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Answer
$7.99\ g/cm^3$
Work Step by Step
If the diameter is 9.40 mm, the radius (half of it) is $4.70\ mm=0.470\ cm$. The volume of the sphere is: $V=\frac43\pi r^3=0.435\ cm^3$ The given mass is 3.475 g, so the density is: $\rho=m/V=7.99\ g/cm^3$
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In last week’s discussion of proofs by contradiction and nonconstructive proofs, we showed:
Theorem: There exist irrational numbers \(x\) and \(y\) with the property that \(x^y\) is rational.
However, our proof was
nonconstructive: it did not pinpoint explicit values for \(x\) and \(y\) that satisfy the condition, instead proving only that such numbers must exist. Would a more constructive proof be more satisfying? Let’s see! I claim \(x=\sqrt{2}\) and \(y=\log_2 9\) work, because \(\sqrt{2}\) we already know to be irrational, \(y=\log_2 9\) can be similarly proved to be irrational (try this!), and $$x^y = \sqrt{2}^{\log_2 9} = \sqrt{2}^{\log_{\sqrt{2}}3}=3,$$ which is rational.
Let’s further discuss why last week’s proof was less satisfying. The following rephrasing of this proof may help shed some light on the situation:
Proof: Assume the theorem were false, so that any time \(x\) and \(y\) were irrational, \(x^y\) would also be irrational. This would imply that \(\sqrt{2}^{\sqrt{2}}\) would be irrational, and by applying our assumption again, \(\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}}\) would also be irrational. But this last number equals 2, which is rational. This contradiction disproves our assumption and thereby proves the theorem, QED.
So perhaps this argument seems less satisfactory simply because it is, at its core, a proof by contradiction. It does not give us evidence for the positive statement “\(x\) and \(y\) exist”, but instead only for the negative statement “\(x\) and \(y\)
don’t not exist.” (Note the double negative.) This distinction is subtle, but a similar phenomenon can be found in the English language: the double negative “not bad” does not mean “good” but instead occupies a hazy middle-ground between the two extremes. And even though we don’t usually think of such a middle-ground existing between logic’s “true” and “false”, proofs by contradiction fit naturally into this haze. In fact, these ideas motivate a whole branch of mathematical logic called Constructive logic that disallows double negatives and proofs by contradiction, instead requiring concrete, constructive justifications for all statements.
But wait; last week’s proof that \(\sqrt{2}\) is irrational used contradiction, and therefore is not acceptable in constructive logic. Can we prove this statement constructively? We must show that \(\sqrt{2}\) is not equal to any rational number; what does it even mean to do this constructively? First, we turn it into a positive statement: we must show that \(\sqrt{2}\)
is unequal to every rational number. And how do we constructively prove that two numbers are unequal? By showing that they are measurably far apart. So, here is a sketch of a constructive proof: \(\sqrt{2}\) is unequal to every rational number \(a/b\) because $$\left|\sqrt{2} – \frac{a}{b}\right| \ge \frac{1}{3b^2}.$$ See if you can verify this inequality! [1]
PS. In case you are still wondering whether \(\sqrt{2}^{\sqrt{2}}\) is rational or irrational: It is irrational (moreover, transcendental), but the only proof that I know uses a very difficult theorem of Gelfond and Schneider.
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Well-Posedness and Ill-Posedness Problems of the Stationary Navier–Stokes Equations in Scaling Invariant Besov Spaces 227 Downloads Abstract
We consider the stationary Navier–Stokes equations in \(\mathbb {R}^n\) for \(n\geqq 3\) in the scaling invariant Besov spaces. It is proved that if \(n<p\leqq \infty \) and \(1\leqq q\leqq \infty \), or \(p=n\) and \(2<q\leqq \infty \), then some sequence of external forces converging to zero in \(\dot{B}^{-3+\frac{n}{p}}_{p,q}\) can admit a sequence of solutions which never converges to zero in \(\dot{B}^{-1}_{\infty ,\infty }\), especially in \(\dot{B}^{-1+\frac{n}{p}}_{p,q}\). Our result may be regarded as showing the borderline case between ill-posedness and well-posedness, the latter of which Kaneko–Kozono–Shimizu proved when \(1\leqq p<n\) and \(1\leqq q\leqq \infty \).
Notes Acknowledgements
The author was partly supported by Grant-in-Aid for JSPS Research Fellow (Grant Number: JP19J11499), Top Global University Project of Waseda University, and by the Research Institute for Mathematical Sciences, an International Joint Usage/Research Center located in Kyoto University.
References 1. 2. 3. 4. 5. 6. 7.Kaneko, K., Kozono, H., Shimizu, S.: Stationary solution to the Navier–Stokes equations in the scaling invariant Besov space and its regularity. Indiana Univ. Math. J.(to appear)Google Scholar 8. 9. 10. 11. 12.
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A quadratic equation can be defined as a equation of a second degree, which implies that it comprises of minimum one term that is squared. The definite form is
ax² + bx + c = 0; where in x is an unknown variable and a,b,c are numerical coefficients.
Quadratics Equation Examples
Beneath are the illustrations of quadratic equations of the form (ax² + bx + c = 0)
x² –x – 9 = 0 5x² – 2x – 6 = 0 3x² + 4x + 8 = 0 -x² +6x + 12 = 0
Examples of a quadratic equation with the absence of a ‘ C ‘- a constant term.
-x² – 9x = 0 x² + 2x = 0 -6x² – 3x = 0 -5x² + x = 0 -12x² + 13x = 0 11x² – 27x = 0
Following are the examples of a quadratic equation in factored form
(x – 6)(x + 1) = 0 [ result obtained after solving is x² – 5x – 6 = 0] –3(x – 4)(2x + 3) = 0 [result obtained after solving is -6x² + 15x + 36 = 0] (x − 5)(x + 3) = 0 [result obtained after solving is x² − 2x − 15 = 0] (x – 5)(x + 2) = 0 [ result obtained after solving is x² – 3x – 10 = 0] (x – 4)(x + 2) = 0 [result obtained after solving is x² – 2x – 8 = 0] (2x+3)(3x – 2) = 0 [result obtained after solving is 6x² + 5x – 6]
Below are the examples of a quadratic equation with an absence of linear co – efficient ‘ bx’
2x² – 64 = 0 x² – 16 = 0 9x² + 49 = 0 -2x² – 4 = 0 4x² + 81 = 0 -x² – 9 = 0 Solving Quadratics by Factoring Begin with a equation of the form ax² + bx + c = 0 Ensure that it is set to adequate zero. Factor the left hand side of the equation by assuming zero on the right hand side of the equation. Assign each factor equal to zero. Now solve the equation in order to determine the values of x.
Suppose if the main coefficient is not equal to one then deliberately, you have to follow a methodology in the arrangement of the factors.
2x²-x-6=0
(2x+3)(x-2)=0
2x+3=0
x=-3/2
x=2
Also, Sridhara wrote down rules for Solving Quadratic Equation, therefore the most common method of finding the roots of the quadratic equation is known as Sridharacharya rule.
For the given Quadratic equation of the form,
\(ax^{2}+bx+ c = 0\)
Therefore the roots of the given equation can be found by:
\(x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\),
where \(\pm\) (one plus and one minus) represent two distinct roots of the given equation.
Lets Work Out:
Solving the quadratic equation using the above method:
\(x= \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\)
\(x = \frac{-(-5)\pm \sqrt{(-5)^{2} -4 \times 3 \times 2}}{2 \times 3}\)
\(x = \frac{5 \pm 1}{6}\)
\(x = \frac{6}{6} \;\; or \;\; \frac{4}{6}\)
or, \(x = 1 \;\; or \;\; \frac{2}{3}\)<
Practice quadratics more by learning quadratic formulas and solving some other examples only at
www.byjus.com
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In Frankel's paper: "Manifolds with positive curvature", he proved the following theorem:
If $M^n$ is complete Riemannian manifold with positive sectional curvature, and $V^r$, $W^s$ are two compact totally geodesic submanifolds. If $r+s\ge n$ then $V$ and $W$ have a non-empty intersection.
The proof is by contradiction. If there is no intersection, then chose $\gamma$ connecting $V$ to $W$ and realized the distance between $V$ and $W$. Due to the dimension reason, he find a parallel vector fileld $X$ along $\gamma$, then apply the second variational formula.
For the boundary term $g(\nabla_X X, \dot{\gamma})$. He claimed this is zero due to the totally geodesic property. My question is: In order to take covariant derivative $\nabla_v Y$, the vector field $Y$ has to be defined at least along one curve $\sigma$ with $\dot{\sigma}=v$, right? But for $\nabla_X X$ in the proof, the vector filed $X$ is only defined along $\gamma$ not along the tangent direction $X$. (My guess is it does not depend on the extension of $X$, but I can't see why, or it's too trivial so Frankel didn't write it down?)
edit: I found that probably, we can just extend $X$ at $\gamma(0)$ and $\gamma(\ell)$ as the tengent vector of geodesic along direction $X$, and extend $X$ at $\gamma(t)$ arbitrary. So this will give the desired boundary condition. Is my claim correct?
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I am currently reading the paper "A Duality Web in 2+ 1 Dimensions and Condensed Matter Physics" by Seiberg et al, and on page 22 they add to the Lagrangian a monopole operator of the form \(\phi^\dagger \mathcal{M}_{\hat b}\). Firstly, is it perhaps a typo that the \(\phi\) is unhatted? Should it be hatted so that it is charged under U(1)\(_{\hat b}\) ? Secondly, how exactly does this operator break the global symmetry whose current is the topological current \(d\hat b\)? I have been trying to understand this under the light of "Generalized Global Symmetries", and if I understand correctly, this would constitute a 1-form global symmetry. However, I could not find in that paper a section which would explain why a monopole of this form would break the symmetry. I would be very grateful if someone could shed a little bit of light on this for me. Thank you!
I find that Nathan Seiberg explaining exactly the same is easier to follow here ( slides 7 - 11 )
1. There is no typo. Physical observables must be (gauged-) charge neutral. The monopole operator $\mathcal{M}_{\hat{b}}$ and $\phi^{\dagger}$ carry the opposite gauged charged so their product is neutral.2. Monopole operator breaks the conservation of the topological current because in the presence of a Dirac magnetic monopole, the gauge field is not globally well-defined anymore. i.e. $d\hat{f}=0$ does not implies that $\hat{f}=d\hat{b}$, where $\hat{f}$ is the field strength.3. I don't think that the second paper "Generalized Global Symmetries" is really related with your questions.
user contributions licensed under cc by-sa 3.0 with attribution required
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I need some help with the following problem:
Let $X$, $Y$, and $Z$ are independent circularly symmetric complex Gaussian random variable with zero mean and unit variance, i.e., $X$, $Y$, and $Z \sim CN(0, 1)$.
These random variables can be represented in polar form as \begin{equation} X=r_1 e^{i \alpha}\\ Y=r_2 e^{i \beta}\\ Z=r_3 e^{i \gamma} \end{equation} where $r_1$, $r_2$ and $r_3$ are the magnitude (Rayleigh distributed) and $\alpha$, $\beta$ and $\gamma$ are the phase of the variables (Uniformly distributed over $[0, 2\pi]$).
How can I represent the random variable $aX+bY+cZ$ (where a,b, c are constant) in polar form as a function of the variables $a$, $b$, $c$, $r_1$, $r_2$, $r_3$, $\alpha$, $\beta$ and $\gamma$. Particularly I am interested to know the phase of $aX+bY+cZ$ in terms of these variables.
Thank you very much
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(A)dS exchanges and partially-massless higher spins
Artikel i vetenskaplig tidskrift, 2008
We determine the current exchange amplitudes for free totally symmetric tensor fields $\vf_{\mu_1 ... \mu_s}$ of mass $M$ in a $d$-dimensional $dS$ space, extending the results previously obtained for $s=2$ by other authors. Our construction is based on an unconstrained formulation where both the higher-spin gauge fields and the corresponding gauge parameters $\Lambda_{\mu_1 >... \mu_{s-1}}$ are not subject to Fronsdal's trace constraints, but compensator fields $\alpha_{\mu_1 ... \mu_{s-3}}$ are introduced for $s > 2$. The free massive $dS$ equations can be fully determined by a radial dimensional reduction from a $(d+1)$-dimensional Minkowski space time, and lead for all spins to relatively handy closed-form expressions for the exchange amplitudes, where the external currents are conserved, both in $d$ and in $(d+1)$ dimensions, but are otherwise arbitrary. As for $s=2$, these amplitudes are rational functions of $(ML)^2$, where $L$ is the $dS$ radius. In general they are related to the hypergeometric functions $_3F_2(a,b,c;d,e;z)$, and their poles identify a subset of the "partially-massless" discrete states, selected by the condition that the gauge transformations of the corresponding fields contain some non-derivative terms. Corresponding results for $AdS$ spaces can be obtained from these by a formal analytic continuation, while the massless limit is smooth, with no van Dam-Veltman-Zakharov discontinuity.
Higher spins
Dam-Veltman-Zakharov discontinuity
AdS space
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Say I have a neural network with some unknown number $N$ of hidden layers. Assume I know the structure (e.g. feedforward, convolutional, or recurrent) of the first $k$ of these hidden layers but know nothing about the remaining $N-k$ layers. (I also don't know $N$.) Assuming the weights in the unknown part are fixed, then if I know the output generated by the full network, is it possible to use the loss function (e.g. MSE) to train only the hidden layers for which I know the structure without knowing anything about the unknown layers?
You are facing the issue that you have a composition of two functions, $f(\cdot)$ being your known part of the neural network and $g(\cdot)$ being the unknown part. The output is then $g(f(x))$ and there is some loss function $\mathcal{L}(g(f(x)), y)$ comparing the network output to the correct answer. To train the network, you need to estimate how much each of the weights contributed to the error, which is done by computing the derivative $\frac{\partial \mathcal L}{\partial w}$.
Since you don't know the analytical form of $g(\cdot)$, you cannot analytically compute this derivative. That leaves you with two options:
Numerical differentiation: instead of computing the derivative analytically, you can perturb a weight $w$ by a small amount and see how it affects $\mathcal L$. Note that if you have many weights, this approach will be extremely slow. Approximate $g(\cdot)$: If you can evaluate $g(\cdot)$, nothing prevents you from creating another neural network $\hat g(\cdot)$ and train it to mimic $g$ simply by feeding random inputs to $g$ and training $\hat g$ to predict the same thing. Then you can use this new neural network as a surrogate for training $f$ which will allow you to evaluate the gradients analytically.
Either way is somewhat impractical.
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As it can be shown, the equations for the irrep with zero mass and helicity 2, -2 respectively can be given in a form $$ \tag 1 \partial^{\dot {b}a}C_{abcd} = 0, \quad \partial^{\dot{b}a}C_{\dot{a}\dot{b}\dot{c}\dot{d}} = 0. $$ Here $C_{abcd} = C_{(abcd)}, C_{\dot{a}\dot{b}\dot{c}\dot{d}} = C_{(\dot{a}\dot{b}\dot{c}\dot{d})}$ are symmetrical spinor tensors or the reps $\left( 2, 0 \right), \left( 0, 2\right)$ of the Lorentz group respectively.
From $C_{abcd}, C_{\dot{a}\dot{b}\dot{c}\dot{d}}$ we may construct 4-tensor $$ C_{\mu \nu \alpha \beta} = (\tilde{\sigma}_{\mu})^{\dot{a}a}(\tilde{\sigma}_{\nu})^{\dot{b}b}(\tilde{\sigma}_{\alpha})^{\dot{c}c}(\tilde{\sigma}_{\beta})^{\dot{d}d}(\varepsilon_{ab}\varepsilon_{cd}C_{\dot{a}\dot{b}\dot{c}\dot{d}} + \varepsilon_{\dot{a}\dot{b}}\varepsilon_{\dot{c}\dot{d}}C_{abcd}) $$ with properties $$ \tag 2 C_{\mu \nu \alpha \beta} = -C_{\nu \mu \alpha \beta} =-C_{\mu \nu \beta \alpha} = C_{\nu \mu \beta \alpha}, \quad C^{\alpha}_{\ \beta \alpha \delta} = 0, \quad \varepsilon^{\alpha \beta \gamma \delta}C_{\alpha \beta \gamma \delta} = 0. $$ In the formal language it is a direct sum of representations with helicity 2 and helicity -2.
From $(1)$ we can also get $$ \tag 3 \varepsilon^{\mu \nu \alpha \beta}\partial_{\nu}C_{\varepsilon \delta \alpha \beta} = 0. $$
$(2), (3)$ gives us the Weyl tensor of the linearized GR.
My question: why we have got the Weyl tensor for massless helicity-2 particles, not the curvature tensor? I.e., why the free gravitons are described by the Weyl tensor, not by the curvature?
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Suppose you have a car traveling on a banked curve with friction. The curve makes an angle $\theta$ with the horizontal and has coefficient of static friction $\mu$.
Suppose we wish to calculate the minimum speed at which the car can travel so that it does not slip down the curve towards the center of the circular racetrack (of radius) $R$.
I know how to calculate this speed, but the following argument seems to reach a contradiction, and I cant tell what I'm doing wrong.
The force of friction keeping the car from slipping down the curve acts opposite the component of gravity parallel to the track. Thus $F_g\sin\theta=F_f$. Also, since the car is on the verge of slipping, $F_f=\mu F_N$ where $F_N$ is the normal force. Since the net force in the direction perpindicular to the car is 0, $F_N=F_g\cos\theta$. Thus $F_g\sin\theta=F_f=\mu F_g\cos\theta$, so $\mu=\tan\theta$.
This makes no sense since it implies the initial conditions must have satisfied this constraint. There is obviously something wrong with this reasoning, and I think I made the incorrect assumption that $F_N=mg\cos\theta$, because the correct method of calculating the minimum speed has a different result for $F_n$.
Please identify the flaw, thank you.
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1. 11, 23, 47, 83, 131, . What is the next number?a. 145
b. 178
c. 176
d. 191
Explanation:
11,23,47,83,131
23–11 = 12
47–23 = 24
83–47 = 36
131–83 = 48
Therefore, 131+60=191
2. A series of book was published at seven year intervals. When the seventh book was published the total sum of publication year was 13, 524. First book was published in?a. 1911
b. 1910
c. 2002
d. 1932
Answer:
Explanation:
Let the years be n, n+7, n+14, ...., n+42. ($\because $ use formula ${T_n} = a + (n - 1)d$ to find nth term)
Sum = ${S_n} = \dfrac{n}{2}\left( {2a + (n - 1)d} \right)$ = $\dfrac{7}{2}\left( {2n + (7 - 1)7} \right)$ = 13,524
$ \Rightarrow 7n + 147 = 13,524$
$ \Rightarrow $ n = 1911
3. Crusoe hatched from a mysterious egg discovered by Angus, was growing at a fast pace that Angus had to move it from home to the lake. Given the weights of Crusoe in its first weeks of birth as 5, 15, 30,135, 405, 1215, 3645. Find the odd weight out.a) 3645
b) 135
c) 30
d) 15
Answer: c
Explanation:
5×3 = 15
15×3 = 45 $ \Rightarrow $ Given as 30
45×3 = 135
135×3 = 405
405×3 = 1215
1215×3 = 3645
4. A can complete a piece of work in 8 hours, B can complete in 10 hours and C in 12 hours. If A,B, C start the work together but A laves after 2 hours. Find the time taken by B and C to complete the remaining work.1) 2 (1/11) hours
2) 4 (1/11) hours
3) 2 (6/11) hours
4) 2 hours
Explanation:
A,B,C's 1 hour work is = $\dfrac{1}{8} + \dfrac{1}{{10}} + \dfrac{1}{{12}}$ = $\dfrac{{15 + 12 + 10}}{{120}} = \dfrac{{37}}{{120}}$
A,B,C worked together for 2 hours, Therefore, 2 hours work is = $\dfrac{{37}}{{120}} \times 2 = \dfrac{{37}}{{60}}$
Remaining work = $1 - \dfrac{{37}}{{60}} = \dfrac{{23}}{{60}}$
(23/60 work is done by B and C together)
B, C's 1 hour work = $\dfrac{1}{{10}} + \dfrac{1}{{12}} = \dfrac{{6 + 5}}{{60}} = \dfrac{{11}}{{60}}$
${\left( {\dfrac{{23}}{{60}}} \right)^{th}}$ part of the work done by B, C in = $\dfrac{{\left( {\dfrac{{23}}{{60}}} \right)}}{{\dfrac{{11}}{{60}}}}$ = $2\dfrac{1}{{11}}$ hours.
5. A tree of height 36m is on one edge of a road broke at a certain height. It fell in such a way that the top of the tree touches the other edge of the road. If the breadth of the road is 12m, then what is the height at which the tree broke?a. 16
b. 24
c. 12
d. 18
Explanation:
Let the tree was broken at x meters height from the ground and 36 - x be the length of other part of the tree.
$ \Rightarrow 1296 - 72x + {x^2} = {x^2} + 144$
$ \Rightarrow 72x = 1296 - 144$
$ \Rightarrow x = 16$
6. The sticks of same length are used to form a triangle as shown below.If 87 such sticks are used then how many triangles can be formed?
Explanation:
First triangle is formed by using 3 sticks, but any subsequent triangle may be formed by using 2 sticks. Therefore, If 1st triangles uses 3 sticks, Remaining sticks = 87 - 3 = 84. With these 84, we can form 42 triangles. So total = 42 + 1 = 43
Shortcut:
To solve questions like these, use formula, 2n + 1 = k. Here n = triangles, k = sticks
2n+1 = 87 $ \Rightarrow $ n = 43.
7. 17 × 8 m rectangular ground is surrounded by 1.5 m width path. Depth of the path is 12 cm. Gravel is filled and find the quantity of gravel required.a. 5.5
b. 7.5
c. 6.05
d. 10.08
Explanation:
Area of the rectangular ground = $17 \times 8 = 136{m^2}$
Area of the big rectangle considering the path width = $\left( {17 + 2 \times 1.5} \right) \times \left( {8 + 2 \times 1.5} \right) = 220{m^2}$
Area of the path = $220 - 136 = 84{m^2}$
Gravel required = $84{m^2} \times \dfrac{{12}}{{100}}m = 10.08{m^3}$
8. A sum of Rs.3000 is distributed among A, B, and C. A gets 2/3 of what B and C got together and c gets 1/3 of what A and B got together, C's share is?Explanation:
Let B+C together got 3 units, then A get 2 units. or $\dfrac{{B + C}}{A} = \dfrac{3}{2}$ - - - (1)
Let A+B together got 3 units, then B get 1 units. or $\dfrac{{A + B}}{C} = \dfrac{3}{1}$ - - - (2)
By using Componendo and Dividendo, we can re-write equations (1) and (2), $\dfrac{{A + B + C}}{A} = \dfrac{{3 + 2}}{2} = \dfrac{5}{2} = \dfrac{{20}}{8}$ and $\dfrac{{A + B + C}}{C} = \dfrac{{3 + 1}}{1} = \dfrac{4}{1} = \dfrac{{20}}{5}$
So A = 8, B = 7, C = 5
C's share = $\dfrac{5}{{(8 + 5 + 7)}} \times 3000 = 750$
9. The numbers 272738 and 232342, when divided by n, a two digit number, leave a remainder of 13 and 17 respectively. Find the sum of the digits of n?a. 7
b. 8
c. 5
d. 4
Explanation:
From the given information, (272738 - 13, 232342 - 17) are exactly divisible by that two digit number.
We have to find the HCF of the given numbers 272725, 232325.
HCF = 25.
So sum of the digits = 7.
10. Assume that f(1)=0 and f(m+n)=f(m)+f(n)+4(9mn-1). For all natural numbers (Integers>0)m and n. What is the value of f(17)?a. 5436
b. 4831
c. 5508
d. 4832
Explanation:
f(1) = 0
f(2) = f(1+1) = f(1)+f(1)+4(9×1×1 – 1) = 0+0+4×8 = 32
f(4) = f(2+2) = f(2)+f(2)+4(9×2×2 – 1) = 32+32+4×35 = 204
f(8) = f(4+4) = f(4)+f(4)+4(9×4×4 – 1) = 204+204+4×143 = 980
f(16) = f(8+8) = f(8)+f(8)+4(9×8×8 – 1) = 980+980+4×575 = 4260
f(17) = f(1+16) = f(16)+f(1)+4(9×16×1 –1) = 4260+0+ 4×143 = 4832
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I have an expectation given as: $\mathbb{E}\left(S_{T}\mathbb{1}_{S_{T}\geq K} \right)$
where $K$ is just an arbitrary number (i.e. the strike price, but that's unimportant) and $S$ can be modelled by the equation $S_{t} = \exp((r-\frac{1}{2})t + \sigma W_{t})$. Also, the expectation is under the $\mathbb{P}$-measure, not the $\mathbb{Q}$-measure, so effectively this expectation is $\mathbb{E}^{\mathbb{P}}\left(S_{T}\mathbb{1}_{S_{T}\geq K} \right)$
Now, when trying to evaluate $\mathbb{E}\left(\mathbb{1}_{S_{T}\geq K} \right)$ under the $\mathbb{Q}$-measure, then solving this expectation is fairly easy, since you can integrate the SDE and take the $\log$ of $S$ to get $\log(S_{T}) = \log(S_{0}) + (r-\frac{1}{2})T + \sigma\sqrt{T}N(0,1)$ (since $W_{t}\approx \sqrt{T}N(0,1)$) and thus we have for $\mathbb{E}\left(\mathbb{1}_{S_{T}\geq K} \right) = \mathbb{P}({S_{T}\geq K})$:
$\log(S_{T}) = (r-\frac{1}{2})T + \sigma\sqrt{T}N(0,1) > \log(K)$ and thus rearranging this equation gives
$N(0,1) > \frac{\log(K/S_{0}) - (r-\frac{1}{2}\sigma^{2})T}{\sigma\sqrt{T}}$
and through a little bit more rearrangement (i.e. knowing that $N(0,1)<-x = 1 - N(x)$ we get finally $d_{2} = \frac{(r-\frac{1}{2}\sigma^{2})T + \log(S_{0}/K)}{\sigma\sqrt{T}}$ and thus $\mathbb{E}\left(\mathbb{1}_{S_{T}\geq K} \right) = N(d_{2})$.
The problem I have now however is that I'm a bit unsure how to find $\mathbb{E}^{\mathbb{P}}\left(S_{T}\mathbb{1}_{S_{T}\geq K} \right)$ (i.e. under the $\mathbb{P}$-measure) since I've never done any measure theory before, so if someone could help me out I'd really appreciate it. Thanks in advance.
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Suppose we have an SU(N) non abelian gauge theory coupled with a multiplet of complex scalar fields $\Phi$. The lagrangian would be
$$ L= - \frac 12 \text{Tr } F_{\mu\nu}F^{\mu\nu} + |D_\mu \Phi|^2 - V(\Phi)$$
where $D_\mu$ is the covariant derivative: $D_\mu \Phi = \partial_\mu \Phi +ig A_\mu^at_a \Phi$, $A_\mu$ is the gauge field and $t_a$ are the generators of $SU(N)$. The kinetic term gives two interaction terms between $\Phi$ $$ \partial_\mu \Phi^* \partial^\mu \Phi + ig A^{\mu a} (\partial_\mu \Phi^* t_a \Phi - \Phi^* t_a \partial_\mu \Phi) + g^2 A_\mu^a A^{\mu b} \Phi^* t_a t_b \Phi$$ I would like to know if I extracted the vertices correctly from the lagrangian. Suppose all the momenta are ingoing. For the $\Phi^*\Phi A_\mu$ vertex I have $$ i g (q+q')^\mu (t^a)_{ij}$$ where $q$,$q'$ are the momenta of $\Phi^*$ and $\Phi$, but I'm not sure about the relative sign of the momenta. For the $\Phi^*\Phi A_\mu^a A_\nu^b$ vertex, I have $$-i g^2 g^{\mu\nu}\{t^a,t^b\}_{ij}$$ ($i,j$ are the component indices for the scalar multiplet).
I would like to underline that this is not homework. Even if the question arises from the preparation of an exam, I'm not asking the solution to an exercise. This is just some speculation following a digression of the lecturer about gauge theories with scalar fields.
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Say that we have a system evolving over discrete timesteps.
The quantity we are interested is X and is given by a distribution $P_X$. This distribution is evolving temporally, and we have a nonlinear equation $F$ that gives us this evolution.
$P_{\Delta X}(x;n) = F\big(~P_X(x;n)~,~P_X(x;n-1)~\big)~~~,~~~n \in \mathbb{N}$
where: $~~~X(x;n+1)=X(x;n)+\Delta X(x;n)$
We also know the initial condition $P_X(x;0) = \delta(x-k)$.
Unfortunately $X$ and $\Delta X$ are correlated.
My question is, what is the appropriate framework to investigate the evolution of $P_X$?
This post imported from StackExchange Physics at 2016-05-21 10:16 (UTC), posted by SE-user Dionysios Gerogiadis
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It took me quite some time to clearly understand the experiment you're describing.
Actually, pouring a full bottle in a container is a quite intriguing thing.
Consider the following starting configuration :
This of course is an unstable situation, as the pressure $P_0$ cannot be at the same time the pressure of the air in the bottle, and the atmospheric pressure since the height of water in the bottle is higher than the level in the container.
So we should quickly get to this configuration instead :
You'll agree that along the red line, the pressure is $P_0$, so what is the pressure $P_1$ ?
Using simple hydrostatics, $ P_1 = P_0 - \rho \, g \, H$
Notice that in the picture as well as in this calculation, we consider the height $H$ to not have changed, i.e. very little water has moved out of the bottle into the container. We'll see why now.
What is now the volume of air in the bottle ?
Using the law of perfect gases $P_0 * V_0 = P_1 * V_1$, hence $$\frac{V_1-V_0}{V_0} = \frac{1}{\frac{P_0}{\rho \, g \, H} - 1} = \frac{1}{\frac{10^5}{10^3 \, 10 \, 10^{-1}} - 1} \approx 1 \% $$
For this numerical estimation I took a water height in the bottle of $10 \, cm$. The variation in volume is so small, it will be hardly noticeable !
The reason why pouring the bottle is intriguing is that it empties itself in bursts. A bubble of air gets in, and water gets out at once. But if you do it in a controlled way, you will end up in the initial configuration I described, and from that point onwards, no air can get in. The variation of volume of the air in the bottle we just obtained obviously corresponds to a volume of water that gets out of the bottle, but again, it is small, and hardly noticeable.
What if you take a longer bottle ?
gigacyan is right, something will happen after a while. Recall that I did the calculation assuming the amount of water exiting the bottle is very small, this assumption is now false. If you have a significant height of water, the pressure will be enough to push out quite a bit of water out of the bottle, in which case the pressure of the air in the bottle will go down, and the level of water in the container go up.
If you consider a very wide container, its level will stay roughly the same, but the level of water in your bottle will go down. A simple calculation leads to:$$P_0-\rho \, g \, h_{final}+\rho \, g \, (H-h_{final})=P_0$$Hence $h_{final} = H/2$, which is the point when the low pressure in the air is able to lift the weight of the water underneath, down to the free surface.
Several interesting remarks can now be made.
To begin with, the pressure in the air keeps on dropping, $P_1 = P_0 - \rho \, g \, h_{final}$. Nothing prevents it from going to negative values, which happens when $h_{final} = \frac{P_0}{\rho \, g} = 10 \, m$. That's where this famous value of 10 meters comes from.
Now, if you think about trees, at first you may imagine they rely on capillary action to carry sap to their leaves, but that can't be the case, as the pressure drops too much after 10 vertical meters against gravity. Any presence of air would make the wood crumple under its own applied pressure.
Which means there is absolutely no air whatsoever in the sap canals of a tree (a.k.a. xylem).
The trees rely principally on another mechanism to pump up sap, known as evaporation. This easily produces (highly) negative pressures in the sap, and the actual limit to the size of a tree is the point when this pressure is small enough that a cavity of water vapor spontaneously appears in its canals, through cavitation. Pull hard enough on water, and you will create two interfaces and evaporate some of the liquid ! This cavitation pressure is around $-120 \, MPa$.
This catastrophic failure is know as embolism, and is also a bad health condition for humans (a gas bubble in a blood vessel).
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One of the main lessons of category theory is that whenever you think about some kind of mathematical gadget, you should also think about maps
between gadgets of this kind. For example, when you think about sets you should also think about functions. When you think about vector spaces you should also think about linear maps. And so on.
We've been talking about various kinds of monoidal preorders. So, let's think about maps
between monoidal preorders.
As I explained in Lecture 22, a monoidal preorder is a crossbreed or hybrid of a
preorder and a monoid. So let's think about maps between preorders, and maps between monoids, and try to hybridize those.
We've already seen maps between preorders: they're called monotone functions:
Definition. A monotone function from a preorder \((X,\le_X)\) to \((Y,\le_Y)\) is a function \(f : X \to Y\) such that
$$ x \le_X x' \textrm{ implies } f(x) \le_Y f(x') $$ for all elements \(x,x' \in X\).
So, these functions preserve what a preorder has, namely the relation \(\le\). A monoid, on the other hand, has an associative operation \(\otimes\) and a unit element \(I\). So, a map between monoids should preserve th0se! That's how this game works.
Just to scare people, mathematicians call these maps "homomorphisms":
Definition. A homomorphism from a monoid \( (X,\otimes_X,I_X) \) to a monoid \( (Y,\otimes_Y,I_Y) \) is a function \(f : X \to Y\) such that:
$$ f(x \otimes_X x') = f(x) \otimes_Y f(x') $$ for all elements \(x,x' \in X\), and
$$ f(I_X) = I_Y .$$ You've probably seen a lot of homomorphisms between monoids. Some of them you barely noticed. For example, the set of integers \(\mathbb{Z}\) is a monoid with addition as \(\otimes\) and the number \(0\) as \(I\). So is the set \(\mathbb{R}\) of real numbers! There's a function that turns each integer into a real number:
$$ i: \mathbb{Z} \to \mathbb{R} . $$It's such a bland function you may never have thought about it: it sends each integer to
itself, but regarded as a real number. And this function is a homomorphism!
What does that mean? Look at the definition. It means you can either add two natural numbers and then regard the result as a real number... or first regard each of them as a real number and then add them... and you get the same answer either way. It also says that integer \(0\), regarded as a real number, is the real number we call \(0\).
Boring facts! But utterly crucial facts. Computer scientists need to worry about these things, because for them integers and real numbers (or floating-point numbers) are different data types, and \(i\) is doing "type conversion".
You've also seen a lot of other, more interesting homomorphisms between monoids.
For example, the whole point of the logarithm function is that it's a homomorphism. It carries multiplication to addition:
$$ \log(x \cdot x') = \log(x) + \log(x') $$ and it carries the identity for multiplication to the identity for addition:
$$ \log(1) = 0. $$ People invented tables of logarithms, and later slide rules, precisely for this reason! They wanted to convert multiplication problems into easier addition problems.
You may also have seen linear maps between vector spaces. A vector space gives a monoid with addition as \(\otimes\) and the zero vector as \(I\); any linear map between vector spaces then gives a homomorphism.
Puzzle 80. Tell me a few more homomorphisms between monoids that you routinely use, or at least know.
I hope I've convinced you: monotone functions between preorders are important, and so are homomorphisms between monoids. Thus, if we hybridize these concepts, we'll get a concept that's likely to be important.
It turns out there are a few different ways! The most obvious way is simply to combine all the conditions. There are other ways, so this way is called "strict":
Definition. A strict monoidal monotone from a monoidal preorder \( (X,\le_X,\otimes_X,I_X) \) to a monoidal preorder \( (Y,\le_Y,\otimes_Y,I_Y) \) is a function \(f : X \to Y\) such that:
$$ x \le_X x' \textrm{ implies } f(x) \le_Y f(x') $$ and
$$ f(x) \otimes_Y f(x') = f(x \otimes_X x') $$ for all elements \(x,x' \in X\), and also
$$ I_Y = f(I_X) . $$ For example, the homomorphism
$$ i : \mathbb{Z} \to \mathbb{R} ,$$ is a strict monoidal monotone: if one integer is \(\le\) another, then that's still true when we regard them as real numbers. So is the logarithm function.
What other definition could we possibly use, and why would we care? It turns out sometimes we want to replace some of the equations in the above definition by inequalities!
Definition. A lax monoidal monotone from a monoidal preorder \((X,\le_X,\otimes_X,I_X)\) to a monoidal preorder \((Y,\le_Y,\otimes_Y,I_Y)\) is a function \(f : X \to Y\) such that:
$$ x \le_X x' \textrm{ implies } f(x) \le_Y f(x') $$ and
$$ f(x) \otimes_Y f(x') \le_Y f(x \otimes_X x') $$ for all elements \(x,x' \in X\), and also
$$ I_Y \le_Y f(I_X). $$Fong and Spivak call this simply a
monoidal monotone, since it's their favorite kind. But I will warn you that others call it "lax".
We could also turn around those last two inequalities:
Definition. An oplax monoidal monotone from a monoidal preorder \((X,\le_X,\otimes_X,I_X)\) to a monoidal preorder \((Y,\le_Y,\otimes_Y,I_Y)\) is a function \(f : X \to Y\) such that:
$$ x \le_X x' \textrm{ implies } f(x) \le_Y f(x') $$ and
$$ f(x) \otimes_Y f(x') \ge_Y f(x \otimes_X x') $$ for all elements \(x,x' \in X\), and also
$$ I_Y \ge_Y f(I_X). $$ You are probably drowning in definitions now, so let me give some examples to show that they're justified. The monotone function
$$ i : \mathbb{Z} \to \mathbb{R} $$ has a right adjoint
$$ \lfloor \cdot \rfloor : \mathbb{R} \to \mathbb{Z} $$which provides the approximation
from below to the nonexistent inverse of \(i\): that is, \( \lfloor x \rfloor \) is the greatest integer that's \(\le x\). It also has a left adjoint
$$ \lceil \cdot \rceil : \mathbb{R} \to \mathbb{Z} $$which is the best approximation
from above to the nonexistent inverse of \(i\): that is, \( \lceil x \rceil \) is the least integer that's \(\ge x\). Puzzle 81. Show that one of the functions \( \lfloor \cdot \rfloor : \mathbb{R} \to \mathbb{Z} \), \( \lceil \cdot \rceil : \mathbb{R} \to \mathbb{Z} \) is a lax monoidal monotone and the other is an oplax monoidal monotone, where we make the integers and reals into monoids using addition.
So, you should be sensing some relation between left and right adjoints, and lax and oplax monoidal monotones. We'll talk about this more! And we'll see why all this stuff is important for resource theories.
Finally, for the bravest among you:
Puzzle 82. Find a function between monoidal preorders that is both lax and oplax monoidal monotone but not strict monoidal monotone.
In case you haven't had enough jargon for today: a function between monoidal preorders that's both lax and oplax monoidal monotone is called
strong monoidal monotone.
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№ 9
All Issues Exact values of the best (α, β) -approximations of classes of convolutions with kernels that do not increase the number of sign changes Abstract
We obtain the exact values of the best $(\alpha , \beta )$-approximations of the classes $K \ast F$ of periodic functions $K \ast f$ such that $f$ belongs to a given rearrangement-invariant set $F$ and $K$ is $2\pi$ -periodic kernel that do not increase the number of sign changes by the subspaces of generalized polynomial splines with nodes at the points $2k\pi /n$ and $2k\pi /n + h, n \in N, k \in Z, h \in (0, 2\pi /n)$. It is shown that these subspaces are extremal for the Kolmogorov widths of the corresponding functional classes.
Citation Example: Parfinovych N. V. Exact values of the best (α, β) -approximations of classes of convolutions with
kernels that do not increase the number of sign changes // Ukr. Mat. Zh. - 2017. - 69, № 8. - pp. 1073-1083. Full text
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Is there a closed-form expression for the distribution of the Sample Kurtosis of data sampled from Gaussian distribution? i.e.,
$P(\hat{K}<a)$ where $\hat{K}$ is the sample kurtosis.
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The exact sampling distribution is tricky to derive; there have been the first few moments (dating back to 1929), various approximations (dating back to the early 1960s), and tables, often based on simulation (dating back to the 1960s).
To be more specific:
Fisher (1929) gives moments of the sampling distribution of the skewness and kurtosis in normal samples, and Pearson (1930) (also) gives the first four moments of the sampling distribution of the skewness and kurtosis and proposes tests based on them.
So for example$^*$:
$E(b_2)=\frac{3(n-1)}{n+1}$
$\text{Var}(b_2)=\frac{24n(n-2)(n-3)}{(n+1)^2(n+3)(n+5)}$
The skewness of $b_2$ is $\frac{216}{n}(1-\frac{29}{n}+\frac{519}{n^2}-\frac{7637}{n^3}+\ldots)$
The excess kurtosis of $b_2$ is $\frac{540}{n}-\frac{20196}{n^2}+\frac{470412}{n^3}+\ldots$.
* Beware - the values for the moments and so on depend on the exact definition of the sample kurtosis being used. If you see a different formula for $E(b_2)$ or $\text{Var}(b_2)$, for example, it will generally be because of a slightly different definition of sample kurtosis.
In this case, the above formulas should apply to $b_2=n\frac{\sum_i (X_i-\bar X)^4}{(\sum_i (X_i-\bar X)^2)^2}$.
Pearson (1963) discusses approximating the sampling distribution of kurtosis in normal samples by a Pearson type IV or a Johnson $S_U$ distribution (doubtless the reason the first four moments were given three decades earlier was in large part to make use of the Pearson family possible).
Pearson (1965) gives tables for percentiles of kurtosis for some values of $n$.
D'Agostino and Tietjen (1971) give more extensive tables of percentiles for kurtosis.
D'Agostino and Pearson (1973) give graphs of percentage points of kurtosis which cover a more extensive range of cases again.
Fisher, R. A. (1929),
"Moments and Product Moments of Sampling Distributions," Proceedings of the London Mathematical Society, Series 2, 30: 199-238.
Pearson, E.S., (1930)
"A further development of tests for normality," Biometrika, 22 (1-2), 239-249.
Pearson, E.S. (1963)
"Some problems arising in approximating to probability distributions, using moments," Biometrika, 50, 95-112
Pearson, E.S. (1965)
"Tables of percentage points of $\sqrt{b_1}$ and $b_2$ in normal samples: A rounding off," Biometrika, 52, 282-285
D'Agostino, R.B. and Tietjen, G.L. (1971),
"Simulation probability points of $b_2$ for small samples," Biometrika, 58, 669-672.
D'Agostino, R.B., and Pearson, E.S. (1973),
"Tests for departure from normality. Empirical results for the distribution of $b_2$ and $\sqrt{b_1}$," Biometrika, 60, 613-622.
The Sample Kurtosis from a normal sample, is approximately distributed as a zero-mean normal with variance $\approx 24/n$, where $n$ is the sample size (naturally, the larger $n$ the better the approximation. More complicated expressions for the variance can be found in the wikipedia page). For Gaussian samples of small size (<40), percentiles have been derived in this paper: Lacher, D. A. (1989). Sampling distribution of skewness and kurtosis. Clinical chemistry, 35(2), 330-331.
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27 February 2012, 05:58
Today’s Theorem comes thanks to Simon, because he is stupid. It is given without proof, because I don’t know how to prove it, and my book doesn’t have a proof. I probably don’t even know what it says.
Theorem: Suppose that a function $f$ is represented by a power series in $x – x_0$ that has a nonzero radius of convergence $R$; that is, \[ f(x) = \sum_{k=0}^{\infty} c_k(x – x_0)^k \qquad (x_0 < x < x_0 + R). \] If the power series representation of $f$ is integrated term by term using an indefinite integral, then the resulting series has radius of converge $R$ and converges to $\int f(x) dx$ on the interval $(x_0 – R, x_0 + R)$; that is, \[ \int f(x)dx = \sum_{k=0}^{\infty} \left[ \int c_k(x – x_0)^k dx \right] + C \qquad (x_0 – R < x < x_0 + R) \] If $\alpha$ and $\beta$ are points in the interval $(x_0 – R, x_0 + R)$, and if the power series representation of $f$ is integrated term by term from $\alpha$ to $\beta$, then the resulting series of numbers converges absolutely on the interval $(x_0 – R, x_0 + R)$ and \[ \int_{\alpha}^{\beta} f(x) dx = \sum_{k=0}^{\infty} \left[\int_{\alpha}^{\beta} c_k(x – x_0)^k dx \right] \]
Posted by Michael Welsh at 05:58.
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My nice nats Collection
context $F$ in ${\bf D}\longrightarrow{\bf C}$ context $G$ in ${\bf C}\longrightarrow{\bf D}$ definiendum $\langle\alpha,\beta\rangle$ in it inclusion $\alpha:FG\xrightarrow{\bullet}1_{\bf C}$ inclusion $\beta:1_{\bf D}\xrightarrow{\bullet}GF$ Discussion
That silly name … I made it up.
The natural transformation $\beta:1_{\bf D}\xrightarrow{\bullet}GF$ squeezes every set $X\in {\bf D}$ into a set $GFX\in {\bf D}$ (although this need not be surjective or injective). The natural transformation $\alpha:FG\xrightarrow{\bullet}1_{\bf C}$ squeezes all sets $FGX$ in the image of $FG$ back into $X$. The latter operation gets rid of lots $FG$'s without changing the structural properties of ${\bf C}$.
The point is that my equivalence of categories and Counit-unit adjunction are two different important special cases of nice nats. In the former case, the two nats actually shift the whole content of a category internally. In the latter case, the two nats end up defining the shifting operations of a monad.
Theorems
Only when the nats are isomorphisms (as in my equivalence of categories) is $F$ fully faithful and dense.
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Cross-posted from my blog.
The purpose of this post is to lay out some of the stuff I wish someone had explained to me when I was first learning about electronics. This is a pretty entry-level introduction, intended by me to ‘fill in the gaps’ with regard to basic operating principles. Opamps are not all that complicated, but for many people they appear as a black box. With so many schematics available online, it can be very tempting to simply ‘cut-and-paste’ subcircuits into your own designs without understanding them. This is okay to a point, and useful for learning, but it may come back to bite you later if your design doesn’t work the way you expect it to.
While this is entry level in terms of concept, there is some significant algebra going on here. I’ve tried to explain each step as I go along, but it does assume you have a basic knowledge of algebra. Unfortunately, there’s really no way around it.
First, let’s look at the schematic symbol for an op-amp and identify the pins and their functions.
V+ is the positive supply and V- is the negative supply. Often these will be omitted in diagrams, especially if the opamp is part of a dual or quad package. For most schematics, it is implied that the opamp is connected to such supply rails. Vin+ is the non-inverting input. Because of the plus sign, you may be tempted to call this the positive input, but that would be something of a misnomer because the signal voltage applied to it can positive or negative with respect to ground. Vin- is the inverting input. Again, the signal voltage applied here can be positive or negative with respect to ground. Vout is the voltage output.
There are two concepts which are typically used to describe how opamps function. The first is: the voltage output of the opamp is the voltage difference between the two inputs multiplied by the gain of the device. The gain of the device is often denoted as
A, and refers to the open-loop gain of the amp. Further, the voltage difference refers to the difference between the voltage at the non-inverting input (Vin+) and the voltage at the inverting input (Vin-). In other words:
[latex]\begin{equation}V_o = A \cdot (V_{in+} – V_{in-})\end{equation}[/latex]
This voltage difference can be considered the input to the opamp. The order of terms in the above subtraction is where the inputs get their names. To explain, let’s assume that the voltage at Vin+ is X, and the voltage at Vin- is 0 (ground). Thus, Vin would be X – 0, which is just X. However, if Vin+ = 0 and Vin- = X, the input voltage would be -X (negative X), thus its sign would be
inverted, hence the term inverting input. This sign inversion would hold true regardless of whether X itself was a positive or negative value (with respect to ground).
So now that we’ve dealt with the Vin terms, let’s talk about
A. A, as mentioned above, is the open-loop gain of the opamp — it is typically very large, on the order or 10,000 or more . This huge gain factor makes an open-loop opamp pretty useless for most applications . There are very few occasions where you would need to multiply a signal 10,000+ times, and even then it would be unwise to do so using an open-loop opamp. The reason is that even though A can be regarded as quite large, it’s very difficult to control exactly how large it will be. This is due to a number of factors, but mostly it’s due to manufacturing variances. The gain of the opamp is directly related to the gains of the individual transistors which make it up. Transistor gain can be difficult to control, particularly in large numbers. While it is possible to create very precise transistors, and thus pretty precise opamps, it’s not at all cheap. Further, it’s completely unnecessary, as we will soon see.
It’s much better to approach opamp design from a different angle. Rather than create a device which has a large but precise gain factor, it’s better to create one with a theoretically infinite gain and provide some other constraint which can be used to control it. This brings us to the second big concept for opamp functionality:
“An opamp tries to keep the voltage across its two inputs at zero.” “An opamp tries to keep the voltage across its two inputs at zero.”
In fact this is the fundamental operating principle of all opamps, but when I was first learning about active devices, I always found this statement to be very confusing. The problem is that very few sources go on to explain
why this is so. It seems a little counter-intuitive. Amplifiers aren’t supposed to make things smaller (or zero), they’re supposed to make them bigger, so why is the opamp trying to mitigate it’s own inputs, and how the heck does it do that?
I’ll answer the second question first. The
how is with a feedback network. The original statement would be more complete if it read: “An opamp tries to keep the voltage across its two inputs zero by using a feedback network to control one of the inputs with the output.“ by using a feedback network to control one of the inputs with the output.“
The construction of the feedback network will determine the extent to which the opamp has to swing the output one direction or another, so that one of the inputs is equal to the other input. The easiest way to explain this is by looking at a simple circuit — the non-inverting amplifier:
In the non-inverting amplifier, the input voltage is applied to the non-inverting (+) input of the amp. The output of the amp is connected, via a feedback resistor, to the inverting (-) input. There are two resistors here which make up the feedback network. The first is the aforementioned feedback resistor (R_feedback or R_f), and the second is a resistor from the inverting input to ground, called R_ground or simply R_g. Let’s ignore the non-inverting input for now and just look at the feedback network — it forms what is known as a voltage divider. Below is an example of a simple voltage divider circuit.
A voltage divider is so named because it provides a fraction of the input voltage as its output. The simplest voltage divider is made with two resistors, R1 and R2. In the figure above, the output voltage (Vout) is equal to the input voltage (Vin) times the ratio of the lower resistance, R2, to the total resistance, R1+R2. The lower point of R2 is tied to ground.
[latex]
\begin{equation}
V_o = V_{in} \cdot \frac{R_2}{R_1+R_2}
\end{equation}
[/latex]
The nice thing about the voltage divider is that it has lots of applications, one of which is in our non-inverting amplifier feedback network. In that case, the input to the voltage divider would be the output of the opamp, and the inverting input of the opamp is the output of the voltage divider. Seems very confusing, I know. Perhaps a diagram will help.
The formula for the voltage divider still applies, and we can substitute the terms from the non-inverting amplifier from above:
[latex]
\begin{equation}
V_{in-} = V_o \cdot \frac{R_g}{R_f+R_g}
\end{equation}
[/latex]
To make things a little clearer, let’s replace the resistor fraction with something less cluttered — the greek letter Beta:
[latex]
\begin{equation}
\beta = \frac{R_g}{R_f+R_g}
\end{equation}
[/latex]
–thus–
[latex]
\begin{equation}
V_{in-} = V_o \cdot \beta
\end{equation}
[/latex]
Ok, now we need to relate this equation to the whole circuit, so that we can come up with the
transfer function for the non-inverting amplifier. A transfer function is an equation that describes the output in terms of the input. It’s usually written as follows:
[latex]
\begin{equation}
\frac{output}{input} = transfer function
\end{equation}
[/latex]
In other words, it’s
a function that describes the ratio of the output to the input. This is an easy form to understand and use, because to find the value of the output, all we need to do is multiply the transfer function by the input value. Remember that the input to the whole circuit is at the non-inverting input, so we are going to need to include that in our function as well. NOTE: From this point on, it’s mostly just a lot of algebra — moving terms around between both sides of the equation until we get a nice Vout/Vin transfer function. If you’re thinking of studying electrical engineering, you should know that this stuff is critical. Being able to do algebra like this quickly is something you should make an effort to get really good at, because you will use it all the time. There aren’t really any shortcuts here — you just have to do it.
One of the problems when you’re studying this stuff is that textbook authors tend to ‘skip steps’ when reducing or reconfiguring equations, because they assume some intermediate steps are self-evident. Sometimes they are, particularly if you have a lot of experience working equations, but for n00bs (or sleepy college students), they are not. As such, I’m going to try to show each and every step, and explain what I’m doing, so you can understand it.
Getting back to the problem at hand, we need to create a transfer function for our opamp+feedback network. In order to do that, we will take equation (5) and substitute it into equation (1), replacing Vin- with the voltage divider equation. Doing that, we get:
[latex]
\begin{equation}
V_o = A \cdot \left(V_{in+} – V_o \cdot \beta\right)
\end{equation}
[/latex]
–or–
[latex]
\begin{equation}
V_o = A \cdot V_{in+} – A \cdot V_o \cdot \beta
\end{equation}
[/latex]
The next step is to get
both Vo terms over on the same side of the equation, so let’s do it:
[latex]
\begin{equation}
V_o + A \cdot V_o \cdot \beta = A \cdot V_{in+}
\end{equation}
[/latex]
Now we
extract the Vo term as a common coefficient of both terms on the left.
[latex]
\begin{equation}
V_o \cdot \left(1 + A \cdot \beta\right) = A \cdot V_{in+}
\end{equation}
[/latex]
Our original mission, of course, was to find the transfer function of Vo/Vin. Vin, in this case, is Vin+, so we want to find Vo/Vin+, so we
divide both sides by Vo.
[latex]
\begin{equation}
1 + A \cdot \beta = \frac {(A \cdot V_{in+})}{V_o}
\end{equation}
[/latex]
Now we’ll divide both sides by
A:
[latex]
\begin{equation}
\frac{\left(1 + A \cdot \beta\right)}{A} = \frac {V_{in+}}{V_o}
\end{equation}
[/latex]
which becomes:
[latex]
\begin{equation}
\frac{1}{A} + \frac{A \cdot \beta}{A} = \frac {V_{in+}}{V_o}
\end{equation}
[/latex]
Here we can use a little trick to simplify things. Recall that
A is very large — theoretically infinite. As such, 1/ A is very small, theoretically zero. We can use this to our advantage, because it means we can get rid of the 1/ A term by equating it to zero. Further, looking at the second term on the left, the A in the numerator and the denominator cancel out, so we can get rid of them too.
[latex]
\begin{equation}
0 + \frac{A \cdot \beta}{A} = \beta = \frac {V_{in+}}{V_o}
\end{equation}
[/latex]
Oh snap! Look how much simpler that is! Now we can substitute back the resistor fraction for Beta:
[latex]
\begin{equation}
\frac{R_g}{R_f+R_g} = \frac {V_{in+}}{V_o}
\end{equation}
[/latex]
and we invert both sides to get the Vout/Vin form we’re aiming for:
[latex]
\begin{equation}
\frac {V_o}{V_{in+}} = \frac{R_f+R_g}{R_g}
\end{equation}
[/latex]
Rg/Rg = 1, so:
[latex]
\begin{equation}
\frac {V_o}{V_{in+}} = \frac{R_f}{R_g} + 1
\end{equation}
[/latex]
To figure out Vo in terms of Vin, we simply multiply:
[latex]
\begin{equation}
{V_o} = V_{in+} \cdot (\frac{R_f}{R_g} + 1)
\end{equation}
[/latex]
And that’s the equation of a non-inverting amplifier. Now that we know where that formula comes from and how it’s derived, it’s a little easier to understand the concept of a feedback network. In order to create a voltage at Vin- which is equal to the voltage at Vin+, the opamp has to produce an output voltage which, when attenuated by the Rf/Rg divider, is equal to Vin+. So, if the voltage divider is set up to output 50% of its input, the opamp has to produce a voltage twice as large as the voltage at Vin+ in order to mitigate it. As such, it is indeed acting as an amplifier with a gain of 2. Likewise, other voltage divider values would produce different gains, as expressed in equation 18.
So now that you know how it works, you should build a couple circuits to try it out. Happy amplifying!
About the equations: the equations here are rendered as alpha-channel PNGs with the WP QuickLaTeX plugin. This is a great plugin which renders LaTeX beautifully and even provides equation numbers. If you want something that will allow you to write nice-looking equations on a blog, you should check it out.
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1. We construct a public-key bit-encryption scheme that is plausibly semantically secure, but is not circular secure. The circular security attack manages to fully recover the private-key. The construction is based on an extension of the Symmetric External Diffie-Hellman assumption (SXDH) from bilinear groups, to $\ell$-multilinear groups of order $p$ where $\ell \geq c \cdot \log p$ for some $c>1$.
While there do exist $\ell$-multilinear groups (unconditionally), for $\ell \geq 3$ there are no known candidates for which the SXDH problem is believed to be hard. Nevertheless, there is also no evidence that such groups do not exist. Our result shows that in order to prove the folklore conjecture, one must rule out the possibility that there exist $\ell$-multilinear groups for which SXDH is hard.
2. We show that the folklore conjecture cannot be proved using a black-box reduction. That is, there is no reduction of circular security of a bit-encryption scheme to semantic security of that very same scheme that uses both the encryption scheme and the adversary as black-boxes.
Both of our negative results extend also to the (seemingly) weaker conjecture that every CCA secure bit-encryption scheme is circular secure.
As a final contribution, we show an equivalence between three seemingly distinct notions of circular security for public-key bit-encryption schemes. In particular, we give a general search to decision reduction that shows that an adversary that distinguishes between encryptions of the bits of the private-key and encryptions of zeros can be used to actually recover the private-key.
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This question focuses on translating/adapting custom defined MATLAB code for creating a CorrelationMatrix to Mathematica.
Software and hardware used: Mathematica Version 10.3.0.0 Computer: Mac Pro (Late 2013) Processor: 3.5 GHz 6-Core Intel Xenon E5 Memory: 32 GB 1866 MHzDDR3 Graphics: AMD FirePro D500 3 GB Background
I have begun the translation of a body of MATLAB code for Pricing Bermudan Swaptions in the LIBOR Market Model, from a dissertation by a Steffen Hippler.
From the paper's introduction:
The fundamental entity modelled in the LIBOR market model is the forward interest rate curve that specifies the simply compounded interest rate at any two given points in time.
...
Bermudan swaptions are interest rate derivatives with early exercise features that are among the most liquidly traded (exotic) interest rate derivative contracts. Consequently, their pricing and risk management is of high practical importance. The pricing of these instruments, however, poses significant conceptual and theoretical difficulties. This is due to the fact that, in general, the pricing in the LIBOR market model has to be carried out via Monte Carlo simulation techniques, which in turn do not lend themselves naturally to the pricing of options with early exercise features.
I hope to eventually post a demonstration of the implementation on the Wolfram Demonstrations Project site.
Note: If
FinancialDerivate[] or
CUDAFinancialDerivative[] enabled such pricing, it would greatly simply my objective.
It does not appear that either function supports
Bermuda style options (options that one can only exercise at defined times) or swaptions, options on swaps of fixed for floating cashflows.
A possible solution to this larger objective (which would likely require specialized knowledge outside the general domain of this forum), could show a way to utilize existing
FinancialDerivate[] or
CUDAFinancialDerivative[] functionality to replicate pricing of Bermuda Swaptions.
As I translate and develop the larger solution, I want to take advantage of Mathematica's functional approach as well as its high-level functionality.
Background specific to the immediate question
Hippler includes the following two custom MatLab functions for building a correlation matrix:
function [corr] = CorrFunc(beta, horizon, i, j) M=horizon; corr=exp(-abs(i-j)/(M-1)*(-log(beta(3))+beta(1)*(i^2+j^2+i*j-3*M*i-3*M*j+3*i+3*j+2*M^2-M-4)*1/((M-2)*(M-3))-beta(2)*(i^2+j^2+i*j-M*i-M*j+3*i+3*j+3*M^2-2)*1/((M-2)*(M-3))));endfunction CORR = CorrelationMatrix(beta, maturities, horizon, startRate, endRate) CORR=zeros(endRate-startRate); for i = startRate:endRate for j = startRate:endRate CORR(i,j) = CorrFunc(beta, horizon, maturities(j), maturities(i)); end endend
Some definitions/term-examples:
beta (β or beta coefficient) - of an investment provides a measure of an instrument's volatility relative to a specific market as a whole. An individual stock or bond may have more or less volatility than the the larger stock or bond markets. horizon :
If you buy a zero-coupon bond and hold it to maturity, the ROR on your investment is the zero rate at which you bought the bond.
If you buy a t-year zero-coupon bond and sell it at time T
maturities - "maturity dates" on which the principal amount of the debt instruments become due and interest payments stop. startRate, endRate - establishes a range of iteration.
The
CorrFunc implements (in MATLAB) a parameterisation due to Schoenmakers and Coffey (J. Schoenmakers and C. Coffey,
Systematic generation of parametric correlation structures for the libor market model, International Journal of Theoretical and Applied Finance (2002)) of the form:
$$\small \begin{align*}\rho_{ij}=\exp\left[-\frac{|i-j|}{M-1}\right.&\left(-\ln\beta_3+\beta_1\frac{i^2+j^2+ij-3Mi-3Mj+3i+3j+2M^2-M-4}{(M-2)(M-3)}\right.\\ &\left.\left.-\beta_2\frac{i^2+j^2+ij-Mi-Mj-3i-3j+3M^2-2}{(M-2)(M-3)}\right)\right]\end{align*}$$
with fitting parameter
β = (β1, β2, β3) and
M denoting the total number of rates under consideration.
I think, the MATLAB
CorrFunc translates pretty directly to Mathematica:
corrFunc[beta_, horizon_, i_, j_] := Module[{M, corr}, M = horizon; corr = Exp[-Abs[i - j]/(M - 1)*(-Log[beta[[3]]] + beta[[1]]*(i^2 + j^2 + i*j - 3*M*i - 3*M*j + 3*i + 3*j + 2*M^2 - M - 4)*1/((M - 2)*(M - 3)) - beta[[2]]*(i^2 + j^2 + i*j - M*i - M*j + 3*i + 3*j + 3*M^2 - 2)*1/((M - 2)*(M - 3)))]]
The MATLAB,
CorrelationMatrix just looks like it simply applies the
CorrFunc to a square matrix of the maturities.
So, it looks like I can simply do something like the following:
beta = {0.5, 0.4, 0.3};horizon = 100;Table[corrFunc[beta, horizon, i, j], {i, 10}, {j, 10}]
Note: I don't have MATLAB so I can't do a direct comparison of output.
Questions:
Does it look like I'm on the right track? Whether, yes or no, can I alternatively do this by other/better means?
LinearModelFit?
GeneralizedLinearModelFit?
Using
Listablefeatures?
As stated in the paper: The entire project of Pricing Bermudan Swaptions,
...poses significant conceptual and theoretical difficulties.
I want/hope to make use of all of Mathematica's advantages in simplifying code and writing efficient performing code.
It occurs to me that writing up the context of a question like this helps me, as an OP, better think about the problem. Not that I have a solution, but I hope it gets me into the area of the solution.
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Difference between revisions of "User:Nikita2"
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m
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I am Nikita Evseev
+
I am Nikita Evseev Novosibirsk, Russia.
My research interests are in [[Mathematical_analysis | Analysis]] and [[Sobolev space|Sobolev spaces]].
My research interests are in [[Mathematical_analysis | Analysis]] and [[Sobolev space|Sobolev spaces]].
Line 54: Line 54:
I'm keen on turning up articles of EoM into better appearance by rewriting formulas and math symbols in TeX.
I'm keen on turning up articles of EoM into better appearance by rewriting formulas and math symbols in TeX.
−
Now there are '''
+
Now there are '''''' (out of 15,890) articles with [[:Category:TeX done]] tag.
<asy>
<asy>
size(0,150);
size(0,150);
−
real tex=
+
real tex=;
real all=15890;
real all=15890;
pair z0=0;
pair z0=0;
Latest revision as of 21:58, 17 June 2018 Pages of which I am contributing and watching
Analytic function | Cauchy criterion | Cauchy integral | Condition number | Continuous function | D'Alembert criterion (convergence of series) | Dedekind criterion (convergence of series) | Derivative | Dini theorem | Dirichlet-function | Ermakov convergence criterion | Extension of an operator | Fourier transform | Friedrichs inequality | Fubini theorem | Function | Functional | Generalized derivative | Generalized function | Geometric progression | Hahn-Banach theorem | Harmonic series | Hilbert transform | Hölder inequality | Lebesgue integral | Lebesgue measure | Leibniz criterion | Leibniz series | Lipschitz Function | Lipschitz condition | Luzin-N-property | Newton-Leibniz formula | Newton potential | Operator | Poincaré inequality | Pseudo-metric | Raabe criterion | Riemann integral | Series | Sobolev space | Vitali theorem |
TeXing
I'm keen on turning up articles of EoM into better appearance by rewriting formulas and math symbols in TeX.
Now there are
3040 (out of 15,890) articles with Category:TeX done tag.
$\quad \rightarrow \quad$ $\sum_{n=1}^{\infty}n!z^n$ Just type $\sum_{n=1}^{\infty}n!z^n$. Today You may look at Category:TeX wanted. How to Cite This Entry:
Nikita2.
Encyclopedia of Mathematics.URL: http://www.encyclopediaofmath.org/index.php?title=Nikita2&oldid=34319
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I have a glmer model that I am performing contrasts on. My model has four factors (emotion, gender, filter, ageGroup). No interactions. I have been using
lsmeans(x, spec, contr='pairwise')) (and)
lsm within
glht) to get lsmeans and to do pairwise contrasts within-factors. I am also interested in comparing the means from one factor against a level from a different factor.
> summary(m7.only_within)Generalized linear mixed model fit by maximum likelihood (Laplace Approximation) ['glmerMod'] Family: binomial ( logit ) Formula: FaceStimulus.RESP.binary ~ emotion + emotion.Sample + gender + gender.Sample + ageGroup + ageGroup.Sample + filter + filter.Sample + (1 | Subject)Fixed effects: Estimate Std. Error z value Pr(>|z|) (Intercept) -0.823451 0.264442 -3.114 0.00185 ** emotiondisgust 1.504896 0.092532 16.264 < 2e-16 ***emotionfear 1.555492 0.091832 16.938 < 2e-16 ***emotion.Sample -0.106948 0.044364 -2.411 0.01592 * gendermale 0.012718 0.074198 0.171 0.86391 gender.Sample 0.007021 0.010046 0.699 0.48462 ageGroupold 0.364239 0.076379 4.769 1.85e-06 ***ageGroup.Sample -0.013519 0.009577 -1.412 0.15808 filtercropAdaptiveThreshold 0.419095 0.075906 5.521 3.37e-08 ***filter.Sample 0.070147 0.029658 2.365 0.01802 * ---
I understand that lsmeans calculates standard errors using variable variances and covariances from vcov() to do pairwise comparisons. Can I just compare one model variable against another? Like so:
summary(glht(m3.only_within, linfct = c("filtercropAdaptiveThreshold - ageGroupold = 0")))Linear Hypotheses: Estimate Std. Error z value Pr(>|z|) filtercropAdaptiveThreshold - ageGroupold == 0 0.05486 0.10798 0.508 1
That contrast is the test of the difference between the betas in the model, not between the lsmeans. Naturally, all my other contrasts compare the lsmeans since I'm using
lsm(). Can I just manually take the difference between the lsmeans, and then use the variances and covariance SE's from the model vcov for the two model effects to get the SE for the
lsmeans difference? Naturally, those SEs are much lower than the SEs reported from the
lsmeans output.
$lsmeans ageGroup lsmean SE df asymp.LCL asymp.UCL young 0.01043529 0.2466025 NA -0.4728967 0.4937673 old 0.37467465 0.2474967 NA -0.1104101 0.8597594$lsmeans filter lsmean SE df asymp.LCL asymp.UCL cropDesat37 -0.01699274 0.2462214 NA -0.49957782 0.4655923 cropAdaptiveThreshold 0.40210268 0.2478032 NA -0.08358268 0.8877880
So it would be: $$ \displaystyle \begin{array}{l} \bar{X_1}-\bar{X_2}= lsmean\_of\_cropAdaptiveThreshold - lsmean\_of\_old \\ = 0.40210268 - 0.37467465 = 0.0274\\ \\ SE_{\bar{X_1}-\bar{X_2}}= \sqrt{SE_1^2 + SE_2^2 + 2*cov}\\ = \sqrt{0.075906^2 + 0.076379^2 + 2*-3.225794e-05}\\ = 0.10798 \end{array} $$
...which is the same SE as the SE in the beta-contrast.
Is it legal to compare cross-factor level lsmeans like this?
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№ 9
All Issues Volume 63, № 2, 2011
Ukr. Mat. Zh. - 2011. - 63, № 2. - pp. 147-155
The best quadrature formula is found for the class of convex-valued functions defined on the interval [0, 1] and monotone with respect to an inclusion.
Ukr. Mat. Zh. - 2011. - 63, № 2. - pp. 156-164
The solvability of the inhomogeneous Dirichlet problem in a bounded domain for scalar improperly elliptic differential equation with complex coefficients is investigated. We study a model case where the unit disk is chosen as a domain and the equation does not contain lowest terms. We prove that the problem has a unique solution in the Sobolev space for special classes of Dirichlet data that are spaces of functions with exponential decrease of the Fourier coefficients.
Ukr. Mat. Zh. - 2011. - 63, № 2. - pp. 165-175
We characterize the least semilattice congruence on the free dimonoid and prove that the free dimonoid is a semilattice of
s-simple subdimonoids each being a rectangular band of subdimonoids. Estimates for the approximate characteristics of the classes $B_{p, \theta}^{\Omega}$ of periodic functions of two variables with given majorant of mixed moduli of continuity
Ukr. Mat. Zh. - 2011. - 63, № 2. - pp. 176-186
Order estimates are obtained for approximation $B_{p, \theta}^{\Omega}$ of the classes of periodic functions of two variables in the space $L_q$ by operators of orthogonal projection as well as by linear operators subjected to some conditions.
Ukr. Mat. Zh. - 2011. - 63, № 2. - pp. 187-199
In the framework of dynamical picture of interacting physical systems, the notion of a spectral measure with quasipoint spectrum is introduced. It is shown that, under conflict interaction with point measures, only quasipoint singularly continuous measures are admitted for the transformation into measures with purely point spectrum.
Group classification of quasilinear elliptic-type equations. II. Invariance under solvable Lie algebras
Ukr. Mat. Zh. - 2011. - 63, № 2. - pp. 200-215
The problem of the group classification of quasilinear elliptic-type equations in a two-dimensional space is considered. The list of all equations of this type, which admit the solvable Lie algebras of symmetry operators, is obtained. The results of this paper along with results obtained by the authors earlier give a complete solution of the problem of the group classification of quasilinear elliptic-type equations.
Ukr. Mat. Zh. - 2011. - 63, № 2. - pp. 216-225
Let $G$ be a group with identity $e$ and let $\mathcal{I}$ be a left-invariant ideal in the Boolean algebra $\mathcal{P}_G$ of all subsets of $G$. A subset $A$ of $G$ is called $\mathcal{I}$-thin if $gA \bigcap A \in \mathcal{I}$ for every $g \in G \ \{e\}$. A subset $A$ of $G$ is called $\mathcal{I}$-sparse if, for every infinite subset $S$ of $G$, there exists a finite subset $F \subset S$ such that $\bigcap_{g \in F}gA \in F$. An ideal $\mathcal{I}$ is said to be thin-complete (sparse-complete) if every $\mathcal{I}$-thin ($\mathcal{I}$-sparse) subset of $G$ belongs to $\mathcal{I}$. We define and describe the thin-completion and the sparse-completion of an ideal in $\mathcal{P}_G$.
Ukr. Mat. Zh. - 2011. - 63, № 2. - pp. 226-236
We establish a criterion of the local linear convexity of sets in the two-dimensional quaternion space $H^2$, that are similar to the bounded Hartogs domains with smooth boundaries in the two-dimensional complex space $C^2$.
Ukr. Mat. Zh. - 2011. - 63, № 2. - pp. 237-269
We consider problems of the linear theory of systems of ordinary differential equations related to the investigation of invariant hyperplanes of these systems, the notion of equivalence for these systems, and the Floquet – Lyapunov theory for periodic systems of linear equations. In particular, we introduce the notion of equivalence of systems of linear differential equations of different orders, propose a new formula of the Floquet form for periodic systems, and present the application of this formula to the introduction of amplitude-phase coordinates in a neighborhood of a periodic trajectory of a dynamical system.
Ukr. Mat. Zh. - 2011. - 63, № 2. - pp. 270-280
The Maxwell - Lorenz system of an electromagnetic field interacting with charged particles (point charges) is considered in the Darwin approximation which is characterized by the Lagrangian and Hamiltonian of the particles both uncoupled with the field. The solution of the equation of motion of the particles with the approximated Darwin Hamiltonian is found on a finite time interval with the use of the Cauchy theorem. Components of this solution are represented as holomorphic functions of time.
Functions of ultraexponential and infralogarithm types and general solution of the Abel functional equation
Ukr. Mat. Zh. - 2011. - 63, № 2. - pp. 281-288
We propose generalized forms of ultraexponential and infralogarithm functions introduced and studied by the author earlier and present two classes of special functions, namely, ultraexponential and infralogarithm $f$-type functions. As a result of present investigation, we obtain general solution of the Abel equation $\alpha (f(x)) = \alpha (x) + 1$ under some conditions on a real function $f$ and prove a new completely different uniqueness theorem for the Abel equation stating that the infralogarithm $f$-type function is its unique solution. We also show that the infralogarithm $f$-type function is an essentially unique solution of the Abel equation. Similar theorems are proved for the ultraexponential $f$-type functions and their functional equation $\beta(x) = f(\beta(x − 1))$ which can be considered as dual to the Abel equation. We also solve certain problem being unsolved before, study some properties of two considered functional equations and some relations between them.
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[This is the 6th post in the current series about Wythoff’s game: see posts #1, #2, #3, #4, and #5.
Caveat lector: this post is a bit more difficult than usual. Let me know what you think in the comments!]
Our only remaining task from last week was to prove the mysterious Covering Theorem: we must show that there is exactly one dot in each row and column of the grid (we already covered the diagonal case). Since the rows and columns are symmetric, let’s focus on columns.
The columns really only care about the
x-coordinates of the points, so let’s draw just these x-coordinates on the number-line. We’ve drawn \(\phi,2\phi,3\phi,\ldots\) with small dots and \(\phi^2,2\phi^2,3\phi^2,\ldots\) with large dots. We need to show that there’s exactly one dot between 1 and 2, precisely one dot between 2 and 3, just one between 3 and 4, and so on down the line. For terminology’s sake, break the number line into length-1 intervals [1,2], [2,3], [3,4], etc., so we must show that each interval has one and only one dot:
Why is this true? One explanation hinges on a nice geometric observation: Take any small dot
s and large dot t on our number-line above, and cut segment st into two parts in the ratio \(1:\phi\) (with s on the shorter side). Then the point where we cut is always an integer! For example, the upper-left segment in the diagram below has endpoints at \(s=2\cdot\phi\) and \(t=1\cdot\phi^2\), and its cutting point is the integer 3:
In general, if
s is the jth small dot—i.e., \(s=j\cdot\phi\)—and \(t=k\cdot\phi^2\) is the kth large dot, then the cutting point between s and t is \(\frac{1}{\phi}\cdot s+\frac{1}{\phi^2}\cdot t = j+k\) (Why?! [1]). But more importantly, this observation shows that no interval has two or more dots: a small dot and a large dot can’t be in the same interval because they always have an integer between them! [2]
So all we have to do now is prove that no interval is
empty: for each integer n, some dot lies in the interval [ n, n+1]. We will prove this by contradiction. What happens if no dot hits this interval? Then the sequence \(\phi,2\phi,3\phi,\ldots\) jumps over the interval, i.e., for some j, the jth dot in the sequence is less than n but the ( j+1)st is greater than n+1. Likewise, the sequence \(\phi^2,2\phi^2,3\phi^2,\ldots\) jumps over the interval: its kth dot is less than n while its ( k+1)st dot is greater than n+1:
By our observation above on segment \(s=j\phi\) and \(t=k\phi^2\), we find that the integer
j+ k is less than n, so \(j+k\le n-1\). Similarly, \(j+k+2 > n+1\), so \(j+k+2 \ge n+2\). But together these inequalities say that \(n\le j+k\le n-1\), which is clearly absurd! This is the contradiction we were hoping for, so the interval [ n, n+1] is in fact not empty. This completes our proof of the Covering Theorem and the Wythoff formula!
It was a long journey, but we’ve finally seen exactly why the Wythoff losing positions are arranged as they are. Thank you for following me through this!
A Few Words on the Column Covering Theorem
Using the
floor function \(\lfloor x\rfloor\) that rounds x down to the nearest integer, we can restate the Column Covering Theorem in perhaps a more natural context. The sequence of integers $$\lfloor\phi\rfloor = 1, \lfloor 2\phi\rfloor = 3, \lfloor 3\phi\rfloor = 4, \lfloor 4\phi\rfloor = 6, \ldots$$ is called the Beatty sequence for the number \(\phi\), and similarly, $$\lfloor\phi^2\rfloor = 2, \lfloor 2\phi^2\rfloor = 5, \lfloor 3\phi^2\rfloor = 7, \lfloor 4\phi^2\rfloor = 8,\ldots$$ is the Beatty sequence for \(\phi^2\). Today we proved that these two sequence are complementary, i.e., together they contain each positive integer exactly once. We seemed to use very specific properties of the numbers \(\phi\) and \(\phi^2\), but in fact, a much more general theorem is true: Beatty’s Theorem: If \(\alpha\) and \(\beta\) are any positive irrational numbers with \(\frac{1}{\alpha}+\frac{1}{\beta}=1\), then their Beatty sequences \(\lfloor\alpha\rfloor, \lfloor 2\alpha\rfloor, \lfloor 3\alpha\rfloor,\ldots\) and \(\lfloor\beta\rfloor, \lfloor 2\beta\rfloor, \lfloor 3\beta\rfloor,\ldots\) are complementary sequences.
Furthermore, our same argument—using \(\alpha\) and \(\beta\) instead of \(\phi\) and \(\phi^2\)—can be used to prove the more general Beatty’s Theorem!
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So what are spin-networks? Briefly, they are graphs with representations ("spins") of some gauge group (generally SU(2) or SL(2,C) in LQG) living on each edge. At each non-trivial vertex, one has three or more edges meeting up. What is the simplest purpose of the intertwiner? It is to ensure that angular momentum is conserved at each vertex. For the case of four-valent edge we have four spins: $(j_1,j_2,j_3,j_4)$. There is a simple visual picture of the intertwiner in this case.
Picture a tetrahedron enclosing the given vertex, such that each edge pierces precisely one face of the tetrahedron. Now, the natural prescription for what happens when a surface is punctured by a spin is to associate the Casimir of that spin $ \mathbf{J}^2 $ with the puncture. The Casimir for spin $j$ has eigenvalues $ j (j+1) $. You can also see these as energy eigenvalues for the quantum rotor model. These eigenvalues are identified with the area associated with a puncture.
In order for the said edges and vertices to correspond to a consistent geometry it is important that certain constraints be satisfied. For instance, for a triangle we require that the edge lengths satisfy the triangle inequality $ a + b \lt c $ and the angles should add up to $ \angle a + \angle b + \angle c = \kappa \pi$, with $\kappa = 1$ if the triangle is embedded in a flat space and $\kappa \ne 1$ denoting the deviation of the space from zero curvature (positively or negatively curved).
In a similar manner, for a classical tetrahedron, now it is the sums of the areas of the faces which should satisfy "closure" constraints. For a quantum tetrahedron these constraints translate into relations between the operators $j_i$ which endow the faces with area.
Now for a triangle giving its three edge lengths $(a,b,c)$ completely fixes the angles and there is no more freedom. However, specifying all four areas of a tetrahedron
does not fix all the freedom. The tetrahedron can still be bent and distorted in ways that preserve the closure constraints (not so for a triangle!). These are the physical degrees of freedom that an intertwiner possesses - the various shapes that are consistent with a tetrahedron with face areas given by the spins, or more generally a polyhedron for n-valent edges.
Some of the key players in this arena include, among others, Laurent Friedel, Eugenio Bianchi, E. Magliaro, C. Perini, F. Conrady, J. Engle, Rovelli, R. Pereira, K. Krasnov and Etera Livine.
I hope this provides some intuition for these structures. Also, I should add, that at present I am working on a review article on LQG for and by "the bewildered". This post imported from StackExchange Physics at 2014-04-01 16:52 (UCT), posted by SE-user user346
I reserve the right to use any or all of the contents of my answers to this and other questions on physics.se in said work, with proper acknowledgements to all who contribute with questions and comments. This legalese is necessary so nobody comes after me with a bullsh*t plagiarism charge when my article does appear :P
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I know that if I have two differentiable functions $f, g$ then the functions $(f + g)$ and $fg$ are also differentiable. I would like to find a way how to argue about the function $h$ where \begin{equation} f(x) = (hg)(x) := h(x)g(x) \quad \text{and } f,g \text{ are differentiable} \end{equation}
For a start I can conclude $h$ is differentiable at all points where $g(x) \neq 0$ since there I can express $h$ as \begin{equation} h = \frac{f}{g} \end{equation}
But for the remaining points I am not sure, my guess is that $h$ is differentiable, any hints how I can make this into a formal argument ? Or am I probably wrong ? In that case, would it help to impose further smoothness on $f$ and $g$, say both are $C^\infty$ ?
Many thanks!
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Consider some (4, say) distinguishable particles. If we wish to distribute them into two exactly similar compartments in an open box, then the priori probability for a particle of going into any one of the compartments will exactly 1/2 as both compartments are identical. If the four particles are named as a , b, c and d and the compartments as compartment 1 and compartment (2), then following table can be made listing all the possible arrangements.
Compartment (1) ↓ Compartment (2) ↓
Distribution Number of Particles Name(Type) of particles Number of Particles Name(Type) of Particles Total Number of Arrangements Distribution (1) 0 – 4 {a,b,c,d} 1 Distribution (2) 1 {a}
{b}
{c}
{d}
3 {b,c,d}
{c,d,a}
{d,a,b}
{a,b,c}
4 Distribution (3) 2 {a,b}
{b,c}
{c,d}
{a,c}
{b,d}
{a,d}
2 {c,d}
{a,d}
{a,b}
{b,d}
{a,c}
{b,c}
6 Distribution (4) 3 {b,c,d}
{c,d,a}
{d,a,b}
{a,b,c}
1 {a}
{b}
{c}
{d}
4 Distribution (5) 4 {a,b,c,d} 0 – 1
Now in the second distribution, only one particle is chosen to be inside compartment (1) and all others in Compartment (2). There are exactly $\binom{4}{1}$ = 4 ways to do so. All these four arrangements are shown in table. Similarly, in the distribution (3) exactly two particles enter in each of the compartment (1) and (2) , i.e., if $a$ and $b$ are placed in the compartment (1), $c$ and $d$ must take place in Compartment (2). There are exactly $\binom{4}{2}$=6 ways to do so. Following the same procedure to complete table by increasing the number of particles one by one in Compartment (1) and decreasing them one by one in Compartment (2).
We have following 5 different distributions for the system of four particles:
Distribution (1) = $d_1$ = (0,4)
Distribution (2) = $d_2$ = (1,3) Distribution (3) = $d_3$ = (2,2) Distribution (4) = $d_4$ = (3,1) Distribution (5) = $d_1$ = (4,0)
These compartment-wise distributions of a system of particles are known as macrostates of the system. There are five macrostates observed corresponding to a system of four particles . In general, for a system of n particles, there are exactly $n+1$ macrostates (for the system of n particles to be distributed in two identical compartments). These macrostates are actually the following distribution of particles:
$d_1= (0,n)$
$d_2= (1,n-1)$ $d_3= (2,n-2)$ $\ldots$ $d_k= (k-1,n-k+1)$ $\ldots$ $d_{n+1} = (n,0)$.
Now, the different possible arrangements for each macrostate are in their own ‘a distribution’, called microstates. For example {{a}, {b,c,d}} is a microstate of macrostate $d_2$ (distribution 2) in the table of system of four particles. From the table, it can be observed that a macrostate may have many corresponding microstates. The distribution $d_1=(0,4)$ has one microstate {{ }{a,b,c,d}}, while the distribution $d_2= (1,3)$ has four, $d_3$ has six , $d_4$ has four and $d_5$ has one.
The total number of microstates for the system of four particles is, therefore, $1+4+6+4+1=16=2^4$. In general, total number of microstates for the system of n-particles in ‘two compartment’ composition is $2^n$. And also, the number of microstates corresponding to a single macrostate with $(r, n-r)$ distribution among $n$ particles is $\binom{n}{r}$.
Thermodynamic Probability
In statistical physics, the number of microstates plays very important role at quantum level. The number of microstates corresponding to any macrostate is called the thermodynamic probability of the macrostate, represented by $W$ or $\Omega$.
Since, for the macrostate $(r,n-r)$ the number of microstates is $$\binom{n}{r} = \frac{n!}{r! \cdot (n-r)!}$$ . Thus, the thermodynamic probability of this macrostate:
$$W_{(r,n-r)} = \binom{n}{r} = \frac{n!}{r! \cdot (n-r)!}$$.
In mathematics, we know that $$\binom{n}{r} = \binom{n}{n-r} =\frac{n!}{r! \cdot (n-r)!}$$. Therefore, $$ W_{(r,n-r)} =W_{(n-r,r)}$$.
Or, $$ W_{(r,n-r)} =\binom{n}{r} = \frac{n!}{r! \cdot (n-r)!}= W_{(n-r,r)}$$ .
That is, for a system of four distinguishable particles, $n=4$, $r=0,1,2,3,4$, provided
$$W_{(0,4)} =\binom{4}{0} = \frac{4!}{0! \cdot (4)!}= W_{(4,0)} =1$$
$$W_{(1,3)} =\binom{4}{1} = \frac{4!}{1! \cdot (3)!}= W_{(3,1)}=4$$
$$W_{(2,2)} =\binom{4}{2} = \frac{4!}{2! \cdot (2)!}= W_{(2,2)}=6$$
All agree with the table.
Thermodynamic probability is not identical to ‘probability’ in mathematics and statistics, but it gives rise to that.
Probability of a Macrostate
The probability of macrostate (i.e., availability of a macrostate) in a system is given by
$$P_m = \frac{\textrm{No. of such microstates in the macrostate}}{\textrm{Total no. of microstates in the system}}$$
Thus, $$P_m = \frac{W}{2^n}$$
Or, $$P_{(r,n-r)} =\frac{W_{(r,n-r)}}{2^n} = \binom{n}{r} \times \frac{1}{2^n} = P_{(n-r,r)} $$
Practically, the probability of finding the macrostate (0,4) in a system of four particles is $$P_{(0,4)} =\frac{W_{(0,4)}}{2^4} = \binom{4}{0} \times \frac{1}{2^4} = \frac{1}{16}$$
Similarly, $P_{(1,3)}= \frac{1}{4}$.
Macrostate with highest probability
The macrostate (r, n-r) has the highest probability only if $$r=\frac{n}{2} \, \textrm{ when n is even}$$ and $$r= \frac{n+1}{2} \, \textrm{or} \frac{n-1}{2} \, \textrm{when n is odd}$$. Then the highest probability is given by:
$$P_{max} = \frac{n!}{\frac{n}{2}! \cdot \frac{n}{2}!} \times \frac{1}{2^n}$$ (n is even)
$$P_{max} = \frac{n!}{ \frac{n-1}{2}! \frac{n+1}{2}!} \times \frac{1}{2^n}$$ (n is odd)
Macrostate with least probability
The macrostate (r, n-r) is least probable when $r$ is either zero or n.
$$P_{min} = \frac{n!}{0! \cdot n!} \times \frac{1}{2^n}$$.
Thus, least probability for a macrostate is $\frac{1}{2^n}$. $\Box$
Remark: The probability of a macrostate (P) is directly proportional to the thermodynamic probability (W).
Feel free to ask questions, send feedback and even point out mistakes. Great conversations start with just a single word.
How to write better comments?
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Consider the Peano axioms. There exists a model for them (namely, the natural numbers with a ordering relation $<$, binary function $+$, and constant term $0$). Therefore, by the model existence theorem, shouldn't this suffice to prove the consistency of first order arithmetic? Why is Gentzen's proof necessary?
The axioms of first-order arithmetic include the induction schema, which says that, for every formula $A(x)$ with free variable $x$, the conjunction of $A(0)$ and $\forall x\,(A(x)\to A(x+1))$ implies $\forall x\,A(x)$. This is, of course, a special case of the well-known and basic induction property of the natural numbers that says the same thing for any property $A(x)$ whatsoever, whether or not it's defined by a first-order formula. For anyone who (1) understands the natural numbers well enough to grasp the general induction principle and (2) believes that (first-order) quantifiers over the natural numbers are meaningful so that first-order formulas $A(x)$ really define properties, it is clear that the natural number system satisfies all of the first-order Peano axioms, and therefore those axioms are consistent.
A difficulty arises if one adopts a very strong constructivist or finitist viewpoint, doubting item (2) above, i.e., questioning the meaning of first-order quantifiers $\forall z$ and $\exists z$ when $z$ ranges over an infinite set (like $\mathbb N$) so that one can't actually check each individual $z$. From such a viewpoint, the formulas $A(x)$ occurring in the induction schema are gibberish (or close to gibberish, or at least not clear enough to be used in mathematical reasoning), and then the proposed consistency proof collapses.
The chief virtue of Gentzen's consistency proof is that it essentially avoids any explicit quantification over infinite sets. It can be formulated in terms of very basic, explicit, computational constructions (technically, in the language of primitive recursive arithmetic). There is, however, a cost for this virtue, namely that one needs an induction principle not just for the usual well-ordering of the natural numbers but for the considerably longer well-ordering $\varepsilon_0$.
Thus, Gentzen uses a much longer well-ordering, but his induction principle is only about primitive recursive properties, not about arbitrary first-order definable properties. There is a trade-off: Length of well-ordering versus quantification.
I believe the trade-off can be made rather precise, but I don't remember the details. Recall that $\varepsilon_0$ is the limit of the sequence of iterated exponentials $\omega(0)=\omega$ and $\omega(n+1)=\omega^{\omega(n)}$. If we weaken PA by limiting the induction principle to formulas $A(x)$ that can be defined with a fixed number $n$ of quantifiers, then the consistency of this weakened theory can be proved using primitive recursive induction up to $\omega(n)$. [I don't guarantee that the last $n$ here is correct; you might need something like $2n$; I hope an expert will come along and edit this answer to fix any error I've made.] In other words, the trade-off is that an additional quantifier in the induction formulas costs an additional exponential [or two?] in the ordinal.
In addition to the reasons Andreas gives, Gentzen's theorem gives additional information that's interesting even if you don't have any qualms about consistency.
In particular, ordinal analysis gives a fairly precise characterization of the provably total computable functions of a theory (and, along with it, a lot of information about the structure of proofs in PA).
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№ 9
All Issues Volume 63, № 9, 2011 Estimates for the norms of fractional derivatives in terms of integral moduli of continuity and their applications
Ukr. Mat. Zh. - 2011. - 63, № 9. - pp. 1155-1168
For functions defined on the real line or a half-line, we obtain Kolmogorov-type inequalities that estimate the $L_p$-norms $(1 \leq p < \infty)$ of fractional derivatives in terms of the Lp-norms of functions (or the $L_p$-norms of their truncated derivatives) and their $L_p$-moduli of continuity and establish their sharpness for $p = 1$. Applications of the obtained inequalities are given.
Laplacian with respect to a measure on a Hilbert space and an L 2-version of the Dirichlet problem for the Poisson equation
Ukr. Mat. Zh. - 2011. - 63, № 9. - pp. 1169-1178
We propose a version of the Laplace operator for functions on a Hilbert space with measure. In terms of this operator, we investigate the Dirichlet problem for the Poisson equation.
Ukr. Mat. Zh. - 2011. - 63, № 9. - pp. 1179-1189
The manifold of symmetric real matrices with fixed multiplicities of eigenvalues was considered for the first time by V. Arnold. In the case of compact real self-adjoint operators, analogous results were obtained by Japanese mathematicians D. Fujiwara, M. Tanikawa, and S. Yukita. They introduced a special local diffeomorphism that maps Arnold's submanifold to a flat subspace. The properties of the indicated diffeomorphism were further studied by Ya. Dymarskii. In the present paper, we describe the smooth structure of submanifolds of finite-dimensional and compact operators of the general form in which a selected eigenvalue is associated with a single Jordan block.
Ukr. Mat. Zh. - 2011. - 63, № 9. - pp. 1190-1205
We propose a regularization of the formal differential expression $$l(y) = i^m y^{(m)}(t) + q(t)y(t),\; t \in (a, b),$$ of order $m \geq 3$ by using quasiderivatives. It is assumed that the distribution coefficient $q$ has an antiderivative $Q \in L ([a, b]; \mathbb{C})$. In the symmetric case $(Q = \overline{Q})$, we describe self-adjoint and maximal dissipative/accumulative extensions of the minimal operator and its generalized resolvents. In the general (nonselfadjoint) case, we establish conditions for the convergence of the resolvents of the considered operators in norm. The case where $m = 2$ and $Q \in L_2 ([a, b]; \mathbb{C})$ was studied earlier.
On modules over integer-valued group rings of locally soluble groups with rank restrictions imposed on subgroups
Ukr. Mat. Zh. - 2011. - 63, № 9. - pp. 1206-1217
We study the $ZG$-module $A$ such that $Z$ is the ring of integers, the group $G$ has infinite section $ p$-rank (or infinite 0-rank), $C_G(A) = 1$, $A$ is not a minimax $Z$-module, and, for every proper subgroup $H$ of infinite section $p$-rank (or infinite 0-rank, respectively), the quotient module $A/C_A(H)$ is a minimax $Z$-module. It is proved that if the group $G$ under consideration is locally solvable, then $G$ is a solvable group. Some properties of a solvable group of this type are obtained.
Structure of a finite commutative inverse semigroup and a finite bundle for which the inverse monoid of local automorphisms is permutable
Ukr. Mat. Zh. - 2011. - 63, № 9. - pp. 1218-1226
For a semigroup $S$, the set of all isomorphisms between subsemigroups of $S$ is an inverse monoid with respect to composition, which is denoted by $P A(S)$ and is called the monoid of local automorphisms of $S$. A semigroup $S$ is called permutable if, for any pair of congruences $p, \sigma$ on $S$, one has $p \circ \sigma = \sigma \circ p$. We describe the structure of a finite commutative inverse semigroup and a finite band whose monoids of local automorphisms are permutable.
On the theory of convergence and compactness for Beltrami equations with constraints of set-theoretic type
Ukr. Mat. Zh. - 2011. - 63, № 9. - pp. 1227-1240
We prove theorems on convergence and compactness for classes of regular solutions of degenerate Beltrami equations with set-theoretic constraints imposed on the complex coefficient and construct variations for these classes.
Ukr. Mat. Zh. - 2011. - 63, № 9. - pp. 1241-1256
For a solution of a reflection problem on a half-line similar to the Skorokhod reflection problem but with possible jump-like exit from zero, we obtain an explicit formula and study its properties. We also construct a Wiener process on a half-line with Wentzell boundary condition as a strong solution of a certain stochastic differential equation.
Ukr. Mat. Zh. - 2011. - 63, № 9. - pp. 1257-1262
We investigate systems of differential equations with essentially infinite-dimensional elliptic operators (of the Laplace - Levy type). For nonlinear systems, we prove theorems on the existence and uniqueness of solutions. For a linear system, we give an explicit formula for the solution.
Ukr. Mat. Zh. - 2011. - 63, № 9. - pp. 1263-1278
Some new oscillation criteria are established for the nonlinear damped differential equation $$(r(t)k_1 (x, x'))' + p (t) k_2 (x, x') x' + q (t) f (x (t)) = 0,\quad t \geq t_0.$$ The results obtained extend and improve some existing results in the literature.
Ukr. Mat. Zh. - 2011. - 63, № 9. - pp. 1279-1278
A finite group $G$ is called an $MSP$-group if all maximal subgroups of the Sylow subgroups of $G$ are Squasinormal in $G$. In this paper, wc give a complete classification of those groups which are not $MSP$-groups but whose proper subgroups are all $MSP$-groups.
Ukr. Mat. Zh. - 2011. - 63, № 9. - pp. 1285-1289
Let $G$ be a bounded domain with a Jordan boundary that is smooth at all points except a single point at which it forms a nonzero angle. We prove Korevaar’s conjecture on the order of polynomial approximation of a conformal mapping of this domain into a disk. We also obtain a pointwise estimate for the error of approximation.
Ukr. Mat. Zh. - 2011. - 63, № 9. - pp. 1290-1296
We consider linear first-order differential equations with shifts of arguments with respect to functions with values in a Banach space. Sufficient conditions for the existence of nontrivial solutions of homogeneous equations are obtained. Ordinary differential equations are constructed for which all solutions defined on an axis are solutions of a given equation with shifts of the argument.
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Vertex degree Function
context $G=\langle V,E,\psi\rangle$ … undirected graph
definiendum $ \mathrm{dom}\ d = V $
range $ u,v\in V, u\neq v $ range $ e\in E $
definiendum $ d(v):=\left|\{e\ |\ \exists u.\ \psi(e)=\{u,v\}\}+2\ \mathrm{card} \{e\ |\ \psi(e)=\{v,v\}\}\right| $ Discussion
The value $d_G(v)$ is the number of neighbours of $v$ with loops counted twice.
It is also equal to the sum over the $v$-row or $v$-column of the adjacency matrix of the graph.
Theorems Because each edge connects twovertices, the number of vertices with odd degree is even, and then so is the sum over all degrees. I.e.
$\left| \{v\ |\ d(v)...\mathrm{odd}\} \right|$ … even
$\sum_{v\in V}d(v)$ … even
For a simple graph, if $A$ is the adjacency matrix, then
$\mathrm{tr}\ A^2=\sum_k(\sum_i A_i^k A_k^i)=\sum_i(\sum_k \delta_{i\ \mathrm{connected\ to }\ k})=d(v_i)$
This also works with $MM^t$, where $M$ is the incidence matrix.
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Search
Now showing items 1-10 of 26
Kaon femtoscopy in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV
(Elsevier, 2017-12-21)
We present the results of three-dimensional femtoscopic analyses for charged and neutral kaons recorded by ALICE in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV. Femtoscopy is used to measure the space-time ...
Anomalous evolution of the near-side jet peak shape in Pb-Pb collisions at $\sqrt{s_{\mathrm{NN}}}$ = 2.76 TeV
(American Physical Society, 2017-09-08)
The measurement of two-particle angular correlations is a powerful tool to study jet quenching in a $p_{\mathrm{T}}$ region inaccessible by direct jet identification. In these measurements pseudorapidity ($\Delta\eta$) and ...
Online data compression in the ALICE O$^2$ facility
(IOP, 2017)
The ALICE Collaboration and the ALICE O2 project have carried out detailed studies for a new online computing facility planned to be deployed for Run 3 of the Large Hadron Collider (LHC) at CERN. Some of the main aspects ...
Evolution of the longitudinal and azimuthal structure of the near-side peak in Pb–Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV
(American Physical Society, 2017-09-08)
In two-particle angular correlation measurements, jets give rise to a near-side peak, formed by particles associated to a higher $p_{\mathrm{T}}$ trigger particle. Measurements of these correlations as a function of ...
J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV
(American Physical Society, 2017-12-15)
We report a precise measurement of the J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector at the LHC. The J/$\psi$ mesons are reconstructed at mid-rapidity ($|y| < 0.9$) ...
Enhanced production of multi-strange hadrons in high-multiplicity proton-proton collisions
(Nature Publishing Group, 2017)
At sufficiently high temperature and energy density, nuclear matter undergoes a transition to a phase in which quarks and gluons are not confined: the quark–gluon plasma (QGP)1. Such an exotic state of strongly interacting ...
K$^{*}(892)^{0}$ and $\phi(1020)$ meson production at high transverse momentum in pp and Pb-Pb collisions at $\sqrt{s_\mathrm{NN}}$ = 2.76 TeV
(American Physical Society, 2017-06)
The production of K$^{*}(892)^{0}$ and $\phi(1020)$ mesons in proton-proton (pp) and lead-lead (Pb-Pb) collisions at $\sqrt{s_\mathrm{NN}} =$ 2.76 TeV has been analyzed using a high luminosity data sample accumulated in ...
Production of $\Sigma(1385)^{\pm}$ and $\Xi(1530)^{0}$ in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV
(Springer, 2017-06)
The transverse momentum distributions of the strange and double-strange hyperon resonances ($\Sigma(1385)^{\pm}$, $\Xi(1530)^{0}$) produced in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV were measured in the rapidity ...
Charged–particle multiplicities in proton–proton collisions at $\sqrt{s}=$ 0.9 to 8 TeV, with ALICE at the LHC
(Springer, 2017-01)
The ALICE Collaboration has carried out a detailed study of pseudorapidity densities and multiplicity distributions of primary charged particles produced in proton-proton collisions, at $\sqrt{s} =$ 0.9, 2.36, 2.76, 7 and ...
Energy dependence of forward-rapidity J/$\psi$ and $\psi(2S)$ production in pp collisions at the LHC
(Springer, 2017-06)
We present ALICE results on transverse momentum ($p_{\rm T}$) and rapidity ($y$) differential production cross sections, mean transverse momentum and mean transverse momentum square of inclusive J/$\psi$ and $\psi(2S)$ at ...
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This was a sub-question in my previous post that I ask separately now.
In
Introduction to Conformal Field Theory by Blumenhagen and Plauschinn (springer link) the Virasoro algebra is introduced the central extension of the Witt algebra. They give the central extension $$\widetilde{\mathfrak{g}} = \mathfrak{g}\oplus \mathbb{C}$$ of a Lie algebra $\mathfrak{g}$ by $\mathbb{C}$ is characterised by the following commutations:
\begin{align}[\widetilde{x},\widetilde{y}]_\widetilde{\mathfrak{g}} &= [x,y]_\mathfrak{g} + c p(x,y),\tag{2.14a}\\ [\widetilde{x}, c]_\widetilde{\mathfrak{g}} &= 0,\tag{2.14b}\\ [c,c]_\widetilde{\mathfrak{g}} &= 0,\tag{2.14c}\end{align}
with $\widetilde{x},\widetilde{y}\in \widetilde{\mathfrak{g}}$, $x,y\in\mathfrak{g}$, $c \in \mathbb{C}$ and $p:\mathfrak{g} \times \mathfrak{g} \rightarrow \mathbb{C}$.
Later on, they write the elements of the central extension as $L_n$, and write the Lie bracket as:
$$[L_m,L_n] = (m-n)L_{m+n} + cp(m,n)\tag{2.15}$$
In the previous section they derived the Witt algebra commutator as: $$[l_m,l_n] = (m-n)l_{m+n}.\tag{2.12}$$
I don't understand why they can write $(m-n)L_{m+n}$ instead of $(m-n)l_{m+n}$ in (2.15). Doesn't (2.14a) tell us that $[x,y]$ is a Lie bracket in $\mathfrak{g}$, while $L$ is in $\widetilde{\mathfrak{g}}$?
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To round off our series on round objects (see the first and second posts), let’s compute the sphere’s surface area. We can compute this in the same way we related the area and circumference of a circle two weeks ago. Approximate the surface of the sphere with lots of small triangles, and connect these to the center of the sphere to create lots of triangular pyramids. Each pyramid has volume \(\frac{1}{3}(\text{area of base})(\text{height})\), where the heights are all nearly \(r\) and the base areas add to approximately the surface area. By using more and smaller triangles these approximations get better and better, so the volume of the sphere is $$\frac{4}{3}\pi r^3 = \frac{1}{3}(\text{surface area})\cdot r,$$ meaning the surface area is \(4\pi r^2\). (This and previous arguments can be made precise with the modern language of integral calculus.)
Here’s an elegant way to rephrase this result: The surface area of a sphere is equal to the area of the curved portion of a cylinder that exactly encloses the sphere. In fact, something very surprising happens here!:
Archimedes’ Hat-Box Theorem: If we draw any two horizontal planes as shown below, then the portions of the sphere and the cylinder between the two planes have the same surface area.
We can prove this with (all!) the methods in the last few posts; here’s a quick sketch. To compute the area of the “spherical band” (usually called a spherical zone), first consider the solid spherical sector formed by joining the spherical zone to the center:
By dividing this into lots of triangular pyramids as we did with the sphere above, we can compute the area of the spherical zone by instead computing the sector’s volume. This volume can be computed by breaking it into three parts: two cones and the spherical segment between the two planes (on the left of the next figure). Compute the volume of the spherical segment by comparing (via Cavalieri’s Principle) to the corresponding part of the vase (from the previous post), which can be expressed with just cylinders and cones.
Last week we saw how to compute the area of a circle from first principles. What about spheres?
To compute the volume of a sphere, let’s show that a hemisphere (with radius \(r\)) has the same volume as the vase shown in the figure below, formed by carving a cone from the circular cylinder with radius and height \(r\). Why this shape? Here’s why: if we cut these two solids at any height \(h\) (between 0 and \(r\)), the areas of the two slices match. Indeed, the slice—usually called cross section—of the sphere is a circle of radius \(\sqrt{r^2-h^2}\), which has area \(\pi(r^2-h^2)\). Similarly, the vase’s cross section is a radius \(r\) circle with a radius \(h\) circle cut out, so its area is \(\pi r^2-\pi h^2\), as claimed.
If we imagine the hemisphere and vase as being made from lots of tiny grains of sand, then we just showed, intuitively, that the two solids have the same number of grains of sand in every layer. So there should be the same number of grains in total, i.e., the volumes should match. This intuition is exactly right:
Cavalieri’s Principle: any two shapes that have matching horizontal cross sectional areas also have the same volume.
So the volumes are indeed equal, and all that’s left is to compute the volume of the vase. But we can do this! Recall that the cone has volume \(\frac{1}{3} (\text{area of base}) (\text{height}) = \frac{1}{3}\pi r^3\) (better yet, prove this too! Hint: use Cavalieri’s Principle again to compare to a triangular pyramid). Likewise, the cylinder has volume \((\text{area of base}) (\text{height}) = \pi r^3\), so the vase (and hemisphere) have volume \(\pi r^3 – \frac{1}{3} \pi r^3 = \frac{2}{3}\pi r^3\). The volume of the whole sphere is thus \(\frac{4}{3}\pi r^3\). Success!
The following visualization illustrates what we have shown, namely $$\text{hemisphere} + \text{cone} = \text{cylinder}.$$ The “grains of sand” in the hemisphere are being displaced horizontally by the stabbing cone, and at the end we have exactly filled the cylinder.
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Abstract
Double quantum dots (DQDs) are a versatile platform for solid-state physics, quantum computation and nanotechnology. The micro-fabrication techniques commonly used to fabricate DQDs are difficult to extend to the atomic scale. Using an alternative approach, which relies on scanning tunneling microscopy and spectroscopy, we prepared a minimal DQD in a wide band-gap semiconductor matrix. It is comprised of a pair of strongly coupled donor atoms that can each be doubly charged. The donor excitation diagram of this system mimicks the charge stability diagram observed in transport measurements of DQDs. We furthermore illustrate how the charge and spin degrees of freedom of the minimal DQD may be used to obtain a single quantum bit and to prepare a Bell state. The results open an intriguing perspective for quantum electronics with atomic-scale structures.
Introduction
Quantum dots (QDs) are artificial nanometer scale structures in which quantum confinement causes the formation of discrete energy levels from continuous electronic bands of a solid. A single QD may be considered as an artificial atom, while a double QD (DQD) can be viewed as a molecule
1. Depending on the strength of the coupling between the QDs, DQDs may be categorized as weakly-coupled or strongly-coupled 1. Strongly-coupled DQDs have attracted much research attention, due to their fundamental properties and their significant applications. The Coulomb staircase, the spin blockade effect, the use as a single spin quantum bit, the realization of a micromaser are just a few of many examples 1,2,3,4. A typical DQD involves source and drain leads, which are coupled to the DQD with tunneling contacts, and two gate electrodes, which individually control the local potential of each QD (Fig. 1a). By varying both gate voltages and measuring the source-drain conductance, the charge states of the DQD may be determined and displayed in the so-called stability diagram (Fig. 1c). The stability diagram may be used to extract the coupling strength and also reveals additional key parameters, such as the on-site binding energy and Coulomb interaction.
To further shrink the dimensions of DQDs to the atomic scale scanning tunneling microscopy and spectroscopy (STM/S) is an obvious choice. Indeed, STM has successfully been used to investigate single impurities in semiconductors like Si or GaAs
5. Examples of recent achievements are the controlled switching of the charge state of a single impurity, the manipulation of individual donor binding energies, the magnetization of individual dopant, and the observation of a valley interference effect in single dopants 6,7,8,9,10,11. However, the lack of a gate electrode is a significant drawback of STM in transport measurements, which are essential for studies of QDs. Here, we have present a data acquisition and analysis method that enables measurements of the stability diagram of atomic-scale DQDs. The DQDs are comprised of dopant dimers.
Monomers and dimers of donors that can carry a single charge have been characterized by STM
6,7,8,9, 12,13,14. However, only donors that can be multiply charged are suitable for implementing a DQD. Therefore, we used the third generation semiconductor material ZnO 15. In ZnO double-donors, which can be charged with up to two electrons, are available 16, 17. From measured donor excitation diagrams (Fig. 1d) we found that a dimer of double-donors can be viewed as a DQD. In order to obtain such data, a set of differential conductance spectra ( dI/ dV) measured along a line crossing two neighboring double-donor atoms is required. The spectra are then represented as a two-dimensional map that displays the conductance vs. the bias voltage and the lateral tip position. The ionization of the donors leads to a peak in dI/ dV that evolves with the position of the tip and separates the map into several regions. The distinct areas of such a map correspond to different occupation numbers ( n 1, n 2) of the two donors, where n 1 ( n 2) is the number of electrons on the left (right) dot.
The donor excitation diagrams from our STM experiment closely resemble the stability diagrams obtained in transport measurements on DQDs. In other words, pairs of double-donors in ZnO represent minimal DQDs, with each QD involving a single donor and the surrounding ZnO lattice.
The clean ZnO(0001) surface displays terraces with triangular islands of adatoms and vacancies
18, 19 as shown in Fig. 1e. Despite the rough surface morphology, the dI/ dV map exhibits sharp rings, similar to what has been observed on other materials, such as GaAs, Bi 2Se 3 and MoSe 2. These rings are due to the ionization of subsurface dopant atoms 6, 20, 21. Importantly, ZnO features concentric double and multiple rings, which result from double or multiple charging of a donor or donor dimer at their centre.
In Fig. 1f, two neighboring elliptical double-rings (yellow lines) are observed. Their centers are ≈10 nm apart and their overlapping contours apparently do not affect each other. These properties are expected from weakly coupled QDs. In addition, the spectroscopic map displays a system exhibiting three concentric rings (red lines). Cross-sectional profiles of the multiple-ring system (Fig. 2a and c), reveal a donor excitation diagram that is typical of a strongly coupled DQD.
Previous works have shown that the binding energy (ionization threshold) of a subsurface donor is largely determined by its depth underneath the surface
22, 23. In other words, donor atoms at the same depth exhibit identical donor excitation diagram and ionization rings in real-space dI/ dV maps. Therefore, a pair of donors at identical depths is expected to generate a symmetric DQD while different depths introduce a degree of asymmetry. The capability of producing both symmetric and asymmetric DQDs is essential for application as a quantum light source 24, 25. In the STM/S data of Figs 2 and 3, we indeed observed both types of DQDs. Using the evolution from a weakly-coupled DQD to the strongly-coupled case (Figs 2e and 3e), we can identify charge occupation numbers ( n 1, n 2). As shown in Figs 3d and 4d, the coherent superposition of particular DQD charge states, e.g. at the boundary between two areas with different ( n 1, n 2), exhibits potential application to quantum electronics.
To describe the dominant features of the donor excitation diagram and to reveal the potential application of our minimal DQD we build an analytic model. The total Hamiltonian is given by
H 1 ( H 2) is the Hubbard Hamiltonian for the left (right) donor atom. H C H T H m H s μ t U stands for the on-site Coulomb repulsion between two electrons at same donor; t 12 is the coherent coupling strength; U m V F V is the bias of the tip; γ ij i induced by TIBB of site j via the coupling between the donor atoms.
In our experiment, the two sites of the DQD are spatially separated (about 6.5 nm), direct coherent tunneling is suppressed by the barrier, so
H T n 1 and n 2 are both good quantum numbers, and the Hamiltonian can be directly diagonalized. By taking into account the position-dependent TIBB (see details in Supplementary Note 1 and Supplementary Fig. 1), we are able to simulate the symmetric DQD donor excitation diagram and deduce all key parameters as listed in Table 1. Specifically, we obtained \({\epsilon }_{d}\) = −200 meV and U ≅ 90 meV for individual donors and U m
By shining microwave radiation with a suitable wavelength into the STM junction
26, 27, coherent tunneling (microwave assisted tunneling) between the two donor atoms may be induced. In such a scenario, it appears possible to build a quantum bit and a Bell state in our minimal DQD by utilizing its charge and spin degrees of freedom. To obtain a quantum bit, the DQD system is first initialized to an occupation (0,1), where the donors carry 0 and 1 electron, respectively. Next, the microwave energy is tuned close to resonance with the (0,1)–(1,0) transition and a coherent coupling between these two states is built up. This process turns the minimal DQD into a minimal single solid-state charge qubit, similar to proposals based on a micro-fabricated QD with larger size 28. To prepare a Bell state, the system is initialized to the (0,2) state. The double donors in our ZnO samples most likely are interstitial Zn atoms 16, 17. When the donor carries two electrons, they occupy the Zn 4 s orbital, thus forming a spin singlet state. Moving one electron to the other donor atom thus produces a fourfold degenerate (1,1) state. Since the coherent tunneling conserves the total spin S 2 and its z-component S z i.e. the Bell state \(|{{\rm{\Psi }}}_{B}^{-}\rangle =\frac{1}{\sqrt{2}}(|\uparrow ,\downarrow \rangle -|\downarrow ,\uparrow \rangle )\), couples to the singlet (0,2) state. By applying the stimulated Raman adiabatic passage protocol (see details in Supplementary Note 2) to the (0,2)–(1,1) transition, the DQD is adiabatically driven from spin-singlet-(0,2) to a spin-singlet-(1,1) state 29, 30. The destination of our DQD is thus in the Bell state \(|{{\rm{\Psi }}}_{B}^{-}\rangle \).
In summary, we have experimentally realized a strongly-coupled DQD which is comprised of a dimer of donor atoms. The electronic levels of the individual donors depend on the depth below the surface, which can break the symmetry between the donors of a pair. As recently demonstrated the depth of an individual donor in ZnO can be manipulated
23 and consequently the electron transport properties of the proposed DQD may be controlled. An analytical model is capable of reproducing the electronic states of the system. We propose that microwave radiation may be used to prepare interesting entangled quantum states of this minimal DQD.
For the time being, STM-based techniques enable probing these electronic properties. Defect densities in ZnO crystals are high and the experiments require a time-consuming search for a suitable sample area. With improved crystal quality, however, it may become possible to more routinely obtain these DQD presented above.
Methods Experiment
All experiments were performed in an ultra-high vacuum system equipped with a home-built STM operated at 5 Kelvin. Single-crystalline ZnO(0001) was prepared by cycles of Ar
+ bombardment and high temperature annealing. Au tips were cut from a polycrystalline wire and in situ annealed prior to transfer to the STM. STM imaging was performed in a constant-current mode with the bias voltage being applied to the sample. A sinusoidal voltage modulation of 10 mV rms and a lock-in amplifier were employed to measure dI/ dV spectra. Theory
We build a total Hamiltonian to describe the strongly-coupled donor dimer. In the semi-classical regime, the Hamiltonian can be directly diagonalized enabling a simulation of the predominant features in the experimental data. In the quantum regime, we used two effective total Hamiltonians to describe the charge quantum bit and spin Bell state. Details of the theory are provided in the Supplementary Figs 1 and 2 and Supplementary Notes 1 and 2.
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Phys. Rev. A 66, 022117 (2002). Acknowledgements
We thank A. Weismann, S. Zhang and Y. Liu for discussions. H.Z. thanks the supports from National Natural Science Foundation of China (No. 11674226), the National Key Research and Development Program of China (2016YFA0300403) and the Young 1000-Talent Program. Financial support by the Deutsche Forschungsgemeinschaft through SFB 855 is acknowledged.
Ethics declarations Competing Interests
The authors declare that they have no competing interests.
Additional information Publisher's note: Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Electronic supplementary material About this article Further reading Chinese Physics B(2019) Surface Review and Letters(2019) Chinese Physics B(2019) Journal of Physics: Condensed Matter(2018) Applied Physics Letters(2017)
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Large deviations for stochastic 3D Leray-$ \alpha $ model with fractional dissipation
1.
School of Mathematical Sciences, Nankai University, 300071 Tianjin, China
2.
School of Mathematics and Statistics, Jiangsu Normal University, 221116 Xuzhou, China
In this paper we establish the Freidlin-Wentzell's large deviation principle for stochastic 3D Leray-$ \alpha $ model with general fractional dissipation and small multiplicative noise. This model is the stochastic 3D Navier-Stokes equations regularized through a smoothing kernel of order $ \theta_1 $ in the nonlinear term and a $ \theta_2 $-fractional Laplacian. The main result generalizes the corresponding LDP result of the classical stochastic 3D Leray-$ \alpha $ model ($ \theta_1 = 1 $, $ \theta_2 = 1 $), and it is also applicable to the stochastic 3D hyperviscous Navier-Stokes equations ($ \theta_1 = 0 $, $ \theta_2\geq\frac{5}{4} $) and stochastic 3D critical Leray-$ \alpha $ model ($ \theta_1 = \frac{1}{4} $, $ \theta_2 = 1 $).
Keywords:Large deviation principle, Leray-$ \alpha $ model, fractional Laplacian, Navier-Stokes equation, weak convergence approach. Mathematics Subject Classification:Primary: 60H15, 60F10; Secondary: 35Q30, 35R11. Citation:Shihu Li, Wei Liu, Yingchao Xie. Large deviations for stochastic 3D Leray-$ \alpha $ model with fractional dissipation. Communications on Pure & Applied Analysis, 2019, 18 (5) : 2491-2509. doi: 10.3934/cpaa.2019113
References:
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D. Barbato, F. Morandin and M. Romito,
Global regularity for a slightly supercritical hyperdissipative Navier-Stokes system,
[5]
V. Barbu and M. Röckner,
An operatorial approach to stochastic partial differential equations driven by linear multiplicative noise,
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H. Bessaih and B. Ferrario,
The regularized 3D Boussinesq equations with fractional Laplacian and no diffusion,
[8]
H. Bessaih, E. Hausenblas and P. A. Razafimandimby,
Strong solutions to stochastic hydrodynamical systems with multiplicative noise of jump type,
[9]
H. Bessaih and A. Millet,
Large deviations and the zero viscosity limit for 2D stochastic Navier-Stokes equations with free boundary,
[10]
H. Bessaih and P. A. Razafimandimby,
On the rate of convergence of the 2-D stochastic Leray-$\alpha$ model to the 2-D stochastic Navier-Stokes equations with multiplicative noise,
[11]
Z. Brzeźniak, B. Goldys and T. Jegaraj,
Large deviations and transitions between equilibria for stochastic Landau-Lifshitz-Gilbert equation,
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S. Cerrai and M. Röckner,
Large deviations for stochastic reaction-diffusion systems with multiplicative noise and non-Lipschitz reaction term,
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V. V. Chepyzhov, E. S. Titi and M. I. Vishik,
On the convergence of solutions of the Leray-$\alpha$ model to the trajectory attractor of the 3D Navier-Stokes system,
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L. Debbi,
Well-posedness of the multidimensional fractional stochastic Navier-Stokes equations on the torus and on bounded domains,
[21]
A. Dembo and O. Zeitouni,
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P. W. Fernando, E. Hausenblas and P. A. Razafimandimby,
Irreducibility and exponential mixing of some stochastic hydrodynamical systems driven by pure jump noise,
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F. Flandoli,
A stochastic view over the open problem of well-posedness for the 3D Navier-Stokes equations. Stochastic analysis: a series of lectures,
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M. I. Freidlin and A. D. Wentzell,
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N. Glatt-Holtz and V. Vicol,
Local and global existence of smooth solutions for the stochastic Euler equations with multiplicative noise,
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[33]
S. Li, W. Liu and Y. Xie, Ergodicity of 3D Leray-$\alpha$ model with fractional dissipation and degenerate stochastic forcing,
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J.-L. Lions,
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W. Liu, M. Röckner and X.-C. Zhu,
Large deviation principles for the stochastic quasi-geostrophic equations,
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D. Martirosyan,
Large deviations for stationary measures of stochastic nonlinear wave equations with smooth white noise,
[43]
E. Olson and E. S. Titi,
Viscosity versus vorticity stretching: global well-posedness for a family of Navier-Stokes-alpha-like models,
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M. Röckner, R.-C. Zhu and X.-C. Zhu,
Local existence and non-explosion of solutions for stochastic fractional partial differential equations driven by multiplicative noise,
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S. S. Sritharan and P. Sundar,
Large deviations for the two-dimensional Navier-Stokes equations with multiplicative noise,
[51]
E. Stein,
Singular Integrals and Differentiability Properties of Functions, Princeton, NJ: Princeton University Press, 1970.
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R. Temam,
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J. Yang and J. Zhai, Asymptotics of stochastic 2D hydrodynamical type systems in unbounded domains,
[58]
show all references
References:
[1] [2] [3] [4]
D. Barbato, F. Morandin and M. Romito,
Global regularity for a slightly supercritical hyperdissipative Navier-Stokes system,
[5]
V. Barbu and M. Röckner,
An operatorial approach to stochastic partial differential equations driven by linear multiplicative noise,
[6] [7]
H. Bessaih and B. Ferrario,
The regularized 3D Boussinesq equations with fractional Laplacian and no diffusion,
[8]
H. Bessaih, E. Hausenblas and P. A. Razafimandimby,
Strong solutions to stochastic hydrodynamical systems with multiplicative noise of jump type,
[9]
H. Bessaih and A. Millet,
Large deviations and the zero viscosity limit for 2D stochastic Navier-Stokes equations with free boundary,
[10]
H. Bessaih and P. A. Razafimandimby,
On the rate of convergence of the 2-D stochastic Leray-$\alpha$ model to the 2-D stochastic Navier-Stokes equations with multiplicative noise,
[11]
Z. Brzeźniak, B. Goldys and T. Jegaraj,
Large deviations and transitions between equilibria for stochastic Landau-Lifshitz-Gilbert equation,
[12] [13] [14]
S. Cerrai and M. Röckner,
Large deviations for stochastic reaction-diffusion systems with multiplicative noise and non-Lipschitz reaction term,
[15] [16]
V. V. Chepyzhov, E. S. Titi and M. I. Vishik,
On the convergence of solutions of the Leray-$\alpha$ model to the trajectory attractor of the 3D Navier-Stokes system,
[17] [18] [19] [20]
L. Debbi,
Well-posedness of the multidimensional fractional stochastic Navier-Stokes equations on the torus and on bounded domains,
[21]
A. Dembo and O. Zeitouni,
[22] [23] [24] [25]
P. W. Fernando, E. Hausenblas and P. A. Razafimandimby,
Irreducibility and exponential mixing of some stochastic hydrodynamical systems driven by pure jump noise,
[26] [27]
F. Flandoli,
A stochastic view over the open problem of well-posedness for the 3D Navier-Stokes equations. Stochastic analysis: a series of lectures,
[28]
M. I. Freidlin and A. D. Wentzell,
[29]
N. Glatt-Holtz and V. Vicol,
Local and global existence of smooth solutions for the stochastic Euler equations with multiplicative noise,
[30] [31] [32]
S. Li, W. Liu and Y. Xie,
[33]
S. Li, W. Liu and Y. Xie, Ergodicity of 3D Leray-$\alpha$ model with fractional dissipation and degenerate stochastic forcing,
[34] [35]
J.-L. Lions,
[36] [37] [38] [39] [40]
W. Liu, M. Röckner and X.-C. Zhu,
Large deviation principles for the stochastic quasi-geostrophic equations,
[41] [42]
D. Martirosyan,
Large deviations for stationary measures of stochastic nonlinear wave equations with smooth white noise,
[43]
E. Olson and E. S. Titi,
Viscosity versus vorticity stretching: global well-posedness for a family of Navier-Stokes-alpha-like models,
[44] [45] [46] [47] [48]
M. Röckner, R.-C. Zhu and X.-C. Zhu,
Local existence and non-explosion of solutions for stochastic fractional partial differential equations driven by multiplicative noise,
[49] [50]
S. S. Sritharan and P. Sundar,
Large deviations for the two-dimensional Navier-Stokes equations with multiplicative noise,
[51]
E. Stein,
Singular Integrals and Differentiability Properties of Functions, Princeton, NJ: Princeton University Press, 1970.
Google Scholar
[52]
R. Temam,
[53] [54]
F.-Y. Wang and L. Xu, Derivative formula and applications for hyperdissipative stochastic Navier-Stokes/Burgers equations,
[55] [56] [57]
J. Yang and J. Zhai, Asymptotics of stochastic 2D hydrodynamical type systems in unbounded domains,
[58]
[1]
Vladimir V. Chepyzhov, E. S. Titi, Mark I. Vishik.
On the convergence of solutions of the Leray-$\alpha $ model to the trajectory attractor of the 3D Navier-Stokes system.
[2]
Phuong Le.
Symmetry of singular solutions for a weighted Choquard equation involving the fractional $ p $-Laplacian.
[3]
Harbir Antil, Mahamadi Warma.
Optimal control of the coefficient for the regional fractional $p$-Laplace equation: Approximation and convergence.
[4]
Anna Amirdjanova, Jie Xiong.
Large deviation principle for a stochastic navier-Stokes equation in
its vorticity form for a two-dimensional incompressible flow.
[5]
Aseel Farhat, M. S Jolly, Evelyn Lunasin.
Bounds on energy and enstrophy for the 3D Navier-Stokes-$\alpha$ and Leray-$\alpha$ models.
[6]
Claudianor O. Alves, Vincenzo Ambrosio, Teresa Isernia.
Existence, multiplicity and concentration for a class of fractional $ p \& q $ Laplacian problems in $ \mathbb{R} ^{N} $.
[7]
Nicholas J. Kass, Mohammad A. Rammaha.
Local and global existence of solutions to a strongly damped wave equation of the $ p $-Laplacian type.
[8]
Peng Mei, Zhan Zhou, Genghong Lin.
Periodic and subharmonic solutions for a 2$n$th-order $\phi_c$-Laplacian difference equation containing both advances and retardations.
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Gabriele Bonanno, Giuseppina D'Aguì.
Mixed elliptic problems involving the $p-$Laplacian with nonhomogeneous boundary conditions.
[13]
Genghong Lin, Zhan Zhou.
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[14]
K. D. Chu, D. D. Hai.
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Newspace parameters
Level: \( N \) = \( 3311 = 7 \cdot 11 \cdot 43 \) Weight: \( k \) = \( 1 \) Character orbit: \([\chi]\) = 3311.h (of order \(2\) and degree \(1\)) Newform invariants
Self dual: Yes Analytic conductor: \(1.65240425683\) Analytic rank: \(0\) Dimension: \(1\) Coefficient field: \(\mathbb{Q}\) Coefficient ring: \(\mathbb{Z}\) Coefficient ring index: \( 1 \) Projective image \(D_{2}\) Projective field Galois closure of \(\Q(\sqrt{-7}, \sqrt{473})\) Artin image size \(8\) Artin image $D_4$ Artin field Galois closure of 4.0.23177.2 Character Values
We give the values of \(\chi\) on generators for \(\left(\mathbb{Z}/3311\mathbb{Z}\right)^\times\).
\(n\) \(904\) \(1893\) \(2927\) \(\chi(n)\) \(-1\) \(-1\) \(-1\)
For each embedding \(\iota_m\) of the coefficient field, the values \(\iota_m(a_n)\) are shown below.
For more information on an embedded modular form you can click on its label.
Label \(\iota_m(\nu)\) \( a_{2} \) \( a_{3} \) \( a_{4} \) \( a_{5} \) \( a_{6} \) \( a_{7} \) \( a_{8} \) \( a_{9} \) \( a_{10} \) 3310.1
−2.00000 0 3.00000 0 0 −1.00000 −4.00000 −1.00000 0
Char. orbit Parity Mult. Self Twist Proved 1.a Even 1 trivial yes 7.b Odd 1 CM by \(\Q(\sqrt{-7}) \) yes 473.d Even 1 RM by \(\Q(\sqrt{473}) \) yes 3311.h Odd 1 CM by \(\Q(\sqrt{-3311}) \) yes
This newform can be constructed as the intersection of the kernels of the following linear operators acting on \(S_{1}^{\mathrm{new}}(3311, [\chi])\):
\(T_{2} \) \(\mathstrut +\mathstrut 2 \) \(T_{3} \)
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In these cases:
$$
r(x,y) = \bigvee \\{a \in A : \; \Delta(a) \leq_{A\times A} (x,y) \\}
$$
$$
l(x,y) = \bigwedge \\{a \in A : \; (x,y) \leq_{A\times A} \Delta(a) \\}
$$
You're assuming that you can just take infima \\(\bigwedge\\) and suprema \\(\bigvee\\). In a simple lattice \\((L, \wedge, \vee)\\) you don't have those operations available.
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AliPhysics 4e47bdd (4e47bdd)
Scripts used in the analysis Algorithms Corrections Monte-carlo code Mid-rapidity tracklet code for dN/deta Tasks Topical
file AliAODForwardMult.h Per-event \( N_{ch}\) per \((\eta,\varphi)\) bin. file AliFMDEventPlaneFinder.h file AliForwardUtil.h Various utilities used in PWGLF/FORWARD. file AliLandauGaus.h Declaration and implementation of Landau-Gauss distributions. file AliLandauGausFitter.h Declaration and implementation of fitter of Landau-Gauss distributions to energy loss spectra.
class AliForwardCreateResponseMatrices class AliForwardTriggerBiasCorrection class AliForwardUtil struct AliForwardUtil::Histos struct AliForwardUtil::RingHistos class AliLandauGaus class AliLandauGausFitter
Code to do the multiplicity analysis in the forward psuedo-rapidity regions
The code in this section defines methods to measure the charged-particle pseudorapidity density over \(|\eta|<2\) using SPD tracklets.
The code in this module constitutes tools for analysing SPD tracklet data for the charged-particle pseudorapidity density. It is based on Ruben's original code (see pwglf_fwd_spd_tracklet_1), but differs in some important aspects.
This code also requires a pass of real-data (AliTrackletAODTask) and simulated (AliTrackletAODMCTask) ESDs, but the output is not near-final histograms but an array of data structures (AliAODTracklet) stored on the output AOD. The data structure contains basic information on each tracklet
Tracklet structures from simulations contain in addition
During the AOD production, no cuts, except those defined for the re-reconstruction are imposed. In this way, the AOD contains the minimum bias information on tracklets for all events,
And for AODs corresponding to simulated data, we can also
In this way, we do a single pass of ESDs for real and simulated data, and we can then process the generated AODs with various cuts imposed. The AODs are generally small enough that they can be processed locally and quickly (for example using ProofLite). This scheme allows for fast turn-around with the largest possible flexibility.
The final charged-particle pseudorapidity density is produced by an external class (AliTrackletdNdeta2).
Other differences to Ruben's code is that the output files are far more structured, allowing for fast browsing of the data and quality assurance.
The analysis requires real data and simulated data, anchored to the real data runs being processed. For both real and simulated data, the analysis progresses through two steps:
A pass over ESD plus clusters to generate an AOD branch containing a TClonesArray of AliAODTracklet objects. In this pass, there are no selections imposed on the events. In this pass, the SPD clusters are reprocesed and the tracklets are re-reconstructed.
In this pass, we also form so-called injection events. In these events, a real cluster is removed and a new cluster put in at some other location in the detector. The tracklets of the event is then reconstructed and stored. This procedure is repeated as many times as possible. The injection events are therefore superpositions of many events - each with a real cluster removed and replaced by a fake cluster. The injection events are used later for background estimates.
When processing simulated data, the tracklets are also inspected for their origin. A tracklet can have three distinct classes of origins:
The last class is the background from combinations of clusters that does not correspond to true particles. This background must be removed from the measurements.
The second class, tracklets from secondaries, also form a background, but these tracklets are suppressed by cuts on the sum-of-square residuals
\[ \Delta = \left[\frac{\Delta_{\theta}^2\sin^2\theta}{\sigma_{\theta}^2}+ \frac{(\Delta_{\phi}-\delta_{\phi})^2}{\sigma_{\phi}^2}\right] \]
\[ \frac{d^2N_X}{d\eta d\mathrm{IP}_z}\quad, \]
where \( X\) is \( M\) or \( I\) for real data, or \( M',I',P',S',C'\) or \( G'\) for simulated data.
For each of these tracklet samples, except \( G'\), we also form the 3-dimensional differential \(\Delta\) distributions
\[ \frac{d^3N_X}{d\eta d\mathrm{IP}_z d\Delta}\quad, \]
which are later used to estimate the background due to wrong combinations of clusters into tracks.
Once both the real and simulated data has passed these two steps, we combine the to data sets into the final measurement. The final measurement is given by
\[ R = \frac{G'}{(1-\beta')M'}(1-\beta)M, \]
where
\[ \beta' = \frac{C'}{M'}\quad\mathrm{and}\quad \beta = k\beta'\quad. \]
Here, \( k\) is some scaling derived from the 3-dimensional differential \(\Delta_M\) and \(\Delta_{M'}\) distributions .
There are classes for containing data, classes that represent analysis tasks, and classes that perform calculations, as well as specialized classes for analysis of simulation (MC) output.
The classes AliAODTracklet and AliAODMCTracklet stores individual tracklet parameters. The difference between the two are that AliAODMCTracklet also stores the PDG code(s) and transverse momentum (momenta) of the mother primary particle (which may be the particle it self).
The pass over the ESD is done by the classes AliTrackletAODTask and AliTrackletAODMCTask. These tasks generated the array of tracklets in the AOD events. The difference between the two is that AliTrackletAODMCTask inspects and groups each tracklet according to it's origin, and create pseudo-tracklets corresponding to the generated primary, charged particles.
\[ \frac{d^2N_X}{d\eta d\mathrm{IP}_z}\quad, \]
and
\[ \frac{d^3N_X}{d\eta d\mathrm{IP}_z d\Delta}\quad. \]
The first task does this for \( X=M\) and \( I\), while the second and thhird tasks does this for \( X=M',I',P',S',C'\) and \( G'\). The third task reweighs all tracklets according to the particle specie(s) and transverse momentum (momenta) of the mother primary particle(s). AliTrackletWeights defines the interface used for reqeighing the data.
To produce the AODs with the tracklet information in, one needs to run a train with a task of the class AliTrackletAODTask (or AliTrackletAODMCTask for simulated data) and a task of the class AliSimpleHeaderTask in it. This is most easily done using the TrainSetup (Using the TrainSetup facility) derived class TrackletAODTrain.
For example for real data from run 245064 of LHC15o using the first physics pass
runTrain --name=LHC15o_245064_fp_AOD \ --class=TrackletAODTrain.C \ --url="alien:///alice/data/2015/LHC15o?run=245064&pattern=pass_lowint_firstphys/*/AliESDs.root&aliphysics=last,regular#esdTree"
or for simulated data from the LHC15k1a1 production anchored to run 245064
runTrain --name=LHC15k1a1_245064_fp_AOD \ --class=TrackletAODTrain.C \ --url=alien:///alice/sim/2015/LHC15k1a1?run=245064&pattern=*/AliESDs.root&aliphysics=last,regular&mc#esdTree
(note the addition of the option "&mc" to the URL argument)
In both cases a sub-directory - named of the name argument - of the current directory is created. In that sub-directory there are scripts for merging the output, downloading results, and downloading the generated AODs.
It is highly recommended to download the generated AODs to your local work station to allow fast second step analysis. To download the AODs, go to the generated sub-directory an run the
DownloadAOD.C script. For example, for the real data analysis of run 245064 of LHC15o, one would do
(cd LHC15o_245064_fp_AOD && root -l -b -q DownloadAOD.C)
and similar for the analysis of the simulated data.
To produce the histograms for the final charged-particle pseudorapidity density , one needs to run a train with a task of the class AliTrackletAODdNdeta (or AliTrackletAODMCdNdeta for simulated data) in it. This is most easily done using the TrainSetup derived class TrackletAODdNdeta.
For example for real data from run 245064 of LHC15o where we store the AODs generated above on the grid
runTrain --name=LHC15o_245064_fp_dNdeta \ --class=TrackletAODdNdeta.C \ --url="alien:///alice/cern.ch/user/a/auser/LHC15o_245064_fp_dNdeta/output?run=245064&pattern=* /AliAOD.root&aliphysics=last,regular#aodTree"
or for simulated data from the LHC15k1a1 production anchored to run 245064
runTrain --name=LHC15k1a1_245064_fp_dNdeta \ --class=TrackletAODdNdeta.C \ --url=alien:///alice/cern.ch/user/a/auser/LHC15k1a1_245064_fp_AOD/output?run=245064&pattern=* /AliAOD.root&aliphysics=last,regular&mc#esdTree
(note the addition of the option "&mc" to the URL argument)
If we had downloaded the AODs, we can use ProofLite to do this step
runTrain --name=LHC15o_245064_fp_dNdeta \ --class=TrackletAODdNdeta.C \ --url="lite:///${PWD}/LHC15o_245064_fp_dNdeta?pattern=AliAOD_*.root#aodTree"
and similar for simulated data
runTrain --name=LHC15k1a1_245064_fp_dNdeta \ --class=TrackletAODdNdeta.C \ --url="lite:///${PWD}/LHC15k1a1_245064_fp_dNdeta?pattern=AliAOD_*.root&mc#aodTree"
(note the addition of the option "&mc" to the URL argument)
The final result is obtained by runnin the class AliTrackletdNdeta2 over the histograms from both real data and simulations. As an example, suppose we ran or histogram production on the Grid and have downloaded the merged results into
LHC15o_245064_fp_dNdeta/root_archive_000245064/AnalysisResult.root (using
LHC15o_245064_fp_dNdeta/Download.C) and LHC15k1a1_245064_fp_dNdeta/root_archive_245064/AnalysisResult.root (using
LHC15k1a1_245064_fp_dNdeta/Download.C). Then we should do
(see AliTrackletdNdeta2::Run for more information on arguments)
Alternatively one can use the script Post.C to do this. The script is used like.
where
simFile is the simulation data input file (or directory)
realFile is the real data input file (or directory)
outDir is the output directory (created)
process are the processing options
visualize are the visualisation options
nCentralities is the maximum number of centrality bins
Processing options:
0x0001 Do scaling by unity
0x0002 Do scaling by full average
0x0004 Do scaling by eta differential
0x0008 Do scaling by fully differential
0x0010 Correct for decay of strange to secondary
0x1000 MC closure test
Visualization options:
0x0001 Draw general information
0x0002 Draw parameters
0x0004 Draw weights
0x0008 Draw dNch/deta
0x0010 Draw alphas
0x0020 Draw delta information
0x0040 Draw backgrounds
0x0100 Whether to make a PDF
0x0200 Whether to pause after each plot
0x0400 Draw in landscape
0x0800 Alternative markers
By default, each plot will be made and the process paused. To advance, simple press the space-bar.
If we had made the histograms using ProofLite, we should do
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I've got a few measurements $\vec{x}$ for some real-world value $\hat{x}$. These measurements have some uncertainty, and are correlated. Given these estimates, and their covariances, I want to take some kind of weighted average of the estimates
$$\bar{x} = \sum_i w_i x_i = \vec{w} \cdot \vec{x}$$
Naturally, I want my weights to sum up to 1:
$$\sum_i w_i = 1 = \vec{w} \cdot \vec{1}$$
I want to choose my weights such that my uncertainty around this estimate is as low as possible, so I want to minimize:
$$\mathbb{E} \left[ \left( \bar{x} - \hat{x} \right)^2 \right] = \sum_i \sum_j w_i w_j C_{i,j} = \vec{w}^T C \vec{w}$$
where $C$ is the covariance matrix of the errors of my measurements: $C_{i,j}=\mathbb{E} \left[ x_i - \hat{x}, x_j - \hat{x} \right]$. Minimizing this under the constraint that the weights sum up to 1 can be done using a Lagrange multiplier, and (assuming I did everything right) results in:
$$ C \cdot \vec{w^*} = \vec{1}$$ $$ \vec{w} = \vec{w^*} / \sum \vec{w^*} $$
Note that the latter step is just rescaling $\vec{w^*}$ to sum up to 1, so I'm mostly interested in the solution of $\vec{w^*}$:
$$ \vec{w^*} = C^{-1} \cdot \vec{1}$$
This procedure results in negative weights. This doesn't make sense for me intuitively. (E.g. if I have just one estimate $x_1$ of 100, then obviously my total estimate $\bar{x}$ will also be 100. How could the introduction of another measurement, of 110, make it so my overall estimate $\bar{x}$
lowers to below 100?) So I suspect that I'm doing something wrong, like using an inconsistent covariance matrix. But is it definite that I am doing something wrong?
Can the above procedure
reasonably return negative weights? How would this make sense in terms of the actual implications of what I'm trying to do (take a weighted average of multiple estimates)?
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I just stumbled upon the definition of the center Z of a group G: $$Z= \{x \in G \mid xz = zx \text{ for all } z \in G\}$$ The name “center” seems to suggest that there is some kind of geometric interpretation of the concept which I fail to see. My question is the following: is there some intuition/motivation behind the choice of naming $Z$ the “center” of a group?
An element is called central if it commutes with everything else...i.e., it does not matter whether you multiply from the left or right, so you can think of such an element as being multiplied in the "center" of any product it is in. Starting from there, it is an easy step to start calling the subgroup of all such elements the center. And from there we call it $Z(G)$, the Z being an abbreviation for the German word for center if I remember right.
By virtue of left and right multiplications, a group $G$ "naturally lives" in $Sym(G)$ (the group of all the bijections of $G$ into itself) in the shape of a
pair of subgroups of $Sym(G)$, say $\Theta$ and $\Gamma$, both of which it is isomorphic to. These subgroups commute, so $\Theta\Gamma$ is also a subgroup of $Sym(G)$. Finally, and this is mostly relevant for your question, $Z(G)$ turns out to be isomorphic to $\Theta \cap \Gamma$. Then, in $Sym(G)$ everything looks symmetric around the "center" $\Theta \cap \Gamma$:
For clarity, I'm not saying this is really the reason why the center was historically named that way. It's just a way I "pictorially" found for myself to accept that such a name actually makes sense.
Likewise, I've given here an interpretation of the wording "inner automorphism" (see Proposition 3 therein and the following comment).
Since
$$xz = zx \iff x = zxz^{-1}$$
$Z$ can also be written as
$$ Z = \{x \in G \mid x = zxz^{-1} \text{ for all } z \in G \}$$
I hope the name is more intuitive now!
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№ 9
All Issues Volume 64, № 3, 2012
Ukr. Mat. Zh. - 2012. - 64, № 3. - pp. 291-306
We study the relationship between the isomorphism of quivers and properties of their spectra. It is proved that two simple strongly connected quivers with at most four vertices are isomorphic to one another if and only if their characteristic polynomials coincide and their left and right normalized positive eigenvectors that correspond to the index can be obtained from one another by the permutation of their coordinates. An example showing that this statement is not true for quivers with five vertices is given.
Ukr. Mat. Zh. - 2012. - 64, № 3. - pp. 307-317
We show that a hyperquadric $M$ in $\mathbb{R}^4_2$ is a Lie group by using the bicomplex number product. For our purpose, we change the definition of tensor product. We define a new tensor product by considering the tensor product surface in the hyperquadric $M$. By using this new tensor product, we classify totally real tensor product surfaces and complex tensor product surfaces of a Lorentzian plane curve and a Euclidean plane curve. By means of the tensor product surfaces of a Lorentzian plane curve and a Euclidean plane curve, we determine a special subgroup of the Lie group M. Thus, we obtain the Lie group structure of tensor product surfaces of a Lorentzian plane curve and a Euclidean plane curve. Morever, we obtain left invariant vector fields of these Lie groups. We consider the left invariant vector fields on these groups, which constitute a pseudo-Hermitian structure. By using this, we characterize these Lie groups as totally real or slant in $\mathbb{R}^4_2$.
Ukr. Mat. Zh. - 2012. - 64, № 3. - pp. 318-343
We study a one-parameter family of positive polynomial operators of one and two variables that approximate the Urysohn operator. In the case of two variables, the integration domain is a "rectangular isosceles triangle". As a special case, Bernstein-type polynomials are obtained. The Stancu asymptotic formulas for remainders are refined.
Ukr. Mat. Zh. - 2012. - 64, № 3. - pp. 344-350
We show that every ballean (equivalently, coarse structure) on a set $X$ can be determined by some group $G$ of permutations of $X$ and some group ideal $\mathcal{I}$ on $G$. We refine this characterization for some basic classes of balleans: metrizable, cellular, graph, locally finite, and uniformly locally finite. Then we show that a free ultrafilter $\mathcal{U}$ on $\omega$ is a $T$-point with respect to the class of all metrizable locally finite balleans on $\omega$ if and only if $\mathcal{U}$ is a $Q$-point. The paper is concluded with а list of open questions.
Ukr. Mat. Zh. - 2012. - 64, № 3. - pp. 351-362
In the spaces $L_{\Psi}(T)$ of periodic functions with metric $\rho(f, 0)_{\Psi} = \int_T \Psi(|f(x)|)dx$, where $\Psi$ is a function of the modulus-of-continuity type, we investigate the inverse Jackson theorems in the case of approximation by trigonometric polynomials. It is proved that the inverse Jackson theorem is true if and only if the lower dilation exponent of the function $\Psi$ is not equal to zero.
Ukr. Mat. Zh. - 2012. - 64, № 3. - pp. 363-383
We consider a nonlinear system in the direct product of a torus and a Euclidean space. For this system, under the conditions of indefinite coercivity and indefinite monotonicity, we establish the existence of a Lipschitzian invariant section.
Ukr. Mat. Zh. - 2012. - 64, № 3. - pp. 384-391
We study Shunkov groups with the following condition: the normalizer of any finite nonunit subgroup has an almost layer-finite periodic part. It is proved that such a group has an almost layer-finite periodic part if it has a strongly imbedded subgroup with almost layer-finite periodic part.
Ukr. Mat. Zh. - 2012. - 64, № 3. - pp. 392-403
We prove the big Picard theorem for holomorphic curves from a punctured disc into $P^n(C)$ with $n + 2$ hypersurfaces. We also prove a theorem on the extension of holomorphic mappings in several complex variables into a submanifold of$P^n(C)$ with several moving hypersurfaces.
Ukr. Mat. Zh. - 2012. - 64, № 3. - pp. 404-414
We introduce the concept of weak $\alpha$-skew Armendariz ideals and investigate their properties. Moreover, we prove that $I$ is a weak $\alpha$-skew Armendariz ideal if and only if $I[x]$ is a weak $\alpha$-skew Armendariz ideal. As a consequence, we show that $R$ is a weak $\alpha$-skew Armendariz ring if and only if $R[x]$ is a weak $\alpha$-skew Armendariz ring.
Ukr. Mat. Zh. - 2012. - 64, № 3. - pp. 415-425
Some relations for quasiunit regular rings and $QB$-rings, as well as for pseudounit regular rings and $QB_{\infty}$-rings, are obtained. In the first part of the paper, we prove that (an exchange ring $R$ is a $QB$-ring) (whenever $x \in R$ is regular, there exists a quasiunit regular element $w \in R$ such that $x = xyx = xyw$ for some $y \in R$) — (whenever $aR + bR = dR$ in $R$, there exists a quasiunit regular element $w \in R$ such that $a + bz = dw$ for some $z \in R$). Similarly, we also give necessary and sufficient conditions for $QB_{\infty}$-rings in the second part of the paper.
Well-posedness of the Dirichlet and Poincare problems for a multidimensional Gellerstedt equation in a cylindric domain
Ukr. Mat. Zh. - 2012. - 64, № 3. - pp. 426-432
We prove the unique solvability of the Dirichlet and Poincare problems for a multidimensional Gellerstedt equation in a ´cylindric domain. We also obtain a criterion for the unique solvability of these problems.
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This is my first study about signal analysing. I'm very confused about filter order. My problem is how can I know whether its 12-th order, or 2nd order like the book says so? I already knew that slope has 2 choises whether its in n multiples of -6n dB/oct or -20n dB/dec. If the slope is -20 dB/dec then the second order should have -40 dB/dec. For example, An n= 4th order is 24dB/octave slope. But, I saw this example below and pretty confuse about determine the slope of those two line and how can the first line is 12-th order. Its not like the slope reaches -240 dB/dec (-20 x 12) and the second line seems not like -40 dB/dec (-20 x 2). Is there anyone could explain this to me? Thank you.
Note that the given slopes of $6n\textrm{ dB}/\textrm{octave}$ and $20n\textrm{ dB}/\textrm{dec}$, respectively, are asymptotic slopes for large frequencies. So you have to extrapolate the lines that are approximated by the magnitude responses at large frequencies.
For the given figure you could do the following. For the second order filter, look at the attenuations at $f=10^2\textrm{Hz}$ and at the next vertical line, which corresponds to $f=2\cdot 10^2\textrm{Hz}$. The ratio between those two frequencies is an octave, and the difference in attenuation is approximately $-22-(-35)\textrm{dB}=13\textrm{dB}$ (should be $12\textrm{dB}$).
For the $12^{th}$order system you could look at the attenuation values at $30\textrm{Hz}$ and $40\textrm{Hz}$, taking into account that the curve flattens towards its knee, and you need to extrapolate the straight line. With this in mind you would arrive at approximately $-31\textrm{dB}$ at $f=40\textrm{Hz}$, and at $-2\textrm{dB}$ at $f=30\textrm{Hz}$, which corresponds to a difference of $29\textrm{dB}$. Solving
$$\left(\frac{4}{3}\right)^x=2\tag{1}$$
for $x$ gives $x\approx 2.41$, i.e., you have to multiply $29\textrm{dB}$ with $2.41$ to obtain the attenuation per octave: $29\cdot 2.41=69.9\textrm{dB}$ (should be $72\textrm{dB}$).
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This question results from the discussion following a previous question: What is the connection between partial least squares, reduced rank regression, and principal component regression?
For principal component analysis, a commonly used probabilistic model is $$\mathbf x = \sqrt{\lambda} \mathbf{w} z + \boldsymbol \epsilon \in \mathbb R^p,$$ where $z\sim \mathcal N(0,1)$, $\mathbf{w}\in S^{p-1}$, $\lambda > 0$, and $\boldsymbol\epsilon \sim \mathcal N(0,\mathbf{I}_p)$. Then the population covariance of $\mathbf{x}$ is $\lambda \mathbf{w}\mathbf{w}^T + \mathbf{I}_p$, i.e., $$\mathbf{x}\sim \mathcal N(0,\lambda \mathbf{w}\mathbf{w}^T + \mathbf{I}_p).$$ The goal is to estimate $\mathbf{w}$. This is known as the spiked covariance model, which is frequently used in the PCA literature. The problem of estimating the true $\mathbf{w}$ can be solved by maximizing $\operatorname{Var} (\mathbf{Xw})$ over $\mathbf{w}$ on the unit sphere.
As pointed out in the answer to the previous question by @amoeba, reduced rank regression, partial least squares, and canonical correlation analysis have closely related formulations,
\begin{align} \mathrm{PCA:}&\quad \operatorname{Var}(\mathbf{Xw}),\\ \mathrm{RRR:}&\quad \phantom{\operatorname{Var}(\mathbf {Xw})\cdot{}}\operatorname{Corr}^2(\mathbf{Xw},\mathbf {Yv})\cdot\operatorname{Var}(\mathbf{Yv}),\\ \mathrm{PLS:}&\quad \operatorname{Var}(\mathbf{Xw})\cdot\operatorname{Corr}^2(\mathbf{Xw},\mathbf {Yv})\cdot\operatorname{Var}(\mathbf {Yv}) = \operatorname{Cov}^2(\mathbf{Xw},\mathbf {Yv}),\\ \mathrm{CCA:}&\quad \phantom{\operatorname{Var}(\mathbf {Xw})\cdot {}}\operatorname{Corr}^2(\mathbf {Xw},\mathbf {Yv}). \end{align}
The question is, what are the probabilistic models behind RRR, PLS, and CCA? In particular, I am thinking about $$(\mathbf{x}^T, \mathbf{y}^T)^T \sim \mathcal N(0, \mathbf{\Sigma}).$$ How does $\mathbf{\Sigma}$ depend on $\mathbf{w}$ and $\mathbf{v}$ in RRR, PLS, and CCA? Moreover, is there a unified probabilistic model (like the spiked covariance model for PCA) for them?
|
In short, my question is why does a spurion analysis work to produce the correct symmetry breaking terms regardless of the high energy physics?
The context that this question arose is from an Effective Field Theory course (for more context, see here, Eq. 5.50). Consider the QCD Lagrangian,
\begin{equation}
{\cal L} _{ QCD} = \bar{\psi} \left( i \gamma^\mu D_\mu - m \right) \psi
\end{equation}
The kinetic part is invariant under a chiral transformation:
\begin{equation}
\psi \rightarrow \left( \begin{array}{cc}
L & 0 \\
0 & R
\end{array} \right) \psi
\end{equation}
however, the mass term is not. Now the claim I don't understand is as follows. Suppose the mass transformed as,
\begin{equation}
m \rightarrow L m R ^\dagger
\end{equation}
In that case the mass term would be invariant under such a transformation. To write down the correct chiral symmetry breaking terms in our Lagrangian we find the terms invariant given this transformation for $ m $ and then make $ m $ a constant again.
The way I understand this physically is that the breaking arises from a high energy spurion field, $ X $, which gets a VEV, $ m $. When we write down all possible chiral symmetry preserving terms using the transforming $m$, we are writing down all the terms that the spurion couples to. The VEV is then inserted and is equal to $ m $. But this procedure assumes that the spurion obeys the chiral symmetry, $ SU(2) _L \times SU(2) _R $, and transforms as, $ X \rightarrow L X R ^\dagger $. How do we know this assumption is true? In fact it seems to fail for the case of QCD since the ``spurion field'' is really the Higgs field, which is a singlet under $ SU(2) _R $.
|
Synchronization for stochastic differential equations with nonlinear multiplicative noise in the mean square sense
School of Mathematics and Statistics, Huazhong University of Science and Technology, Wuhan, Hubei 430074, China
We provide a more clear technique to deal with general synchronization problems for SDEs, where the multiplicative noise appears nonlinearly. Moreover, convergence rate of synchronization is obtained. A new method employed here is the techniques of moment estimates for general solutions based on the transformation of multi-scales equations. As a by-product, the relationship between general solutions and stationary solutions is constructed.
Keywords:Synchronization, stationary solutions, multi-scale, stochastic differential equations, moment estimates. Mathematics Subject Classification:Primary: 60H10; Secondary: 34F05. Citation:Zhen Li, Jicheng Liu. Synchronization for stochastic differential equations with nonlinear multiplicative noise in the mean square sense. Discrete & Continuous Dynamical Systems - B, 2019, 24 (10) : 5709-5736. doi: 10.3934/dcdsb.2019103
References:
[1]
V. S. Afraimovich and H. M. Rodrigues, Uniform dissipativeness and synchronization of nonautonomous equations,
[2]
L. Arnold,
[3]
S. A. Azzawi, J. Liu and X. Liu,
Convergence rate of synchronization of systems with additive noise,
[4]
S. A. Azzawi, J. Liu and X. Liu, The synchronization of stochastic differential equations with linear noise,
[5]
T. Caraballo, I. D. Chueshov and P. E. Kloeden,
Synchronization of a stochastic reaction-diffusion system on a thin two-layer domain,
[6]
T. Caraballo and P. E. Kloeden,
The persistence of synchronization under environmental noise,
[7] [8]
T. Caraballo, P. E. Kloeden and B. Schmalfuss,
Exponentially stable stationary solutions for stochastic evolution equations and their pertubation,
[9] [10]
J. Duan and W. Wei,
[11] [12] [13] [14] [15]
X. Liu, J. Duan, J. Liu and P. E. Kloeden,
Synchronization of systems of Marcus canonical equations driven by $\alpha$-stable noises,
[16]
Y. Liu, X. Wan and E. Wu,
Finite time synchronization of Markovian neural networks with proportional delays and discontinuous activations,
[17]
J. Lu, D. W. C. Ho and Z. Wang,
Pinning stabilization of linearly coupled stochastic neural networks via minimum number of controllers,
[18] [19] [20]
A. Pikovsky, M. Rosenblum and J. Kurths,
Synchronization: A Universal Concept in Nonlinear Sciences, Cambridge Nonlinear Science Series, 12. Cambridge University Press, Cambridge, 2001.
doi: 10.1017/CBO9780511755743.
Google Scholar
[21] [22] [23]
B. Schmalfuss and R. Schneider,
Invariant manifolds for random dynamical systems with slow and fast variables,
[24]
S. Strogatz,
Sync: The Emerging Science of Spontaneous Order, Hyperion Press, New York, 2003.
Google Scholar
[25]
T. Su and X. Yang,
Finite-time synchronization of competitive neural networks with mixed delays,
[26]
X. Yang, J. Lu and D. W. C. Ho,
Synchronization of uncertain hybrid switching and impulsive complex networks,
[27] [28]
W. Zhang, C. Li and T. Huang,
Fixed-time synchronization of complex networks with nonidentical nodes and stochastic noise perturbations,
[29]
C. Zhou, W. Zhang and X. Yang,
Finite-time synchronization of complex-valued neural networks with mixed delays and uncertain perturbations,
show all references
References:
[1]
V. S. Afraimovich and H. M. Rodrigues, Uniform dissipativeness and synchronization of nonautonomous equations,
[2]
L. Arnold,
[3]
S. A. Azzawi, J. Liu and X. Liu,
Convergence rate of synchronization of systems with additive noise,
[4]
S. A. Azzawi, J. Liu and X. Liu, The synchronization of stochastic differential equations with linear noise,
[5]
T. Caraballo, I. D. Chueshov and P. E. Kloeden,
Synchronization of a stochastic reaction-diffusion system on a thin two-layer domain,
[6]
T. Caraballo and P. E. Kloeden,
The persistence of synchronization under environmental noise,
[7] [8]
T. Caraballo, P. E. Kloeden and B. Schmalfuss,
Exponentially stable stationary solutions for stochastic evolution equations and their pertubation,
[9] [10]
J. Duan and W. Wei,
[11] [12] [13] [14] [15]
X. Liu, J. Duan, J. Liu and P. E. Kloeden,
Synchronization of systems of Marcus canonical equations driven by $\alpha$-stable noises,
[16]
Y. Liu, X. Wan and E. Wu,
Finite time synchronization of Markovian neural networks with proportional delays and discontinuous activations,
[17]
J. Lu, D. W. C. Ho and Z. Wang,
Pinning stabilization of linearly coupled stochastic neural networks via minimum number of controllers,
[18] [19] [20]
A. Pikovsky, M. Rosenblum and J. Kurths,
Synchronization: A Universal Concept in Nonlinear Sciences, Cambridge Nonlinear Science Series, 12. Cambridge University Press, Cambridge, 2001.
doi: 10.1017/CBO9780511755743.
Google Scholar
[21] [22] [23]
B. Schmalfuss and R. Schneider,
Invariant manifolds for random dynamical systems with slow and fast variables,
[24]
S. Strogatz,
Sync: The Emerging Science of Spontaneous Order, Hyperion Press, New York, 2003.
Google Scholar
[25]
T. Su and X. Yang,
Finite-time synchronization of competitive neural networks with mixed delays,
[26]
X. Yang, J. Lu and D. W. C. Ho,
Synchronization of uncertain hybrid switching and impulsive complex networks,
[27] [28]
W. Zhang, C. Li and T. Huang,
Fixed-time synchronization of complex networks with nonidentical nodes and stochastic noise perturbations,
[29]
C. Zhou, W. Zhang and X. Yang,
Finite-time synchronization of complex-valued neural networks with mixed delays and uncertain perturbations,
[1]
Thierry Cazenave, Flávio Dickstein, Fred B. Weissler.
Multi-scale multi-profile global solutions of parabolic equations in
$\mathbb{R}^N $.
[2] [3]
Thomas Y. Hou, Pengfei Liu.
Optimal local multi-scale basis functions for linear elliptic equations with rough coefficients.
[4]
Thomas Blanc, Mihai Bostan, Franck Boyer.
Asymptotic analysis of parabolic equations with stiff transport terms by a multi-scale approach.
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Michel Potier-Ferry, Foudil Mohri, Fan Xu, Noureddine Damil, Bouazza Braikat, Khadija Mhada, Heng Hu, Qun Huang, Saeid Nezamabadi.
Cellular instabilities analyzed by multi-scale Fourier series: A review.
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Zhen Wang, Xiong Li, Jinzhi Lei.
Second moment boundedness of linear stochastic delay differential equations.
[8]
Kai Liu.
Stationary solutions of neutral stochastic partial differential equations with delays in the highest-order derivatives.
[9]
Emiliano Cristiani, Elisa Iacomini.
An interface-free multi-scale multi-order model for traffic flow.
[10]
Fuke Wu, George Yin, Le Yi Wang.
Razumikhin-type theorems on moment exponential stability of
functional differential equations involving two-time-scale Markovian switching.
[11]
Jean-Philippe Bernard, Emmanuel Frénod, Antoine Rousseau.
Modeling confinement in Étang de Thau: Numerical simulations and multi-scale aspects.
[12]
Wen-ming He, Jun-zhi Cui.
The estimate of the multi-scale homogenization method
for Green's function on Sobolev space $W^{1,q}(\Omega)$.
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Grigor Nika, Bogdan Vernescu.
Rate of convergence for a multi-scale model of dilute emulsions with non-uniform surface tension.
[14]
Weidong Zhao, Jinlei Wang, Shige Peng.
Error estimates of the $\theta$-scheme for backward stochastic differential
equations.
[15]
Markus Gahn.
Multi-scale modeling of processes in porous media - coupling reaction-diffusion processes in the solid and the fluid phase and on the separating interfaces.
[16]
Miroslava Růžičková, Irada Dzhalladova, Jitka Laitochová, Josef Diblík.
Solution to a stochastic pursuit model using moment equations.
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Yuyun Zhao, Yi Zhang, Tao Xu, Ling Bai, Qian Zhang.
[19]
Tianling Jin, Jingang Xiong.
Schauder estimates for solutions of linear parabolic integro-differential equations.
[20]
Yong Li, Zhenxin Liu, Wenhe Wang.
Almost periodic solutions and stable solutions for stochastic differential equations.
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Please assume that this graph is a highly magnified section of the derivative of some function, say $F(x)$. Let's denote the derivative by $f(x)$.Let's denote the width of a sample by $h$ where $$h\rightarrow0$$Now, for finding the area under the curve between the bounds $a ~\& ~b $ we can a...
@Ultradark You can try doing a finite difference to get rid of the sum and then compare term by term. Otherwise I am terrible at anything to do with primes that I don't know the identities of $\pi (n)$ well
@Silent No, take for example the prime 3. 2 is not a residue mod 3, so there is no $x\in\mathbb{Z}$ such that $x^2-2\equiv 0$ mod $3$.
However, you have two cases to consider. The first where $\binom{2}{p}=-1$ and $\binom{3}{p}=-1$ (In which case what does $\binom{6}{p}$ equal?) and the case where one or the other of $\binom{2}{p}$ and $\binom{3}{p}$ equals 1.
Also, probably something useful for congruence, if you didn't already know: If $a_1\equiv b_1\text{mod}(p)$ and $a_2\equiv b_2\text{mod}(p)$, then $a_1a_2\equiv b_1b_2\text{mod}(p)$
Is there any book or article that explains the motivations of the definitions of group, ring , field, ideal etc. of abstract algebra and/or gives a geometric or visual representation to Galois theory ?
Jacques Charles François Sturm ForMemRS (29 September 1803 – 15 December 1855) was a French mathematician.== Life and work ==Sturm was born in Geneva (then part of France) in 1803. The family of his father, Jean-Henri Sturm, had emigrated from Strasbourg around 1760 - about 50 years before Charles-François's birth. His mother's name was Jeanne-Louise-Henriette Gremay. In 1818, he started to follow the lectures of the academy of Geneva. In 1819, the death of his father forced Sturm to give lessons to children of the rich in order to support his own family. In 1823, he became tutor to the son...
I spent my career working with tensors. You have to be careful about defining multilinearity, domain, range, etc. Typically, tensors of type $(k,\ell)$ involve a fixed vector space, not so many letters varying.
UGA definitely grants a number of masters to people wanting only that (and sometimes admitted only for that). You people at fancy places think that every university is like Chicago, MIT, and Princeton.
hi there, I need to linearize nonlinear system about a fixed point. I've computed the jacobain matrix but one of the elements of this matrix is undefined at the fixed point. What is a better approach to solve this issue? The element is (24*x_2 + 5cos(x_1)*x_2)/abs(x_2). The fixed point is x_1=0, x_2=0
Consider the following integral: $\int 1/4*(1/(1+(u/2)^2)))dx$ Why does it matter if we put the constant 1/4 behind the integral versus keeping it inside? The solution is $1/2*\arctan{(u/2)}$. Or am I overseeing something?
*it should be du instead of dx in the integral
**and the solution is missing a constant C of course
Is there a standard way to divide radicals by polynomials? Stuff like $\frac{\sqrt a}{1 + b^2}$?
My expression happens to be in a form I can normalize to that, just the radicand happens to be a lot more complicated. In my case, I'm trying to figure out how to best simplify $\frac{x}{\sqrt{1 + x^2}}$, and so far, I've gotten to $\frac{x \sqrt{1+x^2}}{1+x^2}$, and it's pretty obvious you can move the $x$ inside the radical.
My hope is that I can somehow remove the polynomial from the bottom entirely, so I can then multiply the whole thing by a square root of another algebraic fraction.
Complicated, I know, but this is me trying to see if I can skip calculating Euclidean distance twice going from atan2 to something in terms of asin for a thing I'm working on.
"... and it's pretty obvious you can move the $x$ inside the radical" To clarify this in advance, I didn't mean literally move it verbatim, but via $x \sqrt{y} = \text{sgn}(x) \sqrt{x^2 y}$. (Hopefully, this was obvious, but I don't want to confuse people on what I meant.)
Ignore my question. I'm coming of the realization it's just not working how I would've hoped, so I'll just go with what I had before.
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@Mathphile I found no prime of the form $$n^{n+1}+(n+1)^{n+2}$$ for $n>392$ yet and neither a reason why the expression cannot be prime for odd n, although there are far more even cases without a known factor than odd cases.
@TheSimpliFire That´s what I´m thinking about, I had some "vague feeling" that there must be some elementary proof, so I decided to find it, and then I found it, it is really "too elementary", but I like surprises, if they´re good.
It is in fact difficult, I did not understand all the details either. But the ECM-method is analogue to the p-1-method which works well, then there is a factor p such that p-1 is smooth (has only small prime factors)
Brocard's problem is a problem in mathematics that asks to find integer values of n and m for whichn!+1=m2,{\displaystyle n!+1=m^{2},}where n! is the factorial. It was posed by Henri Brocard in a pair of articles in 1876 and 1885, and independently in 1913 by Srinivasa Ramanujan.== Brown numbers ==Pairs of the numbers (n, m) that solve Brocard's problem are called Brown numbers. There are only three known pairs of Brown numbers:(4,5), (5,11...
$\textbf{Corollary.}$ No solutions to Brocard's problem (with $n>10$) occur when $n$ that satisfies either \begin{equation}n!=[2\cdot 5^{2^k}-1\pmod{10^k}]^2-1\end{equation} or \begin{equation}n!=[2\cdot 16^{5^k}-1\pmod{10^k}]^2-1\end{equation} for a positive integer $k$. These are the OEIS sequences A224473 and A224474.
Proof: First, note that since $(10^k\pm1)^2-1\equiv((-1)^k\pm1)^2-1\equiv1\pm2(-1)^k\not\equiv0\pmod{11}$, $m\ne 10^k\pm1$ for $n>10$. If $k$ denotes the number of trailing zeros of $n!$, Legendre's formula implies that \begin{equation}k=\min\left\{\sum_{i=1}^\infty\left\lfloor\frac n{2^i}\right\rfloor,\sum_{i=1}^\infty\left\lfloor\frac n{5^i}\right\rfloor\right\}=\sum_{i=1}^\infty\left\lfloor\frac n{5^i}\right\rfloor\end{equation} where $\lfloor\cdot\rfloor$ denotes the floor function.
The upper limit can be replaced by $\lfloor\log_5n\rfloor$ since for $i>\lfloor\log_5n\rfloor$, $\left\lfloor\frac n{5^i}\right\rfloor=0$. An upper bound can be found using geometric series and the fact that $\lfloor x\rfloor\le x$: \begin{equation}k=\sum_{i=1}^{\lfloor\log_5n\rfloor}\left\lfloor\frac n{5^i}\right\rfloor\le\sum_{i=1}^{\lfloor\log_5n\rfloor}\frac n{5^i}=\frac n4\left(1-\frac1{5^{\lfloor\log_5n\rfloor}}\right)<\frac n4.\end{equation}
Thus $n!$ has $k$ zeroes for some $n\in(4k,\infty)$. Since $m=2\cdot5^{2^k}-1\pmod{10^k}$ and $2\cdot16^{5^k}-1\pmod{10^k}$ has at most $k$ digits, $m^2-1$ has only at most $2k$ digits by the conditions in the Corollary. The Corollary if $n!$ has more than $2k$ digits for $n>10$. From equation $(4)$, $n!$ has at least the same number of digits as $(4k)!$. Stirling's formula implies that \begin{equation}(4k)!>\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\end{equation}
Since the number of digits of an integer $t$ is $1+\lfloor\log t\rfloor$ where $\log$ denotes the logarithm in base $10$, the number of digits of $n!$ is at least \begin{equation}1+\left\lfloor\log\left(\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\right)\right\rfloor\ge\log\left(\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\right).\end{equation}
Therefore it suffices to show that for $k\ge2$ (since $n>10$ and $k<n/4$), \begin{equation}\log\left(\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\right)>2k\iff8\pi k\left(\frac{4k}e\right)^{8k}>10^{4k}\end{equation} which holds if and only if \begin{equation}\left(\frac{10}{\left(\frac{4k}e\right)}\right)^{4k}<8\pi k\iff k^2(8\pi k)^{\frac1{4k}}>\frac58e^2.\end{equation}
Now consider the function $f(x)=x^2(8\pi x)^{\frac1{4x}}$ over the domain $\Bbb R^+$, which is clearly positive there. Then after considerable algebra it is found that \begin{align*}f'(x)&=2x(8\pi x)^{\frac1{4x}}+\frac14(8\pi x)^{\frac1{4x}}(1-\ln(8\pi x))\\\implies f'(x)&=\frac{2f(x)}{x^2}\left(x-\frac18\ln(8\pi x)\right)>0\end{align*} for $x>0$ as $\min\{x-\frac18\ln(8\pi x)\}>0$ in the domain.
Thus $f$ is monotonically increasing in $(0,\infty)$, and since $2^2(8\pi\cdot2)^{\frac18}>\frac58e^2$, the inequality in equation $(8)$ holds. This means that the number of digits of $n!$ exceeds $2k$, proving the Corollary. $\square$
We get $n^n+3\equiv 0\pmod 4$ for odd $n$, so we can see from here that it is even (or, we could have used @TheSimpliFire's one-or-two-step method to derive this without any contradiction - which is better)
@TheSimpliFire Hey! with $4\pmod {10}$ and $0\pmod 4$ then this is the same as $10m_1+4$ and $4m_2$. If we set them equal to each other, we have that $5m_1=2(m_2-m_1)$ which means $m_1$ is even. We get $4\pmod {20}$ now :P
Yet again a conjecture!Motivated by Catalan's conjecture and a recent question of mine, I conjecture thatFor distinct, positive integers $a,b$, the only solution to this equation $$a^b-b^a=a+b\tag1$$ is $(a,b)=(2,5).$It is of anticipation that there will be much fewer solutions for incr...
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LaTeX Lambda24 Credit Royalty
Convert latex math notation to lambda functions in Python!
Language
Python 3.x
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API Calls - 1,882 Avg call duration - 1.00sec
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The Algorithm Platform License is the set of terms that are stated in the Software License section of the Algorithmia Application Developer and API License Agreement. It is intended to allow users to reserve as many rights as possible without limiting Algorithmia's ability to run it as a service. Learn More
Run an Example
Input
Output
{ "func": "lambda α,β,ε,λ: sum([(λ**(α))/(β**(1-α)) for i in range(int(1),int(ε)+1)])", "params": [ "α", "β", "ε", "λ" ]}
Install and Use Install
CLI Install Docs
Install the Algorithmia CLI client by running:
curl -sSLf https://algorithmia.com/install.sh | sh
Then authenticate by running:
$ algo auth # When prompted for api endpoint, hit enter # When prompted for API key, enter your key: YOUR_API_KEY
Use
algo run Jeffro/LatexLambda/1.1.2 -d '"\\sum_{i=1}^{\\epsilon}\\dfrac{\\lambda^{\\alpha}}{\\beta^{1-\\alpha}}"' --timeout 300
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I have panel data for counts of new firms in different regions for six years. I am estimating a static poisson regression with multiplicative fixed effects$^*$; I have also tried to estimate a dynamic model by introducing a lagged dependent variable, but could not make that latter model work. Now, I would like to test the residuals from the static model for autocorrelation, so that I have an idea of the importance of the dynamics. However, I cannot find any diagnostic tests for this in a textbook (I've looked at Wooldridge, Cameron & Trivedi, Winkelmann, Greene), and also have not seen such a test in a research paper. Since the individual effects in the model are not identified, I don't know how to compute meaningful residuals in the first place.
Does anyone 1) know how to compute meaningful residuals; and 2) know of any diagnostic tests for these panel fixed effect poisson models?
FYI: I am using Stata (version 12.1) -xtpoisson, fe vce(robust)- command for the static model. Stata's postestimation commands can compute predicted values etc, but only assuming that the individual effects are all zero.
$^*$ The cross-section (or pooled) Poisson regression models the expected number of counts $y$ as $E[y_i|x_i]=\exp(X_i\beta)$, with $\beta$ the coefficients and $X_i$ the variables. A common way to add individual fixed effects with panel data is to let the effects $\alpha_{i}$ enter the model multiplicatively: $E[y_{it}|X_{it},\alpha_i]=\alpha_i\exp(X_{it}\beta)$.
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I have this info but I don't understand anything
The coordinates and why those are the specific angles
You have the same polar co-ordinates for (3, 4) and (-4, 3) - this is incorrect. For (-4, 3), the angle would be more than 90. For (-4, 3), the polar co-ordinates would be (5, 90+53).
The purpose of any co-ordinate system is to be able to describe a point on a plane precisely.
Imagine you had a laser that was pointed East. This light from the laser covers every point on the x-axis, where x is positive. So, to describe any point on the x-axis, all we need to know is the distance from the origin and the direction in which the laser is pointed.
Now, rotate the laser slowly. Once an entire 360 degree rotation is complete, the light would have covered all the points on the plane.
This indicates that all you need to describe a point in a plane is the distance of the point from the origin (r, in polar co-ordinates) and the direction in which the laser is pointed (\theta, in polar co-ordinates).
A point (a, b) in the cartesian system is at a distance of a units from the origin towards the x-axis and at a distance of b units from the origin towards the y-axis.
The distance of this point from the origin is: r=\sqrt{a^2+b^2}. The direction of the laser would such that \tan\theta=\frac{b}{a}; or \theta=\tan^{-1}\left(\frac{b}{a}\right)
Hence, (a, b) in the cartesian system equals \left(\sqrt{a^2+b^2},\ \tan^{-1}\frac{b}{a}\right)in polar co-ordinates.
The most important thing to remember about co-ordinates is that they help you locate an object on a surface or in space. Essentially, they are giving directions to the object from a reference point called the origin.
The co-ordinates allow you to answer the question - Where is the object located? Or How do I get to the object?
They are many ways to answer the question. I will describe two.
Note that knowing one set of directions to arrive at the object, allows you to come up with the second set of directions as well.
Let's take the following example:
Let's describe how to arrive at the black dot on the 2D plane. If we use the rectangular coordinates (or Cartesian coordinates) to describe how to arrive at the black dot, we would say from the origin, walk 3 steps East and then walk 4 steps North (or walk 4 steps North and then walk 3 steps East).
Alternatively, we can use the polar coordinates to describe how to arrive at the dot. We can say start off by facing East, turn \theta \degree = \tan^{-1} \frac{4}{3} counterclockwise and then walk \sqrt{3^2 + 4^2} = 5 steps in the direction you are facing.
Let's use this to figure out how to come up with polar coordinates for the examples you have on your page.
Rectangular (Cartesian) coordinates (2, 0), then the polar coordinates would be (2, 0 \degree) . Do you see how?
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I'm going to go a bit overboard here and give you the sketch of how vectors are geometrically constructed, since I think it's helpful to know. While writing this I found I was phrasing things very carefully, which means: you may need to reread parts of this in a quiet corner if it doesn't all make sense at first.
Suppose that you have a set of scalar fields $\mathbf{S} \subseteq (\mathcal{M} \rightarrow \mathbb R) $ on an underlying set of "points" $\mathcal{M}$. This set
S contains every scalar field that you are interested in. It's helpful to define some jargon: let an $n$-functor be a smooth functions in $\text C^\infty(\mathbb R^n \rightarrow \mathbb R)$. We can of course apply them to real values (which I'll denote with parentheses as $\Phi(\dots)$, but also we can use them to combine scalar fields, which I'll denote with square brackets as $\Phi[a, b, \dots]$ which is the function from any $p \in \mathcal M$ to $\Phi(a(p), b(p), \dots)$. Henceforth I will just call scalar fields scalars, since we're talking about field theory.
When we close
S under $n$-functors for all $n$, we get a lot of stuff: first off we can add and multiply scalars since
+ and
* are 2-functors; but we can even define a topology on $\mathcal M$ with those scalars which makes all of them
continuous (the kernel topology; $s \subseteq \mathcal M$ is closed when $s = \operatorname{ker} \sigma$ for some $\sigma \in \mathbf S$), and we can use this topology to define whether $\mathcal M$ is connected or not, etc. If you go down this rabbit hole far enough, you get to specify criteria by which $\mathcal M$ is a "manifold". But let's get more concrete and specific.
Now in special relativity in particular, there are special fields $w, x, y, z$ which we use as
global coordinates, meaning that every scalar $\phi$ in $\mathbf{S}$ can be written as a 4-functor $\Phi(x_1, x_2, x_3, x_4)$ applied to those scalar fields; $\phi = \Phi[w, x, y, z].$ (In general relativity, you still have coordinates, but you will have to use the topology to define "neighborhoods" of a point and then say that for every point there is a neighborhood enclosing that point with a set of $d$ coordinates which distinguish points in that neighborhood.)
We can take your intuitive 3D notion of "vector fields" and associate them with
derivations $D_v = v^\mu \partial_\mu$, so that we can define vectors (again dropping the word "field") in a purely geometric sense as the set of linear maps $V \in (\mathbf S \rightarrow \mathbf S)$ satisfying the chain rule: given a $k$-functor $\Phi(x_1\dots x_k)$ with partial derivatives $\Phi_{(i)} = \partial \Phi/\partial x_i$, then $$ V [\Phi[s_1\dots s_k]] = \sum_{i=1}^k ~ V[s_i] ~ \Phi_{(i)}[s_1\dots s_k]. $$ We can then identify the components of any vector field by applying it to our coordinates, so you see there is a two-way correspondence between these abstract mathematical operations on S and the lists-of-coordinates that you're used to.
We can write this space of derivations on $\mathbf S$ as $\mathbf S^\bullet$, and make independent copies of it for every symbol that we want to stick up top: this is known as "abstract index notation". We write elements of this set as symbols with a corresponding superscript; so $v^\alpha \in \mathbf S^\alpha$.
Then there are
covectors, like the general gradient of a field, which are linear maps from vectors to scalars. For example, given a scalar $\phi$ there is a covector $\nabla\phi$ defined as $$\nabla\phi~(V) = V[\phi].$$ These we can denote by $\mathbf S_\bullet$ and we copy the space for every symbol. Now you know what $\partial_\mu \phi$ means; it is the copy of $\nabla \phi$ that lives in the space $\mathbf S_\mu$. It is a covector field. (Somewhat important: you don't quite know what $\partial_\mu v^\nu$ means yet. More complicated derivatives of this kind -- derivatives of vectors rather than scalars -- require some added geometric structure called a "connection" on the manifold.)
Finally we have the metric, which is a linear map from a pair of vectors to a scalar; this can be used to convert covectors to vectors.
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I guess by "bosons" you're referring to gauge bosons?
If so then start with some matter field $ \psi(x)$ which transforms under the gauge group. For local gauge transformations the gauge group element $g$ is spacetime dependent $g(x)$, and the transformation is $$\psi(x) \longrightarrow \psi'(x) = g(x)\psi(x).$$
Derivatives would transform as
$$\partial_{\mu}\psi(x) \longrightarrow g(x)\partial_{\mu}\psi(x)+(\partial_{\mu}g(x))\psi(x),$$
i.e. inhomogeneously. We would like a gauge covariant derivative $D_{\mu}$ which transforms homogeneously as
$$D_{\mu}\psi(x) \longrightarrow g(x)D_{\mu}\psi(x).$$
To achieve this, we define
$$D_{\mu}\psi = \partial_{\mu}\psi - A_{\mu}\psi,$$ where $A_{\mu} = \mathbf{A}_{\mu} \cdot\boldsymbol{{\tau}}$ and $\boldsymbol{\tau}$ are the generators of the Lie algebra of the gauge group and $A_{\mu}$ is our bosonic gauge field. This introduction of gauge bosons via the derivative term is sometimes referred to as minimal coupling.
In order to achieve this, $A_{\mu}$ is forced to have the transformation law
$$A_{\mu} \longrightarrow A'_{\mu} = gA_{\mu}g^{-1} + (\partial_{\mu}g)g^{-1}.$$
Just looking at how the $A_{\mu}$ are transforming under the group action (the first term), we recognize the adjoint representation.
Of course, on the global stage, the fields $\psi$ can be interpreted as bundle sections and the gauge fields as bundle connections. $A_\mu$'s transformation law will be recognisable as a transformation of connection coefficients under the action of the bundle's structure group. A good reference is Nakahara, or this link.
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Modeling a Continuous-Time System with Matlab
Many of us are familiar with modeling a continuous-time system in the frequency domain using its transfer function H(s) or H(jω). However, finding the time response can be challenging, and traditionally involves finding the inverse Laplace transform of H(s). An alternative way to get both time and frequency responses is to transform H(s) to a discrete-time system H(z) using the impulse-invariant transform [1,2]. This method provides an exact match to the continuous-time impulse response. Let’s look at how to use the Matlab function impinvar [3] to find H(z).
Consider a 3
rd-order transfer function in s:
$$H(s)=\frac{c_3 s^3 + c_2 s^2 + c_1 s + c_0} {d_3 s^3 + d_2 s^2 + d_1 s + d_0 }$$
We want to transform it into a 3
rd order function in z:
$$H(z)=\frac{b_0 + b_1 z^{-1} + b_2 z^{-2} +b_3 z^{-3} } {1 + a_1 z^{-1} + a_2 z^{-2} + a_3 z^{-3} }$$
Which has a time response:
$$y_n = b_0 x_n + b_1 x_{n-1} + b_2 x_{n-2} + b_3 x_{n-3} - a_1 y_{n-1} -a_2 y_{n-2} -a_3 y_{n-3} $$
Given coefficients [c
3 c 2 c 1 c 0] and [d 3 d 2 d 1 d 0] of H(s), the function impinvar computes the coefficients [b 0 b 1 b 2 b 3] and [1 a 1 a 2 a 3] of H(z).
Example
A 3
rd order Butterworth filter with a -3 dB frequency of 1 rad/s has the transfer function [4,5]
$$H(s)=\frac{1} { s^3 + 2 s^2 + 2 s + 1 } \quad{(1)}$$
Here is the code to calculate the coefficients of H(z) using impinvar:
fs= 4; % Hz % 3rd order butterworth polynomial num= 1; den= [1 2 2 1]; [b,a]= impinvar(num,den,fs) % coeffs of H(z)This gives us the coefficients: b = 0.0000 0.0066 0.0056 a = 1.0000 -2.5026 2.1213 -0.6065So we have
$$H(z)=\frac{.0066 z^{-1} + .0056 z^{-2} } {1 -2.5026 z^{-1} +2.1213 z^{-2} -.6065 z^{-3} }$$
Note we could also use b= [.0066 .0056], making the numerator .0066 + .0056z
-1, which would have the effect of advancing the time response by one sample.
Now we can use the Matlab filter function to calculate the impulse response (Figure 1):
Ts= 1/fs; N= 64; n= 0:N-1; t= n*Ts; x= [1, zeros(1,N-1)]; % impulse x= fs*x; % make impulse response amplitude independent of fs y= filter(b,a,x); plot(t,y,'.'),grid xlabel('seconds')
Let’s compare this discrete-time response to the exact impulse response from the inverse Laplace Transform. The inverse Laplace Transform of H(s) in equation 1 is [6]:
$$h(t)=[e^{-t} - \frac{2}{\sqrt{3}}e^{-t/2}cos(\frac{\sqrt{3}}{2}t + \frac{\pi}{6} )]u(t)$$
Figure 2 plots both responses -- the discrete-time result matches the continuous-time result exactly.
Figure 1. Impulse Response of discrete-time 3
rd
order Butterworth Filter
Figure 2. Impulse Response blue dots = discrete-time filter response
green = continuous-time filter response
We can also look at the step response:
x= ones(1,N); % step y= filter(b,a,x); % filter the step
The continuous-time step response is given by [6]:
$$h(t)=[1 - e^{-t} - \frac{2}{\sqrt{3}}e^{-t/2}sin(\frac{\sqrt{3}}{2}t )]u(t)$$
Figure 3 plots both responses. I had to advance the continuous-time response by T
s/2 to align it with the discrete-time response. The responses don’t exactly match, but are very close – the error is plotted in figure 4. The maximum error is only .0008.
Figure 3. Step Response
blue dots = discrete-time filter response
green = continuous-time filter response
Figure 4. Error of discrete-time step response
Now let’s look at the frequency response, and compare it to the continuous-time frequency response. We use Matlab function freqz for the discrete-time frequency response, and freqs for the continuous-time frequency response:
[h,f]= freqz(b,a,256,fs); % discrete-time freq response H= 20*log10(abs(h)); w= 2*pi*f; [hcont,f]= freqs(num,den,w); % continuous-time freq response Hcont= 20*log10(abs(hcont)); plot(w,H,w,Hcont),grid xlabel('rad/s'),ylabel('dB') axis([0 pi*fs, -80 10])
The results are plotted in Figure 5. The discrete-time response begins to depart from the continuous-time response at roughly f
s/4. To extend the accurate frequency range, we would need to increase the sample rate.
Why use the discrete-time frequency response? When modeling mixed analog/dsp systems, using the discrete-time response allows a single discrete-time model to represent the entire system. You just need to be aware of the valid frequency range of the model.
Figure 5. Discrete-time (blue) and continuous-time (green) frequency responses
note: f
s/2 = 2 Hz = 12.57 rad/s
We can compute the group delay of the filter using the Matlab function grpdelay:
[gd,f]= grpdelay(b,a,256,fs); % samples Group Delay D= gd/fs; % s Group Delay in seconds subplot(211),plot(w,H),grid ylabel('dB') axis([0 5 -50 10]) subplot(212),plot(w,D),grid axis([0 5, 0 3])
The frequency response and group delay are plotted in Figure 6. Note the group delay at 0 rad/s is 2 seconds. This is equal to the delay of the peak of the impulse response in Figure 1.
Finally, note that the impulse-invariant transform is not appropriate for high-pass responses, because of aliasing errors. To deal with this, a discrete-time low-pass filter could be added in cascade with the high-pass system to be modeled. Or we could use the bilinear transform – see Matlab function bilinear [7].
Figure 6. Frequency
Response and Group Delay of the discrete-time filter.
References
1. Oppenheim, Alan V. and Shafer, Ronald W.,
Discrete-Time Signal Processing, Prentice Hall, 1989, section 7.1.1
2. Lyons, Richard G.,
Understanding Digital Signal Processing, 2 nd Ed., Pearson, 2004, section 6.4
5.
Blinchikoff, Herman J. and Zverev, Anatol I.,
Filtering in the Time
and Frequency Domains
, Wiley, p. 110.
6. Blinchikoff and Zverev, p. 116.
% butter_3rd_order.m 6/4/17 nr % Starting with the butterworth transfer function in s, % Create discrete-time filter using the impulse invariance xform and compare % its time and frequency responses to those of the continuous time filter. % Filter fc = 1 rad/s = 0.159 Hz % I. Given H(s), find H(z) using the impulse-invariant transform fs= 4; % Hz sample frequency % 3rd order butterworth polynomial num= 1; den= [1 2 2 1]; [b,a]= impinvar(num,den,fs) % coeffs of H(z) %[b,a]= bilinear(num,den,fs) % II. Impulse Response and Step Response % find discrete-time impulse response Ts= 1/fs; N= 16*fs; n= 0:N-1; t= n*Ts; x= [1, zeros(1,N-1)]; % impulse x= fs*x; % make impulse response amplitude independent of fs y= filter(b,a,x); % filter the impulse plot(t,y,'.'),grid xlabel('seconds'),figure % Continuous-time Impulse response from inverse Laplace transform % Blinchikoff and Zverev, p116 h= exp(-t) - 2/sqrt(3)*exp(-t/2).*cos(sqrt(3)/2*t + pi/6); <p>e= h-y; % error of discrete-time response plot(t,y,'.',t,h),grid xlabel('seconds'),figure % find discrete-time step response x= ones(1,N); % step y= filter(b,a,x); % filter the step % Continuous-time step response. Blinchikoff and Zverev, p116 t= t+Ts/2; % offset t to to align step responses h= 1 - exp(-t) - 2/sqrt(3)*exp(-t/2).*sin(sqrt(3)/2*t); plot(t,y,'.',t,h),grid xlabel('seconds'),figure % III. Frequency Response and Group Delay % Find discrete-time and continuous time frequency responses [h,f]= freqz(b,a,256,fs); % discrete-time freq response H= 20*log10(abs(h)); w= 2*pi*f; [hcont,f]= freqs(num,den,w); % continuous-time freq response Hcont= 20*log10(abs(hcont)); plot(w,H,w,Hcont),grid xlabel('rad/s'),ylabel('dB') axis([0 pi*fs, -80 10]),figure % Find group delay [gd,f]= grpdelay(b,a,256,fs); % samples Group Delay D= gd/fs; % s Group Delay in seconds subplot(211),plot(w,H),grid ylabel('dB') axis([0 5 -50 10]) subplot(212),plot(w,D),grid axis([0 5, 0 3]) xlabel('rad/s'),ylabel('Group Delay (s)')
Neil Robertson June, 2017
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There's No End to It -- Matlab Code Plots Frequency Response above the Unit Circle
Neil, very nice blog!
Hi Rick,
Thanks for the encouragement.
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Gauge theories describe the connectivity of a space with small, symmetric extra dimensions
Start with an infinite cylinder (the direct product of a line and a small circle). The cylinder can be twisted. To avoid appealing to concepts that I'm trying to explain, I'll just say that the cylinder is made of wire mesh: evenly spaced circles soldered to wires running the length of it. The long wires can rotate as a unit, introducing an angular twist between each pair of adjacent circles. It's clear that any such configuration can be continuously deformed into any other: all such cylinders are equivalent from the perspective of the proverbial ant crawling on them.
Replace the line with a closed loop, so that the product is a torus (and think of the torus as a mesh doughnut, even though varying the plane of the small circles like that technically breaks the analogy). Any portion of the doughnut short of the whole thing can be deformed into the same portion of any other doughnut, but the doughnuts as a whole sometimes can't be, because the net twist around the doughnut can't be altered. The classes of equivalent doughnuts are completely characterized by this net twist, which is inherently nonlocal.
Replace the loop (not the small circle) with a manifold of two or more dimensions. It's true, though not obvious, that the physical part of the connection is completely given by the integrated twist around all closed loops (Wilson loops).
$A$ and $F$ quantify the connectivity
In the discrete case, the connection can be described most simply by giving the twist between adjacent circles. In the continuum limit, this becomes a "twist gradient" at each circle. This is $A_\mu$, the so-called vector potential.
Any continuous deformation can be described by a scalar field $\phi$ representing the amount that each circle is twisted (relative to wherever it was before). This alters $A_\mu$ by the gradient of $\phi$, but doesn't change any physical quantity (loop integral).
The description in terms of Wilson loops, $\oint_\gamma A \cdot \, \mathrm dx$, is more elegant because it includes only physically meaningful quantities, but it's nonlocal and highly redundant. If the space is simply connected, you can avoid the redundancy and nonlocality by specifying the twist only around differential loops, since larger loops can be built from them. The so-called field tensor, $\partial_\nu A_\mu - \partial_\mu A_\nu = F_{\mu\nu}$, gives you exactly that.
(If the space is not simply connected, you can still get away with the differential loops plus one net twist for each element of a generating set of the fundamental group. The torus was of course a simple example of this.)
The force comes from the Aharonov–Bohm effect
Consider a scalar field defined over the entire space (unlike the earlier fields, this one takes a value at each point on each circle). The field is zero everywhere except for two narrow beams which diverge from a point and reconverge somewhere else. (Maybe they're reflected by mirrors; maybe the space is positively curved; it doesn't matter.)
Unless the field is constant across the circles, the interference behavior of the beams will depend on the difference in the twist along the two paths. This difference is just the integral around the closed loop formed by the paths.
This is the (generalized) Aharonov–Bohm effect. If you restrict it to differentially differing paths and use $F_{\mu\nu}$ to calculate the effect on the interference, you get the electromagnetic force law.
You can decompose the field into Fourier components. The Fourier spectrum is discrete in the small dimension. The zeroth (constant) harmonic is not affected by the twisting. The second harmonic is affected twice as much as the first. These are the electric charges.
In reality, for unknown reasons, only certain extra-dimensional harmonics seem to exist. If only the first harmonic exists, there's an equivalent description of the field as a single complex amplitude+phase at each point of the large dimensions. The phase is relative to an arbitrary local zero point which is also used by the vector potential. When you compare the phase to the phase at a nearby point, and there is a vector-potential twist of $\mathrm d\theta$ between them, you need to adjust the field value by $i \, \mathrm d\theta$. This is the origin of the gauge covariant derivative.
Circles generalize to other shapes
If you replace the circles with 2-spheres, you get an $\mathrm{SU}(2)$ gauge theory. It is nastier numerically: the symmetry group is noncommutative, so you have to bring in the machinery of Lie algebra. Geometrically, though, nothing much has changed. The connectivity is still described by a net twist around loops.
One unfortunate difference is that the description of charge as extra-dimensional harmonics doesn't quite work any more. Spherical harmonics give you only the integer-spin representations, and all known particles are in the spin-0 or spin-½ representations of the standard model $\mathrm{SU}(2)$, so the particles that are affected by the $\mathrm{SU}(2)$ force at all can't be described this way. There may be a way to work around this problem with a more exotic type of field.
I have nothing insightful to say about the $\mathrm{SU}(3)$ part of the Standard Model gauge group except to point out that the whole SM gauge group can be embedded in $\mathrm{Spin}(10)$, and I think it's easier to visualize a 9-sphere than a shape with $\mathrm{SU}(3)$ symmetry.
General relativity is similar
In general relativity, the Riemann curvature tensor is analogous to the field tensor; it represents the angular rotation of a vector transported around a differential loop. The Aharonov-Bohm effect is analogous to the angular deficit around a cosmic string. Kaluza-Klein theory originally referred to a specific way of getting electromagnetism from general relativity in five dimensions; now it often refers to the broad idea that the Standard Model gauge forces and general relativity are likely to be different aspects of the same thing.
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After the answers by joshphysics and user37496, it seems to me that a last remark remains.
The quantum relevance of the universal covering Lie group in my opinion is (also) due to a fundamental theorem by Nelson. That theorem relates
Lie algebras of symmetric operators with unitary representations of a certain Lie group generated by those operators. The involved Lie group, in this discussion, is always a universal covering.
In quantum theories one often encounters a set of operators $\{A_i\}_{i=1,\ldots, N}$ on a common Hilbert space ${\cal H}$ such that:
(1) They are symmetric (i.e. defined on a dense domain $D(A_i)\subset {\cal H}$ where $\langle A\psi|\phi\rangle = \langle \psi|A\phi\rangle$)
and
(2) they enjoy the
commutation relations of some Lie algebra $\ell$:$$[A_i,A_j]= \sum_{k=1}^N iC^k_{ij}A_k$$on a common invariant domain ${\cal D}\subset {\cal H}$.
As is known, given an abstract Lie algebra $\ell$ there is (up to Lie group isomorphisms) a unique
simply connected Lie group ${\cal G}_\ell$ such that its Lie algebra coincide with $\ell$. ${\cal G}_\ell$ turns out to be the universal covering of all the other Lie groups whose Lie algebra is $\ell$ itself.
All those groups, in a neighbourhood of the identity are isomorphic to a corresponding neighbourhood of the identity of ${\cal G}_\ell$. (As an example just consider the simply connected $SU(2)$ that is the universal covering of $SO(3)$) so that they share the same Lie algebra and are locally identical and differences arise far from the neutral element.
If (1) and (2) hold, the natural question is:
Is there a strongly continuous unitary representation ${\cal G} \ni g \mapsto U_g$ of some Lie group $\cal G$ just admitting $\ell$ as its Lie algebra, such that $$U_{g_i(t)} = e^{-it \overline{A_i}}\:\: ?\qquad (3)$$
Where $t\mapsto g_i(t)$ is the one-parameter Lie subgroup of $\cal G$ generated by (the element $a_i$ of $\ell$ corresponding to) $A_i$ and $\overline{A_i}$ is some self-adjoint extension of $A_i$.
If it is the case, $\cal G$ is a continuous symmetry group for the considered physical system, the self adjoint opertors $\overline{A_i}$ represent physically relevant observables. If time evolution is included in the center of the group (i.e. the Hamiltonian is a linear combination of the $A_i$s and commutes with each of them) all these observables are
conserved quantities.Otherwise the situation is a bit more complicated, nevertheless one can define conserved quantities parametrically depending on time and belonging to the Lie algebra of the representation (think of the boost generator when $\cal G$ is $SL(2,\mathbb C)$).
Well, the fundamental theorem by Nelson has the following statement.
THEOREM (Nelson)
Consider a set of operators $\{A_i\}_{i=1,\ldots, N}$ on a common Hilbert space ${\cal H}$ satisfying (1) and (2) above. If ${\cal D}$ in (2) is a dense subspace such that the symmetric operator$$\Delta := \sum_{i=1}^N A_i^2$$is essentially self-adjoint on $\cal D$ (i.e. its adjoint is self-adjoint or, equivalently, $\Delta$ admits a unique self-adjoint extension, or equivalently its closure $\overline{\Delta}$ is self-adjoint), then:
(a) Every $A_i$ is essentially self-adjoint on $\cal D$,
and
(b) there exists a strongly continuous unitary representation on $\cal H$ of the unique simply connected Lie group ${\cal G}_\ell$ admitting $\ell$ as Lie algebra, completely defined by the requirements:$$U_{g_i(t)} = e^{-it \overline{A_i}}\:\:,$$ where $t\mapsto g_i(t)$ is the one-parameter Lie subgroup of ${\cal G}_\ell$ generated by (the element $a_i$ of $\ell$ corresponding to) $A_i$ and $\overline{A_i}$ is the unique self-adjoint extension of $A_i$ coinciding to $A_i^*$ and with the closure of $A_i$.
Notice that the representation is automatically unitary and not projective unitary: No annoying phases appear.
The simplest example is that of operators $J_x,J_y,J_z$. It is easy to prove that $J^2$ is essentially self adjoint on the set spanned by vectors $|j,m, n\rangle$. The point is that one gets this way unitary representations of $SU(2)$ and not $SO(3)$, since the former is the unique simply connected Lie group admitting the algebra of $J_k$ as its own Lie algebra.
As another application, consider $X$ and $P$ defined on ${\cal S}(\mathbb R)$ as usual. The three symmetric operators $I,X,P$ enjoy the Lie algebra of Weyl-Heisenberg Lie group. Moreover $\Delta = X^2+P^2 +I^2$ is essentially self adjoint on ${\cal S}(\mathbb R)$, because it admits a dense set of analytic vectors (the finite linear combinations of eigenstates of the standard harmonic oscillator). Thus these operators admit unique self-adjoint extensions and are generators of a unitary representation of the (simply connected) Weyl-Heisenberg Lie group. This example holds also replacing $L^2$ with another generic Hilbert space $\cal H$ and $X,P$ with operators verifying CCR on an dense invariant domain where $X^2+P^2$ (and thus also $X^2+P^2 +I^2$) is essentially self adjoint. It is possible to prove that the existence of the unitary rep of the Weyl-Heisenberg Lie group, if the space is irreducible, establishes the existence of a unitary operator from ${\cal H}$ to $L^2$ transforming $X$ and $P$ into the standard operators. Following this way one builds up an alternate proof of Stone-von Neumann's theorem.
As a last comment, I stress that usually ${\cal G}_\ell$ is This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user V. Moretti
not the group acting in the physical space and this fact may create some problem: Think of $SO(3)$ that is the group of rotations one would like to represent at quantum level, while he/she ends up with a unitary representation of $SU(2) \neq SO(3)$. Usually nothing too terrible arises this way, since the only consequence is the appearance of annoying phases as explained by Josh, and overall phases do not affect states. Nevertheless sometimes some disaster takes place: For instance, a physical system cannot assume quantum states that are coherent superpositions of both integer and semi-integer spin. Otherwise an internal phase would take place after a $2\pi$ rotation. What is done in these cases is just to forbid these unfortunate superpositions. This is one of the possible ways to realize superselection rules.
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Complete measure space Set
context $X $
definiendum $ \langle X,\Sigma,\mu\rangle $ … complete measure space over $X$
postulate $ \langle X,\Sigma,\mu\rangle $ … measure space
$\mu(N)=0$ $N'\subseteq N $
postulate $ N'\in\Sigma $ Discussion
In a complete measure space, subsets of null-sets can also be measured (and they then have zero measure as well). This notion is just introduced to prevent some pathologies.
Reference
Wikipedia: Complete measure
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Incorrect theorem: Suppose $A$ is a real matrix, if $\forall x \neq 0,\ x \in \Bbb R^n,\ x^TAx > 0$, then all eigenvalues $> 0$. False proof: If $Ax = \lambda x$, then $$ x^TAx = \lambda x^Tx = \lambda ||x||^2> 0$$
Since $||x||^2> 0$, so $\lambda > 0$ q.e.d.
So I'm pretty sure this isn't true, since we can take a rotation matrix as a counter-example.
However, I can't find the problem with this false proof.
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Let us now take a closer look at the
hex-fractal we sliced last week. Chopping a level 0, 1, 2, and 3 Menger sponge through our slanted plane gives the following:
This suggests an iterative recipe to generate the hex-fractal. Any time we see a hexagon, chop it into six smaller hexagons and six triangles as illustrated below. Similarly, any time we see a triangle, chop it into a hexagon and three triangles like this:
In the limit, each triangle and hexagon in the above image becomes a hex-fractal or a
tri-fractal, respectively. The final hex-fractal looks something like this (click for larger image):
Now we are in a position to answer last week’s question: how can we compute the
Hausdorff dimension of the hex-fractal? Let d be its dimension. Like last week, our computation will proceed by trying to compute the “ d-dimensional volume” of our shape. So, start with a “large” hex-fractal and tri-fractal, each of side-length 1, and let their d-dimensional volumes be h and t respectively. [1]
Break these into “small” hex-fractals and tri-fractals of side-length 1/3, so these have volumes \(h/3^d\) and \(t/3^d\) respectively (this is how “
d-dimenional stuff” scales). Since $$\begin{gather*}(\text{large hex}) = 6(\text{small hex})+6(\text{small tri}) \quad \text{and}\\ (\text{large tri}) = (\text{small hex})+3(\text{small tri}),\end{gather*}$$ we find that \(h=6h/3^d + 6t/3^d\) and \(t=h/3^d+3t/3^d\). Surprisingly, this is enough information to solve for the value of \(3^d\). [2] We find \(3^d = \frac{1}{2}(9+\sqrt{33})\), so $$d=\log_3\left(\frac{9+\sqrt{33}}{2}\right) = 1.8184\ldots,$$ as claimed last week.
As a final thought, why did we choose to slice the Menger sponge on
this plane? Why not any of the (infinitely many) others? Even if we only look at planes parallel to our chosen plane, a mesmerizing pattern emerges: More Information
It takes a bit more work to turn the above computation of the hex-fractal’s dimension into a full proof, but there are a few ways to do it. Possible methods include
mass distributions [3] or similarity graphs [4].
This diagonal slice through the Menger sponge has been proposed as an exhibit at the Museum of Math. Sebastien Perez Duarte seems to have been the first to slice a Menger sponge in this way (see his rendering), and his animated cross section inspired my animation above.
Thanks for reading!
Notes We’re assuming that the hex-fractal and tri-fractal have the same Hausdorff dimension. This is true, and it follows from the fact that a scaled version of each lives inside the other. [↩] There are actually two solutions, but the fact that hand tare both positive rules one out. [↩] Proposition 4.9 in: Kenneth Falconer. Fractal Geometry: Mathematical Foundations and Applications.John Wiley & Sons: New York, 1990. [↩] Section 6.6 in: Gerald Edgar. Measure, Topology, and Fractal Geometry(Second Edition). Springer: 2008. [↩]
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1. The perimeter of a equilateral triangle and regular hexagon are equal. Find out the ratio of their areas?a. 3:2
b. 2:3
c. 1:6
d. 6:1
Correct Option: bExplanation:
Let the side of the equilateral triangle = $a$ units and side of the regular hexagon is $b$ units.
Given that, $3a = 6b$ $ \Rightarrow \dfrac{a}{b} = \dfrac{2}{1}$
Now ratio of the areas of equilateral triangle and hexagon = $ \dfrac{{\sqrt 3 }}{4}{a^2} : \dfrac{{3\sqrt 3 }}{2}{b^2}$
$ \Rightarrow \dfrac{{\sqrt 3 }}{4}{\left( 2 \right)^2} : \dfrac{{3\sqrt 3 }}{2}{\left( 1 \right)^2}$
$ \Rightarrow 2:3$
2. What is the remainder of (32^31^301) when it is divided by 9?
a. 3
b. 5
c. 2
d. 1Correct option: b
Explanation:
See solved example 6 here
$\dfrac{{{{32}^{{{31}^{301}}}}}}{9}$ = $\dfrac{{{5^{{{31}^{301}}}}}}{9}$
Euler totient theorem says that ${\left[ {\dfrac{{{a^{\phi (n)}}}}{n}} \right]_{{\mathop{\rm Re}\nolimits} m}} = 1$
$\phi (n) = n\left( {1 - \dfrac{1}{a}} \right)\left( {1 - \dfrac{1}{b}} \right)...$ here $n = {a^p}.{b^q}...$
Now $\phi (9) = 9\left( {1 - \dfrac{1}{3}} \right) = 6$
Therefore, ${5^6}$ when divided by 9 remainder 1.
Now $\dfrac{{{{31}^{301}}}}{6} = {1^{301}} = 1$
So ${{{31}^{301}}}$ can be written as 6k + 1
$\Rightarrow {5^{{{31}^{301}}}} = {\left( {{5^6}} \right)^K}{.5^1}$
$\dfrac{{{5^{{{31}^{301}}}}}}{9} = \dfrac{{{{\left( {{5^6}} \right)}^K}{{.5}^1}}}{9} = \dfrac{{{1^K}.5}}{9} = 5$
3. Which of the following numbers must be added to 5678 to give a reminder 35 when divided by 460?a. 980
b. 797
c. 955
d. 618
Correct option: b
Explanation:
Let $x$ be the number to be added to 5678.
When you divide 5678 + $x$ by 460 the remainder = 35.
Therefore, 5678 + $x$ = 460k + 35 here $k$ is some quotient.
$ \Rightarrow $ 5643 + $x$ should exactly divisible by 460.
Now from the given options x = 797.
4. A girl entered a store and bought x flowers for y dollars (x and y are integers). When she was about to leave, the clerk said, “If you buy 10 more flowers I will give you all for $\$$2, and you will save 80 cents a dozen”. The values of x and y are:a. (15,1)
b. (10,1)
c. (5,1)
d. Cannot be determined from the given information.
Correct option: c
Explanation:
Given she bought $x$ flowers for $y$ dollars.
So 1 flower cost = $\dfrac{y}{x}$
12 flowers or 1 dozen cost = $\dfrac{{12y}}{x}$
Again, $x$+10 cost = 2 dollars
1 flower cost = $\dfrac{2}{{10 + x}}$
12 flowers or 1 dozen cost = $\dfrac{{2 \times 12}}{{10 + x}} = \dfrac{{24}}{{10 + x}}$
Given that this new dozen cost is 80 cents or 4/5 dollar less than original cost.
$ \Rightarrow \dfrac{{12y}}{x} - \dfrac{{24}}{{10 + x}} = \dfrac{4}{5}$
From the given options, c satisfies this.
5. If a number is divided by 357 the remainder is 5, what will be the remainder if the number is divided by 17?a. 9
b. 3
c. 5
d. 7
Correct option: c
Explanation:
Let $'N'$ be the given number.
$N = 357k + 5$ = $17 \times 21k + 5$
If this number is divided by 17 remainder is 5 as 357k is exactly divided by 17.
6. In how many possible ways can write 3240 as a product of 3 positive integers a,b and c.a. 450
b. 420
c. 350
d. 320
Correct option:
Explanation:
$3450 = {2^3} \times {3^4} \times {5^1} = a \times b \times c$
We have to distribute three 2's to a, b, c in ${}^{3 + 3 - 1}{C_{3 - 1}} = {}^5{C_2} = 10$ ways
We have to distribute four 3's to a, b, c in ${}^{3 + 4 - 1}{C_{3 - 1}} = {}^6{C_2} = 15$ ways
We have to distribute one 5 to a, b, c in 3 ways.
Total ways = $10 \times 15 \times 3 = 450$ ways.
7. On door A - It leads to freedomOn door B - It leads to Ghost houseOn door C - door B leads to Ghost houseThe statement written on one of the doors is wrong.Identify which door leads to freedom.a. A
b. B
c. C
d. None
Correct option: c
Explanation:
Case 1: A, B are true. In this case, Statement C also correct. So contradiction.
Case 2: B, C are true. In this case, B leads to ghost house and C confirms it. Now A is wrong. So door A does not lead to freedom. So Door C leads to freedom.
8. In the given figure, If the sum of the values along each side is equal. Find the possible values a, b, c, d, e, and f.
a. 9, 7, 20, 16, 6, 38
b. 4, 9, 10, 13, 16, 38
c. 4, 7, 20, 13, 6, 38
d. 4, 7, 20, 16, 6, 33
Correct option: c
Explanation:
From the above table, 42 + a + b = 47 + e. Therefore, a + b = 5 + e. Option a, b ruled out.
47 + e = 15 + f. Therefore, 32 + e = f. Option d ruled out.
4 men throw a die each simultaneously. Find the probability that at least 2 people get the same number
a. 5/18
b. 13/18
c. 1/36
d. 1/2
9. 70, 54, 45, 41……. What is the next number in the given series?a. 35
b. 36
c. 38
d. 40
Correct option: d
Explanation:
Consecutive squares are subtracted from the numbers.
70 - 54 = 16
54 - 45 = 9
45 - 41 = 4
So next we have to subtract 1. So answer = 41 - 1 = 40
10. How many positive integers less than 500 can be formed using the numbers 1,2,3,and 5 for digits, each digit being used only once.a. 52
b. 68
c. 66
d. 34
Correct option:
Explanation:
Single digit number = 4
Double digit number = 4$\times$3 = 12
Three digit numbers = 3$\times$3$\times$2= 18 ($\because$ If Hundred's place is 5, then the number is greater than 500)
Total = 34.
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Stochastic calculus is a branch of mathematics that operates on stochastic processes. It allows a consistent theory of integration to be defined for integrals of stochastic processes with respect to stochastic processes. It is used to model systems that behave randomly.
The best-known stochastic process to which stochastic calculus is applied is the Wiener process (named in honor of Norbert Wiener), which is used for modeling Brownian motion as described by Louis Bachelier in 1900 and by Albert Einstein in 1905 and other physical diffusion processes in space of particles subject to random forces. Since the 1970s, the Wiener process has been widely applied in financial mathematics and economics to model the evolution in time of stock prices and bond interest rates.
The main flavours of stochastic calculus are the Itō calculus and its variational relative the Malliavin calculus. For technical reasons the Itō integral is the most useful for general classes of processes but the related Stratonovich integral is frequently useful in problem formulation (particularly in engineering disciplines.) The Stratonovich integral can readily be expressed in terms of the Itō integral. The main benefit of the Stratonovich integral is that it obeys the usual chain rule and does therefore not require Itō's lemma. This enables problems to be expressed in a coordinate system invariant form, which is invaluable when developing stochastic calculus on manifolds other than
R . The dominated convergence theorem does not hold for the Stratonovich integral, consequently it is very difficult to prove results without re-expressing the integrals in Itō form. n
Contents Itō integral 1 Stratonovich integral 2 Applications 3 References 4 Itō integral
The Itō integral is central to the study of stochastic calculus. The integral \int H\,dX is defined for a semimartingale
X and locally bounded predictable process H. Stratonovich integral
The Stratonovich integral of a semimartingale X against another semimartingale
Y can be defined in terms of the Itō integral as \int_0^t X_{s-} \circ d Y_s : = \int_0^t X_{s-} d Y_s + \frac{1}{2} \left [ X, Y\right]_t^c,
where [
X, Y] t denotes the quadratic covariation of the continuous parts of c X and Y. The alternative notation \int_0^t X_s \, \partial Y_s
is also used to denote the Stratonovich integral.
Applications
An important application of stochastic calculus is in quantitative finance, in which asset prices are often assumed to follow stochastic differential equations. In the Black–Scholes model, prices are assumed to follow the geometric Brownian motion.
References
Fima C Klebaner, 2012, Introduction to Stochastic Calculus with Application (3rd Edition). World Scientific Publishing, ISBN 9781848168312 Szabados, T. S.; Székely, B. Z. (2008). "Stochastic Integration Based on Simple, Symmetric Random Walks". Journal of Theoretical Probability 22: 203. Preprint
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Encouraged by Hilmar, I've decided to update the answer with all the steps necessary to calculate the Reverberation Time from a scratch. Presumably, it will be useful for others interested in this area. Obviously, it is the simplest approach because more advanced are definitely beyond a scope.
In the beginning, you must obtain the impulse response of a room. It can be done in various ways:
Firing a starter gun, popping a balloon, etc. - basically recording any impulsive-like signal with broad frequency content and omnidirectional characteristic. This is the simplest method of obtaining impulse response. Sweep Sine measurement with a usage of either linear or exponential sweep. This method is my favorite as it allows you to extract many different parameters of your system at the same time. MLS (Maximum Length Sequence) measurement with this kind of noise and using Hadamard transform to obtain the impulse response. Some comparison with sweep sine technique can be found here.
After getting the impulse response $h(t)$ of your acoustic space you are able to calculate the reverberation time and other speech related parameters such as $C_{80}$, $STI$, $D_{50}$ (Clarity, Speech Transmission Index, Definition), etc.
Very first step is to filter your impulse response in an appropriate frequency band. This is due to fact that reverberation time is a function of position in the room and frequency. Obviously, we assume that sound field is diffused and at each position in a room decay rate is the same. Thus we must still consider dependency on the frequency band. We are using 1/1 octave or 1/3 octave filters. Its desired parameters are defined by
standard, preferably you always wish to use the filter of class 0. For most of the time, Butterworth 3rd order filter is enough (although for frequencies below 125 Hz you might experience some problems with characteristic not meeting specifications). I did the whole implementation on my own with few tweaks, but for usual applications widely used EN 61260 MATLAB implementation is good enough.
Next step after filtering the $h(t)$ and obtaining $h_f(t)$, is to make the response as smooth as possible before conversion to the logarithmic scale. For that purpose, Hilbert Transform is a widely used tool. The goal is to create the analytic signal:
$$h_A(t)=h(t)+j\tilde{h}(t) $$
where $\tilde{h}(t) $ is the HT of your filtered impulse response and $j$ is a complex unit. Now, knowing that your analytic signal is complex, two things are coming from its representation:
$$h_A(t)=A(t)e^{j \psi (t)} $$
where $A(t)$ is the envelope of your signal and $\psi (t)$ is an instantaneous phase. Obviously, we are interested mainly in the envelope (magnitude of the analytic signal). Below you can see overlayed filtered response and its envelope:
One clue for people speaking MATLAB'ish. The
hilbert function returns the analytic signal, so no need of multiplying by $j$ and adding original $h(t)$. Just use
abs(hilbert(h)) to get the envelope.
Now we have the smoothed version of our impulse response, but it must be smoothed further. It is optional and one might consider it unnecessary. It is based on Moving Average filter of appropriate length $M$. Everyone knows the moving average filter, it is basically a low-pass FIR filter with frequency response given by:
$$H(f)=\frac{\sin (\pi f M)}{M\sin (\pi f)} $$
Plots for different lengths $M$ are shown below:
And on the following figure you can observe the effect of smoothing of the previously calculated energy curve ($A(t)$ in logarithmic scale becomes $E(t)$) for two different values of $M$. For sampling frequency of $48\mathtt{kHz}$ I've used $M=5001$.
So now we have the Energy Curve (not a Power) which is simply the smoothed Hilbert envelope in logarithmic scale. Conversion to decibel scale is done by equation:
$$E(t) = 20\log_{10} \frac{A(t)}{\max A(t)} $$
Curve $A(t)$ can be further smoothed for calculations by using the
Schroeder Integration method of your envelope (also known as inversed time integration). This method gives you maximally flat decay curve and is very easy to implement.
$$L(t)=10 \log_{10} \left[ \dfrac{\int_t^{\infty}h^2(\tau )d\tau}{\int_0^{\infty}h^2(\tau )d\tau} \right] $$
Again for MATLAB people:
L(td:-1:1)=(cumsum(hA(td:-1:1))/sum(hA(1:td)));
Mind that limit of integration (
td) is equal to $\infty$. It is true when we do not have any environmental noise. Otherwise, you will see that decaying sound 'dives into noise level'. It is depicted below. The red curve is based on the correct limit, where decay crosses noise floor. If you choose your limit of integration to be too short, then as in green curve, the estimate is not long enough, so you will get under-estimation while doing linear interpolation. On the contrary, the orange curve has too large integration limit and you will get over-estimated. You always want to find the correct limit. If you didn't obtain curve in such way, then please do so.
There are many different methods for estimation of integration limit in Schroeder integral, I think that easiest to implement is the one proposed by
Lundeby et al.
Regarding calculation of RT itself. It must be done by performing linear interpolation of your decay curve (or should I say Schroeder curve) with linear function: $L=A\cdot t+B$ on the correct range (which is described below). When it's done, you calculate the reverberation time from the equation. You can see that point of intercept doesn't really matter (per se), you care about the gradient of your line.
$$RT = \dfrac{-60}{A} $$
where $A$ is a slope coefficient from interpolated line (in dB/s). While you have that, you can also calculate the correlation coefficient of your linear fit (or r-value), that is telling you how good is the fit.
Finally - limits. You cannot really calculate $T_{60}$ having $60 \mathtt{dB}$ of dynamic range as you are trying to do - you always need more (unless you really know what you are doing). So you are doing linear fit on decay curve at following dynamic ranges, and the result is being extrapolated, according to
ISO 3382-2:
$EDT$ (Early Decay Time): upper limit is $0 \mathtt{dB}$ and lower is $-10 \mathtt{dB}$. This parameter correlates well with perceived reverberation time. In practice, though beginning for the sake of algorithms, people are using an interval of $-1 \mathtt{dB}$ and $-10 \mathtt{dB}$ (i.e. in Norsonic analyzers).
$T_{10}$: upper limit must start at $-5 \mathtt{dB}$ to remove any fluctuations and then lower limit is taken to be $-15 \mathtt{dB}$, but it always must be at least $10 \mathtt{dB}$ above the noise floor. So in fact you need at least $25 \mathtt{dB}$ of dynamic range (or INR) to be able to calculate $T_{10}$ ($5+10+10$).
$T_{20}$: upper limit at $-5 \mathtt{dB}$, lower at $-25 \mathtt{dB}$. Minimum dynamic range needed is $35\mathtt{dB}$
$T_{30}$: upper limit of $-5 \mathtt{dB}$, lower at $-35 \mathtt{dB}$, with minimum $45 \mathtt{dB}$ of dynamic range.
Why are people using these values? Well, because you very rarely have $75 \mathtt{dB}$ of dynamic range to be able to estimate $T_{60}$. For very smooth decay these values should be equal. Some more detailed analysis of effects can be found in this publication:
Measuring Room Impulse Responses: Impact of the Decay Range on
Derived Room Acoustic
Parameters
If you will not perform Schroeder integration, then most probably they will diverge a lot. For the very nasty room you can get something like:
In theory: $EDT$ = $T_{10}$ = $T_{20}$ = $T_{30}$. Although they are not necessarily going to be. For example, if your small room is coupled with large one (let's think of small chapel coupled with cathedral by a small doorway), then you will get very sharp decay for smaller one at the beginning of the curve, and very long tail in the end:
To wrap up. According to standards, you should always estimate your reverberation time $T_{N}$ starting from $-5 \mathtt{dB}$ and down to $(-5-N) \ \mathtt{dB}$. Also, ensure that you are at least $10 \mathtt{dB}$ above the noise floor. If not, then you must use another estimate. In your case, I suggest using the $T_{30}$.
Below you have example of $T_{20}$ estimation for a troublesome impulse response. $E(t)$ is the energy decay curve obtained by taking the logarithm of impulse response Hilbert envelope. $L(t)$ is the Schroeder integral. Correlation coefficient between linear fit and $E(t)$ on a given range ($-5 \mathtt{dB}$ - $-25 \mathtt{dB}$) for this case is: $cc=-0.9987$. Calculated RT is $T_{20}=1.15 \mathtt{s}$
I think that should do. If anyone will follow these steps, then he should get an accuracy of $0.1 \mathtt{s} $ against widely used
Dirac. Anyway, it must be remembered that even different software tend to give different results (i.e. EASERA and Dirac), so you should be totally fine. For more complicated impulse responses with artifacts at the end of impulse response one might get worse performance, but anyway, such measurements are not reliable and reflect wrongly conducted an experiment.
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I've been using
Mathematica for a while now, and there are multiple common tasks that I end up doing. I'd like to learn how to convert them into functions so that I can store them in some file and load them at startup.
Most of these common tasks involve taking a set of equations and doing something with them. Thing is, I don't want to fix the variables the equations are written in, and would like behavior similar to
NDSolve where one can specify the functions as well as which variable is which.
For example, take this task:
Pd = ParametricNDSolve[{v'[t] == (v[t] - v[t]^3/3 - ω[t])/0.2, ω'[t] == (1 + v[t])/0.2, ω[0] == b, v[0] == a}, {v[t], ω[t]}, {t, 0, 20}, {a, b}];PVec[x_, y_] := {D[(v[t] /. Pd) [x, y], t] /. t -> 0, D[(ω[t] /. Pd) [x, y], t] /. t -> 0};VectorPlot[PVec[x, y], {x, -2, 1}, {y, -1.2, 0.5}]
Here, I'm parametrically solving the pair of equations
$$ \begin{align} v'(t)&=\frac{-\frac{1}{3} v(t)^3+v(t)-\omega (t)}{0.2}\\ \omega'(t)&=\frac{v(t)+1}{0.2} \end{align} $$
with the parameters being the initial state of $v$ and $\omega$. Now, I take the vector field of their derivatives at $t=0$, and vector plot it. Basically, I want to get a quick map of how the initial trajectories ($(v'(t),\omega'(t))$) change for different initial conditions without having to open
EquationTrekker and do things manually.
So now, I want to take this task and write it as a function akin to
TrajectoryMap[{eqn1,eqn2},{{var1,varmin,varmax},{var2,varmin,varmax}},{{timevar,timepoint},tmin,tmax,tpoint}] where this particular code would run as
TrajectoryMap[{v'[t] == (v[t] - v[t]^3/3 - ω[t])/0.2, ω'[t] == (1 + v[t])/0.2}, {{v[t],-2,1}, {ω[t],-1.2,0.5}}, {{t,0},0,20}]
(or something similar)
Here,
timepoint is the value of
t for which the derivative of the interpolating function is calculated.
I can tell that this will probably need some combination of
Holds and
Evaluates, but I can't seem to get the right one.
How should I go about converting this code (or any arbitrary code of a similar type) into a custom function?
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B̈oni, P.; Mook, H. A.; Martínez, J. L.; Shirane, G.
Title:
Comparison of the Paramagnetic Spin Fluctuations in Nickel with Asymptotic Renormalization-Group Theory
Abstract:
The paramagnetic spin fluctuations in Ni have been investigated at small momentum q and energy transfer $\Elzxh\omega$ by means of inelastic-neutron-scattering techniques. For fixed energy transferred from the neutron to the spin system [S(q,$\omega$=const)], peaks at finite momentum are observed. Their positions are very sensitive to the low-intensity parts of the magnetic cross section. The results are compared with an analytical expression for the dynamical correlation function for an isotropic ferromagnet introduced by Iro. The agreement with the theory is good near Tc for excitation energies 1 meV $\leq\Elzxh\omega{}<$kBTc$\simeq$50 meV. Further away from Tc serious discrepancies occur, mainly because the spin fluctuations slow down more quickly than expected on the basis of mode-mode coupling or renormalization-group theory. If the experimentally determined scaling function is used, then the results can be well parametrized. The additional slowing down, which affects also the line shape at Tc, is most likely caused by the interaction of the 3d moments with the conduction electrons, i.e., by an itinerant effect. We have also extracted from our data the asymptotic behavior of the scattering function at very large energy transfers, i.e., SI(q,$\omegåightarrow훜fty$)$\propto\omega$-(z+4)/z at Tc and the results are in reasonable agreement with renormalization-group theory. No indications for a breakdown of dynamical scaling has been found within the q and E ranges investigated.
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Difference between revisions of "Mean field models"
(I can't believe there was not an entry about this... work in progress)
(→Mean field solution of the Ising model)
Line 3: Line 3:
==Mean field solution of the Ising model==
==Mean field solution of the Ising model==
−
A well-known mean field solution of the [[Ising model]] goes as follows. From the original
+
A well-known mean field solution of the [[Ising model]]goes as follows.
−
:<math>
+
From the original ,
+
:<math> U = - \sum_iS_i \S_j , </math>
suppose we may approximate
suppose we may approximate
−
:<math> \
+
:<math> \S_j \approx \bar{s}, </math>
−
where <math>
+
where <math></math> is the number of neighbors of site <math>i</math> (e.g. 4 in a 2-D lattice), and <math>\bar{s}</math> is the (unknown) magnetization:
:<math> \bar{s}=\frac{1}{N} \sum_i S_i . </math>
:<math> \bar{s}=\frac{1}{N} \sum_i S_i . </math>
+ + + + + + + + + + + + + + + Revision as of 16:17, 3 May 2010
A
mean field model, or a mean field solution of a model, is an approximation to the actual solution of a model in statistical physics. The model is made exactly solvable by treating the effect of all other particles on a given one as a mean field (hence its name). It appear in different forms and different contexts, but all mean field models have this feature in common. Mean field solution of the Ising model
A well-known mean field solution of the Ising model, known as the
Bragg-Williams approximation goes as follows.From the original Hamiltonian,
suppose we may approximate
where is the number of neighbors of site (e.g. 4 in a 2-D square lattice), and is the (unknown) magnetization:
Therefore, the Hamiltonian turns to
as in the regular Langevin theory of magnetism: the spins are independent, but coupled to a constant field of strength
The magnetization of the Langevin theory is
Therefore:
This is a
self-consistent expression for . There exists a critical temperature, defined by
At temperatures higher than this value the only solution is . Below it, however, this solution becomes unstable (it corresponds to a maximum in energy), whereas two others are stable. Slightly below ,
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I'm working through example 3.2 of Zangwill's Modern Electrodynamics and have come across a change in integration variables that I just can't seem to get. The example has two different change distributions related by $ \rho_b(\lambda \mathbf{r}) = \rho_a(\mathbf{r})$ and asks for the relationship between the electric potentials. According to the book, $$ \int d^3 r'= \frac{1}{\lambda^2} \int d^3 (\lambda r'). $$ When I try to work through the Jacobian, I get $$ d^3 r' = \begin{vmatrix} \frac{1}{\lambda} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{vmatrix} d^3(\lambda r') = \frac{1}{\lambda} d^3(\lambda r'),$$ where I'm assuming changing from $ r'$ to $ \lambda r'$ is like going from $\vec{x}$ to $\vec{y}$ where $$ y_1 = \lambda x_1, \hskip{1cm} y_2 = x_2\hskip{0.5cm} \text{and} \hskip{0.5cm} y_3 = x_3 . $$ What am I missing here?
closed as off-topic by Aaron Stevens, stafusa, Kyle Kanos, Jon Custer, user191954 Feb 16 at 9:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
"Homework-like questions should ask about a specific physics conceptand show some effortto work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Aaron Stevens, stafusa, Kyle Kanos, Jon Custer, Community
If $$ \boldsymbol r = \begin{vmatrix} x \\ y \\ z \\ \end{vmatrix} = \begin{vmatrix} r \sin \theta \cos \phi \\ r \sin \theta \sin \phi \\ r \cos \theta \\ \end{vmatrix} $$ the associated volume element is $$ d^3 \boldsymbol r = \begin{vmatrix} r \sin \theta \cos \phi & r \cos \theta \cos \phi & - r \sin \theta \sin \phi \\ r \sin \theta \sin \phi & r \cos \theta \sin \phi & r \sin \theta \cos \phi \\ r \cos \theta & - r \sin \theta & 0 \\ \end{vmatrix} dr d \theta d \phi = r^2 \sin \theta dr d \theta d \phi $$ Similarly, if $$ \lambda \boldsymbol r = \begin{vmatrix} \lambda x \\ \lambda y \\ \lambda z \\ \end{vmatrix} = \begin{vmatrix} \lambda r \sin \theta \cos \phi \\ \lambda r \sin \theta \sin \phi \\ \lambda r \cos \theta \\ \end{vmatrix} $$ the associated volume element is $$ d^3 ( \lambda \boldsymbol r ) = \begin{vmatrix} \lambda r \sin \theta \cos \phi & \lambda r \cos \theta \cos \phi & - \lambda r \sin \theta \sin \phi \\ \lambda r \sin \theta \sin \phi & \lambda r \cos \theta \sin \phi & \lambda r \sin \theta \cos \phi \\ \lambda r \cos \theta & - \lambda r \sin \theta & 0 \\ \end{vmatrix} dr d \theta d \phi = ( \lambda r)^2 \sin \theta dr d \theta d \phi = \lambda^2 ( r^2 \sin \theta dr d \theta d \phi ) $$ So, comparing the two expressions above for $d^3 \boldsymbol r$ and $d^3 ( \lambda \boldsymbol r )$, you get $$ d^3 ( \lambda \boldsymbol r ) = \lambda^2 d^3 \boldsymbol r $$ or $$ d^3 r = \frac{1}{\lambda^2} d^3 ( \lambda \boldsymbol r ) $$
I think you forget pieces of the integral that gives the potential :
${{d}^{3}}(\lambda \overrightarrow{r'})={{\lambda }^{3}}{{d}^{3}}(\overrightarrow{r'})$ as indicated in the comment.
But : $\int{{{d}^{3}}(\overrightarrow{r'})\frac{\rho (\lambda \overrightarrow{r'})}{\left| \overrightarrow{r}-\overrightarrow{r'} \right|}}=\frac{1}{{{\lambda }^{2}}}\int{{{d}^{3}}(\lambda \overrightarrow{r'})\frac{\rho (\lambda \overrightarrow{r'})}{\left| \lambda \overrightarrow{r}-\lambda \overrightarrow{r'} \right|}}$
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I don't see why Gauss' Law holds for volumes that contain no source or sink. I am trying to understand Gauss' Law as a general effect of any vector field, so I would appreciate if any answers do not centrally focus on electrostatics.
For an arbitrary 3-dimensional region $R$ placed in a vector field
F, the divergence theorem states that the flux integrated across the surface of the volume $R$ is equal to the divergence of the vector field F integrated over the entire volume of $R$:
$$\begin{equation} \oint_S \textbf{F}\cdot\hat{\textbf{n}}\>dS = \int_V \nabla\cdot\textbf{F}\>dV \end{equation}$$
This much makes sense to me, intuitively and mathematically. But, Gauss' Law then states that if the region $R$ does
not contain a source or sink, then the left-hand side of the equation above is zero (and hence the right hand side is zero as well).
But this doesn't make sense to me. For example, consider a rectangular region $\bar{R}$ in a two dimensional vector field
k such that
$$\textbf{k} = x\hat{x}$$
All boundaries of $\bar{R}$ are parallel to either the $x$ or $y$ axis. I will refer to the "left" and "right" sides of the region $\bar{R}$ as those parallel to the $y$ axis, and the "top" and "bottom" sides as those parallel to the $x$ axis.
The magnitude of
k is increasing in the $+x$ direction, so any rectangle $\bar{R}$ drawn in the $+x$ region will have a larger positive flux on its right side, as compared to a smaller negative flux on its left side, and zero flux on the top and bottom.
This region contains no sources or sinks (nor does any region at all in this field), but I don't see how the divergence integrated across the area of $\bar{R}$ could be zero, or how the value of
k integrated across the boundary of the shape could be zero either. It seems that in either case it should be positive.
As a physical example, I don't understand how a volume hovering above the Earth's surface could possibly have zero net gravitational flux (which is implied by Gauss' law, since the volume does not contain the source of the field - the Earth)
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Improved Three Bin Exact Frequency Formula for a Pure Real Tone in a DFT
Introduction
This is an article to hopefully give a better understanding of the Discrete Fourier Transform (DFT) by extending the exact two bin formulas for the frequency of a real tone in a DFT to the three bin case. This article is a direct extension of my prior article "Two Bin Exact Frequency Formulas for a Pure Real Tone in a DFT"[1]. The formulas derived in the previous article are also presented in this article in the computational order, rather than the indirect order they were derived in.
Martin Vicanek also has a three bin formula, which is an extension of his two bin formula. It used to be in an earlier version of his two bin formula paper[2], but it seems to be removed in his latest version as of the date of this blog article. His three bin formula is used as a comparison below, but not presented.
These formulas represent a significant improvement over my original three bin formulas[3] which is the discovery I made that prompted my study of DSP and more specifically the DFT.
A System of Equations In Vector Form
The system of equations derived in the two bin case are extended to the three bin case. Instead of two arbitrary bins, $j$ and $k$, three consecutive bins are used, $k-1$, $k$, and $k+1$. Three arbitrary bins could have been used, and the equations can be easily formulated as such, but in practice the three bins surrounding the peak should be used.
The complex bin values are split into real and imaginary parts: $$ Z_k = x_k + i \cdot y_k \tag {1} $$ The same applies for $k-1$ and $k+1$.
The following vectors then need to be constructed using the bin values. Vector $\vec C$ does not depend on the bin values. $$ \vec A = \begin{bmatrix} ( x_k - x_{k-1} ) / \sqrt{2} \\ ( x_k - x_{k+1} ) / \sqrt{2} \\ y_{k-1}\\ y_k \\ y_{k+1} \end{bmatrix} \tag {2} $$ $$ \vec B = \begin{bmatrix} [ \cos( \beta_k ) \cdot x_k - \cos( \beta_{k-1} ) \cdot x_{k-1} ] / \sqrt{2} \\ [ \cos( \beta_k ) \cdot x_k - \cos( \beta_{k+1} ) \cdot x_{k+1} ] / \sqrt{2} \\ \cos( \beta_{k-1} ) \cdot y_{k-1} \\ \cos( \beta_k ) \cdot y_k \\ \cos( \beta_{k+1} ) \cdot y_{k+1} \end{bmatrix} \tag {3} $$ $$ \vec C = \begin{bmatrix} [ \cos( \beta_k ) - \cos( \beta_{k-1} ) ] / \sqrt{2} \\ [ \cos( \beta_k ) - \cos( \beta_{k+1} ) ] / \sqrt{2} \\ \sin( \beta_{k-1} ) \\ \sin( \beta_k ) \\ \sin( \beta_{k+1} ) \end{bmatrix} \tag {4} $$ Where $$ \beta_k = k \cdot \frac{ 2\pi }{ N } \tag {5} $$ The same $\sqrt{2}$ adjustment done in the two bin case is made to the equations derived from the difference of two source equations to make the variance of the difference of the random variables have the same value as the variance of a single variable. Results with, and without, this adjustment are shown below.
Solving for the Frequency Term
Once the formulas are in vector form, the three bin case is identical to the two bin case. Here are the formulas derived in the two bin case presented in computational order.
First, a unit vector is made from $\vec C$. This calculation can be done ahead of time. $$ \vec P_{k} = \frac{ \vec C }{ \| \vec C \| } = \frac{ \vec C }{ \sqrt{ \vec C\cdot \vec C } } \tag {6} $$ Second, a compromise vector in the general direction of $\vec A$ and $\vec B$ is produced. This is the simplest version. Either $\vec A$ or $\vec B$, or any linear combination (other than zeroing) will give close to the same results because the two vectors are nearly collinear and its length doesn't matter. $$ \vec D = \vec A + \vec B \tag {7} $$ Third, the projection of the compromise vector onto the plane orthogonal to $\vec C$ is found. $$ \vec K = \vec D - ( \vec D \cdot \vec P_{k} ) \vec P_{k} \tag {8} $$ Finally the actual frequency, in units of cycles per frame, can be calculated. $$ f = \cos^{-1} \left( \frac{ \vec K \cdot \vec B }{ \vec K \cdot \vec A } \right) \cdot \frac{ N }{ 2\pi } \tag {9} $$
Some Numerical Results
These tests consist of taking a sweep of frequencies from the low end of a bin to the high end. For each frequency a real pure tone is generated. The phase is swept from zero to $2\pi$. The formulas are evaluated using the same data sets. The Unadjusted column does not have the $\sqrt{2}$ rescaling.
The numbers that are shown are the average and standard deviation of the distribution of frequency errors. The Target Noise Level is the standard deviation of the added noise. The error levels are shown multiplied by 100 for better presentation.
The sample count is 100 and the run size is 4000 Errors are shown at 100x actual value Target Noise Level = 0.100 Freq Original Vicanek Unadjusted Improved ---- ------------- ------------- ------------- ------------- 4.0 -0.008 1.019 -0.014 1.036 -0.011 1.108 -0.013 1.042 4.1 0.009 0.998 -0.006 1.025 -0.002 1.052 -0.004 0.990 4.2 -0.019 1.017 -0.037 1.007 -0.032 1.002 -0.034 0.944 4.3 0.025 1.102 -0.008 0.996 -0.010 0.963 -0.016 0.906 4.4 0.011 1.231 -0.020 0.995 -0.007 0.944 -0.004 0.892 4.5 0.004 1.394 -0.048 0.990 -0.044 0.947 -0.047 0.900 4.6 0.031 1.484 0.042 0.979 0.056 1.011 0.053 0.942 4.7 0.005 1.321 0.012 1.009 0.018 1.070 0.017 0.997 4.8 -0.010 1.146 -0.003 1.003 -0.002 1.090 -0.000 1.018 4.9 0.016 1.057 0.019 1.015 0.022 1.090 0.021 1.029
Similar to the two bin case, the new three bin formulas do best when the frequency is between two bins. In constrast, my original formula does best near the bins and worse between the two bins. This is because the original formula has a fixed weighting of the three bins and the new formulas are essentially weighted by the magnitude of the bins. So when the frequency is close to the center, the original formula weighs the third bin, which has a lower signal value and thus a higher SNR, too heavily.
Conclusion
Another solid formula for the calculation of the frequency of a single pure tone from DFT bin values has been presented. Like the two bin case, there is no guarantee that it is the optimal one.
References
[1] Dawg, Cedron, Two Bin Exact Frequency Formulas for a Pure Real Tone in a DFT
[2] Vicanek, Martin, Frequency Estimation from two DFT Bins
[3] Dawg, Cedron, Exact Frequency Formula for a Pure Real Tone in a DFT
Previous post by Cedron Dawg:
Two Bin Exact Frequency Formulas for a Pure Real Tone in a DFT
Next post by Cedron Dawg:
An Alternative Form of the Pure Real Tone DFT Bin Value Formula
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Main question: I've been reading about communication games a lot, and I'm wondering if there are good criteria to select between two separating-ish equilibria. I think of a separating equilibria as coordination equilibria among types. So, if we grant that these types successfully coordinate, why wouldn't we grant that they coordinate to a sender-optimal (in a Pareto efficient among senders sense) equilibrium? That is, suppose there is a single sequential equilibrium where all senders do strictly better than in the remaining equilibria. What arguments are there for selecting this equilibrium?
Consider the following communication game. Receiver payoffs are the second number in the pair. There are six types of senders, with payoffs given as the first element of the pairs. I will show there is a pooling equilibrium and at least two partial separations. I'm wondering what kind of techniques can be used to argue in favor of either separating equilibrium. One is sender-optimal and the other is receiver-optimal.
$$\begin{array}{l*{6}{c}r} & Action \;B & Action \;L & Action\;R& Action\:LL & Action\;RR & \\ \hline type\;B & (0,3) & (1,2) & (1,2) & (2,1)& (2,1) \\ type\;L & (0,2) & (1,3) & (1,2) & (2,0) & (2,2.25) \\ type\;R & (0,2) & (1,2) & (1,3) & (2,2.25) & (2,0) \\ type\;LL & (0,1) & (1,2) & (1,0) & (2,3) & (2,1) \\ type\;RR & (0,1) & (1,0) & (1,2) & (2,1) & (2,3) \\ type\;H & (0,0) & (1,0.9) & (1,0.9) & (2,3.1) & (2,3.1) \\ \end{array}$$
Let their be a prior distribution on types $\pi$ where $$\pi(B)=.3,\pi(L)=\pi(R)=.2, \pi(LL)=\pi(RR)=.1, \pi(H)=.1.$$
In a pooling equilibrium, the receiver will take action $B$ for expected payoff $EU_2(B) = .3(3) + .4(2) + .2(1)=1.9$, edging out $EU_2(L)= .3(2) + .2(3) + .2(2) + .1(2) + .1(.9)=1.89$.
However, there are partially separating equilibria.
Separation 1Let types $L,LL$ "ask" for action $L$, types $R$ and $RR$ "ask" for $R$ and then $B$ and $H$ mix 50/50 between the two signals. Let the messages be $l$ and $r$ with the natural interpretation.
So $EU_2(L\mid l)\Pr(l)=.15(2) + .2(3) + .1(2) + .025(1)=1.125=EU_2(R\mid r)\Pr(r)$
So the receiver earns $2.25$ in expectation. The senders are better off too.
Separation 2But let's consider another kind of separation. Types $R$ and $LL$ always send a message $ll$, "asking" for action $LL$. Types $L$ and $RR$ send $rr$, asking for action $RR$. Again, $B$ and $H$ randomize evenly.
Then, $EU_2(RR\mid rr)\Pr(rr)= .15(1) + .2(2.25) + .1(3) + .025(3.1) = .9775=EU_2(LL\mid ll)\Pr(ll).$ The expected payoff is 1.955 because each message is received half the time.
Responding to $rr$ with action $R$ and $ll$ with $L$ yields a lower payoff of, so the separation, being jumbled with types $L$ and $RR$ pooling, isn't useful for taking the "correct" actions $L$ or $R$ as the receiver would like.
It seems to me that this last equilibrium is more robust. There are two separating equilibrium, which require coordination. Granting that senders can coordinate, why wouldn't they coordinate in the sender-optimal way?
I'm wondering if any methods exist that would refine the equilibrium set to exclude the receiver-optimal separation. The first pooling equilibrium might be said not to be neologism proof.
Neologism proofness is defined in section 3 of this paper. Roughly, there must not be an additional (off path) message such, that if observed, the receiver could form beliefs and a rational strategy based on those beliefs such that all who sent the message are strictly better off relative to the proposed equilibrium and those who didn't weakly prefer the proposed equilibrium outcome. I'm guessing that won't work here, because you have to consider
two neologisms ($ll$ and $rr$) at once to eliminate separation 1, which requires collusion essentially. But are there any other ideas?
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I believe there are some big misunderstandings in the question...
First: usually error terms are , additive not multiplicative.
There are the so-called
multiplicative error models (MEM), but I don't think it's the case here, because usually in MEM the multiplicative error is time-dependent (i.e., $\epsilon_{t} \in [0, \infty)$), while in the OP's question the error term is withdrawn from a fixed interval (i.e, $\epsilon \in [0.5, 1.5]$). A good introduction to MEM can be found here.
Second: This said, it seems that we cannot find for this "model" because it is actually possible to find confidence intervals (sorry if this term doesn't exist, maybe I've just created it...). certainty intervals
Consider, for instance, the original model:
$$f(x)=\dfrac{1}{e^{(x-2)^2}} + \dfrac{1}{e^{(x+2)^2}}\epsilon,$$s.t. $\epsilon \in [0.5, 1.5]$.
If we make a small change to the original model it will be easier to understand how to draw the "certainty intervals". Consider the following modification:
$$f(x)=\dfrac{1}{e^{(x-2)^2}}\epsilon_{1} + \dfrac{1}{e^{(x+2)^2}}\epsilon_{2}.$$
If we set $\epsilon_{1} \in [1, 1]$ and $\epsilon_{2} \in [0.5, 1.5]$ we have the original model.
Now we can show that the
are dependent upon the certainty intervals ($\epsilon_{1}$ and $\epsilon_{2}$) error terms . intervals
Manipulate[
Module[{f, g, h, y},
{f[y_] := error1*Exp[-(y - 2)^2],
g[y_] := (error1 + error2)/2*Exp[-(y - 2)^2],
h[y_] := error2*Exp[-(y - 2)^2],
G1 = Plot[{f[y], g[y], h[y]}, {y, -4, 4}, Filling -> {1 -> {3}},
PlotRange -> {{-4, 4}, {0, 2}}]}];
Module[{f, g, h, y}, {f[y_] := error3*Exp[-(y + 2)^2],
g[y_] := (error3 + error4)/2*Exp[-(y + 2)^2],
h[y_] := error4*Exp[-(y + 2)^2],
G2 = Plot[{f[y], g[y], h[y]}, {y, -4, 4}, Filling -> {1 -> {3}},
PlotRange -> {{-4, 4}, {0, 2}}]}];
sim = Table[{x,Exp[-(x - 2)^2]*RandomReal[{error1, error2}] +
Exp[-(x + 2)^2]*RandomReal[{error3, error4}]},
{x,RandomReal[{-4, 4}, Points]}];
G3 = ListPlot[sim, PlotStyle -> Thick];
Show[If[bands, {G3, G2, G1}, G3], PlotRange -> {{-4, 4}, {0, 2.2}}],
{{Points, 500}, 100, 1500, Appearance -> "Labeled"},
{{error1, 1}, 0, 1, Appearance -> "Labeled"},
{{error2, 1}, 1, 2, Appearance -> "Labeled"},
{{error3, 0.5}, 0, 1, Appearance -> "Labeled"},
{{error4, 1.5}, 1, 2, Appearance -> "Labeled"},
Button["new sim", {Clear@sim, sim}, ImageSize -> 100],
{{bands, False, Style["Show \"confidence\" bands?", Bold, Red,
FontSize -> 16]}, {True, False}}]
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In general relativity and for its Einstein-Hilbert action, we usually ask that the metric variations $\delta g_{\mu \nu}$ cancel on the boundary $\partial \, \Omega$ of some region $\Omega$ of the spacetime manifold ($\delta g_{\mu \nu} = 0$). But if the variations are of
compact support, why can't we also ask that the derivatives cancel at the boundary ; $\partial_{\lambda} \, \delta g_{\mu \nu} = 0$ on $\partial \, \Omega$ ?
Why is this too restrictive, or too abusive ? Why would this be
too easy ? (I know about the surface term problem in General Relativity)
Under the variational principle, the variations are arbitrary and don't have to obey the equations of motion (the variations are not
on shell). Under the variational method, only $g_{\mu \nu}$ and $\partial_{\lambda} \, g_{\mu \nu}$ are supposed to obey some equation of motion (the Einstein equation in this case).
There's something important that I do not understand here, and I need to clear this up !
EDIT 1 :
Maybe the question is badly formulated and something is missing. The variational method is usually (?) used to find the
classical equations of motion (no quantum mechanics here). In this case, the variational method is just a recipe to find the equations. The variation $\delta g_{\mu \nu}$ are then fully arbitrary but are of compact support, and all its derivatives cancels too on the boundary. Is that right ?
If that is the case, then we don't need to care much about the surface terms during some arbitrary variations, and we don't need to add any surface counter-term to the action (aka
Gibbons-Hawking-York surface integral). This is what I need to clear up.
Maybe I'm simply confusing this classical recipe with some other uses of the variational principle ?
EDIT 2 :
Under the action integral variation, is the perturbed field $\phi'(x) = \phi(x) + \delta \phi(x)$ "on-shell" or "off-shell" ?
As far as I know, the
perturbed field is "off-shell" ; it doesn't necessarily satisfy the field equation, contrary to the unperturbed $\phi(x)$ (which is "on-shell", by definition). Is that right ?
It is possible that there's another interpretation of the
stationary action principle (to be confirmed) ; the variation $\delta \phi(x)$ may correspond to a , that the real field should satisfy. In this case, I guess that the perturbed field $\phi'(x)$ should also satisfy the field equation (the perturbation stay "on-shell"). modification of the boundary conditions Someone confirm this interpretation ?
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Per cent is derived from the Latin word ‘per centum’ meaning ‘per hundred’. It is represented by the symbol % and means 100. If we say 1% that means 1 out of 100 or one part of a complete quantity. In chapter 7 of ICSE Class 8 Maths, you will learn all the concepts related to Percent and Percentage. Also, you will solve the slightly advanced problems related to the application of percentages in the exercise.
We have provided the ICSE Class 8 Maths Selina Solutions Chapter 7 Percent and Percentage in PDF format to help you in your studies. These questions are solved by the experts at BYJU’S and covers the detailed explanation. In case you have missed the class then refer to this pdf to know the solution. To download the pdf visit the link below;
ICSE Class 8 Maths Chapter 7 Percent and Percentage has only one exercise i.e 7 (A) consisting of a total of 15 questions. The answers to all these questions have also been shown below;
ICSE Class 8 Maths Selina Solutions Chapter 7 Percent and Percentage – Exercise 7 (A) Question 1 Evaluate:
(i) 55% of 160 + 24% of 50 – 36% of 150
Solution:
Equating them in the following form \( =\frac{55 \times 160}{100}+\frac{24 \times 50}{100}-\frac{36 \times 150}{100} \) \( =11 \times 8+12-18 \times 3=88+12-54=46\)
(ii) 9.3% of 500 – 4.8% of 250 – 2.5% of 240
Solution:
Equating them in the following form\( =\frac{9.3 \times 500}{100}-\frac{4.8 \times 250}{100}-\frac{2.5 \times 240}{100}\) \( =9.3 \times 5-1.2 \times 10-0.5 \times 12\)
=46.5-12-6=46.5-18=28.5
Question 2.
(i) A number is increased from 125 to 150; find the percentage increase.
Solution:
Original value = 125
New value = 150
Increase = (150-125) =25
Increase \( \%=\frac{25}{125} \times 100=20 \% \)
(ii) A number is decreased from 125 to 100; find the percentage decrease.
Solution:
Original value = 125,
New value=100
Decrease =(125-100)=25
Decrease \( \%=\frac{25}{125} \times 100=20 \%\)
Question 3.
Find:
(i) 45 is what percent of 54?
Solution:
et 45=x percent of \( 54=\frac{54 \times x}{100} \) \( \Rightarrow x=\frac{45 \times 100}{54}=\frac{5 \times 100}{6}\) \( =\frac{250}{3}=83 \frac{1}{3} \%\)
∴ Required percentage\( =83 \frac{1}{3} \%\)
Find (ii) 2.7 is what percent of 18?
Solution:
Let 2.7=x percent of 18\( =\frac{18 \times x}{100}\) \( ∴ x=\frac{2.7 \times 100}{18}=\frac{270}{18}=\frac{30}{2}=15\)
∴ Required percentage =15%
Question 4.
(i) 252 is 35% of a certain number, find the number.
Solution:
(i) Let the number be x
By the given condition\( 252=\frac{x \times 35}{100}=\frac{x \times 7}{20}\) \( ∴ \quad x=\frac{252 \times 20}{7}=36 \times 20=720\)
Hence, the required number = 720
(ii) If 14% of a number is 315; find the number.
Solution:
Let the number be x
By the given condition\( 315=\frac{x \times 14}{100}\) \( ∴ x=\frac{315 \times 100}{14}=\frac{45 \times 100}{2}=45 \times 50=2250\)
Hence the required number = 2250.
Question 5.
Find the percentage change, when a number is changed from:
(i) 80 to 100
Solution:
Original number = 80
New number = 100,
Change = (100 – 80) = 20
∴ Percentage change (increase)\( =\frac{20}{80} \times 100=25 \%\)
(ii) 100 to 80
Solution:
Original number = 100
New number = 80
Change (100 – 80) = 20
∴ Percentage change (decrease) \( =\frac{20}{100} \times 100=20 \%\)
(iii) 6.25 to 7.50
Solution:
Original number = 6.25
New number = 7.50
Change (increase) = (7.50 – 6.25) = 1.25
∴ Increase \( =\frac{1.25}{6.25} \times 100=20 \% \)
Question 6.
An auctioneer charges 8% for selling a house. If a house is sold for Rs.2,30,500; find the charges of the auctioneer.
Solution:
Selling price of the house = Rs.2,30,500
Rate of charges of the auctioneer = 8% of selling price
∴ Charges of the auctioneer = 8% of 2,30,500,\( =\frac{8}{100} \times 2,30,500=R s .18,440\)
Question 7.
Out of 800 oranges, 50 are rotten. Find the percentage of good oranges.
Solution:
Rotten oranges = 50
Number of good oranges = 800 – 50 = 750
Percentage of good oranges \( =\frac{750}{800} \times 100\) \( =\frac{750}{8}=\frac{375}{4}=93 \frac{3}{4} \%\)
Question 8.
A cistern contains 5 thousand liters of water. If 6% water is leaked. Find how many liters of water are left in the cistern.
Solution:
Water in the cistern = 5000 liters
Quantity of water leaked \( =\frac{6}{100} \times 500=300 liters\)
Quantity of water left in the cistern
= (5000 – 300) liters = 4700 liters
Question 9.
A man spends 87% of his salary. If he saves Rs. 325; find his salary.
Solution:
Let salary =Rs x
∴ Expenditure \( =\frac{87}{100} \text { of } x\) \( =\mathrm{Rs} \cdot \frac{87 \mathrm{x}}{100}\)
Saving =Rs.325\( ∴ x-\frac{87 x}{100}=325\) \( \frac{100 x-87 x}{100}=325 \Rightarrow \frac{13 x}{100}=325\) \( x=\frac{325 \times 100}{13} \Rightarrow x=\frac{32500}{13}\)
x=2500
∴ Salary =Rs.2500
Question 10.
(i) A number 3.625 is wrongly read as 3.265; find the percentage error.
Solution:
Correct number =3.625
Number wrongly read as =3.265
Error =3.625-3.265=0.360\( \% \text { Error }=\frac{0.360}{3 \cdot 625} \times 100\) \( =\frac{360}{3625} \times 100=\frac{36000}{3625}=9.93 \%\)
(ii) A number \( 5.78 \times 10^{3}\) is wrongly written as \( 5.87 \times 10^{3}\) , find the percentage e
Solution:
Correct number \( =5.78 \times 10^{3} \)
Number wrongly written as \( =5.87 \times 10^{3}\)
Error\( =5.87 \times 10^{3}-5.78 \times 10^{3} \) \( =0.09 \times 10^{3}\) \( \% \text { Error }=\frac{0.09 \times 10^{3}}{5 \cdot 78 \times 10^{3}} \times 100\) \( =\frac{0.09}{5 \cdot 78} \times 100=\frac{9}{578} \times 100=\frac{900}{578} \%\)
=1.56%
Question 11.
In an election between two candidates, one candidate secured 58% of the votes polled and won the election by 18,336 votes.Find the total number of votes polled and the votes secured by each candidate.
Solution:
Since, winning candidate secured 58% of the votes polled.
∴ Losing candidate secured = (100-58)% of the votes polled = 42% of the votes polled
Difference of votes =58-42 =16% of the votes polled
We are given:
16% of votes polled = 18,336\( \frac{16}{100} \) of votes polled = 18,336
⇒ Votes polled \( =18,336 \times \frac{100}{16}\)
⇒ Votes polled \( =\frac{18,33,600}{16}\)
⇒ Votes polled =1,14,600
∴ Votes secured by winning candidate\( =\frac{58}{100} \times 1,14,600=66,468\)
Votes secured by losing candidate\( =\frac{42}{100} \times 1,14,600=48,132\)
Votes polled =1,14,600
Votes secured by winning candidate =66,468
Votes secured by losing candidate =48,132
Question 12.
In an election between two candidates one candidate secured 47% of votes polled and lost the election by 12,366 votes. Find the total votes and die votes secured by the winning candidate.
Solution:
Since, the losing candidate secured 47% of the votes polled
Winning candidate secures votes = (100-47)% of the votes polled
= 53 % of the votes polled
Difference of votes =53-47 =6 % of the votes polled
We are given:
6% of the votes polled =12,366\( \frac{6}{100} of the votes polled =12,366 \)
Votes polled \( =12,366 \times \frac{100}{6}=\frac{1236600}{6}=2,06,100\)
Votes secured by winning candidate\( =\frac{53}{100} \times 2,06,100=1,09,233\)
∴ Votes polled =2,06,100
Votes secured by winning candidate = 1,09,233
Question 13.
The cost of a scooter depreciates every year by 15% of its value at the beginning of the year. If the present cost of the scooter is 8,000; find its cost:
(i) After one year
(ii) After 2 years
Solution:
Present cost of scooter = Rs.8000
The cost of scooter depreciates by 15% every year
(i) Cost of scooter after one year\( =\frac{(100-15)}{100} \times 8000=\frac{85}{100} \times 8000=R s .6800\)
(ii) Cost of scooter after 2 year\( =\frac{(100-15)}{100} \times 6800=\frac{85}{100} \times 6800=\text { Rs. } 5780\)
Question 14.
In an examination, the pass mark is 40%. If a candidate gets 65 marks and fails by 3 marks; find the maximum marks.
Solution:
Marks obtained by the candidate =65
Fails by = 3 marks
Pass marks = 65 + 3 = 68
% of Pass marks = 40 %
∴ Required maximum marks \( =\frac{100}{40} \times 68\) \(=10 \times 17\)
=170
Question 15.
In an examination, a candidate secured 125 marks and failed by 15 marks. If the pass percentage was 35%. Find the maximum marks.
Solution:
Total marks secured = 125
Failed by 15 marks
∴Pass marks =125+15=140
Let maximum marks = x\( ∴ \frac{x \times 35}{100}=140\) \( \Rightarrow \quad x=\frac{140 \times 100}{35}=4 \times 100=400\)
Hence maximum marks = 400
To get the Selina Solutions for all the chapters of class 8 Maths, visit the ICSE class 8 Maths Selina Solutions page.
ICSE Class 8 Maths Selina Solutions Chapter 7 – Percent and Percentage
When a result is released, you must have heard people saying that the topper had scored 580 marks out of 600 and thus scored 96.6% in the board exam. Also, your parents may have compared your marks with your classmates. Have you thought about how they have calculated the percentage and then finally came to the conclusion that whether you have performed well or your classmate. Well, the answers to all these questions you will find in this chapter. So, go through the theory part of this chapter and then start to solve the exercise questions.
Students can access the ICSE Class 8 Selina Solutions for all questions from Maths, Physics, Chemistry and Biology subjects at one place.
We hope this information on “ICSE Class 8 Maths Selina Solutions Chapter 7 Percent and Percentage” is useful for students. Keep learning and stay tuned for further updates on ICSE and other competitive exams. To access interactive Maths and Science Videos download BYJU’S App and subscribe to YouTube Channel.
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Stupid Question: If I'm given a non-euclidean space, such as $S^n $ (n-dimensional unit sphere), with the the uniform measure $\mu$ on it.
I'm also given a function $f:S^n \to \mathbb{R} $ that is bounded, by a constant $C$ . As an analogous to the Lebesgue measure case, can I say that: $ \int_{S^n } f d\mu \leq C \mu ( S^n) = C $?
Hope you'll be able to help
Thanks !
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Getting Started¶
MathJax allows you to include mathematics in your web pages, either using TeX and LaTeX notation, or as MathML. To use MathJax, you will need to do the following things:
Obtain a copy of MathJax and make it available on your server. Configure MathJax to suit the needs of your site. Link MathJax into the web pages that are to include mathematics. Put mathematics into your web pages so that MathJax can display it.
Each of these steps is described briefly below, with links to more detailed explanations. This page gives the quickest and easiest ways to get MathJax up and running on your web site, but you may want to read the details in order to customize the setup for your pages.
Obtaining and Installing MathJax¶
The easiest way to set up MathJax is to obtain the
MathJax-v1.0.1.zip archive from the MathJax download page. This includes both the MathJax codeand the MathJax webfonts, so this is the only file you need. (This isdifferent from the beta releases, which had the fonts separate fromthe rest of the code).
Unpack the
MathJax-v1.0.1.zip archive and place theresulting MathJax folder onto your web server at a convenientlocation where you can include it into your web pages. For example,making
MathJax a top-level directory on your server would beone natural way to do this. That would let you refer to the mainMathJax file via the URL
/MathJax/MathJax.js from within any pageon your server.
Note: While this is the easiest way to set up MathJax initially, there is a better way to do it if you want to be able to keep your copy of MathJax up-to-date easily. That uses the Git version control system, and is described in the Installing MathJax document. If you prefer using Subversion, we also maintain an SVN mirror (see Installing MathJax via SVN).
Once you have MathJax set up on your server, you can test it using thefiles in the
MathJax/test directory. Load them in your browserusing its web address rather than opening them locally (i.e., use an
http:// URL rather than a
file:// URL). When you view the
index.html file, after a few moments you should see a message thatMathJax appears to be working. If not, check that the files have beentransferred to the server completely and that the permissions allowthe server to access the files and folders that are part of theMathJax directory. (Be sure to verify the MathJax folder’s permissionsas well.) Check the server log files for any errors that pertain tothe MathJax installation; this may help locate problems in thepermission or locations of files.
Configuring MathJax¶
When you include MathJax into your web pages as described below, itwill load the file
config/MathJax.js (i.e., the file named
MathJax.js in the
config folder of the main
MathJaxfolder). This file contains the configuration parameters thatcontrol how MathJax operates. There are comments in it thatexplain each of the parameters, and you can edit the file to suityour needs.
The default settings are appropriate for pages that use TeX as theinput language, but you might still want to adjust some settings; forexample, you might want to include some additional extensions such asthe
AMSmath and
AMSsymbols extensions. The comments in thefile should help you do this, but more detailed instructions areincluded in the Configuring MathJax document.There are also ways to configure MathJax other than by using the
config/MathJax.js file; these are descibed on that page as well.
Putting mathematics in a web page¶
To put mathematics in your web page, you can use either TeX and LaTeX notation, or MathML notation (or both); the configuration file tells MathJax which you want to use, and how you plan to indicate the mathematics when you are using TeX notation. The following sections tell you how to use each of these formats.
TeX and LaTeX input¶
To process mathematics that is written in TeX or LaTeXformat, include
"input/TeX" in your configuration’s jax array,and add
"tex2jax.js" to the extensions array so that MathJaxwill look for TeX-style math delimiters to identify the mathematics onthe page.
extensions: ["tex2math.js"],jax: ["input/TeX", "output/HTML-CSS"]
Note that the default math delimiters are
$$...$$ and
\[...\]for displayed mathematics, and
\(...\) for in-line mathematics.In particular, the
$...$ in-line delimiters are
not used bydefault. That is because dollar signs appear too often innon-mathematical settings, which could cause some text to be treatedas mathematics unexpectedly. For example, with single-dollardelimiters, ”... the cost is $2.50 for the first one, and $2.00 foreach additional one ...” would cause the phrase “2.50 for the firstone, and” to be treated as mathematics since it falls between dollarsigns. For this reason, if you want to use single-dollars for in-linemath mode, you must enable that explicitly in your configuration:
tex2jax: {inlineMath: [['$','$'], ['\\(','\\)']]}
See the
config/MathJax.js file, or the tex2jax configurationoptions page, for additional configurationparameters that you can specify for the
tex2jax preprocessor.
Here is a complete sample page containing TeX mathematics (whichassumes that
config/MathJax.js is configured as described above):
<html><head><title>MathJax TeX Test Page</title><script type="text/javascript" src="/MathJax/MathJax.js"></script></head><body>When \(a \ne 0\), there are two solutions to \(ax^2 + bx + c = 0\) and they are$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$</body></html>
There are a number of extensions for the TeX input processor that you might want to add to the extensions array. These include:
TeX/AMSmath.js, which defines the AMS math environments and macros, TeX/AMSsymbols.js, which defines the macros for the symbols in the msam10 and msbm10 fonts, TeX/noErrors.js, which shows the original TeX code rather than an error message when there is a problem processing the TeX, and TeX/noUndefined.js, which prevents undefined macros from producing an error message, and instead shows the macro name in red.
For example,
extensions: ["tex2math.js","TeX/noErrors.js","TeX/noUndefined.js", "TeX/AMSmath.js","TeX/AMSsymbols.js"]
loads all four extensions, in addition to the
tex2mathpreprocessor.
MathML input¶
To process mathematics written in MathML, include
"input/MathML" in your configuration’s jax array, and add
"mml2jax.js" to the extensions array so that MathJax willlocate the
<math> elements in the page automatically.
extensions: ["mml2jax.js"],jax: ["input/MathML", "output/HTML-CSS"]
With this configuration, you would mark your mathematics usingstandard
<math> tags, where
<math display="block"> representsdisplayed mathematics and
<math display="inline"> or just
<math> represents in-line mathematics.
Note that this will work in HTML files, not just XHTML files (MathJaxworks with both), and that the web page need not be served with anyspecial MIME-type. Also note that, unless you are using XHTML ratherthan HTML, you should not include a namespace prefix for your
<math> tags; for example, you should not use
<m:math> exceptin a file where you have tied the
m namespace to the MathML DTD.
Here is a complete sample page containing MathML mathematics (whichassumes that
config/MathJax.js is configured as described above):
<html><head><title>MathJax MathML Test Page</title><script type="text/javascript" src="/MathJax/MathJax.js"></script></head><body>When <math><mi>a</mi><mo>≠</mo><mn>0</mn></math>,there are two solutions to <math> <mi>a</mi><msup><mi>x</mi><mn>2</mn></msup> <mo>+</mo> <mi>b</mi><mi>x</mi> <mo>+</mo> <mi>c</mi> <mo>=</mo> <mn>0</mn></math> and they are<math mode="display"> <mi>x</mi> <mo>=</mo> <mrow> <mfrac> <mrow> <mo>−</mo> <mi>b</mi> <mo>±</mo> <msqrt> <msup><mi>b</mi><mn>2</mn></msup> <mo>−</mo> <mn>4</mn><mi>a</mi><mi>c</mi> </msqrt> </mrow> <mrow> <mn>2</mn><mi>a</mi> </mrow> </mfrac> </mrow> <mtext>.</mtext></math></body></html>
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Transvection A transvection is a linear mapping $f$ of a (right) vector space $V$ over a skew-field $K$ with the properties
$$\def\rk{\textrm{rk}\;}\rk(f-E) = 1\quad\textrm{and}\quad \textrm{Im}(f - E)\subseteq \ker(f-E),$$ where $E$ is the identity linear transformation. A transvection can be represented in the form
$$\def\a{\alpha} f(x) = x+a\a(x),$$ where $a\in V$, $\a\in V^*$ and $\a(a) = 0$.
The transvections of a vector space $V$ generate the special linear, or unimodular, group $\def\SL{\textrm{SL}}\SL(V)$. It coincides with the commutator subgroup of $\def\GL{\textrm{GL}}\GL(V)$, with the exception of the cases when $\dim V = 1$ or $\dim V = 2$ and $K$ is the field of two elements. If $K$ is a field, then $\SL(V)$ is the group of matrices with determinant 1. In the general case (provided that $\dim V \ne 1$), $\SL(V)$ is the kernel of the epimorphism
$$\GL(V) \to K^*/[K^*,K^*],$$ which is called the Dieudonné determinant (cf. Determinant).
In the projective space $\def\P{\mathbb{P}}\P(V)$, whose points are the $1$-dimensional subspaces of $V$, a transvection $f$ as above induces a (projective) transvection with $aK$ as centre and $\ker(f-E)$ as axis. If one takes $\ker(f-E)$ to be a hyperplane at infinity in $\P(V)$, such a transvection induces a translation $x\mapsto x+b$ in the remaining affine space (interpreted as a linear space). See also Shear.
References
[Di] J.A. Dieudonné, "La géométrie des groupes classiques", Springer (1955) MR0072144 Zbl 0067.26104 How to Cite This Entry:
Transvection.
Encyclopedia of Mathematics.URL: http://www.encyclopediaofmath.org/index.php?title=Transvection&oldid=21541
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Let us now take a closer look at the
hex-fractal we sliced last week. Chopping a level 0, 1, 2, and 3 Menger sponge through our slanted plane gives the following:
This suggests an iterative recipe to generate the hex-fractal. Any time we see a hexagon, chop it into six smaller hexagons and six triangles as illustrated below. Similarly, any time we see a triangle, chop it into a hexagon and three triangles like this:
In the limit, each triangle and hexagon in the above image becomes a hex-fractal or a
tri-fractal, respectively. The final hex-fractal looks something like this (click for larger image):
Now we are in a position to answer last week’s question: how can we compute the
Hausdorff dimension of the hex-fractal? Let d be its dimension. Like last week, our computation will proceed by trying to compute the “ d-dimensional volume” of our shape. So, start with a “large” hex-fractal and tri-fractal, each of side-length 1, and let their d-dimensional volumes be h and t respectively. [1]
Break these into “small” hex-fractals and tri-fractals of side-length 1/3, so these have volumes \(h/3^d\) and \(t/3^d\) respectively (this is how “
d-dimenional stuff” scales). Since $$\begin{gather*}(\text{large hex}) = 6(\text{small hex})+6(\text{small tri}) \quad \text{and}\\ (\text{large tri}) = (\text{small hex})+3(\text{small tri}),\end{gather*}$$ we find that \(h=6h/3^d + 6t/3^d\) and \(t=h/3^d+3t/3^d\). Surprisingly, this is enough information to solve for the value of \(3^d\). [2] We find \(3^d = \frac{1}{2}(9+\sqrt{33})\), so $$d=\log_3\left(\frac{9+\sqrt{33}}{2}\right) = 1.8184\ldots,$$ as claimed last week.
As a final thought, why did we choose to slice the Menger sponge on
this plane? Why not any of the (infinitely many) others? Even if we only look at planes parallel to our chosen plane, a mesmerizing pattern emerges: More Information
It takes a bit more work to turn the above computation of the hex-fractal’s dimension into a full proof, but there are a few ways to do it. Possible methods include
mass distributions [3] or similarity graphs [4].
This diagonal slice through the Menger sponge has been proposed as an exhibit at the Museum of Math. Sebastien Perez Duarte seems to have been the first to slice a Menger sponge in this way (see his rendering), and his animated cross section inspired my animation above.
Thanks for reading!
Notes We’re assuming that the hex-fractal and tri-fractal have the same Hausdorff dimension. This is true, and it follows from the fact that a scaled version of each lives inside the other. [↩] There are actually two solutions, but the fact that hand tare both positive rules one out. [↩] Proposition 4.9 in: Kenneth Falconer. Fractal Geometry: Mathematical Foundations and Applications.John Wiley & Sons: New York, 1990. [↩] Section 6.6 in: Gerald Edgar. Measure, Topology, and Fractal Geometry(Second Edition). Springer: 2008. [↩]
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First, the critical dimension. There are many ways (seemingly inequivalent ways but ultimately bound to give the same result) to calculate $D=10$ for the superstring that mirror the methods to calculate $D=26$ for the bosonic string.
For the bosonic string, one may use a conformally invariant world sheet theory. Because of the residual conformal symmetry, it has to have $bc$ ghosts. The central charge of the $bc$ system is $c=1-3k^2$ where $k=2J-1$ where $J$ is the dimension of the $b$ antighost, in this case $J=2$. You see that my formulae imply $k=3$ and $c=1-27=-26$ so one has to add 26 bosons, i.e. 26 dimensions of spacetime, to get $c=0$ in total.
Now, for the superstring, the local symmetries on the world sheet are enhanced from the ordinary conformal group to the $N=1$ superconformal group. One needs to add the $\beta\gamma$ (bosonic) ghosts for the new (fermionic) generators. Their dimension is $J=3/2$, different from $J=2$ of $bc$ by $1/2$, as usual for the spin difference of things related by supersymmetry. You see that $k=2J-1=2$ and $3k^2-1=12-1=11$. Now, the central charge of $\beta\gamma$ is $3k^2-1$ and not $1-3k^2$, the sign is the opposite one, because they are bosons.
So the $bc$ and $\beta\gamma$ have $c=-26+11=-15$. This minus fifteen must be compensated by 10 bosonic fields and 10 fermionic fields (whose $c=1/2$ per dimension: note that a fermion is half a boson) and $10+10/2=15$ so that the total $c=0$. If some of the steps aren't understandable above, it's almost certainly because the reader isn't familiar with basics of conformal field theory and it is not possible to explain conformal field theory without conformal field theory. It's a whole subject, not something that should be written as one answer on this server.
In this formalism with the new world sheet fermions $\psi^\mu$ transforming as spacetime vectors, one has to protect the spin-statistics relationship. Vector-like fermions violate it so they are only allowed in pairs. This is achieved by the GSO projection – well, there are actually two GSO projections, one separate for left-movers and one for right-movers. Only 1/4 of the states are kept in the spectrum. The projection is a flip side of having four sectors – the left-moving and right-moving fermions may independently be periodic or antiperiodic. I wrote about the GSO projection a month ago:
http://motls.blogspot.com/2012/11/david-ian-olive-1937-2012.html?m=1
Again, if anything is incomprehensible and incomplete, it's because it's not really one isolated insight that a layman may understand from one sentence. It's one of many technical results that follows from a large subject – string theory – that has to be systematically studied if one wants to understand it.This post imported from StackExchange Physics at 2014-03-12 15:28 (UCT), posted by SE-user Luboš Motl
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A classical question on signals and systems, for which I provide you one method of solution.
Now given an LTI system, and its input $x[n]$ and output $y[n]$, the fundamental approach to determine the impulse response $h[n]$ involves the utilisation of a frequency domain approach to assert $H(e^{jw}) = Y(e^{jw})/X(e^{jw})$
In this problem however, I would choose a different approach which involves more of an analysis by inspection rather than pure mathematical operations, because of the fact that the input $x[n]$ involves a cosine term whose DTFT is includes impulses and will be problematic in the division process.
The method proceeds by drawing the DTFTs of input and output:The input $x[n]$ involves two terms one of which is $$5 \sin(0.4 \pi n)/ \pi n$$ whose DTFT is a zero phase ideal lowpass filter of cutoff frequency $W_c = 0.4 \pi$ with a gain of 5.
And the second term of input is $$10 \cos (0.5 \pi n)$$ whose DTFT is modeled as two impulses at $$ 5 \pi \delta(\omega + 0.5 \pi) + 5 \pi \delta(\omega - 0.5 \pi) $$ Because of linearity property of DTFT, these two DTFT are added to represent the DTFT of the input $x[n]$.
Then, look at the DTFT of the output $$y[n] = \frac {10 \sin(0.3 \pi (n-10) )}{\pi (n-10)} $$ which is again an ideal low pass filter of gain = 10 and $W_c = 0.3 \pi$ with a phase term of $e^{-10 \omega}$ hence $Y(e^{j\omega}) =10 e^{-j10 \omega}$ when $|\omega| < 0.3 \pi$ and zero elsewhere.
Now we want to find a frequency response $H(e^{j\omega})$ such that when it is multiplied by the DTFT $X(e^{j\omega})$ of the input $x[n]$, it should produce the DTFT $Y(e^{j\omega})$ of the output $y[n]$.
The observation of the input and output DTFTs yields the conclusion that one such $H(e^{j \omega})$ is a linear phase ideal lowpass filter of gain = 2, $W_c = 0.3\pi$ and phase term of $e^{-j10 \omega}$, whose time domain correspondence is found by inspection as $$h[n] = \frac {2 \sin(0.3 \pi (n-10))}{\pi (n-10)}$$
The justification of this solution can be made by observing the effect of this filter on its input: it will completely supress any signal above the frequency $\omega = 0.3 \pi$ and it will multiply any signal below that frequency by 2, while adding a phase shift of -10 radians which reflects itself as a group delay of 10 samples in (n-10) term.
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I will give two explanations for why games crop up so fruitfully, one explanation from mathematics and the other arising from evolutionary biology.
The mathematical explanation for why games crop up in mathematics is that the truth of a complex statement
$$\forall x_0 \exists y_0 \forall x_1 \exists y_1 \cdots \varphi(x_0,y_0,x_1,y_1,\ldots)$$
is easily thought of as a game, where the first player (called Adam for $\forall$) plays $x_0$, challenging Eve (for $\exists$) to respond with a value $y_0$, for which Adam plays $x_1$, to which Eve responds with $y_1$ and so on, with Eve winning when $\varphi(x_0,y_0,x_1,y_1,\ldots)$. The point is that(in the finite case) the statement above is true exactly when Eve has a winning strategy allowing her to defeat all plays of Adam. Thus, the truth of the statement with alternating quantifiers can be thought of in terms of these games and whether they have strategies.
Such a perspective is fruitful even when there are only a few quantifiers, as our calculus students learn when studying continuity or when seeing $\epsilon,\delta$ proofs for the first time ("you pick $\epsilon$, and I can respond with $\delta$").
Thus, the answer to your question about why games are so fruitful is that the existence of strategies for those games is intimately connected with the truth of complex statements. Since it is these statements that we are really interested in, it is fruitful to investigate them particularly with the game perspective.
One can prove using these ideas that all finite length games are determined, in the sense that they have a winning strategy for one of the players. That is, either Eve has a strategy that allows her to defeat all plays by Adam, or else Adam has a winning strategy allowing him to successfully challenge any play of Eve.
The natural generalization of these ideas lead to the Axiom of Determinacy, which asserts even that infinitely long games are determined. In this context, one imagines an infinite game, with Adam playing $x_n$ and Even playing $y_n$ for all natural numbers $n$, and then Eve wins if the play $(x_0,y_0,x_1,y_1,\ldots)$ is in the payoff set $A$, a set of infinite sequences. The Axiom of Determancy asserts that for every set A, either Adam or Eve has a winning strategy, and this axiom can be thought of as an infinitary de Morgan law:
$$\neg[\forall x_0 \exists y_0 \forall x_1 \exists y_1 \cdots A(x_0,y_0,x_1,y_1,\ldots)] \iff\exists x_0 \forall y_0 \exists x_1 \forall y_1 \cdots \neg A(x_0,y_0,x_1,y_1,\ldots)$$
which expresses the idea that if Eve does not have a winning strategy, then Adam does have a winning strategy. Synatically, this equivalence looks quite natural, since we are used to pushing $\neg$ through all the quantifiers and changing them to the dual quantifier. But in the infinitary context, this equivalence is not a matter of logic, and actually contradicts the axiom of choice. But it does suggest AD as a natural axiom. From AD one can prove that every set of reals is Lebesgue measurable and many other natural regularity properties, such as the property of Baire and the perfect set property. The consistency of AD over ZF is equivalent to the consistency of infinitely many Woodin cardinals, a large cardinal property.
Evolutionary biology. But let me come to another more speculative explanation of why games crop up so frequently and fruitfully. This reason has to do with evolutionary biology. The fact is that the truth of a statement involving many alternations of quantifiers is amazingly complex mathematically. Even definitions involving five or six quantifiers are rather complex, and give rise to very subtle distinctions in mathematics, such as the difference between uniform continuity and continuity, or families of continuous functions versus equicontinuous families of functions. Nevertheless, because of the way that we humans evolved, we are used to thinking strategically, in making decisions that are provisional based on the future actions of other individuals. Thus, the strategic way of thinking allows us more easily to understand the meaning of these complex statements. I would say that this facility with strategic thinking means in a sense that we may have a hard-wired kind of inherent ability to reason about extremely complex mathematic statements, since these statements can be equivalently expressed as strategic games.
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question_answer1) Suppose that the division \[x\div 5\] leaves a remainder 4 and the division \[x\div 2\] leaves a remainder 1. Find the ones digit of x. play_arrow
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question_answer2) Find the number of digits in the square root of 4489. (Without any calculation). play_arrow
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question_answer5) How many vertices are there of a sphere? play_arrow
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question_answer7) 160 \[{{m}^{3}}\]of water is to be used to irrigate a rectangular field whose area is 800\[{{m}^{2}}\]. What will be the height of the water level in the field? play_arrow
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question_answer8) If \[117\frac{1}{3}\]m long rope is cut into equal pieces measuring \[7\frac{1}{3}\]m each. How many such small pieces are there? play_arrow question_answer9)
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Factorise the following: (a) \[\frac{{{x}^{2}}}{4}+2x+4\] (b) \[16{{x}^{2}}+40x+25\] play_arrow
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question_answer10) Find the side of a square whose area is equal to the area of a rectangle with sides 6.4 m and 2.5 m. play_arrow
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question_answer11) A colour TV is available for Rs. 26880 inclusive of VAT. If the original cost of the TV is Rs. 24,000, find the rate of VAT. play_arrow
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question_answer12) The length and breadth of a rectangle are \[3{{x}^{2}}-2\text{ }and\text{ }2x+5\]respectively. Find its area. play_arrow question_answer13)
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(a) Find the area of rectangular park which is \[36\frac{3}{5}\]m long and \[16\frac{2}{3}\]m broad. (b) Write the name of property for any rational numbers \[\frac{a}{b}\] and \[\frac{c}{d}\], we have \[\left( \frac{a}{b}\times \frac{c}{d} \right)=\left( \frac{c}{d}\times \frac{a}{b} \right)\] play_arrow
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question_answer14) If 756 x is divisible by 11, where x is a digit find the value of x. play_arrow
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question_answer15) A volleyball court is in a rectangular shape and its dimensions are directly proportional to the dimensions of the swimming pool given below. Find the width of the pool. play_arrow
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question_answer16) The denominator of a rational number is greater then its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is \[\frac{3}{2}\] . Find the rational number. play_arrow
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question_answer17) If \[\frac{{{5}^{m}}\times {{5}^{3}}\times {{5}^{-2}}}{{{5}^{-5}}}={{5}^{12}},\] find m. play_arrow
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question_answer18) Construct a quadrilateral ABCD in which AB = 5-6 cm, BC = 4.1 cm, CD = 4.4 cm, AD = 3.3 cm and \[\angle A\text{ }=75{}^\circ .\] play_arrow
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question_answer19) Find the area to be painted in the following block with a cylindrical hole. Given that length is 15 cm, weight 12 cm, and radius of the hole 2.8 cm. play_arrow
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question_answer20) Prove that if x number is doubled then its cube is 8 times cube of the given number. play_arrow question_answer21) Factorise the following :
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(a) \[{{a}^{3}}-4{{a}^{2}}+12-3a\] (b) \[4{{x}^{2}}-20x+25\] play_arrow
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question_answer22) Vishakha offers a discount of 20% on all the items at her shop and still makes a profit of 12%. What is the cost price of an article marked at Rs. 280? play_arrow question_answer23)
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Factorise and divide the following: (a)\[\left( {{x}^{2}}-22x+117 \right)\div \left( x-13 \right)~~~~\] (b) \[\left( 9{{x}^{2}}-4 \right)\div \left( 3x+2 \right)\] play_arrow question_answer24)
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It is given that I varies directly as m. (a) Write an equation which relates 1 and m. (b) Find the constant of proportion (k), when 1 is 6 then m is 18. (c) Find 1, when m is 33. (d) Find m when 1 is 18. play_arrow
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question_answer25) The product of two rational numbers is\[\frac{-28}{75}\]. If one of the numbers is\[\frac{14}{25}\]. Find the other. play_arrow question_answer26)
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(a) A cylindrical tank has a capacity of 5632\[{{m}^{3}}\]. If the diameter of its base is 16 m. Find its depth. (b) If side of square is 14 cm, then find the area of semi-circle as shown in the figure. play_arrow
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question_answer27) Lakshmi is a cashier in a bank. She has currency notes of denominations Rs.100,Rs. 50 and Rs. 10 respectively. The ratio of the number of these notes is 2: 3: 5. The total cash with Lakshmi is Rs. 400,000. How many notes of each denomination does she have? play_arrow question_answer28) In a quadrilateral ABCD, DO and CO are the bisectors of \[\angle D\] and \[\angle C\] respectively.
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Prove that \[\angle COD=\frac{1}{2}[\angle A+\angle B].\] play_arrow
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question_answer29) A shopkeeper bought two TV sets at Rs. 10,000 each. He sold one at a profit 10% and the other at a loss of 10%. Find whether he made an overall profit or loss. play_arrow
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question_answer30) Divide 63 \[\left( {{p}^{4}}+5{{p}^{3}}-24{{p}^{2}} \right)\text{ }by\text{ }9p\left( p+8 \right)\] play_arrow
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I am studying for my qualifiers, and I ran into this question from a previous year's exam.
$\textbf{Question:}$
Consider a two-consumer two-good pure exchange economy. Both preferences are locally non-satiated and convex. Prove or disprove the following statement: if $(x_1,x_2)$ and $(\hat{x}_1,\hat{x}_2)$ are two different pareto optimal allocations, then the convex combination, $(\alpha x_1+(1-\alpha)\hat{x}_1,\alpha x_2+(1-\alpha)\hat{x}_2$ MUST also be pareto optimal for any $\alpha\in(0,1)$.
I believe the statement is true, and here is the work for my proof below.
$\textbf{My Proof:}$ By pareto optimality of $x_i$ and $\hat{x}_i$: $$\not\exists\; x_i^\star\; s.t.\; u_i(x_i^\star)\geq u_i(x_i)\;\forall i\; \text{and}\; u_i(x_i^\star)> u_i(x_i)\;\text{for at least one }i \;\text{or}\;u_i(x_i^\star)\geq u_i(\hat{x}_i)\;\forall i\; \text{and}\; u_i(x_i^\star)> u_i(\hat{x}_i)\;\text{for at least one }i$$ $$\implies\;u_i(\alpha x_i)\geq u_i(\alpha x_i^\star)\;\forall i\;\text{and}\;u_i((1-\alpha)\hat{x}_i)\geq u_i((1-\alpha)x_i^\star)\;\forall i$$ $$\implies\not\exists\;x_i^\star\;s.t.\;u_i(\alpha x_i^\star+(1-\alpha)x_i^\star)\geq u_i(\alpha x_i+(1-\alpha)\hat{x}_i)\;\forall i$$ $$\text{and}\;u_i(\alpha x_i^\star+(1-\alpha)x_i^\star)> u_i(\alpha x_i+(1-\alpha)\hat{x}_i)\;\text{for at least one}\;i$$ $$\implies (\alpha x_i+(1-\alpha)\hat{x}_i)\;\text{is pareto optimal}$$ $$\blacksquare$$
This proof seemed almost too easy, so I'm wondering if it is correct/rigorous.
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First off, I want to warn you just a little bit about putting battery systems in parallel. It's usually not a good idea because often the two batteries (or battery systems) don't have exactly the same voltage. If they are different, then the one with the larger voltage will supply some current into the battery with the lower voltage and this often isn't a good thing. It also messes up your experiment, somewhat, because it adds another complication to it.
In this case, you are curious and imagine that two batteries in parallel can supply more current. So it doesn't serve your purposes to use only one in your experiment because it doesn't test your assumptions. So you have to do it the way you did. But I just want you to also realize that there is another unknown (to you) factor that you aren't accounting for in your experimental design. But it's not enough to worry about, for now.
So set that aside...
Let me suggest a new idea for you to consider. Suppose that the green LED you have
requires exactly \$1.9\:\text{V}\$ in order to "turn on" and that it also has, unknown to you, an internal resistor of exactly \$50\:\Omega\$. You can't get inside the LED to see these things. But let's say, as a thought experiment, that this is the way this particular LED works.
Also, let's assume that your battery systems provide exactly \$2.9\:\text{V}\$ each. If you put them in series then you are applying \$5.8\:\text{V}\$ to the circuit. If you put them in parallel then you are applying \$2.9\:\text{V}\$ to the circuit. The only difference here may be the current compliance (the ability to supply more or less current to a load, if required.)
Your assumption is that if the current compliance is more, then the current is more. But that may be true sometimes and not others. So, for now, let's use my above idea about the LED and see where that takes us.
Your series circuit includes also a \$100\:\Omega\$ resistor. Taken together with my hypothetical internal \$50\:\Omega\$ resistor inside the LED, there is a total series resistance in the circuit of \$150\:\Omega\$ (let's just assume I'm right for now.) Plus, the LED itself (the one inside the package that you cannot actually touch) also requires (subtracts) \$1.9\:\text{V}\$ from the applied voltage before we can compute the current. (You can see that the LED is ON in both cases, so this must be true if my assertion is correct.)
So in the batteries-in-parallel case you have \$I_\text{parallel}=\frac{2.9\:\text{V}-1.9\:\text{V}}{150\:\Omega}\approx 6.7\:\text{mA}\$ and in the batteries-in-series case you have \$I_\text{series}=\frac{2\cdot 2.9\:\text{V}-1.9\:\text{V}}{150\:\Omega}\approx 26\:\text{mA}\$.
This would appear to predict your measurements within a reasonably small error.
So which idea do you think works better here? Your thoughts about two parallel battery systems doubling the current? Or my suggestion about how an LED may behave? Do you have still further ideas that you may want to consider? How might you test or validate my above suggestion? Can you think of another way to change your circuit that might put my suggestion to another test to see if it still holds up? Or can you think of another voltage measurement or current measurement you might try to test it?
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E.g. if I know that my topology is that of a hyperboloid, how much freedom do I have left for my choice of the metric? And the other way around: if my metric is some conformal factor times the unit hyperboloid how much freedom do I have left in my choice for the topology? Am I forced to conclude that the topology of this (Lorentzian) spacetime needs to be that of a hyperboloid?
Let a smooth manifold $M$ be given, namely a suitably non-pathological topological space along with a smooth structure. A metric $g$ on $M$ is any symmetric, nondegenerate $(0,2)$ tensor field on $M$ having constant index. The pair consisting of $(M,g)$ is then called a semi-riemannian manifold.
So suppose you hand someone the a topological space like a hyperboloid $H$, and suppose that someone turns this hyperboloid into a smooth manifold by equipping it with a smooth structure, then you can still choose any metric on the hyperboloid you wish and make it a perfectly acceptable semi-riemannian manifold, so needless to say, there's a whole lot of freedom.
Note, however, that (especially in physics literature on, say, AdS/CFT) a hyperboloid $H$, as a smooth manifold, is often defined as an embedded submanifold of some higher dimensional smooth manifold $N$. Moreover, as it often happens, this higher dimensional smooth manifold already has a specified metric $g_N$. In this case, there is a particularly natural choice for the metric on $H$; it's called the
pullback metric or the induced metric on $H$ depending on who you're talking to. The pullback metric is defined as follows. If $f:M\to N$ is the embedding through which $H$ was defined (in other words $H$ is just the image of $M$ under $f$), then the pullback metric is usually denoted $f^*g^N$ and is defined as follows:$$ (f^*g^N)_p(v,w) = g^N_{f(p)}(df_p(v), df_p(w))$$for all $p\in M$ and $v,w\in T_pM$. Intuitively, the pullback metric is precisely the notion of distance inherited by $H$ from $N$, a smooth manifold that already has a notion of distance defined on it. I can't remember the last time I encountered a situation in the (high energy theoretical) physics literature in which someone considered an embedded submanifold with a metric other than the induced metric.
Now, for the second question which is basically the reverse of the first, we need to be careful. When we use the word "metric" as defined above, implicit in the use of this terminology is the existence of a smooth manifold on which the metric is defined, so in this sense, the reverse direction doesn't make much sense.
However, I think (and correct me if I'm wrong) you're second question can be reformulated as follows:
For each $\mu,\nu = 1, \dots, n$, I have a matrix $g=(g_{\mu\nu})$ of functions $g_{\mu\nu} = g_{\mu\nu}(x_1, \dots, x_n)$ defined on some open subset of $\mathbb R^n$ that is nondegenerate and has constant index, then does there exist a realization of $g$ as the local coordinate representation of the induced metric of some embedded submanifold $M$ of some ambient manifold $N$? If so, is this realization unique?
If this is not a correct reformulation of your question, then let me know in the comments below.
I don't know the definitive answer to these questions, but I strongly suspect that the answers are yes and no respectively. If such a realization exists, then, as Jerry Schirmer basically points out in his comment, you can just, for example, chop $M$ into two, disconnected pieces to obtain another realization.
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I have a discrete time stochastic process, where at each time the state of the system $X_t$ is given by: $$ X_t = f_\theta(X_{t-1},\epsilon_t), \; \; \text{for} \; t = 1,\dots,T $$ and, for example, the perturbations $\epsilon_t$ are i.i.d. $N(0,\sigma_e)$. So the state of the system at each time is a function (parametrized by $\theta$) of the state at the previous time step $X_{t-1}$ and of the noise $\epsilon_t$.
Suppose that I have observed $X^o_{1:T}$ and that I want to do inference about $\theta$ (by $X^o_{1:T}$ in mean $X^o_1,\dots,X^o_T$). In particular if, instead of observing $X^o_{1:T}$ directly, I was observing $Y_{1:T}$ given by: $$ Y_{t} = X^o_t + Z_t\; \; \text{for} \; t = 1,\dots,T. $$ where the observational noise $z_t$ are i.i.d. from some kernel density $K_{\sigma_o}$ (for example $Z_t \sim N(0,\sigma_o)$), then I would be dealing with a state space model with observations $Y_{1:T}$ and hidden states $X^o_{1:T}$. This means that I could use all the methodologies that have been developed to do inference for state space models (such as particle filters for example).
But the observational noise $Z_{1:T}$ is simply not there in my case, so there are two ways to go:
1) Fit the data as if the observational noise $Z_{1:T}$ was there (that is as if I was observing $Y_{1:T}$, while I'm really observing $X^o_{1:T}$).
2) Add to the the states $X^o_{1:T}$ randomly generated noise $Z_{1:T}$, to obtain the observations $Y_{1:T}$. Then $Y_{1:T}$ has been really generated according to a (artificial) state space model.
Let's consider the first approach. In this case I'm fitting the wrong model, because I'm fitting a model that includes observational noise $Z_{1:T} \sim K_{\sigma_o}$ which is not present in reality. In particular, by treating the model as a state space model, we obtain the likelihood:
$$ \hat{p}_1(X^o_{1:T}|\theta) = \int p(X^o_{1:T},X_{1:T}|\theta) d X_{1:T} = \int K_{\sigma_o}(X^o_{1:T}-X_{1:T})p(X_{1:T}|\theta) d X_{1:T} $$
so $\hat{p}_1(X^o_{1:T}|\theta) = (K_{\sigma_o} \star p(\cdot |\theta))(X^o_{1:T}) \neq p(X^o_{1:T}|\theta)$. The likelihood that we obtain is the convolution of kernel $K_{\sigma_o}()$ with $ p(\cdot |\theta)$ and it is different from the true likelihood $p(X^o_{1:T}|\theta)$. So the resulting ML estimates for $\theta$ are likely to be biased.
The second approach looks more promising. If I add some noise $Z_{1:T}$, distributed according to $K_{\sigma_o}$, to the data then I'm justified to fit state space model. This is because if I define $Y_{1:T} = X^o_{1:T} + Z_{1:T}$, then $X^o_{1:T}$ is really a hidden state. So I would expect the resulting estimates for $\theta$ to be unbiased, even though some efficiency will be lost because I'm adding noise (and thus losing information).
Lets call $\hat{p}_2(X^o_{1:T}|\theta)$ the likelihood given by this procedure. I'm interested in knowing what is it's expected value, conditionally on the observed $X^o_{1:T}$. The expected likelihood over all the possible simulated vectors $Z_{1:T}$, is given by:
$$ E(\hat{p}_2(X^o_{1:T}|\theta)|X^o_{1:T}) = \int \int K_{\sigma_o}(Y_{1:T}-X_{1:T})p(X_{1:T}|\theta) d X_{1:T} K_{\sigma_o}(Z_{1:T}) dZ_{1:T} $$ $$ = \int \bigg ( \int K_{\sigma_o}(Z_{1:T} + X^o_{1:T}-X_{1:T})\, K_{\sigma_o}(Z_{1:T})\, dZ_{1:T} \bigg ) \, p(X_{1:T}|\theta)\, d X_{1:T} $$ $$ = \int K_{\sigma_o} \star K_{\sigma_o}( X^o_{1:T}-X_{1:T}) p(X_{1:T}|\theta)\, d X_{1:T}. $$ $$ = (K_{\sigma_o} \star K_{\sigma_o}) \star \, p(X_{1:T}|\theta)( X^o_{1:T}). $$
So we have the same problem as with the first approach! So if we add noise (generated according to $K_{\sigma_o}$) to the data $X^o_{1:T}$ and we fit $Y_{1:T}$ as a space state model the expected likelihood that we get out is the same that we get I we fit the data without adding noise, using the kernel $K_{\sigma_o} \star K_{\sigma_o}$. This result has to be wrong because we are fitting the right model if we add the noise $Z_{1:T}$!
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Python is available on any Linux/Unix machine including department machines and Macs.
You can download and install python on your own computer by following instructions at http://www.python.org
You can use the Python interpreter interactively by typing
python at a terminal window.Ipython is a nicer front end to python that is invoked with ipython
To quit, type control-d
To run python code in a file
code.py, either type run code.py
in
ipython, or type python code.py
at the unix command line.
When in
ipython, you may type python statements or expressionsthat are evaluated, or ipython commands. See theVideotutorial on using ipython, in five parts by Jeff Rush, for helpgetting started with ipython.
Documentation is immediately available for many things. For example:
> ipython asa:~$ ipython Python 2.7.3 (v2.7.3:70274d53c1dd, Apr 9 2012, 20:52:43) Type "copyright", "credits" or "license" for more information. IPython 0.13.2 -- An enhanced Interactive Python. ? -> Introduction and overview of IPython's features. %quickref -> Quick reference. help -> Python's own help system. object? -> Details about 'object', use 'object??' for extra details. In [1]: list? Type: type Base Class: <type 'type'> String Form: <type 'list'> Namespace: Python builtin Docstring: list() -> new list list(sequence) -> new list initialized from sequence's items In [2]: help(list) Help on class list in module __builtin__: class list(object) | list() -> new list | list(sequence) -> new list initialized from sequence's items | | Methods defined here: . . . | append(...) | L.append(object) -- append object to end | . . . | | sort(...) | L.sort(cmp=None, key=None, reverse=False) -- stable sort *IN PLACE*; | cmp(x, y) -> -1, 0, 1
What is the value of $(100\cdot 2 - 12^2) / 7 \cdot 5 + 2\;\;\;$?
In [301]: (100*2 - 12**2) / 7*5 + 2 Out[301]: 42
In order to compute something like $\sin(\pi/2)$ we first need to
import the math module: In [303]: import math In [304]: math.sin(math.pi/2) 1.0
How do I find out what other mathematical functions are available?
help("math")
Let's get on to that all important step of visualizing data. We will be using the matplotlib Python package for that. Let's start by plotting the function $f(x) = x^2$.
First, let's generate the numbers. Well, there are tons of ways to do so. First, using a
for loop.
In [3]: f = [] In [4]: for i in range(10) : ...: f.append(i**2) ...: In [5]: f Out[5]: [0, 1, 4, 9, 16, 25, 36, 49, 64, 81]
To plot the data, first import the
pyplot module.
In [6]: import matplotlib.pyplot as plt In [7]: plt.plot(range(10), f) Out[7]: [<matplotlib.lines.Line2D at 0x10549b590>]
In order to actually see the plot you need to do:
In [8]: plt.show()
As an alternative, you can put matplotlib in interactive mode before plotting using the command
plt.ion().
Python has some nifty syntax for generating lists. Watch this! A list comprehension!!
In [9]: f = [i**2 for i in range(10)] In [10]: f Out[10]: [0, 1, 4, 9, 16, 25, 36, 49, 64, 81]
There's an alternative way of doing this using
numpy:
In [11]: import numpy as np In [12]: f = np.arange(10)**2 In [13]: f Out[13]: array([ 0, 1, 4, 9, 16, 25, 36, 49, 64, 81])
Note that plotting functions to accept either lists or
numpy arrays, so a fast way of doing our plot is
In [14]: plt.plot(np.arange(10), np.arange(10)**2)
For a smoother plot:
In [14]: x = np.arange(10, 0.1) In [15]: plt.plot(x, x**2, 'ob') Out[15]: [<matplotlib.lines.Line2D at 0x1054162d0>]
We can add a second plot to the same axes by calling
plot again: In [16]: plt.plot(x, x, 'dr') Out[16]: [<matplotlib.lines.Line2D object at 0x3608990>]
Can I work with vectors and matrices in python?
Of course! No data analysis tool is worth the bytes it burns if itdoesn't. The
numpy package provides the required magic.Let's create an array that represents the following matrix:\[\left ( \begin{array}{cc} 1 & 2\\ 3 & 4\\ 5 & 6 \end{array} \right ) \]by doing
In [17]: import numpy as np In [18]: m = np.array([[1,2], [3,4], [5,6]]) In [19]: m Out[19]: array([[1, 2], [3, 4], [5, 6]])
Let's construct the matrices \[a = \left ( \begin{array}{cc} 2 & 2 & 2\\ 2 & 2 & 2\\ 2 & 2 & 2 \end{array} \right ) \] and \[b = \left ( \begin{array}{cc} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{array} \right ) \]
In [16]: a = np.ones((3,3)) * 2 In [17]: a Out[17]: array([[ 2., 2., 2.], [ 2., 2., 2.], [ 2., 2., 2.]]) In [18]: b = np.resize(np.arange(9)+1,(3,3)) In [19]: b Out[19]: array([[1, 2, 3], [4, 5, 6], [7, 8, 9]])
What is the value of $a * b$?
In [20]: a * b Out[21]: array([[ 2, 4, 6], [ 8, 10, 12], [14, 16, 18]])
The
* operator does a component-wise multiplication. Use the
numpy function
dot to do matrix multiplication.
In [22]: np.dot(a,b) Out[22]: array([[24, 30, 36], [24, 30, 36], [24, 30, 36]])
An array is transposed by
In [23]: b.transpose() Out[23]: array([[1, 4, 7], [2, 5, 8], [3, 6, 9]]) In [24]: b.T Out[24]: array([[1, 4, 7], [2, 5, 8], [3, 6, 9]])
Elements and sub-matrices are easily extracted:
In [25]: b Out[25]: array([[1, 2, 3], [4, 5, 6], [7, 8, 9]]) In [26]: b[0,0] Out[26]: 1 In [27]: b[0,1] Out[27]: 2 In [28]: b[0:2, 1:3] Out[28]: array([[2, 3], [5, 6]])
Let's multiply the first row of a $a$ by the second column of $b$.
In [29]: np.dot(a[0], b[:,1]) Out[29]: 30.0 In [30]: np.dot(a[0],b.T[1]) Out[30]: 30.0
How do I find the inverse of a matrix?
In [2]: z = np.array([[2,1,1],[1,2,2],[2,3,4]]) In [3]: z Out[3]: array([[2, 1, 1], [1, 2, 2], [2, 3, 4]]) In [4]: np.linalg.inv(z) Out[4]: array([[ 0.66666667, -0.33333333, 0. ], [ 0. , 2. , -1. ], [-0.33333333, -1.33333333, 1. ]]) In [5]: np.dot(z, np.linalg.inv(z)) Out[5]: array([[ 1., 0., 0.], [ 0., 1., 0.], [ 0., 0., 1.]])
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Hello, I've never ventured into char before but cfr suggested that I ask in here about a better name for the quiz package that I am getting ready to submit to ctan (tex.stackexchange.com/questions/393309/…). Is something like latex2quiz too audacious?
Also, is anyone able to answer my questions about submitting to ctan, in particular about the format of the zip file and putting a configuration file in $TEXMFLOCAL/scripts/mathquiz/mathquizrc
Thanks. I'll email first but it sounds like a flat file with a TDS included in the right approach. (There are about 10 files for the package proper and the rest are for the documentation -- all of the images in the manual are auto-generated from "example" source files. The zip file is also auto generated so there's no packaging overhead...)
@Bubaya I think luatex has a command to force “cramped style”, which might solve the problem. Alternatively, you can lower the exponent a bit with f^{\raisebox{-1pt}{$\scriptstyle(m)$}} (modify the -1pt if need be).
@Bubaya (gotta go now, no time for followups on this one …)
@egreg @DavidCarlisle I already tried to avoid ascenders. Consider this MWE:
\documentclass[10pt]{scrartcl}\usepackage{lmodern}\usepackage{amsfonts}\begin{document}\noindentIf all indices are even, then all $\gamma_{i,i\pm1}=1$.In this case the $\partial$-elementary symmetric polynomialsspecialise to those from at $\gamma_{i,i\pm1}=1$,which we recognise at the ordinary elementary symmetric polynomials $\varepsilon^{(n)}_m$.The induction formula from indeed gives\end{document}
@PauloCereda -- okay. poke away. (by the way, do you know anything about glossaries? i'm having trouble forcing a "glossary" that is really an index, and should have been entered that way, into the required series style.)
@JosephWright I'd forgotten all about it but every couple of months it sends me an email saying I'm missing out. Oddly enough facebook and linked in do the same, as did research gate before I spam filtered RG:-)
@DavidCarlisle Regarding github.com/ho-tex/hyperref/issues/37, do you think that \textNFSSnoboundary would be okay as name? I don't want to use the suggested \textPUnoboundary as there is a similar definition in pdfx/l8uenc.def. And textnoboundary isn't imho good either, as it is more or less only an internal definition and not meant for users.
@UlrikeFischer I think it should be OK to use @, I just looked at puenc.def and for example \DeclareTextCompositeCommand{\b}{PU}{\@empty}{\textmacronbelow}% so @ needs to be safe
@UlrikeFischer that said I'm not sure it needs to be an encoding specific command, if it is only used as \let\noboundary\zzznoboundary when you know the PU encoding is going to be in force, it could just be \def\zzznoboundary{..} couldn't it?
@DavidCarlisle But puarenc.def is actually only an extension of puenc.def, so it is quite possible to do \usepackage[unicode]{hyperref}\input{puarenc.def}. And while I used a lot @ in the chess encodings, since I saw you do \input{tuenc.def} in an example I'm not sure if it was a good idea ...
@JosephWright it seems to be the day for merge commits in pull requests. Does github's "squash and merge" make it all into a single commit anyway so the multiple commits in the PR don't matter or should I be doing the cherry picking stuff (not that the git history is so important here) github.com/ho-tex/hyperref/pull/45 (@UlrikeFischer)
@JosephWright I really think I should drop all the generation of README and ChangeLog in html and pdf versions it failed there as the xslt is version 1 and I've just upgraded to a version 3 engine, an dit's dropped 1.0 compatibility:-)
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I came across John Duffield Quantum Computing SE via this hot question. I was curious to see an account with 1 reputation and a question with hundreds of upvotes.It turned out that the reason why he has so little reputation despite a massively popular question is that he was suspended.May I ...
@Nelimee Do we need to merge? Currently, there's just one question with "phase-estimation" and another question with "quantum-phase-estimation". Might we as well use just one tag? (say just "phase-estimation")
@Blue 'merging', if I'm getting the terms right, is a specific single action that does exactly that and is generally preferable to editing tags on questions. Having said that, if it's just one question, it doesn't really matter although performing a proper merge is still probably preferable
Merging is taking all the questions with a specific tag and replacing that tag with a different one, on all those questions, on a tag level, without permanently changing anything about the underlying tags
@Blue yeah, you could do that. It generally requires votes, so it's probably not worth bothering when only one question has that tag
@glS "Every hermitian matrix satisfy this property: more specifically, all and only Hermitian matrices have this property" ha? I though it was only a subset of the set of valid matrices ^^ Thanks for the precision :)
@Nelimee if you think about it it's quite easy to see. Unitary matrices are the ones with phases as eigenvalues, while Hermitians have real eigenvalues. Therefore, if a matrix is not Hermitian (does not have real eigenvalues), then its exponential will not have eigenvalues of the form $e^{i\phi}$ with $\phi\in\mathbb R$. Although I'm not sure whether there could be exceptions for non diagonalizable matrices (if $A$ is not diagonalizable, then the above argument doesn't work)
This is an elementary question, but a little subtle so I hope it is suitable for MO.Let $T$ be an $n \times n$ square matrix over $\mathbb{C}$.The characteristic polynomial $T - \lambda I$ splits into linear factors like $T - \lambda_iI$, and we have the Jordan canonical form:$$ J = \begin...
@Nelimee no! unitarily diagonalizable matrices are all and only the normal ones (satisfying $AA^\dagger =A^\dagger A$). For general diagonalizability if I'm not mistaken onecharacterization is that the sum of the dimensions of the eigenspaces has to match the total dimension
@Blue I actually agree with Nelimee here that it's not that easy. You get $UU^\dagger = e^{iA} e^{-iA^\dagger}$, but if $A$ and $A^\dagger$ do not commute it's not straightforward that this doesn't give you an identity
I'm getting confused. I remember there being some theorem about one-to-one mappings between unitaries and hermitians provided by the exponential, but it was some time ago and may be confusing things in my head
@Nelimee if there is a $0$ there then it becomes the normality condition. Otherwise it means that the matrix is not normal, therefore not unitarily diagonalizable, but still the product of exponentials is relatively easy to write
@Blue you are right indeed. If $U$ is unitary then for sure you can write it as exponential of an Hermitian (time $i$). This is easily proven because $U$ is ensured to be unitarily diagonalizable, so you can simply compute it's logarithm through the eigenvalues. However, logarithms are tricky and multivalued, and there may be logarithms which are not diagonalizable at all.
I've actually recently asked some questions on math.SE on related topics
@Mithrandir24601 indeed, that was also what @Nelimee showed with an example above. I believe my argument holds for unitarily diagonalizable matrices. If a matrix is only generally diagonalizable (so it's not normal) then it's not true
also probably even more generally without $i$ factors
so, in conclusion, it does indeed seem that $e^{iA}$ unitary implies $A$ Hermitian. It therefore also seems that $e^{iA}$ unitary implies $A$ normal, so that also my argument passing through the spectra works (though one has to show that $A$ is ensured to be normal)
Now what we need to look for is 1) The exact set of conditions for which the matrix exponential $e^A$ of a complex matrix $A$, is unitary 2) The exact set of conditions for which the matrix exponential $e^{iA}$ of a real matrix $A$ is unitary
@Blue fair enough - as with @Semiclassical I was thinking about it with the t parameter, as that's what we care about in physics :P I can possibly come up with a number of non-Hermitian matrices that gives unitary evolution for a specific t
Or rather, the exponential of which is unitary for $t+n\tau$, although I'd need to check
If you're afraid of the density of diagonalizable matrices, simply triangularize $A$. You get $$A=P^{-1}UP,$$ with $U$ upper triangular and the eigenvalues $\{\lambda_j\}$ of $A$ on the diagonal.Then$$\mbox{det}\;e^A=\mbox{det}(P^{-1}e^UP)=\mbox{det}\;e^U.$$Now observe that $e^U$ is upper ...
There's 15 hours left on a bountied question, but the person who offered the bounty is suspended and his suspension doesn't expire until about 2 days, meaning he may not be able to award the bounty himself?That's not fair: It's a 300 point bounty. The largest bounty ever offered on QCSE. Let h...
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Let $(\Omega,\mathcal F, P)$ be such a probability space that for every event $A\in\mathcal F$, the events that are independent to $A$ form an algebra $\mathcal B_A$ ($\Omega\in\mathcal B_A$). Independence to $A$ implying (I hope I get this right) for every finite $\mathcal C\subseteq\mathcal B_A$ $$P\left (A\cap\bigcap_{C\in\mathcal C}C\right ) = P(A)\prod_{C\in\mathcal C} P(C)\tag{1}\label{eq:1} $$
Let $A_1,A_2,A_3\in\mathcal F$ be pairwise independent events. Show that they are in fact independent events. So it remains to verify $$P(A_1\cap A_2\cap A_3) = P(A_1)P(A_2)P(A_3). $$
A quick (perhaps insignificant) observation is that $A\notin\mathcal B_A$ (because $P(A) = P(A\cap A)\neq P(A)P(A)$).
An idea was to use the fact that $\Omega\in \mathcal B_{A_1\cap A_2}$ : by additivity$$P(A_1\cap A_2\cap (A_3\cup A_3^c)) = P(A_1\cap A_2\cap A_3)+P(A_1\cap A_2\cap A_3^c) = P(A_1)P(A_2) $$Similarly we obtain:
$$P(A_1\cap A_2\cap A_3)+P(A_1\cap A_2^c\cap A_3) = P(A_1)P(A_3)\\ P(A_1\cap A_2\cap A_3)+P(A_1^c\cap A_2\cap A_3) = P(A_2)P(A_3) $$ We now could use additivity again and compute the probability of $$(A_1^c\cap A_2\cap A_3)\cup (A_1\cap A_2^c\cap A_3)\cup (A_1\cap A_2\cap A_3^c) $$ my set arithmetics is complete garbage, so I don't know if we even obtain what we need here. I haven't actually invoked $\mathcal B_{A}$ being an algebra anywhere, so I'm kind of stuck.
Are there any ideas, any hints on how to either proceed or rethink my strategy?
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The Chaplygin sleigh problem is very popular in nonholonomic mechanics (more details can be seen in Bloch,
Nonholonomic Mechanics and Control, Springer, New York, 2015, section 1.7, pp 29-30),
The Chaplygin sleigh is a rigid body moving on two sliding posts and one knife edge
The equations are:
$\qquad v =\dot{x}\cos\theta+\dot{y}sin\theta,$
$\qquad \ddot{\theta} =\dot{\omega}=-\frac{ma}{1+ma^2}v\omega.$
where $v$ is the velocity in the direction of motion, and
$\qquad \dot{v}=a(\cos^2\theta+\sin^2\theta)\dot{\theta}^2=a\dot{\theta}^2=a\omega^2.$
In Mathematica, I can plot the trajectory in $vω$ space, but how can I plot the Chaplygin sleigh model trajectory in $xy$ space?
My code as follows (maybe something wrong):
Clear["Global`*"]m = 1; a = 1;sol = NDSolve[ {w'[t] == -v[t]*w[t]*m*a/(1 + m*a^2), v'[t] == a*w[t]^2, w[0] == 0.2, v[0] == 0.1}, {w, v}, {t, -4, 4}]ParametricPlot[Evaluate[{w[t], v[t]} /. sol], {t, -4, 4}]
The result in the referenced text is:
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@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.
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Problem Statement:-
Find the set of real values of $x$ for which $$x^{(\log_{10}x)^2-3\log_{10}x+1}\gt1000$$
Correct Approach:-
Domain of $x^{(\log_{10}x)^2-3\log_{10}x+1}\implies x\gt0$
Now, $$x^{(\log_{10}x)^2-3\log_{10}x+1}\gt1000$$
Taking $\log_{10}$ on both sides, we get
$$((\log_{10}x)^2-3\log_{10}x+1)(\log_{10}x)\gt3$$
Let $z=\log_{10}{x}$
Hence, $(z^2-3z+1)z\gt3\implies z^3-3z^2+z-3\gt0$
Fortunately for factorising the polynomial we dont have to go too far we get the first factor at $z=3$. So, on factorising, we get $$(z-3)(z^2+1)\gt0$$
As, $\forall z\in\Bbb{R}.(z^2+1\gt0)$, hence $$z-3\gt0\implies z\gt3\implies \log_{10}x\gt3\implies x\gt1000$$
Incorrect Approach:-
Domain of $x^{(\log_{10}x)^2-3\log_{10}x+1}\implies x\gt0$
We also see that for $x=1$ the inequality is not satisfied, so we would not be eliminating any values from the interval satisfying the inequality by putting the condition $x\neq1$
Now, take $\log_{x}$ on both sides to get
$$((\log_{10}x)^2-3\log_{10}x+1)\gt\dfrac{3}{\log_{10}x}\\ \implies ((\log_{10}x)^2+1)\gt\dfrac{3}{\log_{10}x}+3\log_{10}x$$
Using $A.M.\ge G.M.$, we get $$((\log_{10}x)^2+1)\gt6\implies (\log_{10}x)^2\gt5\implies -\sqrt{5}\lt\log_{10}x\lt\sqrt5\\ \implies 10^{-\sqrt{5}}\lt x\lt10^{\sqrt5}$$
But, as we had excluded $1$ as a solution, so the interval satisfying the inequality comes out to be $x\in(10^{-\sqrt5},10^{\sqrt5})-\{1\}$
I have tried two different approaches to the problem, while one does provide the correct answer and the other doesn't but what is wrong with the incorrect approach.
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History and Feedback Christian Lawson-Perfect 2 years, 2 months ago
Saved a checkpoint:
I'd like a description of how you decide if a root is a surd or not, like "$\sqrt{a}$ is a surd because there is no whole number $b$ such that $b^2 = a$".
Similarly for part
b), describe the strategy: find a square number which divides $a$, and rewrite as $\sqrt{b^2} \times \sqrt{c}$. (Or, do what I did and multiply out the $a\sqrt{b}$ forms - much easier!)
Otherwise, this looks good.
Bradley Bush 2 years, 3 months ago
This question was really hard to criticise, I only came up with a few pedantic points.
It might have been nice to see more random variables in part
a)and maybe part b)so the student can repeat this question for practice.
You may have missed a fullstop in the advice for part bi), and I'm not sure you need fullstops after each equation in the advice to part b, maybe using a comma instead would be better?
Chris Graham 2 years, 3 months ago
Where you are using
\sqrt, you should have braces around the argument
\sqrt{...}. Note the difference: $\sqrt100$ and $\sqrt{100}$.
Also the nth root can be written
\sqrt[n]{...}, e.g.
\sqrt[3]{100}, $\sqrt[3]{100}$.
If this is the first question on surds when we put questions together, then I think an example surd would be beneficial in the statement.
Christian Lawson-Perfect 2 years, 3 months ago
The prompt for part
bis hopelessly generic! "Match each product with the equivalent surd" might do, but that's not quite right either. Also think about how the numbers are laid out - would it be better to put the products down the side and collected surds on top? As it is, I can rule out a few combinations without thinking, like $\sqrt{44}$ is obviously not equal to $\sqrt{30}$.
Part
c: I would repeat $\sqrt{n} = $ before the gap-fill. As stated, how is "$\sqrt{300} = \sqrt{100} \times \sqrt{3}$, so $\sqrt{300} = ?? \times \sqrt{3}$" different to $\sqrt{100} = ??$". Maybe just $\sqrt{300} = \text{[gapfill]} \times \sqrt{3}$ would do.
Part
d: for "simplify" to make sense here, I think you need to give the hint that they can be simplified to whole numbers.
Part
e: instead of [1] and [2], write $a$ and $b$. I got $\frac{\sqrt{6}}{\sqrt{3}}$, which it expected me to write as $\frac{\sqrt{18}}{3}$. I'd write that as just $\sqrt{2}$. Can you set it up so there's only one way of writing it? Making sure the top and bottom of the original fraction are coprime might do it.
Part
f: I had to reduce the fraction - you should say that in the prompt. "Rationalise the denominator of this expression and reduce to lowest terms".
In parts
g, hand i, the big expression is REALLY big - have you formatted it as a header? I don't think you need the "= ?" bit, or "and select the correct answer from the list of options".
None of the parts of this question really lead on from each other - you could split this into several smaller questions.Advice
Some numbers not in LaTeX in part
a. "Roots are necessary but not sufficient conditions for surds" isn't completely clear - I'd say "all surds are roots, but not all roots are surds".
Part
e: this might be incredibly pedantic, but does "the denominator is $\sqrt{6}$" sound better than "\sqrt{6} is the denominator" to you?
A few sentence of a similar form to "This gives the final answer as:"; I would say "So the final answer is:".
Some wordy explanation of what's happening at each step, or a description of the plan of attack, in part
fwouldn't go amiss.
Part
g: It's not true that you can'tmultiply by those things, it just doesn't help. Get someone else to read that paragraph and have a go at rewriting it: it's a bit long and easy to get lost in. There's an unhelpful "simply" in there, too.
I have a few problems with this sentence: "To be able to do a question like $\frac{2}{\sqrt2+\sqrt8}+\frac{1}{3}$ which requires you to add fractions, you need to have the same denominator."
$\frac{2}{\sqrt2+\sqrt8}+\frac{1}{3}$ isn't a question, it's an expression. "you need to have the same denominator" isn't a complete phrase - the same as what?
Try:
To add $\frac{2}{\sqrt2+\sqrt8}$ to $\frac{1}{3}$, you must put write both terms with the same denominator."
Likewise, "the question is now $\frac{2}{3\sqrt2}+\frac{1}{3}$" isn't right. You could say "the expression is now ...".
Name Status Author Last Modified Surds simplification Ready to use Lauren Richards 01/08/2017 14:09 Inbbavathie's copy of Surds simplification draft Inbbavathie Ravi 24/07/2017 03:48 LHS Surds simplification draft Harry Flynn 27/09/2018 10:33 Simon's copy of Surds simplification draft Simon Thomas 06/06/2019 10:08 Katy's copy of Surds simplification draft Katy Dobson 03/09/2019 08:29
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This question is used in the following exams:
Summer Project (Optimisation + Methods of proof) by Inbbavathie Ravi in Inbbavathie's workspace. summer project 2 ( egyptian fractions+complex numbers) by Inbbavathie Ravi in Inbbavathie's workspace. Summer Project (IGCSE) by Inbbavathie Ravi in Inbbavathie's workspace. Surds by Christian Lawson-Perfect in Transition to university. maths practice-revision by David Martin in David's workspace. Surds by Mark Hodds in Tutoring. Nick's copy of Surds by Nick Walker in Nick's workspace. Summer Project 5 - assessed by Inbbavathie Ravi in Lee's Newcastle Maths & Stats Summer Project 2017. Surds [L0 Randomised] by Matthew James Sykes in CHY1205. PRE-LECTURE SKILLS by Matthew James Sykes in CHY1205. NUMBAS - Surds by Katy Dobson in Katy's workspace. Summer Project 2 - practice by Inbbavathie Ravi in Lee's Newcastle Maths & Stats Summer Project 2017. Summer Project 3 - practice by Inbbavathie Ravi in Lee's Newcastle Maths & Stats Summer Project 2017. Summer Project 4 - practice by Inbbavathie Ravi in Lee's Newcastle Maths & Stats Summer Project 2017. Surds by Jo-Ann Lyons in Jo-Ann's workspace. Portfolio - Logs and Indices by Paul Finley in Paul's workspace. Possible applicable questions from previous year by Tom Bold in Transition to university maths.
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S Bhattacharyya
Articles written in Pramana – Journal of Physics
Volume 60 Issue 1 January 2003 pp 47-52 Reasearch Articles
Analysis of a part of the meteorite which fell at Dergaon (India) on March 2, 16.40 local time (2001) is presented with the help of FTIR, absorption and atomic spectra. The FTIR spectrum exhibits prominent absorption bands in the region 800–1100 cm
−1, originating from the valence vibration of SiO 4, a basic component of the silicate lattice.
Volume 62 Issue 6 June 2004 pp 1299-1301
An emission band system in the region 5700–6700 Å from Dergaon stoney iron meteorite which fell at Dergaon, India on March 2, 16.40 local time (2001) was excited with the help of a continuous 500 mW Ar
+ laser. The band system is attributed to silicate (olivine), a major component of the meteorite.
Volume 71 Issue 3 September 2008 pp 599-610 Research Articles
AC impedance spectroscopy technique has been used to study electrical properties of Bi
3.25La 0.75Ti 3O 12 (BLT) ceramic. Complex impedance plots were fitted with three depressed semicircles, which are attributed to crystalline layer, plate boundary and grain boundary and all three were found to comprise of universal capacitance nature $[C = C_{0}w^{n−1}]$. Grain boundary resistance and capacitance evaluated from complex impedance plots have larger values than that of plate boundary and crystalline layer. The activation energies ($E_{a}$) for DC-conductance in grain boundary, plate boundary and crystalline layer are 0.68 eV, 0.89 eV and 0.89 eV, respectively. Relaxation activation energies calculated from impedance plots showed similar values, 0.81 eV and 0.80 eV for crystalline layer and plate boundary, respectively. These activation energy values are found to be consistent with the $E_{a}$ value of oxygen vacancies in perovskite materials. A mechanism is offered to explain the generation of oxygen vacancies in BLT ceramic and its role in temperature dependence of DC-conductance study.
Volume 82 Issue 3 March 2014 pp 585-605 Research Articles
A K Tickoo R Koul R C Rannot K K Yadav P Chandra V K Dhar M K Koul M Kothari N K Agarwal A Goyal H C Goyal S Kotwal N Kumar P Marandi K Venugopal K Chanchalani N Bhatt S Bhattacharyya C Borwankar N Chouhan S R Kaul A K Mitra S Sahaynathan M Sharma K K Singh C K Bhat
The TeV atmospheric Cherenkov telescope with imaging camera (TACTIC) 𝛾-ray telescope has been in operation at Mt. Abu, India since 2001 to study TeV 𝛾-ray emission from celestial sources. During the last 10 years, apart from consistently detecting a steady signal from the Crab Nebula above ∼1.2 TeV energy, at a sensitivity level of ∼5.0𝜎 in ∼25 h, the telescope has also detected flaring activity from Mrk 421 and Mrk 501 on several occasions. Although we used Crab Nebula data partially, in some of the reported results, primarily for testing the validity of the full data analysis chain, the main aim of this work is to study the long term performance of the TACTIC telescope by using consolidated data collected between 2003 and 2010. The total on-source data, comprising ∼402 h, yields an excess of ∼(3742±192) 𝛾-ray events with a statistical significance of ∼19.9𝜎 . The off-source data, comprising ∼107 h of observation, is found to be consistent with a no-emission hypothesis, as expected. The resulting 𝛾-ray rate for the onsource data is determined to be ∼(9.31±0.48) h
-1. A power law fit (d𝛷/d𝐸 = $f_0E^{−\Gamma}$) with $f_0 \tilde (2.66 \pm 0.29) \times 10^{−11}$ cm -2 s -1 TeV -1 and $\Gamma \tilde$ 2.56 ± 0.10 is found to provide reasonable fit to the inferred differential spectrum within statistical uncertainties. The spectrum matches reasonably well with that obtained by other groups. A brief summary of the improvements in the various subsystems of the telescope carried out recently, which has resulted in a substantial improvement in its detection sensitivity (viz., ∼5𝜎 in an observation period of ∼13 h as compared to ∼25 h earlier) are also presented in this paper. Encouraged by the detection of strong 𝛾-ray signals from Mrk 501 and Mrk 421 on several occasions, there is considerable scope for the TACTIC telescope to monitor similar TeV 𝛾-ray emission activity from other active galactic nuclei on a long-term basis.
Current Issue
Volume 93 | Issue 5 November 2019
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Find the value of $P[\Pi_{i=1}^{10}X_i > C]$ for $C=5$, where $X_{10\times 1}$ is a random vector with $10$ dimensional Normal Distribution having location parameter $\mu_{10\times 1} = (1,1,\dots,1)$ and the scatter parameter $\Sigma = I_{10}$, where $I_{10}$ is the $10 \times 10$ identity matrix.
My question is related to this question. In the previously asked question main interest was Cauchy Distribution. But my problem is for normal distribution. How I can handle this problem. I am able to use Monte Carlo simulation. So, I get value of this probability with Monte Carlo Simulation. Now, I am interested to compare that value with exact probability.
Update
I mainly use
Probability[] but can't solve the problem with this function
c = 5;n = 1;Probability[Product[x[i], {i, 1, n}] > c, Table[x[i] \[Distributed] NormalDistribution[a, b], {i, 1, n}]](* Out[20]= 1/2 (1-Erf[2 Sqrt[2]]))c = 5;n = 2;Probability[Product[x[i], {i, 1, n}] > c, Table[x[i] \[Distributed] NormalDistribution[1, 1], {i, 1, n}]]
This is taking huge amount of time..
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In order to analyze the magnitude of earthquakes or compare the magnitudes of two different earthquakes, we need to be able to convert between logarithmic and exponential form. For example, suppose the amount of energy released from one earthquake were 500 times greater than the amount of energy released from another. We want to calculate the difference in magnitude. The equation that represents this problem is [latex]{10}^{x}=500[/latex], where
x represents the difference in magnitudes on the Richter Scale. How would we solve for x?
We have not yet learned a method for solving exponential equations. None of the algebraic tools discussed so far is sufficient to solve [latex]{10}^{x}=500[/latex]. We know that [latex]{10}^{2}=100[/latex] and [latex]{10}^{3}=1000[/latex], so it is clear that
x must be some value between 2 and 3, since [latex]y={10}^{x}[/latex] is increasing. We can examine a graph to better estimate the solution.
Estimating from a graph, however, is imprecise. To find an algebraic solution, we must introduce a new function. Observe that the graph above passes the horizontal line test. The exponential function [latex]y={b}^{x}[/latex] is
one-to-one, so its inverse, [latex]x={b}^{y}[/latex] is also a function. As is the case with all inverse functions, we simply interchange x and y and solve for y to find the inverse function. To represent y as a function of x, we use a logarithmic function of the form [latex]y={\mathrm{log}}_{b}\left(x\right)[/latex]. The base b logarithm of a number is the exponent by which we must raise b to get that number.
We read a logarithmic expression as, “The logarithm with base
b of x is equal to y,” or, simplified, “log base b of x is y.” We can also say, “ b raised to the power of y is x,” because logs are exponents. For example, the base 2 logarithm of 32 is 5, because 5 is the exponent we must apply to 2 to get 32. Since [latex]{2}^{5}=32[/latex], we can write [latex]{\mathrm{log}}_{2}32=5[/latex]. We read this as “log base 2 of 32 is 5.”
We can express the relationship between logarithmic form and its corresponding exponential form as follows:
Note that the base
b is always positive.
Because logarithm is a function, it is most correctly written as [latex]{\mathrm{log}}_{b}\left(x\right)[/latex], using parentheses to denote function evaluation, just as we would with [latex]f\left(x\right)[/latex]. However, when the input is a single variable or number, it is common to see the parentheses dropped and the expression written without parentheses, as [latex]{\mathrm{log}}_{b}x[/latex]. Note that many calculators require parentheses around the
x.
We can illustrate the notation of logarithms as follows:
Notice that, comparing the logarithm function and the exponential function, the input and the output are switched. This means [latex]y={\mathrm{log}}_{b}\left(x\right)[/latex] and [latex]y={b}^{x}[/latex] are inverse functions.
A General Note: Definition of the Logarithmic Function
A
logarithm base b of a positive number x satisfies the following definition.
For [latex]x>0,b>0,b\ne 1[/latex],
where,
we read [latex]{\mathrm{log}}_{b}\left(x\right)[/latex] as, “the logarithm with base bof x” or the “log base bof x.” the logarithm yis the exponent to which bmust be raised to get x.
Also, since the logarithmic and exponential functions switch the
x and y values, the domain and range of the exponential function are interchanged for the logarithmic function. Therefore, the domain of the logarithm function with base [latex]b \text{ is} \left(0,\infty \right)[/latex]. the range of the logarithm function with base [latex]b \text{ is} \left(-\infty ,\infty \right)[/latex]. Q & A Can we take the logarithm of a negative number? No. Because the base of an exponential function is always positive, no power of that base can ever be negative. We can never take the logarithm of a negative number. Also, we cannot take the logarithm of zero. Calculators may output a log of a negative number when in complex mode, but the log of a negative number is not a real number. How To: Given an equation in logarithmic form [latex]{\mathrm{log}}_{b}\left(x\right)=y[/latex], convert it to exponential form. Examine the equation [latex]y={\mathrm{log}}_{b}x[/latex] and identify b, y, and x. Rewrite [latex]{\mathrm{log}}_{b}x=y[/latex] as [latex]{b}^{y}=x[/latex]. Example 1: Converting from Logarithmic Form to Exponential Form
Write the following logarithmic equations in exponential form.
[latex]{\mathrm{log}}_{6}\left(\sqrt{6}\right)=\frac{1}{2}[/latex] [latex]{\mathrm{log}}_{3}\left(9\right)=2[/latex] Solution
First, identify the values of
b, y, and x. Then, write the equation in the form [latex]{b}^{y}=x[/latex]. [latex]{\mathrm{log}}_{6}\left(\sqrt{6}\right)=\frac{1}{2}[/latex]
Here, [latex]b=6,y=\frac{1}{2},\text{and } x=\sqrt{6}[/latex]. Therefore, the equation [latex]{\mathrm{log}}_{6}\left(\sqrt{6}\right)=\frac{1}{2}[/latex] is equivalent to [latex]{6}^{\frac{1}{2}}=\sqrt{6}[/latex].
[latex]{\mathrm{log}}_{3}\left(9\right)=2[/latex]
Here,
b= 3, y= 2, and x= 9. Therefore, the equation [latex]{\mathrm{log}}_{3}\left(9\right)=2[/latex] is equivalent to [latex]{3}^{2}=9[/latex]. Try It 1
Write the following logarithmic equations in exponential form.
a. [latex]{\mathrm{log}}_{10}\left(1,000,000\right)=6[/latex]
b. [latex]{\mathrm{log}}_{5}\left(25\right)=2[/latex]
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Multiplicity Results for the Biharmonic Equation with Singular Nonlinearity of Super Exponential Growth in ℝ 4 21 Downloads Abstract
We consider the following elliptic problem of exponential-type growth posed in an open bounded domain with smooth boundary
B 1 (0) ⊂ ℝ 4: \((P_\lambda)\begin{cases}\Delta^{2}u = \lambda(u^{-\delta}+h(u)e^{u^{\alpha}}),\;\;u>0\;in\;B_{1}(0),\\\;\;\;\;\;u=\Delta{u}=0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;on\;\partial{B}_{1}(0).\end{cases}\) Here Δ 2(.):= −Δ(−Δ)(.) denotes the biharmonic operator, 1 < α < 2, 0 < δ < 1, λ > 0, and h( t) is assumed to be a smooth “perturbation” of \({e^{{t^\alpha }}}\) as t→∞ (see (H1)–(H4) below). We employ variational methods in order to show the existence of at least two distinct (positive) solutions to the problem (P λ) in \({H^2} \cap H_0^1({B_1}(0))\). Keywordsbiharmonic equation multiple solutions super exponential growth Dirichlet boundary conditions Preview
Unable to display preview. Download preview PDF.
References 1. 2. 3. 4. 5. 6. 7. 8. 9.R. Dhanya, J. Giacomoni, S. Prashanth, and K. Saoudi, “Multiplicity results for elliptic equations with singular nonlinearity of super exponential growth in ℝ 2,” Advances in Differential Equations, 17(3–4), (March/April) (2012).Google Scholar 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21.P. L. Lions, “The concentration compactness principale in calculus of variations Part 1,” Revista Mathematica Iberoamericana 1, 185–201 (1985).Google Scholar 22. 23. 24. 25. 26. 27. 28. 29.
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The key point here is that without an air gap an inductor will saturate if you try to put any current through it so inductance will fall and you can't store any energy.
The term "Flyback Transformer" is a little misleading and its more useful to consider it as coupled inductors rather than a transformer because the action is quite different with a conventional transformer energy is going into the primary and out of the secondary at the same time it does not store energy. With a "Flyback" transformer energy is first stored then released.
Taking some things we know about inductors
$$v = L \frac{di}{dt} = N A \frac {dB}{dt}$$
Where v is voltage, i is current, N is turns, B is flux density and A is the effective magnetic area.
Also
$$H = \frac {N \ i}{l} \Rightarrow i = \frac {H \ l}{N} $$
where H is the magnetic field strength, N is turns and l is magnetic path length
Finally permiability
$$ \mu = \frac {B}{H} \Rightarrow H = \frac {B}{\mu} $$
Thus
$$i = \frac{B \ l}{\mu \ N}$$
Now we can calculate Energy
$$$$\begin{align}Energy & = \int{i \ v} \ dt\\ & = \int{\left( \frac{B \ l}{\mu \ N} \right) \ \left( N A \frac {dB}{dt} \right)} \ dt\\ & = \frac {A \ l}{\mu}\int{B} \ dB\\ & = \frac {A \ l}{\mu}\frac{B^2}{2}\\\end{align}$$$$
The energy storage is therefore only possible in the air gap and is proportional to be air gap volume and the square of the flux density.
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