Galois group of $x^4-2x^2-2$ is dihedral group of order 8.I already have an element of order 2, I have to get an element of order 4 from $Gal(\mathbb{Q}(\sqrt{1+\sqrt{3}},\sqrt{2}i)/\mathbb{Q}(\sqrt{2}i))$ which is becoming more cumbersome.
Any help would be appreciated. Thank You.
|
Homoclinic orbits for a class of asymptotically quadratic Hamiltonian systems
School of Mathematics and Statistics, Southwest University, Chongqing 400715, China
$ \ddot{q}(t)-\lambda q(t)+\nabla W(t,q(t)) = 0 $
$ \lambda>0 $
$ \frac{|\nabla W(t,x)|}{|x|} $
$ |x|\rightarrow\infty $
$ |t|\rightarrow\infty $ Keywords:Homoclinic orbit, Hamiltonian systems, asymptotically quadratic, variational method, concentration-compactness principle. Mathematics Subject Classification:Primary: 37J45, 37K05; Secondary: 58E05. Citation:Ying Lv, Yan-Fang Xue, Chun-Lei Tang. Homoclinic orbits for a class of asymptotically quadratic Hamiltonian systems. Communications on Pure & Applied Analysis, 2019, 18 (5) : 2855-2878. doi: 10.3934/cpaa.2019128
References:
[1]
C. O. Alves, P. C. Carriao and O. H. Miyagaki,
Existence of homoclinic orbits for asymptotically periodic systems involving Duffing-like equation,
[2]
G. Arioli and A. Szulkin, Homoclinic solution for a class of systems of second order differential equtions, Tech. Rep. 5, Dept. of Math., Univ. Stockholm, Sweden, 1995. doi: 10.12775/TMNA.1995.040. Google Scholar
[3]
P. Bartolo, V. Benci and D. Fortunato,
Abstract critical point theorems and applications to some nonlinear problems with ``strong" resonance at infinity,
[4] [5]
P. C. Carriao and O. H. Miyagaki,
Existence of homoclinic solutions for a class of time-dependent Hamiltonian systems,
[6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20]
Y. Lv and C. L. Tang,
Homoclinic orbits for second-order Hamiltonian systems with subquadratic potentials,
[21] [22] [23] [24]
P. H. Rabinowitz, Minimax methods in critical point theory with applications to differential equations, in
[25] [26] [27] [28]
E. Serra, M. Tarallo and S. Terracini,
Subharmonic solutions to second-order differential equations with periodic nonlinearities,
[29]
J. Sun, H. Chen and Juan J. Nieto,
Homoclinic solutions for a class of subquadratic second-order Hamiltonian systems,
[30] [31] [32]
Z. Zhang and R. Yuan,
Homoclinic solutions for a class of non-autonomous subquadratic second-order Hamiltonian systems,
[33]
Q. Zhang and X. H. Tang,
Existence of homoclinic solutions for a class of asymptotically quadratic non-autonomous Hamiltonian systems,
[34] [35]
show all references
References:
[1]
C. O. Alves, P. C. Carriao and O. H. Miyagaki,
Existence of homoclinic orbits for asymptotically periodic systems involving Duffing-like equation,
[2]
G. Arioli and A. Szulkin, Homoclinic solution for a class of systems of second order differential equtions, Tech. Rep. 5, Dept. of Math., Univ. Stockholm, Sweden, 1995. doi: 10.12775/TMNA.1995.040. Google Scholar
[3]
P. Bartolo, V. Benci and D. Fortunato,
Abstract critical point theorems and applications to some nonlinear problems with ``strong" resonance at infinity,
[4] [5]
P. C. Carriao and O. H. Miyagaki,
Existence of homoclinic solutions for a class of time-dependent Hamiltonian systems,
[6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20]
Y. Lv and C. L. Tang,
Homoclinic orbits for second-order Hamiltonian systems with subquadratic potentials,
[21] [22] [23] [24]
P. H. Rabinowitz, Minimax methods in critical point theory with applications to differential equations, in
[25] [26] [27] [28]
E. Serra, M. Tarallo and S. Terracini,
Subharmonic solutions to second-order differential equations with periodic nonlinearities,
[29]
J. Sun, H. Chen and Juan J. Nieto,
Homoclinic solutions for a class of subquadratic second-order Hamiltonian systems,
[30] [31] [32]
Z. Zhang and R. Yuan,
Homoclinic solutions for a class of non-autonomous subquadratic second-order Hamiltonian systems,
[33]
Q. Zhang and X. H. Tang,
Existence of homoclinic solutions for a class of asymptotically quadratic non-autonomous Hamiltonian systems,
[34] [35]
[1]
Jun Wang, Junxiang Xu, Fubao Zhang.
Homoclinic orbits for a class of Hamiltonian systems
with superquadratic or asymptotically quadratic potentials.
[2] [3]
Oksana Koltsova, Lev Lerman.
Hamiltonian dynamics near nontransverse homoclinic orbit to saddle-focus equilibrium.
[4]
Zheng Yin, Ercai Chen.
The conditional variational principle for maps with the pseudo-orbit tracing property.
[5] [6] [7]
Shiwang Ma.
Nontrivial periodic solutions for asymptotically linear hamiltonian systems at resonance.
[8]
Paolo Gidoni, Alessandro Margheri.
Lower bound on the number of periodic solutions for asymptotically linear planar Hamiltonian systems.
[9] [10] [11]
Juntao Sun, Jifeng Chu, Zhaosheng Feng.
Homoclinic orbits for first order periodic
Hamiltonian systems with spectrum point zero.
[12] [13]
Zheng Yin, Ercai Chen.
Conditional variational principle for the irregular set in some nonuniformly hyperbolic systems.
[14] [15]
D. Bartolucci, L. Orsina.
Uniformly elliptic Liouville type equations: concentration compactness and a priori estimates.
[16]
S. Aubry, G. Kopidakis, V. Kadelburg.
Variational proof for hard Discrete breathers in
some classes of Hamiltonian dynamical systems.
[17] [18]
Dong-Lun Wu, Chun-Lei Tang, Xing-Ping Wu.
Existence and nonuniqueness of homoclinic solutions for second-order Hamiltonian systems with mixed nonlinearities.
[19]
Amadeu Delshams, Pere Gutiérrez.
Exponentially small splitting for whiskered tori in Hamiltonian systems: continuation of transverse homoclinic orbits.
[20]
Li-Li Wan, Chun-Lei Tang.
Existence and multiplicity of homoclinic orbits for second order Hamiltonian systems without (
2018 Impact Factor: 0.925
Tools Metrics Other articles
by authors
[Back to Top]
|
Just a question for personal comprehension. Consider the following statement:
It is NP-hard to approximate Set-Cover within a $(1 - \epsilon) \log n$ factor for any $0 < \epsilon < 1$.
Now, NP-hardness refers to decision problems.So what is NP-Hard exactly here ?
My guess is that the statement is equivalent to saying that the following problem is NP-hard (but I really ain't sure):
Given a set cover instance $S$ and an integer $k$ with the guarantee that either $S$ admits a cover of size at most $k$, or $S$ has a cover of size at least $k \cdot (1 - \epsilon) \log n$, decide if $S$ has a cover of size at most $k$.
Note that I took set cover as an example, but my question is on problems that are said 'hard to approximate' in general.
|
How many positive integer values of n are there such that $2^n + 7^n$ is a perfect square?
I am not sure how to approach this question given that there are two different bases 2 and 7
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community
How many positive integer values of n are there such that $2^n + 7^n$ is a perfect square?
I am not sure how to approach this question given that there are two different bases 2 and 7
The only instance is $n=1$.
Let $a_n=2^n+7^n$.
Of course $a_1$ is a square. Assume that $n>1$.
We see that $a_n\equiv (-1)^n \pmod 4$ so $n$ must be even (this is where we use $n>1$).
Similarly we see that $a_n\equiv 2^{n+1}\pmod 5$ which implies that $n$ is odd, and we are done
Set $2^n+7^n=x^2$. Since all perfect squares are congruent to $0$ or $1$ modulo $3$ . On the other hand $7^n\equiv 1\pmod 3$ and $2^n\equiv 1\pmod 3$ or $2^n\equiv 2\pmod 3$ .This means that $n$ must be odd(why?). Proof by Contradiction: We wish to show that the only value of $n$ is $1$. Assume that $n>2$. Then consider the equation $2^n=x^2-7^n$. let $x=2a+1$. We have $2^n=4a^2+4a+1-7^n$. Dividing by $4$, $2^ {n-2}=a^2+a+(1-7^n)/4$. the entire LHS is an integer, and so are $a^2$ and $a$. Thus, $\dfrac {1}{4} (1 - 7^n)$ must be an integer. Let $\dfrac {1}{4} (1 - 7^n) = k$.
Then we have $1- 7^n = 4k$. This means that $1- 7^n \equiv 0 \pmod {4}$. Thus, $n$ is even. However, it has already been shown that $n$ must be odd. This is a contradiction. Therefore, $n$ is not greater than or equal to $ 2$, and must hence be less than $2$. The only positive integer less than $2$ is $1$
If $n=2m$, $2^{2m}+7^{2m}$ is too close to the square of $7^m$ to be a square itself. This gives that $n$ is odd.
If we consider $n=2m+1$ for $m\geq 1$ (the case $n=1$ leads to a trivial solution) we have that $$ 2^n+7^n = 2\cdot 4^m + (8-1)^{2m+1} \equiv (-1)^{2m+1} \equiv -1\pmod{4}, $$ so $2^n+7^n$ cannot be a square (a square $\!\!\pmod{4}$ is either $0$ or $1$). It follows that the trivial solution is the only solution.
This the first attempted solution not only on this post but also my overall first:I can only give you a start with this answer.
N can take either an even value or an odd value. Im attempting only one part-easier one.
Case I(N is even)
The last digit of powers of 2 and 7 repeat in groups of four i.e (2,4,8,6) and (7,9,3,1).For N=4k+2 and N=4k both the last digit of the overall expression turns out to be (4+9=)3 and (6+1=)7,both of which can never be last digits of a perfect square.
(or)
Simply take modulo 3 when n is even,which tells us that the whole expression is 2(mod 3),never a perfect square.
Ill try to complete it soon.(Sorry for not using LaTex as my knowledge of it is non existent.)
|
Tool to calculate Double Integral. The calculation of two consecutive integral makes it possible to compute areas for functions with two variables to integrate over a given interval.
Double Integral - dCode
Tag(s) : Functions, Symbolic Computation
dCode is free and its tools are a valuable help in games, maths, geocaching, puzzles and problems to solve every day!
A suggestion ? a feedback ? a bug ? an idea ? Write to dCode!
Sponsored ads
Tool to calculate Double Integral. The calculation of two consecutive integral makes it possible to compute areas for functions with two variables to integrate over a given interval.
The calculation of double integral is equivalent to a calculation of two consecutive integrals, from the innermost to the outermost.
$$ \iint f(x,y) \text{d}x \text{d}y = \int_{(y)} \left(\int_{(x)} f(x,y) \text{d}x \right) \text{d}y $$
Example: Calculate the integral of $ f(x,y)=x+y $ over $ x \in [0,1] $ and $ y \in [0,2] $ $$ \int_{0}^{2} \int_{0}^{1} x+y \text{ d}x\text{ d}y = \int_{0}^{2} \frac{1}{2}y^2+y \text{ d}y = 3 $$
Enter the function on dCode with the upper and lower bounds for each variable and the calculator will return the resultat automatically.
It is possible to use variables in the bounds of the integrals:
$$ \iint (x+y) \text{ d}x \text{ d}y = \int_0^1 \left( \int_0^{y} (x+y) \text{ d}x \right) \text{ d}y $$
Polar coordinates are useful for performing area calculations via double integration by variable change:
$$ \iint f(x,y) \text{ d}x \text{ d}y = \iint (r\cos(\theta),r\sin(\theta))r\text{ d}r \text{ d}\theta $$
dCode retains ownership of the source code of the script Double Integral online. Except explicit open source licence (indicated Creative Commons / free), any algorithm, applet, snippet, software (converter, solver, encryption / decryption, encoding / decoding, ciphering / deciphering, translator), or any function (convert, solve, decrypt, encrypt, decipher, cipher, decode, code, translate) written in any informatic langauge (PHP, Java, C#, Python, Javascript, Matlab, etc.) which dCode owns rights will not be released for free. To download the online Double Integral script for offline use on PC, iPhone or Android, ask for price quote on contact page !
|
Difference between revisions of "Lower attic"
From Cantor's Attic
Line 1: Line 1: −
{{DISPLAYTITLE:The
+
{{DISPLAYTITLE: The }}
[[File:SagradaSpiralByDavidNikonvscanon.jpg | right | Sagrada Spiral photo by David Nikonvscanon]]
[[File:SagradaSpiralByDavidNikonvscanon.jpg | right | Sagrada Spiral photo by David Nikonvscanon]]
Revision as of 09:32, 29 December 2011
Welcome to the lower attic, where the countably infinite ordinals climb ever higher, one upon another, in an eternal self-similar reflecting ascent.
$\omega_1$, the first uncountable ordinal, and the other uncountable cardinals of the middle attic stable ordinals The ordinals of infinite time Turing machines, including $\omega_1^x$ admissible ordinals Church-Kleene $\omega_1^{ck}$, the supremum of the computable ordinals $\Gamma$ $\epsilon_0$ and the hierarchy of $\epsilon_\alpha$ numbers $\omega+1$, $\omega\cdot 2$, $\omega^3$, $\omega^{\omega^\omega}$ and the other small countable ordinals below $\epsilon_0$ Hilbert's hotel $\omega$, the smallest infinity down to the subattic, containing very large finite numbers
|
Wave energy converters in coastal structures: verschil tussen versies
(→Application for wave energy converters)
(→Application for wave energy converters)
Regel 80: Regel 80:
where <math>m_n</math>
where <math>m_n</math>
−
represents the spectral moment of order n. An equation similar to that describing the power of regular waves is then obtained :
+
represents the spectral moment of order n. An equation similar to that describing the power of regular waves is then obtained :
<p>
<p>
<br>
<br>
Versie van 3 sep 2012 om 12:04 Introduction
Fig 1: Construction of a coastal structure.
Coastal works along European coasts are composed of very diverse structures. Many coastal structures are ageing and facing problems of stability, sustainability and erosion. Moreover climate change and especially sea level rise represent a new danger for them. Coastal dykes in Europe will indeed be exposed to waves with heights that are greater than the dykes were designed to withstand, in particular all the structures built in shallow water where the depth imposes the maximal amplitude because of wave breaking.
This necessary adaptation will be costly but will provide an opportunity to integrate converters of sustainable energy in the new maritime structures along the coasts and in particular in harbours. This initiative will contribute to the reduction of the greenhouse effect. Produced energy can be directly used for the energy consumption in harbour area and will reduce the carbon footprint of harbours by feeding the docked ships with green energy. Nowadays these ships use their motors to produce electricity power on board even if they are docked. Integration of wave energy converters (WEC) in coastal structures will favour the emergence of the new concept of future harbours with zero emissions.
Inhoud Wave energy and wave energy flux
For regular water waves, the time-mean wave energy density E per unit horizontal area on the water surface (J/m²) is the sum of kinetic and potential energy density per unit horizontal area. The potential energy density is equal to the kinetic energy
[1] both contributing half to the time-mean wave energy density E that is proportional to the wave height squared according to linear wave theory [1]:
(1)
[math]E= \frac{1}{8} \rho g H^2[/math]
g is the gravity and [math]H[/math] the wave height of regular water waves. As the waves propagate, their energy is transported. The energy transport velocity is the group velocity. As a result, the time-mean wave energy flux per unit crest length (W/m) perpendicular to the wave propagation direction, is equal to
[1]:
(2)
[math] P= Ec_{g}[/math]
with [math]c_{g}[/math] the group velocity (m/s). Due to the dispersion relation for water waves under the action of gravity, the group velocity depends on the wavelength λ (m), or equivalently, on the wave period T (s). Further, the dispersion relation is a function of the water depth h (m). As a result, the group velocity behaves differently in the limits of deep and shallow water, and at intermediate depths:
[math](\frac{\lambda}{20} \lt h \lt \frac{\lambda}{2})[/math]
Application for wave energy convertersFor regular waves in deep water:
[math]c_{g} = \frac{gT}{4\pi} [/math] and [math]P_{w1} = \frac{\rho g^2}{32 \pi} H^2 T[/math]
The time-mean wave energy flux per unit crest length is used as one of the main criteria to choose a site for wave energy converters.
For real seas, whose waves are random in height, period (and direction), the spectral parameters have to be used. [math]H_{m0} [/math] the spectral estimate of significant wave height is based on zero-order moment of the spectral function as [math]H_{m0} = 4 \sqrt{m_0} [/math] Moreover the wave period is derived as follows
[2].
[math]T_e = \frac{m_{-1}}{m_0} [/math]
where [math]m_n[/math] represents the spectral moment of order n. An equation similar to that describing the power of regular waves is then obtained
[3]:
[math]P_{w1} = \frac{\rho g^2}{64 \pi} H_{m0}^2 T_e[/math]
If local data are available ([math]H_{m0}^2, T_e [/math]) for a sea state through in-situ wave buoys for example, satellite data or numerical modelling, the last equation giving wave energy flux [math]P_{w1}[/math] gives a first estimation. Averaged over a season or a year, it represents the maximal energetic resource that can be theoretically extracted from wave energy. If the directional spectrum of sea state variance F (f,[math]\theta[/math]) is known with f the wave frequency (Hz) and [math]\theta[/math] the wave direction (rad), a more accurate formulation is used:
[math]P_{w2} = \rho g\int\int c_{g}(f,h)F(f,\theta) dfd \theta[/math]
Fig 2: Time-mean wave energy flux along
West European coasts
[4] .
It can be shown easily that equations (5 and 6) can be reduced to (4) with the hypothesis of regular waves in deep water. The directional spectrum is deduced from directional wave buoys, SAR images or advanced spectral wind-wave models, known as third-generation models, such as WAM, WAVEWATCH III, TOMAWAC or SWAN. These models solve the spectral action balance equation without any a priori restrictions on the spectrum for the evolution of wave growth.
From TOMAWAC model, the near shore wave atlas ANEMOC along the coasts of Europe and France based on the numerical modelling of wave climate over 25 years has been produced
[5]. Using equation (4), the time-mean wave energy flux along West European coasts is obtained (see Fig. 2). This equation (4) still presents some limits like the definition of the bounds of the integration. Moreover, the objective to get data on the wave energy near coastal structures in shallow or intermediate water requires the use of numerical models that are able to represent the physical processes of wave propagation like the refraction, shoaling, dissipation by bottom friction or by wave breaking, interactions with tides and diffraction by islands.
The wave energy flux is therefore calculated usually for water depth superior to 20 m. This maximal energetic resource calculated in deep water will be limited in the coastal zone:
at low tide by wave breaking; at high tide in storm event when the wave height exceeds the maximal operating conditions; by screen effect due to the presence of capes, spits, reefs, islands,...
Technologies
According to the International Energy Agency (IEA), more than hundred systems of wave energy conversion are in development in the world. Among them, many can be integrated in coastal structures. Evaluations based on objective criteria are necessary in order to sort theses systems and to determine the most promising solutions.
Criteria are in particular:
the converter efficiency : the aim is to estimate the energy produced by the converter. The efficiency gives an estimate of the number of kWh that is produced by the machine but not the cost. the converter survivability : the capacity of the converter to survive in extreme conditions. The survivability gives an estimate of the cost considering that the weaker are the extreme efforts in comparison with the mean effort, the smaller is the cost.
Unfortunately, few data are available in literature. In order to determine the characteristics of the different wave energy technologies, it is necessary to class them first in four main families
[4].
An interesting result is that the maximum average wave power that a point absorber can absorb [math]P_{abs} [/math](W) from the waves does not depend on its dimensions
[6]. It is theoretically possible to absorb a lot of energy with only a small buoy. It can be shown that for a body with a vertical axis of symmetry (but otherwise arbitrary geometry) oscillating in heave the capture (or absorption) width [math]L_{max}[/math](m) is as follows [6]:
[math]L_{max} = \frac{P_{abs}}{P_{w}} = \frac{\lambda}{2\pi}[/math] or [math]1 = \frac{P_{abs}}{P_{w}} \frac{2\pi}{\lambda}[/math]
Fig 4: Upper limit of mean wave power
absorption for a heaving point absorber.
where [math]{P_{w}}[/math] is the wave energy flux per unit crest length (W/m). An optimally damped buoy responds however efficiently to a relatively narrow band of wave periods.
Babarit et Hals propose
[7] to derive that upper limit for the mean annual power in irregular waves at some typical locations where one could be interested in putting some wave energy devices. The mean annual power absorption tends to increase linearly with the wave power resource. Overall, one can say that for a typical site whose resource is between 20-30 kW/m, the upper limit of mean wave power absorption is about 1 MW for a heaving WEC with a capture width between 30-50 m.
In order to complete these theoretical results and to describe the efficiency of the WEC in practical situations, the capture width ratio [math]\eta[/math] is also usually introduced. It is defined as the ratio between the absorbed power and the available wave power resource per meter of wave front times a relevant dimension B [m].
[math]\eta = \frac{P_{abs}}{P_{w}B} [/math]
The choice of the dimension B will depend on the working principle of the WEC. Most of the time, it should be chosen as the width of the device, but in some cases another dimension is more relevant. Estimations of this ratio [math]\eta[/math] are given
[7]: 33 % for OWC, 13 % for overtopping devices, 9-29 % for heaving buoys, 20-41 % for pitching devices. For energy converted to electricity, one must take into account moreover the energy losses in other components of the system.
Civil engineering
Never forget that the energy conversion is only a secondary function for the coastal structure. The primary function of the coastal structure is still protection. It is necessary to verify whether integration of WEC modifies performance criteria of overtopping and stability and to assess the consequences for the construction cost.
Integration of WEC in coastal structures will always be easier for a new structure than for an existing one. In the latter case, it requires some knowledge on the existing coastal structures. Solutions differ according to sea state but also to type of structures (rubble mound breakwater, caisson breakwaters with typically vertical sides). Some types of WEC are more appropriate with some types of coastal structures.
Fig 5: Several OWC (Oscillating water column) configurations (by Wavegen – Voith Hydro).
Environmental impact
Wave absorption if it is significant will change hydrodynamics along the structure. If there is mobile bottom in front of the structure, a sand deposit can occur. Ecosystems can also be altered by change of hydrodynamics and but acoustic noise generated by the machines.
Fig 6: Finistere area and locations of
the six sites (google map).
Study case: Finistere area
Finistere area is an interesting study case because it is located in the far west of Brittany peninsula and receives in consequence the largest wave energy flux along the French coasts (see Fig.2). This area with a very ragged coast gathers moreover many commercial ports, fishing ports, yachting ports. The area produces a weak part of its consumption and is located far from electricity power plants. There are therefore needs for renewable energies that are produced locally. This issue is important in particular in islands. The production of electricity by wave energy will have seasonal variations. Wave energy flux is indeed larger in winter than in summer. The consumption has peaks in winter due to heating of buildings but the consumption in summer is also strong due to the arrival of tourists.
Six sites are selected (see figure 7) for a preliminary study of wave energy flux and capacity of integration of wave energy converters. The wave energy flux is expected to be in the range of 1 – 10 kW/m. The length of each breakwater exceeds 200 meters. The wave power along each structure is therefore estimated between 200 kW and 2 MW. Note that there exist much longer coastal structures like for example Cherbourg (France) with a length of 6 kilometres.
(1) Roscoff (300 meters) (2) Molène (200 meters) (3) Le Conquet (200 meters) (4) Esquibien (300 meters) (5) Saint-Guénolé (200 meters) (6) Lesconil (200 meters) Fig.7: Finistere area, the six coastal structures and their length (google map).
Wave power flux along the structure depends on local parameters: bottom depth that fronts the structure toe, the presence of caps, the direction of waves and the orientation of the coastal structure. See figure 8 for the statistics of wave directions measured by a wave buoy located at the Pierres Noires Lighthouse. These measurements show that structures well-oriented to West waves should be chosen in priority. Peaks of consumption occur often with low temperatures in winter coming with winds from East- North-East directions. Structures well-oriented to East waves could therefore be also interesting even if the mean production is weak.
Fig 8: Wave measurements at the Pierres Noires Lighthouse.
Conclusion
Wave energy converters (WEC) in coastal structures can be considered as a land renewable energy. The expected energy can be compared with the energy of land wind farms but not with offshore wind farms whose number and power are much larger. As a land system, the maintenance will be easy. Except the energy production, the advantages of such systems are :
a “zero emission” port industrial tourism test of WEC for future offshore installations.
Acknowledgement
This work is in progress in the frame of the national project EMACOP funded by the French Ministry of Ecology, Sustainable Development and Energy.
See also Waves Wave transformation Groynes Seawall Seawalls and revetments Coastal defense techniques Wave energy converters Shore protection, coast protection and sea defence methods Overtopping resistant dikes
References Mei C.C. (1989) The applied dynamics of ocean surface waves. Advanced series on ocean engineering. World Scientific Publishing Ltd Vicinanza D., Cappietti L., Ferrante V. and Contestabile P. (2011) : Estimation of the wave energy along the Italian offshore, journal of coastal research, special issue 64, pp 613 - 617. Citefout: Onjuist label
<ref>; er is geen tekst opgegeven voor referenties met de naam
ref2
Mattarolo G., Benoit M., Lafon F. (2009), Wave energy resource off the French coasts: the ANEMOC database applied to the energy yield evaluation of Wave Energy, 10th European Wave and Tidal Energy Conference Series (EWTEC’2009), Uppsala (Sweden) Benoit M. and Lafon F. (2004) : A nearshore wave atlas along the coasts of France based on the numerical modeling of wave climate over 25 years, 29th International Conference on Coastal Engineering (ICCE’2004), Lisbonne (Portugal), pp 714-726. De O. Falcão A. F. (2010) Wave energy utilization: A review of the technologies. Renewable and Sustainable Energy Reviews, Volume 14, Issue 3, April 2010, pp. 899–918. Babarit A. and Hals J. (2011) On the maximum and actual capture width ratio of wave energy converters – 11th European Wave and Tidal Energy Conference Series (EWTEC’2011) – Southampton (U-K).
|
So I was recently discussing the transitions in Egyptian Blue ($\ce{CaCu[Si4O10]}$) with some of my students, who had to prepare this compound. What I like in particular in this case is how, at least in a simplified view you can show, that the blue is not simply due to one single transition with the complementary color to blue but actually composed of three possible transitions in the visible region that cause an absorption by all colors but blue. I also looked for a paper so they could give a reference in their report and this paper summarizes the transitions pretty well.
But when I thought about it for a while some questions came up in my mind on the actual transitions. So there are three transitions, either from the $\mathrm{a_{1g}}$, the set of $\mathrm{e_{g}}$, or the $\mathrm{b_{2g}}$ into the $\mathrm{b_{1g}}$. As the square plane has an inversion center we are dealing with the same problem as in octahedral geometries, the odd / even parity rule upon excitation.
Then I found this line in a text on symmetry rules in electronic transitions
In cases where transitions coincide with vibrations of the initial or final state, the electronic transition moment R needs to be replaced by the transition moment Rv. [...] Here Ψv denotes a vibronic wave function. In complete analogy with electronic transitions, we could derive the following set of selection rules for vibronic transitions: $$\Gamma(\psi_v') \otimes \Gamma(\psi_v'') = \Gamma(T_x)$$
At first I misread that line and thought that the dipole moment would change to vibrational modes as well, so I could choose from a larger set of possible symmetry elements to do the transitions, but then going through the examples at the bottom of the page it seems more like the vibration would cause a lower symmetry and change the point group into one where the transition may be allowed. It further says:
The total symmetry of a system can be expressed as a conjunction of the symmetry of the electronic states Γ(Ψ) and the symmetry of the vibration Γ(Ψv). $$\Gamma(\psi_{ev}) = \Gamma(\psi) \otimes \Gamma(\psi_v)$$
And this is the line that I don't understand anymore. So I looked at the examples And for the one with the point group $\mathrm{c_{2v}}$ for the transition between $\mathrm{b_{1}}$ and $\mathrm{b_{2}}$ it says:
The picture changes if we account for vibrational modes too. An asymmetric stretching causes the molecule to get from $\mathrm{c_{2v}}$ to $\mathrm{c_{s}}$ and only the molecular plane remains as symmetry element. State $\mathrm{B_{2}}$ becomes $\mathrm{A'}$ and state $\mathrm{B_{1}}$ $\mathrm{A''}$. Consequently, the transition dipole moment is of symmetry $\mathrm{A''}$ and perpendicular to the plane of the molecule. Obviously, reduced symmetry increases the number of possible transitions.
I also found a correlation table for the point group $\mathrm{c_{2v}}$ that links it to it's sub-groups. For $\mathrm{c_{2v}}$ the irreducible representation for the vibrational modes should give: $$\Gamma_{vib} = 2 A_1 + B_2$$
So for $\mathrm{B_{2}}$ we have this case, mentioned in the link, where we have to change to a vibrational mode. But how do I determine what sub-group is created by this vibration? In their example they showed that $\mathrm{B_{2}}$ becomes $\mathrm{A'}$. In the correlation table this would be $\mathrm{c_{s} (σ_{yz})}$. And if I then go back to the character table for $\mathrm{c_{2v}}$ the only thing that I could find would be that for this entry, $\mathrm{c_{s} (σ_{yz})}$, $\mathrm{B_{1}}$ would return -1 while $\mathrm{B_{2}}$ remains unchanged.
So does this give me any hint on how to transform my point group to its sub-group by a vibrational mode? As I have never had any lecture on point groups and symmetry I can unfortunately only look for tables, texts or similarities since I have no clue about point groups at all. This means the above-stated ideas could be terribly wrong. I tried to understand the text I quoted above as well as possible and this would be my interpretation.
This means that in my Egyptian Blue case, I would determine the symmetry element of the transition dipole moment. If the transition is not allowed I consider the vibrational normal modes and if either the initial and or final state has the same symmetry element, then I need to find out (using a correlation table) how the point group can be changed to a sub-group, where the transition would be allowed.
Therefore, I finally remain with the question: How do I determine which sub-group is chosen by the actional of a vibrational mode in this specific case?
|
There are six power rules in exponentiation and here is the list of the formulas in algebraic form.
$\large {(b^m)}^n \,=\, b^{mn}$
$b^0 \,=\, 1$
$\large b^{-m} \,=\, \dfrac{1}{b^m}$
$\large b^{\frac{1}{n}} \,=\, \sqrt[\displaystyle n]{b}$
$\large b^{\frac{m}{n}} \,=\, \sqrt[\displaystyle n]{b^m}$
$\large b^1 \,=\, b$
Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.
|
Let $c \in \mathbb{R}$ be such that $\{f,f_c\}$ is a basis for $V$, where $f_c(x)=f(x+c)$. Then by definition there are unique functions $a,b\colon \mathbb{R} \to \mathbb{R}$ such that $f_t=a(t)f+b(t)f_c$ for all $t \in \mathbb{R}$. Now for any $x_1, x_2 \in \mathbb{R}$, let $$M(x_1,x_2)=\left[\begin{matrix} f(x_1) & f_c(x_1) \\ f(x_2) & f_c(x_2)\end{matrix}\right]$$There must exist $x_1,x_2$ such that $M(x_1,x_2)$ is nonsingular, since otherwise $f$ would be an exponential function and $V$ would only have dimension one. Then for such $x_1$ and $x_3$, $\left[\begin{matrix} a(t) \\ b(t)\end{matrix}\right]=(M(x_1,x_2))^{-1}\left[\begin{matrix}f_t(x_1) \\ f_t(x_2)\end{matrix}\right]=(M(x_1,x_2))^{-1}\left[\begin{matrix}f_{x_1}(t) \\ f_{x_2}(t)\end{matrix}\right]$, so $a$ and $b$ can be written as linear combinations of $f_{x_1}$ and $f_{x_2}$. Since $f$ and its translates are differentiable, this implies that $a$ and $b$ are differentiable.
Now taking $\frac{\partial}{\partial t}$ of both sides of $f_t=a(t)f+b(t)f_c$ at $t=0$, we get $f'=a'(0)f+b'(0)f_c$. This is a linear combination of $f$ and $f_c$, so $f' \in V$. Note that $f$ and $f'$ are linearly independent (since otherwise $f$ would be an exponential and $V$ would have dimension one), so $\{f, f'\}$ span $V$. Since $f', f'_c \in V$, we also have $f'' = a'(0)f'+b'(0)f_c' \in V$, so $f'' \in \text{span}\{f, f''\}$. Therefore $f$ satisfies some ODE of the form
$$ \frac{d^{2}f}{dx^{2}}+B\frac{df}{dx}+Cf=0 \:\:\:\:\:B,C\text{ constants }$$
As T.A.E. pointed out, the solutions to this ODE are all solutions to to the question (excluding $f(x)=c e^{rx}$, since in this case $V$ only has dimension one), so the solutions to the question are:
$f(x)= c_1e^{r_1 x}+c_2 e^{r_2 x}\:\:\:$ ($c_1, c_2 \neq 0$ and $r_1 \neq r_2$)
$f(x) = (c_1+c_2 x)e^{rx}\:\:\:\:\,\,$ ($c_2 \neq 0$)
$f(x) = ce^{rx}\sin(sx+t)\,\,$ ($c, s \neq 0$)
(Postscript: Note that we did not need to initially assume that $V$ was a vector space, just that $V$ contains all $af_b$ for $a,b \in \mathbb{R}$ and is contained in $\text{span}\{f, f_c\}$ for some translate $f_c$ which is linearly independent of $f$. For then it still follows that $f', f'' \in \text{span}\{f, f_c\}$, so $f$ obeys the ODE. It is then an observed property of the solutions that $V$ is indeed a vector space.)
|
Defining parameters
Level: \( N \) = \( 12 = 2^{2} \cdot 3 \) Weight: \( k \) = \( 6 \) Character orbit: \([\chi]\) = 12.a (trivial) Character field: \(\Q\) Newforms: \( 0 \) Sturm bound: \(12\) Trace bound: \(0\) Dimensions
The following table gives the dimensions of various subspaces of \(M_{6}(\Gamma_0(12))\).
Total New Old Modular forms 13 0 13 Cusp forms 7 0 7 Eisenstein series 6 0 6
|
Assume two identical solid blocks of copper with molar heat capacity at constant volume $C_{V,m}=24.5 \text{ J$\cdot$K$^{-1}$mol$^{-1}$}$ . Assume the heat capacity to be constant in the temperature range of the experiment. Block 1 has a temperature $T_1$ and block 2 has a temperature $T_2=4T_1$. The copper blocks are thermally isolated from the surrounding. Only energy exchange between the two blocks is possible.
a) What is the final temperature $T_{final}$, both blocks will eventually reach, when they are in thermal contact?
b) Calculate the entropy change $\Delta S$ for each of the blocks. What is the total change in entropy? Would it have been possible to predict the sign (minus or plus) of the entropy change (give arguments!).
c) Now, instead of direct thermal contact, a reversible thermodynamic machine is placed between both copper blocks at their initial temperatures $T_1$ and $T_2$. What is the total entropy change now (give arguments)?
d) Calculate the entropy change $\Delta S$ for each of the blocks. Use the result to show, that for the case of the reversible thermodynamic machine $T_{final}=2T_1$.
e) Show, that the work done by the reversible thermodynamic machine is: $w= C_{V,m}T_1$
I tried to answer this question as part of an exam preparation (no homework) and got stuck at part C. Part A and B are quite trivial. You can show that for part A, $T_{final}=2.5T_1$. For part B we see that
$$\Delta S = \int_{T_1}^{T_{final}}\frac{C_{V,m}\mathrm{d}T}{T}+\int_{T_2}^{T_{final}}\frac{C_{V,m}\mathrm{d}T}{T}=C_{V,m}\ln \frac{25}{16}\approx 10.9 \text{ J$\cdot$K$^{-1}$mol$^{-1}$}$$ Now I do not quite get C and D (E follows very easily from the first law and the answer found at part D, i.e. $T_{final}=2T_1$). What is exactly meant by a reversible thermodynamic machine, and moreover, how do I calculate part C and D? I can see that the total entropy change will be lower when some of the thermal energy is used to do work, but I don't know how to quantitatively calculate the new entropy change.
|
What is Induction?
Induction is also known as inductance and is defined for a conductor such that the magnetic field is proportional to the rate of change of the magnetic field. L is used to represent the inductance and Henry is the SI unit of inductance.
1 Henry is defined as the amount of inductance required to produce an emf of 1 volt in a conductor when the current change in the conductor is at the rate of 1 Ampere per second.
Factors Affecting Inductance
Following are the factors that affect the inductance:
The number of turns of the wire used in the inductor. The material used in the core. The shape of the core.
Electromagnetic Induction law was given by Faraday which states that by varying the magnetic flux electromotive force is induced in the circuit. From Faraday’s law of electromagnetic induction, the concept of induction is derived. Inductance can be defined as the electromotive force generated to oppose the change in current in a particular time duration.
According to Faraday’s Law:
Electromotive force = – L \( \frac {\Delta I}{\Delta t}\)
Unit of Inductance = \( \frac {Volt ~Second }{Ampere}\)=Henry
Types of Inductance
Two types of inductance are there:
Self Induction Mutual Induction
What is Self Induction?
When there is a change in the current or magnetic flux of the coil, an opposed induced electromotive force is produced. This phenomenon is termed as Self Induction. When the current starts flowing through the coil at any instant, it is found that that the magnetic flux becomes directly proportional to the current passing through the circuit. The relation is given as:
\( \phi \)= I
\( \phi \) = L I
Where L is termed as self-inductance of the coil or the coefficient of self-inductance. The self-inductance depends on the cross-sectional area, the permeability of the material or the number of turns in the coil.
The rate of change of magnetic flux in the coil is given as,
e = – \( \frac {d \phi}{dt} \) = – \( \frac {d (LI) }{dt} \)
or e = – L \( \frac {dI}{dt} \)
What is Mutual Induction?
We take two coils, and they are placed close to each other. The two coils are P- coil (Primary coil) and S- coil (Secondary coil). To the P-coil, a battery, and a key is connected wherein the S-coil a galvanometer is connected across it. When there is a change in the current or magnetic flux linked with two coils an opposing electromotive force is produced across each coil, and this phenomenon is termed as Mutual Induction. The relation is given as:
\( \phi \) = I
\( \phi \) = M I
Where M is termed as the mutual inductance of the two coils or the coefficient of the mutual inductance of the two coils.
The rate of change of magnetic flux in the coil is given as,
e = – \( \frac {d \phi}{dt} \) = – \( \frac {d (MI)}{dt}\)
e = – M \( \frac {dI}{dt} \)
Know more about the inductance and other related concepts like the uses of the inductor by visiting BYJU’S.
|
Equivalence of Definitions of Local Basis/Neighborhood Basis of Open Sets Implies Local Basis for Open Sets Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let $x$ be an element of $S$.
every neighborhood of $x$ contains a set in $\mathcal B$. Then $\mathcal B$ satisfies: $\forall U \in \tau: x \in U \implies \exists H \in \mathcal B: H \subseteq U$ Proof
Let $U \in \tau$ such that $x \in U$.
By assumption, there exists $H \in \mathcal B$ such that $H \subseteq U$.
The result follows.
$\blacksquare$
|
Example of Converting from Slope Intercept to Standard Form
Multiply by the least common denominator of the fractions (if any)
The only fraction is $$ \frac { 5}{ \red 4} $$ so you can multiply everything by 4.
$ \red 4 \cdot y = \red 4 \cdot \big( \frac { 5}{ \red 4}x +5 \big) \\ 4y = 5x + 20 $ Use our Calculator
You can use the calculator below to find the equation of a line from any two points. Just type numbers into the boxes below and the calculator (which has its own page here) will automatically calculate the equation of line in standard and slope intercept forms
Answer:
$ $
( Try this 'equation from 2 points' calculator on its own page here . )
Practice Problems
Multiply by the least common denominator of the fractions
The only fraction is $$ \frac{2}{\red 3 } $$ so you can multiply everything by 3.
$$ \red 3 \cdot y = \red 3 \big( \frac{2}{\red 3 } -4 \big) \\ 3y = 2x-12 $$
Multiply by the least common denominator of the fractions (if any)
The only fraction is $$ \frac{1}{\red 2}$$ so you can multiply everything by 2.
$$ \red 2 \cdot y =\red 2 \cdot \big( \frac{1}{\red 2} x + 5 \big) \\2y = x + 10 $$
Unlike the prior examples, this problem has two fractions $$ \big( \frac 2 3 \text{ and } \frac 5 9 \big) $$ so you can multiply everything by their common denominator of $$ \red 9$$ .
$$ \red 9 \cdot y = \red 9 \big( \frac 2 3 x + \frac 5 9 \big) \\ 9y = 6x + 5 $$
This problem only has the fraction $$ \frac{4}{\red 7 } $$ so we must multiply everything by 7.
$$ \red 7 \cdot y = \red 7 \big( 5x + \frac 4 7 \big) \\ 7y = 35x + 4 $$
This problem only has the fraction $$ \frac{b}{ \red c } $$ so we must multiply everything by $$ \red c $$.
$$ \red c \cdot y = \red c \big( ax + \frac b c \big) \\ cy = cax + b $$ .
|
The value of secant function when angle of right triangle equals to $45^°$ is called secant of angle $45$ degrees. It is written as $\sec{(45^°)}$ as per sexagesimal system in mathematics.
$\sec{(45^°)} \,=\, \sqrt{2}$
The value of sec of angle $45$ degrees in fraction is $\sqrt{2}$ exactly. It is an irrational number and equal to $1.4142135624\ldots$ in decimal form but the approximate value of secant of angle $45$ degrees is considered as $1.4142$ in mathematics. Generally, the value of $\sec{(45^°)}$ is called as trigonometric function or trigonometric ratio for standard angle.
$\sec{(45^°)}$ is written in alternative form as $\sec{\Big(\dfrac{\pi}{4}\Big)}$ in circular system and also written as $\sec{(50^g)}$ in centesimal system.
$(1) \,\,\,$ $\sec{\Big(\dfrac{\pi}{4}\Big)}$ $\,=\,$ $\sqrt{2}$ $\,=\,$ $1.4142135624\ldots$
$(2) \,\,\,$ $\sec{(50^g)}$ $\,=\,$ $\sqrt{2}$ $\,=\,$ $1.4142135624\ldots$
You know the exact value of secant of angle $45$ degrees in both decimal and fraction from and it is time to learn how to evaluate the exact value of $\sec{\Big(\dfrac{\pi}{4}\Big)}$ in trigonometry.
Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.
|
There are three basic notable properties in a right triangle when its angle equals to $30$ degrees.
$\Delta POQ$ is a right triangle and its angle is $30^°$. Assume, the length of hypotenuse is equal to $d$.
Then, the length of opposite side is exactly equal to half of the length of the hypotenuse.
$Length \, of \, Opposite \, side$ $\,=\,$ $\dfrac{Length \, of \, Hypotenuse}{2}$
$\implies$ $Length \, of \, Opposite \, side$ $\,=\,$ $\dfrac{d}{2}$
The lengths of opposite side and hypotenuse are known. They can be used to calculate the length of adjacent side (base) mathematically by Pythagorean Theorem.
${OP}^2 \,=\, {PQ}^2 + {OQ}^2$
$\implies d^2 \,=\, {\Bigg(\dfrac{d}{2}\Bigg)}^2 + OQ^2$
$\implies d^2 \,-\, {\Bigg(\dfrac{d}{2}\Bigg)}^2 \,=\, OQ^2$
$\implies OQ^2 \,=\, d^2 \,-\, {\Bigg(\dfrac{d}{2}\Bigg)}^2$
$\implies OQ^2 \,=\, d^2 \,-\, \dfrac{d^2}{4}$
$\implies OQ^2 \,=\, d^2 \Bigg[1 \,-\, \dfrac{1}{4}\Bigg]$
$\implies OQ^2 \,=\, d^2 \Bigg[\dfrac{3}{4}\Bigg]$
$\implies OQ^2 \,=\, \Bigg[\dfrac{3}{4}\Bigg]d^2$
$\implies OQ \,=\, \sqrt{\Bigg[\dfrac{3}{4}\Bigg]d^2}$
$\implies OQ \,=\, \dfrac{\sqrt{3}}{2}d$
$\implies OQ \,=\, \dfrac{\sqrt{3}}{2} \times d$
$\implies OQ \,=\, \dfrac{\sqrt{3}}{2} \times OP$
$\therefore \,\, Length \, of \, Adjacent \, side \,=\, \dfrac{\sqrt{3}}{2} \times Length \, of \, Hypotenuse$
The properties of right angled triangle can also be proved geometrically by constructing an example triangle with an angle of $30$ degrees.
Let us study the geometrical relations between sides of a right triangle when its angle is $30^°$.
The length of opposite side (perpendicular) equals to half of the length of hypotenuse when angle equals to $30$ degrees.
Now, take a ruler and measure the length of the opposite side ($\small \overline{ST}$). You observe that the length of opposite side $\small \overline{ST}$ is $5 \, cm$ exactly.
Actually, it is half of the length of the hypotenuse. It is possible only when the angle of a right triangle equals to $30^°$.
So, remember that the length of opposite side is always equal to half of the length of the hypotenuse when the angle of right angled triangle is $30^°$.
The length of adjacent side (base) equals to $\dfrac{\sqrt{3}}{2}$ times of the length of hypotenuse when angle equals to $30$ degrees.
Measure the length of the adjacent side $\small \overline{RT}$ by ruler and you observe that the length of the adjacent side is equal to $8.65 \, cm$. Later, calculate the length of the adjacent side in theoretical method to compare difference between them.
$Length \, of \, Adjacent \, side$ $\,=\,$ $\dfrac{\sqrt{3}}{2} \times Length \, of \, Hypotenuse$
$\implies$ $Length \, of \, Adjacent \, side \,=\, \dfrac{\sqrt{3}}{2} \times 10$
$\implies$ $Length \, of \, Adjacent \, side \,=\, 8.66025\ldots$
$\,\,\, \therefore \,\,\,\,\,\,$ $Length \, of \, Adjacent \, side \,\approx\, 8.66 \, cm$
Theoretically, the length of adjacent side is $8.66 \, cm$ approximately and geometrically, it is $8.65 \, cm$. They both are equal approximately in mathematics.
It proves that the length of adjacent side is equal to $\dfrac{\sqrt{3}}{2}$ times of length of hypotenuse when angle of right triangle is equal to $30^{°}$.
Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.
|
Risk and Return
Risk-averse investors attempt to maximize the return they earn per unit of risk. Ratios such as Sharpe ratio, Treynor’s ratio, Sortino ratio, etc. and coefficient of variation measure return per unit of investment risk.
It is important to analyze and attempt to quantify the relationship between risk and return. Modern portfolio theory and the capital asset pricing model offers a framework for analysis of risk-return trade-off.
Risk in equity investments is broadly classified into unique risk (also called unsystematic risk) and systematic risk.
Total Risk = Unique Risk + Systematic Risk
Unique risk vs systematic risk
Unique risk is the risk that arises from investment-specific factors. By adding more investments to a portfolio, unsystematic risk can be eliminated, hence, it is also called diversifiable risk. Unsystematic risk can be further classified into business risk and financial risk. Business risk is the risk of loss in business while financial risk is the risk of default due to the company taking on too much debt.
The systematic risk, on the other hand, is the risk of the whole economy and financial market performing poorly due to economy-wide factors. Its component risks are interest rate risk, inflation risk, regulatory risk, exchange rate risk, etc.
Since an individual investment is not diversified, it carries total risk and the appropriate measure of risk is the standard deviation. On the other hand, the risk of a fully diversified portfolio is best measured by its beta coefficient.
Individual investment risk and return Expected return of individual investment
Expected return of an individual investment can be estimated by creating a distribution of likely return and their associated likelihood using the following formula:
$$ \text{E}(\text{R})=\text{r} _ \text{1}\times \text{p} _ \text{1}+\text{r} _ \text{2}\times \text{p} _ \text{2}+\text{...}+\text{r} _ \text{n}\times \text{p} _ \text{n} $$
Where,
E(R) is the expected return on individual asset, r 1, r2 and rn are the first, second and n threturn outcomes, and pare the associated probabilities. 1, p 2and p n
The percentage return on an individual investment can be calculating using the following holding period return formula:
$$ \text{r}=\frac{\text{P} _ \text{t}-\text{P} _ {\text{t}-\text{1}}+\text{i}}{\text{P} _ {\text{t}-\text{1}}} $$
Where,
P t is the ending value of investment, Pis the beginning value of investment, and t-1 iis its income. Variance, standard deviation, and beta of individual investment
Historical standard deviation of an investment on a standalone basis can be estimated by using the following general formula:
$$ \sigma=\sqrt{\frac{\sum{(\text{x} _ \text{i}\ -\ \mu)}^\text{2}}{\text{n}}} $$
Where,
σ is standard deviation, x i is return value, µis the mean return value, and nis the total number of observations.
Forward-looking standard deviation is usually calculated based on the expected return and their associated probability distribution using the following formula:
$$ \sigma=\sqrt{\sum{(\text{r} _ \text{n}\ -\ \text{E}(\text{R}))}^\text{2}\times \text{p} _ {\ \text{n}}} $$
Where,
σ is standard deviation, r n is the nth return outcome, E(R)is the expected return i.e. the probability-weighted average return, and pis the associated probability. n
Variance of an individual investment simply equals squared standard deviation. In most cases standard deviation is a better measure.
The ratio of standard deviation of investment to its return is called coefficient of variation and it gives us an indication of total risk per unit of return.
Portfolio risk and return Expected return of a portfolio of investments
Expected return of a portfolio is calculated as the weighted average of the expected return on individual investments using the following formula:
$$ \text{E} _ \text{r}=\text{w} _ \text{1}\times \text{R} _ \text{1}+\text{w} _ \text{2}\times \text{R} _ \text{2}+\text{....}+\text{w} _ \text{n}\times \text{R} _ \text{n} $$
Where,
E r is the portfolio expected return, wis the weight of first asset in the portfolio, 1 Ris the expected return on the first asset, 1 wis the weight of second asset, and 2 Ris the expected return on the second asset and so on. 2 Portfolio risk: variance, standard deviation & beta
Standard deviation of a portfolio depends on the weight of each asset in the portfolio, standard deviation of individual investments and their mutual correlation coefficient. If two and more assets have correlation of less than 1, the portfolio standard deviation is lower than the weighted average standard deviation of the individual investments. This is because less than perfect correlation causes certain unique company-specific risks to cancel out each other. Portfolio standard deviation is calculated as follows:
$$ \sigma _ \text{P}=\sqrt{{\text{w} _ \text{A}}^\text{2}{\sigma _ \text{A}}^\text{2}{+\text{w} _ \text{A}}^\text{2}{\sigma _ \text{A}}^\text{2}+\text{2}\times \text{w} _ \text{A} \text{w} _ \text{B}\sigma _ \text{A}\sigma _ \text{B}\rho} $$
Where,
σ P = portfolio standard deviation, w= weight of asset A in the portfolio, A w= weight of asset B in the portfolio, A σ= standard deviation of asset A, A σ= standard deviation of asset B, and B ρ= correlation coefficient between returns on asset A and asset B.
Portfolio variance simply standard deviation raised to the power of 2.
Portfolio beta is the weighted average beta coefficient of the portfolio's constituent securities.
Diversification ratio
Diversification ratio ® is the extent of diversification of an investment portfolio. It is calculated by dividing the weighted average volatility (standard deviation) of the constituent investments divided by portfolio standard deviation.
$$ \text{Diversification Ratio}=\frac{\text{w} _ \text{A}\times\sigma _ \text{A}+\text{w} _ \text{B}\times\sigma _ \text{B}+\text{...}+\text{w} _ \text{n}\times\sigma _ \text{n}}{\sigma _ \text{P}} $$
Since the portfolio standard deviation in a diversified portfolio is lower than the weighted average of individual investment standard deviations, the ratio is greater than 1. A higher ratio is better.
by Obaidullah Jan, ACA, CFA and last modified on
|
How do we add a matrix to a LaTeX document?
Ash's answer typesets the matrix inline with the text. A (perhaps) nicer way to do this is to use the
smallmatrix environment in the
amsmath package. Add to the document preamble:
\usepackage{amsmath}
And then you can do:
$M = \begin{smallmatrix} a&b\\ c&d \end{smallmatrix}$
If you want to bracket the matrix you can also do:
$M = \left( \begin{smallmatrix} a&b\\ c&d \end{smallmatrix} \right)$
The
amsmath package also offers the shortcut
matrix environments which default to centered alignment for their columns:
matrix: unbracketed matrix
pmatrix: matrix surrounded by parentheses
bmatrix: matrix surrounded by square brackets
vmatrix: matrix surrounded by single vertical lines
Vmatrix: matrix surrounded by double vertical lines
This info is found in "The LaTeX Companion", and the
amsmath manual section 4.
First: if you intend to do math in LaTeX, you SHOULD learn and use AMS LaTeX. The best reference is the Short Math Guide for LaTeX. In this guide, you will learn that there are many different matrix macros available when you use the
amsmath package (e.g.,
\usepackage{amsmath} ).
To quote the document,
4.4. MatricesThe environments
pmatrix,
bmatrix,
Bmatrix,
vmatrixand
Vmatrixhave (respectively) ( ), [ ], { }, | |, and || || delimiters built in. There is also a
matrixenvironment sans delimiters, and an
arrayenvironment that can be used to obtain left alignment or other variations in the column specs. [ed. To produce a matrix with parenthesis around it, use:]
\begin{pmatrix} \alpha & \beta^{*}\\ \gamma^{*} & \delta \end{pmatrix}
To produce a small matrix suitable for use in text, there is a
smallmatrixenvironment [ed. here was a matrix appropriate for text mode] that comes closer to fitting within a single text line than a normal matrix. This example was produced by
\bigl( \begin{smallmatrix} a & b\\ c & d \end{smallmatrix} \bigr)
To produce a row of dots in a matrix spanning a given number of columns, use \hdotsfor. For example,
\hdotsfor{3}in the second column of a four-column matrix will print a row of dots across the final three columns.
Note. The plain TeX form
\matrix{...\cr...\cr}and the related commands
\pmatrix,
\casesshould be avoided in LaTeX (and when the
amsmathpackage is loaded they are disabled).
Finally, I'd like to mention that, while it is possible to set matrices without AMS LaTeX, just use it. It offers so many benefits that until you get the hang of LaTeX, it's the best single macro package for math.
For a matrix of the form:
M = x y z w
use the LaTeX code:
$M =\begin{array}{cc}x & y \\z & w \\\end{array}$
For flexible typesetting of matrices with color, lines and justification done by formatting parameters see An extension to amsmath matrix environments.
$M = \left[ {\begin{array}{*{20}c} x & y \\ z & w \\ \end{array} } \right]$$M = \left( {\begin{array}{*{20}c} x & y \\ z & w \\ \end{array} } \right)$$M = \left| {\begin{array}{*{20}c} x & y \\ z & w \\ \end{array} } \right|$
|
Search
Now showing items 1-10 of 24
Production of Σ(1385)± and Ξ(1530)0 in proton–proton collisions at √s = 7 TeV
(Springer, 2015-01-10)
The production of the strange and double-strange baryon resonances ((1385)±, Ξ(1530)0) has been measured at mid-rapidity (|y|< 0.5) in proton–proton collisions at √s = 7 TeV with the ALICE detector at the LHC. Transverse ...
Forward-backward multiplicity correlations in pp collisions at √s = 0.9, 2.76 and 7 TeV
(Springer, 2015-05-20)
The strength of forward-backward (FB) multiplicity correlations is measured by the ALICE detector in proton-proton (pp) collisions at s√ = 0.9, 2.76 and 7 TeV. The measurement is performed in the central pseudorapidity ...
Inclusive photon production at forward rapidities in proton-proton collisions at $\sqrt{s}$ = 0.9, 2.76 and 7 TeV
(Springer Berlin Heidelberg, 2015-04-09)
The multiplicity and pseudorapidity distributions of inclusive photons have been measured at forward rapidities ($2.3 < \eta < 3.9$) in proton-proton collisions at three center-of-mass energies, $\sqrt{s}=0.9$, 2.76 and 7 ...
Rapidity and transverse-momentum dependence of the inclusive J/$\mathbf{\psi}$ nuclear modification factor in p-Pb collisions at $\mathbf{\sqrt{\textit{s}_{NN}}}=5.02$ TeV
(Springer, 2015-06)
We have studied the transverse-momentum ($p_{\rm T}$) dependence of the inclusive J/$\psi$ production in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV, in three center-of-mass rapidity ($y_{\rm cms}$) regions, down to ...
Measurement of pion, kaon and proton production in proton–proton collisions at √s = 7 TeV
(Springer, 2015-05-27)
The measurement of primary π±, K±, p and p¯¯¯ production at mid-rapidity (|y|< 0.5) in proton–proton collisions at s√ = 7 TeV performed with a large ion collider experiment at the large hadron collider (LHC) is reported. ...
Two-pion femtoscopy in p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV
(American Physical Society, 2015-03)
We report the results of the femtoscopic analysis of pairs of identical pions measured in p-Pb collisions at $\sqrt{s_{\mathrm{NN}}}=5.02$ TeV. Femtoscopic radii are determined as a function of event multiplicity and pair ...
Measurement of charm and beauty production at central rapidity versus charged-particle multiplicity in proton-proton collisions at $\sqrt{s}$ = 7 TeV
(Springer, 2015-09)
Prompt D meson and non-prompt J/$\psi$ yields are studied as a function of the multiplicity of charged particles produced in inelastic proton-proton collisions at a centre-of-mass energy of $\sqrt{s}=7$ TeV. The results ...
Charged jet cross sections and properties in proton-proton collisions at $\sqrt{s}=7$ TeV
(American Physical Society, 2015-06)
The differential charged jet cross sections, jet fragmentation distributions, and jet shapes are measured in minimum bias proton-proton collisions at centre-of-mass energy $\sqrt{s}=7$ TeV using the ALICE detector at the ...
Centrality dependence of high-$p_{\rm T}$ D meson suppression in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Springer, 2015-11)
The nuclear modification factor, $R_{\rm AA}$, of the prompt charmed mesons ${\rm D^0}$, ${\rm D^+}$ and ${\rm D^{*+}}$, and their antiparticles, was measured with the ALICE detector in Pb-Pb collisions at a centre-of-mass ...
K*(892)$^0$ and $\Phi$(1020) production in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(American Physical Society, 2015-02)
The yields of the K*(892)$^0$ and $\Phi$(1020) resonances are measured in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV through their hadronic decays using the ALICE detector. The measurements are performed in multiple ...
|
I'll preface the examples I am trying to understand with the definitions of weak and weak-* convergence in a Banach space $X$:
"A sequence $(x_n)_{n=1}^\infty$ converges weakly to $x$ in $X$ if for all $f\in X'$, $f(x_n)\to f(x)$ in $\mathbb F$ as $n\to\infty.$ A sequence $(f_n)_{n=1}^\infty\subset X'$ converges weak-* to $f$ in $X'$ if for all $x\in X$, $f_n(x)\to f(x)$ in $\mathbb F$ as $n\to\infty.$"
In my notes I am given the following two examples, although, I am having trouble following them completely. In all examples $(e_n)_{n=1}^\infty$ denotes the infinite sequence with zero in every entry apart from its $n$-th entry, which is one.
Example $1$:
Let $(e_n)_{n=1}^\infty\subset\ell^2$. Then $(e_n)_{n=1}^\infty$ does not converge strongly anywhere. But since $(\ell ^2)'$ is isometrically isomorphic to $\ell^2$ we can represent every $f\in(\ell^2)'$ as the action of some $a\in \ell^2$ on another $x\in\ell^2$. Letting $f:=f_a$ we have that then $f_a(x)=\sum_{k=1}^\infty a_kx_k$. We then have that $f(e_n)=a_n$. My notes then go on to say that if $a\in\ell^2$ then $a_n\to0$ as $n\to\infty$ and $x_n\rightharpoonup x$ in $\ell^2$.
It is at this last sentence that I am lost. I understand that $f(e_n)=a_n$, but how does the fact that $a\in\ell^2$ allow us to deduce that $a_n\to0$ as $n\to\infty$ and then in turn allow us to establish the weak convergence?
Example $2$:
Take $(e_n)_{n=1}^\infty\subset\ell^1$. Then for every $f\in(\ell^1)'$ we can define similarly to what we done above $f:=f_a$ for some $a\in\ell^\infty$. But then my notes say that $f_a(e_n)=a_n$ does not have to go to zero as $n\to\infty$. From which point it then goes on to say that it is not true that $e_n\rightharpoonup 0$ in $\ell^1$.
Again, similar to the last example, I am completely lost on the latter part of the example. Why in this case for $a\in\ell^\infty$ does $f_a(e_n)=a_n$ not have to converge strongly to zero as $n$ increases?
|
@Tsemo Aristide is absolutely correct, you can follow that link and also The Proof for your specific case here. However, this is a different kind of explanation for what you have, which is not a proof but I think it might help you better grasp the concept.
At the first glance, this looks very similar to the Vandermonde matrix; except we have
$$ M = \left[ {\begin{array}{ccccc} x_1 & x_2 & x_3 & \cdots & x_n \\x_1^2 & x_2^2 & x_3^2 &\cdots & x_n^2\\ \vdots & & \ddots & & \vdots \\ x_1 ^n & x_2^n & x_3^n & \cdots &x_n^n \\ \end{array} } \right]$$Now let us consider the Vandermonde's matrix which is
$$V = \left[ {\begin{array}{ccccc} 1 & x_1 & x_1^2& \cdots & x_1^{n-1}\\ 1 & x_2 & x_2^2 &\cdots & x_2^{n-1}\\ \vdots & & \ddots & & \vdots \\ 1 & x_n& x_n & \cdots &x_n^{n-1}\\ \end{array} } \right]$$As we can see, there are very closely related; so we want to find the matrix $X$ so that $$VX=M$$ and in that case the determinant of $M$ will be $$\det(M) = \det(V) \cdot \det(X)$$Now let us look at an example of a $2 \times 2$ by $2 \times 2$ matrices, so assume that we have :$$\left[ {\begin{array}{cc} a & 0\\ 0 & b \\ \end{array} } \right] \times \left[ {\begin{array}{cc} 1 & 1\\ 2& 2 \\ \end{array} } \right] = \left[ {\begin{array}{cc} a & a\\ 2b& 2b \\ \end{array} } \right] $$ So we can see that by multiplying this diagonal matrix by a the square matrix, we can multiply each component of the first row by $a$ and second row by $b$. Now this is exactly what we want in order to go from our the Vandermonde matrix to the given matrix; furthermore this is very beneficial since we know the determinant of the diagonal matrix is the multiplication of the diagonals and we know the Vandermonde's determinant. So to get $M$ we will have
$VX = N $ where $$X = \left[ {\begin{array}{ccccc} x_1 & 0 & 0& \cdots & 0\\ 0 & x_2 & 0 &\cdots & 0\\ \vdots & & \ddots & & \vdots \\ 0&0 &0 & \cdots &x_n\\ \end{array} } \right]$$Now$$VX = \left[ {\begin{array}{ccccc} x_1 & x_1^2 & x_1^3 & \cdots & x_1^n\\ x_2 & x_2^2 & x_2^2 &\cdots & x_2^n\\ \vdots & & \ddots & & \vdots \\ x_n & x_n^2 & x_n^3 & \cdots &x_n^n\\ \end{array} } \right] = N $$ This is almost what we want except we have the transpose of this matrix, so $(VX)^T = M$. Also, for a square matrix, the determinant of the transpose is the same as the determinant of the matrix, we will have : $$\det(X)\cdot\det(V) = \prod_{1\leq i \leq n}x_i \cdot \prod_{1 \leq i \leq j \leq n} (x_j - x_i) = \det(M^T) = \det(M)$$
Again this is not a proof, but I hope this clarified some stuff from both of the links.
|
Quadratic Formula Calculator and Solver
The Discriminant
The Work
The Actual Solutions
The Graph
The calculator uses the quadratic formula to find solutions to any quadratic equation.
The formula is: $ \frac{ -b \pm \sqrt{b^2 -4ac}}{2a } $
The quadratic formula calculator below will solve any quadratic equation that you type in. Simply type in a number for 'a', 'b' and 'c' then hit the 'solve' button.
The Discriminant
The Work
The Actual Solutions
The Graph
The calculator on this page shows how the quadratic formula operates, but if you have access to a graphing calculator you should be able to solve quadratic equations, even ones with imaginary solutions.
|
A fundamental question about mathematics
This answer was reorganized after the OP gave more precisions as tothe meaning and intent of his question. I also comment other answershere, as it is awkward to do so in the usual comment format. Commenting them also gives extra insight into the relevant issues. In a nutshell
Your intuition is quite correct that the empty string plays a specialrole in the study of strings and formal languages, and that is thereason why it is often given a special name or notation. Strings overa given set of symbols form an algebraic structure called a monoid,with the concatenation operation which has a neutral element: theempty string. See the answer by J.-E. Pin.
You are also correct that there could be many other notations orrepresentation for it. The choice of representation is dictated byconvenience, perspicuity and simplification of discourse, reasoningand computation.
One such convenience, as you rightfully wonder, is having a uniformnotation for all strings, including the empty string. This can beachieved in several ways, whether on paper or in thecomputer. Terminating strings with a special symbol supposed not tobelong to the set of symbols included in the strings is one way ofdoing it. I guess this is what you suggest with EOL. This was donesome 45 years ago by Denis Ritchie for the programming language C,except that he used the byte 0, also noted NUL or ^@, rather than EOL.
In text it can be done with surrounding quotes, or with a finalturnstyle $\dashv$. Note however that while the $\dashv$ alone willdenote the empty string, it terminates then all strings, which is notthe case for the use of the letter ε. They do not play exactly thesame syntactic role.
In principle, such a termination symbol as EOL, ^@ or $\dashv$ cannotbe also a symbol belonging to a string, unless you add more complexrepresentation mechanisms.
In the computer, the null reference string could be used to representthe empty string. Otherwise it is only a programming concept that hasnothing to do with the abstract concept of string.
However your question was a bit confusing and not too wellstated. Talking of a "
separate concept" hints at semantic issuesrather than syntactic reresentation. And you were mixing textual,printed representations, which use εbut not EOL, with computerrepresentation which do the opposite. With many more details
This is a strange question. In its way, it also raises one or two fundamentalissues about mathematics.
Understanding such issues is non obvious, as witnessed by theinadequacies of some answers given by obviously competent users, andthe inadequacies of the question itself. This is what attracted me tothis question.
These two issues are concerned with :
The second issue, which has to do with semantics, has probably beenaddressed by logicians, and possibly by historians of science. But Ido not recall seeing it formally addressed (or possibly I did notrecognized it).
A confusion between syntax and semantics probably arose from the factthat the OP talks of a "
separate concept" where he should rathertalk of a " separate notation". Such a mistake is probably fair inhis case as he is trying to understand issues. But it further confused someusers who answered, clearly Yuval Filmus and myself, as we took the word"concept" for what it is supposed to mean. About Semantics
I realize now that the next paragraph is not about the question you intended; but it is the question you wrote, and which is to be understood as semantics, and was by several people, while you meant syntax (to be addressed in the syntax part below).
Let's start with yourquestion "
Why do you need a separate concept, that of 'empty string'?", which I understood as: "could we use strings, in theory and in programming, without ever considering the empty string?", as apparently did Yuval Filmus.
The fact is, we often do not need the empty string, but it is generally more convenient to have it. Most of the theory could probably be developedwithout ever considering empty strings. After all, a lot ofarithmetics was developed by the Greeks without consideringzero as a number. Zero was introduced syntactically and semanticallyonly a few centuries later in India. Extending the number system is not just introducing newconcepts, but also a way of simplifying the understanding and use ofold concepts. Introducing zero and the negative numbers made it easierto understand the properties of the natural positive numbers, and soon. Some properties of functions on the reals (such as convergence ofseries) are much easier to analyze and understand when you considerthe extension to complex numbers.
So introducing new concepts and extensions in mathematics is often agood way of making theories simpler (and usually more powerful forexpressing problems).
Introducing the empty string along with "natural strings" willsimplify theories built on strings, and that is good enough a reason.Typically, as stated in other answers, having the empty string enablesus to consider strings as representatives (models) of well knownalgebraic structures (monoids), and apply directly all known resultsabout such structures. Indeed, as noted by J.-E. Pin, the empty stringis directly related to the concatenation operation on strings (and Iwould add, in the same way that zero is related to the addition ofintegers).
We do not or may not need the empty string, but it is a lot moreconvenient to do mathematics with it than without it. And this is alsotrue of programming (which is a form of mathematics aiming at producingconstructive proofs).
A matter of consistency
However
I disagree with the answer of Yuval Filmus regarding theeffect of not allowing for the concept of an empty string, in the sameway that the Greeks would not consider a number zero. Introducing zeroas a new number would not have been acceptable if it had changed theknown results of arithmetics. At best it would have been considered adifferent theory, with its own purpose.
Similarly, a theory of strings should give consistent results whether it allows for the empty string or not. But both approaches should useconsistent definitions for that to be apparent and meaningful, and Yuval Filmus didnot do that.
When the empty string is allowed, the usual definition of prefix is:
A string u is a
prefix of a string v iff there is a string w such that
u.w=v
where the dot denotes the string concatenation. This allows for astring being a prefix of itself by taking w=ε (the empty string). Then you can define:
A string u is a
proper prefix of a string v iff it is a
prefix of v and not equal to v.
However,
when the empty string is not allowed, you have to state thesedefinitions consistently, but differently. For example:
A string u is a
proper prefix of a string v iff there is a
string w such that u.w=v
Note that w must have at least one symbol. Then you can define:
A string u is a
prefix of a string v iff u is a proper prefix of
v or u=v.
With such consistent definitions, a word remains a prefix of itself,even when the empty string is not allowed in the theory.
So the point to be made is not that not allowing the emptystring changes the properties of strings (at least not in such atrivial way) as asserted by Yuval Filmus. The point is much more that it makes the study ofstrings more complicated, in the same way that arithmetics is morecomplicated when you cannot talk of zero.
About Syntax
The second issue is syntactic. How should strings be represented, onpaper or in the computer. In particular, assuming we agree that it is useful to have the concept of an empty string, how should it be represented syntactically so that we can talk or write about it.
The question actually arises for all mathematical concepts: how shouldthey be represented so that we can talk or write about them, and do soas conveniently as possible. Much of the evolution of mathematics isalso related to improvement of syntax, of the representation ofconcepts. A trivial example is the awkwardness of doing arithmeticswith the ancient Roman representation of integers.
A first answer regarding the empty string is that you may want that tobe consistent with the representation of other strings. Typically, therepresentation of a string will include the sequence of symbols in thestrings plus some additional notation, such as quotes : "
gattaca" forexample. It then becomes quite natural to represent the empty stringas "".
If you rather represent the above example as
gattaca$\dashv$, then thenatural representation for the empty string is $\dashv$ (as noted implicitly by David Richerby).
So the question about the need to introduce a
separate notation(rather than a separate concept, as actually written) has a negativeanswer. No, it is not needed. Uniform notation, uniformrepresentation, is possible for all strings, including the emptystring.
However, if you simply represent the string by the sequence of includedsymbols, such as
gattaca, with no other characters, then the empty string would becomeinvisible syntactically, which is rather inconvenient. Then it isnecessary to introduce some specific notation, such as the Greekletter ε or some other name.
Similarly, when studying strings abstractedly, it is a bit awkward to use "" torepresent the empty string, if only because it does not make for niceand clear sentences in oral speech, when scientists talk to eachother, which is supposed to happen on occasion. Hence it is nicer togive it a name. Saying
empty string might do, but is awkward inwriting. Hence the habit of using a single letter symbol as is oftendone in mathematics to denote entities of specific relevance,
The suggestion to represent the empty word by EOL is essentially thesame as representing it by $\dashv$. It is simply a representation ofstrings with a special terminating character. EOL is just a special character"somehow available in computers".
As noted above for Roman integer arithmetics, the choice of a representation should bedictated by convenience, expecially in an algorithmicenvironment. There are many way to represent strings in general, andthe empty string in particular, in the computer. From a theoreticalstandpoint, it does not matter much which you choose. From a practicalstandpoint, it is essential to choose one that will make stringoperations and manipulation more efficient. This is a basic issue inany class on algorithms and data structures.
On the confusion of syntax and semantics
The answer by David Richerby is interesting for its confusion ofsyntax and semantics.
He tries to introduce the syntactic use of EOL suggested in the question, which hereplaces by the symbol $\dashv$, but he strangely mixes it with thedefinition of the semantic domain of strings, making what is supposedto be only a notation part of that semantic domain.
His second definition should actually have been the following:
An
alphabet is a finite set $\Sigma$ of symbols. A
string $s$ over the alphabet $\Sigma$ is a finite
sequence of $\ell$ symbols $s_i$, where $0\le\ell$, $1\le i\le \ell$ and
$s_i\in\Sigma$ for all values of $i$. It is noted $s_1\dots
s_\ell\dashv$ where $\dashv$ is a special character not denoting a
symbol in $\Sigma$. We write $|s|$ for the length of
$s$, defined by $|s_1\dots s_\ell\dashv|=\ell$. A substring of $s_1\dots
s_\ell\dashv$ is any string $s_i\dots s_j\dashv$, where $1\leq i\leq
j\leq \ell$. The concatenation of strings $s_1\dots s_\ell\dashv$
and $t_1\dots t_m\dashv$ is the string $s_1\dots s_\ell t_1\dots
t_m\dashv$ of length $\ell+m$.
Note that as a consequence, the unique string of length
zero is denoted $\dashv$.
This definition is just a notational variant of the conventionaldefinition given by David Richerby. It does not introduce anycomplexity or "
extra fiddliness" and changes nothing to automatatheory, for the simple reason that $\dashv$ is part of the notation,not a symbol in the strings. And it does give a uniform notation forall strings, including the empty one.
Yuval Filmus makes a similar error in his second remark, since EOL isintended as a syntactic notational device for representing strings,not as a symbol in strings, while $\{0,1\}$ concerns the list ofsymbols that can constitute strings, semantically.
To summarize answers
J.-E. Pin's answer is quite correct, but it addresses only one part of thequestion, regarding the importance of the empty string. It does not addressthe possibility of a uniform notation.
The answers of Yuval Filmus and David Richerby confusing syntax andsemantics, thus erroneously rejecting the suggestion of the OPśquestion to use EOL. Also Yuval Filmus'argument to assert the semanticimportance of the empty string is very disputable. Though it deos make some sense, David Richerby's remark on the use of the null reference is also somewhat unwarranted: it could well be used to represent the empty string, provided the code is written accordingly.
The answer by Pseudonym is theoretical overkill regarding theimportance of the empty string in formal language, but does notactually discuss the issues raised by the question.
As for my own answer, I can only hope it addresses adequately the issuesand contains no error, but it is far far too long.
|
I am trying to figure out where I went wrong on the following problem:
The two batteries are identical, and each has an open-circuit voltage of 1.5V. The lamp has a resistance of 5\$\Omega\$ when lit. With the switch closed, 2.5V is measured across the lamp. What is the internal resistance of each battery?
(Problem 2.1 in Agarwal and Lang's,
Foundations of Analog and Digital Electronic Circuits). Note the answer printed in the back of the book: 0.5\$\Omega\$ . Here's my solution:Step 1
Use element law to find current, \$ {i}_{1} \$, through bulb. $$ v=iR \rightarrow {i}_{1} = \frac{v}{{R}_{bulb}}=\frac{2.5V}{5\Omega}=\frac{1}{2}A. $$Step 2
Model the internal resistance of each battery as a resistor. State the equivalent resistance of the two
resistors in series.$${R}_{eq}={R}_{1}+{R}_{2}=2{R}_{n}$$
By Kirchoff's Voltage Law, the potential difference across the two batteries must be equal and opposite to the potential difference across the lamp. I combine the element law with the above expression in the following manner: $$ v={i}_{2}{R}_{eq} \rightarrow {R}_{n}=\frac{1}{2}\frac{v}{{i}_{2}} (eqn. 1) $$Step 4
By Kirchoff's Current Law, the sum of currents at any node is zero. $$ {i}_{1}-{i}_{2}=0 \rightarrow {i}_{2}={i}_{1} (eqn.2) $$Step 5
Combine eqns. 1 & 2 to find \${R}_{n}\$, the internal resistance of a single battery. $$ {R}_{n}=\frac{1}{2}\frac{v}{{i}_{1}}=2.5\Omega $$
Conclusion
After reflecting on the problem statement, especially the open-circuit voltage part, I know that I am committing some logical fallacy. However I just cannot see it on my own. Where did I go wrong? Should I not imagine that the internal resistance of the batteries can be modeled as a resistor? Would an energy / power approach be better suited to this problem?
|
If you want coefficients of your expression, you'd better use Coefficient.Here are the functions for which you want to determine the coefficients:functions = Cases[{exp1, exp2, exp3},Exp[x_]|Sin[x_]|Cos[x_], Infinity] // Union$\{ e^{-\frac{t}{x^2+y^2}}\,, \, \cos\Big[\frac{2\pi t}{T}\Big]\,,\, \sin\Big[\frac{2\pi t}{T}\Big] \}$Then the ...
ExplanationThe problem is that pattern matching in Mathematica is purely structural - this means that e.g. the pattern a_ + b_ * f[_] will not match 3 (even though you could set b to 0 from a mathematical point of view):split = a_ + b_*f[_] :> {a, b};1 + 3 f[x] /. split3 /. split(* {1, 3} *)(* 3 *)To circumvent this, we can use Optional to ...
If you are calculating the expectation of an expression, use Expectation or other built-in statistics functions rather than replacements which can be error-prone (which is why you ended up asking the question).expr = x + x^2;Expectation[expr, x \[Distributed] BinomialDistribution[k, p]] //Simplify(* k p (2 + (-1 + k) p) *)Alternatively, using ...
This is the purpose of Formal Symbols. They are special symbols that can not be set to a value so that they will not clash with variables in your code. For formal lowercase x you enter Esc$xEsc.foo[a_] :=Probability[\[FormalX] > 0, \[FormalX] \[Distributed] NormalDistribution[a, 1]]Which looks like the image below in a notebook.With this the ...
One way around this could be to make local x inside your module. Like thisfoo[a_] := Module[{x},Probability[x > 0, x \[Distributed] NormalDistribution[a, 1]]]And now you can dofoo[-1] // NAndfoo[x] /. x -> -1
I am a bit confused but if you use your definition of Riccil1 and make the replacement without defining m[u_] it would look like the following. First I remove the power to the 1/4 on each side and multiple both sides by negative one.Ricci11 /. a12[u]^2 - a11[u] a22[u] -> -Det[m[u]]Hope this helps.
Perhaps this will handle your first problem.Simplify[{4*i*sigma*t1-i*B,-4*i*sigma*t1- -i*B, 8*i*sigma*t1-2*i*B,-8*i*sigma*t1--2*i*B,etc},4*i*sigma*t1-i*B==i*gamma]instantly returns{gamma*i, -(gamma*i), 2*gamma*i, -2*gamma*i, etc}Carefully read the documentation for Simplify and in particular how you can add an extra argument specifying assumptions ...
|
Return to Glossary.
Formal Definition
Let $H$ be a subgroup of a group $G$. The number of left cosets of $H$ in $G$ is the
index $(G:H)$ of $H$ in $G$. Informal Definition
The index is the number of cosets which can be enumerated by dividing the number of elements in the group by the number of elements in the subgroup.
Example(s)
Find the index of $\langle 3 \rangle$ in the group $\mathbb{Z_{24}}$.
$\langle 3 \rangle = \{1,3,6,9,12,15,18,21\}$ has 8 elements, so its index is $24/8 = 3$ Non-example(s) Claim: $A_4$ has no subgroup of order 6. $|A_4| = 12$ Proof: Suppose $A_4$ does have a subgroup of order 6, call it $H$. Then the cosets of $H$ partition $A_4$ into 2 cells. $A_4$ has 8 elements of order 3. $(a_1 a_2 a_3)$. How many distinct elements of this form are there? ($4 \times 3 \times 2 = 24$) but $(a_1 a_2 a_3) = (a_2 a_3 a_1) = (a_3 a_1 a_2)$ where $24/3 = 8$. So we let $a$ be an element of order 3. Then either $a\in H$ or $a\notin H$. If $a\notin H$, then $A_4 = H \cup aH$. So $a^2 \in H$ or $a^2 \in aH$. If $a \in H$ then $(a^2)^{-1} \in H$ but $(a^2)^{-1} = a$ so $a \in H$. If $a^2 \in aH$ then $a^2 =ah$ for some $h \in H$ \implies a^3=a^2h$ (by multiplication on the left) $\implies 1 = a^2h \in H$ $\implies a^2 = h^{-1}$ $\implies a^2 \in H$ Therefore there is no subgroup of order 6. Additional Comments
The index $(G:H)$ just defined may be finite or infinite. If $G$ is finite, then obviously $(G:H)$ is finite and $(G:H)=|G|/|H|$,since every coset of $H$ contains $|H|$ elements.
|
Consider the CFT that corresponds to a gauge-fixed closed bosonic string.
Ground level string states are described by vertex operators such as $$V(p) = :\exp(i p_{\mu} X^{\mu}(z, \bar{z})):$$ which are conformal primaries with weight $$ h = \bar{h} = \frac{\alpha'}{4} p^2. $$
The physical states of the strings must have $h = \bar{h} = 1$, therefore, the physical ground state is the tachyon.
Consider the 3-point function of three ground state operators:
$$ G_{p_1, p_2, p_3}(z_1, z_2, z_3; \bar{z}_1, \bar{z}_2, \bar{z}_3) = \left< V(p_1)(z_1, \bar{z}_1) V(p_2)(z_2, \bar{z}_2) V(p_3)(z_3, \bar{z}_3) \right>. $$
Because we're dealing with a free quantum field theory, it isn't hard to calculate this function exactly. Afaik this is called the "Coulomb gas" representation, and the expression is
$$ G_{p_1, p_2, p_3}(z_1, z_2, z_3; \bar{z}_1, \bar{z}_2, \bar{z}_3) = |z_{12}|^{\alpha' p_1 \cdot p_2} \cdot |z_{13}|^{\alpha' p_1 \cdot p_3} \cdot |z_{23}|^{\alpha' p_2 \cdot p_3}, $$
where $z_{ij} = z_i - z_j = -z_{ji}$, and $|z_{ij}| = (z_{ij} \cdot \bar{z}_{ij})^{1/2} = |z_{ji}|$.
However, I expect a general 3-point function to be completely fixed by global conformal symmetries – the Mobius group $SL(2, \mathbb{C})$. The general form for three primary fields with weights $h_i, \bar{h}_i$ is:
$$ \left< \phi_1(z_1, \bar{z}_1) \phi_2(z_2, \bar{z}_2) \phi_3(z_3, \bar{z}_3) \right> = C_{123} z_{12}^{h_3-h_1-h_2} z_{13}^{h_2-h_1-h_3} z_{23}^{h_1-h_2-h_3} \bar{z}_{12}^{\bar{h}_3-\bar{h}_1-\bar{h}_2} \bar{z}_{13}^{\bar{h}_2-\bar{h}_1-\bar{h}_3} \bar{z}_{23}^{\bar{h}_1-\bar{h}_2-\bar{h}_3}, $$
where $C_{123}$ is the 3-point structure constant of the CFT.
For any physical string state, $h = \bar{h} = 1$, therefore Mobius invariance requires that
$$ \left< \phi_1(z_1, \bar{z}_1) \phi_2(z_2, \bar{z}_2) \phi_3(z_3, \bar{z}_3) \right> = C_{123} z_{12}^{-1} z_{13}^{-1} z_{23}^{-1} \bar{z}_{12}^{-1} \bar{z}_{13}^{-1} \bar{z}_{23}^{-1} = C_{123} |z_{12}|^{-2} |z_{13}|^{-2} |z_{23}|^{-2}. $$
This seems to suggest that for the physical ground states participating in scattering (tachyons),
$$\forall i, j, i \neq j: \quad \alpha' p_i \cdot p_j = -2. $$
This restriction seems very odd to me, I've never seen anything like it before.
My questions are:
Have I missed something crucial? If not, what is the origin of the constraint on the values of tachyon momenta?
|
There is a normality assumption when it comes to consider OLS models and that is that the errors be normally distributed. I have been browsing through Cross Validated and it sounds like Y and X don't have to be normal in-order for errors to be normal. My question is why when we have non-normally distributed errors is the validity of our significance statements compromised? Why will confidence intervals be too wide or narrow?
why when we have non-normally distributed errors is the validity of our significance statements compromised? Why will confidence intervals be too wide or narrow?
The confidence intervals are based on the way that the numerator and denominator are distributed in a t-statistic.
With normal data the numerator of a t-statistic has a normal distribution and the distribution of the square of the denominator (which is then a variance) is a particular multiple of a chi-squared distribution. When the numerator and denominator are also independent (as will only be the case with normal data, given the observations themselves are independent), the whole statistic has a t-distribution.
This then means that a t-statistic like $\frac{\hat \beta - \beta}{s_{\hat\beta}}$ will be a
pivotal quantity (its distribution doesn't depend on what the true slope coefficient is, and it's a function of the unknown $\beta$), which makes it suitable for constructing confidence intervals ... and these intervals will then use $t$-quantiles in their construction to get the desired coverage.
If the data were from some other distribution, the statistic wouldn't have a t-distribution. For example, if it were heavy tailed, the t-distribution would tend to be a bit lighter tailed (the outlying observations affect the denominator more than the numerator). Here's an example. In both cases, the histogram is for 10,000 regressions:
The histogram on the left is for when the data are conditionally normal, n=30 (and where in this case, $\beta=0$). The distribution looks as it should. The histogram on the right is for the case when the conditional distribution is right skewed and heavy-tailed, and the histogram shows very few values outside $(-2,2)$ - the distribution isn't much like the theoretical distribution for normal data, because the statistic no longer has the t-distribution.
A 95% t-interval (which should include 95% of the slopes in our sample) runs from -2.048 to 2.048. For the normal data, it actually included 95.15% of the 10000 sample slopes. For the skewed data it includes 99.91%.
|
Travelling Wave Line Model Contents Introduction
A travelling wave on a transmission line is a transient disturbance that that moves along the line at a constant speed yet maintains its shape (see Figure 1). Examples include lightning surges, switching transients, faults, etc.
In this article, we will derive various time-domain models for travelling waves on transmission lines, starting from the simplest case of the lossless single-phase line, then building to a generic lossy single-phase line. The derivation of the model is also looked at starting from the frequency-domain, showing the symmetry of the frequency and time-domain models. Finally, the considerations required to extend the model to a multi-conductor line is discussed.
This article provides the theoretical background for the travelling wave model. Actual implementations of the model suitable for use in computer simulations (i.e. discrete-time) can be found in the following articles:
Lossless Single-Phase Lines
Consider the small segment of length [math] \Delta x [/math] metres from a lossless single-phase transmission line shown in Figure 2.
Notice that the model is similar to that of the classical distributed parameter line model, except that in this model, the voltages and currents are functions of both distance [math]x[/math]
and time [math]t[/math] and are real, not phasor quantities. In the classical distributed parameter line model, the voltages and currents are evaluated at steady-state and are only functions of distance [math]x[/math]. Derivation of Voltage and Current Equations [math] V(x + \Delta x, t) = V(x,t) - L \Delta x \frac{\partial I(x + \Delta x,t)}{\partial t} \, [/math] [math] I(x,t) = I(x + \Delta x, t) + C \Delta x \frac{\partial V(x,t)}{\partial t} \, [/math] Notes: 1. The voltage across an inductor is [math]v = L \frac{di}{dt} [/math] and the current through a capacitor is [math]i = C \frac{dv}{dt} [/math]. 2. Partial derivatives are used because the voltages and currents are functions of both time and distance. Re-arranging the above equations, we get: [math] \frac{V(x + \Delta x, t) - V(x,t)}{\Delta x} = - L \frac{\partial I(x + \Delta x,t)}{\partial t} \, [/math] [math] \frac{I(x + \Delta x, t) - I(x,t)}{\Delta x} = - C \frac{\partial V(x,t)}{\partial t} \, [/math] Taking the limit as [math]\Delta x \to 0 \, [/math] yields the well known Telegrapher's equations: [math] \frac{\partial V(x, t)}{\partial x} = - L \frac{\partial I(x,t)}{\partial t} \, [/math] ... Equ. (1) [math] \frac{\partial I(x,t)}{\partial x} = - C \frac{\partial V(x,t)}{\partial t} \, [/math] ... Equ. (2) Differentiating the voltage equation with respect to x and the current equation with respect to t: [math] \frac{\partial^{2} V(x, t)}{\partial x^{2}} = - L \frac{\partial^{2} I(x,t)}{\partial x \partial t} \, [/math] ... Equ. (3) [math] \frac{\partial^{2} I(x,t)}{\partial x \partial t} = - C \frac{\partial^{2} V(x,t)}{\partial t^{2}} \, [/math] ... Equ. (4) Now substituting Equ. (4) into Equ. (3), we get: [math] \frac{\partial^{2} V(x, t)}{\partial x^{2}} = LC \frac{\partial^{2} V(x,t)}{\partial t^{2}} \, [/math] ... Equ. (5)
Similarly, we could have differentiated the current equation with respect to x and the voltage equation with respect to t and then solved for the current equation:
[math] \frac{\partial^{2} I(x, t)}{\partial x^{2}} = LC \frac{\partial^{2} I(x,t)}{\partial t^{2}} \, [/math] ... Equ. (6)
The pair of equations (5) and (6) above are referred to as the
transmission line wave equations.
The general solution to either of these equations can be found using d'Alembert's formula:
[math] I(x, t) = i^{+}(x - vt) + i^{-}(x + vt) \, [/math] [math] V(x, t) = v^{+}(x - vt) + v^{-}(x + vt) \, [/math]
where [math] v = \frac{1}{\sqrt{LC}} \, [/math] is the velocity of propagation (m/s).
[math] i^{+}(t) \, [/math] and [math] i^{-}(t) \, [/math] are arbitrary current functions of time. [math] v^{+}(t) \, [/math] and [math] v^{-}(t) \, [/math] are arbitrary voltage functions of time.
Using the current equation above, we can solve for the voltage equation as follows (see here for the full derivation):
[math] v(x, t) = Z_{c} \left[ i^{+}(x - vt) - i^{-}(x+vt) \right] \, [/math]
where [math]Z_{c} = \frac{1}{vC} = \frac{\sqrt{LC}}{C} = \sqrt{\frac{L}{C}} [/math] is the characteristic impedance (Ohms)
Significance of + and - Equations
So we have the pair of voltage and current equations:
[math] I(x, t) = i^{+}(x - vt) + i^{-}(x + vt) \, [/math] [math] V(x, t) = Z_{c} \left[ i^{+}(x - vt) - i^{-}(x+vt) \right] \, [/math]
What is the significance of [math] i^{+}(x - vt) \, [/math] and [math] i^{-}(x + vt) \, [/math]?
Consider that time t=0s, the function [math] i^{+}(x) \, [/math] represents an arbitrary waveform / shape distributed in space along the transmission line. For example, it could look like this:
Now consider what happens at time t=1s, the function [math] i^{+}(x - v) \, [/math] has the same shape as the function above, but is translated to the right by [math]v \, [/math] metres:
What we can conclude is that the function [math] i^{+}(x - vt) \, [/math] is an arbitrary waveform that maintains its shape but moves in the direction of the receiving end with increasing time at speed [math]v \, [/math] m/s. This is called a
forward travelling wave.
We can use similar logic to conclude that the function [math] i^{-}(x + vt) \, [/math] is an arbitrary waveform that moves in the opposite direction (i.e. in the direction of the sending end). This is referred to as a
backward travelling wave. Generic Lossy Single-Phase Lines
Consider the generic single-phase distributed parameter line model of length [math]l \,[/math] metres shown in Figure 5. Analysing the circuit using Kirchhoff's laws as we did previously in the lossless case, we can get:
[math] \frac{\partial V(x, t)}{\partial x} = - I(x,t) R - L \frac{\partial I(x,t)}{\partial t} \, [/math] [math] \frac{\partial I(x,t)}{\partial x} = -V(x,t) G - C \frac{\partial V(x,t)}{\partial t} \, [/math]
Where [math] G = \frac{1}{R_{sh}} \, [/math] is the shunt conductance.
Differentiating the voltage equation with respect to x and the current equation with respect to t:
[math] \frac{\partial^{2} V(x, t)}{\partial x^{2}} = - R \frac{\partial I(x,t)}{\partial x} - L \frac{\partial^{2} I(x,t)}{\partial x \partial t} \, [/math] ... Equ. (7) [math] \frac{\partial^{2} I(x,t)}{\partial x \partial t} = - G \frac{\partial V(x,t)}{\partial t} - C \frac{\partial^{2} V(x,t)}{\partial t^{2}} \, [/math] ... Equ. (8)
Substituting Equ. (8) into Equ. (7), we get:
[math] \frac{\partial^{2} V(x, t)}{\partial x^{2}} = - R \frac{\partial I(x,t)}{\partial x} + LG \frac{\partial V(x,t)}{\partial t} + LC \frac{\partial^{2} V(x,t)}{\partial t^{2}} \, [/math] ... Equ. (9)
If we were to set R = G = 0 in the equation above, then we would get the same transmission line wave equation as Equ. (5) in the lossless case.
However, while there is a general solution to the lossless line using d'Alembert's formula (as shown earlier), the generic lossy line
does not have a neat, closed-form solution. Frequency Domain Derivation
In the preceding sections, the travelling wave model was developed purely in the time-domain. Another way of looking at the travelling wave model is to derive it first in the frequency domain and then convert it to the time-domain. Recall from the steady-state (frequency domain) distributed parameter model that the sending end voltage and current can be related to the receiving end quantities as follows:
[math] \left[ \begin{matrix} \boldsymbol{V_{s}} \\ \\ \boldsymbol{I_{s}} \end{matrix} \right] = \left[ \begin{matrix} \cosh (\boldsymbol{\gamma} l) & \boldsymbol{Z}_{c} \sinh(\boldsymbol{\gamma} l) \\ \\ \frac{1}{\boldsymbol{Z}_{c}} sinh (\boldsymbol{\gamma} l) & \cosh(\boldsymbol{\gamma} l) \end{matrix} \right] \left[ \begin{matrix} \boldsymbol{V_{r}} \\ \\ \boldsymbol{I_{r}} \end{matrix} \right] \, [/math]
where [math] \boldsymbol{\gamma} = \sqrt{\boldsymbol{zy}} [/math] is the propagation constant ([math]m^{-1}[/math])
[math] \boldsymbol{Z}_{c} = \sqrt{\boldsymbol{\frac{z}{y}}} [/math] is the characteristic impedance ([math]\Omega[/math]) [math] \boldsymbol{z} = R + j \omega l [/math] and [math] \boldsymbol{y} = G + j \omega C [/math] are the series impedance and shunt admittance respectively
With some straightforward algebraic manipulation, the steady-state equations above can be represented in the following form:
[math] \boldsymbol{V_{r}} + \boldsymbol{Z}_{c} \boldsymbol{I_{r}} = (\boldsymbol{V_{s}} + \boldsymbol{Z}_{c} \boldsymbol{I_{s}}) e^{-\boldsymbol{\gamma} l} [/math] ... Equ. (10)
The time-domain model can be derived by calculating the inverse Fourier transform of Equ. (10).
In the
lossless case (i.e. R = G = 0.), the propagation constant and characteristic impedance are as follows: [math] \boldsymbol{\gamma} = \sqrt{\boldsymbol{zy}} = \sqrt{(j \omega L)(j \omega C)} = j \omega \sqrt{LC} [/math] [math] \boldsymbol{Z}_{c} = \sqrt{\boldsymbol{\frac{z}{y}}} = \sqrt{\frac{(j \omega L)}{(j \omega C)}} = \sqrt{\frac{L}{C}}[/math]
A property of Fourier transforms is that multiplying by a complex exponential in the frequency domain results in a time shift in the time domain. Therefore, taking the inverse Fourier transform of Equ. (10) for the lossless case, we get:
[math] v_{r}(t) + \sqrt{\frac{L}{C}} i_{r}(t) = v_{s}(t - \sqrt{LC} l) + \sqrt{\frac{L}{C}} i_{s}(t - \sqrt{LC} l) \, [/math]
The lossless case above shows that the receiving end quantities are essentially time-shifted (past) versions of the sending end quantities, as was derived earlier using d'Alembert's formula.
In the
general lossy case, the characteristic impedance is not a scalar constant (but rather a function of frequency [math] \boldsymbol{Z}_{c}(\omega) \, [/math]) and the propagation constant does not lead to a simple time shift in the time-domain. Therefore, the inverse Fourier transform of Equ. (10) leads to a series of messy convolutions: [math] v_{r}(t) + z_{c}(t) * i_{r}(t) = \left[ v_{s}(t) + z_{c}(t) * i_{s}(t) \right] * a(t) \, [/math]
where [math] a(t) \, [/math] is the inverse Fourier transform of [math] e^{-\boldsymbol{\gamma}(\omega) l} \, [/math]
[math] z_{c}(t) \, [/math] is the inverse Fourier transform of [math] \boldsymbol{Z}_{c}(\omega) \, [/math]
As can be imagined, the convolution integrals in the equation above are not neatly solvable.
Multi-conductor Lines
The travelling wave line model for the single-phase lossless line can be extended to a multi-conductor line of n conductors by replacing the voltage and current quantities with [math]n \times 1 [/math] vectors, e.g. for a three-phase, three-conductor line:
[math] \boldsymbol{V}(x,t) = \left[ \begin{matrix} \boldsymbol{V_{a}}(x,t) \\ \boldsymbol{V_{b}}(x,t) \\ \boldsymbol{V_{c}}(x,t) \end{matrix} \right] \, [/math], [math] \boldsymbol{I}(x,t) = \left[ \begin{matrix} \boldsymbol{I_{a}}(x,t) \\ \boldsymbol{I_{b}}(x,t) \\ \boldsymbol{I_{c}}(x,t) \end{matrix} \right] \, [/math]
And the inductance and capacitance (per unit length) with [math]n \times n [/math] matrices to capture the mutual coupling between phases, e.g. for a three-phase, three-conductor line:
[math] [L] = \left[ \begin{matrix} L_{aa} & L_{ab} & L_{ac} \\ L_{ba} & L_{bb} & L_{bc} \\ L_{ca} & L_{cb} & L_{cc} \end{matrix} \right] \, [/math] [math] [C] = \left[ \begin{matrix} C_{aa} & C_{ab} & C_{ac} \\ C_{ba} & C_{bb} & C_{bc} \\ C_{ca} & C_{cb} & C_{cc} \end{matrix} \right] \, [/math]
The transmission line wave equations for the single-phase model (see equations (5) and (6) above) can now be re-written for the multi-conductor model as follows:
[math] \frac{\partial^{2} \boldsymbol{V}(x, t)}{\partial x^{2}} = [L][C] \frac{\partial^{2} \boldsymbol{V}(x,t)}{\partial t^{2}} \, [/math] ... Equ. (10) [math] \frac{\partial^{2} \boldsymbol{I}(x, t)}{\partial x^{2}} = [C][L] \frac{\partial^{2} \boldsymbol{I}(x,t)}{\partial t^{2}} \, [/math] ... Equ. (11)
Unlike the single-phase case, there is no general solution to the second-order differential equations (10) and (11) since the matrix product [L][C] is typically full (i.e. off-diagonals are non-zero). To solve these differential equations, a
modal transformation is required to decouple the phases (refer to this section of the distributed parameter line article for more details on the modal transformation).
The voltage and current vectors are transformed from the phase domain to the modal domain by the [math]n \times n [/math] transformation matrices [math][T_{v}] \, [/math] and [math][T_{i}] \, [/math], such that:
[math]\boldsymbol{V}(x,t) = [T_{v}] \boldsymbol{V'}(x,t)[/math] [math]\boldsymbol{I}(x,t) = [T_{i}] \boldsymbol{I'}(x,t)[/math] where [math] \boldsymbol{V'}(x,t)[/math] and [math] \boldsymbol{I'}(x,t)[/math] are the modal voltage and current vectors, i.e. [math] \boldsymbol{V'}(x,t) = \left[ \begin{matrix} \boldsymbol{V_{0}}(x) \\ \boldsymbol{V_{1}}(x) \\ \boldsymbol{V_{2}}(x) \end{matrix} \right] \, [/math], [math] \boldsymbol{I'}(x,t) = \left[ \begin{matrix} \boldsymbol{I_{0}}(x) \\ \boldsymbol{I_{1}}(x) \\ \boldsymbol{I_{2}}(x) \end{matrix} \right] \, [/math]
The transformation matrices [math][T_{v}] \, [/math] and [math][T_{i}] \, [/math] are selected such that they are the eigenvectors of [L][C] and [C][L] respectively (note that they have the same eigenvalues).
In the modal domain, equations (10) and (11) can be re-written as sets of decoupled equations, i.e.
[math] \left[ \begin{matrix} \frac{d^{2} \boldsymbol{V_{0}}(x,t)}{dx^{2}} \\ \\ \frac{d^{2} \boldsymbol{V_{1}}(x,t)}{dx^{2}} \\ \\ \frac{d^{2}\boldsymbol{V_{2}}(x,t)}{dx^{2}} \end{matrix} \right] = \left[ \begin{matrix} \lambda_{0} & & \\ & \lambda_{1} & \\ & & \lambda_{2} \end{matrix} \right] \left[ \begin{matrix} \frac{d^{2} \boldsymbol{V_{0}}(x,t)}{dt^{2}} \\ \\ \frac{d^{2} \boldsymbol{V_{1}}(x,t)}{dt^{2}} \\ \\ \frac{d^{2}\boldsymbol{V_{2}}(x,t)}{dt^{2}} \end{matrix} \right] \, [/math] ... Equ. (12) [math] \left[ \begin{matrix} \frac{d^{2} \boldsymbol{I_{0}}(x,t)}{dx^{2}} \\ \\ \frac{d^{2} \boldsymbol{I_{1}}(x,t)}{dx^{2}} \\ \\ \frac{d^{2}\boldsymbol{I_{2}}(x,t)}{dx^{2}} \end{matrix} \right] = \left[ \begin{matrix} \lambda_{0} & & \\ & \lambda_{1} & \\ & & \lambda_{2} \end{matrix} \right] \left[ \begin{matrix} \frac{d^{2} \boldsymbol{I_{0}}(x,t)}{dt^{2}} \\ \\ \frac{d^{2} \boldsymbol{I_{1}}(x,t)}{dt^{2}} \\ \\ \frac{d^{2}\boldsymbol{I_{2}}(x,t)}{dt^{2}} \end{matrix} \right] \, [/math] ... Equ. (13)
These modal differential equations can then be solved individually using d'Alembert's formula. For example, the solution for mode 0 is:
[math] I_{0}(x, t) = i_{0}^{+}(x - v_{0}t) + i_{0}^{-}(x + v_{0}t) \, [/math] [math] V_{0}(x, t) = v_{0}^{+}(x - v_{0}t) + v_{0}^{-}(x + v_{0}t) \, [/math]
where [math] v_{0} = \frac{1}{\sqrt{\lambda_0}} \, [/math] is the velocity of propagation for mode 0 (m/s)
[math] i^{+}(t) \, [/math] and [math] i^{-}(t) \, [/math] are current functions for mode 0 [math] v^{+}(t) \, [/math] and [math] v^{-}(t) \, [/math] are voltage functions for mode 0
Once modal domain quantities are calculated, then the phase domain quantities can be found by applying the inverse transformation.
Finally, it should be noted that the general solution using the modal transformation only applies to
lossless multi-conductor lines. The generic lossy line suffers from the same issues as described earlier for the lossy single-phase line, i.e. multiple cross-coupling terms that render d'Alembert's formula unusable. Frequency Dependency of Line Parameters
So far, we have described the per unit length inductance (L), capacitance (C), resistance (R) and conductance (G) parameters of the line as fixed and constant for the full range of frequencies. However, this is not a valid assumption, particularly at high frequencies. For example, the skin effect leads to higher resistances at higher frequencies. It is therefore more correct to characterise the line parameters as having frequency dependencies, i.e. [math]L(\omega) \, [/math], [math]C(\omega) \, [/math], [math]R(\omega) \, [/math] and [math]G(\omega) \, [/math].
|
I'm a physics student starting grad school, and I figured I'd read up on manifolds since they pop up so much. However, one thing that continues to elude me is why tangent spaces have such involved definitions. Given that the tangent space of an $n$ dimensional manifold at any point is diffeomorphic to $\mathbb{R}^n$, why do we bother with all the long definitions involving derivations and equivalence classes of curves in the first place?
We don't just want to have a vector space to call the "tangent space". We want to
do geometric things with the tangent space, and we can't do those things if it's just an arbitrary vector space of the right dimension. We need it to more specifically be a vector space that encodes the parts of the geometry of our manifold that we care about.
For instance, here's one geometric thing we'd like to be able to do: given a smooth curve $\gamma:\mathbb{R}\to M$ on a manifold $M$, we'd like to define a "tangent vector" at each point of the curve. For each $t\in\mathbb{R}$, this should give us a vector "$\gamma'(t)$" which is in the tangent space of $M$ at the point $\gamma(t)$. If the tangent space is just defined as some arbitrary $n$-dimensional vector space, there isn't going to be any natural way to define $\gamma'(t)$. But for the usual definition, there is: we just take the equivalence class of the curve $\gamma$.
Another very useful thing we like to do is differentiate functions between manifolds: if $f:M\to N$ is a smooth function between manifolds and $p\in M$, there should be a "derivative" $df$ which is a linear map from $T_pM\to T_{f(p)}N$. Again, there's not any natural way to get such a map if the tangent spaces are just some arbitrary vector spaces.
To put it another way, the tangent space to a manifold at a point is
not merely a vector space. It is a vector space together with a bunch of extra structure relating it to other geometric features of the manifold. So we care about it not just as a vector space up to isomorphism, but as a "vector space plus extra structure" up to isomorphism. It's possible to axiomatize this extra structure, and then you could define the tangent space as just any abstract gadget satisfying those axioms, rather than restricting yourself to the usual definition by equivalence classes of curves. This isn't commonly done though because (as far as I know) there isn't any nice axiomatization that isn't essentially just saying "a vector space equipped with an isomorphism to this particular concrete construction", so you might as well just use the concrete construction itself.
If they are “just vector spaces” at each point, then every set would be a manifold.
The whole point is to link the topological structure of the manifold (using coordinate patches that fit together well) to linear algebra that goes on in the tangent spaces.
Another reason to give many different definitions is that they all expose some aspect of what a manifold does. In particular, I’ve found D. Holm’s chapter on manifolds in
Geometric mechanics and symmetry very helpful, as it relates several equivalent definitions.
Finally, another contribution to your angst might be the books you’re reading. Some math expositions by physicists can be extremely frustrating for mathematicians, due to different styles in the two disciplines (to put it charitably), so if you’re reading books by such authors it may contribute to what you’re describing.
|
I read some proofs that show that the outer measure $m^*(I)$ of an interval is equal to its length $l(I)$, i.e. $m^*(I)=l(I)$, where for an interval $I=[a,b]$, we have $l(I)=b-a=m^*(I)$.
I understand the part that $m^*(I) \leq l(I)$, but for the other direction $m^*(I) \geq l(I)$, I could not see why the proofs really wanted to use the compactness property of $I$ (being bounded and closed). From what I read, outer measure of an interval $I$ is:
$$ m^*(I) = \inf \bigg\{\sum_{j\in J} l(j) \bigg\} $$
where $J$ forms an open covering of $I$, and $j$ refers to any open interval inside the open covering $J$ - so that the outer measure gets the infimum of the sum above for all open coverings of $I$.
Since we have (for sure) that $I \subseteq J$, shouldn't it hold trivially that $m^*(I) \geq l(I)$ ? given that whether $J$ is finite or infinite countable, it should be able to cover all elements of $I$.
So why do we need to guarantee (using compactness and the Heine Borel theorem) that there is a $J$ with finite cardinality $|J| \neq \infty$ that covers $I$ to show that $m^*(I) \geq l(I)$ ?
|
In this calculus problem, the limit of the quotient of $\cos{x}$ by $\dfrac{\pi}{2}-x$ should have to evaluate as $x$ approaches $\dfrac{\pi}{2}$. In this problem, the $x$ represents a variable and also represents an angle of a right triangle.
$\displaystyle \large \lim_{x \,\to\, \Large \frac{\pi}{2}}{\normalsize \dfrac{\cos{x}}{\dfrac{\pi}{2}-x}}$
Now. let’s try to evaluate the limit of the algebraic trigonometric function as $x$ approaches $\dfrac{\pi}{2}$ by using direct substitution method.
$= \,\,\,$ $\dfrac{\cos{\Big(\dfrac{\pi}{2}\Big)} }{\dfrac{\pi}{2}-\dfrac{\pi}{2}}$
$= \,\,\,$ $\dfrac{0}{0}$
It is evaluated that the limit of the given function is indeterminate as the input $x$ approaches $\dfrac{\pi}{2}$. So, it should be evaluated in another method.
Take $y = \dfrac{\pi}{2}-x$, then $x = \dfrac{\pi}{2}-y$.
$(1) \,\,\,$ If $x \to \dfrac{\pi}{2}$ then $x-\dfrac{\pi}{2} \to \dfrac{\pi}{2}-\dfrac{\pi}{2}$. Therefore, $x-\dfrac{\pi}{2} \to 0$.
$(2) \,\,\,$ If $x-\dfrac{\pi}{2} \to 0$ then $-\Big(\dfrac{\pi}{2}-x\Big) \to 0$.
In this step, we have taken that $y = \dfrac{\pi}{2}-x$. Therefore, $-y \to 0$.
$(3) \,\,\,$ If $-y \to 0$ then $-y \times (-1) \to 0 \times (-1)$. Therefore, $y \to 0$.
Therefore, it is cleared that if $x$ approaches $\dfrac{\pi}{2}$, then $y$ approaches $0$. Now, write the whole given problem in terms of $y$ by eliminating the $x$.
$\implies$ $\displaystyle \large \lim_{x \,\to\, \Large \frac{\pi}{2}}{\normalsize \dfrac{\cos{x}}{\dfrac{\pi}{2}-x}}$ $\,=\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\cos{\Big(\dfrac{\pi}{2}-y\Big)}}{y}}$
The angle $\dfrac{\pi}{2}-y$ represents a variable angle, which belongs to the first quadrant. In the first quadrant, $\cos{\Big(\dfrac{\pi}{2}-y\Big)} = \sin{y}$.
Therefore, the function in the numerator $\cos{\Big(\dfrac{\pi}{2}-y\Big)}$ can be replaced by its equivalent value $\sin{y}$.
$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{y}}{y}}$
According to the limit rules of trigonometric functions, the limit of sinx/x is equal to one as x approaches 0. Therefore, the limit of the $\dfrac{\sin{y}}{y}$ as $y$ tends to $0$ is also equal to $1$.$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{y}}{y}}$
$= \,\,\,$ $1$
Therefore, it is evaluated that the limit of the ratio of $\cos{x}$ to $\dfrac{\pi}{2}-x$ is equal to $1$ as $x$ approaches $\dfrac{\pi}{2}$.
Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.
|
Let $S_4$ denote the group of permutations of $\{1,2,3,4\}$ and let $H$ be a sub group of $S_4$ of order $6$ . Show that $\exists~ i \in \{1,2,3,4\}$ which is fixed by each element of $H$.
Attempt: As per the given question, $H$ is a sub group of $S_4$ of order $6$ .
We have to prove that $\exists~ i \in \{1,2,3,4\} ~ s.t. ~ \alpha(i) = i ~\forall ~\alpha \in H$ . So, If i prove that $ H \subseteq Stab_G (i)$ , then I can prove the same.
Now, by
orbit stabilizer theorem, we have $|G| = |Stab_G (i)|~|Orb_G(i)|$ .
{ So, if $G$ is a group of permutation of a set $S$. For $\forall ~s \in S$, $orb_G(s)= \{\phi(s)~ | ~~\phi \in G .$ The set $orb_G(s)$ is a subset of $S$ called the orbit of $s$ under $G$. }
We also know that $Stab_G(i)$ is a subgroup of $G$.
IF i show that $\exists ~orb_G(i)$ such that $| orb_G(i) |=4$ under $G$, then we are done because
$Stab_G (i)$ is already a subgroup of $G$.
So, my problem boils down to finding an $i$ such that $| orb_G(i) |=4$
EDIT : I also know that Since $S_4$ has no elements of order $6$, by Lagrange's theorem the elements of H must have orders $2$ or $3 $ which means they can be which are $2$ cycled or $3$ cycled
How do I find such an $i$. Do i now write all the elements of $S_4$? Writing down everything seems tedious and impractical though ? Help will be really appreciated.
Thank you.
|
Usage is fairly complex but there is a nice tutorial here:
http://www.forkosh.c...extutorial.html
Use the following syntax:
[math]f(x)=\int_{-\infty}^x e^{-t^2}dt[/math]
...which yields
[math]f(x)=\int_{-\infty}^x e^{-t^2}dt[/math]
Jump to content
Posted 04 May 2006 - 03:12 AM
[math]f(x)=\int_{-\infty}^x e^{-t^2}dt[/math]
Posted 04 May 2006 - 03:15 AM
Posted 04 May 2006 - 03:17 AM
Posted 04 May 2006 - 06:08 AM
As a side note I'll mention that the forum also supports a [noparse][/noparse] tag so that example BBCode can be posted. This allows you to actually post BBCode code that can literally be copy and pasted like
[ latex ]f(x)=int_{-\infty}^x e^{-t^2}dt[ /latex ]
Posted 04 May 2006 - 02:18 PM
Posted 04 May 2006 - 02:21 PM
Posted 04 May 2006 - 02:36 PM
Posted 04 May 2006 - 03:41 PM
Posted 04 May 2006 - 03:42 PM
Posted 11 May 2006 - 08:15 PM
Posted 13 May 2006 - 08:14 PM
Posted 09 November 2006 - 06:30 PM
Posted 09 November 2006 - 07:23 PM
I understand how to show sqrt-root but how do I show cube-root or evern higher powers of this function?
Posted 09 November 2006 - 07:26 PM
Excellent, thank you so much my friend.............................Infy
\sqrt[3]{X} renders as [math]\sqrt[3]{X}[/math]
Though there are many excellent LaTeX references, I usually refer to http://meta.wikimedi...ki/Help:Formula . I don’t think it’s complete, but it’s yet to fail to answer any of my questions.
|
Searching for just a few words should be enough to get started. If you need to make more complex queries, use the tips below to guide you.
Purchase individual online access for 1 year to this journal.
Impact Factor 2019: 0.808
The journal
Asymptotic Analysis fulfills a twofold function. It aims at publishing original mathematical results in the asymptotic theory of problems affected by the presence of small or large parameters on the one hand, and at giving specific indications of their possible applications to different fields of natural sciences on the other hand. Asymptotic Analysis thus provides mathematicians with a concentrated source of newly acquired information which they may need in the analysis of asymptotic problems.
Article Type: Research Article
Abstract: In this article we study the homogenization, of a particular example, of degenerate elliptic equations of second order in the setup of viscosity solutions. These results are an attempt to extend the corresponding results of Evans [8] to degenerate situations.
Citation: Asymptotic Analysis, vol. 36, no. 3,4, pp. 187-201, 2003
Authors: Vakulenko, S.A.
Article Type: Research Article
Abstract: The paper is devoted to localized reaction fronts in a viscous fluid. A rigorous analytical theory describing small front deformations is suggested, and the dependence of the front orientation on the gravity force is studied. Estimates for perturbation of the front velocity are given. The critical Rayleigh number when the convective instability appears is found.
Citation: Asymptotic Analysis, vol. 36, no. 3,4, pp. 203-239, 2003
Article Type: Research Article
Abstract: We consider the semilinear parabolic system with absorption terms in a bounded domain Ω of $\mathbb{R}^{N}$ \[\left\{\begin{array}{l@{\quad}l}u_{t}-\Delta u+\vert v\vert ^{p}\vert u\vert ^{k-1}u=0,&\hbox{in }\varOmega\times(0,\infty),\\v_{t}-\Delta v+\vert u\vert ^{q}\vert v\vert ^{\ell -1}v=0,&\hbox{in }\varOmega\times(0,\infty),\\u(0)=u_{0},\quad v(0)=v_{0},&\hbox{in }\varOmega,\end{array}\right.\] where p,q>0 and k,ℓ≥0, with Dirichlet or Neuman conditions on $\curpartial\varOmega\times(0,\infty)$ . We study the existence and uniqueness of the Cauchy problem when the initial data are L1 functions or bounded measures. We find invariant regions when u0 , v0 are nonnegative, and give sufficient conditions for positivity or extinction in finite time.
Citation: Asymptotic Analysis, vol. 36, no. 3,4, pp. 241-283, 2003
Article Type: Research Article
Abstract: A phase‐field model of Penrose–Fife type for diffusive phase transitions with conserved order parameter is introduced. A Cauchy–Neumann problem is considered for the related parabolic system which couples a nonlinear Volterra integro‐differential equation for the temperature $\teta$ with a fourth order relation describing the evolution of the phase variable χ. The latter equation contains a relaxation parameter μ related to the speed of the transition process, which happens to be very small in the applications. Existence and uniqueness for this model as μ>0 have been recently proved by the first author. Here, the asymptotic behaviour of the model …is studied as μ is let tend to zero. By a priori estimates and compactness arguments, the convergence of the solutions is shown. The approximating initial data have to be properly chosen. The problem obtained at the limit turns out to couple the original energy balance equation with an elliptic fourth order inclusion. Show more
Keywords: phase transition, Penrose–Fife model, singular limit, Neumann problem, memory kernel
Citation: Asymptotic Analysis, vol. 36, no. 3,4, pp. 285-301, 2003
Article Type: Research Article
Abstract: We prove an existence and uniqueness result for solutions of some bilateral problems with right‐hand side in L1 by using an approach involving increasing powers.
Keywords: strongly nonlinear elliptic problem in $L^{1}$, truncations, obstacle problem
Citation: Asymptotic Analysis, vol. 36, no. 3,4, pp. 303-317, 2003
Article Type: Research Article
Abstract: The framework of this paper is given by the mixed boundary‐value problem \[\left\{\begin{array}{@{}l@{\quad}l}\Delta u(x)=0,&x;\in \Omega,\\u(x)=0,&x;\in \Gamma _{0},\\\frac{\partial u}{\partial n}(x)=q(x),&x;\in \Gamma _{1},\end{array}\right.\] where Ω is a plane domain bounded by a regular curve composed by two arcs Γ0 and Γ1 . Assuming that |Γ1 |=ε and denoting by u[ε] the solution to this problem, we study some asymptotic expansions in terms of ε which are related to u[ε]. Some connections are presented among these expansions, on one hand, and the geometry of the domain Ω, on the other. In addition, a systematic way is found for computing …at the boundary the Ghizzetti's integral that solves the problem. Show more
Citation: Asymptotic Analysis, vol. 36, no. 3,4, pp. 319-343, 2003
Authors: Almog, Y.
Article Type: Research Article
Abstract: The creeping motion of a Newtonian fluid around a particle in a smooth domain is studied. It is proved that the force distribution on the surface of the particle can be approximated, in the limit where the ratio between particle size and the domain's radius of curvature tends to zero, by the force distribution on the same particle near a flat wall. This result is then utilized to show that the velocities of the particle in mobility problems, or the forces acting on it in resistance problems can be approximated by replacing the domain with a flat wall.
Citation: Asymptotic Analysis, vol. 36, no. 3,4, pp. 345-357, 2003
Inspirees International (China Office)
Ciyunsi Beili 207(CapitaLand), Bld 1, 7-901 100025, Beijing China Free service line: 400 661 8717 Fax: +86 10 8446 7947 china@iospress.cn
For editorial issues, like the status of your submitted paper or proposals, write to editorial@iospress.nl
如果您在出版方面需要帮助或有任何建, 件至: editorial@iospress.nl
|
Tool for Second Derivative calculation f''. The second derivative is the application of the derivation tool to the (first) derivative of a function, a double derivation on the same variable.
Second Derivative - dCode
Tag(s) : Functions
dCode is free and its tools are a valuable help in games, maths, geocaching, puzzles and problems to solve every day!
A suggestion ? a feedback ? a bug ? an idea ? Write to dCode!
Sponsored ads
Tool for Second Derivative calculation f''. The second derivative is the application of the derivation tool to the (first) derivative of a function, a double derivation on the same variable.
Example: $$ f(x) = x^2+\sin(x) \Rightarrow f'(x) = 2 x+\cos(x) \Rightarrow f´´'x) = 2 - \sin(x) $$
In physics the
second derivative is usually used for acceleration calculations
A
second derivative can be written $ f´´ (x) $ or $ f^{(2)} (x) $ or $ \frac{d^2f}{dx^2} $. On dCode prefer f ' ' which is the most used notation.
The
second derivative is used to know the variation of the slope of the curve representing the function. For a given interval:
- a positive
second derivative means an increase of the slope (convex function)
- a negative
second derivative signifies a decrease in thought (concave function)
- a zero
second derivative means a straight / straight curve
For a given point:
- a
second derivative canceling with a change of sign means a point of inflection, the curvature of the graphical representation changes and is reversed. It can be a maximum of the function or a minimum of the function.
dCode retains ownership of the source code of the script Second Derivative online. Except explicit open source licence (indicated Creative Commons / free), any algorithm, applet, snippet, software (converter, solver, encryption / decryption, encoding / decoding, ciphering / deciphering, translator), or any function (convert, solve, decrypt, encrypt, decipher, cipher, decode, code, translate) written in any informatic langauge (PHP, Java, C#, Python, Javascript, Matlab, etc.) which dCode owns rights will not be released for free. To download the online Second Derivative script for offline use on PC, iPhone or Android, ask for price quote on contact page !
|
Not sure if this approach has any gotcha's.
\documentclass{article}
\usepackage[nomessages]{fp}
\newcommand\FPuse[1]{\FPeval{\result}{#1}{\result}}
\begin{document}
$\cos(\pi)=\FPuse{clip(cos(pi))}$\
$\sin(\pi/3)=\FPuse{sin(pi/3)}$
$\sin(\pi/3)=\FPuse{round(sin(pi/3),3)}$
\end{document}
In the comments below, jfbu and I discuss why I grouped
{\result} at the end of the
\FPuse definition. First, let's see what happens if I ungroup it:
\newcommand\FPuse[1]{\FPeval{\result}{#1}\result}
The result on the first operation is
What we see is that
\result sets itself as
{} - 1, using a binary minus sign. The conclusion is that
\FPeval{}{} creates a
\bgroup...\egroup quantity that, in math mode, causes the subsequent minus sign to act in a binary fashion. Thus, the only way to eliminate this problem (without changing the
fp package), is to isolate the final
\result in its own group, as I did in my original code.
While jfbu has probed a bit into the guts of
fp, I am no expert to know if the
fp code can be revised to use
\begingroup...\endgroup instead (which is truly transparent in math mode) or not. I do know that
fp has a few issues, for example, a stray space is introduced via
\FPpow which has to be
\unskiped after its use.
See jfbu's comments below for more information on the group issue.
|
As I mentioned in other blogs, we can still use a classically derived test known as the generalized log-likelihood ratio as a way of simply ranking different A-B combinations against each other according to how interesting they are. Even without being able to interpret the statistical score as a statistical test, we get useful results in practice.
The generalized log-likelihood ratio most commonly used in these situations is derived assuming we have two binomial observations. This test can be extended to compare two multinomial conditions for independence, but this is rarely done, if only because comparing two binomials is so darned useful.
With the binomial test, we look at the number of positive observations out of some total number of observations for each condition. In some situations, it is much more natural to talk about the number of positive observations not as a fraction of all observations, but as a fraction of the duration of the condition. For instance, we might talk about the number of times we noticed a particular kind of network error under different conditions. In such a case, we probably can say how long we looked for the errors under each condition, but it can be very hard to say how many observations there were without an error.
Count $$\Delta t$$
$$A$$
$$k_1$$
$$\,\,\,\, t_1 \,\,\,\,$$
$$\neg A$$
$$k_2$$
$$t_2$$
We can investigate whether the Poisson distribution the same under both conditions using the generalized log-likelihood ratio test. Such a test uses $\lambda$, the generalized log-likelihood ratio,
\[
\lambda = \frac{
\max_{\theta_1 = \theta_2 = \theta_0} p(k_1 | \theta_0)p(k_2 | \theta_0)
}{
\max_{\theta_1, \theta_2} p(k_1 | \theta_1)p(k_2 | \theta_2)
}
\]
According to Wilks\cite{wilks1938} and also later Chernoff\cite{Chernoff1954}, the quantity $-2 \log \lambda$ is asymptotically $\chi^2$ distributed with one degree of freedom.
For the Poisson distribution, \[ p(k | \theta, t) = \frac{(\theta t)^k e^{-\theta t}}{k!} \\ \log p(k|\theta, t) = k \log \theta t - \theta t - \log k! \] The maximum likelihood estimator $\hat\theta$ can be computed by maximizing the log probability
For the Poisson distribution,
\[
p(k | \theta, t) = \frac{(\theta t)^k e^{-\theta t}}{k!} \\
\log p(k|\theta, t) = k \log \theta t - \theta t - \log k!
\]
The maximum likelihood estimator $\hat\theta$ can be computed by maximizing the log probability
\[
\max_\theta \frac{(\theta t)^k e^{-\theta t}}{k!} = \max_\theta \log \frac{(\theta t)^k e^{-\theta t}}{k!} \\ \log \frac{(\theta t)^k e^{-\theta t}}{k!} = k \log \theta t - \theta t - \log k! \\ \frac{\partial \log L(k | \theta, t)}{\partial \theta} = \frac{k}{ \theta} - t =0 \\ \hat \theta = k \] Returning to the log-likelihood ratio test, after some cancellation we get
\max_\theta \frac{(\theta t)^k e^{-\theta t}}{k!} = \max_\theta \log \frac{(\theta t)^k e^{-\theta t}}{k!} \\
\log \frac{(\theta t)^k e^{-\theta t}}{k!} = k \log \theta t - \theta t - \log k! \\
\frac{\partial \log L(k | \theta, t)}{\partial \theta} = \frac{k}{ \theta} - t
=0 \\
\hat \theta = k
\]
Returning to the log-likelihood ratio test, after some cancellation we get
\[
-\log \lambda = k_1 \log k_1 + k_2 \log k_2 - k_1 \log \frac{k_1+k_2}{t_1+t_2} t_1 - k_2 \log \frac{k_1+k_2}{t_1+t_2} t_2 \]
-\log \lambda =
k_1 \log k_1 +
k_2 \log k_2
- k_1 \log \frac{k_1+k_2}{t_1+t_2} t_1 - k_2 \log \frac{k_1+k_2}{t_1+t_2} t_2
\]
Some small rearrangement gives the following preferred form that is very reminiscent of the form most commonly to compute the log-likelihood ratio test for binomials and multinomials
\[ -2 \log \lambda = 2 \left( k_1 \log \frac{k_1}{t_1} + k_2 \log \frac{k_2}{t_2} - (k_1+k_2) \log \frac{k_1+k_2}{t_1+t_2} \right) \]
\[
-2 \log \lambda = 2 \left( k_1 \log \frac{k_1}{t_1} +
k_2 \log \frac{k_2}{t_2} - (k_1+k_2) \log \frac{k_1+k_2}{t_1+t_2}
\right)
\]
|
Mentor: Bjoern Muetzel
If you are an undergraduate interested in a reading course, independent study or working on a research project, feel free to contact me. I am particularly interested in the following topics.
The hyperbolic plane is a space of constant negative curvature minus one, where different rules than in Euclidean space apply for geodesics, the geometry of polygons and the area of disks. A hyperbolic surface can be seen as a polygon in the hyperbolic plane with identified sides. We call such a surface a Riemann surface. Many questions about Riemann surfaces are still open or under study. Hyperbolic geometry is used in the theory of special relativity, particularly Minkowski spacetime.
A systole of a surface is a shortest non-contractible loop on a surface. Every surface has a genus \( g \), where informally \( g \) denotes the number of holes. Surprisingly given any surface of fixed genus \( g \) and area one, the systole can not take a value larger than \(c \cdot \frac{\log(g)}{ \sqrt{g}} \), where \( c \) is a constant. A large number of families of short curves on surfaces satisfy this upper bound and example surfaces can be found among the hyperbolic Riemann surfaces.
Advisor: Prof. Orellana
I have a number of projects accessible to undergraduate students in Combinatorics, Algebra and Graph Theory. These projects can lead to a senior thesis for honors or high honors. The ideal student should have taken math 24 and preferably (although not required) Math 28, 31 (71), 38 and have some programming skills. For more details schedule an appointment.
Advisor: Prof. Voight
Classical unsolved problems often serve as the genesis for the formulation of a rich and unified mathematical fabric. Diophantus of Alexandria first sought solutions to algebraic equations in integers almost two thousand years ago. For instance, he stated that if a two numbers can be written as the sum of squares, then their product is also a sum of two squares: since $5=2^2+1^2$ and $13=3^2+2^2$, then also $13\cdot 5=65$ can be written as the sum of two squares, indeed $65=8^2+1^2$. Equations in which only integer solutions are sought are now called Diophantine equations in his honor.
Diophantine equations may seem perfectly innocuous, but in fact within them can be found the deep and wonderously complex universe of number theory. Pierre de Fermat, a seventeenth century French lawyer and mathematician, famously wrote in his copy of Diophantus’ treatise “Arithmetica” that “it is impossible to separate a power higher than two into two like powers”, i.e., if $n>2$ then the equation $x^n+y^n=z^n$ has no solution in integers $x,y,z\ge 1$; provocatively, that he had “discovered a truly marvelous proof of this, which this margin is too narrow to contain.” This deceptively simple mathematical statement known as “Fermat’s last ‘theorem’” remained without proof until the pioneering work of Andrew Wiles, who in 1995 (building on the work of many others) used the full machinery of modern algebra to exhibit a complete proof. Over 300 years, attempts to prove there are no solutions to this innocent equation gave birth to some of the great riches of modern number theory.
Even before the work of Wiles, mathematicians recognized that geometric properties often govern the behavior of arithmetic objects. For example, Diophantus may have asked if there is a cube which is one more than a square, i.e., is there a solution in integers x,y to the equation $E : x^3-y^2=1$? This equation describes a curve in the plane called an elliptic curve, and a property of elliptic curves known as modularity was the central point in Wiles’s proof. One sees visibly the solution $(x,y)=(1,0)$ to the equation—but are there any others? What happens if 1 is replaced by 2 or another number?
Computational tools provide a means to test conjectures and can sometimes furnish partial solutions; for example, one can check in a fraction of a second on a desktop computer that there is no integral point on E other than $(1,0)$ with the coordinates x,y at most a million. Although this experiment does not furnish a proof, it is strongly suggestive. (Indeed, one can prove there are no other solutions following an argument of Leonhard Euler.) At the same time, theoretical advances fuel dramatic improvements in computation, allowing us to probe further into the Diophantine realm.
My research falls into this area of computational arithmetic geometry: I am concerned with algorithmic aspects of the problem of finding rational and integral solutions to polynomial equations, and I investigate the arithmetic of moduli spaces and elliptic curves. My work blends number theory with the explicit methods in algebra, analysis, and geometry in the exciting context of modern computation. This research is primarily theoretical, but it has potential applications in the areas of cryptography and coding theory. The foundation of modern cryptography relies upon the apparent difficulty of certain computational problems in number theory, in particular, the factorization of integers (in RSA) or the discrete logarithm problem (in elliptic curve cryptography).
I have several problems in the area of computational and explicit methods in number theory suitable for experimentation and possible resolution by motivated students. These problems can be tailored to the student based on interests, background, and personality, so there is little need to present the details here; but they all will feature a explicit mathematical approach and, very likely, some computational aspects. Mathematical maturity and curiosity is essential; some background (at the level of MATH 71) is desirable.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.