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Perm - Maple Help
Home : Support : Online Help : Mathematics : Discrete Mathematics : Combinatorics : Permutations : Perm
construct a permutation class
Perm( L )
Perm( LL )
list(posint) : a list of positive integers that forms a permutation of 1..n, for some n
list(list(posint)) : a list of lists of positive integers representing disjoint cycles
\left\{1,2,\dots ,n\right\}
n
n
\left[{c}_{1},{c}_{2},\dots ,{c}_{k}\right]
{c}_{i}
\left[{i}_{1},{i}_{2},\dots ,{i}_{m}\right]
{i}_{1}↦{i}_{2}↦{i}_{m}↦{i}_{1}
The Perm constructor creates a permutation, given a specification of its disjoint cycle structure in the form of a list of lists. You can also use a permutation list, which is just the representation of the permutation as a list L of points in which L[ i ] specifies the image of i under the permutation. In particular, the identity permutation is represented by the expression Perm([]).
The Permutation Operations in GroupTheory page lists commands that operate on permutation objects and are part of the GroupTheory package.
Note that the non-commutative multiplication operator . can be used to multiply permutations.
a≔\mathrm{Perm}\left([[1,2],[3,4,5]]\right)
\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{≔}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\right)\left(\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}\right)
a[1]
\textcolor[rgb]{0,0,1}{2}
a[2]
\textcolor[rgb]{0,0,1}{1}
a[3]
\textcolor[rgb]{0,0,1}{4}
a[4]
\textcolor[rgb]{0,0,1}{5}
a[5]
\textcolor[rgb]{0,0,1}{3}
b≔\mathrm{Perm}\left([[1,3],[2,6]]\right)
\textcolor[rgb]{0,0,1}{b}\textcolor[rgb]{0,0,1}{≔}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\right)\left(\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{6}\right)
In the following examples, the PermDegree() and PermProduct() commands are part of the GroupTheory package. They operate on permutation objects constructed by Perm.
\mathrm{with}\left(\mathrm{GroupTheory}\right):
\mathrm{PermDegree}\left(a\right)
\textcolor[rgb]{0,0,1}{5}
\mathrm{PermDegree}\left(b\right)
\textcolor[rgb]{0,0,1}{6}
\mathrm{PermProduct}\left(a,b\right)
\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{6}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}\right)
The Perm command was introduced in Maple 17.
Permutation Operations in GroupTheory |
Structural, Electrical, and Ethanol-Sensing Properties of Nanoparticles
Nguyen Thi Thuy, Dang Le Minh, Ho Truong Giang, Nguyen Ngoc Toan, "Structural, Electrical, and Ethanol-Sensing Properties of Nanoparticles", Advances in Materials Science and Engineering, vol. 2014, Article ID 685715, 5 pages, 2014. https://doi.org/10.1155/2014/685715
Nguyen Thi Thuy,1 Dang Le Minh,2 Ho Truong Giang,3 and Nguyen Ngoc Toan3
1Physics Department, Hue University’s College of Education, Hue, Vietnam
2Faculty of Physics, Hanoi University of Science, VNU, Hanoi, Vietnam
3Institute of Material Science, Institute of Technology and Science, Hanoi, Vietnam
The nanocrystalline () powders with orthorhombic perovskite phase were prepared by sol-gel method. The average crystallite sizes of powders are about 20 nm. The resistance and gas-sensing properties of the based sensors were investigated in the temperature range from 160 to 300°C. The results demonstrated that the resistance and response of the perovskite thick films changed with the increase of Nd content.
There has been much interest in perovskite structured compounds (of general formula ABO3) because of their catalytic activity, colossal magnetoresistance effects, thermoelectric effects, gas-sensing properties, and so forth [1–8]. Specially perovskite oxides with AFeO3 structure (A: rare earth) have shown the good gas-sensing properties such as LaFeO3, , , , and . Among the modified perovskites, showed the best ethanol gas-sensing characteristics; its response to 100 ppm ethanol was more than 80% in the temperature range from 140 to 240°C; it was also found that the based sensor had the best response and selectivity to ethanol gas; the response to 500 ppm ethanol is 128 at 220°C or the highest response to 500 ppm ethanol gas reaches 57.8 at 260°C for sensor and so forth [9–12]. Numerous perovskites show p-type semiconductor properties in air. Oxygen adsorption enhances the conductivity of these materials on account of the increased concentration of holes, which are the main charge carrier species in p-type semiconductors. Furthermore, their resistance increases by applying reducing gases, such as ethanol. Interaction between the reducing gas and the oxygen adsorbed on the metal oxide surface leads to a change in conductance [13–16]. Perovskite powder AFeO3, used in thick film gas sensors, can be manufactured by different chemical methods: coprecipitation method, sol-gel method, and hydrothermal method. They are used broadly due to their advantage in which precursors can be admixed at atomic scale. So, the products are pure and homogeneous. The products also have small grain size and great surface area and are compatible in metal oxide semiconductor (MOS) gas sensors.
In this paper, perovskite oxides were prepared by a citrate-gel method. The influence of Nd doping on the A site of the crystalline structure of LaFeO3 and also on their ethanol-sensing characteristics has been investigated in detail.
Nanopowders of were prepared by a sol-gel (citrate-gel) method, which is based on the chelation of the metal cations by citric acid in a solution of water. The specified amount of Fe(NO3)3·9H2O; La(NO3)3·6H2O; and Nd(NO3)3·6H2O was first dissolved in citric acid solution and then mixture was stirred slowly and kept at a temperature of 70°C until the reaction mixture became clear. To completely create compound matters, ammonium solution was added drop by drop at a time until the pH reached 6 and 7. The complete dissolution of the salts resulted in a transparent solution. After continuously stirring for 2 hours the brown semitransparent sol was produced, and then the solution containing La, Fe, and Nd cations was homogenized; the solution became more viscous as the temperature was continuously kept at 70°C, without showing any visible phase separation. This resin was placed in a furnace and dried to 120°C for 4 h in air to pulverize into powders. The crystalline phase was obtained by heating the powder 500°C for 10 h in air.
Structural characterization was performed by means of X-ray diffraction using a D5005 diffractometer with Cu K radiation and with varied in the range of 10–70° at a step size of 0.02°. The particle size and morphology of the calcined powders were examined by SEM (-4800), Hitachi-Japan.
The fabrication of thick films, structure of sensor prototypes, and measuring conditions were described in [17]. In order to improve their stability and repeatability, the thick film sensors were calcined at 400°C for 2 h in air. The gas sensitivity of sensors was measured in a temperature range of 100°C–300°C. Their resistance was measured in air with test gas equipment. The response, , was defined by the following equation: where is the resistance of sensor measured in air and is the resistance of sensors measured in the test gas equipment.
XRD patterns of the samples were shown in Figure 1. All of them are single phase, with orthorhombic structure (space group Pnma). The wide diffraction peaks (in position of 2 about 32-33°) show that the samples have small grain size. The a-cell parameter versus Nd content is presented in Figure 2, and it can be seen that the a-cell parameter of the samples decreases with the increase of Nd doping concentration. The lattice distortion may be caused by the radius of (0.127 Å) that is smaller than one of (0.136 Å). It leads to the decrease of the lattice parameters with increase of the Nd concentration (Figure 2).
XRD patterns of nanoparticles after annealing in air at 500°C for 10 hours.
a-Cell parameter versus Nd content.
The crystalline sizes (nm) of the samples are calculated by Scherrer formula: where is the average size of crystalline particle, assuming that particles are spherical, , is the wavelength of X-ray radiation, is full width at half maximum of the diffracted peak, and is angle of diffraction.
The cell parameters and the crystalline sizes of powdersare shown in Table 1. These small grain sizes of the nanopowders are favourable for preparing the thick film sensors.
Compounds (Å) (Å) (Å) (Å)3 (nm)
0 LaFeO3 5.5656 5.2544 7.5659 221.256 20.31
0.15 La0.85Nd0.15FeO3 5.5538 5.2432 7.5498 219.850 19.62
0.30 La0.7Nd0.3FeO3 5.54500 5.2350 7.5379 218.811 21.24
0.50 La0.5Nd0.5FeO3 5.5406 5.2308 7.5319 218.280 19.34
1 NdFeO3 5.5046 5.1967 7.4829 214.050 17.40
The cell parameters and crystallite sizes of powders.
The thick film sensors were prepared by using the nanopowder and their ethanol-sensing characters were studied. The resistance of these sensors was examinated with the different temperatures and ethanol concentrations. Figure 3 presents the temperature dependence of resistance of thick film sensors based on the nanosized in the temperature range from 160°C to 300°C in air. It is suggested that the electrical conductivity mechanism is small polaron hopping process [18, 19] following the equation where is constant relating to carrier concentration, is the temperature, is the Boltzmann constant, and is activation energy. Figure 4 shows the temperature dependent on conductivity and Figure 5 demonstrates the Arrhenius plots of conductivities of the samples. From Figure 5 the activation energy can be calculated (Table 2).
(0 ≤ ≤ 1.0) = 0.0 = 0.15 = 0.3 = 0.5 = 1.0
(kJ mol−1) 27 28 26 27 20
The Activation energy () of the electrical conduction process.
Resistance versus temperature of () measured in air.
Electrical conductivity versus temperature of () measured in air.
Arrhenius plots of electrical conductivity for ().
It is noted that the resistance was decreased with increasing temperature due to an intrinsic characteristic of a semiconductor. This would result from the ionization of oxygen vacancies. LaFeO3 and doped-LaFeO3 are the kind of p-type semiconductive material [20].
When the sensor is exposed to ethanol, the ethanol reacts with the chemisorbed oxygen, releasing electrons back to the valence band, decreasing the holes concentration, and increasing resistance [16]. Figure 6 depicts the response and recovery curve of when exposed to 0.25 mg/L ethanol at 212°C. The response and recovery times of this sensor are relatively short. The doping at A site caused a disorder in structure and oxygen deficiency can occur during heating sample at high temperature. On the other hand, interacts with the oxygen, by transferring the electrons from the valence band to adsorbed oxygen atoms, forming ionic species such as or . The electron transferring from the valence band to the chemisorbed oxygen results in an increase in holes concentration and a reduction in resistance of these sensors.
Response and recovery curve of when exposed to 0.25 mg/L ethanol at 212°C.
The temperature dependence of the sensor responses to 0.25 mg/L ethanol is shown in Figure 7. We found that the sensors’ sensitivity increases with Nd replaced concentration. On the other hand, the temperature, at which sensor responses reach maximum value, decreases with increasing Nd replaced concentrations. All sensors showed excellent ethanol-sensing characteristics. The response of was positive; this suggests that the semiconductivity is p-type behavior. Mechanism of gas-sensing is based on the oxidation-reduction on the surface of the material. The absorbed accelerates the reaction: This should give an increase in and thus increase the sensitivity of these sensors [9–13].
Temperature dependence of the response in 0.25 mg/L ethanol of sensors.
Figure 8 presents the dependence of the response upon the concentration of ethanol at 182°C for the sensor. The change of electric resistance of the sensor is strongly affected by an increase in ethanol gas concentration.
Ethanol concentration dependence of response of at 182°C.
The perovskite compounds with orthorhombic perovskite structure were prepared successfully by gel-citrate method. With increasing of the Nd replaced concentrations, both the particle size and a-cell parameter of the samples decrease. The nanocrystallite materials were manufactured thick film sensors and studied ethanol-sensing characters. All sensors showed excellent ethanol-sensing characteristics. The lattice structure of is strongly distorted, and this leads to the change of the ethanol-sensing characters as function of replaced Nd concentrations.
This work was supported by Vietnam’s National Foundation for Science and Technology Development (NAFOSTED) with the project code “103.03.69.09.”
L. Zhang, J. Hu, P. Song, H. Qin, and M. Jiang, “Electrical properties and ethanol-sensing characteristics of perovskite La1−XPbxFeO3,” Sensors and Actuators B Chemical, vol. 114, no. 2, pp. 836–840, 2006. View at: Publisher Site | Google Scholar
V. Caignaert, A. Maignan, and B. Raveau, “Up to 50 000 per cent resistance variation in magnetoresistive polycrystalline perovskites Ln23Sr13MnO3 (Ln=Nd; Sm),” Solid State Communications, vol. 95, no. 6, pp. 357–359, 1995. View at: Publisher Site | Google Scholar
N. Gayathri, A. K. Raychaudhuri, S. K. Tiwary, R. Gundakaram, A. Arulraj, and C. N. R. Rao, “Electrical transport, magnetism, and magnetoresistance in ferromagnetic oxides with mixed exchange interactions: a study of the La0.7Ca0.3Mn1–xCoxO3 system,” Physical Review B, vol. 56, no. 3, pp. 1345–1353, 1997. View at: Publisher Site | Google Scholar
H. Taguchi, M. Nagao, and M. Shimada, “Mechanism of metal-insulator transition in the systems ( Ln1-xCax)MnO 3-δ( Ln: La, Nd, and Gd) and (Nd 0.1Ca 0.9-ySr y)MnO 2.97,” Journal of Solid State Chemistry, vol. 97, no. 2, pp. 476–480, 1992. View at: Google Scholar
Md. A. Choudhury, S. Akhter, D. L. Minh, N. D. Tho, and N. Chau, “Large magnetic-entropy change above room temperature in the colossal magnetoresistance La0.7Sr0.3Mn1-xNixO3 materials,” Journal of Magnetism and Magnetic Materials, vol. 272–276, pp. 1295–1297, 2004. View at: Google Scholar
K. Iwasaki, T. Ito, M. Yoshino, T. Matsui, T. Nagasaki, and Y. Arita, “Power factor of La1-xSrxFeO3 and LaFe1-yNiyO3,” Journal of Alloys and Compounds, vol. 430, no. 1-2, pp. 297–301, 2007. View at: Publisher Site | Google Scholar
M.-H. Hung, M. D. M. Rao, and D.-S. Tsai, “Microstructures and electrical properties of calcium substituted LaFeO3 as SOFC cathode,” Materials Chemistry and Physics, vol. 101, no. 2-3, pp. 297–302, 2007. View at: Publisher Site | Google Scholar
D. Bayraktar, F. Clemens, S. Diethelm, T. Graule, J. Van herle, and P. Holtappels, “Production and properties of substituted LaFeO3-perovskite tubular membranes for partial oxidation of methane to syngas,” Journal of the European Ceramic Society, vol. 27, no. 6, pp. 2455–2461, 2007. View at: Publisher Site | Google Scholar
X. Liu, B. Cheng, J. Hu, H. Qin, and M. Jiang, “Semiconducting gas sensor for ethanol based on LaMgxFe1−xO3 nanocrystals,” Sensors and Actuators B: Chemical, vol. 129, no. 1, pp. 53–58, 2008. View at: Publisher Site | Google Scholar
H. Suo, F. Wu, Q. Wang et al., “Study on ethanol sensitivity of nanocrystalline La0.7Sr0.3FeO3-based gas sensor,” Sensors and Actuators B, vol. 45, no. 3, pp. 245–249, 1997. View at: Publisher Site | Google Scholar
L. Chena, J. Hua, S. Fanga et al., “Ethanol-sensing properties of SmFe1−xNixO3 perovskite oxides,” Sensors and Actuators B, vol. 139, pp. 407–410, 2009. View at: Google Scholar
N. N. Toan, S. Saukko, and V. Lantto, “Gas sensing with semiconducting perovskite oxide LaFeO3,” Physica B: Condensed Matter, vol. 327, no. 2–4, pp. 279–282, 2003. View at: Publisher Site | Google Scholar
J. Xu, J. Han, Y. Zhang, Y. Sun, and B. Xie, “Studies on alcohol sensing mechanism of ZnO based gas sensors,” Sensors and Actuators, B: Chemical, vol. 132, no. 1, pp. 334–339, 2008. View at: Publisher Site | Google Scholar
J. R. Stetter, W. R. Penrose, and S. Yao, “Sensors, chemical sensors, electrochemical sensors, and ECS,” Journal of the Electrochemical Society, vol. 150, no. 2, pp. S11–S16, 2003. View at: Publisher Site | Google Scholar
H. Suo, J. Wang, E. Wu, G. Liu, B. Xu, and M. Zhao, “Influence of Sr content on the ethanol sensitivity of nanocrystalline La1−xSrxFeO3,” Journal of Solid State Chemistry, vol. 130, pp. 152–153, 1997. View at: Publisher Site | Google Scholar
L. Zhang, J. Hu, P. Song, H. Qin, and M. Jiang, “Electrical properties and ethanol-sensing characteristics of perovskite
{La}_{1- x}{Pb}_{x}{FeO}_{3}
,” Sensors and Actuators B, vol. 114, pp. 836–840, 2006. View at: Publisher Site | Google Scholar
H. T. Giang, H. T. Duy, P. Q. Ngan, G. H. Thai, D. T. A. Thu, and N. N. Toan, “Hydrocarbon gas sensing of nano-crystalline perovskite oxides LnFeO3 (Ln = La, Nd and Sm),” Sensors and Actuators B: Chemical, vol. 158, no. 1, pp. 246–251, 2011. View at: Publisher Site | Google Scholar
M. Hung, M. V. M. Rao, and D. Tsai, “Microstructures and electrical properties of calcium substituted LaFeO3 as SOFC cathode,” Materials Chemistry and Physics, vol. 101, no. 2-3, pp. 297–302, 2007. View at: Publisher Site | Google Scholar
S. Komine and E. Iguchi, “Dielectric properties in LaFe0.5 Ga0.5 O3,” Journal of Physics and Chemistry of Solids, vol. 68, no. 8, pp. 1504–1507, 2007. View at: Publisher Site | Google Scholar
Copyright © 2014 Nguyen Thi Thuy et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. |
\begin{array}{cccccc}Ten-Crores& Crores& Ten-Lacs& Lacs& Ten-Thousands& Thousands\\ 6& 8& 9& 7& 4& 5\end{array}\begin{array}{ccc}Hundreds& Tens& Units\\ 1& 3& 2\end{array}
6×{10}^{8}
{5}^{n}
{2}^{n}
975436×{5}^{4}
\frac{9754360000}{{2}^{4}}
\left({x}^{n}-{a}^{n}\right)
\left({x}^{n}-{a}^{n}\right)
\left({x}^{n}+{a}^{n}\right)
\frac{n}{2}\left[2a+\left(n-1\right)d\right]
\frac{n}{2}\left[a+l\right]
\left(1+2+3+...+n\right)=\frac{1}{2}n\left(n+1\right)
\left({1}^{2}+{2}^{2}+{3}^{2}+...+{n}^{2}\right)=\frac{1}{6}n\left(n+1\right)\left(2n+1\right)
\left({1}^{3}+{2}^{3}+{3}^{3}+...+{n}^{3}\right)=\frac{1}{4}{n}^{2}\left(n+1\right){ }^{2}
a,ar,a{r}^{2},a{r}^{3},...
a{r}^{n-1}
\left\{\begin{array}{ll}\frac{a\left(1-{r}^{n}\right)}{\left(1-r\right)}& where r<1\\ \frac{a\left({r}^{n}-1\right)}{\left(r-1\right)}& where r>1\end{array}\right\
Let the two parts be (54 - x) and x.
Then, 10 (54 - x) + 22x = 780
Bigger part = (54 - x) = 34.
The denominator of a fraction is 3 more than the numerator. If the numerator as well as the denominator is increased by 4, the fraction becomes 4/5. What was the original fraction ?
Answer & Explanation Answer: B) 8/11
Let the numerator be x Then, denominator = x + 3.
Now. (x + 4)/(x + 3) +4 = 4/5 <=> 5 (x + 4) = 4(x + 7)
The fraction is 8/11.
What is the common ratio of the following geometric sequence?
4, 2, 1, 0.5, 0.25, 0.125,...
A) 0.5 B) -1
C) 1.5 D) -0.5
Herein the given sequence, 4, 2, 1, 0.5, 0.25, 0.125,...
Common Ratio r = 2/4 = 1/2 = 0.5/1 = 0.25/0.5 = 0.125/0.25 == 0.5.
Let the three integers be x, x + 2 and x + 4. Then, 3x = 2 (x + 4) + 3 <=> x = 11.
The difference between two numbers is 1365. When larger number is divided by the smaller one, the quotient is 6 and the remainder is 15. The smaller number is ?
Let the numbers be x and 1365+x
Then 1365+x = 6x+15
Let the numbers be x and y. Then, 2x + 3y = 39...(i)and 3x + 2y = 30 ...(ii)
On solving (i) and (ii), we get : x = 6 and y = 9.
larger number = 9
A) 476190476 B) 48617
C) 47619 D) 4587962
By hit and trail, we find that
7) 333333 (47619
And 476190476 x 7 = 3333333333 but smallest number is 47619.
The sum of the squares of two numbers is 3341 and the diference of their squares is 891. The numbers are :
Answer & Explanation Answer: A) 35 and 46
Let the numbers be x and y. Then,
x^2+y^2=3341......(1)
x^2-y^2=891......(2)
Adding (i) and (ii), we get : 2x^2=4232 or x^2= 2116 or x =46
Subtracting (ii) from (i), we get : 2y^2= 2450 or y^2 = 1225 or y = 35
So, the numbers are 35 and 46 |
Equivariant fundamental classes in $\mathrm{RO}(C_2)$–graded cohomology with $\underline{\mathbb{Z}/2}$–coefficients
{C}_{2}
denote the cyclic group of order
2
. Given a manifold with a
{C}_{2}
–action, we can consider its equivariant Bredon
RO\left({C}_{2}\right)
–graded cohomology. We develop a theory of fundamental classes for equivariant submanifolds in
RO\left({C}_{2}\right)
–graded cohomology with constant
ℤ∕2
–coefficients. We show the cohomology of any
{C}_{2}
–surface is generated by fundamental classes, and these classes can be used to easily compute the ring structure. To define fundamental classes we are led to study the cohomology of Thom spaces of equivariant vector bundles. In general, the cohomology of the Thom space is not just a shift of the cohomology of the base space, but we show there are still elements that act as Thom classes, and cupping with these classes gives an isomorphism within a certain range.
equivariant cohomology, equivariant homotopy theory, Bredon cohomology
Primary: 55N91, 55P91
https://www.math.ucla.edu/~chazel/ |
I want to solve these Integrals 1. \int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\sqrt{2}}x}dx 2. \int_0^{\frac{\pi}{2}}\frac{1}{(\sqrt{2}\cos^2x+\sin^2x)^2}
I want to solve these Integrals
{\int }_{0}^{\frac{\pi }{2}}\frac{1}{1+{\mathrm{tan}}^{\sqrt{2}}x}dx
{\int }_{0}^{\frac{\pi }{2}}\frac{1}{{\left(\sqrt{2}{\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x\right)}^{2}}
macalpinee3
For the second integral:
t=\mathrm{tan}x
in order to get:
{\int }_{0}^{\frac{\pi }{2}}\frac{1}{{\left(\sqrt{2}{\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x\right)}^{2}}dx={\int }_{0}^{\mathrm{\infty }}\frac{{t}^{2}+1}{{\left(\sqrt{2}+{t}^{2}\right)}^{2}}
d\left(-\frac{t}{{t}^{2}+\sqrt{2}}=\frac{{t}^{2}-\sqrt{2}}{\left({t}^{2}+\sqrt{2}{\right)}^{2}}\right)
{t}^{2}+1
as the following sum:
a\left({t}^{2}+\sqrt{2}\right)+b\left({t}^{2}-\sqrt{2}\right)
It easy to calculate that
a=\frac{\sqrt{2}+1}{2\sqrt{2}}
b=\frac{\sqrt{2}-1}{2\sqrt{2}}
, from the system of equations
a+b=1
a-b=\frac{1}{\sqrt{2}}
Using the previous, we get
{\int }_{0}^{\mathrm{\infty }}\frac{{t}^{2}+1}{{\left(\sqrt{2}+{t}^{2}\right)}^{2}}dt=a{\int }_{0}^{\mathrm{\infty }}\frac{1}{\sqrt{2}+{t}^{2}}dt+b{\int }_{0}^{\mathrm{\infty }}\frac{{t}^{2}-\sqrt{2}}{{\left(\sqrt{2}+{t}^{2}\right)}^{2}}
{\int }_{0}^{\mathrm{\infty }}\frac{{t}^{2}+1}{\left(\sqrt{2}+{t}^{2}{\right)}^{2}}dt=a\frac{\mathrm{arctan}\frac{t}{\sqrt[4]{2}}}{\sqrt[4]{2}}{|}_{0}^{\mathrm{\infty }}-b\frac{t}{{t}^{2}+\sqrt{2}}{|}_{0}^{\mathrm{\infty }}
{\int }_{0}^{\mathrm{\infty }}\frac{{t}^{2}+1}{\left(\sqrt{2}+{t}^{2}{\right)}^{2}}dt=a\frac{\frac{\pi }{2}}{\sqrt[4]{2}}
{\int }_{0}^{\mathrm{\infty }}\frac{{t}^{2}+1}{\left(\sqrt{2}+{t}^{2}}dt=\frac{\sqrt{2}+1}{4\sqrt{2}}\frac{\pi }{\sqrt[4]{2}}
I={\int }_{0}^{\frac{\pi }{2}}\frac{1}{1+{\mathrm{tan}}^{\sqrt{2}}x}dx
{\int }_{0}^{a}f\left(x\right)dx={\int }_{0}^{a}f\left(a-x\right)dx
I={\int }_{0}^{\frac{\pi }{2}}\frac{1}{1+{\mathrm{cot}}^{\sqrt{2}}x}dx={\int }_{0}^{\frac{\pi }{2}}\frac{{\mathrm{tan}}^{\sqrt{2}}x}{1+{\mathrm{tan}}^{\sqrt{2}}x}dx
2I={\int }_{0}^{\frac{\pi }{2}}1\cdot dx=\frac{\pi }{2}⇒I=\frac{\pi }{4}
\begin{array}{}I\left(a,b,n\right)={\int }_{0}^{\frac{\pi }{2}}\frac{1}{\left(a{\mathrm{cos}}^{2}x+b{\mathrm{sin}}^{2}x{\right)}^{n}}dx\\ \frac{dI\left(a,b,n\right)}{da}=-n{\int }_{0}^{\frac{\pi }{2}}\frac{{\mathrm{cos}}^{2}x}{\left(a{\mathrm{cos}}^{2}x+b{\mathrm{sin}}^{2}x{\right)}^{n+1}}\\ \frac{dI\left(a,b,n\right)}{db}=-n{\int }_{0}^{\frac{\pi }{2}}\frac{{\mathrm{sin}}^{2}x}{\left(a{\mathrm{cos}}^{2}x+b{\mathrm{sin}}^{2}x{\right)}^{n+1}}\\ \text{Therefore}\\ \frac{dI\left(a,b,n\right)}{da}+\frac{dI\left(a,b,n\right)}{db}=-nI\left(a,b,n+1\right)\\ \text{we have}\\ I\left(a,b,n+1\right)=-\frac{1}{n}\left(\frac{dI\left(a,b,n\right)}{da}+\frac{dI\left(a,b,n\right)}{db}\right)\\ \text{Now we should compute}\text{ }I\left(a,b,1\right)\\ I\left(a,b,1\right)={\int }_{0}^{\frac{\pi }{2}}\frac{1}{a{\mathrm{cos}}^{2}x+b{\mathrm{sin}}^{2}x}dx={\int }_{0}^{\frac{/pi}{2}}\frac{1+{\mathrm{tan}}^{2}x}{a+b{\mathrm{tan}}^{2}x}dx\\ \text{Set}\text{ }u=\mathrm{tan}x,\text{ }\text{thus}\\ I\left(a,b,1\right)={\int }_{0}^{\mathrm{\infty }}\frac{1}{a+b{u}^{2}}dx=\frac{\pi }{2\sqrt{ab}}\\ \text{so}\\ \frac{dI\left(a,b,1\right)}{da}=\frac{\pi \sqrt{b}}{4a\sqrt{a}b}\\ \text{similarly}\\ \frac{dI\left(a,b,1\right)}{db}=\frac{\pi \sqrt{a}}{4b\sqrt{b}a}\\ \text{apply}\\ I\left(a,b,2\right)=-\left(\frac{dI\left(a,b,1\right)}{da}+\frac{dI\left(a,b,1\right)}{db}\right)=\frac{\pi \left(a+b\right)}{4ab\sqrt{ab}}\\ \text{set}a=\sqrt{2}\text{ }\text{and}\text{ }b=1,\text{finally}\\ {\int }_{0}^{\frac{\pi }{2}}\frac{1}{\left(\sqrt{2}{\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x{\right)}^{2}}=\frac{\pi \left(1+\sqrt{2}\right)}{4\sqrt[4]{8}}\end{array}
\int \frac{1}{{\left({x}^{2}+1\right)}^{2}}dx
Evaluate the line integral, where C is the given curve. integral C
x{e}^{yzds}
C is the line segment from
\left(0,0,0\right)
\left(1,2,3\right)
Solve this equation please
\int \frac{{e}^{8x}}{\left({e}^{16x}+36\right)}
How to add two compound fractions with fractions in numerator like this one:
\frac{\frac{1}{x}}{2}+\frac{\frac{2}{3x}}{x}
or fractions with fractions in denominator like this one:
\frac{x}{\frac{2}{x}}+\frac{\frac{1}{x}}{x}
Evaluate the line integral
{\int }_{C}27{x}^{2}yzds
with respect to s along the
C:x=t,y={t}^{3},z=\frac{2}{3}{t}^{3}\left(0\le t\le 10
{\int }_{0}^{\mathrm{\infty }}{e}^{-{x}^{2}}dx=\frac{\sqrt{\pi }}{2} |
EUDML | Exponentially convergent parallel discretization methods for the first order evolution equations. EuDML | Exponentially convergent parallel discretization methods for the first order evolution equations.
Exponentially convergent parallel discretization methods for the first order evolution equations.
Gavrilyuk, I., and Makarov, V.. "Exponentially convergent parallel discretization methods for the first order evolution equations.." Computational Methods in Applied Mathematics 1.4 (2001): 333-355. <http://eudml.org/doc/225155>.
@article{Gavrilyuk2001,
author = {Gavrilyuk, I., Makarov, V.},
journal = {Computational Methods in Applied Mathematics},
keywords = {evolution equation; strongly -positive operator; parallel computation; initial value problem; Banach space; Dunford-Cauchy integral; Sinc quadrature formula; strongly -positive operator},
publisher = {Institute of Mathematics of the National Academy of Sciences of Belarus},
title = {Exponentially convergent parallel discretization methods for the first order evolution equations.},
AU - Gavrilyuk, I.
TI - Exponentially convergent parallel discretization methods for the first order evolution equations.
PB - Institute of Mathematics of the National Academy of Sciences of Belarus
KW - evolution equation; strongly -positive operator; parallel computation; initial value problem; Banach space; Dunford-Cauchy integral; Sinc quadrature formula; strongly -positive operator
evolution equation, strongly
P
-positive operator, parallel computation, initial value problem, Banach space, Dunford-Cauchy integral, Sinc quadrature formula, strongly
P
-positive operator
Spectral, collocation and related methods
Abstract parabolic equations
Equations with linear operators (do not use )
Articles by Gavrilyuk
Articles by Makarov |
EUDML | Riesz basis property of Timoshenko beams with boundary feedback control. EuDML | Riesz basis property of Timoshenko beams with boundary feedback control.
Riesz basis property of Timoshenko beams with boundary feedback control.
Feng, De-Xing; Xu, Gen-Qi; Yung, Siu-Pang
Feng, De-Xing, Xu, Gen-Qi, and Yung, Siu-Pang. "Riesz basis property of Timoshenko beams with boundary feedback control.." International Journal of Mathematics and Mathematical Sciences 2003.28 (2003): 1807-1820. <http://eudml.org/doc/50199>.
author = {Feng, De-Xing, Xu, Gen-Qi, Yung, Siu-Pang},
keywords = {-semigroups; exponential decay; -semigroups},
title = {Riesz basis property of Timoshenko beams with boundary feedback control.},
AU - Feng, De-Xing
AU - Xu, Gen-Qi
AU - Yung, Siu-Pang
TI - Riesz basis property of Timoshenko beams with boundary feedback control.
KW - -semigroups; exponential decay; -semigroups
{C}_{0}
-semigroups, exponential decay,
{C}_{0}
(Generalized) eigenfunction expansions; rigged Hilbert spaces
Accretive operators, dissipative operators, etc.
Higher-order hyperbolic systems
Articles by Feng
Articles by Yung |
Find the solution of {f}{''}{\left({x}\right)}={8}{x}+ \sin{{x}}
f\left(x\right)=8x+\mathrm{sin}x
\frac{{d}^{2}f\left(x\right)}{{dx}^{2}}=8x+\mathrm{sin}\left(x\right):
\frac{df\left(x\right)}{dx}=\int \left(8x+\mathrm{sin}\left(x\right)\right)dx=4{x}^{2}-\mathrm{cos}\left(x\right)+{c}_{1}
{c}_{1}
Take the integral:
\int \left(8x+\mathrm{sin}\left(x\right)\right)dx
Integrate the sum term by term and factor out constants:
=\int \mathrm{sin}\left(x\right)dx+8\int xdx
The integral of
\mathrm{sin}\left(x\right)
-\mathrm{cos}\left(x\right)
=-\mathrm{cos}\left(x\right)+8\int xdx
The integral of x is
\frac{{x}^{2}}{2}:
=4{x}^{2}-\mathrm{cos}\left(x\right)+\text{constant}
f\left(x\right)=\int \left(4{x}^{2}-\mathrm{cos}\left(x\right)+{c}_{1}\right)dx=\frac{4{x}^{3}}{3}-\mathrm{sin}\left(x\right)+x{c}_{1}+{c}_{2}
{c}_{2}
\int \left({c}_{1}+4{x}^{2}-\mathrm{cos}\left(x\right)\right)dx
={c}_{1}\int 1dx-\int \mathrm{cos}\left(x\right)dx+4\int {x}^{2}dx
The integral of 1 is x:
={c}_{1}x-\int \mathrm{cos}\left(x\right)dx+4\int {x}^{2}dx
\mathrm{cos}\left(x\right)
\mathrm{sin}\left(x\right)
=-\mathrm{sin}\left(x\right)+{c}_{1}x+4\int {x}^{2}dx
{x}^{2}
\frac{{x}^{3}}{3}
={c}_{1}x+\frac{4{x}^{3}}{3}-\mathrm{sin}\left(x\right)+\text{constant}
x\left(\frac{dy}{dx}\right)+3\left(y+{x}^{2}\right)=\frac{\mathrm{sin}x}{x}
y{}^{″}-{y}^{\prime }-12y=0
y{}^{″}+4y=\mathrm{cos}x\mathrm{sin}x
Use logarithmic differentiation to find
\frac{dy}{dx}y=x\sqrt{{x}^{2}+48}
Solve the following IVP for the second order linear equations
{y}^{″}-4{y}^{\prime }+9y=0,\text{ }\text{ }y\left(0\right)=0,\text{ }\text{ }{y}^{\prime }\left(0\right)=-8
Determine whether three series converges or diverges. if it converges, find its sum
\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{n}}{{4}^{n}}
\sqrt{16-{x}^{2}}
between -4 and 0 |
Option Greeks - Vega | Brilliant Math & Science Wiki
This is an advanced topic in option theory. Please refer to this wiki Options Glossary if you do not understand any of the terms.
Vega is one of the option Greeks, and it measures the rate of change of the price of the option with respect to volatility. Specifically, the vega of an option tells us by how much the price of an option would increase when volatility increases by 1%.
Note that vega isn't an actual greek letter. It is often represented by nu
(\nu)
, which looks like a "v".
The vega of an option is the sensitivity of the option to a change in volatility:
\nu = \frac{ \partial V } { \partial \sigma },
\nu
is the vega of the option,
V
is the price of the option, and
\sigma
is the symbol for volatility.
As with the other Greeks, the units of vega are often ignored/unstated. It has a unit of
\frac{\$}{\sigma}
Vega of Option
Implications of Put-Call Parity on Vega
Graph of Vega
Vega Changes over Volatility
Vega Changes over Time
The vega of an option tells us how much the price of an option would increase by when volatility increases by 1%. It allows us to make predictions about how much the option value would change as volatility changes.
When the stock is trading at $45, the call option on the $45 strike with 25 days to expiry is worth $3.48 at an implied volatility of 62. If the vega of the option is 0.056, what would be the price of the option when implied volatility is 70?
The volatility has increased by
70 - 62 = 8
vol points. Since the vega of the option is 0.056, our best guess of the option value is that it has increased by
8 \times 0.056 = 0.448
\$3.48 + \$0.448 = \$3.93.\ _\square
The above example shows how knowing the vega of an option allows us to calculate the price change which results from a volatility change. This would be accurate as a first-order approximation. In the above example, because the vega of an ATM option is mostly constant, the approximation is extremely accurate.
The call option on the 49 strike is currently worth $4.50 and has a vega of 0.11.
How much would the call option be worth if volatility increases by 5%?
The vega of an option is always positive. We know that this is instinctively true, since when volatility goes up, we should be increasing the price of protection/insurance which the options offer.
The price will increase. The price will decrease. The price will increase for ATM and decrease for OTM. The price will decrease for ATM and increase for OTM.
As volatility increases, what happens to the price of an option?
C - P = S - K e^{-rt}
. Let us differentiate this equation with respect to volatility:
\frac { \partial } { \partial \sigma } ( C - P ) = \nu_C - \nu_P
, which is the vega of the call minus the vega of the put.
\frac{ \partial } { \partial \sigma} ( S - K e^{-rt} )
Since the underlying price
S
and the present value of the strike
K e^{-rt}
are independent of volatility, the RHS is 0. Thus, we obtain
\nu_C - \nu_P = 0 \Longrightarrow \nu_C = \nu_P.
This tells us that the vega of the call and the put on the same strike and expiration is the same. Thus, to know the vega of an option on a strike, we can consider either the call or the put option, or even consider the case of the straddle!
For example, let's consider the vega of a straddle. As volatility increases, we are likely to see larger moves in the underlying, which will result in a higher payoff as the underlying moves away from the strike. Thus, the vega of the straddle is positive, which implies that the vega of the individual options is positive. This backs up the observation made in the previous section.
-0.8 -0.2 0.2 0.8 Cannot be determined
If the vega of the call on the 30 strike is 0.2, what is the vega of the put on the 30 strike of the same expiry?
To better understand how vega changes with respect to the underlying, see the wiki Vanna.
Let's consider the graph of vega against the underlying:
Put on the 100 strike Put on the 120 strike Call on the 80 strike Call on the 120 strike
Which of the following options (on the same expiry) has the largest vega when the stock is trading at 100?
Explanation for characteristics of the above graph:
To think about the vega of an option, we look at the option value. Since the intrinsic value is constant as volatility changes, we should focus on the extrinsic value.
When the stock is far away from the strike, the extrinsic value is low and an increase in volatility would not affect the payoffs by much. Hence the extrinsic value will not increase significantly, so the vega is low.
When the stock is near the strike, an increase in volatility has a direct effect on the payoffs. Hence the extrinsic value will increase significantly, so the vega is higher.
Let's consider the graph of vega against volatility:
Differentiating the straddle approximation formula with respect to volatility, we see that the ATM vega is pretty constant. (Actually, it is slowly decreasing as volatility increases, but not noticeably so). This is represented by the red line above.
For the other options, when volatility is 0, the extrinsic value is clearly 0. As volatility increases slightly but not sufficiently enough to affect the payoff which is far away, there would be little change in the extrinsic value, hence a low vega. After a while when the stock is volatile enough to result in a payoff, the extrinsic value would start to increase and thus vega becomes larger.
Of course, the further away an option is from ATM, the higher the volatility will have to be before this effect takes place. This explains the difference between the green and blue curves.
For a given volatility, the ATM option has the largest vega, and this sets a maximum limit on the vega of other options.
5
\int_0^{10} \nu \, d vol
5 + \int_0^{5} \nu \, d vol
5 + \int_0^{10} \nu \, d vol
10 + \int_0^{5} \nu \, d vol
10 + \int_0^{10} \nu \, d vol
The stock is trading at 50. For the 55 strike, suppose that we are given the graph of option's vega against volatility. The option has a volatility of 10.
What is the price of the 55 put?
Let's consider how vega changes over time:
The blue curve represents an option with more time to expiry, and the red curve represents an option on the same strike with less time to expiry.
By the straddle approximation formula, the ATM vega is equal to
\frac{ S \sqrt{t} }{2000}
. Hence, as the time to expiry decreases, the ATM vega decreases.
A similar effect happens in the wings, since the underlying has less time to move and thus is less likely to affect the extrinsic value.
Call on the 100 strike with 1 month to expiry Call on the 80 strike with 1 months to expiry Call on the 80 strike with 3 months to expiry Call on the 100 strike with 3 months to expiry
When the stock is trading at 100, which of the following options have the largest vega?
Cite as: Option Greeks - Vega. Brilliant.org. Retrieved from https://brilliant.org/wiki/option-greeks-vega/ |
RiemannWindow - Maple Help
Home : Support : Online Help : Science and Engineering : Signal Processing : Windowing Functions : RiemannWindow
multiply an array of samples by a Riemann windowing function
RiemannWindow(A)
The RiemannWindow(A) command multiplies the Array A by the Riemann windowing function and returns the result in an Array having the same length.
The Riemann windowing function
w\left(k\right)
N
{\begin{array}{cc}1& k=\frac{N}{2}\\ \frac{\mathrm{sin}\left(\left(\frac{2k}{N}-1\right)\mathrm{\pi }\right)}{\left(\frac{2k}{N}-1\right)\mathrm{\pi }}& \mathrm{otherwise}\end{array}
The SignalProcessing[RiemannWindow] command is thread-safe as of Maple 18.
\mathrm{with}\left(\mathrm{SignalProcessing}\right):
N≔1024:
a≔\mathrm{GenerateUniform}\left(N,-1,1\right)
{\textcolor[rgb]{0,0,1}{\mathrm{_rtable}}}_{\textcolor[rgb]{0,0,1}{36893628315422794084}}
\mathrm{RiemannWindow}\left(a\right)
{\textcolor[rgb]{0,0,1}{\mathrm{_rtable}}}_{\textcolor[rgb]{0,0,1}{36893628315267284988}}
c≔\mathrm{Array}\left(1..N,'\mathrm{datatype}'='\mathrm{float}'[8],'\mathrm{order}'='\mathrm{C_order}'\right):
\mathrm{RiemannWindow}\left(\mathrm{Array}\left(1..N,'\mathrm{fill}'=1,'\mathrm{datatype}'='\mathrm{float}'[8],'\mathrm{order}'='\mathrm{C_order}'\right),'\mathrm{container}'=c\right)
{\textcolor[rgb]{0,0,1}{\mathrm{_rtable}}}_{\textcolor[rgb]{0,0,1}{36893628315267260652}}
u≔\mathrm{`~`}[\mathrm{log}]\left(\mathrm{FFT}\left(c\right)\right):
\mathbf{use}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathrm{plots}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathbf{in}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathrm{display}\left(\mathrm{Array}\left(\left[\mathrm{listplot}\left(\mathrm{ℜ}\left(u\right)\right),\mathrm{listplot}\left(\mathrm{ℑ}\left(u\right)\right)\right]\right)\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathbf{end use}
The SignalProcessing[RiemannWindow] command was introduced in Maple 18. |
EUDML | Coverings of foliations anf associated C*-algebras. EuDML | Coverings of foliations anf associated C*-algebras.
Coverings of foliations anf associated C*-algebras.
Moto O'Uchi
O'Uchi, Moto. "Coverings of foliations anf associated C*-algebras.." Mathematica Scandinavica 58 (1986): 69-76. <http://eudml.org/doc/166967>.
@article{OUchi1986,
author = {O'Uchi, Moto},
keywords = {smooth manifold; transverse submanifold; holonomy groupoid; action of a group; -algebras; homogeneous covering map; crossed product; Anosov foliations},
title = {Coverings of foliations anf associated C*-algebras.},
AU - O'Uchi, Moto
TI - Coverings of foliations anf associated C*-algebras.
KW - smooth manifold; transverse submanifold; holonomy groupoid; action of a group; -algebras; homogeneous covering map; crossed product; Anosov foliations
smooth manifold, transverse submanifold, holonomy groupoid, action of a group,
{C}^{*}
-algebras, homogeneous covering map, crossed product, Anosov foliations
{C}^{*}
{W}^{*}
{C}^{*}
Dynamics of group actions other than
𝐙
𝐑
, and foliations
Articles by Moto O'Uchi |
Find the derivatives of the functions. h(x)=e^{2 x^{2}-x+1 / x}
Find the derivatives of the functions.
h\left(x\right)={e}^{2{x}^{2}-x+1\text{ }x2F;x}
Want to know more about Derivatives?
{h}^{\prime }\left(x\right)=\left(4x-1-\frac{1}{{x}^{2}}\right){e}^{2{x}^{2}}-x+\frac{1}{x}
Use the given graph to estimate the value of each derivative.(Round all answers to one decimal place.)Graph uploaded below.
(a) f ' (0)1
(b) f ' (1)2
(c) f ' (2)3
(d) f ' (3)4
(e) f ' (4)5
(f) f ' (5)6
What is the Mixed Derivative Theorem for mixed second-order partial derivatives? How can it help in calculating partial derivatives of second and higher orders? Give examples.
f\left(x\right)=\frac{{2}^{x}}{{2}^{x}+1}
Derivatives Find the derivative of the following functions.
y=\mathrm{ln}2{x}^{8}
f\left(x\right)=7{x}^{2}+7x+2
g\left(x\right)={x}^{3}+6{x}^{2}+7x+4
, find the derivative of the compostion f(g(x)).
Write 10 usages of derivatives.
Second-order derivatives Find y" for the following functions
y=\mathrm{cos}0\mathrm{sin}0 |
An analysis of variance produces SS_(total)=40 and SS_(within)=10. For rhis anal
An analysis of variance produces SS_(total)=40 and SS_(within)=10. For rhis analysis, what is SS_(between)?a) 30b) 400c) Cannot be determined grom the information givend) 50
An analysis of variance produces
S{S}_{total}=40
S{S}_{within}=10
. For rhis analysis, what is
S{S}_{between}
c) Cannot be determined grom the information given
Nathalie Redfern
S{S}_{between}=S{S}_{total}-S{S}_{within}
S{S}_{between}=40-10
S{S}_{between}=30
Geographical Analysis (Oct. 2006) published a study of a new method for analyzing remote-sensing data from satellite pixels in order to identify urban land cover. The method uses a numerical measure of the distribution of gaps, or the sizes of holes, in the pixel, called lacunarity. Summary statistics for the lacunarity measurements in a sample of 100 grassland pixels are
\overline{x}=225\text{ }and\text{ }s=20s=20
. It is known that the mean lacunarity measurement for all grassland pixels is 220. The method will be effective in identifying land cover if the standard deviation of the measurements is 10% (or less) of the true mean (i.e., if the standard deviation is less than 22).
a. Give the null and alternative hypotheses for a test to determine whether, in fact, the standard deviation of all grassland pixels is less than 22.
b. A MINITAB analysis of the data is provided below. Locate and interpret the p-value of the test. Use
\alpha =0.10
. Test for One Standard Deviation Method Null hypothesis
\mathrm{\Sigma }=22
Method Alternative hypothesis
\mathrm{\Sigma }\le 22
The standard method is only for the normal distribution. Statistics NStDevVariance 10020.0400 Tests
Two runners start a race at the same time and finish in a tie. Prove that at some time during the race they have the same speed. [Hint: Consider
f\left(t\right)=g\left(t\right)-h\left(t\right)
, where and are the position functions of the two runners.
Which graphs represent functions that have inverse functions?
Find the exact area enclosed by the following functions. (do not give decimals in the answer)
see (int1) for the functions
f\left(x\right)={x}^{2}
g\left(x\right)=-3x+4
$3450 shares were bought and than sold for $6100 after a while. What is the capital gain as a percent of the original purchace price? What is the gross capital gain?
y={e}^{x}\mathrm{sin}h\left(x\right)
Find the maximum and minimum values attained by the function falong the path
c\left(t\right)
f\left(x,y\right)=xy;\text{ }c\left(t\right)=\left(\mathrm{cos}\left(t\right),\text{ }\mathrm{sin}\left(t\right)\right);\text{ }0\le t\le 2\pi |
Determine whether the given matrices are inverses of each other. A=begin{bmatrix
Determine whether the given matrices are inverses of each other. A=begin{bmatrix} 8 & 3 &-4 -6 & -2 &3-3&1&1 end{bmatrix} text{ and } B=begin{bmatrix} -1 & -1 &-1 3 & 4 &00&1&-2 end{bmatrix}
Determine whether the given matrices are inverses of each other.
A=\left[\begin{array}{ccc}8& 3& -4\\ -6& -2& 3\\ -3& 1& 1\end{array}\right]\text{ and }B=\left[\begin{array}{ccc}-1& -1& -1\\ 3& 4& 0\\ 0& 1& -2\end{array}\right]
Here we are given two matrices:
AB=\left[\begin{array}{ccc}8& 3& -4\\ -6& -2& 3\\ -3& 1& 1\end{array}\right]\left[\begin{array}{ccc}-1& -1& -1\\ 3& 4& 0\\ 0& 1& -2\end{array}\right]
To show that the given matrices are multiplicative inverses of each other.
Multiply AB and BA and if both products equal the identity, then the two matrices are inverses of each other:
Find AB and BA:
AB=\left[\begin{array}{ccc}8& 3& -4\\ -6& -2& 3\\ -3& 1& 1\end{array}\right]\left[\begin{array}{ccc}-1& -1& -1\\ 3& 4& 0\\ 0& 1& -2\end{array}\right]
=\left(\begin{array}{ccc}8\left(-1\right)+3\cdot 3+\left(-4\right)\cdot 0& 8\left(-1\right)+3\cdot 4+\left(-4\right)\cdot 1& 8\left(-1\right)+3\cdot 0+\left(-4\right)\left(-2\right)\\ \left(-6\right)\left(-1\right)+\left(-2\right)\cdot 3+3\cdot 0& \left(-6\right)\left(-1\right)+\left(-2\right)\cdot 4+3\cdot 1& \left(-6\right)\left(-1\right)+\left(-2\right)\cdot 0+3\cdot \left(-2\right)\\ \left(-3\right)\left(-1\right)+1\cdot 3+1\cdot 0& \left(-3\right)\left(-1\right)+1\cdot 4+1\cdot 1& \left(-3\right)\left(-1\right)+1\cdot 0+1\cdot \left(-2\right)\end{array}\right)
=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 6& 9& 1\end{array}\right)
Find the product BA:
BA=\left[\begin{array}{ccc}-1& -1& -1\\ 3& 4& 0\\ 0& 1& -2\end{array}\right]\left[\begin{array}{ccc}8& 3& -4\\ -6& -2& 3\\ -3& 1& 1\end{array}\right]
=\left[\begin{array}{ccc}\left(-1\right)\cdot 8+\left(-1\right)\left(-6\right)+\left(-1\right)\left(-3\right)& \left(-1\right)\cdot 3+\left(-1\right)\left(-2\right)+\left(-1\right)\cdot 1& \left(-1\right)\left(-4\right)+\left(-1\right)\cdot 3+\left(-1\right)1\\ 3\cdot 8+4\left(-6\right)+0\left(-3\right)& 3\cdot 3+4\left(-2\right)+0\cdot 1& 3\left(-4\right)+4\cdot 3+0\cdot 1\\ 0\cdot 8+1\left(-6\right)+\left(-2\right)\left(-3\right)& 0\cdot 3+1\left(-2\right)+\left(-2\right)\cdot 1& 0\left(-4\right)+1\cdot 3+\left(-2\right)1\end{array}\right]
=\left[\begin{array}{ccc}1& -2& 0\\ 0& 1& 0\\ 0& -4& 1\end{array}\right]
So, the product of A and B matrices are not identity matrix. so, that the matrices are not inverse of each other.
2×2
\text{Basis }=\left\{\left[\begin{array}{cc}& \\ & \end{array}\right],\left[\begin{array}{cc}& \\ & \end{array}\right]\right\}
Show that B is the multiplicative inverse of A, where:
A=\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]\text{ and }B=\left[\begin{array}{cc}1& -1\\ -1& 2\end{array}\right]
Solve the system of linear equations using matrices.
Construct a
3×3
matrix A, with nonzero entries, and a vector b in
{\mathbb{R}}^{3}
such that b is not in the set spanned by the columns of A.
Explain the term Comparable matrices?
Find the products AB and BA to determine whether B is the multiplicative inverse of A.
A=\left[\begin{array}{cc}-4& 0\\ 1& 3\end{array}\right],B=\left[\begin{array}{cc}-2& 4\\ 0& 1\end{array}\right] |
A technique for estimating geometrical synchronization of biomedical signals | JVE Journals
Inga Timofejeva1 , Rollin McCraty2 , Mike Atkinson3 , Alfonsas Vainoras4 , Minvydas Ragulskis5
1, 5Department of Mathematical Modelling, Kaunas University of Technology, 51368, Kaunas, Lithuania
2, 3HeartMath Institute, Boulder Creek, CA 95006, USA
4Cardiology Institute, Lithuanian University of Health Sciences, 44307, Kaunas, Lithuania
Received 7 September 2019; accepted 15 September 2019; published 26 September 2019
Copyright © 2019 Inga Timofejeva, et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
A technique for the evaluation of geometrical synchronization between data signals based on optimal attractor reconstruction is demonstrated and validated using coupled chaotic logistic maps. The measure is then applied to estimate the degree of synchronization between human heart rate variability and Earth’s local geomagnetic activity.
Keywords: synchronization, attractor embedding, heart rate variability, magnetic field.
Evaluation of synchronization between data signals is a broadly discussed concept among researchers since it can be applied in the analysis of a wide range of phenomena. Several examples include analysis of biomedical signals in health sciences and biological systems, investigation of coupled circuits or laser systems in electronics and optics [1-3].
In [4, 5] we proposed a technique capable of estimating the degree of geometrical synchronization via near-optimal chaotic attractor embedding. In this paper, the measure is demonstrated and validated using the example of two coupled chaotic logistic maps. The technique is then applied to assess the impact of Earth’s local magnetic field on individual’s biomedical parameters.
2. Estimation of geometrical synchronization between two time series
Geometrical similarity between two time series can be estimated via the algorithms developed and validated in [4, 5]. A brief overview of those algorithms is presented below. The feasibility of the discussed approach is demonstrated by considering two logistic maps with diffusive coupling:
\begin{array}{c}{x}_{k+1}={ax}_{k}\left(1-{x}_{k}\right)\left(1-\epsilon \right)+\epsilon {y}_{k},\\ {y}_{k+1}=b{y}_{k}\left(1-{y}_{k}\right)\left(1-\epsilon \right)+\epsilon {x}_{k},\end{array}
a=b=\text{4}
(such parameter values result in chaotic behavior);
0\le \epsilon \le \text{1}
is the coupling parameter (low values of
\epsilon
result in low synchronization between maps and vice versa). Initial conditions are set to
{x}_{0}=\text{0.3}
{y}_{0}=\text{0.6}
Two resulting trajectories
x=\left({x}_{1},\dots ,{x}_{N}\right)
y=\left({y}_{1},\dots ,{y}_{N}\right)
N =
6000 are sampled using Eq. (1) after transient processes die down. Fig. 1(a) illustrates the evolution of
x
y
and the difference
x-y
when two logistic maps are uncoupled (
\epsilon =0
). Fig. 1(b) depicts
x
y
x-y
\epsilon =\text{0.13}
and Fig. 1(c) corresponds to
\epsilon =\text{0.16}
. It can be seen that the similarity between trajectories
x
y
increases as the coupling parameter
\epsilon
increases; however, the difference
x-y
remains chaotic even at
\epsilon =\text{0.16}
Fig. 1. The trajectories of coupled logistic maps (see Eq. (1)) for different values of the coupling parameter
\epsilon
. Parts a), b) and c) illustrate
x,y
x-y
\epsilon =0
\epsilon =\text{0.13}
\epsilon =\text{0.16}
In order to evaluate geometrical synchronization between
x
y
, both data signals are split into
m=\text{20}
equal-sized segments of size
n=\text{300}
. The resulting segments
{x}_{i}^{\left(s\right)}
{y}_{i}^{\left(s\right)}
i=1,\dots ,m
are then mapped onto integers
{\tau }_{i}^{*x}
{\tau }_{i}^{*y}
i=1,\dots ,m
denoted as optimal time lags using the steps below:
z=\left({z}_{1},\dots ,{z}_{n}\right)
be a data signal of size
n
(corresponding to one of the segments) (see Fig. 2(a)).
2) Embed data signal
z
into a 2D delay coordinate space using parameter
\tau \in \left\{1,\dots ,n-1\right\}
{z}_{i}\to \left({z}_{i},{z}_{i+\tau }\right), i=1,\dots ,n-\tau .
The obtained set of the embedded points is called an attractor (see Fig. 2(b)).
3) Compute the area of reconstructed attractor using the following formula for each
\tau =1,\dots ,200
{S}_{\tau }=\frac{1}{\sqrt{2}\left(n-\tau \right)}{\sum }_{k=1}^{n-\tau }\sqrt{{z}_{k}^{2}+{z}_{k+\tau }^{2}}.
4) Determine the optimal time lag
{\tau }^{*}
resulting in the largest area of the attractor (see Fig. 2(c)):
{\tau }^{*}=\mathrm{arg}\underset{1\le \tau \le 200}{\mathrm{max}}{S}_{\tau }.
It was shown in [4, 5] that the optimal time lag
{\tau }^{*}
can be used as a scalar feature representing the geometrical properties of the analyzed data signal.
Fig. 2. Part a) depicts one segment of data signal
x
displayed in Fig. 1(b). Parts b) and c) illustrate the corresponding reconstructed attractors at
\tau =\text{4}
{\tau }^{*}=\text{63}
(optimal time lag), respectively
Geometrical synchronization between data signals
x
y
is then estimated as the Pearson correlation coefficient between optimal time lag vectors
{\tau }_{ }^{*x}
{\tau }_{ }^{*y}
. Optimal time lags for
x
y
are represented in Fig. 3 for three values of coupling parameter
\epsilon
. It can be seen that the correlation between sequences of optimal time lags is
\rho =
\rho =
\rho =
–0.9934 at
\epsilon =
\epsilon =
\epsilon =
0.16 respectively. Obtained results show that described geometrical synchronization estimation algorithm based on the optimal attractor embedding is able to detect the similarity between two chaotic data signals in an effective and efficient way.
Fig. 3. Optimal time lag vectors
{\tau }_{ }^{*x}
(solid line) and
{\tau }_{ }^{*y}
(dashed line) corresponding to
\epsilon =0
\epsilon =\text{0.13}
\epsilon =\text{0.16}
(parts a), b), c) respectively)
3. Estimation of synchronization between human heart rate variability and Earth’s local magnetic field
In order to assess the impact of magnetic field on human’s biomedical parameters the following experiment was conducted:
1) RR interval data was gathered from a group of 20 Lithuanian students that continuously wore heart rate monitors for a period of 11 days (2015.02.28 - 2015.03.10). Consequently, a total of 20 RR interval data series (
{X}^{\left(i\right)}, i=1,\dots ,20
) were collected from all 20 individuals.
2) During the same eleven-day period of time the intensity of local magnetic field was
measured using the magnetometer located in Lithuania. Spectral power of the local magnetic field (
M
) was then computed using obtained intensity values by summing the spectrogram (evaluated for one-second intervals) over the physically relevant frequency range [0.2; 3.5] Hz [4].
3.2. Application of the synchronization estimation algorithm
The geometrical synchronization measure presented in the previous section was applied to the experiment data to estimate the degree of synchronization between RR interval time series and the spectral power of local magnetic field. Note, that parameter
n
, denoting the length of data segments for the attractor embedding was selected to correspond to 5 minutes of data since it is the standard time span for the analysis of human heart rate variability (HRV). Fig. 4 displays mean synchronization values between all participants’ HRV and magnetic field power computed for each day of the experiment. Obtained synchronization values can be further analyzed with regards to magnetic field activity at a given date in order to investigate the impact of the geomagnetic field on human heart activity.
Fig. 4. Mean synchronization between RR data series and magnetic field power for each day of the experiment
A technique for the estimation of geometrical synchronization between two data series using optimal attractor embedding is validated in this paper via the example of two coupled chaotic logistic maps. Presented technique is then applied to the real data in order to evaluate the degree of synchronization between human HRV and Earth’s geomagnetic activity.
Dörfler F., Bullo F. Synchronization in complex networks of phase oscillators: a survey. Automatica, Vol. 50, Issue 6, 2014, p. 1539-1564. [Publisher]
Quiroga R. Q., Kraskov A., Kreuz T., Grassberger P. Performance of different synchronization measures in real data: a case study on electroencephalographic signals. Physical Review E, Vol. 65, Issue 4, 2002, p. 041903. [Search CrossRef]
González-Miranda J. M. Amplitude envelope synchronization in coupled chaotic oscillators. Physical Review E, Vol. 65, Issue 3, 2002, p. 036232. [Publisher]
Timofejeva I., McCraty R., Atkinson M., Joffe R., Vainoras A., Alabdulgader A., Ragulskis M. Identification of a group’s physiological synchronization with earth’s magnetic field. International Journal of Environmental Research and Public Health, Vol. 14, Issue 9, 2017, p. 998. [Publisher]
Timofejeva I., Poskuviene K., Cao M., Ragulskis M. Synchronization measure based on a geometric approach to attractor embedding using finite observation windows. Complexity, Vol. 2018, 2018, p. 8259496. [Publisher] |
Solve nonnegative linear least-squares problem - MATLAB lsqnonneg - MathWorks France
\underset{x}{\mathrm{min}}{‖C\cdot x-d‖}_{2}^{2},\text{ where }x\ge 0.
lsqnonneg applies only to the solver-based approach. For a discussion of the two optimization approaches, see First Choose Problem-Based or Solver-Based Approach.
x = lsqnonneg(C,d,options) minimizes with the optimization options specified in the structure options. Use optimset to set these options.
x = lsqnonneg(problem) finds the minimum for problem, a structure described in problem.
\mathrm{min}||Cx-d||
\mathrm{min}||Cx-d||
\mathrm{min}||Cx-d||
\mathrm{min}||Cx-d||
\ne
\underset{x}{\mathrm{min}}{‖Cx-d‖}_{2}^{2}.
\underset{x}{\mathrm{min}}{‖Cx-d‖}_{2}^{2}.
For problems where d has length over 20, lsqlin might be faster than lsqnonneg. When d has length under 20, lsqnonneg is generally more efficient.
To convert between the solvers when C has more rows than columns (meaning the system is overdetermined),
The only difference is that the corresponding Lagrange multipliers have opposite signs: lambda = -lambda_lsqlin.ineqlin.
mldivide | lsqlin | optimset | Optimize |
Classification loss for observations not used in training - MATLAB - MathWorks América Latina
Estimate k-Fold Cross-Validation Classification Error
Specify Custom Classification Loss
Find Good Lasso Penalty Using k-fold Classification Loss
kfoldLoss returns a different value for a model with a nondefault cost matrix
Classification loss for observations not used in training
L = kfoldLoss(CVMdl) returns the cross-validated classification losses obtained by the cross-validated, binary, linear classification model CVMdl. That is, for every fold, kfoldLoss estimates the classification loss for observations that it holds out when it trains using all other observations.
L contains a classification loss for each regularization strength in the linear classification models that compose CVMdl.
L = kfoldLoss(CVMdl,Name,Value) uses additional options specified by one or more Name,Value pair arguments. For example, indicate which folds to use for the loss calculation or specify the classification-loss function.
To obtain estimates, kfoldLoss applies the same data used to cross-validate the linear classification model (X and Y).
Folds — Fold indices to use for classification-score prediction
Fold indices to use for classification-score prediction, specified as the comma-separated pair consisting of 'Folds' and a numeric vector of positive integers. The elements of Folds must range from 1 through CVMdl.KFold.
'mincost' is appropriate for classification scores that are posterior probabilities. For linear classification models, logistic regression learners return posterior probabilities as classification scores by default, but SVM learners do not (see predict).
Let n be the number of observations in X and K be the number of distinct classes (numel(Mdl.ClassNames), Mdl is the input model). Your function must have this signature
L — Cross-validated classification losses
Cross-validated classification losses, returned as a numeric scalar, vector, or matrix. The interpretation of L depends on LossFun.
If Mode is 'average', then L is a 1-by-R vector. L(j) is the average classification loss over all folds of the cross-validated model that uses regularization strength j.
Otherwise, L is an F-by-R matrix. L(i,j) is the classification loss for fold i of the cross-validated model that uses regularization strength j.
Estimate the average of the out-of-fold, classification error rates.
ce = kfoldLoss(CVMdl)
ce = 7.6017e-04
Alternatively, you can obtain the per-fold classification error rates by specifying the name-value pair 'Mode','individual' in kfoldLoss.
Load the NLP data set. Preprocess the data as in Estimate k-Fold Cross-Validation Classification Error, and transpose the predictor data.
Cross-validate a binary, linear classification model using 5-fold cross-validation. Optimize the objective function using SpaRSA. Specify that the predictor observations correspond to columns.
CVMdl = fitclinear(X,Ystats,'Solver','sparsa','KFold',5, ...
CVMdl is a ClassificationPartitionedLinear model. It contains the property Trained, which is a 5-by-1 cell array holding a ClassificationLinear models that the software trained using the training set of each fold.
Create an anonymous function that measures linear loss, that is,
L=\frac{\sum _{j}-{w}_{j}{y}_{j}{f}_{j}}{\sum _{j}{w}_{j}}.
{w}_{j}
is the weight for observation j,
{y}_{j}
is response j (-1 for the negative class, and 1 otherwise), and
{f}_{j}
is the raw classification score of observation j. Custom loss functions must be written in a particular form. For rules on writing a custom loss function, see the LossFun name-value pair argument. Because the function does not use classification cost, use ~ to have kfoldLoss ignore its position.
linearloss = @(C,S,W,~)sum(-W.*sum(S.*C,2))/sum(W);
Estimate the average cross-validated classification loss using the linear loss function. Also, obtain the loss for each fold.
ce = kfoldLoss(CVMdl,'LossFun',linearloss)
ce = -8.0982
ceFold = kfoldLoss(CVMdl,'LossFun',linearloss,'Mode','individual')
ceFold = 5×1
To determine a good lasso-penalty strength for a linear classification model that uses a logistic regression learner, compare test-sample classification error rates.
Load the NLP data set. Preprocess the data as in Specify Custom Classification Loss.
1{0}^{-6}
1{0}^{0.5}
Cross-validate binary, linear classification models using 5-fold cross-validation, and that use each of the regularization strengths. Optimize the objective function using SpaRSA. Lower the tolerance on the gradient of the objective function to 1e-8.
Extract a trained linear classification model.
Mdl1 is a ClassificationLinear model object. Because Lambda is a sequence of regularization strengths, you can think of Mdl as 11 models, one for each regularization strength in Lambda.
Choose the indexes of the regularization strength that balances predictor variable sparsity and low classification error. In this case, a value between
1{0}^{-4}
1{0}^{-1}
\sum _{j=1}^{n}{w}_{j}=1.
L=\sum _{j=1}^{n}{w}_{j}\mathrm{log}\left\{1+\mathrm{exp}\left[-2{m}_{j}\right]\right\}.
L=\sum _{j=1}^{n}{w}_{j}{c}_{{y}_{j}{\stackrel{^}{y}}_{j}},
{\stackrel{^}{y}}_{j}
{c}_{{y}_{j}{\stackrel{^}{y}}_{j}}
{\stackrel{^}{y}}_{j}
L=\sum _{j=1}^{n}{w}_{j}I\left\{{\stackrel{^}{y}}_{j}\ne {y}_{j}\right\},
L=-\sum _{j=1}^{n}\frac{{\stackrel{˜}{w}}_{j}\mathrm{log}\left({m}_{j}\right)}{Kn},
{\stackrel{˜}{w}}_{j}
L=\sum _{j=1}^{n}{w}_{j}\mathrm{exp}\left(-{m}_{j}\right).
L=\sum _{j=1}^{n}{w}_{j}\mathrm{max}\left\{0,1-{m}_{j}\right\}.
L=\sum _{j=1}^{n}{w}_{j}\mathrm{log}\left(1+\mathrm{exp}\left(-{m}_{j}\right)\right).
{\gamma }_{jk}={\left(f{\left({X}_{j}\right)}^{\prime }C\right)}_{k}.
{\stackrel{^}{y}}_{j}=\underset{k=1,...,K}{\text{argmin}}{\gamma }_{jk}.
L=\sum _{j=1}^{n}{w}_{j}{c}_{j}.
L=\sum _{j=1}^{n}{w}_{j}{\left(1-{m}_{j}\right)}^{2}.
R2022a: kfoldLoss returns a different value for a model with a nondefault cost matrix
If you specify a nondefault cost matrix when you train the input model object, the kfoldLoss function returns a different value compared to previous releases.
The kfoldLoss function uses the observation weights stored in the W property. Also, the function uses the cost matrix stored in the Cost property if you specify the LossFun name-value argument as "classifcost" or "mincost". The way the function uses the W and Cost property values has not changed. However, the property values stored in the input model object have changed for a model with a nondefault cost matrix, so the function can return a different value.
ClassificationPartitionedLinear | ClassificationLinear | kfoldPredict | loss |
h is related to one of the parent functions described in this chapter. Describe the sequen
h is related to one of the parent functions described in this chapter. Describe the sequence of transformations from f to h. h(x)=−1/3x^3
h\left(x\right)=-\frac{1}{3}{x}^{3}
ensojadasH
We are starting with the parent function
f\left(x\right)=-\left(1/3\right){x}^{3}
STEP 1: Reflect the graph across x-axis, to get
y=-{x}^{3}
STEP 2: Vertically compress the graph by a factor of 1/3, to get
y=-\left(1/3\right){x}^{3}
, which is the required function h(x)
g is related to one of the parent functions described in Section 1.6. Describe the sequence of transformations from f to g. g(x)
Given that f(x) = 3x - 7 and that (f + g)(x) = 7x + 3, find g(x).
{x}^{2}+8x+15
By using the transformation of function
y={x}^{2}
sketch the function of y=
\frac{{\left(x-2\right)}^{2}}{3}+4
Graph f and g in the same rectangular coordinate system. Use transformations of the graph of f to obtain the graph of g. Graph and give equations of all asymptotes Use the graphs to determine each functions
An observational study is retrospective if it considers only existing data. It is prospective if the study design calls for data to be collected as time goes on. Tell which of the following observational studies are retrospective and which are prospective. Out of curiosity, Mimi looks up the ages of all the Best Actress winners in the years they won their Oscars, and then gathers the same data for the Best Actor winners, to sec whether male or female winners were, on the average younger
h is related to one of the six parent functions. (a) Identify the parent function f. (b) Describe the sequence of transformations from f to h. (c) Sketch the graph of h by hand. (d) Use function notation to write h in terms of the parent function f.
h\left(x\right)=\mid 2x+8\mid -1
Upper level algebra |
Students arrive for the much dreaded JEE Advanced according to a Poisson process with rate lambda. For sanitization proc
Students arrive for the much dreaded JEE Advanced according
Students arrive for the much dreaded JEE Advanced according to a Poisson process with rate lambda. For sanitization process they must stand in a queue,each student can take different time or same time compared with some other student for sanitization. Let us denote the time taken by i-th student as
{X}_{i}
{X}_{i}
are independent identically distributed random variables. We assume that
{X}_{i}
takes integer values in range 1,2, ... , n, with probabilities
{p}_{1},{p}_{2},\dots ,{p}_{n}
. Find the PMF for
{N}_{t}
, the number of students in sanitization queue at time t.
Given: Students arrive for the much dreaded JEE advanced according to Poisson process with rate Lamda
\left(\lambda \right)
{X}_{1},{X}_{2},\dots \dots
Xnbe the time taken by students.
{X}_{i}
are independent and identically distributed and it takes integer values in range 1,2,3....n with probability
{p}_{1},{p}_{2},{p}_{3}\dots .pn
The Poisson process is used in happening of repeatedly and independently in several settings
{N}_{t}
be the number of students in sanitization queue at time [0,t].
{N}_{t}
is increasing integer valued, continuous time random process.
suppose the time [0,t]is divided into subinterval of width
\mathrm{△}t=\frac{t}{n}
1) Probability of happing two events in subinterval is negligible, that is probability of happing of event in subinterval is either success or failure it gives only two outcomes hence it is Bernoulli trial.
2) Two events can not happen simultaneously either it give success or failure hence subinterval is independent of outcome in each interval, hence Bernoulli trials are independent.
The above two assumption shows that the counting process Nt can be binomial process (by result : The sum of independent Bernoulli trial approximated to Binomial distribution)
If the probability of number of students sanitize in queue in each subinterval is fixed P. then expected value of event happen in time [0,t] is given by binomial distribution is np. but event occurred at rate Lamda.
\left(\lambda t\right)=\left(np\right)
n⇒\mathrm{\infty },P⇒0\text{ }whi\le \text{ }\lambda t=np
is fixed ,the binomial approaches to Poission distribution and its PMF is given by
N\left(t{N}_{t}\right)
P\left[N\left(t\right)=k\right]=\frac{{e}^{\lambda t}{\left(\lambda t\right)}^{k}}{k!}k=0,1,\dots
e=2.71828
\lambda t=
mean of distribution and
k=
number of arrival in each subinterval
Answer: The PMF is given by,
P\left[N\left(t\right)=k\right]=\frac{{e}^{\lambda t}{\left(\lambda t\right)}^{k}}{k!}\text{ }k=0,1,\dots
Introduction to Poisson distribution:
A random variable X is said to have a Poisson distribution with a parameter
\lambda
if its probability mass function (p.m.f) is as given below:
P\left(x,\lambda \right)=\left\{\begin{array}{ccccc}\frac{{e}^{-\lambda }{\lambda }^{x}}{x!}& & x=012\dots \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\lambda >0& & \\ 0& & otherwise.& & \end{array}
Here, students arrive for the much dreaded JEE advanced according to a Poisson process with the rate
\lambda
Denote the time taken by the student i as
{X}_{i}
{X}_{i}^{\prime }s
are independent and identically distributed with probabilities
{p}_{1},{p}_{2},\cdots ,pn
Find the PMF of
{N}_{t}
Here, Nt is the number of students in the sanitization queue at time t.
\lambda >0
is fixed. The counting process
\left\{N\left(t\right),t\in \left[0,\mathrm{\infty }\right)\right\}
is called a Poisson process with rate
\lambda
under the below given conditions:
N\left(0\right)=0
2. N(t) has independent increments;
3. The number of arrivals in any interval of length
\tau >0
has Poisson
\left(\lambda \tau \right)
The PMF of Nt is as given below:
N\left(t\right)=
Number of students in the sanitization queue at time t
N\left(t\right)\sim B\left(n,p\right)
n=\frac{t}{\delta }
p={p}_{1}+{p}_{2}+\cdots {p}_{n}
p=\lambda \delta
np=\frac{t}{\delta }×\lambda \delta
=\lambda
The PMF of N(t) converges to a Poisson distribution with parameter
\lambda
P\left(N\left(t\right)\right)=\left\{\begin{array}{ccccc}\frac{{e}^{-\lambda t}{\left(\lambda t\right)}^{x}}{x!}& & x>0& & \\ 0& & otherwise.& & \end{array}
If a seed is planted, it has a 90% chance of growing into a healthy plant.
If 7 seeds are planted, what is the probability that exactly 2 don't grow?
It is conjectured that an impurity exists in 30% of all drinking wells in a certain rural community. In order to gain some insight on this problem, it is determined that some tests should be made. It is too expensive to test all of the many wells in the area, so 10 were randomly selected for testing. a. Using the binomial distribution, what is the probability that exactly three wells have the impurity assuming that the conjecture is correct? b. What is the probability that more than three wells are impure?
According to a 2014 Gallup poll. 56% of uninsured Americans who plan to get health insurance say they will do so through a government health insurance exchange.
a) What is the probability that in a random sample of 10 people exactly 6 plan to get health insurance through a goverment health insurance exchange?
b) What is the probability that in a random sample of 1000 people exactly 600 plan to get health insurance through a government health insurance exchange?
c) What are the expected value and the variance of X?
d) What is the probability that less than 600 people plan to get health insurance through a government health insurance exchange?
A Bernoulli process has 4 trials and probability of success 0.31. Find the following probabilities.
1. Exactly 2 successes.
2. Exactly 2 failures.
3. At most 1 success.
4. At least 1 success.
5. At least 1 success and 1 failure.
A 10 question multiple choice exam is given and each question has 5 possible answers. A student takes this exam and guesses at every question. What is the probability they get at least 9 questions correct?
Suppose a certain breed of piglets have a sex-ratio of 0.54 (that is the probability of male piglets). If 5 piglets were born, what is the probability that 3 of them are male? |
Filter - Maple Help
Home : Support : Online Help : Science and Engineering : Signal Processing : Filtering : Filter
Filter( A, a, b )
Array of real numeric values; the Array of IIR coefficients
(optional) Array of real numeric values; the Array of FIR coefficients
The Filter( A, a, b ) command filters the signal (sample) in the array A using the IIR coefficients in the array a and the FIR coefficients in the array b. If b is omitted, no FIR filter is applied (there is only a single coefficient, 1).
Before the code performing the computation runs, the input Arrays are converted to datatype float[8] if they do not have that datatype already. For this reason, it is most efficient if the input Arrays have this datatype beforehand.
The SignalProcessing[Filter] command is thread-safe as of Maple 17.
\mathrm{with}\left(\mathrm{SignalProcessing}\right):
\mathrm{with}\left(\mathrm{plots}\right):
A≔\mathrm{GenerateTone}\left(128,1,0.05,3.0\right):
a≔\mathrm{Array}\left(1..3,[\frac{1}{3},\frac{1}{3},\frac{1}{3}],\mathrm{datatype}=\mathrm{float}[8],\mathrm{order}=\mathrm{C_order}\right):
b≔\mathrm{Array}\left(1..1,[1],\mathrm{datatype}=\mathrm{float}[8],\mathrm{order}=\mathrm{C_order}\right):
B≔\mathrm{Filter}\left(A,a,b\right):
\mathrm{display}\left(\mathrm{Array}\left([\mathrm{SignalPlot}\left(A\right),\mathrm{SignalPlot}\left(B\right)]\right)\right)
The SignalProcessing[Filter] command was introduced in Maple 17. |
Convex preferences - Wikipedia
In economics, convex preferences are an individual's ordering of various outcomes, typically with regard to the amounts of various goods consumed, with the property that, roughly speaking, "averages are better than the extremes". The concept roughly corresponds to the concept of diminishing marginal utility without requiring utility functions.
5 Relation to indifference curves and utility functions
Comparable to the greater-than-or-equal-to ordering relation
{\displaystyle \geq }
for real numbers, the notation
{\displaystyle \succeq }
below can be translated as: 'is at least as good as' (in preference satisfaction).
{\displaystyle \succ }
can be translated as 'is strictly better than' (in preference satisfaction), and Similarly,
{\displaystyle \sim }
can be translated as 'is equivalent to' (in preference satisfaction).
Use x, y, and z to denote three consumption bundles (combinations of various quantities of various goods). Formally, a preference relation
{\displaystyle \succeq }
on the consumption set X is called convex if for any
{\displaystyle x,y,z\in X}
{\displaystyle y\succeq x}
{\displaystyle z\succeq x}
{\displaystyle \theta \in [0,1]}
{\displaystyle \theta y+(1-\theta )z\succeq x}
i.e., for any two bundles that are each viewed as being at least as good as a third bundle, a weighted average of the two bundles is viewed as being at least as good as the third bundle.
A preference relation
{\displaystyle \succeq }
is called strictly convex if for any
{\displaystyle x,y,z\in X}
{\displaystyle y\succeq x}
{\displaystyle z\succeq x}
{\displaystyle y\neq z}
{\displaystyle \theta \in (0,1)}
{\displaystyle \theta y+(1-\theta )z\succ x}
i.e., for any two distinct bundles that are each viewed as being at least as good as a third bundle, a weighted average of the two bundles (including a positive amount of each bundle) is viewed as being strictly better than the third bundle.[1][2]
Use x and y to denote two consumption bundles. A preference relation
{\displaystyle \succeq }
is called convex if for any
{\displaystyle x,y\in X}
{\displaystyle y\succeq x}
{\displaystyle \theta \in [0,1]}
{\displaystyle \theta y+(1-\theta )x\succeq x}
That is, if a bundle y is preferred over a bundle x, then any mix of y with x is still preferred over x.[3]
A preference relation is called strictly convex if for any
{\displaystyle x,y\in X}
{\displaystyle y\sim x}
{\displaystyle x\neq y}
{\displaystyle \theta \in (0,1)}
{\displaystyle \theta y+(1-\theta )x\succ x}
{\displaystyle \theta y+(1-\theta )x\succ y}
That is, for any two bundles that are viewed as being equivalent, a weighted average of the two bundles is better than each of these bundles.[4]
1. If there is only a single commodity type, then any weakly-monotonically-increasing preference relation is convex. This is because, if
{\displaystyle y\geq x}
, then every weighted average of y and ס is also
{\displaystyle \geq x}
2. Consider an economy with two commodity types, 1 and 2. Consider a preference relation represented by the following Leontief utility function:
{\displaystyle u(x_{1},x_{2})=\min(x_{1},x_{2})}
This preference relation is convex. Proof: suppose x and y are two equivalent bundles, i.e.
{\displaystyle \min(x_{1},x_{2})=\min(y_{1},y_{2})}
. If the minimum-quantity commodity in both bundles is the same (e.g. commodity 1), then this implies
{\displaystyle x_{1}=y_{1}\leq x_{2},y_{2}}
. Then, any weighted average also has the same amount of commodity 1, so any weighted average is equivalent to
{\displaystyle x}
{\displaystyle y}
. If the minimum commodity in each bundle is different (e.g.
{\displaystyle x_{1}\leq x_{2}}
{\displaystyle y_{1}\geq y_{2}}
), then this implies
{\displaystyle x_{1}=y_{2}\leq x_{2},y_{1}}
{\displaystyle \theta x_{1}+(1-\theta )y_{1}\geq x_{1}}
{\displaystyle \theta x_{2}+(1-\theta )y_{2}\geq y_{2}}
{\displaystyle \theta x+(1-\theta )y\succeq x,y}
. This preference relation is convex, but not strictly-convex.
3. A preference relation represented by linear utility functions is convex, but not strictly convex. Whenever
{\displaystyle x\sim y}
, every convex combination of
{\displaystyle x,y}
is equivalent to any of them.
4. Consider a preference relation represented by:
{\displaystyle u(x_{1},x_{2})=\max(x_{1},x_{2})}
This preference relation is not convex. Proof: let
{\displaystyle x=(3,5)}
{\displaystyle y=(5,3)}
{\displaystyle x\sim y}
since both have utility 5. However, the convex combination
{\displaystyle 0.5x+0.5y=(4,4)}
is worse than both of them since its utility is 4.
Relation to indifference curves and utility functions[edit]
A set of convex-shaped indifference curves displays convex preferences: Given a convex indifference curve containing the set of all bundles (of two or more goods) that are all viewed as equally desired, the set of all goods bundles that are viewed as being at least as desired as those on the indifference curve is a convex set.
Convex preferences with their associated convex indifference mapping arise from quasi-concave utility functions, although these are not necessary for the analysis of preferences.
^ Hal R. Varian; Intermediate Microeconomics A Modern Approach. New York: W. W. Norton & Company. ISBN 0-393-92702-4
^ Mas-Colell, Andreu; Whinston, Michael; & Green, Jerry (1995). Microeconomic Theory. Oxford: Oxford University Press. ISBN 978-0-19-507340-9
^ Board, Simon (October 6, 2009). "Preferences and Utility" (PDF). Econ 11. Microeconomic Theory. Autumn 2009. University of California, Los Angeles.
^ Sanders, Nicholas J. "Preference and Utility - Basic Review and Examples" (PDF). College of William & Mary. Archived from the original (PDF) on March 20, 2013.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Convex_preferences&oldid=1067152117" |
To calculate: To convert the radical expression \sqrt[4]{3^{2}} to rat
To calculate:To convert the radical expression \sqrt[4]{3^{2}} to rat
To convert the radical expression
\sqrt[4]{{3}^{2}}
to rational exponent form and simplify.
\sqrt[4]{{3}^{2}}
According to exponential rules,
\sqrt[n]{a}={a}^{\frac{1}{n}}
Applying the above rule: where
a={3}^{2},n=4
\sqrt[4]{{3}^{2}}=\left({3}^{2}{\right)}^{\frac{1}{4}}
According to the power rule,
{\left({a}^{m}\right)}^{n}={a}^{mn}
a=3.m=2,n=\frac{1}{4}
\sqrt[4]{{3}^{2}}={3}^{\left(2\right)\left(\frac{1}{4}\right)}
\sqrt[4]{{3}^{2}}={3}^{\frac{1}{2}}
A stunt man whose mass is 70 kg swings from the end ofa 4.0 m long rope along thearc of a vertical circle. Assuming that he starts from rest whenthe rope is horizontal, find the tensions in the rope that are required to make him follow his circular path at each of the following points.
(a) at the beginning of his motion N
(b) at a height of 1.5 m above the bottom of the circular arc N
(c) at the bottom of the arc N
{25.0}^{\circ }
A 95.0kg person stands on a scale in an elevator. What is the apparent weight when the elevator is (a) accelerating upward with an acceleration of 1.80
\frac{m}{{s}^{2}}
, (b) moving upward at a constant speed and (c) accelerating downward with an acceleration of 1.30
\frac{m}{{s}^{2}}
Find the probabilities for the standard normal random variable z: P(-1.43 < z < 0.68)
If X is a normal random variable with parameters
\mu =10
{\sigma }^{2}=36
, compute P[4 < X < 16]
“CVP analysis is both simple and simplistic. If you want realistic analysis to underpin your decisions, look beyond CVP analysis.” Do you agree? Explain. |
Defaults - Maple Help
Home : Support : Online Help : System : Security : Defaults
retrieve the standard security settings
This routine is used to retrieve the standard security settings. These are the security settings used when Maple is started with the -z option. These settings are returned in the same format as those returned by Security,Config.
\mathrm{Security}:-\mathrm{Defaults}\left(\right)
\textcolor[rgb]{0,0,1}{\mathrm{SECURE_READ_LIST}}\textcolor[rgb]{0,0,1}{=}[\textcolor[rgb]{0,0,1}{"/usr/local/maple/lib/*"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"/usr/local/maple/java"}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{SECURE_NOREAD_LIST}}\textcolor[rgb]{0,0,1}{=}[]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{SECURE_WRITE_LIST}}\textcolor[rgb]{0,0,1}{=}[]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{SECURE_NOWRITE_LIST}}\textcolor[rgb]{0,0,1}{=}[]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{SECURE_EXTCALL_LIST}}\textcolor[rgb]{0,0,1}{=}[\textcolor[rgb]{0,0,1}{"/usr/local/maple/bin.X86_64_LINUX/*"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"/usr/local/maple/toolbox/Database/bin.X86_64_LINUX/*"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"/usr/local/maple/toolbox/NAG/bin.X86_64_LINUX/*"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"/usr/local/maple/toolbox/Finance/bin.X86_64_LINUX/*"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"/usr/local/maple/toolbox/GlobalOptimization/bin.X86_64_LINUX/*"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"/usr/local/maple/toolbox/LinearSystem/bin.X86_64_LINUX/*"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"/usr/local/maple/java/*"}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{SECURE_NOEXTCALL_LIST}}\textcolor[rgb]{0,0,1}{=}[\textcolor[rgb]{0,0,1}{"/usr/local/maple/bin.X86_64_LINUX/libmsock.so"}]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{SECURE_SYSCALL_ENABLED}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{false}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{SECURE_MODE}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{true}} |
Find the slope, if it exists, of the line containing the pair of points (5, -1) and (5, -1
Find the slope, if it exists, of the line containing the pair of points (5, -1) and (5, -10).
Use the slope formula:
m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}
\left({x}_{1},{y}_{1}\right)=\left(5,-1\right)
\left({x}_{2},{y}_{2}\right)=\left(5,-10\right)
m=\frac{-10-\left(-1\right)}{5-5}=-\frac{9}{0}\to
f\left(x\right)=\sqrt{1-x}
a=0
and use it to approximate the numbers
\sqrt{0.9}
\sqrt{0.99}
Determine whether each of these functions is a bijection from R to R.
f\left(x\right)=-3x+4
f\left(x\right)=-3{x}^{2}+7
f\left(x\right)=\frac{x+1}{x+2}
d\right)f\left(x\right)={x}^{5}+1
A baseball team plays in a stadium that holds 55,000 spectators. With ticket prices at 10, the average attendance had been 27,000. When ticket prices were lowered to10,the average attend ance had been 27,000.When ticket prices were lowered to 8, the average attendance rose to 33,000. How should ticket prices be set to maximize revenue?
Using a directrix of y = −2 and a focus of (2, 6), what quadratic function is created?
f\left(x\right)=-{\left(x-2\right)}^{2}-2
f\left(x\right)={\left(x-2\right)}^{2}+2
f\left(x\right)={\left(x-2\right)}^{2}-2
f\left(x\right)=-{\left(x+2\right)}^{2}-2
|2\left(x-1\right)+4|\le 8
\text{If X is uniformly distributed over }\left(-1,1\right),\text{ find (a) }P|X|>\frac{1}{2}
(b) the density function of the random variable |X|.
Use the definition to show that each function is
O\left({n}^{2}\right).
O is big O notation
F\left(n\right)=\mathrm{ln}\left(n\right)+5{n}^{2} |
Refit generalized linear mixed-effects model - MATLAB - MathWorks Deutschland
glmenew
Refit Model to New Response Vector
Refit generalized linear mixed-effects model
glmenew = refit(glme,ynew)
glmenew = refit(glme,ynew) returns a refitted generalized linear mixed-effects model, glmenew, based on the input model glme, using a new response vector, ynew.
ynew — New response vector
n-by-1 vector of scalar values
New response vector, specified as an n-by-1 vector of scalar values, where n is the number of observations used to fit glme.
For an observation i with prior weights wip and binomial size ni (when applicable), the response values yi contained in ynew can have the following values.
\left\{0,\frac{1}{{w}_{i}^{p}{n}_{i}},\frac{2}{{w}_{i}^{p}{n}_{i}},.\dots ,1\right\}
wip and ni are integer values > 0
\left\{0,\frac{1}{{w}_{i}^{p}},\frac{2}{{w}_{i}^{p}},\cdots ,1\right\}
wip is an integer value > 0
Gamma (0,∞) wip ≥ 0
InverseGaussian (0,∞) wip ≥ 0
Normal (–∞,∞) wip ≥ 0
You can access the prior weights property wip using dot notation.
glme.ObservationInfo.Weights
glmenew — Generalized linear mixed-effects model
Generalized linear mixed-effects model, returned as a GeneralizedLinearMixedModel object. glmenew is an updated version of the generalized linear mixed-effects model glme, refit to the values in the response vector ynew.
For properties and methods of this object, see GeneralizedLinearMixedModel.
{\text{defects}}_{ij}\sim \text{Poisson}\left({\mu }_{ij}\right)
\mathrm{log}\left({\mu }_{ij}\right)={\beta }_{0}+{\beta }_{1}{\text{newprocess}}_{ij}+{\beta }_{2}{\text{time}\text{_}\text{dev}}_{ij}+{\beta }_{3}{\text{temp}\text{_}\text{dev}}_{ij}+{\beta }_{4}{\text{supplier}\text{_}\text{C}}_{ij}+{\beta }_{5}{\text{supplier}\text{_}\text{B}}_{ij}+{b}_{i},
{\text{defects}}_{ij}
i
j
{\mu }_{ij}
i
i=1,2,...,20
j
j=1,2,...,5
{\text{newprocess}}_{ij}
{\text{time}\text{_}\text{dev}}_{ij}
{\text{temp}\text{_}\text{dev}}_{ij}
i
j
{\text{newprocess}}_{ij}
i
j
{\text{supplier}\text{_}\text{C}}_{ij}
{\text{supplier}\text{_}\text{B}}_{ij}
i
j
{b}_{i}\sim N\left(0,{\sigma }_{b}^{2}\right)
i
Use random to simulate a new response vector from the fitted model.
rng(0,'twister'); % For reproducibility
ynew = random(glme);
Refit the model using the new response vector.
glme = refit(glme,ynew)
Name Estimate SE tStat DF pValue
{'(Intercept)'} 1.5738 0.18674 8.4276 94 4.0158e-13
{'newprocess' } -0.21089 0.2306 -0.91455 94 0.36277
{'time_dev' } -0.13769 0.77477 -0.17772 94 0.85933
{'temp_dev' } 0.24339 0.84657 0.2875 94 0.77436
{'supplier_C' } -0.12102 0.07323 -1.6526 94 0.10175
{'supplier_B' } 0.098254 0.066943 1.4677 94 0.14551
1.203 1.9445
-1.676 1.4006
-0.26642 0.024381
-0.034662 0.23117
You can use refit and random to conduct a simulated likelihood ratio test or parametric bootstrap.
GeneralizedLinearMixedModel | fitted | residuals | designMatrix |
They are gases that have been produced with the process of production according to the Kyoto Protocol [17]
measured using tons of CO2 [22]
Processing of waste (i.e. undesirable or futile materials) back to the material cycle with the goal that contamination of the earth is limited [18]
measured using a percentage of materials reused out of the total materials used in the industry through one whole year [22]
Anything that adds adverse effects to the
an environment without adding value [19]
Measured using the tons of resources discarded or unused resources throughout the year.
Energy could be utilized over and over [20]
Measured using the megawatts of energy generated from renewable resources.
measures productivity by uncovering how a lot of benefits an organization creates with the cash investors have contributed [21]
Measured as a percentage of net income after tax to total equity [21]
\text{ROE}=\frac{\text{net}\text{\hspace{0.17em}}\text{income}}{\text{Total}\text{\hspace{0.17em}}\text{Equity}} |
Solving quadratically constrained quadratic programming (QCQP) problems | nag
Solving quadratically constrained quadratic programming (QCQP) problems
Quadratic functions are a powerful modelling construct in mathematical programming and appear in various disciplines such as statistics, machine learning (Lasso regression), finance (portfolio optimization), engineering (OPF) and control theory. At Mark 27.1 of the NAG Library, NAG introduced two new additions to the NAG Optimization Modelling Suite to help users easily define quadratic objective functions and/or constraints, seamlessly integrate them with other constraints and solve the resulting problems using compatible solvers without the need of a reformulation or any extra effort.
Formally, a QCQP problem can be written in its pure form with only quadratic functions in both objective and constraints as follows:
\[ \begin{array}{cc}\underset{x\in {\Re }^{n}}{\text{minimize}}\hfill & \frac{1}{2}{x}^{T}{Q}_{0}x+{r}_{0}^{T}x+{s}_{0}\hfill \\ & \\ \text{subject to}\hfill & \frac{1}{2}{x}^{T}{Q}_{i}x+{r}_{i}^{T}x+{s}_{i}\le 0,\phantom{\rule{1em}{0ex}}i=1,\dots ,p,\hfill \end{array} \]
{Q}_{i}\in {\Re }^{n×n},i=0,\dots ,p
, are symmetric matrices,
{r}_{i}\in {\Re }^{n},i=0,\dots ,p
, are vectors and
{s}_{i}
scalars. However, in practice it will often be stated with linear constraints and simple bounds as well. The two new routines are handle_set_qconstr (e04rs) which defines quadratic functions in assembled form as above and handle_set_qconstr_fac (e04rt) which uses factored form
\frac{1}{2}{x}^{T}{F}_{i}^{T}{F}_{i}x+{r}_{i}^{T}x+{s}_{i}
. The latter form is useful in many applications where the factors
{F}_{i}
appear naturally, such as in data fitting
\parallel Fx-b{\parallel }_{2}
The key question is if the problem is convex or non-convex as it determines if the problem can be solved via conic optimization (second-order cone programming, SOCP) or only by generic nonlinear programming (NLP). The problem is convex if all
{Q}_{i},i=0,\dots ,p
, are positive semidefinite (the factored form is positive semidefinite by definition).
QCQP problems were solvable even without the new additions in this release but a nontrivial knowledge of reformulation techniques was required if an SOCP solver was to be used or callbacks covering the quadratic functions had to be introduced in the general case. To remove the unnecessary burden from practitioners, the new routines provide the quadratic data directly to the solvers with an appropriate automatic reformulation as is required by the chosen solver from the Suite.
For non-convex QCQP, NAG will use the input data to automatically assembly first and second derivatives that are used by the nonlinear programming solver (such as handle_solve_ipopt (e04st)).
Even though convex QCQP problems can also be solved via nonlinear programming, we generally recommend the second-order cone programming solver (handle_solve_socp_ipm (e04pt)) due to its computational efficiency and ability to detect infeasibility. In that case, NAG will take care of the reformulation of quadratic functions to cones, including the factorization of all
{Q}_{i}
matrices in an efficient and numerically robust way. This is particularly helpful in the case of singular or close-to-singular matrices.
By maintaining the consistency of the interface of the solvers within the NAG Optimization Modelling Suite, e04rs and e04rt become part of the suite that simplifies building models and enhances NAG’s offering in mathematical optimization. For examples and further reading visit our GitHub Local optimisation page. |
GenerateSimilar - Maple Help
Home : Support : Online Help : Programming : Random Objects : RandomTools package : Commands : GenerateSimilar
create a random expression similar to the one given
GenerateSimilar( expr )
integer, float, polynomial, or general expression
The GenerateSimilar command produces a new expression that is similar to the given expression expr. The structure and variables in the original expression are preserved, but new constants and coefficients replace the initial ones. The random floating-point numbers and integers in the new expression are the same magnitude as in the original.
Trig functions are not always preserved: sin and cos can switch, sec and csc can switch, and tan and cot can switch. The inverse trig functions are paired and can switch in the same way.
Inputting a polynomial that can be factored and has integer roots returns a polynomial of the same order that can be factored and has integer roots.
If a polynomial is input without integer roots then a polynomial up to the same degree as the input polynomial with random coefficients up to the same order of magnitude as the largest integer in the input expression is returned.
Numerators and denominators of input rational expressions are replaced in the same way as polynomials with the additional feature that if the numerator and denominator have a common root then the returned rational expression also has a shared root in its numerator and denominator.
If a singular square matrix is input then a singular square matrix is output. Inputting upper triangular, lower triangular, diagonal, symmetric, hermitian, or antisymmetric matrices returns the same type. Otherwise the elements of the output matrix are generated individually from each element of the input matrix.
Rational and integer exponents of functions or expressions (other than exponential functions) are preserved.
If an integral is passed into the function then the resultant integral will be solvable using the same integration technique (u substitution, partial fractions, trig substitution, integration by parts).
If an equation is passed into GenerateSimilar then a similar equation is output: radical equations produce radical equations, polynomial equations produce polynomial equations, logarithmic equations produce logarithmic equations, and exponential equations produce exponential equations.
Sums over values of binomial or poisson probability distributions produce sums over values of binomial or poisson probability distributions respectively.
Expectation value of a function over a binomial or poisson probability distribution returns an expectation value of a similar function over a binomial or poisson probability distribution respectively.
Integrals over values of exponential or gaussian probability distributions return integrals over values of exponential or gaussian probability distributions respectively.
Expectation value of a function over an exponential or gaussian probability distribution returns an expectation value of a similar function over an exponential or gaussian probability distribution respectively.
Parametrizations of circles, ellipses and cycloids return parametrizations of circles, ellipses and cycloids.
Differential equations should not be input into GenerateSimilar; for differential equations use GenerateSimilarODE.
{\textstyle \mathrm{with}\left(\mathrm{RandomTools}\right)\:}
{\textstyle \mathrm{GenerateSimilar}\left(x\right)}
{\textstyle \textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{2}}
The cube on the sin function is preserved.
{\textstyle \mathrm{GenerateSimilar}\left({\mathrm{sin}\left(x\right)}^{3}+3\mathrm{exp}\left({x}^{2}\right)\right)}
{\textstyle \textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{Typesetting}}\textcolor[rgb]{0,0,1}{:-}\textcolor[rgb]{0,0,1}{\mathrm{\_Hold}}\textcolor[rgb]{0,0,1}{}\left(\left[\textcolor[rgb]{0,0,1}{\mathrm{\%cos}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{7}\right)\right]\right)}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{Typesetting}}\textcolor[rgb]{0,0,1}{:-}\textcolor[rgb]{0,0,1}{\mathrm{\_Hold}}\textcolor[rgb]{0,0,1}{}\left(\left[\textcolor[rgb]{0,0,1}{\mathrm{\%exp}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{\+}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\right)\right]\right)}
Inputting a factorable polynomial returns a factorable polynomial..
{\textstyle \mathrm{factor}\left({r}^{2}+r-6\right)}
{\textstyle \left(\textcolor[rgb]{0,0,1}{r}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{3}\right)\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{r}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}\right)}
{\textstyle \mathrm{poly1}≔\mathrm{GenerateSimilar}\left({r}^{2}+r-6\right)}
{\textstyle \textcolor[rgb]{0,0,1}{\mathrm{poly1}}\textcolor[rgb]{0,0,1}{≔}{\textcolor[rgb]{0,0,1}{r}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{6}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{r}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{16}}
{\textstyle \mathrm{factor}\left(\mathrm{poly1}\right)}
{\textstyle \left(\textcolor[rgb]{0,0,1}{r}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{2}\right)\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{r}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{8}\right)}
A polynomial without integer roots returns a polynomial with random coefficients
{\textstyle \mathrm{solve}\left({r}^{2}-2r+2=0\right)}
{\textstyle \textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{I}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{I}}}
{\textstyle \mathrm{GenerateSimilar}\left({r}^{2}-2r+2\right)}
{\textstyle {\textcolor[rgb]{0,0,1}{r}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{r}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{2}}
Factorable numerators and denominators remain factorable and if a factor is shared between the numerator and denominator then the resultant rational function will share a factor between numerator and denominator.
{\textstyle \mathrm{factor}\left(\frac{{y}^{2}-1}{y-3}\right)}
{\textstyle \frac{\left(\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\right)\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\right)}{\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{3}}}
{\textstyle \mathrm{rational1}≔\mathrm{GenerateSimilar}\left(\frac{{y}^{2}-1}{y-3}\right)}
{\textstyle \textcolor[rgb]{0,0,1}{\mathrm{rational1}}\textcolor[rgb]{0,0,1}{≔}\frac{\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{3}}{\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{4}}}
{\textstyle \mathrm{factor}\left(\mathrm{rational1}\right)}
{\textstyle \frac{\left(\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\right)\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{3}\right)}{\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{4}}}
{\textstyle \mathrm{factor}\left(\frac{{y}^{2}-1}{y+1}\right)}
{\textstyle \textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}} |
Pointwise convergence - formulasearchengine
In mathematics, pointwise convergence is one of various senses in which a sequence of functions can converge to a particular function.[1][2]
4 Almost everywhere convergence
Suppose { fn } is a sequence of functions sharing the same domain and codomain (for the moment, defer specifying the nature of the values of these functions, but the reader may take them to be real numbers). The sequence { fn } converges pointwise to f, often written as
{\displaystyle \lim _{n\rightarrow \infty }f_{n}=f\ {\mbox{pointwise}},}
{\displaystyle \lim _{n\rightarrow \infty }f_{n}(x)=f(x).}
for every x in the domain.
This concept is often contrasted with uniform convergence. To say that
{\displaystyle \lim _{n\rightarrow \infty }f_{n}=f\ {\mbox{uniformly}}}
{\displaystyle \lim _{n\rightarrow \infty }\,\sup\{\,\left|f_{n}(x)-f(x)\right|:x\in {\mbox{the domain}}\,\}=0.}
That is a stronger statement than the assertion of pointwise convergence: every uniformly convergent sequence is pointwise convergent, to the same limiting function, but some pointwise convergent sequences are not uniformly convergent. For example we have
{\displaystyle \lim _{n\rightarrow \infty }x^{n}=0\ {\mbox{pointwise}}\ {\mbox{on}}\ {\mbox{the}}\ {\mbox{interval}}\ [0,1),\ {\mbox{but}}\ {\mbox{not}}\ {\mbox{uniformly}}\ {\mbox{on}}\ {\mbox{the}}\ {\mbox{interval}}\ [0,1).}
The pointwise limit of a sequence of continuous functions may be a discontinuous function, but only if the convergence is not uniform. For example,
{\displaystyle f(x)=\lim _{n\rightarrow \infty }\cos(\pi x)^{2n}}
takes the value 1 when x is an integer and 0 when x is not an integer, and so is discontinuous at every integer.
The values of the functions fn need not be real numbers, but may be in any topological space, in order that the concept of pointwise convergence make sense. Uniform convergence, on the other hand, does not make sense for functions taking values in topological spaces generally, but makes sense for functions taking values in metric spaces, and, more generally, in uniform spaces.
Pointwise convergence is the same as convergence in the product topology on the space YX, where X is the domain and Y is the codomain. If the codomain Y is compact, then, by Tychonoff's theorem, the space YX is also compact.
In measure theory, one talks about almost everywhere convergence of a sequence of measurable functions defined on a measurable space. That means pointwise convergence almost everywhere. Egorov's theorem states that pointwise convergence almost everywhere on a set of finite measure implies uniform convergence on a slightly smaller set.
hu:Függvénysorozatok konvergenciája#Pontonkénti konvergencia
Retrieved from "https://en.formulasearchengine.com/index.php?title=Pointwise_convergence&oldid=228714" |
EUDML | -Correspondences: The inclusion . EuDML | -Correspondences: The inclusion .
L
-Correspondences: The inclusion
{L}^{p}\left(\mu ,X\right)\subset {L}^{q}\left(\upsilon ,Y\right)
Dawson, C.Bryan
Dawson, C.Bryan. "-Correspondences: The inclusion .." International Journal of Mathematics and Mathematical Sciences 19.4 (1996): 723-726. <http://eudml.org/doc/47646>.
author = {Dawson, C.Bryan},
keywords = {Lebesgue-Bochner spaces; measurable point mapping; inclusions; -correspondence; equimeasurability theorem; -correspondence},
title = {-Correspondences: The inclusion .},
AU - Dawson, C.Bryan
TI - -Correspondences: The inclusion .
KW - Lebesgue-Bochner spaces; measurable point mapping; inclusions; -correspondence; equimeasurability theorem; -correspondence
Lebesgue-Bochner spaces, measurable point mapping, inclusions,
L
-correspondence, equimeasurability theorem,
L
{L}^{p}
Articles by Dawson |
Sound | Brilliant Math & Science Wiki
Ashish Menon, Samara Simha Reddy, Rushikesh Jogdand, and
Sound is a mechanically propagating pressure wave in a material medium, and is a typical example of longitudinal wave. When in a certain range of frequency, it causes the sensation of hearing.
Vibrations in a Stretched String
Musical Sound and Noise
Sound energy, like light energy, obeys the laws of reflection:
Its angle of incidence is equal to the angle of reflection.
Incident wave, reflected wave, and the normal lie in the same plane.
The following experiment demonstrates the laws of reflection for sound waves:
Take a smooth polished large wooden board and mount it vertically on a table, at right angles to the wooden screen. On each side of the screen place a long, narrow, and highly polished tube from inside. Place a clock at the end of one tube. Move the other tube slightly left or right, till a distinct tick of the clock is heard. Measure the angles of incidence and reflection, and you will find that they are equal. Thus, the experiment illustrates the law of reflection.
Practical Applications of Sound
Megaphone or speaking tube:
When you have to call someone at a far-off distance (say 100 m), what do you do? You cup your hands and call the person with the maximum sound you can produce. Why do you cup your hands? It is because the hands prevent the sound energy from spreading in all directions, much the same way people use horn-shaped metal tubes, commonly called megaphones, while addressing a group of people in fairs or tourist spots. Similarly, the loud-speakers used at the public address system have horn-shaped openings. In all such devices, sound energy is prevented from spreading out by successive reflections from the horn-shaped tubes.
Ear trumpet or hearing aid:
Ear trumpet is a device which is used by people who is hard of hearing. Its shape is like a trumpet. The narrow end of it is kept in the earhole of the person who is hard of hearing, whereas the wider end faces towards the speaker. The waves received by the wider end of the trumpet are reflected into the narrower end, thus increasing its intensity. So, the person who is hard of hearing can hear easily.
Sound boards:
Sound waves obey the laws of reflection on plane as well as curved reflecting surfaces. In order to spread sound evenly in big halls or auditoriums, the speaker is fixed at the focus of a concave reflector which is commonly called sounding board. The sound waves on striking the sounding board gets reflected parallel to the principal axis. So, everyone can hear clearly.
The phenomenon, due to which the repetition of sound is heard after reflection from a distant object (such as a hillock or a high building) after the original sound from a given source dies off, is called an echo.
It has been found that the sensation of any sound lasts for
\frac{1}{10}
of a second. This time is called the persistence of audibility or persistence of hearing. Thus, it is clear that if any sound reaches back to the ear in less than the mentioned time, we cannot make out when the original sound died and the reflected sound reached the ear. In other words, no echo is heard. But the observations is different in a contrary situation.
Relation between speed of sound, distance of a reflecting body from a source of sound, and time for hearing an echo:
t
is the time after which an echo is heard,
d
is the distance between the source of sound and the reflecting body, and
v
is the speed of sound, then the total distance travelled by the sound is
2d
Hence , in time
the distance travelled by the sound is
2d.
Hence, in 1 second the distance travelled by the sound is
\frac {2d}{t},
which is "the speed of sound."
v=\frac {2d}{t}.
Conditions for formation of echoes:
The minimum distance between the source of sound and the reflecting body should be
17\text{ m}
normally because the speed of sound in air is normally
332 \text{ m/s}.
The wavelength of sound should be less than the height of the reflecting body.
The intensity of sound should be sufficient enough that it can be heard after reflection.
When a series of reflections fall on the ear from various reflectors one after the another in a closed room, thereby forming a continuous rolling sound, reverberation is said to be caused.
Auditoriums are made in such a manner that each person receives the sound signals.
Numerical problems on echoes:
A boy stands
83 \text{ m}
in front of a high wall and then blows a whistle. Calculate the time interval when he hears an echo. Speed of sound is
332\text{ m/s}
Distance between wall and the boy:
d = 83\text{ m}
v= 332 \text{ m/s}
\frac {2d}{v} = \frac{2\times 83}{332} = 0.5
_\square
A man stands between two parallel cliffs and fires a gun. He hears two successive echoes after
2
4.5
seconds. What is the distance between the two cliffs? The speed of sound in air is
332\text{ m/s}
Speed of sound in air:
v = 332\text{ m/s}
Time taken to hear first echo:
t_1= 2\text{ s}
Distance of the man from the nearer cliff:
d_1=\frac {v\times t}{2}=\frac{332\times 2}{2}=332\text{ m}
Time taken to hear second echo:
t_2= 4.5\text{ s}
Distance of the man from the farther cliff:
d_2=\frac{v\times t}{2}=\frac{332\times 4.5}{2}=747\text{ m}
So, the distance between the cliffs is
d_1+d_2=332 + 1079 = 1079
_\square
A boy standing in front of a cliff on the other side of a river explodes a powerful cracker. He hears an echo after 4 seconds. Then he moves
125\text{ m}
backwards and again explodes another powerful cracker. This time, he hears an echo after 5 seconds. Calculate the width of the river (in metres).
The free vibrations produced in a body, on being slightly disturbed from its mean position, are called free vibrations of natural vibrations. It can also be defined as the periodic vibrations of a body of constant amplitude in the absence of any external force. In the absence of any resistance (such as air, etc.), the amplitude of free vibrations remains constant and so does its frequency. Theoretically, such vibrations are possible only in vacuum. However, practically, such vibrations are not possible because of the presence of a medium.
Natural time period: The time period of a body executing natural vibrations is called natural time period.
Natural frequency: The number of vibrations executed per second by a freely vibrating body is called natural frequency.
Examples of free vibrations:
a freely suspended pendulum vibrating about its mean position
a metal blade clamped at one end being gently disturbed
a tuning fork on being struck on a rubber pad
When a simple pendulum is set in motion, it is observed that its amplitude continuously decreases, till it comes to rest. Similarly, the amplitude of any freely vibrating body decreases with the passage of time. However, it is noticed that the frequency or time period of the vibrating body remains the same, and thus the motion is periodic in nature.The damping (decrease in amplitude) occurs due to frictional force, which the surrounding medium exerts on the vibrating body.
Definition: The periodic vibrations of continuously decreasing amplitude in the presence of the resistive force are called damped vibrations.
The frictional force is found proportional to the velocity of the body and the nature of the surrounding material, such as its density, viscosity, etc. Due to frictional force, the vibrating system loses energy, with the result that its amplitude gradually decreases, till it comes to rest. The energy lost by the vibrating body gradually changes into heat energy and is dissipated in the surrounding medium.
Examples of damped vibrations:
vibrations of tuning fork in air
vibration of stringed instruments in air
vibration of simple pendulum in air.
Definition: The phenomenon of damping does not allow a freely vibrating body to maintain its amplitude of vibration. If the amplitude of vibration has to be maintained, then an external periodic force has to be applied. The vibrations, so produced by the external periodic force, are called forced vibrations.
Characteristics of forced vibrations:
The body on which the external periodic force is applied does not vibrate with its natural frequency, but with the frequency of an external periodic force.
The amplitude of vibrations remain constant with time, but its magnitude depends upon the frequency of the driving force.
If the frequency of the external force is much different from the natural frequency of the body, the amplitude of oscillations is very small.
If the frequency of the external force is exactly equal to, or is an integral multiple of, the natural frequency of the vibrating body, the amplitude of oscillations is very large.
Examples of forced vibrations:
When the handle of a vibrating tuning fork is pressed against a table top, a loud sound is heard. It is because the vibrating tuning fork forces the table top to vibrate with its own natural frequency. As the table top is very large, the forced vibrations produce larger sound.
When the wire of a sitar or a guitar is plucked, its board and wind box make forced vibrations.
All stringed instruments are provided with a wind box. When the vibrations produced by strings are impressed on the enclosed air, they produce forced vibrations and a loud sound is heard.
When the needle of a gramophone player moves in the grooves of a record, it produces forced vibrations.
Definition: It is defined as a phenomenon when the frequency of an applied external force is equal to, or is an integral multiple of, the natural frequency of the body on which the force is applied, such that it readily takes up the vibrations and begins to vibrate with an increased amplitude.
Conditions for the phenomenon of resonance:
The natural frequency of the given body must be equal to, or is an integral multiple of, the frequency of the external force.
The vibrating body must have sufficient force so that it can set the other body into vibrations.
Examples of resonance:
It is a common experience that the frame of a motor cycle starts vibrating violently when driven at some particular speed. The reason is that at this time the natural frequency of the frame matches with the frequency of the piston of the engine, causing resonance.
Soldiers are often asked to break their steps while crossing a bridge. This precaution is taken to prevent any sudden collapse of the bridge, which may occur by the matching of the frequency of the impressed force due to footsteps of the soldiers with the natural frequency of bridge.
When the radio station broadcasts at some particular wavelength or frequency, these waves are received by the antenna in which they produce forced vibrations. When the person at the receiving end tunes his radio, he is in a way changing the natural frequency of the radio set. When the natural frequency of the radio set corresponds to the frequency of the broadcasting station, resonance takes place. Thus, the radio signals are amplified. These signals are further processed in the radio, and hence the sound is amplified.
Resonance is a special case of forced vibrations. When the frequency of an externally applied periodic force on a body is equal to its natural frequency, the body readily begins to vibrate with an increased amplitude. This phenomenon is known as resonance. The vibrations of large amplitude are called resonant vibrations.
With which of the following frequencies does a tuning fork of
256\text{ Hz}
resonate?
Whenever a string is plucked, it starts vibrating with a specific frequency. The frequency of the note depends upon the following factors:
Law of length: The frequency of a note produced by a stretched string is inversely proportional to its length.
Law of tension: The frequency of a note produced by a stretched string is directly proportional to the square root of the tension in the string.
Law of mass: The frequency of a note produced in a stretched string is inversely proportional to the square root of the mass per unit length of the string.
Node: Node is a point on the vibrating string with maximum tension and minimum displacement.
Antinode: Antinode is the point on the vibrating string with maximum displacement and zero tension.
Principal note or first harmonic or fundamental note: If a wire is made to vibrate in
2
3
antinodes, then the note produced is called pricipal note or fundamental note or first harmonic.
Second harmonic or first octave: If a wire is made to vibrate in
3
2
antinodes, then the note produced is called principal note or second harmonic or first octave. Second harmonic
Two strings of guitar are of length
80\text{ cm}
60\text{ cm}
, respectively, are made from the same material, and are of the same thickness and under the same tension. If the frequency of the longer wire is
300\text{ Hz}
, find the frequency of the smaller wire.
f \propto \frac {1}{l},
\frac {f_1}{f_2}= \frac {l_2}{l_1}.
Substituting the values,
\begin{aligned} \dfrac {300}{f_2} &= \dfrac {60}{80}\\ f_2 &= \dfrac {300×80}{60}\\ &= 400\text{ (Hz)}.\ _\square \end{aligned}
Sound waves which produce pleasant sensation in our ears and are acceptable are called musical sounds. Sound waves which produce unpleasant sensation in our ears and are unacceptable are called noise. However, it is difficult to differentiate between musical sound and noise because a particular sound may be pleasant and acceptable to someone, but it may be unpleasant and unacceptable to others. Generally, sound waves which are produced by regular periodic vibrations are musical in character, but those produced by irregular non-periodic vibrations of a body are non-musical in character or noise.
This characteristic of musical sound enable us to differentiate between two sounds of equal loudness, coming from different sources and having different frequencies:
\begin{aligned} \text {Pitch} &\propto \text {Frequency}\\\\ \text {Pitch} &\propto \dfrac{1}{\text {Wavelength}}. \end{aligned}
Intensity of sound is the time rate at which the sound energy flows through a unit area. Loudness is the sensation produced in the brain by the combined effect of intensity and response of the ear:
\begin{aligned} \text {Intensity} &\propto {\text {Amplitude}}^2\\\\ \text {Intensity} &\propto \dfrac {1}{{\text {Distance}}^2}\\\\\\ \text {Intensity} &\propto \text {Density of the medium}. \end{aligned}
Unit for measuring loudness of sound: The loudness of sound is measured in decibels (dB). One decibel is defined as the change in loudness of sound when the intensity increases by
26\%
The notes of different musical instruments are distinguished by this characteristic. It is because different waveforms are produced by different musical instruments. Sound produced by musical instruments differ in wavelength, loudness and waveform. The waveforms differ because a musical instrument produces a number of subsidiary notes other than the main note and this difference depends upon the shape, size and material of the instruments. This is the same reason why few people are really singers because the quality of sound produced by them is far superior and has a more pleasant sensation on our ears.
Waveforms of different musical instruments
Cite as: Sound. Brilliant.org. Retrieved from https://brilliant.org/wiki/sound/ |
RealRootCounting - Maple Help
Home : Support : Online Help : Mathematics : Factorization and Solving Equations : RegularChains : SemiAlgebraicSetTools Subpackage : RealRootCounting
number of distinct real solutions of a semi-algebraic system
RealRootCounting(F, N, P, H, R)
The command RealRootCounting(F, N, P, H, R) returns the number of distinct real solutions of the system whose equations, inequations, positive polynomials, and non-negative polynomials are given by F, H, P and N respectively.
This computation assumes that the polynomial system given by F and H (as equations and inequations respectively) has finitely many complex solutions.
The base field of R is meant to be the field of rational numbers.
The algorithm is described in the paper by Xia, B., Hou, X.: "A complete algorithm for counting real solutions of polynomial systems of equations and inequalities." Computers and Mathematics with applications, Vol. 44 (2002): pp.633-642.
\mathrm{with}\left(\mathrm{RegularChains}\right):
\mathrm{with}\left(\mathrm{SemiAlgebraicSetTools}\right):
R≔\mathrm{PolynomialRing}\left([y,x]\right):
F≔[{x}^{2}-1,{y}^{2}+2xy+1]
\textcolor[rgb]{0,0,1}{F}\textcolor[rgb]{0,0,1}{≔}[{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}]
Compute the number of nonnegative solutions.
N≔[x,y];
P≔[];
H≔[]
\textcolor[rgb]{0,0,1}{N}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}]
\textcolor[rgb]{0,0,1}{P}\textcolor[rgb]{0,0,1}{≔}[]
\textcolor[rgb]{0,0,1}{H}\textcolor[rgb]{0,0,1}{≔}[]
\mathrm{RealRootCounting}\left(F,N,P,H,R\right)
\textcolor[rgb]{0,0,1}{0}
R≔\mathrm{PolynomialRing}\left([c,z,y,x]\right)
\textcolor[rgb]{0,0,1}{R}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{polynomial_ring}}
F≔[1-cx-x{y}^{2}-x{z}^{2},1-cy-y{x}^{2}-y{z}^{2},1-cz-z{x}^{2}-z{y}^{2},8{c}^{6}+378{c}^{3}-27]
\textcolor[rgb]{0,0,1}{F}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{z}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{c}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{z}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{c}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{c}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{8}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{c}}^{\textcolor[rgb]{0,0,1}{6}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{378}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{c}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{27}]
Require c to be positive here.
N≔[];
P≔[c];
H≔[]
\textcolor[rgb]{0,0,1}{N}\textcolor[rgb]{0,0,1}{≔}[]
\textcolor[rgb]{0,0,1}{P}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{c}]
\textcolor[rgb]{0,0,1}{H}\textcolor[rgb]{0,0,1}{≔}[]
\mathrm{RealRootCounting}\left(F,N,P,H,R\right)
\textcolor[rgb]{0,0,1}{4} |
Nazca Plate - Wikipedia
Oceanic tectonic plate in the eastern Pacific Ocean basin
The Nazca Plate or Nasca Plate,[2] named after the Nazca region of southern Peru, is an oceanic tectonic plate in the eastern Pacific Ocean basin off the west coast of South America. The ongoing subduction, along the Peru–Chile Trench, of the Nazca Plate under the South American Plate is largely responsible for the Andean orogeny. The Nazca Plate is bounded on the west by the Pacific Plate and to the south by the Antarctic Plate through the East Pacific Rise and the Chile Rise respectively. The movement of the Nazca Plate over several hotspots has created some volcanic islands as well as east-west running seamount chains that subduct under South America. Nazca is a relatively young plate both in terms of the age of its rocks and its existence as an independent plate having been formed from the break-up of the Farallon Plate about 23 million years ago. The oldest rocks of the plate are about 50 million years old.[3]
1.1 East Pacific and Chile Rise
1.2 Peru–Chile Trench
2 Intraplate features
2.2 Aseismic ridges
2.3 Fracture zones
3 Plate motion
East Pacific and Chile Rise[edit]
Main articles: East Pacific Rise and Chile Rise
A triple junction, the Chile Triple Junction,[4] occurs on the seafloor of the Pacific Ocean off Taitao and Tres Montes Peninsula at the southern coast of Chile. Here, three tectonic plates meet: the Nazca Plate, the South American Plate, and the Antarctic Plate.
Peru–Chile Trench[edit]
See also: Peru–Chile Trench
The eastern margin is a convergent boundary subduction zone under the South American Plate and the Andes Mountains, forming the Peru–Chile Trench. The southern side is a divergent boundary with the Antarctic Plate, the Chile Rise, where seafloor spreading permits magma to rise. The western side is a divergent boundary with the Pacific Plate, forming the East Pacific Rise. The northern side is a divergent boundary with the Cocos Plate, the Galapagos Rise.
The subduction of the Nazca plate under southern Chile has a history of producing massive earthquakes, including the largest ever recorded on earth, the moment magnitude 9.5 1960 Valdivia earthquake.
Intraplate features[edit]
Hotspots[edit]
Main articles: Easter hotspot, Juan Fernández hotspot, and Galápagos hotspot
A second triple junction occurs at the northwest corner of the plate where the Nazca, Cocos, and Pacific Plates all join off the coast of Colombia. Yet another triple junction occurs at the southwest corner at the intersection of the Nazca, Pacific, and Antarctic Plates off the coast of southern Chile. At each of these triple junctions an anomalous microplate exists, the Galapagos Microplate at the northern junction and the Juan Fernandez Microplate at the southern junction. The Easter Island Microplate is a third microplate that is located just north of the Juan Fernandez Microplate and lies just west of Easter Island.
Aseismic ridges[edit]
Main articles: Nazca Ridge, Juan Fernández Ridge, and Carnegie Ridge
The Carnegie Ridge is a 1,350-kilometre-long (840 mi) and up to 300-kilometre-wide (190 mi) feature on the ocean floor of the northern Nazca Plate that includes the Galápagos archipelago at its western end. It is being subducted under South America with the rest of the Nazca Plate.
Fracture zones[edit]
The Darwin Gap is the area between the Nazca Plate and the coast of Chile, where Charles Darwin experienced the earthquake of 1835. It is expected that this area will be the epicenter of a major quake in the near future.[5]
Plate motion[edit]
The absolute motion of the Nazca Plate has been calibrated at 3.7 cm/year (1.5 in/year) east motion (88°), one of the fastest absolute motions of any tectonic plate. The subducting Nazca Plate, which exhibits unusual flat slab subduction, is tearing as well as deforming as it is subducted (Barzangi and Isacks). The subduction has formed, and continues to form, the volcanic Andes Mountain Range. Deformation of the Nazca Plate even affects the geography of Bolivia, far to the east (Tinker et al.). The 1994 Bolivia earthquake occurred on the Nazca Plate; this had a magnitude of 8.2
{\displaystyle M_{w}}
, which at that time was the strongest instrumentally recorded earthquake occurring deeper than 300 km (190 mi).
Aside from the Juan Fernández Islands, this area has very few other islands that are affected by the earthquakes that are a result of complicated movements at these junctions.
See also: Farallon Plate
The precursor of the Nazca Plate, Juan de Fuca Plate, and the Cocos Plate was the Farallon Plate, which split in the late Oligocene, about 22.8 Mya, a date arrived at by interpreting magnetic anomalies. Subduction under the South American continent began about 140 Mya, although the formation of the high parts of the Central Andes and the Bolivian orocline did not occur until 45 Mya. It has been suggested that the mountains were forced up by the subduction of the older and heavier parts of the plate, which sank more quickly into the mantle.[6]
^ "Sizes of Tectonic or Lithospheric Plates". About.com Geology. Retrieved 4 January 2016.
^ Oxford Atlas Of The World 26th Ed. New York, NY: Oxford University Press. 2019. p. 74. ISBN 978-0-19-006581-2.
^ Dutch, Steven (10 August 2009). "Sea Floor Spreading in the Pacific (Plate Boundaries Shown)". University of Wisconsin - Green Bay. Archived from the original on 17 March 2010.
^ Kelly McGuire (8 April 2004). "Tectonics of South America: Chile Triple Junction" (PDF). Retrieved 27 February 2016.
^ Kate Ravilious (30 Jan 2011). "Darwin Gap quake will shake Chile again". New Scientist. Retrieved 8 Feb 2011.
^ "Mountains on a plate form the Andes". No. 214. University World News. 25 March 2012. Retrieved 8 February 2016.
Extreme Science site: "A Lesson in Plate tectonics" The basics explained.
James, D. (1978). "Subduction of the Nazca plate beneath Central Peru." Geology 6 (3) pp 174–178
Tectonic plates of Central America (Pacific Plate–North American Plate–Caribbean Plate Convergence Zone)
Atl Fault
Atoyac Fault
Ballenas Fault
Carmen Fault
Cerro Prieto Fault
Chapala Tula Fault Zone
Donaji Fault
Guaymas Fault
Gulf of California Rift Zone
Farallon Fault
Oaxaca Fault
Pedro Miguel Fault
Pescadero Fault
Rivera Transform Fault
Polochic Fault - Motagua Fault
San Lorenzo Fault
Tamayo Fault
Trenches and
Alarcon Basin
Consag Basin
Delfin Basin
Middle America Trench (Acapulco/Guatemala)
San Pedro Mártir Basin
Wagner Basin
Rivera Triple Junction
List of faults in Costa Rica
List of seismic faults in Mexico
Retrieved from "https://en.wikipedia.org/w/index.php?title=Nazca_Plate&oldid=1078213147"
Natural history of South America |
A uniform classification of discrete series representations of affine Hecke algebras
Dan Ciubotaru and Eric Opdam
We give a new and independent parametrization of the set of discrete series characters of an affine Hecke algebra
{\mathsc{ℋ}}_{v}
, in terms of a canonically defined basis
{\mathsc{ℬ}}_{gm}
of a certain lattice of virtual elliptic characters of the underlying (extended) affine Weyl group. This classification applies to all semisimple affine Hecke algebras
\mathsc{ℋ}
, and to all
v\in \mathsc{Q}
\mathsc{Q}
denotes the vector group of positive real (possibly unequal) Hecke parameters for
\mathsc{ℋ}
. By analytic Dirac induction we define for each
b\in {\mathsc{ℬ}}_{gm}
a continuous (in the sense of Opdam and Solleveld (2010)) family
{\mathsc{Q}}_{b}^{reg}:={\mathsc{Q}}_{b}\setminus {\mathsc{Q}}_{b}^{sing}\ni v\to {Ind}_{D}\left(b;v\right)
ϵ\left(b;v\right){Ind}_{D}\left(b;v\right)
(for some
ϵ\left(b;v\right)\in \left\{±1\right\}
) is an irreducible discrete series character of
{\mathsc{ℋ}}_{v}
{\mathsc{Q}}_{b}^{sing}\subset \mathsc{Q}
is a finite union of hyperplanes in
\mathsc{Q}
In the nonsimply laced cases we show that the families of virtual discrete series characters
{Ind}_{D}\left(b;v\right)
are piecewise rational in the parameters
v
. Remarkably, the formal degree of
{Ind}_{D}\left(b;v\right)
in such piecewise rational family turns out to be rational. This implies that for each
b\in {\mathsc{ℬ}}_{gm}
there exists a universal rational constant
{d}_{b}
determining the formal degree in the family of discrete series characters
ϵ\left(b;v\right){Ind}_{D}\left(b;v\right)
. We will compute the canonical constants
{d}_{b}
, and the signs
ϵ\left(b;v\right)
. For certain geometric parameters we will provide the comparison with the Kazhdan–Lusztig–Langlands classification.
https://projecteuclid.org/ant
Affine Hecke algebra, graded affine Hecke algebra, Dirac operator, discrete series representation
Secondary: 22D25, 43A30 |
Hurwitz's_automorphisms_theorem Knowpia
In mathematics, Hurwitz's automorphisms theorem bounds the order of the group of automorphisms, via orientation-preserving conformal mappings, of a compact Riemann surface of genus g > 1, stating that the number of such automorphisms cannot exceed 84(g − 1). A group for which the maximum is achieved is called a Hurwitz group, and the corresponding Riemann surface a Hurwitz surface. Because compact Riemann surfaces are synonymous with non-singular complex projective algebraic curves, a Hurwitz surface can also be called a Hurwitz curve.[1] The theorem is named after Adolf Hurwitz, who proved it in (Hurwitz 1893).
Hurwitz's bound also holds for algebraic curves over a field of characteristic 0, and over fields of positive characteristic p>0 for groups whose order is coprime to p, but can fail over fields of positive characteristic p>0 when p divides the group order. For example, the double cover of the projective line y2 = xp −x branched at all points defined over the prime field has genus g=(p−1)/2 but is acted on by the group SL2(p) of order p3−p.
Interpretation in terms of hyperbolicityEdit
One of the fundamental themes in differential geometry is a trichotomy between the Riemannian manifolds of positive, zero, and negative curvature K. It manifests itself in many diverse situations and on several levels. In the context of compact Riemann surfaces X, via the Riemann uniformization theorem, this can be seen as a distinction between the surfaces of different topologies:
X a sphere, a compact Riemann surface of genus zero with K > 0;
X a flat torus, or an elliptic curve, a Riemann surface of genus one with K = 0;
and X a hyperbolic surface, which has genus greater than one and K < 0.
While in the first two cases the surface X admits infinitely many conformal automorphisms (in fact, the conformal automorphism group is a complex Lie group of dimension three for a sphere and of dimension one for a torus), a hyperbolic Riemann surface only admits a discrete set of automorphisms. Hurwitz's theorem claims that in fact more is true: it provides a uniform bound on the order of the automorphism group as a function of the genus and characterizes those Riemann surfaces for which the bound is sharp.
Statement and proofEdit
{\displaystyle X}
be a smooth connected Riemann surface of genus
{\displaystyle g\geq 2}
. Then its automorphism group
{\displaystyle \mathrm {Aut} (X)}
has size at most
{\displaystyle 84(g-1)}
Proof: Assume for now that
{\displaystyle G=\mathrm {Aut} (X)}
is finite (we'll prove this at the end).
{\displaystyle X\to X/G}
{\displaystyle G}
acts by holomorphic functions, the quotient is locally of the form
{\displaystyle z\to z^{n}}
{\displaystyle X/G}
is a smooth Riemann surface. The quotient map
{\displaystyle X\to X/G}
is a branched cover, and we will see below that the ramification points correspond to the orbits that have a non trivial stabiliser. Let
{\displaystyle g_{0}}
be the genus of
{\displaystyle X/G}
By the Riemann-Hurwitz formula,
{\displaystyle 2g-2\ =\ |G|\cdot \left(2g_{0}-2+\sum _{i=1}^{k}\left(1-{\frac {1}{e_{i}}}\right)\right)}
where the sum is over the
{\displaystyle k}
ramification points
{\displaystyle p_{i}\in X/G}
for the quotient map
{\displaystyle X\to X/G}
. The ramification index
{\displaystyle e_{i}}
{\displaystyle p_{i}}
is just the order of the stabiliser group, since
{\displaystyle e_{i}f_{i}=\deg(X/\,X/G)}
{\displaystyle f_{i}}
the number of pre-images of
{\displaystyle p_{i}}
(the number of points in the orbit), and
{\displaystyle \deg(X/\,X/G)=|G|}
. By definition of ramification points,
{\displaystyle e_{i}\geq 2}
{\displaystyle k}
ramification indices.
Now call the righthand side
{\displaystyle |G|R}
{\displaystyle g\geq 2}
{\displaystyle R>0}
. Rearranging the equation we find:
{\displaystyle g_{0}\geq 2}
{\displaystyle R\geq 2}
{\displaystyle |G|\leq (g-1)}
{\displaystyle g_{0}=1}
{\displaystyle k\geq 1}
{\displaystyle R\geq 0+1-1/2=1/2}
{\displaystyle |G|\leq 4(g-1)}
{\displaystyle g_{0}=0}
{\displaystyle k\geq 3}
{\displaystyle k\geq 5}
{\displaystyle R\geq -2+k(1-1/2)\geq 1/2}
{\displaystyle |G|\leq 4(g-1)}
{\displaystyle k=4}
{\displaystyle R\geq -2+4-1/2-1/2-1/2-1/3=1/6}
{\displaystyle |G|\leq 12(g-1)}
{\displaystyle k=3}
then write
{\displaystyle e_{1}=p,\,e_{2}=q,\,e_{3}=r}
. We may assume
{\displaystyle 2\leq p\leq q\ \leq r}
{\displaystyle p\geq 3}
{\displaystyle R\geq -2+3-1/3-1/3-1/4=1/12}
{\displaystyle |G|\leq 24(g-1)}
{\displaystyle p=2}
{\displaystyle q\geq 4}
{\displaystyle R\geq -2+3-1/2-1/4-1/5=1/20}
{\displaystyle |G|\leq 40(g-1)}
{\displaystyle q=3}
{\displaystyle R\geq -2+3-1/2-1/3-1/7=1/42}
{\displaystyle |G|\leq 84(g-1)}
{\displaystyle |G|\leq 84(g-1)}
{\displaystyle G}
is finite, note that
{\displaystyle G}
acts on the cohomology
{\displaystyle H^{*}(X,\mathbf {C} )}
preserving the Hodge decomposition and the lattice
{\displaystyle H^{1}(X,\mathbf {Z} )}
In particular, its action on
{\displaystyle V=H^{0,1}(X,\mathbf {C} )}
gives a homomorphism
{\displaystyle h:G\to \mathrm {GL} (V)}
with discrete image
{\displaystyle h(G)}
In addition, the image
{\displaystyle h(G)}
preserves the natural non degenerate Hermitian inner product
{\textstyle (\omega ,\eta )=i\int {\bar {\omega }}\wedge \eta }
{\displaystyle V}
. In particular the image
{\displaystyle h(G)}
is contained in the unitary group
{\displaystyle \mathrm {U} (V)\subset \mathrm {GL} (V)}
which is compact. Thus the image
{\displaystyle h(G)}
is not just discrete, but finite.
It remains to prove that
{\displaystyle h:G\to \mathrm {GL} (V)}
has finite kernel. In fact, we will prove
{\displaystyle h}
is injective. Assume
{\displaystyle \phi \in G}
acts as the identity on
{\displaystyle V}
{\displaystyle \mathrm {fix} (\phi )}
is finite, then by the Lefschetz fixed-point theorem,
{\displaystyle |\mathrm {fix} (\phi )|=1-2\mathrm {tr} (h(\phi ))+1=2-2\mathrm {tr} (\mathrm {id} _{V})=2-2g<0.}
This is a contradiction, and so
{\displaystyle \mathrm {fix} (\phi )}
is infinite. Since
{\displaystyle \mathrm {fix} (\phi )}
is a closed complex sub variety of positive dimension and
{\displaystyle X}
is a smooth connected curve (i.e.
{\displaystyle \dim _{\mathbf {C} }(X)=1}
), we must have
{\displaystyle \mathrm {fix} (\phi )=X}
{\displaystyle \phi }
is the identity, and we conclude that
{\displaystyle h}
{\displaystyle G\cong h(G)}
is finite. Q.E.D.
Corollary of the proof: A Riemann surface
{\displaystyle X}
{\displaystyle g\geq 2}
{\displaystyle 84(g-1)}
automorphisms if and only if
{\displaystyle X}
is a branched cover
{\displaystyle X\to \mathbf {P} ^{1}}
with three ramification points, of indices 2,3 and 7.
The idea of another proof and construction of the Hurwitz surfacesEdit
By the uniformization theorem, any hyperbolic surface X – i.e., the Gaussian curvature of X is equal to negative one at every point – is covered by the hyperbolic plane. The conformal mappings of the surface correspond to orientation-preserving automorphisms of the hyperbolic plane. By the Gauss–Bonnet theorem, the area of the surface is
A(X) = − 2π χ(X) = 4π(g − 1).
In order to make the automorphism group G of X as large as possible, we want the area of its fundamental domain D for this action to be as small as possible. If the fundamental domain is a triangle with the vertex angles π/p, π/q and π/r, defining a tiling of the hyperbolic plane, then p, q, and r are integers greater than one, and the area is
A(D) = π(1 − 1/p − 1/q − 1/r).
Thus we are asking for integers which make the expression
1 − 1/p − 1/q − 1/r
strictly positive and as small as possible. This minimal value is 1/42, and
1 − 1/2 − 1/3 − 1/7 = 1/42
gives a unique (up to permutation) triple of such integers. This would indicate that the order |G| of the automorphism group is bounded by
A(X)/A(D) ≤ 168(g − 1).
However, a more delicate reasoning shows that this is an overestimate by the factor of two, because the group G can contain orientation-reversing transformations. For the orientation-preserving conformal automorphisms the bound is 84(g − 1).
Hurwitz groups and surfaces are constructed based on the tiling of the hyperbolic plane by the (2,3,7) Schwarz triangle.
To obtain an example of a Hurwitz group, let us start with a (2,3,7)-tiling of the hyperbolic plane. Its full symmetry group is the full (2,3,7) triangle group generated by the reflections across the sides of a single fundamental triangle with the angles π/2, π/3 and π/7. Since a reflection flips the triangle and changes the orientation, we can join the triangles in pairs and obtain an orientation-preserving tiling polygon. A Hurwitz surface is obtained by 'closing up' a part of this infinite tiling of the hyperbolic plane to a compact Riemann surface of genus g. This will necessarily involve exactly 84(g − 1) double triangle tiles.
The following two regular tilings have the desired symmetry group; the rotational group corresponds to rotation about an edge, a vertex, and a face, while the full symmetry group would also include a reflection. The polygons in the tiling are not fundamental domains – the tiling by (2,3,7) triangles refines both of these and is not regular.
order-7 triangular tiling
Wythoff constructions yields further uniform tilings, yielding eight uniform tilings, including the two regular ones given here. These all descend to Hurwitz surfaces, yielding tilings of the surfaces (triangulation, tiling by heptagons, etc.).
From the arguments above it can be inferred that a Hurwitz group G is characterized by the property that it is a finite quotient of the group with two generators a and b and three relations
{\displaystyle a^{2}=b^{3}=(ab)^{7}=1,}
thus G is a finite group generated by two elements of orders two and three, whose product is of order seven. More precisely, any Hurwitz surface, that is, a hyperbolic surface that realizes the maximum order of the automorphism group for the surfaces of a given genus, can be obtained by the construction given. This is the last part of the theorem of Hurwitz.
Examples of Hurwitz groups and surfacesEdit
The small cubicuboctahedron is a polyhedral immersion of the tiling of the Klein quartic by 56 triangles, meeting at 24 vertices.[2]
The smallest Hurwitz group is the projective special linear group PSL(2,7), of order 168, and the corresponding curve is the Klein quartic curve. This group is also isomorphic to PSL(3,2).
Next is the Macbeath curve, with automorphism group PSL(2,8) of order 504. Many more finite simple groups are Hurwitz groups; for instance all but 64 of the alternating groups are Hurwitz groups, the largest non-Hurwitz example being of degree 167. The smallest alternating group that is a Hurwitz group is A15.
Most projective special linear groups of large rank are Hurwitz groups, (Lucchini, Tamburini & Wilson 2000). For lower ranks, fewer such groups are Hurwitz. For np the order of p modulo 7, one has that PSL(2,q) is Hurwitz if and only if either q=7 or q = pnp. Indeed, PSL(3,q) is Hurwitz if and only if q = 2, PSL(4,q) is never Hurwitz, and PSL(5,q) is Hurwitz if and only if q = 74 or q = pnp, (Tamburini & Vsemirnov 2006).
Similarly, many groups of Lie type are Hurwitz. The finite classical groups of large rank are Hurwitz, (Lucchini & Tamburini 1999). The exceptional Lie groups of type G2 and the Ree groups of type 2G2 are nearly always Hurwitz, (Malle 1990). Other families of exceptional and twisted Lie groups of low rank are shown to be Hurwitz in (Malle 1995).
There are 12 sporadic groups that can be generated as Hurwitz groups: the Janko groups J1, J2 and J4, the Fischer groups Fi22 and Fi'24, the Rudvalis group, the Held group, the Thompson group, the Harada–Norton group, the third Conway group Co3, the Lyons group, and the Monster, (Wilson 2001).
Automorphism groups in low genusEdit
The largest |Aut(X)| can get for a Riemann surface X of genus g is shown below, for 2≤g≤10, along with a surface X0 with |Aut(X0)| maximal.
Largest possible |Aut(X)|
Aut(X0)
2 48 Bolza curve GL2(3)
3 168 (Hurwitz bound) Klein quartic PSL2(7)
4 120 Bring curve S5
7 504 (Hurwitz bound) Macbeath curve PSL2(8)
In this range, there only exists a Hurwitz curve in genus g=3 and g=7.
(2,3,7) triangle group
^ Technically speaking, there is an equivalence of categories between the category of compact Riemann surfaces with the orientation-preserving conformal maps and the category of non-singular complex projective algebraic curves with the algebraic morphisms.
^ (Richter) Note each face in the polyhedron consist of multiple faces in the tiling – two triangular faces constitute a square face and so forth, as per this explanatory image.
Hurwitz, A. (1893), "Über algebraische Gebilde mit Eindeutigen Transformationen in sich", Mathematische Annalen, 41 (3): 403–442, doi:10.1007/BF01443420, JFM 24.0380.02.
Lucchini, A.; Tamburini, M. C. (1999), "Classical groups of large rank as Hurwitz groups", Journal of Algebra, 219 (2): 531–546, doi:10.1006/jabr.1999.7911, ISSN 0021-8693, MR 1706821
Lucchini, A.; Tamburini, M. C.; Wilson, J. S. (2000), "Hurwitz groups of large rank", Journal of the London Mathematical Society, Second Series, 61 (1): 81–92, doi:10.1112/S0024610799008467, ISSN 0024-6107, MR 1745399
Malle, Gunter (1990), "Hurwitz groups and G2(q)", Canadian Mathematical Bulletin, 33 (3): 349–357, doi:10.4153/CMB-1990-059-8, ISSN 0008-4395, MR 1077110
Malle, Gunter (1995), "Small rank exceptional Hurwitz groups", Groups of Lie type and their geometries (Como, 1993), London Math. Soc. Lecture Note Ser., vol. 207, Cambridge University Press, pp. 173–183, MR 1320522
Tamburini, M. C.; Vsemirnov, M. (2006), "Irreducible (2,3,7)-subgroups of PGL(n,F) for n ≤ 7", Journal of Algebra, 300 (1): 339–362, doi:10.1016/j.jalgebra.2006.02.030, ISSN 0021-8693, MR 2228652
Wilson, R. A. (2001), "The Monster is a Hurwitz group", Journal of Group Theory, 4 (4): 367–374, doi:10.1515/jgth.2001.027, MR 1859175, archived from the original on 2012-03-05, retrieved 2015-09-04 |
Angular Kinematics | Brilliant Math & Science Wiki
Matt DeCross contributed
Angular kinematics is the study of rotational motion in the absence of forces. The equations of angular kinematics are extremely similar to the usual equations of kinematics, with quantities like displacements replaced by angular displacements and velocities replaced by angular velocities. Just as kinematics is routinely used to describe the trajectory of almost any physical system moving linearly, the equations of angular kinematics are relevant to most rotating physical systems.
Basic Equations of Angular Kinematics
Derivation of Equations of Angular Kinematics
Rotating Systems in Physics
In purely rotational (circular) motion, the equations of angular kinematics are:
v = r\omega, \qquad a_c = -r\omega^2, \qquad a = r\alpha
The tangential velocity
v
describes the velocity of an object tangent to its path in rotational motion at angular frequency
\omega
r
. This is the velocity an object would follow if it suddenly broke free of rotational motion and traveled along a straight line. The rate of change of this velocity is the tangential acceleration
a
. The centripetal acceleration
a_c
is a second acceleration experienced by rotating objects, because changing the direction of a velocity vector requires an acceleration. Since the direction of the velocity vector changes constantly in rotational motion, rotating objects must be continuously accelerated towards the axis of rotation by some force providing a centripetal acceleration.
From the above equations, the usual kinematic equations hold in angular form. If an object undergoes constant angular acceleration
\alpha
, the total angular displacement is:
\theta - \theta_0 = \omega_0 t + \frac12 \alpha t^2
\theta_0
is the initial angle and
\omega_0
is the initial angular velocity. Similarly, the angular velocity changes according to:
\omega^2 = \omega_0^2 + 2\alpha (\theta - \theta_0)
in terms of the angular displacement, or
\omega = \omega_0 + \alpha t
Though the above derivation gives the magnitudes of angular quantities correctly, it does not capture the fact that angular quantities are also vector quantities. The direction in which the angular velocity points can be found from the right-hand rule: curving the fingers of your right hand along the direction of rotation, your thumb points in the direction of the angular velocity vector, along the axis of rotation. This is true by definition; although it seems strange since the vector is perpendicular to the rotation, this definition turns out to be the only way to formulate a consistent vector theory of rotational forces.
12\pi^2
2\pi\sqrt{3}
3\pi
\pi\sqrt{3}
A turntable is spun from rest with a constant angular acceleration of
\frac{\pi}{2} \text{ rad}/\text{s}^2
. After completing six full revolutions, what is its angular velocity in
\text{rad}/\text{s}
\frac{4\pi}{5}
\frac{3\pi}{10}
\frac{\pi}{4}
\frac{\pi}{8}
A merry-go-round has a radius of about
8 \text{ m}
. A child sitting on a horse on the outer edge sees his parents, standing still at the entrance, every
20 \text{ s}
. If the child's horse could break free of its merry-go-round restraints and continue forward tangentially without accelerating, how fast would it be traveling in
\text{m}/\text{s}
In a rotating frame of reference, it is often more convenient to use polar coordinates than Cartesian coordinates. Similarly, for vectors it is more convenient to use the radial and tangential vectors
\hat{r}
\hat{\theta}
in place of the usual Cartesian basis vectors
\hat{x}
\hat{y}
. The radial vector to an object is defined so that it always points towards the object:
\hat{r} = \cos \theta\, \hat{x} + \sin \theta\, \hat{y}.
x
y
components of the radial vector are just the usual polar coordinates, normalized to one. Similarly, the tangential vector
\hat{\theta}
is defined so that it is always orthogonal to the radial vector and tangent to the circle on which the radial vector lies:
\hat{\theta} = -\sin \theta\,\hat{x} + \cos \theta\, \hat{y}.
The polar basis vectors
\hat{r}
\hat{\theta}
explained graphically.
\frac{d\hat{r}}{dt} = \dot{\theta} \hat{\theta}
\frac{d\hat{\theta}}{dt} = -\dot{\theta} \hat{r}
, where dots indicate time derivatives.
Note that in Cartesian coordinates, the derivatives of the basis vectors
\hat{x}
\hat{y}
, etc. always vanish, because these basis vectors are fixed. The polar basis vectors, however, rotate in time so that they are always pointing radially and tangentially along some trajectory. Computing the derivative from the definitions above using the chain rule:
\begin{aligned} \frac{d\hat{r}}{dt} &= \frac{d}{dt} \left( \cos \theta\, \hat{x} + \sin \theta\, \hat{y}\right) = -\dot{\theta} \sin \theta\,\hat{x} + \dot{\theta} \cos \theta\,\hat{y} = \dot{\theta} \hat{\theta} \\ \frac{d\hat{\theta}}{dt} &= \frac{d}{dt} \left( -\sin \theta\, \hat{x} + \cos \theta\, \hat{y}\right) = -\dot{\theta} \cos \theta\,\hat{x} - \dot{\theta} \sin \theta\,\hat{y} = -\dot{\theta} \hat{r} \end{aligned}
From the definitions above, all of the laws of angular kinematics are straightforward to derive. Suppose that the vector pointing to some rotating object is
\vec{r} = r\hat{r}
in polar coordinates, where
r
is the magnitude of the distance from the origin. The velocity of the object is then:
\frac{d\vec{r}}{dt} = \dot{r} \hat{r} +r\dot{\theta} \hat{\theta}.
The velocity has two components, as can be seen above. The first term,
\dot{r} \hat{r}
, describes the radial velocity of the object away from the origin. The second term is the tangential velocity. Denoting
\dot{\theta} = \omega
as the angular velocity, the tangential velocity is just
v = r\omega
Taking another derivative allows identification of the different terms contributing to the acceleration of the object:
\begin{aligned} \frac{d^2 \vec{r}}{dt} &= \frac{d}{dt} \frac{d\vec{r}}{dt} = \frac{d}{dt} \left( \dot{r} \hat{r} +r\dot{\theta} \hat{\theta} \right) \\ &=(\ddot{r} - r\dot{\theta}^2)\hat{r} + (r\ddot{\theta} + 2\dot{r} \dot{\theta}) \hat{\theta} \end{aligned}
The radial terms
\ddot{r}
r\dot{\theta}^2 = r\omega^2
describe the radial acceleration outward from the origin and the centripetal acceleration towards the origin, respectively. The tangential terms are the tangential acceleration
a = r\ddot{\theta} = r\alpha
\alpha = \dot{\omega}
is the angular acceleration, and
2\dot{r} \dot{\theta} = 2\dot{r} \omega
, the Coriolis acceleration.
A straight line in an inertial frame becomes a curved line in the rotating frame when there is no force to provide Coriolis acceleration [2].
Remarkably, this derivation proves without reference to any forces at all that an object in circular motion with angular velocity
\omega
must accelerate radially inward with the centripetal acceleration
a_c
given above.
Countless rotating systems in physics can be analyzed using the laws of angular kinematics; a few are explored in the following examples.
As most kids learn, if you quickly rotate a tube containing a ball, the ball can be made to "slingshot" out the end at very high speeds. The same effect is visible in rotating a rod with a bead around it and many other physical scenarios. How fast does the velocity of the ball/bead increase if the tube/rod is rotated at constant angular velocity
\omega
Since there is no radial force on the ball/bead, the total radial acceleration is zero according to Newton's second law. In polar coordinates this means that:
\ddot{r} - r\omega^2 = 0.
This differential equation in
r
is solved by:
r(t) = Ae^{\omega t} + Be^{-\omega t},
A
B
depending on initial conditions. If the ball/bead starts from rest at radius
r_0
, these constants are fixed to be:
A = B = \frac{r_0}{2},
so the solution for the velocity of the ball/bead is found by differentiating to be:
\dot{r}(t) = \frac{r_0 \omega}{2} e^{\omega t} - \frac{r_0 \omega}{2} e^{-\omega t}.
The velocity of the ball/bead grows exponentially, since the second term damps to zero quickly.
A bead is lodged in a wheel that rolls without slipping with constant velocity
V
. Show that the trajectory of the bead traces a cycloid in the lab frame.
In the wheel reference frame, the bead is in uniform circular motion. If the wheel is of radius
r
, the bead has coordinates and velocity with respect to the center of the wheel:
\begin{aligned} \vec{r} &= r\hat{r} \\ \vec{v} &= r \omega \hat{\theta} = V \hat{\theta} \end{aligned}
In the lab reference frame, the center of the wheel moves with velocity
V
, so it is located at:
\vec{R} = Vt \hat{x} + r\hat{y} = r\omega t \hat{x} + r\hat{y},
keeping in mind that the center of the wheel is a height
r
above the ground. The position of the bead in the lab frame is therefore:
\vec{R} + \vec{r} = r\omega t \hat{x} + r\hat{y} + r\hat{r}
Now if the wheel rolls clockwise (i.e., to the right) without slipping, the angle the bead has traveled is:
\theta(t) = -\omega t
assuming the bead starts at
\theta = 0
. So the position of the bead is:
r\omega t \hat{x} + r\hat{y} + r\hat{r} = r(\omega t +\cos (\omega t)) \hat{x} + r(1-\sin (\omega t)) \hat{y}.
Below is a plot of the position above for
r = \omega = 1
to verify that it is indeed a cycloid:
Cycloidal motion of a bead at the edge of a wheel. :
The fact that the Earth rotates means that everywhere on the surface of the earth is a rotating reference frame. This has observable effects from the Coriolis acceleration, most notably in the precession of the axis of rotation of a sufficiently large pendulum. This experimental apparatus is usually called a Foucault pendulum.
Precession of the axis of rotation of a Foucault pendulum [3].
Derive the precession of the Foucault pendulum assuming the Earth rotates with angular frequency
\Omega
The precession results from the fact that the plane in which the pendulum oscillates rotates with the rotation of the Earth. However, at higher latitudes
\varphi
, this precession is slower than at the equator. In two dimensions, the Coriolis acceleration is then:
\begin{aligned} \ddot{x} &= \Omega \sin \varphi \dot{y} \\ \ddot{y} &= \Omega \sin \varphi \dot{x} \end{aligned}
Small oscillations of a pendulum at frequency
\omega
obey Hooke's law. Using this fact and Newton's second law gives the following equations of motion:
\begin{aligned} \ddot{x} &= -\omega^2 x + 2\Omega \sin \varphi \dot{y} \\ \ddot{y} &= -\omega^2 y - 2\Omega \sin \varphi \dot{x} \end{aligned}
The solution for the complex coordinate
z= x+iy
can be found by matrix ODE methods to be:
z = e^{-i\Omega \sin \varphi t} \left(A e^{i\omega t} + Be^{-i \omega t} \right).
A
B
to be determined by initial conditions. The leading prefactor
e^{-i\Omega \sin \varphi t}
describes the
z
coordinate as rotating over time. Since
z = x+iy
, the axis of the pendulum thus rotates in the
x,y
plane with frequency
\Omega \sin \varphi
\Omega = 2\pi \text{ rads}/\text{day}
, over the course of a single day, the pendulum oscillation precesses by an angle
-2\pi \sin \varphi
[1] D. Kleppner and R. Kolenkow, An Introduction to Mechanics. McGraw-Hill, 1973.
[2] Image from https://en.wikipedia.org/wiki/Coriolis_force#/media/File:Corioliskraftanimation.gif under Creative Commons licensing for reuse with modification.
[3] Image from https://en.wikipedia.org/wiki/Foucault_pendulum#/media/File:Foucault-rotz.gif under Creative Commons licensing for reuse with modification.
Cite as: Angular Kinematics. Brilliant.org. Retrieved from https://brilliant.org/wiki/angular-kinematics-problem-solving/ |
Peak gain of dynamic system frequency response - MATLAB getPeakGain - MathWorks í•œêµ
gpeak = getPeakGain(sys,tol,fband) returns the peak gain in the frequency interval fband = [fmin,fmax] with 0 ≤ fmin < fmax. This syntax also takes into account the negative frequencies in the band [–fmax,–fmin] for models with complex coefficients.
sys=\frac{90}{{s}^{2}+1.5s+90}.
sys=\frac{90}{{s}^{2}+1.5s+90}.
sys=\left(\frac{1}{{s}^{2}+0.2s+1}\right)\left(\frac{100}{{s}^{2}+s+100}\right).
sys=\left(\frac{1}{{s}^{2}+0.2s+1}\right)\left(\frac{100}{{s}^{2}+s+100}\right).
Now compute the peak gain in the frequency interval [–50,–1] ∪ [1,50]. To do so, specify fband = [1,50].
Frequency interval in which to calculate the peak gain, specified as a 1-by-2 vector of positive real values. Specify fband as a row vector of the form [fmin,fmax] with 0 ≤ fmin < fmax.
For models with complex coefficients, getPeakGain calculates the peak gain in the range [–fmax,–fmin]∪[fmin,fmax]. As a result, the function can return a peak at a negative frequency. |
Cup products of line bundles on homogeneous varieties and generalized PRV components of multiplicity one
Ivan Dimitrov and Mike Roth
X=G∕B
{L}_{1}
{L}_{2}
be two line bundles on
X
. Consider the cup-product map
{H}^{{d}_{1}}\left(X,{L}_{1}\right)\otimes {H}^{{d}_{2}}\left(X,{L}_{2}\right)\underset{}{\overset{\cup }{\to }}{H}^{d}\left(X,L\right),
L={L}_{1}\otimes {L}_{2}
d={d}_{1}+{d}_{2}
. We answer two natural questions about the map above: When is it a nonzero homomorphism of representations of
G
? Conversely, given generic irreducible representations
{V}_{1}
{V}_{2}
, which irreducible components of
{V}_{1}\otimes {V}_{2}
may appear in the right hand side of the equation above? For the first question we find a combinatorial condition expressed in terms of inversion sets of Weyl group elements. The answer to the second question is especially elegant: the representations
V
appearing in the right hand side of the equation above are exactly the generalized PRV components of
{V}_{1}\otimes {V}_{2}
of stable multiplicity one. Furthermore, the highest weights
\left({\lambda }_{1},{\lambda }_{2},\lambda \right)
corresponding to the representations
\left({V}_{1},{V}_{2},V\right)
fill up the generic faces of the Littlewood–Richardson cone of
G
of codimension equal to the rank of
G
. In particular, we conclude that the corresponding Littlewood–Richardson coefficients equal one.
Homogeneous variety, Littlewood–Richardson coefficient, Borel–Weil–Bott theorem, PRV component. |
Rewrite each of the expressions below in a simpler form using exponents.
4 · 4 · 5 · 5 · 5
Notice how many 4's there are and how many 5's there are. The exponent refers to how many times a number is multiplied by itself.
Since there are two 4's and three 5's:
4^{2} · 5^{3}
3 · 3 · 3 · 3 · 3 · y · y
Use the same method as in (a).
(6x)(6x)(6x)(6x)
\left(6x\right)^{4} |
Find laplace transform of the following functions. f(t)=t \sin h(2t)
Find laplace transform of the following functions
f\left(t\right)=t\mathrm{sin}h\left(2t\right)
The given function is in the form of
f\left(t\right)={t}^{k}g\left(t\right)
. The Laplace transform of
{t}^{k}g\left(t\right)\text{ }\text{ is }\text{ }{\left(-1\right)}^{k}\frac{{d}^{k}}{d{s}^{k}}L\left[g\left(t\right)\right]
L\left[{t}^{k}g\left(t\right)\right]={\left(-1\right)}^{k}\frac{{d}^{k}}{d{s}^{k}}\left(L\left[g\left(t\right)\right]\right)
Compare the function
f\left(t\right)=t\mathrm{sin}h\left(2t\right)
f\left(t\right)={t}^{k}g\left(t\right)
we found,
g\left(t\right)=\mathrm{sin}h\left(2t\right)
L\left[{t}^{k}g\left(t\right)\right]={\left(-1\right)}^{k}\frac{{d}^{k}}{d{s}^{k}}\left(L\left[g\left(t\right)\right]\right)
L\left[t\mathrm{sin}h2t\right]={\left(-1\right)}^{1}\frac{d}{ds}\left(L\left[\mathrm{sin}h2t\right]\right)\stackrel{˙}{s}\left(1\right)
L\left[\mathrm{sin}h\left(at\right)\right]=\frac{a}{{s}^{2}-{a}^{2}}
, So use this find
L\left[\mathrm{sin}h\left(2t\right)\right]
L\left[\mathrm{sin}h\left(2t\right)\right]=\frac{2}{{s}^{2}-{2}^{2}}
=\frac{2}{{s}^{2}-4}\stackrel{˙}{s}\left(2\right)
From (1) and (2) ,
L\left[t\mathrm{sin}\left(2t\right)\right]=-\frac{d}{ds}\left[\frac{2}{{s}^{2}-4}\right]
\frac{d}{ds}\left[\frac{2}{{s}^{2}-4}\right]
\frac{d}{ds}\left[\frac{2}{{s}^{2}-4}\right]=\frac{-2}{{\left({s}^{2}-4\right)}^{2}}\cdot 2s
=-\frac{4s}{{\left({s}^{2}-4\right)}^{2}}
y\left({x}^{2}+xy-2{y}^{2}\right)dx+x\left(3{y}^{2}-xy-{x}^{2}\right)dy=0;\text{ }when\text{ }x=1,y=1
When talking about boundary conditions for partial Differential equations, what does an open boundary mean?
\left(t+2\right)dx=2{x}^{2}dt
Solve the given LDE:
{y}^{6}-64y=0
{L}^{-1}\left(F\right)
F\left(s\right)=\frac{6}{{s}^{2}+2s+2}
Find the differntial of each function.
y=\mathrm{tan}\sqrt{7t}
y=\frac{3-{v}^{2}}{3+{v}^{2}}
{\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{sin}\left(t\right)}{t}dt=\frac{\pi }{2}
by using Laplace Transform method. I know that
L\left\{\mathrm{sin}\left(t\right)\right\}={\int }_{0}^{\mathrm{\infty }}{e}^{-st}\mathrm{sin}\left(t\right)dt=\frac{1}{{s}^{2}+1} |
p(x) = x^3 + 3x^2 + 4x - 8
\color{#D61F06}{a}
\color{#3D99F6}{b}
\color{#69047E}{c}
\color{#D61F06}{a}^2 \big(1 + \color{#D61F06}{a}^2\big) + \color{#3D99F6}{b}^2 \big(1 + \color{#3D99F6}{b}^2\big) + \color{#69047E}{c}^2 \big(1 + \color{#69047E}{c}^2\big)?
\alpha , \beta , \gamma
x^{3} + 3x + 9 = 0,
\alpha^{9} + \beta^9 + \gamma^{9}.
Consider all pairs of non-zero integers
(a,b)
( ax-b)^2 + (bx-a)^2 = x
has at least one integer solution.
The sum of all (distinct) values of
x
which satisfy the above condition can be written as
\frac{ m}{n}
m
n
m + n
by Jordi Bosch
x_1,x_2,\ldots,x_{2015}
x^{2015} + x^{2014} + x^{2013} + \ldots + x^2 + x + 1 =0.
\frac1{1-x_1} + \frac1{1-x_2} + \ldots + \frac1{1-x_{2015}}.
by Ravi Dwivedi
n \leq 1000
(a, b)
n = \frac{ a^2 + b^2 } { ab - 1 } ?
(a,b) = (0,0)
\frac{ 0^2 + 0^2 } { 0 \times 0 - 1 } = 0
0 . |
\frac { 2 } { 3 } \text { of } \frac { 3 } { 7 }
Draw a diagram to represent three sevenths. A rectangle divided into 7 equal vertical parts, with the first 3 parts shaded.
Since the green shaded region is three pieces, you can see that two thirds of this region would be the orange shaded region. The first 2 parts change from being shaded green to being shaded orange.
\frac{2}{7}
\quad \frac { 1 } { 2 } \text { of } \frac { 3 } { 5 }
At right is a diagram representing one half of three fifths. Use it to help you find the answer to this problem. |
Q What is a flat spiral spring Describe an experiment to determine the modulus of rigidity of a - Physics - Work Energy And Power - 10807477 | Meritnation.com
Flat spiral springs are also known as spiral torsion, clock springs or brush springs . They are characterized by the requirement that the coil contact is minimized during operation. 3
1) Measure the diameter of the wire of the spring by using the micrometer.
2) Measure the diameter of spring coils by using the vernier caliper
3) Count the number of turns. 4) Insert the spring in the spring testing machine and load the spring by a suitable weight and note the corresponding axial deflection in tension or compression.
5) Increase the load and take the corresponding axial deflection readings.
6) Plot a curve between load and deflection. The shape of the curve gives the stiffness of the spring.
modulus of rigidity is given by:\phantom{\rule{0ex}{0ex}}C= \frac{8WD{m}^{3}n }{\delta {d}^{4}}\phantom{\rule{0ex}{0ex}}where W= load\phantom{\rule{0ex}{0ex}}{D}_{m}=mean diameter of the sprin \left(spring diameter - diameter of wire\right)\phantom{\rule{0ex}{0ex}}\delta is deflection\phantom{\rule{0ex}{0ex}}d is the diameter of the spring\phantom{\rule{0ex}{0ex}}n is the no of turns.
This is the complete solution it is not necessary that each and every answer must contain derivation and diagram.
If you have any doubt do let us know we will help you out. |
Maria and Jorge were trying to simplify the expression
1 \frac { 2 } { 5 } \cdot ( - \frac { 3 } { 4 } ) \cdot ( - \frac { 4 } { 3 } )
. Maria started by rewriting
1 \frac { 2 } { 5 }
\frac { 7 } { 5 }
. Her work is below.
\left. \begin{array} { c } { 1 \frac { 2 } { 5 } \cdot ( - \frac { 3 } { 4 } ) \cdot ( - \frac { 4 } { 3 } ) } \\ { \frac { 7 } { 5 } \cdot ( - \frac { 3 } { 4 } ) \cdot ( - \frac { 4 } { 3 } ) } \\ { ( - \frac { 21 } { 20 } ) \cdot ( - \frac { 4 } { 3 } ) } \end{array} \right.
\frac { 84 } { 60 } = 1 \frac { 24 } { 60 } = 1 \frac { 2 } { 5 }
Jorge had a different idea. He multiplied
( - \frac { 3 } { 4 } ) \cdot ( - \frac { 4 } { 3 } )
1 \frac { 2 } { 5 } \cdot ( - \frac { 3 } { 4 } ) \cdot ( - \frac { 4 } { 3 } )
using Jorge's method. Is your answer equal to Maria's?
What do you get when you multiply
\left(-\frac{3}{4}\right) \cdot \left(-\frac{4}{3}\right)
Does this answer make sense and match Maria's answer?
Since you are multiplying by reciprocal fractions, the fractions reduce to
1
\text{Yes, }1\frac{2}{5} \cdot \left(-\frac{3}{4}\right) \cdot \left(-\frac{4}{3}\right) = 1\frac{2}{5}
Why might Jorge have decided to multiply
( - \frac { 3 } { 4 } ) \cdot ( - \frac { 4 } { 3 } )
first?
As you found in part (a),
\left(-\frac{3}{4}\right)\left(-\frac{4}{3}\right)
1
, due to the multiplicative property.
In a multiplication problem, the factors can be grouped together in different ways. This is called the Associative Property of Multiplication. Read the Math Notes box for this lesson, then show that
( \frac { 3 } { 4 } \cdot \frac { 2 } { 5 } ) \cdot ( - 2 ) = \frac { 3 } { 4 } \cdot ( \frac { 2 } { 5 } \cdot ( - 2 ) )
Solve each equation. What do you get?
Are the answers equal?
-\frac{12}{20} = -\frac{12}{20}
Simplify each expression. First, decide if you want to group some factors together.
\left(-9\right)\cdot\frac{1}{9}\cdot\frac{3}{8}
-\frac{5}{12}\cdot\frac{3}{7}\cdot\frac{4}{9}
-8.1·5·2
i. Like numbers in the numerator and denominator cancel out.
ii. Multiply the last two values. Do you see anything you can cancel out?
iii. Multiply two terms together to get an easier number to deal with.
ii. -\frac{5}{60} |
y=3x-5 find intersept in y
Aneeka Hunt 2020-11-26 Answered
The y intercept of a line is the point where the line crosses the y-axis. You can find the y intercept of a line multiple ways. If a point touches the y-axis, the value of x of that certain point will be 0. Therefore, you can replace x in the equation with 0.
Once you do this, you will see that the y intercept for this equation is -5.
A college fraternity house spent $670 for an order of 85 pizzas.
The order consisted of cheese pizzas, which cost $5 each, and Supreme pizzas, which cost $12 each.
Find the number of each kind of pizza ordered.
According to the U.S. Bureau of Labor Statistics, you will devote 37 years to sleeping and watching TV. The number of years sleeping will exceed the number of years watching TV by 19. Over your lifetime, how many years will you spend each of these activities?
solve given system of equation
2x+5y=-18
y=-3x-7
Solve the equations and inequalities. Write the solution sets to the inequalities in interval notation.
\mid 5x-1\mid =\mid 3-4x\mid
Find the side lengths of the triangle if the hypotenuse of a right triangle is 5 cm long, and the shorter leg is 1 cm shorter than the longer leg.
Customers of a phone company can choose between two service plans for long distance calls. The first plan has a $9 monthly fee and charges an additional $0.13 for each minute of calls. The second plan has a $25 monthly fee and charges an additional $0.09 for each minute of calls.
For how many minutes of calls will the costs of the two plans be equal?
Solve the following equations and inequalities:
-{x}^{2}+2x>1
|1-3x|=5
|4x-3|\le 2 |
What Is An Electrical Charge? – Cute Lava
Cute Lava All About Electrical and Electronics What Is An Electrical Charge?
An electron is the smallest particle that exhibits negative electrical charge. When an excess of electrons exists in a material, there is a net negative electrical charge. When a deficiency of electrons exists, there is a net positive electrical charge.
The charge of an electron and that of a proton are equal in magnitude. Electrical charge is an electrical property of matter that exists because of an excess or deficiency of electrons. The letter Q symbolizes charge. Static electricity is the presence of a net positive or negative charge in a material.
Materials with charges of opposite polarity are attracted to each other, and materials with charges of the same polarity are repelled. As shown in figure 1, a force acts between charges, as evidenced by the attraction or repulsion. This force is called an electric field.
Figure 1. Attraction and repulsion of electrical charges
Coulomb's Law states:
A force (F) exists between two point-source charges (
// <![CDATA[ Q_{1},Q_{2} // ]]>
) that is directly proportional to the product of the two charges and inversely proportional to the square of the distance (d) between the charges.
2.0 Coulomb: The Unit of Charge
Electrical charge (
// <![CDATA[ Q // ]]>
) is measured in coulombs, symbolized by
// <![CDATA[ C // ]]>
One coulomb is the total charge possessed by
// <![CDATA[ 6.25\times10^{18} // ]]>
electrons. A single electron has a charge of
// <![CDATA[ 1.6\times10^{- 19}\text{C} // ]]>
. The total charge
// <![CDATA[ Q // ]]>
, expressed in coulombs, for a given number of electrons is sated in the following formula:
Consider a neutral atom -- that is, one that has the same number of electrons and protons and thus has no net charge. As you know, when a valence electron is pulled away from the atom by the application of energy, the atom is left with a net positive charge (more protons than electrons) and becomes a positive ion. If an atom acquires an extra electron in its outer shell, it has a negative charge and becomes a negative ion.
The energy required to free a valence electron is related to the number of electrons in the outer shell. An atom can have up to eight valance electrons. The more complete the outer shell, the more stable the atom and thus the more energy is required to remove an electron.
What is the symbol for charge?
What is the unit of charge, and what is the unit symbol?
What causes a positive and negative charge?
How much charge, in coulombs, is there in
// <![CDATA[ 10\times10^{12} // ]]>
electrons?
This project is the "Hello World!" of Arduino. In this project, we are going to learn about physical outputs and two ele
Back to All About Electrical and Electronics |
Mnemonic for understanding orientation of vectors in 3D space
This article is about three-dimensional vector geometry. For the maze-solving technique, see Wall follower. For the traffic rule, see Priority to the right.
Find sources: "Right-hand rule" – news · newspapers · books · scholar · JSTOR (November 2017) (Learn how and when to remove this template message)
In mathematics and physics, the right-hand rule is a common mnemonic for understanding orientation of axes in three-dimensional space.
Most of the various left-hand and right-hand rules arise from the fact that the three axes of three-dimensional space have two possible orientations. One can see this by holding one's hands outward and together, palms up, with the fingers curled, and the thumb out-stretched. The curl of the fingers represents a movement from the first (x axis) to the second (y axis), then the third (z axis) can point along either thumb. Left-hand and right-hand rules arise when dealing with coordinate axes. The rule can be used to find the direction of the magnetic field, rotation, spirals, electromagnetic fields, mirror images, and enantiomers in mathematics and chemistry.
The sequence is often: index finger, then middle, then thumb. However, two other sequences also work because they preserve the cycle:
Middle finger, then thumb, then index finger.
Thumb, then index finger, then middle (e.g., see the ninth series of the Swiss 200-francs banknote).
1 Curve orientation and normal vectors
3.1 A rotating body
3.2 Helices and screws
4.1 Ampère's right-hand grip rule
5 Cross products
Curve orientation and normal vectors[edit]
In vector calculus, it is often necessary to relate the normal to a surface to the curve bounding it. For a positively-oriented curve C bounding a surface S, the normal to the surface n̂ is defined such that the right thumb points in the direction of n̂, and the fingers curl along the orientation of the bounding curve C.
Right-hand rule for curve orientation.
Left-handed coordinates on the left,
right-handed coordinates on the right.
For right-handed coordinates use right hand.
For left-handed coordinates use left hand.
Axis or vector
Two fingers and thumb
x, 1, or A First or index Fingers extended
y, 2, or B Second finger or palm Fingers curled 90°
z, 3, or C Thumb Thumb
Coordinates are usually right-handed.
For right-handed coordinates the right thumb points along the z axis in the positive direction and the curl of the fingers represents a motion from the first or x axis to the second or y axis. When viewed from the top or z axis the system is counter-clockwise.
For left-handed coordinates the left thumb points along the z axis in the positive direction and the curled fingers of the left hand represent a motion from the first or x axis to the second or y axis. When viewed from the top or z axis the system is clockwise.
Interchanging the labels of any two axes reverses the handedness. Reversing the direction of one axis (or of all three axes) also reverses the handedness. (If the axes do not have a positive or negative direction then handedness has no meaning.) Reversing two axes amounts to a 180° rotation around the remaining axis.[1]
A rotating body[edit]
Conventional direction of the axis of a rotating body
In mathematics, a rotating body is commonly represented by a pseudovector along the axis of rotation. The length of the vector gives the speed of rotation and the direction of the axis gives the direction of rotation according to the right-hand rule: right fingers curled in the direction of rotation and the right thumb pointing in the positive direction of the axis. This allows some easy calculations using the vector cross product. No part of the body is moving in the direction of the axis arrow. By coincidence, if the thumb is pointing north, Earth rotates in a prograde direction according to the right-hand rule. This causes the Sun, Moon, and stars to appear to revolve westward according to the left-hand rule.
Helices and screws[edit]
Left- and right-handed screws
A helix is a curved line formed by a point rotating around a center while the center moves up or down the z axis. Helices are either right- or left-handed, curled fingers giving the direction of rotation and thumb giving the direction of advance along the z axis.
The threads of a screw are a helix and therefore screws can be right- or left-handed. The rule is this: if a screw is right-handed (most screws are) point your right thumb in the direction you want the screw to go and turn the screw in the direction of your curled right fingers.
When electricity (conventional current) flows in a long straight wire it creates a circular or cylindrical magnetic field around the wire according to the right-hand rule. The conventional current, which is the opposite of the actual flow of electrons, is a flow of positive charges along the positive z axis. The conventional direction of a magnetic line is given by a compass needle.
Electromagnet: The magnetic field around a wire is quite weak. If the wire is coiled into a helix all the field lines inside the helix point in the same direction and each successive coil reinforces the others. The advance of the helix, the non-circular part of the current and the field lines all point in the positive z direction. Since there is no magnetic monopole, the field lines exit the +z end, loop around outside the helix, and reenter at the −z end. The +z end where the lines exit is defined as the north pole. If the fingers of the right hand are curled in the direction of the circular component of the current, the right thumb points to the north pole.
Lorentz force: If a positive electric charge moves across a magnetic field it experiences a force according Lorentz force, with the direction given by the right-hand rule. If the curl of the right fingers represents a rotation from the direction the charge is moving to the direction of the magnetic field then the force is in the direction of the right thumb. Because the charge is moving, the force causes the particle path to bend. The bending force is computed by the vector cross product. This means that the bending force increases with the velocity of the particle and the strength of the magnetic field. The force is maximum when the particle direction and magnetic fields are at right angles, is less at any other angle and is zero when the particle moves parallel to the field.
Ampère's right-hand grip rule[edit]
Prediction of direction of field (B), given that the current I flows in the direction of the thumb
Finding direction of magnetic field (B) for an electrical coil
Ampère's right-hand grip rule[2] (also called right-hand screw rule, coffee-mug rule or the corkscrew-rule) is used either when a vector (such as the Euler vector) must be defined to represent the rotation of a body, a magnetic field, or a fluid, or vice versa, when it is necessary to define a rotation vector to understand how rotation occurs. It reveals a connection between the current and the magnetic field lines in the magnetic field that the current created.
André-Marie Ampère, a French physicist and mathematician, for whom the rule was named, was inspired by Hans Christian Ørsted, another physicist who experimented with magnet needles. Ørsted observed that the needles swirled when in the proximity of an electric current-carrying wire, and concluded that electricity could create magnetic fields.
This rule is used in two different applications of Ampère's circuital law:
An electric current passes through a straight wire. When the thumb is pointed in the direction of conventional current (from positive to negative), the curled fingers will then point in the direction of the magnetic flux lines around the conductor. The direction of the magnetic field (counterclockwise instead of clockwise when viewing the tip of the thumb) is a result of this convention and not an underlying physical phenomenon.
An electric current passes through a solenoid, resulting in a magnetic field. When wrapping the right hand around the solenoid with the fingers in the direction of the conventional current, the thumb points in the direction of the magnetic north pole.
Cross products[edit]
Illustration of the right-hand rule on the ninth series of the Swiss 200-francs banknote.
The cross product of two vectors is often taken in physics and engineering. For example, in statics and dynamics, torque is the cross product of lever length and force, while angular momentum is the cross product of distance and linear momentum. In electricity and magnetism, the force exerted on a moving charged particle when moving in a magnetic field B is given by:
{\displaystyle \mathbf {F} =q\mathbf {v} \times \mathbf {B} }
The direction of the cross product may be found by application of the right hand rule as follows:
The index finger points in the direction of the velocity vector v.
The middle finger points in the direction of the magnetic field vector B.
The thumb points in the direction of the cross product F.
For example, for a positively charged particle moving to the north, in a region where the magnetic field points west, the resultant force points up.[1]
The right-hand rule is in widespread use in physics. A list of physical quantities whose directions are related by the right-hand rule is given below. (Some of these are related only indirectly to cross products, and use the second form.)
For a rotating object, if the right-hand fingers follow the curve of a point on the object, then the thumb points along the axis of rotation in the direction of the angular velocity vector.
A torque, the force that causes it, and the position of the point of application of the force.
A magnetic field, the position of the point where it is determined, and the electric current (or change in electric flux) that causes it.
A magnetic field in a coil of wire and the electric current in the wire.
The force of a magnetic field on a charged particle, the magnetic field itself, and the velocity of the object.
The vorticity at any point in the field of flow of a fluid
The induced current from motion in a magnetic field (known as Fleming's right-hand rule).
The x, y and z unit vectors in a Cartesian coordinate system can be chosen to follow the right-hand rule. Right-handed coordinate systems are often used in rigid body and kinematics.
^ a b Watson, George (1998). "PHYS345 Introduction to the Right Hand Rule". udel.edu. University of Delaware.
^ IIT Foundation Series: Physics – Class 8, Pearson, 2009, p. 312.
Wikimedia Commons has media related to Right-hand rule.
Right and Left Hand Rules - Interactive Java Tutorial National High Magnetic Field Laboratory
A demonstration of the right-hand rule at physics.syr.edu
Weisstein, Eric W. "Right-hand rule". MathWorld.
Dr. Johannes Heidenhain : Right Hand Rule - Heidenhain TNC Training : heidenhain.de
Christian Moser : right-hand-rule : wpftutorial.net
Clorpt
List of physics mnemonics
Mnemonics in trigonometry
Taxonomy mnemonic
Thermodynamic square
Retrieved from "https://en.wikipedia.org/w/index.php?title=Right-hand_rule&oldid=1088150490" |
Parallel-plate transmission line - MATLAB - MathWorks France
rfckt.parallelplate
Parallel-plate transmission line
Use the parallelplate class to represent parallel-plate transmission lines that are characterized by line dimensions and optional stub properties.
A parallel-plate transmission line is shown in cross-section in the following figure. Its physical characteristics include the plate width w and the plate separation d.
h = rfckt.parallelplate
h = rfckt.parallelplate(Name,Value)
h = rfckt.parallelplate returns a parallel-plate transmission line object whose properties are set to their default values.
h = rfckt.parallelplate(Name,Value) sets properties using one or more name-value pairs. For example, rfckt.parallelplate('LineLength',0.045) creates a parallel-plate transmission line object with a physical length of 0.045 meters. You can specify multiple name-value pairs. Enclose each property name in a quote. Properties not specified retain their default values.
\epsilon
{\epsilon }_{0}
. The default value is 2.3.
LineLength — Physical length of parallel-plate transmission line
Physical length of parallel-plate transmission line, specified as a scalar in meters. The default value is 0.01.
scalar in Siemens per meter
Relative permeability of dielectric, specified as a scalar. The ratio of permeability of dielectric,
\mu
, to the permeability in free space,
{\mu }_{0}
. The default value is 1.
Thickness of the dielectric separating the plates, specified as a scalar in meters. The default value is 1.0e-3.
Width — Physical width of parallel-plate transmission line
Physical width of parallel-plate transmission line, specified as a scalar in meters. The default value is 6.0e-4.
Create a parallel plate transmission line using rfckt.parallelplate.
tx1=rfckt.parallelplate('LineLength',0.045)
rfckt.parallelplate with properties:
Separation: 1.0000e-03
Name: 'Parallel-Plate Transmission Line'
The analyze method treats the parallel-plate line as a 2-port linear network and models the line as a transmission line with optional stubs. The analyze method computes the AnalyzedResult property of the line using the data stored in the rfckt.parallelplate object properties as follows:
\begin{array}{l}A=\frac{{e}^{kd}+{e}^{-kd}}{2}\\ B=\frac{{Z}_{0}*\left({e}^{kd}-{e}^{-kd}\right)}{2}\\ C=\frac{{e}^{kd}-{e}^{-kd}}{2*{Z}_{0}}\\ D=\frac{{e}^{kd}+{e}^{-kd}}{2}\end{array}
\begin{array}{c}{Z}_{0}=\sqrt{\frac{R+j2\pi fL}{G+j2\pi fC}}\\ k={k}_{r}+j{k}_{i}=\sqrt{\left(R+j2\pi fL\right)\left(G+j2\pi FC\right)}\end{array}
\begin{array}{l}R=\frac{2}{w{\sigma }_{cond}{\delta }_{cond}}\\ L=\mu \frac{d}{w}\\ G=\omega {\epsilon }^{″}\frac{w}{d}\\ C=\epsilon \frac{w}{d}\end{array}
w is the plate width.
d is the plate separation.
1/\sqrt{\pi f\mu {\sigma }_{cond}}
\begin{array}{c}A=1\\ B=0\\ C=1/{Z}_{in}\\ D=1\end{array}
\begin{array}{c}A=1\\ B={Z}_{in}\\ C=0\\ D=1\end{array}
rfckt.amplifier | rfckt.cascade | rfckt.coaxial | rfckt.cpw | rfckt.datafile | rfckt.delay | rfckt.hybrid | rfckt.hybridg | rfckt.mixer | rfckt.microstrip | rfckt.passive | rfckt.parallel | rfckt.rlcgline | rfckt.series | rfckt.seriesrlc | rfckt.shuntrlc | rfckt.twowire | rfckt.txline |
ERRATA—Volume VI
258 i 3 Bremner, Robert: for 1763 read 1762
263 i 15 Brent, Sir Nathaniel: for 1639 read 1630
ii 5 f.e. for Latin read Italian
266 i 14 f.e. Brenton, Sir Jahleel: after Sheerness insert He was colonel of marines 1825-30
268 i 8-10 Brereton, John: omit John Brereton . . . 1592-3
ii 23 f.e. Brereton, Owen S.: after Lloyd insert He was M.P. for Liverpool from 1724 till his death in 1756, changing his name to Salusbury some years before
18 f.e. omit about the year 1756
269 i 5 after 1780 insert supporting Lord North
ii 3 f.e.
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \\\ \end{matrix}}\right\}\,}}
Brereton, Thomas: for Queen's square read Queen square
271 i 12 f.e. Brereton, Sir William (1604-61): for 1639-40 read to the short and to the long parliaments in 1640
273 i 4 Brereton, Sir William (1789-1864): after constabulary insert and in April 1864 was made colonel commandant of the royal artillery
274 i 15 f.e. Brerewood, Sir Robert: after law insert He was M.P. in the same year for Chester in the short parliament
282 i 32 Brett, Henry: for was for a short time member read was in three parliaments— from 1701-8—the tory member
283 ii 1-2 Brett, John W.: for but he did not live to see it accomplished read and an electric telegraph cable across the Atlantic was completed in 1858, though it failed to work after the transmission of a few messages
4 for Trim read Trym
284 i 9 f.e. Brett, Sir Peircy: for He became a vice-admiral read He was colonel of marines 1760-2 and M.P. for Queenborough 1754-74. He became a vice-admiral of the blue on 18 Oct. and of the white
285 i 7 f.e. Brett, Thomas: for 1743 read 1744
286 i 21 for 1743 read 1743-4
299 i 14 f.e. Brewster, Abraham: for home secretary read first lord of the admiralty
300 ii 13 Brewster, Sir David: for that university read Aberdeen University
301 ii 17 f.e. for In 1838 read On 7 Dec. 1837
11 f.e. after St. Andrews insert He held the post till 1859. From that year till his death he was principal of Edinburgh University
302 i 38 for Professor Blackie read Professor Blaikie
303 i 14 for Montrose read Melrose
26-27 Brewster, Sir Francis: for (fl. 1674-1702) read (d. 1704)
29 after 1674 insert He was M.P. in the Irish house of commons for Tuam 1692-1703, and for Doneraile 1703-4
17 f.e. after colleagues insert Brewster died in 1704
304 ii 2-1 f.e. Brewster, William: for but he mentions . . . nor read He matriculated from Peterhouse in 1580, but apparently failed to graduate
305 i 1 omit the line
314 ii 14 Brideoake, Ralph: omit his deanery of Salisbury
8 f.e. Bridge, Bewick: after 1811 insert He was proctor in 1800
318 i 3-2 f.e. Bridgeman, Sir Orlando: for the long parliament read both the short and the long parliaments
l.l. for in the read on 27 Nov. 1643
ii 1 omit same year
320 ii 23 Bridges, John (d. 1618): after Kent insert From 1565 to 1610 he was prebendary of Winchester
321 ii 25 Bridges, John (1626-1724): for 1711 read 1711-2
325 ii 6 Bridport, Giles of: for Buckinghamshire read Berkshire
327 ii 15 Briggs, Henry P.: for 1793 read 1791?
16 after painter insert son of John Hobart Briggs
17-18 for in 1793; he was . . . Opie the artist read about 1791; he was descended from Vincent Perronet [q. v.] whose daughter Ellizabeth married William Briggs of the Customs House, secretary to the Wesleys
11 f.e. after 1844 insert He married Eliza Alderson, by whom he had one son who died young, and a daughter who married John Carr, rector of Adisham, Kent
7 f.e. for Redgraves' read Redgrave's
328 ii 22 Briggs, John (1785-1875) : after major-general insert of lieut.-general 1851 and full general 6 Feb. 1861 |
Data Presentation - Tables Practice Problems Online | Brilliant
The above table shows the percentage of citizens in a given country that are in a certain age group. If
38 \%
of the population are aged 40 and over, what would be the value of
P
The above table shows the percentage of citizens in a given country that are in a certain age group. What percentage of the population is aged 30 and above?
62 \%\%
47 \%\%
53 \%\%
35 \%\%
In the above table, students recorded down the number of peapods which had a certain number of peas in them. It was discovered that no pea pod had 5 or more peas. How many of the peapods had 2 or fewer peas in them?
The above table represents the number of days that workers in a company drive to work in the past 5 work days.
How many workers are there in the company?
Stella polled her class of 43 students and asked them for the time that they spent watching television each day.
How many students spent 2 to 3 hours watching television? |
Radar target - MATLAB - MathWorks í•œêµ
ScatteringMatrixSource
MeanRCSSource
MeanRCS
Compute Reflected Signal from a Non-fluctuating Radar Target
The RadarTarget System object™ models how a signal is reflected from a radar target. The quantity that determines the response of a target to incoming signals is called the radar target cross-section (RCS). While all electromagnetic radar signals are polarized, you can sometimes ignore polarization and process them as if they were scalar signals. To ignore polarization, specify the EnablePolarization property as false. To utilize polarization, specify the EnablePolarization property as true. For non-polarized processing, the radar cross section is encapsulated in a single scalar quantity called the MeanRCS. For polarized processing, specify the radar cross-section as a 2-by-2 scattering matrix in the ScatteringMatrix property. For both polarization processing types, there are several Swerling models available that can generate random fluctuations in the RCS. Choose these models using the Model property. The SeedSource and Seed properties control the random fluctuations.
The properties that you can use to model the radar cross-section or scattering matrix depend upon the polarization type.
EnablePolarization Value
Use These Properties
To compute the signal reflected from a radar target:
Define and set up your radar target. See Construction.
Call step to compute the reflected signal according to the properties of phased.RadarTarget. The behavior of step is specific to each object in the toolbox.
H = phased.RadarTarget creates a radar target System object, H, that computes the reflected signal from a target.
H = phased.RadarTarget(Name,Value) creates a radar target object, H, with each specified property set to the specified value. You can specify additional name-value pair arguments in any order as (Name1,Value1,...,NameN,ValueN).
Allow polarized signals
Set this property to true to allow the target to simulate the reflection of polarized radiation. Set this property to false to ignore polarization.
Target scattering mode
Target scattering mode specified as one of 'Monostatic' or 'Bistatic'. If you set this property to 'Monostatic', the reflected signal direction is opposite to its incoming direction. If you set this property to 'Bistatic', the reflected direction of the signal differs from its incoming direction. This property applies when you set the EnablePolarization property to true.
Default: 'Monostatic'
Sources of mean scattering matrix of target
Source of mean scattering matrix of target specified as one of 'Property' or 'Input port'. If you set the ScatteringMatrixSource property to 'Property', the target’s mean scattering matrix is determined by the value of the ScatteringMatrix property. If you set this property to 'Input port', the mean scattering matrix is determined by an input argument of the step method. This property applies only when you set the EnablePolarization property to true. When the EnablePolarization property is set to false, use the MeanRCSSource property instead, together with the MeanRCS property, if needed.
Mean radar scattering matrix for polarized signal
Mean radar scattering matrix specified as a complex–valued 2-by-2 matrix. This matrix represents the mean value of the target's radar cross-section. Units are in square meters. The matrix has the form [s_hh s_hv;s_vh s_vv]. In this matrix, the component s_hv specifies the complex scattering response when the input signal is vertically polarized and the reflected signal is horizontally polarized. The other components are defined similarly. This property applies when you set the ScatteringMatrixSource property to 'Property' and the EnablePolarization property to true. When the EnablePolarization property is set to false, use the MeanRCS property instead, together with the MeanRCSSource property. This property is tunable.
Default: [1 0;0 1i]
Source of mean radar cross section
Specify whether the mean RCS value of the target comes from the MeanRCS property of this object or from an input argument in step. Values of this property are:
'Property' The MeanRCS property of this object specifies the mean RCS value(s).
'Input port' An input argument in each invocation of step specifies the mean RCS value.
When EnablePolarization property is set to true, use the ScatteringMatrixSource property together with the ScatteringMatrix property.
Mean radar cross section
Specify the mean value of the target's radar cross section as a nonnegative scalar or as a 1-by-M real-valued, nonnegative row vector. Units are in square meters. Using a vector lets you simultaneously process multiple targets. The quantity M is the number of targets. This property is used when MeanRCSSource is set to 'Property'. This property is tunable.
When EnablePolarization property is set to true, use the ScatteringMatrix property together with the ScatteringMatrixSource.
Target statistical model
Specify the statistical model of the target as one of 'Nonfluctuating', 'Swerling1', 'Swerling2', 'Swerling3', or 'Swerling4'. If you set this property to a value other than 'Nonfluctuating', you must use the UPDATERCS input argument when invoking step. You can set the mean value of the radar cross-section model by specifying MeanRCS or use its default value.
Default: 'Nonfluctuating'
Specify the carrier frequency of the signal you are reflecting from the target, as a scalar in hertz.
The random numbers are used to model random RCS values. This property applies when the Model property is 'Swerling1', 'Swerling2','Swerling3', or 'Swerling4'.
reset Reset states of radar target object
step Reflect incoming signal
Create a simple signal and compute the value of the reflected signal from a target having a radar cross section of
10{m}^{2}
. Set the radar cross section using the MeanRCS property. Set the radar operating frequency to 600 MHz.
x = ones(10,1);
'MeanRCS',10,...
'OperatingFrequency',600e6);
y = target(x);
disp(y(1:3))
This value agrees with the formula
y=\sqrt{G}x
G=4\mathrm{Ï}\mathrm{Ï}/{\mathrm{λ}}^{2}
For a narrowband nonpolarized signal, the reflected signal, Y, is
Y=\sqrt{G}â
X,
X is the incoming signal.
G is the target gain factor, a dimensionless quantity given by
G=\frac{4\mathrm{Ï}\mathrm{Ï}}{{\mathrm{λ}}^{2}}.
σ is the mean radar cross-section (RCS) of the target.
λ is the wavelength of the incoming signal.
The incident signal on the target is scaled by the square root of the gain factor.
For narrowband polarized waves, the single scalar signal, X, is replaced by a vector signal, (EH, EV), with horizontal and vertical components. The scattering matrix, S, replaces the scalar cross-section, σ. Through the scattering matrix, the incident horizontal and vertical polarized signals are converted into the reflected horizontal and vertical polarized signals.
\left[\begin{array}{c}{E}_{H}^{\left(scat\right)}\\ {E}_{V}^{\left(scat\right)}\end{array}\right]=\sqrt{\frac{4\mathrm{Ï}}{{\mathrm{λ}}^{2}}}\left[\begin{array}{cc}{S}_{HH}& {S}_{VH}\\ {S}_{HV}& {S}_{VV}\end{array}\right]\left[\begin{array}{c}{E}_{H}^{\left(inc\right)}\\ {E}_{V}^{\left(inc\right)}\end{array}\right]=\sqrt{\frac{4\mathrm{Ï}}{{\mathrm{λ}}^{2}}}\left[S\right]\left[\begin{array}{c}{E}_{H}^{\left(inc\right)}\\ {E}_{V}^{\left(inc\right)}\end{array}\right]
For further details, see Mott, [1] or Richards, [2] .
phased.FreeSpace | phased.Platform | phased.BackscatterRadarTarget | phased.BackscatterSonarTarget | phased.WidebandBackscatterRadarTarget | backscatterPedestrian (Radar Toolbox) |
Simulation of chip-formation by a single grain of pyramid shape | JVE Journals
S. A. Voronov1 , Weidong Ma2
The article focuses on the analysis of thermomechanical modeling of cutting by single abrasive micro-size grain while the process of grinding. In this research, applying the Johnson-Cook” material model, which relates the intensity of stress with strain rate, temperature and the accumulated plastic strain. Using arbitrary Lagrangian-Eulerian (ALE) approach makes it possible avoid the distortion of finite elements in the simulation of chip-formation under large deformations. The simulation allows predicting cutting force for processing workpiece of titanium alloy Ti6Al4V by single grain grinding.
Keywords: grinding, abrasive grain, FEM, approach ALE, cutting forces, chip formation.
Grinding is a finishing machining operation, which is used to obtain the desired surface roughness and specified precision of shape characteristics. In the process, a large amount of abrasive grains, distributed on the outer cylindrical surface of the grinding wheel, play the role of cutting tools and remove unwanted material. Thermomechanical behavior of the material of workpiece is determined by its interaction with the cutting tool during the formation of chips. The cutting process takes place at high pressure and temperature in the zone of contact of grain and material [1]. The periodical grain entrance in machined material leads to the dynamical workpiece and tool behavior that can decrease the surface quality and the accuracy of processing. The arising vibration have the mechanism of forced vibrations and the regenerative mechanism due to cutting by each grain the surface that was generated by the previous grain [2]. The determination of cutting forces under the specified cutting conditions is required for the vibration analysis. It can be done by modeling of single grain and machined material interaction while cutting.
Therefore, the mechanism of micro material removal with a single grain and the process of chip formation are necessary to investigate in details, as they are determined by the properties of workpiece material, cutting thickness, cutting speed, geometry of cutting edge, and etc. Simulation of cutting process with single grain is conducted by using “Abaqus/Explicit”, which allows to predict the shape of the chips, cutting forces, the surface quality, to estimate the residual stresses in workpiece, and determine the effect of operating parameters on the quality of the treated surface, in order to optimize primary operational parameters and geometry of cutting tools [3].
2. Description of cutting model with a single grain
Real abrasive grain usually has shape of pyramid with flat faces. Grain enters into the material like a cutting wedge with a negative rake angle. In this case, the chip material easily slides along the rake face of the grain, after been separated from the workpiece material. In order to avoid infinity in the calculation of FEM model, the side four sharp edges of pyramid are chamfered while the geometry modeling of pyramidal shape grain. Taking into account of grain deterioration in grinding process, the bottom of grain is covered by spherical top with a given radius
r=
10 μm. Geometric scheme of pyramid-shaped grain is shown in Fig. 1.
In fact, each abrasive grain on grinding wheel, entering into the material, moves along a path close to a circle. For simplicity, the trajectory is replaced by the path shown in Fig. 2, consisting of three straight segments. This designed scheme of grain cutting can reduce time of computation and allows explore the cutting forces acting on single grain. We assume that, the process of exit from cutting of grain from the material in the third stage doesn't greatly change characteristics of the cutting forces, so the results of the third stage are not essential in simulation. Thus modeling is limited to two stages, which can significantly reduce the time of calculation. Besides, all the dimensions are in micrometers.
Fig. 1. Geometric scheme of pyramidal shaped grain
Fig. 2. Simulation path of process single grain cutting
We assume that, the movement of grain is specified kinematically and doesn’t be influenced by the interaction between tool and workpiece material. Grain penetrates into the workpiece material with a constant cutting speed (specified
\stackrel{\to }{V}=\mathrm{ }
5 m/s). At step-1, grain enters into the material with the depth, which varies linearly from 0 to
{h}_{cu}
, and passes along horizontal direction through 100 μm. Next, at step-2, the grain moves with constant speed and the depth of
{h}_{cu}
, and passes on at 200 μm. Modeling was performed for a given cutting thickness
{h}_{cu}=
40 μm.
Zone of deformed surface, where grain penetrates into the material, is several degrees less than the dimensions of the part. Therefore, the actual shape and dimensions of the whole part it is of no importance, and the workpiece is considered as a cuboid. Due to the large temperature gradient and plastic deformations in the contact area, it requires a fine mesh of finite elements. Using uniform grid for workpiece is not efficient, besides it significantly increases the amount of computation.
In the simulation approach ALE is used for finite element mesh construction with the help of built-in procedures of software “ABAQUS”, in order to get a normal chip formation without distortion of finite elements. Fig. 3 shows construction of grid.
Fig. 3. Construction of grid: a) partition of blocks, b) initial mesh, c) the grinding part
3. Approach ALE for the formation of chips
The three main approaches are widely used in the field of metal cutting FEM simulation: Lagrangian, Eulerian, and Arbitrary Lagrangian-Eulerian (ALE).
Approach ALE combines peculiarity of both approaches Lagrangian and Eulerian. The ALE approach FE mesh is neither spatially fixed nor attached to the material. The mesh moves with the flow of material, determination of displacement is performed applying the Lagrangian approach, meanwhile, at each iteration, the grid is rebuilt, determination of velocities is performed applying Euler approach.
The idea of modeling used in metal cutting consists in following way. In the zone near the edge of the cutting tool, Lagrange approach is used, while to analyze the areas where there is a free-boundary flow of chip material, Euler approach is used. This eliminates significant distortion of the grid elements [4].
Fig. 4 shows a scheme of ALE approach and the boundary conditions for workpiece in the simulation. Workpiece is modeled by using element of type C3D8RT, which is an 8-node thermally coupled brick, considering 3-D displacement and temperature, reduces integration time, and controls element parasitic form generation. The number of elements is 210540. The grain is considered as a rigid body.
Fig. 4. Scheme of ALE approach and boundary conditions of workpiece
4. Material model and properties
The behavior of the workpiece material during deformation beyond the elasticity limit is considered by using thermo-elastic-plastic “Johnson-Cook” constitutive model. This model relates the intensity of the stress with strain rate, temperature and the accumulated plastic strain [5]:
\sigma =\left(A+B{\stackrel{-}{{\epsilon }_{pl}}}^{n}\right)\bullet \left[1+C\mathrm{l}\mathrm{n}\left(\frac{\stackrel{˙}{\stackrel{-}{\epsilon }}}{\stackrel{˙}{\stackrel{-}{{\epsilon }_{0}}}}\right)\right]\bullet \left[1-{\left(\frac{T-{T}_{ref}}{{T}_{melt}-{T}_{ref}}\right)}^{m}\right],
\sigma
– equivalent stress,
\stackrel{-}{{\epsilon }_{pl}}
– equivalent plastic deformation,
\stackrel{˙}{\stackrel{-}{\epsilon }}
– actual strain rate,
\stackrel{˙}{\stackrel{-}{{\epsilon }_{0}}}
– basic effective strain rate,
\stackrel{˙}{\stackrel{-}{{\epsilon }_{0}}}=
1.0 s-1,
{T}_{ref}
– room temperature,
{T}_{melt}
– melting temperature of the material.
A
– characteristic coefficient, having dimension of stress,
A={\sigma }_{T}
(yield strength of material),
B
– characteristic coefficient, having the dimension of stress,
n
– the exponent of hardening effect of plastic deformation,
C
– coefficient of influence of the strain rate (dimensionless),
m
– material constant considering heat softening (dimensionless).
Coefficients of model “Johnson-Cook” for material Ti6Al4V are shown in Table 1 [6].
Table 1. Coefficients of model “Johnson-Cook” for material Ti6Al4V
A
(МPа)
B
n
C
m
In the simulation the deformation of the material is supposed to use model of cumulative damage summation
D
D=\frac{1}{{\epsilon }_{f}}\sum _{i}∆{\epsilon }_{p}^{i},
∆{\epsilon }_{p}^{i}
– increment of effective plastic strain in FE on
i
th step of integration time.
The value of critical accumulated strain
{\epsilon }_{f}
is used as the criteria of material damage, which is determined by the “Johnson-Cook” destruction model:
{\epsilon }_{f}=\left[{D}_{1}+{D}_{2}\bullet \mathrm{exp}\left({D}_{3}\frac{p}{\stackrel{-}{\sigma }}\right)\right]\bullet \left[1+{D}_{4}\mathrm{ln}\left(\frac{\stackrel{˙}{\stackrel{-}{\epsilon }}}{\stackrel{˙}{\stackrel{-}{{\epsilon }_{0}}}}\right)\right]\bullet \left[1+{D}_{5}\left(\frac{T-{T}_{ref}}{{T}_{melt}-{T}_{ref}}\right)\right],
p
– the hydrostatic pressure,
\stackrel{-}{\sigma }
– the equivalent von Mises stress,
{D}_{1}
{D}_{5}
– material damage parameters, that show the influence of strain, strain rate and temperature on the material damage.
To describe fracture of material the “Johnson-Cook” model in the package “Abaqus” is applied which states that, when the damage parameter of FE reaches 1, then destruction of the FE occurs. The parameters of the damage law in “Johnson-Cook” material model for Ti6Al4V are given in Table 2 [7].
Mechanical, physical and thermal parameters of the workpiece material are given in Table 3 [8].
Table 2. parameters of the damage law in “Johnson-Cook” material model for Ti6Al4V
{D}_{1}
{D}_{2}
{D}_{3}
{D}_{4}
{D}_{5}
Table 3. Mechanical, physical and thermal parameters of the workpiece material Ti6Al4V
\rho
{C}_{p}
(J/kg°C)
E
\lambda
(W/m°C)
\nu
{\alpha }_{L}
(µm/m°C)
{T}_{melt}
{T}_{ref}
5. Finite element simulation results
Heavily distorted elements during simulation of metal cutting can be removed from the mesh by setting the damage variables (SDEG), so as to reduce the possibility of termination program. Element deletion occurs when the degradation value (SDEG) calculated at a specified increment time reaches 1, ultimately element failure occurs via element deletion technique. Fig. 5 shows distribution of the damage variable
D
(SDEG) in the simulation.
The process of chip formation during the immersion of grain into material is shown in Fig. 6. Several pieces of material are distributed out of workpiece into air. This phenomenon is related to the fact that; material is separated from workpiece in the region around the edge of pyramidal shaped grain. Chips are formed during the extrusion of the material under the grain, and slide along the rake face of grain.
The von Mises stress distribution through the simulation path is illustrated with contour lines in Fig. 6. It is very obvious to detect stress change during process. The maximum stress is achieved at the area near the end of step 1, and near the beginning of step 2, also at the region where the grain changes the direction of its motion. It’s interesting that, the stress decreases, along the path of grain immersion into the material. But if determine the temperature distribution, which is shown in Fig. 7, we can see that maximum temperature is reached at the contact area of grain and material. This coincides with the fact that, metal cutting is the thermomechanical process with high temperature and high stress. Besides, cutting heat will soften metal, consequently influences on stress distribution in workpiece.
Fig. 5. Scalar element degradation parameter
D
(SDEG) during simulation of chip-formation
Fig. 6. The von Mises stress distribution during simulation of chip-formation (Semi-sectional view)
Fig. 7. Temperature distribution during simulation of chip-formation (Semi-sectional view)
Fig. 8. Configuration of cross-section (
{A}_{c}
– cross-sectional area of the contact grain and material)
Fig. 8 shows the configuration of grain and material cross-section. Obviously, the immersion of grain into material, formed groove, side flow, side burrs, as shown in Fig. 9(i).
Finally, the cutting forces of chip-formation with a single grain are shown in Fig. 9. At step-1, cutting forces increase, when the grain moves forward, meanwhile, the cutting thickness also increases linearly with the position of the grain. In most cases of approximate calculation, the cutting force can be seen as proportional to the cutting thickness.
At Step-2, cutting forces have some obvious waves with 4 peaks and 3 troughs, which are consistent with the appearance of chip separation into pieces in the simulation, as shown in Fig. 7. Comparing to the change of cross-sectional area of the contact grain and material (
{A}_{c}
), as shown in Fig. 10, we can assume that, there is some correlation between cutting forces and
{A}_{c}
. In this paper only the results of the simulated case for cutting thickness
{h}_{cu}=
40 μm are presented. The results are not enough to find the correlation between cutting conditions and forces. It’s necessary to add several simulations with different cutting thickness
{h}_{cu}
in future and investigate in details. It will allow construct mechanistic model which can be used for vibration analysis under grinding.
Fig. 9. The dependence of cutting forces (
{F}_{z}
{F}_{x}
) and cross-sectional area of the contact grain and material (
{A}_{c}
) on the position of the grain
FEM model of metal cutting is analyzed; it confirms that approach ALE is very convenient to investigate chip-formation. The results of FEM simulation provide stress and temperature distribution, chip formation during the implantation of single grain. Configuration of cross-section is observed, cross-sectional area of the contact grain and material
\text{(}{A}_{c}\text{)}
is measured, and dependence of cutting forces on grain path is also given.
Astakhov V. P., Shvets S. The assessment of plastic deformation in metal cutting. Journal of Materials Processing Technology, Vol. 146, 2004, p. 193-202. [Search CrossRef]
Altintas Y., Weck M. Chatter stability in metal cutting and grinding. Annals of the CIRP, Vol. 53, Issue 2, 2004, p. 619-642. [Search CrossRef]
Li Xuekun Modeling and Simulation of Grinding Processes Based on a Virtual Wheel Model and Microscopic Interaction Analysis. Dissertation, 2010, p. 4-12. [Search CrossRef]
Kilicaslan Cenk Modelling and Simulation of Metal Cutting by Finite Element Method. Dissertation, 2009, p. 22-24. [Search CrossRef]
Ratchev S. M., Afazov S. M. Mathematical modelling and integration of micro-scale residual stresses into axisymmetric FE models of Ti6Al4V alloy in turning. CIRP Journal of Manufacturing Science and Technology, Vol. 4, 2011, p. 80-89. [Search CrossRef]
Leseur D. R. Experimental Investigations of Material Models for Ti-6Al-4V Titanium and 2024-T3 Aluminum. Technical Report, US Department of Transportation, Federal Aviation Administration, Vol. 9, 2000. [Search CrossRef]
Kay Gregory Failure Modeling of Titanium 6Al-4V and Aluminum 2024-T3 with the Johnson-Cook Material Model. US William J. Hughes Technical Center, Washington, Vol. 9, 2003. [Search CrossRef]
Bragov A., Konstantinov A. Experimental and numerical analysis of high strain rate response of Ti-6Al-4V titanium alloy. DYMAT International Conferences EDP Sciences, Vol. 7, 2009. [Search CrossRef] |
Symmetry in Trigonometric Graphs | Brilliant Math & Science Wiki
Mei Li, Pranjal Jain, Omkar Kulkarni, and
The trigonometric functions cosine, sine, and tangent satisfy several properties of symmetry that are useful for understanding and evaluating these functions.
Symmetry in Angles
The cosine and sine functions satisfy the following properties of symmetry:
\begin{aligned} \cos(-\theta) &= \cos(\theta) \\ \sin(-\theta) &= -\sin(\theta). \end{aligned}
From the definition of cosine and sine in the unit circle,
x= \cos \theta \quad \text{ and } \quad y= \sin \theta.
We can see that for both
\theta
-\theta
x
remains the same. Thus,
\cos \theta=\cos (-\theta)
Similarly, we can see that the
y
in two cases are additive inverse of each other. Thus,
\sin (-\theta)=-\sin\theta.\ _\square
Now that we have the above identities, we can prove several other identities, as shown in the following example.
\begin{aligned} \tan(-\theta) &= -\tan(\theta)\\ \cot(-\theta) &= -\cot(\theta)\\ \csc(-\theta) &= -\csc(\theta)\\ \sec(-\theta) &= \sec(\theta). \end{aligned}
\begin{aligned} \tan(-\theta) &=\frac{\sin(-\theta)}{\cos(-\theta)}=\frac{-\sin(\theta)}{\cos(\theta)}=-\tan(\theta)\\ \cot(-\theta) &=\frac{1}{\tan(-\theta)}=\frac{1}{-\tan(\theta)}=-\cot(\theta)\\ \csc(-\theta) &=\frac{1}{\sin(-\theta)}=\frac{1}{-\sin(\theta)}=-\csc(\theta)\\ \sec(-\theta) &=\frac{1}{\cos(-\theta)}=\frac{1}{\cos(\theta)}=\sec(\theta).\ _\square \end{aligned}
Using the properties of symmetry above, we can show that sine and cosine are special types of functions.
f(x)
is an even function if and only if for all real values of
x
f(-x)=f(x)
. In other words, the graph is symmetric about
y
f(x)
is an odd function if and only if for all real values of
x
f(-x)=-f(x)
. In other words, the graph is symmetric about origin.
f(-0)=-f(0)\implies f(0)=0
. That is, an odd function must pass through the origin.
From this definition, the cosine function is an even function and the sine function is an odd function.
What symmetry is there between the angles
\theta
(\theta + \pi)?
If we plug in a few values for
\theta
, how do the basic trigonometric functions change?
By the properties of symmetry, we can write
\sin (\theta + \pi)
\sin(\theta)
\sin (\theta+\pi)=-\sin(\theta).
Similarly, we can write
\cos (\theta + \pi)
\cos(\theta)
\cos (\theta+\pi)=-\cos(\theta).
Determine whether the function
f(x) = \tan^2(x) + \cos(x)
is an odd function, an even function, or neither.
The function satisfies
f(-x) = \tan^2(-x) + \cos(-x) = \tan^2(x) + \cos(x) = f(x)
\cos(x)
is an even function. Therefore,
f(x)
is an even function.
_\square
Find a relationship between
\tan (\theta + \pi)
\tan (\theta ).
Solution 1: We have
\begin{aligned} \tan (\theta + \pi) & = \frac{\sin(\theta + \pi ) }{\cos(\theta + \pi) } \\ &= \frac{-\sin(\theta)}{-\cos(\theta)}\\ &= \frac{\sin(\theta)}{\cos(\theta)}\\ &= \tan (\theta ). \end{aligned}
\tan (\theta + \pi) = \tan (\theta ).
\tan(x + \pi) = \frac{\tan(x) + \tan(\pi)}{ 1 - \tan(x) \tan(\pi)} = \frac{\tan(x) + 0}{ 1 - \tan(x) \cdot 0} = \tan(x).
This shows the period of the tangent function is at most
\pi
_\square
Since the cosine function satisfies
\cos(-\theta) = \cos(\theta)
, the graph of the function
\cos(x)
is symmetric about the
y
-axis. What is the symmetry satisfied by the graph of
\sin(x)?
\sin(x)
\sin(-x) = \sin(x)
\sin(x)
is symmetric about the origin.
_\square
In general, for any even function
f(x)
, the the graph of
f(x)
y
-axis; for any odd function
g(x)
g(x)
See Sine and Cosine graphs for more properties of the sine and cosine graphs.
Cite as: Symmetry in Trigonometric Graphs. Brilliant.org. Retrieved from https://brilliant.org/wiki/symmetry-in-trigonometric-graphs/ |
Consider a binomial experiment with 15 trials and probability 0.45
Consider a binomial experiment with 15 trials and probability 0.45 of success on
Consider a binomial experiment with 15 trials and probability 0.45 of success on a single trial.
(a) Use the binomial distribution to find the probability of exactly 10 successes. (Round your answer to three decimal places.)
(b) Use the normal distribution to approximate the probability of exactly 10 successes. (Round your answer to three decimal places.)
Solution: It is given here that a random variable say x follows the binomial distribution with parameters
n=15\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }p=0.45
The binomial probability function is:
P\left(X=x\right)=\frac{n!}{\left(n-x\right)!x!}{p}^{x}{\left(1-p\right)}^{n-x};x=0,1,2,..,n
(a) Use the binomial distribution to find the probability of exactly 10 successes.
Answer: It is required to find:
P\left(x=10\right)
Using the binomial distribution function:
P\left(x=10\right)=\frac{15!}{\left(15-10\right)!10!}{0.45}^{10}{\left(1-0.45\right)}^{15-10}
=3003×0.000340506×0.050328438
=0.051
Therefore, the probability of exactly 10 successes is 0.051
(b) Use the normal distribution to approximate the probability of exactly 10 successes.
The mean and standard deviation of the random variable x is:
\mu =np=15×0.45=6.75
\sigma =\sqrt{np\left(1-p\right)}=\sqrt{15×0.45\left(1-0.45\right)}=1.92678
It is required to find:
P\left(x=10\right)
Using the continuity correction factor, the above probability can be written as:
P\left(x=10\right)=P\left(10-0.5<x<10+0.5\right)
=P\left(9.5<x<10.5\right)
Using the z-score formula:
P\left(9.5<x<10.5\right)=P\left(\frac{9.5-6.75}{1.92678}<\frac{x-\mu }{\sigma }<\frac{10.5-6.75}{1.92678}
=P\left(1.4272<z<1.9462\right)
=P\left(z<1.9462\right)-P\left(z<1.4272\right)
Now using the excel functions:
P\left(9.5<x<10.5\right)=P\left(z<1.9462\right)-P\left(z<1.4272\right)=0.9742-0.9232=0.051
The excel functions are:
=NORMSDIST\left(1.4272\right)=0.9232
=NORMSDIST\left(1.9462\right)=0.9742
Therefore, Using the normal distribution to approximate the probability of exactly 10 successes is 0.051
The number of products manufactured in a factory in a day are 3500 and the probability that some pieces are defected is 0.55 then the mean of binomial probability distribution is
A binomial probaity is given. Write the probably in words. Then, use a continuity correction to convert the binomial probability to a normal distributon probability.
P\left(x=138\right)
Write the probability in words.
The probability of getting ? 138 successes.
Which of the fallowing is the normal probability statement that corresponds to the binomial probability statement?
P\left(x>137.5\right)
P\left(137.5<x<138.5\right)
P\left(x<137.5\right)
P\left(x>138.5\right)
P\left(x<138.5\right)
Given a sequence of Bernoulli trials with
n=6
, k is at least 1, and
p=0.25
, find the binomial probability.
Suppose that a family has 7 children. Also, suppose that the probability of having a girl is 1/2. Find the probability that the family has at least 6 girls.
Let X be a binomial random variable with
p=0.3
n=15
P\left(5\le X\le 10\right)
How can I get the probability of a binomial distribution if the values are between 6 and 9, and the number of trials is 19? |
Coplanar waveguide transmission line - MATLAB - MathWorks France
rfckt.cpw
Use the cpw object to represent coplanar waveguide transmission lines that are characterized by line dimensions, stub type, and termination.
A coplanar waveguide transmission line is shown in cross-section in the following figure. Its physical characteristics include the conductor width (w), the conductor thickness (t), the slot width (s), the substrate height (d), and the permittivity constant (ε).
h = rfckt.cpw
h = rfckt.cpw(Name,Value)
h = rfckt.cpw returns a coplanar waveguide transmission line object whose properties are set to their default values.
h = rfckt.cpw(Name,Value) sets properties using one or more name-value pairs. For example, rfckt.cpw('ConductorWidth',0.3) creates an RF coplanar waveguide transmission line with a width of 0.3 meters. You can specify multiple name-value pairs. Enclose each property name in a quote. Properties not specified retain their default values.
Computed S-parameters, noise figure, OIP3, and group delay values, specified as an rfdata.data object. For more information refer, Algorithms.
ConductorWidth — Physical width of conductor
Physical width of conductor, specified as a scalar in meters. By default, the value is 0.6e-4.
\epsilon
{\epsilon }_{0}
. By default, the value is 9.8.
Dielectric thickness or physical height of the conductor, specified as a scalar in meters. The default value is 0.635e-4.
Physical length of transmission, specified as a scalar in meters. The default value is 0.01.
Loss angle tangent of dielectric, specified as a scalar. The default value is 0.
Name — Name of coplanar waveguide transmission line object
Name of coplanar waveguide transmission line object, specified as a 1-by-N character array.
Physical width of slot, specified as a scalar in meters. The default value is 0.2e-4.
Thickness — Physical thickness of conductor
Physical thickness of conductor, specified as a scalar in meters. The default value is 0.005e-6.
Create a coplanar waveguide transmission line using rfckt.cpw.
tx=rfckt.cpw('Thickness',0.0075e-6)
rfckt.cpw with properties:
Name: 'Coplanar Waveguide Transmission Line'
The analyze method treats the transmission line as a 2-port linear network. It computes the AnalyzedResult property of a stub or as a stub less line using the data stored in the rfckt.cpw object properties as follows:
\begin{array}{l}A=\frac{{e}^{kd}+{e}^{-kd}}{2}\\ B=\frac{{Z}_{0}*\left({e}^{kd}-{e}^{-kd}\right)}{2}\\ C=\frac{{e}^{kd}-{e}^{-kd}}{2*{Z}_{0}}\\ D=\frac{{e}^{kd}+{e}^{-kd}}{2}\end{array}
Z0 and k are vectors whose elements correspond to the elements of f, the vector of frequencies specified in the analyze input argument freq. Both can be expressed in terms of the specified conductor strip width, slot width, substrate height, conductor strip thickness, relative permittivity constant, conductivity and dielectric loss tangent of the transmission line, as described in [1].
\begin{array}{c}A=1\\ B=0\\ C=1/{Z}_{in}\\ D=1\end{array}
\begin{array}{c}A=1\\ B={Z}_{in}\\ C=0\\ D=1\end{array}
rfckt.amplifier | rfckt.cascade | rfckt.coaxial | rfckt.datafile | rfckt.delay | rfckt.hybrid | rfckt.hybridg | rfckt.mixer | rfckt.microstrip | rfckt.passive | rfckt.parallel | rfckt.parallelplate | rfckt.rlcgline | rfckt.series | rfckt.seriesrlc | rfckt.shuntrlc | rfckt.twowire | rfckt.txline |
What are the three cube roots of -1? Not sure if
A canoe has a velocity of 0.40 m/s southeast relative
A canoe has a velocity of 0.40 m/s southeast relative to the earth. The canoe is on a river that is flowing 0.50 m/s east relative to the earth. Find the velocity (magnitude and direction) of the canoe relative to the river.
The velocity (magnitude and direction) of the canoe to the river is
0.4{\mathrm{cos}45}^{\circ }=0.5-{v}_{crE}
{v}_{crE}=0.5-0.4{\mathrm{cos}45}^{\circ }=0.22\text{ }\frac{m}{s}
{v}_{crS}=0.4{\mathrm{sin}45}^{\circ }-0.28\text{ }\frac{m}{s}
{v}_{cr}=\sqrt{{v}_{crE}^{2}+{v}_{crS}^{2}}=0.36\text{ }\frac{m}{s}
\theta =\mathrm{arctan}\frac{{v}_{crS}}{{v}_{crE}}={52}^{\circ }
S of E
0.4\mathrm{cos}{45}^{\circ }=0.5-{v}_{crE}
{v}_{crE}=0.5-0.4\mathrm{cos}{45}^{\circ }=0.22\text{ }m/s
{v}_{crS}=0.4\mathrm{sin}{45}^{\circ }-0.28\text{ }m/s
{v}_{cr}=\sqrt{{v}_{crE}^{2}+{v}_{crS}^{2}}=0.36\text{ }m/s
\theta =\mathrm{arctan}\frac{{v}_{crS}}{{v}_{crE}}={52}^{\circ }S\text{ }of\text{ }E
A 0.75-kg ball is attached to a 1.0-m rope and whirled in avertical circle. The rope will break when the tension exceeds 450 N. What is the maximum speed theball can have at the bottom of the circle without breaking the rope?
Two small charged objects repel each other with a force F whenseparated by a distance d. if the charge on each object is reducedto one-fourth of its original value and the distance between themis reduced to d/2, the force becomes:
\frac{F}{16}
\frac{F}{8}
\frac{F}{4}
\frac{F}{2}
In Figure, a stationary block explodes into two pieces and R that slide across a frictionless floor and then into regions with friction, where they stop. Piece L, with a mass of 2.6 kg, encounters a coefficient of kinetic friction
\mu L=0.40
and slides to a stop in distance deal = 0.15 m. Piece R encounters a coefficient of kinetic friction
\mu R=0.50
and slides to a stop in distanced R=0.30 m. What was the mass of the original block?
The drawing shows a version of the loop-the-loop trick for a small car. If the car is given an initial speed of 4 m/s, what is the largest value that the radius r can have if the car is to remain in contact with the circular track at all times?
Individual A has a red die and B has a green die (both fair).If they each roll until they obtain five "doubles" (1-1,....,6-6),what is the pmf of X= the total number of times a die is rolled?What are E(X) and V(X)?
Apply Newton’s Method using the given initial guess, and explain why the method fails.
y={x}^{3}-2x-2,\text{ }{x}_{1}=0
A rancher has 400 feet of fencing with which to enclose two adjacent rectangular corrals. What dimensions should be used so that the enclosed area will be a maximum? |
Source coding mu-law or A-law compressor or expander - MATLAB compand - MathWorks France
Source coding mu-law or A-law compressor or expander
out = compand(in,param,v)
out = compand(in,param,v,method)
out = compand(in,param,v) performs mu-law compression on the input data sequence. The param input specifies the mu-law compression value and must be set to a mu value for mu-law compressor computation (a mu-law value of 255 is used in practice). v specifies the peak magnitude of the input data sequence.
out = compand(in,param,v,method) performs mu-law or A-law compression or expansion on the input data sequence. param specifies the mu-law compander or A-law compander value (a mu-law value of 255 and an A-law value of 87.6 are used in practice). method specifies the type of compressor or expander computation for the function to perform on the input data sequence.
When transmitting signals with a high dynamic range, quantization using equal length intervals can result in loss of precision and signal distortion. Companding is a operation that applies a logarithmic computation to compress the signal before quantization on the transmit side and applies an inverse operation to expand the signal to restore it to full scale on the receive side. Companding avoids signal distortion without the need to specify many quantization levels. Compare distortion when using 6-bit quantization on an exponential signal with and without companding. Plot the original exponential signal, the quantized signal and the expanded signal.
Create an exponential signal and calculate its maximum value.
sig = exp(-4:0.1:4);
V = max(sig);
Quantize the signal by using equal-length intervals. Set partition and codebook values, assuming 6-bit quantization. Calculate the mean square distortion.
partition = 0:2^6 - 1;
codebook = 0:2^6;
[~,qsig,distortion] = quantiz(sig,partition,codebook);
Compress the signal by using the compand function configured to apply the mu-law method. Apply quantization and expand the quantized signal. Calculate the mean square distortion of the companded signal.
mu = 255; % mu-law parameter
csig_compressed = compand(sig,mu,V,'mu/compressor');
[~,quants] = quantiz(csig_compressed,partition,codebook);
csig_expanded = compand(quants,mu,max(quants),'mu/expander');
distortion2 = sum((csig_expanded - sig).^2)/length(sig);
Compare the mean square distortion for quantization versus combined companding and quantization. The distortion for the companded and quantized signal is an order of magnitude lower than the distortion of the quantized signal. Equal-length intervals are well suited to the logarithm of an exponential signal but not well suited to an exponential signal itself.
[distortion, distortion2]
Plot the original exponential signal, the quantized signal, and the expanded signal. Zoom in on axis to highlight the quantized signal error at lower signal levels.
plot([sig' qsig' csig_expanded']);
title('Comparison Between Original, Quantized, and Expanded Signals');
ylabel('Apmlitude');
legend('Original','Quantized','Expanded','location','nw');
in — Input data sequence
Input data sequence, specified as a row vector. This input specifies the data sequence for the function to perform compression or expansion.
param — mu or A value of compander
positive scalar | 255 | 87.6
mu or A value of the compander, specified as a positive scalar. The prevailing values used in practice are µ = 255 and A = 87.6.
method — Type of compressor or expander computation
mu/compressorr | mu/expander | A/compressor | A/expander
Type of compressor or expander computation for the function to perform on the input data sequence, specified as one of these values.
mu/compressor
mu/expander
A/compressor
A/expander
v — Peak magnitude of input data sequence
Peak magnitude of the input data sequence, specified as a positive scalar.
out — Compressed or expanded signal
positive row vector
Compressed or expanded signal, returned as a positive row vector. The size of out matches that of input argument in.
In certain applications, such as speech processing, using a logarithmic computation (called a compressor) before quantizing the input data is common. The inverse operation of a compressor is called an expander. The combination of a compressor and expander is called a compander.
For a given signal, x, the output of the (µ-law) compressor is
y=\frac{\mathrm{log}\left(1+\mu |x|\right)}{\mathrm{log}\left(1+\mu \right)}\mathrm{sgn}\left(x\right).
µ is the µ-law parameter of the compander, log is the natural logarithm, and sgn is the signum function (sign in MATLAB®).
µ-law expansion for input signal x is given by the inverse function y-1,
{y}^{-1}=\mathrm{sgn}\left(y\right)\left(\frac{1}{\mu }\right)\left({\left(1+\mu \right)}^{|y|}-1\right)\text{ for -1}\le y\le 1
For a given signal, x, the output of the (A-law) compressor is
y=\left\{\begin{array}{cc}\begin{array}{c}\frac{A|x|}{1+\mathrm{log}A}\mathrm{sgn}\left(x\right)\\ \frac{\left(1+\mathrm{log}\left(A|x|\right)\right)}{1+\mathrm{log}A}\mathrm{sgn}\left(x\right)\end{array}& \begin{array}{c}\text{for }0\le |x|\le \frac{1}{A}\\ \text{for }\frac{1}{A}<|x|\le 1\end{array}\end{array}
A is the A-law parameter of the compander, log is the natural logarithm, and sgn is the signum function (sign in MATLAB).
A-law expansion for input signal x is given by the inverse function y-1,
{y}^{-1}=\mathrm{sgn}\left(y\right)\left\{\begin{array}{cc}\begin{array}{c}\frac{|y|\left(1+\mathrm{log}\left(A\right)\right)}{A}\\ \frac{\mathrm{exp}\left(|y|\left(1+\mathrm{log}\left(A\right)\right)-1\right)}{A}\end{array}& \begin{array}{c}\text{for }0\le |y|<\frac{1}{1+\mathrm{log}\left(A\right)}\\ \text{for }\frac{1}{1+\mathrm{log}\left(A\right)}\le |y|<1\end{array}\end{array}
[1] Sklar, Bernard. Digital Communications: Fundamentals and Applications. Englewood Cliffs, NJ: Prentice-Hall, 1988.
lloyds | quantiz | dpcmenco | dpcmdeco | huffmanenco | huffmandeco |
Polynomial sample rate converter with arbitrary conversion factor - MATLAB - MathWorks Australia
The decimation factor is now only 13. The lower the decimation factor, the more flexibility in input size. The output rate is within the range OutputSampleRate
±
1%. |
Ordinary, weak, repulsive magnetism that all materials possess
Diamagnetism was first discovered when Anton Brugmans observed in 1778 that bismuth was repelled by magnetic fields.[1] In 1845, Michael Faraday demonstrated that it was a property of matter and concluded that every material responded (in either a diamagnetic or paramagnetic way) to an applied magnetic field. On a suggestion by William Whewell, Faraday first referred to the phenomenon as diamagnetic (the prefix dia- meaning through or across), then later changed it to diamagnetism.[2][3]
A simple rule of thumb is used in chemistry to determine whether a particle (atom, ion, or molecule) is paramagnetic or diamagnetic:[4] If all electrons in the particle are paired, then the substance made of this particle is diamagnetic; If it has unpaired electrons, then the substance is paramagnetic.
2.1 Curving water surfaces
3.1 Langevin diamagnetism
3.2 In metals
Notable diamagnetic materials[5]
χm [× 10−5 (SI units)]
Superconductor −105
Pyrolytic carbon −40.9
Bismuth −16.6
Neon −6.74
Mercury −2.9
Silver −2.6
Carbon (diamond) −2.1
Carbon (graphite) −1.6
Copper −1.0
Water −0.91
Diamagnetism is a property of all materials, and always makes a weak contribution to the material's response to a magnetic field. However, other forms of magnetism (such as ferromagnetism or paramagnetism) are so much stronger that, when multiple different forms of magnetism are present in a material, the diamagnetic contribution is usually negligible. Substances where the diamagnetic behaviour is the strongest effect are termed diamagnetic materials, or diamagnets. Diamagnetic materials are those that some people generally think of as non-magnetic, and include water, wood, most organic compounds such as petroleum and some plastics, and many metals including copper, particularly the heavy ones with many core electrons, such as mercury, gold and bismuth. The magnetic susceptibility values of various molecular fragments are called Pascal's constants.
Diamagnetic materials, like water, or water-based materials, have a relative magnetic permeability that is less than or equal to 1, and therefore a magnetic susceptibility less than or equal to 0, since susceptibility is defined as χv = μv − 1. This means that diamagnetic materials are repelled by magnetic fields. However, since diamagnetism is such a weak property, its effects are not observable in everyday life. For example, the magnetic susceptibility of diamagnets such as water is χv = −9.05×10−6. The most strongly diamagnetic material is bismuth, χv = −1.66×10−4, although pyrolytic carbon may have a susceptibility of χv = −4.00×10−4 in one plane. Nevertheless, these values are orders of magnitude smaller than the magnetism exhibited by paramagnets and ferromagnets. Because χv is derived from the ratio of the internal magnetic field to the applied field, it is a dimensionless value.
In rare cases, the diamagnetic contribution can be stronger than paramagnetic contribution. This is the case for gold, which has a magnetic susceptibility less than 0 (and is thus by definition a diamagnetic material), but when measured carefully with X-ray magnetic circular dichroism, has an extremely weak paramagnetic contribution that is overcome by a stronger diamagnetic contribution.[6]
Transition from ordinary conductivity (left) to superconductivity (right). At the transition, the superconductor expels the magnetic field and then acts as a perfect diamagnet.
Superconductors may be considered perfect diamagnets (χv = −1), because they expel all magnetic fields (except in a thin surface layer) due to the Meissner effect.[7]
Curving water surfaces[edit]
If a powerful magnet (such as a supermagnet) is covered with a layer of water (that is thin compared to the diameter of the magnet) then the field of the magnet significantly repels the water. This causes a slight dimple in the water's surface that may be seen by a reflection in its surface.[8][9]
Main article: Magnetic levitation § Diamagnetic levitation
A live frog levitates inside a 32 mm (1.26 in) diameter vertical bore of a Bitter solenoid in a magnetic field of about 16 teslas at the Nijmegen High Field Magnet Laboratory.[10]
Diamagnets may be levitated in stable equilibrium in a magnetic field, with no power consumption. Earnshaw's theorem seems to preclude the possibility of static magnetic levitation. However, Earnshaw's theorem applies only to objects with positive susceptibilities, such as ferromagnets (which have a permanent positive moment) and paramagnets (which induce a positive moment). These are attracted to field maxima, which do not exist in free space. Diamagnets (which induce a negative moment) are attracted to field minima, and there can be a field minimum in free space.
A thin slice of pyrolytic graphite, which is an unusually strongly diamagnetic material, can be stably floated in a magnetic field, such as that from rare earth permanent magnets. This can be done with all components at room temperature, making a visually effective and relatively convenient demonstration of diamagnetism.
The Radboud University Nijmegen, the Netherlands, has conducted experiments where water and other substances were successfully levitated. Most spectacularly, a live frog (see figure) was levitated.[11]
In September 2009, NASA's Jet Propulsion Laboratory (JPL) in Pasadena, California announced it had successfully levitated mice using a superconducting magnet,[12] an important step forward since mice are closer biologically to humans than frogs.[13] JPL said it hopes to perform experiments regarding the effects of microgravity on bone and muscle mass.
Recent experiments studying the growth of protein crystals have led to a technique using powerful magnets to allow growth in ways that counteract Earth's gravity.[14]
A simple homemade device for demonstration can be constructed out of bismuth plates and a few permanent magnets that levitate a permanent magnet.[15]
The electrons in a material generally settle in orbitals, with effectively zero resistance and act like current loops. Thus it might be imagined that diamagnetism effects in general would be common, since any applied magnetic field would generate currents in these loops that would oppose the change, in a similar way to superconductors, which are essentially perfect diamagnets. However, since the electrons are rigidly held in orbitals by the charge of the protons and are further constrained by the Pauli exclusion principle, many materials exhibit diamagnetism, but typically respond very little to the applied field.
The Bohr–Van Leeuwen theorem proves that there cannot be any diamagnetism or paramagnetism in a purely classical system. However, the classical theory of Langevin for diamagnetism gives the same prediction as the quantum theory.[16] The classical theory is given below.
Langevin diamagnetism[edit]
Paul Langevin's theory of diamagnetism (1905)[17] applies to materials containing atoms with closed shells (see dielectrics). A field with intensity B, applied to an electron with charge e and mass m, gives rise to Larmor precession with frequency ω = eB / 2m. The number of revolutions per unit time is ω / 2π, so the current for an atom with Z electrons is (in SI units)[16]
{\displaystyle I=-{\frac {Ze^{2}B}{4\pi m}}.}
The magnetic moment of a current loop is equal to the current times the area of the loop. Suppose the field is aligned with the z axis. The average loop area can be given as
{\displaystyle \scriptstyle \pi \left\langle \rho ^{2}\right\rangle }
{\displaystyle \scriptstyle \left\langle \rho ^{2}\right\rangle }
is the mean square distance of the electrons perpendicular to the z axis. The magnetic moment is therefore
{\displaystyle \mu =-{\frac {Ze^{2}B}{4m}}\langle \rho ^{2}\rangle .}
If the distribution of charge is spherically symmetric, we can suppose that the distribution of x,y,z coordinates are independent and identically distributed. Then
{\displaystyle \scriptstyle \left\langle x^{2}\right\rangle \;=\;\left\langle y^{2}\right\rangle \;=\;\left\langle z^{2}\right\rangle \;=\;{\frac {1}{3}}\left\langle r^{2}\right\rangle }
{\displaystyle \scriptstyle \left\langle r^{2}\right\rangle }
is the mean square distance of the electrons from the nucleus. Therefore,
{\displaystyle \scriptstyle \left\langle \rho ^{2}\right\rangle \;=\;\left\langle x^{2}\right\rangle \;+\;\left\langle y^{2}\right\rangle \;=\;{\frac {2}{3}}\left\langle r^{2}\right\rangle }
{\displaystyle n}
is the number of atoms per unit volume, the volume diamagnetic susceptibility in SI units is[18]
{\displaystyle \chi ={\frac {\mu _{0}n\mu }{B}}=-{\frac {\mu _{0}e^{2}Zn}{6m}}\langle r^{2}\rangle .}
In atoms, Langevin susceptibility is of the same order of magnitude as Van Vleck paramagnetic susceptibility.
In metals[edit]
The Langevin theory is not the full picture for metals because there are also non-localized electrons. The theory that describes diamagnetism in a free electron gas is called Landau diamagnetism, named after Lev Landau,[19] and instead considers the weak counteracting field that forms when the electrons' trajectories are curved due to the Lorentz force. Landau diamagnetism, however, should be contrasted with Pauli paramagnetism, an effect associated with the polarization of delocalized electrons' spins.[20][21] For the bulk case of a 3D system and low magnetic fields, the (volume) diamagnetic susceptibility can be calculated using Landau quantization, which in SI units is
{\displaystyle \chi =-\mu _{0}{\frac {e^{2}}{12\pi ^{2}m\hbar }}{\sqrt {2mE_{\rm {F}}}},}
{\displaystyle E_{\rm {F}}}
is the Fermi energy. This is equivalent to
{\displaystyle -\mu _{0}\mu _{\rm {B}}^{2}g(E_{\rm {F}})/3}
, exactly
{\textstyle -1/3}
times Pauli paramagnetic susceptibility, where
{\displaystyle \mu _{\rm {B}}=e\hbar /2m}
is the Bohr magneton and
{\displaystyle g(E)}
is the density of states (number of states per energy per volume). This formula takes into account the spin degeneracy of the carriers (spin ½ electrons).
In doped semiconductors the ratio between Landau and Pauli susceptibilities may change due to the effective mass of the charge carriers differing from the electron mass in vacuum, increasing the diamagnetic contribution. The formula presented here only applies for the bulk; in confined systems like quantum dots, the description is altered due to quantum confinement.[22][23] Additionally, for strong magnetic fields, the susceptibility of delocalized electrons oscillates as a function of the field strength, a phenomenon known as the De Haas–Van Alphen effect, also first described theoretically by Landau.
Diamagnetic inequality – Mathematical inequality relating the derivative of a function to its covariant derivative
^ Gerald Küstler (2007). "Diamagnetic Levitation – Historical Milestones". Rev. Roum. Sci. Techn. – Électrotechn. Et Énerg. 52, 3: 265–282.
^ Jackson, Roland (21 July 2014). "John Tyndall and the Early History of Diamagnetism". Annals of Science. 72 (4): 435–489. doi:10.1080/00033790.2014.929743. PMC 4524391. PMID 26221835.
^ "diamagnetic, adj. and n". OED Online. Oxford University Press. June 2017.
^ "Magnetic Properties". Chemistry LibreTexts. 2 October 2013. Retrieved 21 January 2020.
^ Nave, Carl L. "Magnetic Properties of Solids". Hyper Physics. Retrieved 9 November 2008.
^ Motohiro Suzuki, Naomi Kawamura, Hayato Miyagawa, Jose S. Garitaonandia, Yoshiyuki Yamamoto, and Hidenobu Hori (24 January 2012). "Measurement of a Pauli and Orbital Paramagnetic State in Bulk Gold Using X-Ray Magnetic Circular Dichroism Spectroscopy". Physical Review Letters. 108 (4): 047201. Bibcode:2012PhRvL.108d7201S. doi:10.1103/PhysRevLett.108.047201. PMID 22400883. {{cite journal}}: CS1 maint: multiple names: authors list (link)
^ Poole, Jr., Charles P. (2007). Superconductivity (2nd ed.). Amsterdam: Academic Press. p. 23. ISBN 9780080550480.
^ Beatty, Bill (2005). "Neodymium supermagnets: Some demonstrations—Diamagnetic water". Science Hobbyist. Retrieved 26 September 2011.
^ Quit007 (2011). "Diamagnetism Gallery". DeviantART. Retrieved 26 September 2011.
^ "Diamagnetic Levitation". High Field Laboratory. Radboud University Nijmegen. 2011. Retrieved 26 September 2020.
^ "The Real Levitation". High Field Laboratory. Radboud University Nijmegen. 2011. Retrieved 26 September 2011.
^ Liu, Yuanming; Zhu, Da-Ming; Strayer, Donald M.; Israelsson, Ulf E. (2010). "Magnetic levitation of large water droplets and mice". Advances in Space Research. 45 (1): 208–213. Bibcode:2010AdSpR..45..208L. doi:10.1016/j.asr.2009.08.033.
^ Choi, Charles Q. (9 September 2009). "Mice levitated in lab". Live Science. Retrieved 26 September 2011.
^ Kleiner, Kurt (10 August 2007). "Magnetic gravity trick grows perfect crystals". New Scientist. Retrieved 26 September 2011.
^ "Fun with diamagnetic levitation". ForceField. 2 December 2008. Archived from the original on 12 February 2008. Retrieved 26 September 2011.
^ a b Kittel, Charles (1986). Introduction to Solid State Physics (6th ed.). John Wiley & Sons. pp. 299–302. ISBN 978-0-471-87474-4.
^ Langevin, Paul (1905). "Sur la théorie du magnétisme". Journal de Physique Théorique et Appliquée (in French). 4 (1): 678–693. doi:10.1051/jphystap:019050040067800. ISSN 0368-3893.
^ Kittel, Charles (2005). "Chapter 14: Diamagnetism and Paramagnetism". Introduction to Solid State Physics (8 ed.). John Wiley & Sons. ISBN 978-0471415268.
^ Landau, L. D. "Diamagnetismus der metalle." Zeitschrift für Physik A Hadrons and Nuclei 64.9 (1930): 629-637.
^ Chang, M. C. "Diamagnetism and paramagnetism" (PDF). NTNU lecture notes. Retrieved 24 February 2011.
^ Drakos, Nikos; Moore, Ross; Young, Peter (2002). "Landau diamagnetism". Electrons in a magnetic field. Retrieved 27 November 2012.
^ Lévy, L.P.; Reich, D.H.; Pfeiffer, L.; West, K. (1993). "Aharonov-Bohm ballistic billiards". Physica B: Condensed Matter. 189 (1–4): 204–209. Bibcode:1993PhyB..189..204L. doi:10.1016/0921-4526(93)90161-x.
^ Richter, Klaus; Ullmo, Denis; Jalabert, Rodolfo A. (1996). "Orbital magnetism in the ballistic regime: geometrical effects". Physics Reports. 276 (1): 1–83. arXiv:cond-mat/9609201. Bibcode:1996PhR...276....1R. doi:10.1016/0370-1573(96)00010-5. S2CID 119330207.
Media related to Diamagnetism at Wikimedia Commons
Diamagnetic Levitation (YouTube)
Diamagnetism of water (YouTube, in Japanese)
Retrieved from "https://en.wikipedia.org/w/index.php?title=Diamagnetism&oldid=1058559375" |
Missing sock - Wikipedia
Single sock in a pair of socks known or perceived to be missing
Not to be confused with One Sock Missing.
An illustration of a singular sock from 1840, published in Godey's lady's book[1]
A missing sock, lost sock, or odd sock (primarily British English)[2][3] is a single sock in a pair of socks known or perceived to be permanently or temporarily missing. According to popular media articles regarding missing socks, people almost always report losing one sock in a pair, and hardly ever the entire pair of two socks. Socks are usually perceived to be lost immediately before, during, or immediately after doing laundry. Various explanations or theories—some scientific or pseudo-scientific and others humorous or facetious—have been proposed to show how or why single socks go missing or are perceived to have gone missing. The terms odd sock or mismatched sock may refer to the remaining "orphaned" sock in a pair where the other matching sock is missing or lost.
1 Possible causes and explanations
Possible causes and explanations[edit]
Two common plausible explanations for missing socks are that they are lost in transit to or from the laundry, or that they are trapped inside, between, or behind components of ("eaten by") washing machines and/or clothes dryers. Due to the high rotational speeds of modern front-loading washing machines and dryers, it may be possible for small clothes items such as socks to slip through any holes or tears in the rubber gasket between either machine's spinning drums and their outer metal or plastic cases.[4] Socks may also bunch up or unravel and get caught in the water drain pipe of washing machines or in the lint trap of dryers.[4]
Some explanations for missing socks seem to imply socks' propensity for going missing is—or is related to—a physical property of the universe. For example, in the 1996 book The Nature of Space and Time by physicist Stephen Hawking and mathematician Roger Penrose, they posited spontaneous black holes are responsible for lost socks.[2]
In 2008, American science educator and writer George B. Johnson proposed six hypotheses for why socks go missing, some plausible and others fanciful:
an "intrinsic property" of the socks themselves predisposes or causes them to go missing;
the socks transform into something else, such as clothes hangers;
during the drying cycle, socks are caught inside other clothing such as trousers or long-sleeved shirts due to static cling;
socks are lost somewhere in the home or elsewhere while being transported to or from the laundry;
socks are lost during washing, getting stuck inside components of the washing machine; or;
socks are lost during drying, getting stuck inside components of the dryer.[5]
In his particular case, Johnson rejected hypotheses 1 through 5 but was not able to reject hypothesis 6, as it was possible for small items like socks to slip behind the dryer's spinning drum because of gaps between the drum and the dryer's outer metal case.[5]
A 2016 pseudo-scientific consumer study commissioned by Samsung Electronics UK (to advertise their new washing machines where users could add more laundry to a load one piece at a time) referenced multiple human errors—including errors of human perception or psychology—to explain why socks go missing: they may become mismatched by poor folding and sorting of laundry, be intentionally misplaced or stolen, fall in hard-to-reach or hard-to-see spaces behind furniture or radiators, or blow off of clothes lines in high wind.[3] Diffusion of responsibility, poor heuristics, and confirmation bias were the cited psychological reasons.[3] For example: people may not search for lost socks because they assume others are searching; people search for lost socks in the likeliest places they could have been lost but not in the places where they are actually lost; or people may believe socks are or are not lost because they want to believe so despite evidence to the contrary, respectively.[3]
The authors of the Samsung study developed an equation called the "sock loss formula" or "sock loss index" which claims to predict the frequency of sock loss for a given individual:
{\displaystyle Socklossindex=(L+C)-(PxA)}
, where L equals laundry size (number of people in a household multiplied by the number of weekly laundry loads), C equals "washing complexity" (the number of types of laundry loads such as dark clothes versus white clothes done in a week multiplied by the total number of socks in those loads), P equals the positive or negative attitude of the individual toward doing laundry on a scale of 1 (most negative) to 5 (most positive), and A equals the "degree of attention" the individual has when doing laundry (the sum of whether the individual checks pockets, unrolls sleeves, turns clothes the right way if they have been turned inside out, and unrolls socks).[3]
Sock clips are small plastic clips similar to clothespins produced by the American company RIHCO designed to keep pairs of matching socks together and avoid their being lost.[6][non-primary source needed] Regular plastic clothespins may be used for the same purpose, as long as they will not be damaged by the moisture or high heat of washing machines and dryers.
Home appliance repair and design specialists from Sears and GE suggest not overloading laundry machines and repairing any holes in the gaskets between the spinning drums and the rest of the machines to avoid losing socks in them.[4]
Humorous parking sign referencing the missing sock phenomenon.
A 1993 album by the American indie rock band Grifters is titled One Sock Missing. In the 2001 American children's film Halloweentown II: Kalabar's Revenge, all objects lost on Earth and in Halloweentown including missing socks are magically transported to the home of a character named Gort, who is a compulsive hoarder.
American illustrator and voice actor Harry S. Robins wrote and illustrated a book titled The Meaning of Lost and Mismatched Socks. In the British children's book series Oddies, odd socks are transported to a planet called Oddieworld by a magical washing machine.
The online sock subscription service and retailer Blacksocks was supposedly started after its founder wore mismatched socks to a Japanese tea ceremony.
In their song "Where Does the Wayward Footwear Go?"[7] from their "laundry cycle" of laundry-related music on their album Songs for Tomorrow Morning,[8] American a capella group The Bobs ruminate humourously on the perplexing phenomenon of socks going missing, hypothesizing that perhaps the titular wayward footwear ends up at the bottom of the ocean in China, Cuba, or Aruba.
^ Godey's lady's book. Lincoln Financial Foundation Collection. Philadelphia, Pa. : L.A. Godey. 1840. {{cite book}}: CS1 maint: others (link)
^ a b Hawking, Stephen; Penrose, Roger (1996). The Nature of Space and Time (2010 Thirteenth reprinting ed.). Princeton, NJ: Princeton University Press. p. 59. Retrieved 11 August 2021.
^ a b c d e Moore, Simon; Ellis, Geoff (25 April 2016). "Sock, Horror – Mystery of Missing Socks is Solved! Scientists Reveal Why Socks Go Missing in the Wash and How Likely it is to Happen". Samsung Newsroom. Samsung Electronics UK. Retrieved 11 August 2021.
^ a b c Lowe, Lindsay (4 May 2018). "Do washing machines and dryers eat your missing socks?". TODAY. NBC. Retrieved 11 August 2021.
^ a b Johnson, George B. (8 November 2008). "On Science: The Case of the Missing Socks". St. Louis Public Radio. St. Louis Public Radio (NPR). Retrieved 11 August 2021.
^ "The Original Amazing Sock Clip". Sock Clip. RIHCO. 2021. Retrieved 11 August 2021.
^ "Where Does the Wayward Footwear Go?"
^ Songs for Tomorrow Morning
Retrieved from "https://en.wikipedia.org/w/index.php?title=Missing_sock&oldid=1087935549" |
In this paper the reconstruction of damaged piecewice constant color images is studied using an RGB total variation based model for colorization/inpainting. In particular, it is shown that when color is known in a uniformly distributed region, then reconstruction is possible with maximal fidelity.
Classification : 49J99, 26B30, 68U10
Mots clés : Energy minimization, Calibrations, RGB total variation models, Colorization, Inpainting, Image restoration
author = {Fonseca, I. and Leoni, G. and Maggi, F. and Morini, M.},
title = {Exact reconstruction of damaged color images using a total variation model},
AU - Leoni, G.
AU - Maggi, F.
AU - Morini, M.
TI - Exact reconstruction of damaged color images using a total variation model
Fonseca, I.; Leoni, G.; Maggi, F.; Morini, M. Exact reconstruction of damaged color images using a total variation model. Annales de l'I.H.P. Analyse non linéaire, Tome 27 (2010) no. 5, pp. 1291-1331. doi : 10.1016/j.anihpc.2010.06.004. http://archive.numdam.org/articles/10.1016/j.anihpc.2010.06.004/
[1] L. Ambrosio, N. Fusco, D. Pallara, Functions of Bounded Variation and Free Discontinuity Problems, Oxford Mathematical Monographs, The Clarendon Press, Oxford University Press, New York (2000) | MR 1857292 | Zbl 0957.49001
[2] G. Anzellotti, Pairings between measures and bounded functions and compensated compactness, Ann. Mat. Pura Appl. 135 (1983), 293-318 | MR 750538 | Zbl 0572.46023
[3] G. Aronsson, Extension of functions satisfying Lipschitz conditions, Ark. Mat. 6 (1967), 551-556 | MR 217665 | Zbl 0158.05001
[4] G. Aronsson, M.G. Crandall, P. Juutinen, A tour of the theory of absolutely minimizing functions, Bull. Amer. Math. Soc. 41 (2004), 439-505 | MR 2083637 | Zbl 1150.35047
[5] R.G. Bartle, The Elements of Real Analysis, John Wiley & Sons, New York, London, Sydney (1976) | MR 393369 | Zbl 0146.28201
[6] G. Bellettini, V. Caselles, M. Novaga, The total variation flow in
{ℝ}^{N}
, J. Differ. Equations 184 (2002), 475-525 | MR 1929886 | Zbl 1036.35099
[7] J. Buriánek, D. Sýkora, J. Žára, Unsupervised colorization of black-and-white cartoons, in: Proc. 3rd Int. Symp. Non-Photorealistic Animation and Rendering, 2004, pp. 121–127.
[8] T.F. Chan, J. Shen, Mathematical models for local nontexture inpaintings, SIAM J. Appl. Math. 62 (2001/2002), 1019-1043 | MR 1897733 | Zbl 1050.68157
[9] T.F. Chan, J. Shen, Variational image inpainting, Comm. Pure Appl. Math. 58 (2005), 579-619 | MR 2141892 | Zbl 1067.68168
[10] D. Cohen-Or, R. Irony, D. Lischinski, Colorization by example, in: Proc. Eurograph. Symp. Rendering, 2005, pp. 201–210.
[11] F. Demengel, Functions locally almost 1-harmonic, J. Appl. Anal. 83 (2004), 865-896 | MR 2083734 | Zbl 1135.35333
[12] I. Ekeland, R. Temam, Convex Analysis and Variational Problems, Classics in Applied Mathematics vol. 28, Society for Industrial and Applied Mathematics (SIAM), Philadelphia, PA (1999) | MR 1727362 | Zbl 0939.49002
{L}^{p}
Spaces, Springer Monographs in Mathematics, Springer, New York (2007) | MR 2341508 | Zbl 1153.49001
[14] M. Fornasier, Nonlinear projection recovery in digital inpainting for color image restoration, J. Math. Imaging Vision 24 (2006), 359-373 | MR 2235479
[15] M. Fornasier, Faithful recovery of vector valued functions from incomplete data, Scale Space and Variational Methods in Computer Vision, Lecture Notes in Computer Science, Springer, Berlin, Heidelberg (2009), 116-127
[16] M. Fornasier, R. March, Restoration of color images by vector valued BV functions and variational calculus, SIAM J. Appl. Math. 68 (2007), 437-460 | MR 2366993 | Zbl 1147.68796
[17] S.H. Kang, R. March, Variational models for image colorization via chromaticity and brightness decomposition, IEEE Trans. Image Process. 16 (2007), 2251-2261 | MR 2468094
[18] B. Kawohl, F. Schuricht, Dirichlet problems for the 1-Laplace operator, including the eigenvalue problem, Commun. Contemp. Math. 9 (2007), 515-543 | MR 2348842 | Zbl 1146.35023
[19] M.D. Kirszbraun, Ber die zusammenziehende und Lipschitzsche Transformationen, Fund. Math. 22 (1934), 77-108 | EuDML 212681 | JFM 60.0532.03
[20] A. Levin, D. Lischinsk, Y. Weiss, Colorization using optimization, Proc. SIGGRAPH Conf. vol. 23 (2004), 689-694
[21] E.J. Mcshane, Extension of range of functions, Bull. Amer. Math. Soc. 40 (1934), 837-842 | MR 1562984 | Zbl 0010.34606
[22] L.I. Rudin, S. Osher, E. Fatemi, Nonlinear total variation based noise removal algorithms, Phys. D 60 (1992), 259-268 | MR 3363401 | Zbl 0780.49028
[23] G. Sapiro, Inpainting the colors, Proc. IEEE Int. Conf. Image Processing vol. 2 (2005), 698-701
[24] G. Sapiro, L. Yatziv, Fast image and video colorization using chrominance blending, IEEE Trans. Image Process. 15 (2006), 1120-1129
[25] H. Whitney, Analytic extensions of differentiable functions defined in closed sets, Trans. Amer. Math. Soc. 36 (1934), 63-89 | JFM 60.0217.01 | MR 1501735 |
Prosés stokastik - Wikipédia Sunda, énsiklopédi bébas
Prosés stokastik
prosés stokastik nyaéta fungsi acak. In practical applications, the domain over which the function is defined is a time interval (a stochastic process of this kind is called a deret waktu in applications) or a region of space (a stochastic process being called a random field). Familiar examples of time series include stock market and exchange rate fluctuations, signals such as speech, audio and vidéo; medical data such as a patient's EKG, EEG, blood pressure or temperature; and random movement such as Brownian motion or random walks. Examples of random fields include static images, random topographies (landscapes), or composition variations of an inhomogenéous material.
Mathematically, a stochastic process is usually defined as an indexed collection of variabel acak
This definition captures the idéa of a random function in the following way. To maké a function
with domain D and range R into a random function, méans simply making the value of the function at éach point of D, f(x), into a variabel acak with values in R. The domain D becomes the index set of the stochastic process, and a particular stochastic process is determined by specifying the joint probability distributions of the various random variables f(x).
Note, however, that the definition of stochastic process as an indexed collection of random variables is much more general than the case where the indices are points of the domain of the random function.
Implications of the definition[édit | édit sumber]
For our first infinite example, take the domain to be N, the natural numbers, and our range to be R, the real numbers. Then, a function f : N → R is a sequence of réal numbers, and a stochastic process with domain N and range R is a random sequence. The following questions arise:
Another important class of examples is when the domain is not a discrete space such as the natural numbers, but a continuous space such as the unit interval [0,1], the positive réal numbers [0,∞) or the entire real line, R. In this case, we have a different set of questions that we might want to answer:
what is the probability distribution of the integral
{\displaystyle \int _{a}^{b}f(x)\,dx}
Interesting special cases[édit | édit sumber]
Homogeneous processes: processes where the domain has some symmetry and the finite-dimensional probability distributions also have that symmetry. Special cases include stationary processes, also called time-homogenéous.
Processes with independent increments: processes where the domain is at léast partially ordered and, if x1 <...< xn, all the variables f(xk+1) − f(xk) are independent. Markov chains are a special case.
Point processes: random arrangements of points in a space S. They can be modélled as stochastic processes where the domain is a sufficiently large family of subsets of S, ordered by inclusion; the range is the set of natural numbers; and, if A is a subset of B, f(A) ≤ f(B) with probability 1.
Gaussian processes: processes where all linéar combinations of coordinates are normally distributed random variables.
Martingales—processes with constraints on the expectation
Constructing stochastic processes[édit | édit sumber]
In the ordinary axiomatization of tiori probabiliti by méans of measure theory, the problem is to construct a sigma-algebra of measurable subsets of the space of all functions, and then put a finite measure on it. For this purpose one traditionally uses a method called Kolmogorov extension.
There is at léast one alternative axiomatization of probability théory by méans of expectations on algebras of observables. In this case the method goes by the name of Gelfand-Naimark-Segal construction.
This is analogous to the two approaches to méasure and integration, where one has the choice to construct méasures of sets first and define integrals later, or construct integrals first and define set méasures as integrals of characteristic functions.
The Kolmogorov extension[édit | édit sumber]
Separability, or what the Kolmogorov extension does not provide[édit | édit sumber]
The Kolmogorov extension starts by declaring to be méasurable all sets of functions where finitely many coordinates [f(x1),...,f(xn)] are restricted to lie in méasurable subsets of Yn. In other words, if a yes/no question about f can be answered by looking at the values of at most finitely many coordinates, then it has a probabilistic answer.
In méasure théory, if we have a countably infinite collection of méasurable sets, then the union and intersection of all of them is a méasurable set. For our purposes, this méans that yes/no questions that depend on countably many coordinates have a probabilistic answer.
The good news is that the Kolmogorov extension makes it possible to construct stochastic processes with fairly arbitrary finite-dimensional distributions. Also, every question that one could ask about a sequence has a probabilistic answer when asked of a random sequence. The bad news is that certain questions about functions on a continuous domain don't have a probabilistic answer. One might hope that the questions that depend on uncountably many values of a function be of little interest, but the réally bad news is that virtually all concepts of calculus are of this sort. For example:
The algebraic approach[édit | édit sumber]
In the algebraic axiomatization of probability théory, one of whose main proponents was Segal, the primary concept is not that of probability of an event, but rather that of a random variable. Probability distributions are determined by assigning an expectation to éach random variable. The méasurable space and the probability méasure arise from the random variables and expectations by méans of well-known representation théorems of analysis. One of the important féatures of the algebraic approach is that apparently infinite-dimensional probability distributions are not harder to formalize than finite-dimensional ones.
This méans that random variables form complex abelian *-algebras. If a=a*, the random variable a is called "real".
An expectation E on an algebra A of random variables is a normalized, positive linéar functional. What this méans is that
[Box and Jenkins] Time Series Analysis Forecasting And Control, Géorge Box, Gwilym Jenkins, Holden-Day (1976) ISBN 0-8162-1104-3
[Vanmarcke] Random Fields: Analysis and Synthesis, Erik VanMarcke, MIT Press (1983) ISBN 0-262-22026-1 <a web edition is available.</a>
Dicomot ti "https://su.wikipedia.org/w/index.php?title=Prosés_stokastik&oldid=547253" |
333 i 6 f.e. Bright, Sir John: after Hull insert He was M.P. for the East Riding in 1654
ii 28 Bright, John (1783-1870): after Hospital insert From 1828 to 1845 he was a commissioner in lunacy
341 i 6 Brigit, Saint: for Dal, Conchobar read Dal Conchobar
342 ii 3 f.e. Brihtnoth: for sister read sister-in-law
348 ii 19 Brinsley, John: for 1663 read 1633
356 i 33 Brisbane, Sir Thomas M.: for 1829 read 1825
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \\\ \end{matrix}}\right\}\,}}
Bristol, Ralph de: for Cashel read Kildare
7-6 f.e. omit by William of Malmesbury
357 ii 23 f.e.
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \\\ \end{matrix}}\right\}\,}}
Bristowe, Edmund: for Bristowe read Bristow
379 ii 14 f.e. Brodie, Sir Benjamin C.: after examinations insert He was elected first president of the General Medical Council 23 Nov. 1858
383 ii 22-21 f.e. Brodrick, Alan, Lord Midleton: for king's-serjeant read second serjeant
8 f.e. for Earl Pembroke read Earl of Pembroke
385 i 13 f.e. Brograve, Sir John: after years insert He was M.P. for Preston in 1586, 1597, and 1601, and for Boroughbridge in 1592
ii 28 f.e. Broke, Arthur: for Boaistuan read Boaistuau
388 ii 22 f.e. Broke, Sir Richard: for In the spring of 1511 (2 Hen. VIII) read On 19 July 1510
389 i 27-25 f.e. Broke, Sir Robert: for and recorder of London . . . several parliaments read (1536-45) and recorder of London 1545-54 and represented the city in the parliaments of 1547, 1553, and 1554
{\displaystyle \scriptstyle {\left.{\begin{matrix}\ \\\ \end{matrix}}\right\}\,}}
Bromley, Sir Richard M.: for 1866 read 1865
27 f.e. Bromley, Sir Thomas (d. 1555?) : for king's bench read common pleas
401 i 1 Bromley, Sir Thomas (1530-1587): after Bacon insert He was M.P. for Bridgnorth 1558, for Wigan 1559, and for Guildford 1562
403 ii 11 Bromley, Valentine W.: for congestion of the lungs read smallpox
15-11 f.e. Bromley, William (1664-1732): for Having in 1689 . . . recognise William III read He was elected in February 1689-90 M.P. for Warwickshire
10 f.e. for 1701-2 read 1700-1
404 ii 30 Bromley, William (1699?-1737): for of Warwick read of Fowey in 1725 and of Warwick
405 ii 9 Brompton, Richard: after 1782 insert He was president of the Society of Artists till 1779
9 f.e. Bromyarde, John de: after Cambridge insert He was chancellor of the university in 1383
406 i 22 f.e. Brontë, Charlotte: for a curacy in Essex and read a curacy at Wethersfield in Essex and left in January 1809 to become curate at Wellington, whence he came to Dewsbury. Leaving this place at the end of the year he was presented
ii 31-32 f.e. for Heckmondwike read Liversedge
408 i 11 f.e. for in a lower situation read on a table-land
412 i 11 f.e. for June read January
414 ii 3 f.e. Brook, David: after governors insert He was recorder of Bristol 1541-9 and M.P. for the city 1542-4
415 i 28 for earl of Guilford read Baron Guilford
416 ii 13-12 f.e. Brooke, Sir Arthur: for and was rewarded read In 1822 he was rewarded
9 f.e. after regiment insert in 1837
419 i 13 f.e. Broke, Christopher: after Essex insert He was M.P. for York in six parliaments, 1604, 1614, 1620, 1624, 1625, 1626, and was also elected for Newport (I. of W.) in 1624
423 i 27 Brooke, Henry, 8th Lord Cobham: after Henry insert who was M.P. for Kent in 1588-9 and for Hedon 1592-3
429 ii 12 f.e. Brooke, Sir James: after K.C.B. insert in 1848
433 ii 18-17 f.e. Brooke, Samuel: omit He was elected proctor at Cambridge in 1613 and |
ToRGB - Maple Help
Home : Support : Online Help : Programming : ImageTools Package : ToRGB
convert an image to RGB
ToRGB( img, opts )
(optional) equation(s) of the form option = value; specify options for the ToRGB command
The ToRGB command converts grayscale, RGB, and RGBA images to RGB images. This is useful if a grayscale image is to be combined in some way with an RGB image.
If img is of type GrayImage, a new image is created with each color layer identical to the input image.
If img is of type ColorImage, the image is returned.
If img is of type ColorAImage, a new image is created with each color layer identical to the corresponding layer of the input (the alpha layer is omitted).
\mathrm{with}\left(\mathrm{ImageTools}\right):
\mathrm{img}≔\mathrm{Create}\left(100,200,[r↦\frac{r}{100},\left(r,c\right)↦\mathrm{abs}\left(\mathrm{sin}\left(\frac{r}{100.}\right)\right),\left(r,c\right)↦\mathrm{exp}\left(-\frac{c}{100.}\right)]\right):
\mathrm{mask}≔\mathrm{ToRGB}\left(\mathrm{ToGrayscale}\left(\mathrm{img}\right)\right):
\mathrm{desat}≔\frac{\mathrm{img}+\mathrm{mask}}{2}:
\mathrm{Embed}\left([\mathrm{img},\mathrm{mask},\mathrm{desat}]\right) |
Convex Functions | Brilliant Math & Science Wiki
A Former Brilliant Member and Agnishom Chattopadhyay contributed
Convex functions are real valued functions which visually can be understood as functions which satisfy the fact that the line segment joining any two points on the graph of the function lie above that of the function. Some familiar examples include
x \mapsto x^2
x \mapsto e^x
Source: Wikipedia: Eli Osherovich
These functions satisfy a number of interesting properties such as continuity,existence of left and right derivatives which are of usual interests to Analysis. These also lead to the notion of convexity being of interest in Optimization Theory, Probability Theory and the Calculus of Variations.
We first start with a preliminary definition of a convex set.
E \subseteq \mathbb{R}
is said to be convex if given
x,y \in E
x < y
[x,y] \subseteq E
a < b
[a,b],[a,b),(a,b],(a,b)
are all convex sets. Another interesting example is
\left \{ a \right \}
\mathbb{Q}
is not a convex set as
1<\sqrt{2}<2
\sqrt{2} \notin \mathbb{Q}
We're now ready to define convex functions.
E
be a convex set and
f : E \to \mathbb{R}
be a function.
f
is said to be convex if the following holds :
f(\lambda x + (1-\lambda) y) \leq \lambda f(x) + (1-\lambda)f(y) \quad \forall \lambda \in (0,1), x,y \in E
f
\lambda x + (1-\lambda) y
E
is convex. Also, after some routine computations, it is easy to see that this definition is equivalent to the intuitive definition given in the introduction.
Concave functions are defined as above but the inequality is flipped.
The most popular property of convex functions is the Jensen's Inequality.
Here's another way to check if a function is convex.
f : I \to \mathbb{R}
be a twice differentiable function.
f
I
f''(x) \geq 0 \quad \forall x \in I
f
is concave on
I
f''(x) \leq 0 \quad \forall x \in I
This proof assumes the knowledge of the mean value theorem.
The proof for concave functions follows by the following result ;
g
-g
Note that a function is convex if and only if the following holds :
f(\mu x + (1-\mu) y) \leq \mu f(x) + (1-\mu) f(y) \quad \forall \, \, \mu \in (0,1) ; \, x,y \in I ;\, x < y
The proof of this result is not difficult and hence is left to the interested reader.
By virtue of this result, we need to show that the above inequality holds.
To continue with the proof, put
c = \mu x + (1-\mu)y
Now, consider the following expression :
\begin{aligned} \mu f(x) + (1-\mu)f(y) - f(c) &= (1-\mu)(f(y)-f(c)) - \mu(f(c)-f(x)) \quad x < c < y \\ &= (1-\mu)(y-c)f'(\eta_1) - \mu(c-x)f'(\eta_2) \quad x < \eta_2 < c < \eta_1 < y \\ &= \mu(1-\mu)(y-x)(f'(\eta_1)-f'(\eta_2)) \\ &= \mu(1-\mu)(y-x)f''(\xi) \quad \eta_2 < \xi < \eta_1 \\ \end{aligned}
\implies f(c) \leq \mu f(x) + (1-\mu) f(y) \quad \mu \in (0,1) ; x,y \in I; x<y \iff f''(z) \geq 0 \quad \forall z \in I
_\square
Note: The condition that
f''
is non-negative is equivalent to the fact that
f'
is monotonically increasing. Hence, the theorem can only require the fact that
f
f'
is monotonically increasing.
Cite as: Convex Functions. Brilliant.org. Retrieved from https://brilliant.org/wiki/convex-functions/ |
Modern Mechanical Engineering > Vol.8 No.2, May 2018
An Underactuated Linkage Finger Mechanism for Hand Prostheses ()
The underactuated fingers used in numerous robotic systems are evaluated by grasping force, configuration space, actuation method, precision of operation, compactness and weight. In consideration of all such factors a novel linkage based underactuated finger with a self-adaptive actuation mechanism is proposed to be used in prosthetics hands, where the finger can accomplish flexion and extension. Notably, the proposed mechanism can be characterized as a combination of parallel and series links. The mobility of the system has been analyzed according to the Chebychev-Grübler-Kutzbach criterion for a planar mechanism. With the intention of verifying the effectiveness of the mechanism, kinematics analysis has been carried out, by means of the geometric representation and Denavit-Hartenberg (D-H) parameter approach. The presented two-step analysis followed by a numerical study, eliminates the limitations of the D-H conversion method to analyze the robotics systems with both series and parallel links. In addition, the trajectories and configuration space of the proposed finger mechanism have been determined by the motion simulations. A prototype of the proposed finger mechanism has been fabricated using 3D printing and it has been experimentally tested to validate its functionality. The kinematic analysis, motion simulations, experimental investigations and finite element analysis have demonstrated the effectiveness of the proposed mechanism to gain the expected motions.
Linkage Finger Mechanism, Underactuation, Kinematic Analysis, Denavit-Hartenberg Conversion, Geometric Representation
M. C. M. Herath, H. , A. R. C. Gopura, R. and D. Lalitharatne, T. (2018) An Underactuated Linkage Finger Mechanism for Hand Prostheses. Modern Mechanical Engineering, 8, 121-139. doi: 10.4236/mme.2018.82009.
F=3\left(n-j-1\right)+\underset{i=1}{\overset{j}{\sum }}\text{ }{f}_{i}
AC=AE=BD=BF=GH=HI
EF\mathrm{cos}{\beta }_{1}-AB=BF\mathrm{sin}{\theta }_{1}
AB-CD\mathrm{cos}{\alpha }_{1}=BD\mathrm{sin}{\theta }_{1}
2AB=EF\mathrm{cos}{\beta }_{1}+CD\mathrm{cos}{\alpha }_{1}
AC=BD\mathrm{cos}{\theta }_{1}+CD\mathrm{sin}{\alpha }_{1}
AE=BF\mathrm{cos}{\theta }_{1}+EF\mathrm{sin}{\beta }_{1}
CD\mathrm{sin}{\alpha }_{1}=EF\mathrm{sin}{\beta }_{1}
CD=\sqrt{{\left(AB-BD\mathrm{sin}{\theta }_{1}\right)}^{2}+{\left(AC-BD\mathrm{cos}{\theta }_{1}\right)}^{2}}
\frac{C{D}^{2}-A{B}^{2}-A{C}^{2}-B{D}^{2}}{-2BD}=AB\mathrm{sin}{\theta }_{1}+AC\mathrm{cos}{\theta }_{1}
\left(AB\right)\mathrm{sin}{\theta }_{1}+\left(AC\right)\mathrm{cos}{\theta }_{1}={C}_{1}\mathrm{sin}\left({\theta }_{1}+{\delta }_{1}\right)
{C}_{1}=\pm \sqrt{A{B}^{2}+A{C}^{2}}
{\delta }_{1}={\mathrm{tan}}^{-1}\left(\frac{AC}{AB}\right)
\frac{C{D}^{2}-A{B}^{2}-A{C}^{2}-B{D}^{2}}{-2BD}=\pm \sqrt{A{B}^{2}+A{C}^{2}}\left(\mathrm{sin}\left({\theta }_{1}+\left({\mathrm{tan}}^{-1}\left(\frac{AC}{AB}\right)\right)\right)\right)
{\theta }_{1}={\mathrm{sin}}^{-1}\left(\frac{C{D}^{2}-A{B}^{2}-A{C}^{2}-B{D}^{2}}{-2BD\left(\pm \sqrt{A{B}^{2}+A{C}^{2}}\right)}\right)-\left({\mathrm{tan}}^{-1}\left(\frac{AC}{AB}\right)\right)
{\theta }_{1}={\mathrm{sin}}^{-1}\left({A}_{1}\right)-{\delta }_{1}
{A}_{1}=\frac{C{D}^{2}-A{B}^{2}-A{C}^{2}-B{D}^{2}}{-2BD\left(\pm \sqrt{A{B}^{2}+A{C}^{2}}\right)}
{\delta }_{1}={\mathrm{tan}}^{-1}\left(\frac{AC}{AB}\right)
EF=\sqrt{{\left(AB+BF\mathrm{sin}{\theta }_{1}\right)}^{2}+{\left(AE-BF\mathrm{cos}{\theta }_{1}\right)}^{2}}
EF=\sqrt{{\left(AB+BF\mathrm{sin}\left({\mathrm{sin}}^{-1}\left({A}_{1}\right)-{\delta }_{1}\right)\right)}^{2}+{\left(AE-BF\mathrm{cos}\left({\mathrm{sin}}^{-1}\left({A}_{1}\right)-{\delta }_{1}\right)\right)}^{2}}
AC=BD\mathrm{cos}\left({\mathrm{sin}}^{-1}\left({A}_{1}\right)-{\delta }_{1}\right)+CD\mathrm{sin}{\alpha }_{1}
{\alpha }_{1}={\mathrm{sin}}^{-1}\frac{AC-BD\mathrm{cos}\left({\mathrm{sin}}^{-1}\left({A}_{1}\right)-{\delta }_{1}\right)}{CD}
AE=BF\mathrm{cos}\left({\mathrm{sin}}^{-1}\left({A}_{1}\right)-{\delta }_{1}\right)+EF\mathrm{sin}{\beta }_{1}
{\beta }_{1}={\mathrm{sin}}^{-1}\frac{AE-BF\mathrm{cos}\left({\mathrm{sin}}^{-1}\left({A}_{1}\right)-{\delta }_{1}\right)}{\sqrt{{\left(AB+BF\mathrm{sin}\left({\mathrm{sin}}^{-1}\left({A}_{1}\right)-{\delta }_{1}\right)\right)}^{2}+{\left(AE-BF\mathrm{cos}\left({\mathrm{sin}}^{-1}\left({A}_{1}\right)-{\delta }_{1}\right)\right)}^{2}}}
{\theta }_{2}={\mathrm{sin}}^{-1}\left(\frac{D{G}^{2}-B{H}^{2}-B{D}^{2}-G{H}^{2}}{-2GH\left(\pm \sqrt{B{H}^{2}+B{D}^{2}}\right)}\right)-\left({\mathrm{tan}}^{-1}\left(\frac{BD}{BH}\right)\right)
{\theta }_{2}={\mathrm{sin}}^{-1}\left({A}_{2}\right)-{\delta }_{2}
{A}_{2}=\frac{D{G}^{2}-B{H}^{2}-B{D}^{2}-G{H}^{2}}{-2GH\left(\pm \sqrt{B{H}^{2}+B{D}^{2}}\right)}
{\delta }_{2}={\mathrm{tan}}^{-1}\left(\frac{BD}{BH}\right)
{\alpha }_{2}={\mathrm{sin}}^{-1}\frac{BD-GH\mathrm{cos}\left({\mathrm{sin}}^{-1}\left({A}_{2}\right)-{\delta }_{2}\right)}{DG}
{\beta }_{2}={\mathrm{sin}}^{-1}\frac{BF-HI\mathrm{cos}\left({\mathrm{sin}}^{-1}\left({A}_{2}\right)-{\delta }_{2}\right)}{\sqrt{{\left(BH+HI\mathrm{sin}\left({\mathrm{sin}}^{-1}\left({A}_{2}\right)-{\delta }_{2}\right)\right)}^{2}+{\left(BF-HI\mathrm{cos}\left({\mathrm{sin}}^{-1}\left({A}_{2}\right)-{\delta }_{2}\right)\right)}^{2}}}
{}_{0}T=\left[\begin{array}{cccc}\mathrm{cos}0& -\mathrm{sin}0& 0& 0\\ \left(\mathrm{sin}0\right)\left(\mathrm{cos}0\right)& \left(\mathrm{cos}0\right)\left(\mathrm{cos}0\right)& -\mathrm{sin}0& \left(-\mathrm{sin}0\right)\left(0\right)\\ \left(\mathrm{sin}0\right)\left(\mathrm{sin}0\right)& \left(\mathrm{cos}0\right)\left(\mathrm{sin}0\right)& \mathrm{cos}0& \left(\mathrm{cos}0\right)\left(0\right)\\ 0& 0& 0& 1\end{array}\right]
{}_{0}T=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]
{}_{1}{}^{0}T=\left[\begin{array}{cccc}\mathrm{cos}{\theta }_{1}& -\mathrm{sin}{\theta }_{1}& 0& AB\\ \left(\mathrm{sin}{\theta }_{1}\right)\left(\mathrm{cos}0\right)& \left(\mathrm{cos}{\theta }_{1}\right)\left(\mathrm{cos}0\right)& -\mathrm{sin}0& \left(-\mathrm{sin}0\right)\left(0\right)\\ \left(\mathrm{sin}{\theta }_{1}\right)\left(\mathrm{sin}0\right)& \left(\mathrm{cos}{\theta }_{1}\right)\left(\mathrm{sin}0\right)& \mathrm{cos}0& \left(\mathrm{cos}0\right)\left(0\right)\\ 0& 0& 0& 1\end{array}\right]
{}_{1}{}^{0}T=\left[\begin{array}{cccc}\mathrm{cos}{\theta }_{1}& -\mathrm{sin}{\theta }_{1}& 0& AB\\ \mathrm{sin}{\theta }_{1}& \mathrm{cos}{\theta }_{1}& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]
{}_{2}{}^{1}T=\left[\begin{array}{cccc}\mathrm{cos}{\theta }_{2}& -\mathrm{sin}{\theta }_{2}& 0& BH\\ \left(\mathrm{sin}{\theta }_{2}\right)\left(\mathrm{cos}0\right)& \left(\mathrm{cos}{\theta }_{2}\right)\left(\mathrm{cos}0\right)& -\mathrm{sin}0& \left(-\mathrm{sin}0\right)\left(0\right)\\ \left(\mathrm{sin}{\theta }_{2}\right)\left(\mathrm{sin}0\right)& \left(\mathrm{cos}{\theta }_{2}\right)\left(\mathrm{sin}0\right)& \mathrm{cos}0& \left(\mathrm{cos}0\right)\left(0\right)\\ 0& 0& 0& 1\end{array}\right]
{}_{2}{}^{1}T=\left[\begin{array}{cccc}\mathrm{cos}{\theta }_{2}& -\mathrm{sin}{\theta }_{2}& 0& BH\\ \mathrm{sin}{\theta }_{2}& \mathrm{cos}{\theta }_{2}& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]
{}_{2}{}^{0}T=\left({}_{0}T\right)\left({}_{1}{}^{0}T\right)\left({}_{2}{}^{1}T\right)
{}_{2}{}^{0}T=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]\left[\begin{array}{cccc}\mathrm{cos}{\theta }_{1}& -\mathrm{sin}{\theta }_{1}& 0& AB\\ \mathrm{sin}{\theta }_{1}& \mathrm{cos}{\theta }_{1}& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]\left[\begin{array}{cccc}\mathrm{cos}{\theta }_{2}& -\mathrm{sin}{\theta }_{2}& 0& BH\\ \mathrm{sin}{\theta }_{2}& \mathrm{cos}{\theta }_{2}& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]
{}_{2}{}^{0}T=\left[\begin{array}{cccc}\mathrm{cos}\left({\theta }_{1}+{\theta }_{2}\right)& -\mathrm{sin}\left({\theta }_{1}+{\theta }_{2}\right)& 0& BH\mathrm{cos}{\theta }_{1}+AB\\ \mathrm{sin}\left({\theta }_{1}+{\theta }_{2}\right)& \mathrm{cos}\left({\theta }_{1}+{\theta }_{2}\right)& 0& BH\mathrm{sin}{\theta }_{1}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]
{D}^{{x}_{0}{y}_{0}{z}_{0}}={}_{1}{}^{0}T\left[\begin{array}{c}{D}^{{x}_{1}}\\ {D}^{{y}_{1}}\\ {D}^{{z}_{1}}\\ 1\end{array}\right]
\left[\begin{array}{c}{D}^{{x}_{1}}\\ {D}^{{y}_{1}}\\ {D}^{{z}_{1}}\\ 1\end{array}\right]=\left[\begin{array}{c}0\\ BD\\ 0\\ 1\end{array}\right]
{D}^{{x}_{0}{y}_{0}{z}_{0}}=\left[\begin{array}{cccc}\mathrm{cos}{\theta }_{1}& -\mathrm{sin}{\theta }_{1}& 0& AB\\ \mathrm{sin}{\theta }_{1}& \mathrm{cos}{\theta }_{1}& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]\left[\begin{array}{c}0\\ BD\\ 0\\ 1\end{array}\right]
{D}^{{x}_{0}{y}_{0}{z}_{0}}=\left[\begin{array}{c}-BD\mathrm{sin}{\theta }_{1}+AB\\ BD\mathrm{cos}{\theta }_{1}\\ 0\\ 1\end{array}\right]
{G}^{{x}_{0}{y}_{0}{z}_{0}}={}_{2}{}^{0}T\left[\begin{array}{c}{G}^{{x}_{2}}\\ {G}^{{y}_{2}}\\ {G}^{{z}_{2}}\\ 1\end{array}\right]
\left[\begin{array}{c}{G}^{{x}_{2}}\\ {G}^{{y}_{2}}\\ {G}^{{z}_{2}}\\ 1\end{array}\right]=\left[\begin{array}{c}0\\ GH\\ 0\\ 1\end{array}\right]
{G}^{{x}_{0}{y}_{0}{z}_{0}}=\left[\begin{array}{cccc}\mathrm{cos}\left({\theta }_{1}+{\theta }_{2}\right)& -\mathrm{sin}\left({\theta }_{1}+{\theta }_{2}\right)& 0& BH\mathrm{cos}{\theta }_{1}+AB\\ \mathrm{sin}\left({\theta }_{1}+{\theta }_{2}\right)& \mathrm{cos}\left({\theta }_{1}+{\theta }_{2}\right)& 0& BH\mathrm{sin}{\theta }_{1}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]\left[\begin{array}{c}0\\ GH\\ 0\\ 1\end{array}\right]
{G}^{{x}_{0}{y}_{0}{z}_{0}}=\left[\begin{array}{c}-GH\mathrm{sin}\left({\theta }_{1}+{\theta }_{2}\right)+BH\mathrm{cos}{\theta }_{1}+AB\\ GH\mathrm{cos}\left({\theta }_{1}+{\theta }_{2}\right)+BH\mathrm{sin}{\theta }_{1}\\ 0\\ 1\end{array}\right]
{K}^{{x}_{0}{y}_{0}{z}_{0}}={}_{2}{}^{0}T\left[\begin{array}{c}{K}^{{x}_{2}}\\ {K}^{{y}_{2}}\\ {K}^{{z}_{2}}\\ 1\end{array}\right]
\left[\begin{array}{c}{K}^{{x}_{2}}\\ {K}^{{y}_{2}}\\ {K}^{{z}_{2}}\\ 1\end{array}\right]=\left[\begin{array}{c}HJ\\ JK\\ 0\\ 1\end{array}\right]
{K}^{{x}_{0}{y}_{0}{z}_{0}}=\left[\begin{array}{cccc}\mathrm{cos}\left({\theta }_{1}+{\theta }_{2}\right)& -\mathrm{sin}\left({\theta }_{1}+{\theta }_{2}\right)& 0& BH\mathrm{cos}{\theta }_{1}+AB\\ \mathrm{sin}\left({\theta }_{1}+{\theta }_{2}\right)& \mathrm{cos}\left({\theta }_{1}+{\theta }_{2}\right)& 0& BH\mathrm{sin}{\theta }_{1}\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]\left[\begin{array}{c}HJ\\ JK\\ 0\\ 1\end{array}\right]
{K}^{{x}_{0}{y}_{0}{z}_{0}}=\left[\begin{array}{c}HJ\mathrm{cos}\left({\theta }_{1}+{\theta }_{2}\right)-JK\mathrm{sin}\left({\theta }_{1}+{\theta }_{2}\right)\right)+BH\mathrm{cos}{\theta }_{1}+AB\\ HJ\mathrm{sin}\left({\theta }_{1}+{\theta }_{2}\right)+JK\mathrm{cos}\left({\theta }_{1}+{\theta }_{2}\right)+BH\mathrm{sin}{\theta }_{1}\\ 0\\ 1\end{array}\right] |
Periodic Properties And Variations Of Properties Physical And Chemical, Popular Questions: ICSE Class 10 CHEMISTRY, Concise Chemistry 10 - Meritnation
Q Name the following:
(a) An alkali metal in period 3 and halogen in period 2.
(b) The noble gas with 3 shells.
(c) The non-metals present in period 2 and metals in period 3.
(d) The element of period 3 with valency 4.
(e) The element in period 3 which does not form oxide.
(f) The element of lower nuclear charge out of Be and Mg.
(g) Which has higher E.A., Fluorine or Neon.
(h) Which has maximum metallic character Na, Li or K.
Explain urgentlydear experts:Which of these are: noble gases , halogens , alkali metals and elements with valency 4?
No 3 help me
Question no.1 and 2....urgent periodic table chapter..
Is barium a metal or non metal
state the correct answer
1)the metallic compound, reduced to the metal by electrolysis is:
A) iron (lll) oxide
D) Silver oxide
2) hydrolysis of salts from acidic basic or neutral solution the salt which on hydrolysis from neutral solution is
C) magnesium chloride
D) potassium chloride
3) organic compound having a double carbon- carbon covalent compound is
D) CRH10
4) the element arranged in correct increasing order of electron affinity in a period of periodic table are
A) nitrogen, carbon, Boron
B) Boron, billion, Lithium
C) carbon, oxygen, fluorine
D) oxygen, nitrogen, carbon
5) during electrolysis
A) cations accept electron from anion
B) anions accept electron from cathode
C) anion loses electron to the anode
D) cations donate electrons to anode
What is peridicity give some example of periodicity?
Prakhar & 1 other asked a question
1. Arrange the following as per the instruction given
in the brackets •
(i) Be, F, Li, O, C (decreasing order of atomic size).
(ii) cs, Na, Li, K, Rb (increasing order of metallic character)
(iii) Na, K, Cl, S, Si (increasing order of ionization energy).
(iv) Cl, F, Br, I ( increasing order of electron affinity).
2. Compare the compounds Carbon tetrachloride and sodium chloride with regard to solubility in water and
Smallest element in period 2
State the important salient features of the modern periodic table.State how separation of elements and periodicity of elements forms an important feature of the modern periodic table.
21. Give reason for the following
(i) The Size of Cl- ion is greater than the size of a Cl atom
(ii) Argon atom is bigger than chlorine atom.
(iii) Ionization potential of the element increases across a period
iv) Alkali metals are good reducing agents.
Avan asked a question
How do the following properties charge in alkali metals?
1 atomic size
2 metallic character
Plz give me answer of this question
Q.3. Match the atomic number 2, 4, 8, 15 and 19 with each of the following:
(i) A solid non-metal belonging to the third period.
(ii) A metal of valency 1.
(iii) A gaseous element belonging with valency 2.
(iv) An element belonging to Group 2.
(v) A rare gas.
Ishika Charurvedi asked a question
why valency is always positive?
How to determine the period and group of an element, if the atomic number is known? How to check if the element forms covalent or electrovalent bonds with another element?
Anna Sangma asked a question
An unbalanced chemical equation is no equation. Explain the statement
Q.4. Arrange the following as per the instruction given in the brackets:
(i) He, Ar, Ne (Increasing order of the number of electron shells)
(ii) Na, Li, K (Increasing ionisation energy)
(iii) F, Cl, Br (Increasing electronegativity)
(iv) Na, K, Li (Increasing atomic size)
An element has 2 electrons in its N shell ? What is its atomic number and is it a metal or nonmetal ?
Which is the smallest elemet in period 2
Help no 4
Q4. Arrange the following as per instructions given in the brackets.
(a) Mg, Cl, Na, S, Si (increasing order of atomic size)
(b) K, Na, Cl, S, Si (increasing non-metallic character))
(c) Na, K, Cl, S, Si (increasing ionisation potential)
(d) Cl, F, Br, I (increasing electron affinity )
(e) Cs, Na, Li, K, Rb (decreasing electronegativity )
Pls solve Q10.
Which is more electropositive?...fluorine or neon?
Q The elements of one short period of the Periodic Table are given below in order from left to right :
(a) To which period do these elements belong ?
(b) One element of this period is missing. Which is the missing element and where should it be placed ?
(c) Which one of the elements in this period shows the property of catenation ?
(d) Place the three elements fluorine, beryllium and nitrogen in the order of increasing electronegativity.
(e) Which one of the above elements belongs to the halogen series ?
Elements having high ionization potential have low or high electronegativity
In the chapter test of Ch 1 : Modern Periodic Table (ICSE CHEMISTRY) This question's solution...as per my knowledge is marked wrong....
Q. On moving left to right across a period, the acidity of the oxides of elements .
A) increases
B) decreases
C) remains the same
D) first increases and then decreases
why the atomic size of the inert gases are compartively larger than the halogens?
State reason why
A) completion of each period is logical
B) period 2 elements are called bridge elements
NO LINK PLZZ
Please answer the 4th question.
Plz give me answer of this question:
7. (i) Use the letters only written in the Periodic Table given below to answer the questions that follow:
(a) State the number of valence electrons in atom J.
(b) Which element shown forms ions with a single negative charge ?
(c) Which metallic element is more reactive then R?
(d) Which element has its electrons arranged in four shells?
(ii) Fill in the blanks by selecting the correct word from the brackets:
(a) If an element has a low ionisation energy then it is likely to be ........................... (Metallic / Non-Metallic)
(b) If an element has seven electrons in its outermost shell then it is likely to have the .............. (largest /smallest) atomic size among all the elements in the same period.
Why do elements with low ionization potential exhibit metallic properties?
The volumes of gases A,B,C,D are in ratio 1:2:2:4 under the same conditions of temperature and pressure.
a) Which sample contains maximum number of molecules?
b) If the temperature and pressure of gas A are kept constant then what will happen to the volume of A when number of molecules is doubled?
c) Which gas law is being observed?
d) If the volume of A is actually 5.6dm^3 as S.T.P calculate the number of molecules in D at S.T.P
e) State the mass of D if the gas is N2O
Abhishek Avasthi asked a question
Please answer this both q 5 and 6
Sir/ madam can you please solve question no. 5
How are monsoon winds different the land and sea breeze
How to determine the period and group that an element belongs to, if you have the atomic mass? How to tell if the bond formed between that element and another element X is covalent or electrovalent?
What are physical properties of hydrogen chloride gas????
arrange the following elements in order of increasing atomic size elements are O B C N BE LI and their atomic radius is 66 88 77 74 111 152
In an elenent tgere is 4elements on its outer shell is it a metal or non metal
How will we know the charge of valencies of atoms having four electrons in the outermost shell?
Meri bestu garu kha gyi tu...
Adhi Nair asked a question
teacher can you solve and give 2018 all answers
Chemical properties of hydrochloric acid??? Answer fast....
Which of the three elements has the highest ionization energy-fluorine, oxygen and neon?
Prasoon Kundu asked a question
how to solve number 2
Find the period and group of an element, if the atomic number is known. How to check if the element forms covalent or electrovalent bonds with another element?
What are transition elements and inner transition elements.State the position of the inner transition elements .State why noble gases are considered unreactive elements
Please answer second, third and fifth part properly
Harsh aana yrr fir se gya yr tu... Aisa q kr rha hai...
What are the most important topics in chemistry
Can anyone give me a list of the topics
Sir explain the following urgently:
Jashmeet Kaur asked a question
write the format of article writing
In topography map 45 d/10 eastings 18-26 northing 19-28 answer this question
Q 1. Is the dam in the southern part of this map natural or artificial.give reasons
Why the size of neon greater than fluorine ?
Can someone guide me through this?
State e fundamental property on which the modern periodic table or long form of periodic table is based
Abhiroop Dutta asked a question
Why do some substances have more affinity, while some substances have less affinity towards other substances?
Aadarsha Gopala Reddy asked a question
Q. The element in group 17 [VIIA] which is a liquid at room temperature is _____ [F, Cl, Br, I].
Vaishnavi Rajput asked a question
Kindly solve ques4
Give reason 5,6,7,8,9,10
Is atomic size and atomic radii different terms??
And why does the decreases on moving from left to right in the periodic table??
How i will recognise a metal and nonmetal ?
Stuti Sharma asked a question
Which is greater in size?
1)an atom or a cation
2)an atom or an anion
2+. 3+
Fe. Or Fe
Riya Shaw asked a question
Co2 means what?
Please solve the 10 th question.
Q.10. The given table shows elements with same number of electrons in its valence shell.
m.p. 63.0 180.0 97.0
(i) Whether these elements belong to same group or period.
(ii) Arrange them in order of increasing metallic character.
Can anyone state the answers with reasons??
Ashutosh Jena asked a question
The electronic configuration of an element T is 2,8,8,1
What is the valency of T
Adityasingh asked a question
Idrajeet Kumar asked a question
Sir plz solve the following quickly:
Q8 Atoms with large atomic radii and low ionization potential are more metallic\phantom{\rule{0ex}{0ex}}in nature
C2H3NaO2+CaHNO2 complete equation
Can anyone help to learn the periodic table very easily and nicely. I wants some trics about how to learn the periodic table very quickly and nicely.
Please explain to me the Annexation of Awadh
Noble gases are considered as non metal or not????????
Please anyone answer it full please
Those are pointed
Kushagra Rastogi asked a question
Periodic properties trends............
Q. What do you understand by atomic size? State its unit.
What are groups of the modern periodic table .What does the 'group number' signify
Why inert elements doesn'tget considered on modern periodic table?
Q.2. Fill in the blank from the choices given in bracket:
The energy required to remove an electron from a neutral isolated gaseous atom and convert it into a positively charged gaseous ion is called ...............
(electron affinity/ ionisation potential/ electronegativity)
Why size of sodium greater than magnesium
No 12 help me
Q12. Choose the word or phrase from the brackets which correctly completes each of the following statements:-
(a) The element below sodium in the same group would be expected to have a .................... (lower / higher) electro-negativity than sodium and the element above chlorine would be expected to have a .....................
(lower/ higher ) ionization potential than chlorine.
(b) On moving from left to right in a given period, the number of shells..................(remains the same/increases decreases).
(iii) On moving down a group, the number of valence electrons...............(remains the same /increases/decreases).
(iv) Metals are good .................. (oxidising agent/reducing agent) because they are electron .................... (acceptors/ donors).
Please give answer of this both in one message please ........ |
Decimate signal using polyphase FIR halfband filter - Simulink - MathWorks Switzerland
h\left(n\right)=\frac{1}{2\pi }{\int }_{-\pi /2}^{\pi /2}{e}^{j\omega n}d\omega =\frac{\mathrm{sin}\left(\frac{\pi }{2}n\right)}{\pi n}.
g\left(n\right)=\frac{1}{2\pi }{\int }_{-\pi }^{-\pi /2}{e}^{j\omega n}d\omega +\frac{1}{2\pi }{\int }_{\pi /2}^{\pi }{e}^{j\omega n}d\omega .
g\left(n\right)=\frac{\mathrm{sin}\left(\pi n\right)}{\pi n}-\frac{\mathrm{sin}\left(\frac{\pi }{2}n\right)}{\pi n}.
{\ell }^{\infty }
w\left(n\right)=\frac{{I}_{0}\left(\beta \sqrt{1-{\left(\frac{n-N/2}{N/2}\right)}^{2}}\right)}{{I}_{0}\left(\beta \right)},\text{ }\text{ }0\le n\le N,
\beta =\left\{\begin{array}{ll}0.1102\left(\alpha -8.7\right),\hfill & \alpha >50\hfill \\ 0.5842{\left(\alpha -21\right)}^{0.4}+0.07886\left(\alpha -21\right),\hfill & 50\ge \alpha \ge 21\hfill \\ 0,\hfill & \alpha <21\hfill \end{array}
n=\frac{\alpha -7.95}{2.285\left(\Delta \omega \right)}
{H}_{0}\left(z\right)=\sum _{n}h\left(2n\right){z}^{-n}.
{H}_{1}\left(z\right)=\sum _{n}h\left(2n+1\right){z}^{-n}.
H\left(z\right)={H}_{0}\left({z}^{2}\right)+{z}^{-1}{H}_{1}\left({z}^{2}\right). |
Convert symbolic expressions to function handle for ODE solvers - MATLAB odeFunction
Create Function Handle for ODE Solvers and Solve DAEs
Function Handles for System Containing Symbolic Parameters
Write Function Handles to File with Comments
Convert symbolic expressions to function handle for ODE solvers
f = odeFunction(expr,vars)
f = odeFunction(expr,vars,p1,...,pN)
f = odeFunction(___,Name,Value)
f = odeFunction(expr,vars) converts a system of symbolic algebraic expressions to a MATLAB® function handle. This function handle can be used as input to the numerical MATLAB ODE solvers, except for ode15i. The argument vars specifies the state variables of the system.
f = odeFunction(expr,vars,p1,...,pN) specifies the symbolic parameters of the system as p1,...,pN.
f = odeFunction(___,Name,Value) uses additional options specified by one or more Name,Value pair arguments.
Convert a system of symbolic differential algebraic equations to a function handle suitable for the MATLAB ODE solvers. Then solve the system by using the ode15s solver.
Create the following second-order differential algebraic equation.
eqn = diff(y(t),t,2) == (1-y(t)^2)*diff(y(t),t) - y(t);
Use reduceDifferentialOrder to rewrite that equation as a system of two first-order differential equations. Here, vars is a vector of state variables of the system. The new variable Dy(t) represents the first derivative of y(t) with respect to t.
[eqs,vars] = reduceDifferentialOrder(eqn,y(t))
diff(Dyt(t), t) + y(t) + Dyt(t)*(y(t)^2 - 1)
Set initial conditions for y(t) and its derivative Dy(t) to 2 and 0 respectively.
initConditions = [2 0];
Find the mass matrix M of the system and the right sides of the equations F.
[ -1, 0]
- y(t) - Dyt(t)*(y(t)^2 - 1)
M and F refer to the form
M\left(t,x\left(t\right)\right)\stackrel{˙}{x}\left(t\right)=F\left(t,x\left(t\right)\right).
. To simplify further computations, rewrite the system in the form
\stackrel{˙}{x}\left(t\right)=f\left(t,x\left(t\right)\right)
f = M\F
- Dyt(t)*y(t)^2 - y(t) + Dyt(t)
Convert f to a MATLAB function handle by using odeFunction. The resulting function handle is input to the MATLAB ODE solver ode15s.
odefun = odeFunction(f,vars);
ode15s(odefun, [0 10], initConditions)
Convert a system of symbolic differential equations containing both state variables and symbolic parameters to a function handle suitable for the MATLAB ODE solvers.
vars = [x1(t) x2(t)];
Find the mass matrix M and vector of the right side F for this system. M and F refer to the form
M\left(t,x\left(t\right)\right)\stackrel{˙}{x}\left(t\right)=F\left(t,x\left(t\right)\right).
b*x2(t)^2 + a*x1(t)
r(t)^2 - x1(t)^2 - x2(t)^2
Use odeFunction to generate MATLAB function handles from M and F. The function handle F contains symbolic parameters.
M = odeFunction(M,vars)
F = odeFunction(F,vars,a,b,r(t))
@(t,in2)reshape([1.0,0.0,0.0,0.0],[2,2])
@(t,in2,param1,param2,param3)[param1.*in2(1,:)+...
param2.*in2(2,:).^2;param3.^2-in2(1,:).^2-in2(2,:).^2]
r = @(t) cos(t)/(1+t^2);
Create the reduced function handle F.
F = @(t,Y) F(t,Y,a,b,r(t));
yp0 = [a*y0(1) + b*y0(2)^2; 1.234];
Create an option set that contains the mass matrix M of the system and vector yp0 of initial conditions for the derivatives.
opt = odeset('mass',M,'InitialSlope',yp0);
Now, use ode15s to solve the system of equations.
ode15s(F, [t0, 1], y0, opt)
Write the generated function handles to files by using the File option. When writing to files, odeFunction optimizes the code using intermediate variables named t0, t1, .… Include comments the files by specifying the Comments option.
Define the system of differential equations. Find the mass matrix M and the right side F.
eqs = [diff(x(t),t)+2*diff(y(t),t) == 0.1*y(t), ...
x(t)-y(t) == cos(t)-0.2*t*sin(x(t))];
vars = [x(t) y(t)];
[M,F] = massMatrixForm(eqs,vars);
Write the MATLAB code for M and F to the files myfileM and myfileF. odeFunction overwrites existing files. Include the comment Version: 1.1 in the files You can open and edit the output files.
M = odeFunction(M,vars,'File','myfileM','Comments','Version: 1.1');
function expr = myfileM(t,in2)
%MYFILEM
% EXPR = MYFILEM(T,IN2)
expr = reshape([1.0,0.0,2.0,0.0],[2, 2]);
F = odeFunction(F,vars,'File','myfileF','Comments','Version: 1.1');
function expr = myfileF(t,in2)
%MYFILEF
% EXPR = MYFILEF(T,IN2)
expr = [y.*(1.0./1.0e1);-x+y+cos(t)-t.*sin(x).*(1.0./5.0)];
Specify consistent initial values for x(t) and y(t) and their first derivatives.
xy0 = [2; 1]; % x(t) and y(t)
xyp0 = [0; 0.05*xy0(2)]; % derivatives of x(t) and y(t)
Create an option set that contains the mass matrix M, initial conditions xyp0, and numerical tolerances for the numerical search.
opt = odeset('mass', M, 'RelTol', 10^(-6),...
'AbsTol', 10^(-6), 'InitialSlope', xyp0);
Solve the system of equations by using ode15s.
ode15s(F, [0 7], xy0, opt)
Use the name-value pair argument 'Sparse',true when converting sparse symbolic matrices to MATLAB function handles.
Create the system of differential algebraic equations. Here, the symbolic functions x1(t) and x2(t) represent the state variables of the system. Specify the equations and state variables as two symbolic vectors: equations as a vector of symbolic equations, and variables as a vector of symbolic function calls.
M\left(t,x\left(t\right)\right)\stackrel{˙}{x}\left(t\right)=F\left(t,x\left(t\right)\right).
- (3*x1(t))/5 - x2(t)^2/10
cos(t)^2/(t^2 + 1)^2 - x1(t)^2 - x2(t)^2
Generate MATLAB function handles from M and F. Because most of the elements of the mass matrix M are zeros, use the Sparse argument when converting M.
M = odeFunction(M,vars,'Sparse',true)
F = odeFunction(F,vars)
@(t,in2)sparse([1],[1],[1.0],2,2)
@(t,in2)[in2(1,:).*(-3.0./5.0)-in2(2,:).^2./1.0e+1;...
cos(t).^2.*1.0./(t.^2+1.0).^2-in2(1,:).^2-in2(2,:).^2]
opt = odeset('mass',M,'InitialSlope', yp0);
Solve the system of equations using ode15s.
expr — System of algebraic expressions
System of algebraic expressions, specified as a vector of symbolic expressions.
Parameters of the system, specified as symbolic variables, functions, or function calls, such as f(t). You can also specify parameters of the system as a vector or matrix of symbolic variables, functions, or function calls. If expr contains symbolic parameters other than the variables specified in vars, you must specify these additional parameters as p1,...,pN.
Example: odeFunction(expr,vars,'File','myfile')
Path to the file containing generated code, specified as a character vector. The generated file accepts arguments of type double, and can be used without Symbolic Math Toolbox™. If the value is empty, odeFunction generates an anonymous function. If the character vector does not end in .m, the function appends .m.
By default, odeFunction with the File argument generates a file containing optimized code. Optimized means intermediate variables are automatically generated to simplify or speed up the code. MATLAB generates intermediate variables as a lowercase letter t followed by an automatically generated number, for example t32. To disable code optimization, use the Optimize argument.
By default, odeFunction with the File argument generates a file containing optimized code. Optimized means intermediate variables are automatically generated to simplify or speed up the code. MATLAB generates intermediate variables as a lowercase letter t followed by an automatically generated number, for example t32.
odeFunction without the File argument (or with a file path specified by an empty character vector) creates a function handle. In this case, the code is not optimized. If you try to enforce code optimization by setting Optimize to true, then odeFunction throws an error.
Flag that switches between sparse and dense matrix generation, specified as true or false. When you specify 'Sparse',true, the generated function represents symbolic matrices by sparse numeric matrices. Use 'Sparse',true when you convert symbolic matrices containing many zero elements. Often, operations on sparse matrices are more efficient than the same operations on dense matrices. See Sparse Matrices.
f — Function handle that is input to numerical MATLAB ODE solvers, except ode15i
Function handle that can serve as input argument to all numerical MATLAB ODE solvers, except for ode15i, returned as a MATLAB function handle.
odeFunction returns a function handle suitable for the ODE solvers such as ode45, ode15s, ode23t, and others. The only ODE solver that does not accept this function handle is the solver for fully implicit differential equations, ode15i. To convert the system of equations to a function handle suitable for ode15i, use daeFunction.
findDecoupledBlocks | daeFunction | decic | incidenceMatrix | isLowIndexDAE | massMatrixForm | matlabFunction | ode15i | ode15s | ode45 | ode23t | reduceDAEIndex | reduceDAEToODE | reduceDifferentialOrder | reduceRedundancies |
Conservation of Momentum | Brilliant Math & Science Wiki
In physics, the systems of interest are often full of dynamics, and in Newtonian mechanics, that is certainly the case. When everything is changing from moment to moment: forces, positions, velocities, et cetera, it may come as a surprise that there exist some quantities that never change. There are however a few special quantities that remain constant even as the components of a system move around and explore the space of possible arrangements. One of these is the total momentum, whose conservation is implied by Newton's laws of motion.
n
particle case
Momentum conservation in the real world
Systems that conserve momentum are closed
As we showed before, if we consider two particles interacting with one another in free space, A and B, particle A feels the force
F_\textrm{AB}
from B, while B feels the force
F_\textrm{BA}
from A. In Newton's third law, we showed that
F_\textrm{AB}=-F_\textrm{BA}
From the second law, we know that the change in momentum of particle A per unit time is given by
\Delta p_\textrm{A} / \Delta t = F_\textrm{AB}
\Delta p_\textrm{B} / \Delta t = F_\textrm{BA}
. Now, the total momentum of the system is given by
p_\textrm{A} + p_\textrm{B}
and the rate at which the total momentum change is given by
\frac{\Delta p_\textrm{A}}{\Delta t} + \frac{\Delta p_\textrm{B}}{\Delta t} = \frac{F_\textrm{AB} + F_\textrm{BA}}{\Delta t} = \frac{0}{\Delta t} = 0
If something changes at a rate of zero, it isn't changing at all. Hence, the total momentum of the two particles remains constant!
n
This calculation can be easily extended to systems of
n
particles interacting with each other. The change in the total momentum of a system of
n
particles is given by
\displaystyle \sum_i \frac{\Delta p_i}{\Delta t}
Now the change in momentum of each particle
\displaystyle \frac{\Delta p_i}{\Delta t}
is given by the sum of its interactions with all the other particles
\Delta p_i = \Delta t \sum_j F_{ij}
, so that the change in the total momentum of the system of particles is given by
\Delta t \sum_{ij} F_{ij} = \Delta t \frac{1}{2}\sum_{ij} \left(F_{ij} + F_{ji}\right)
Because the forces of interaction between any two particles are balanced
\left(F_{ij} =- F_{ji}\right)
, each term
\left(F_{ij} + F_{ji}\right)
\Delta p = \Delta t \frac{1}{2}\sum_{ij} 0 = 0
and the total momentum of the system is constant!
The conservation of momentum is useful whenever we analyze collisions, whether they be a bat and a ball, two cars in a crash, or subatomic particles colliding in a particle accelerator, giving rise to exotic new forms of matter. In each one of these cases the conservation of momentum aids us in our investigation.
Momentum conservation in problems
In the case of two cars of mass
M_1
M_2
colliding, we have:
M_1v_1^i + M_2v_2^i = M_1v_1^f + M_2v_2^f
If you study particle physics, the annihilation between an electron and positron to form two photons the conservation of momentum becomes:
p_{e^-}^\mu + p_{e^+}^\mu = p_\gamma^\mu + p+\gamma^\mu
\mu
indicates that these are the four-momentum, the generalization of ordinary momentum in relativity.
The crucial property of the system that leads to this result is the fact that it is closed. In other words, the particles under consideration interact with each other and nothing else. A practical definition of a closed system in classical mechanics is one that we can contain within a surface, across which no forces act, i.e.
F_\textrm{ext} = 0
, and no matter passes. For example, in a space where two particles are the only things in existence, we can draw an imaginary ball around them. No forces act across the ball because there is nothing else in the universe.
In the real world, this is of course never the case. No matter how far one gets away from other matter and energy in the universe, there is always some remnant that spoils the "closed'' nature of any real system. However, such external forces and fields can usually be taken to be very small compared to the magnitude of the interactions being studied, so that for all practical purposes, the system is "closed'', and the conservation of momentum still applies, approximately.
Were the system to be acted upon by a external force, or some kind of field, we'd have expand our system to include the source of the external forces if we wanted to preserve the conservation of momentum.
Finally, we point out that momentum depends on the frame of reference of the observer. If someone sees a bullet of mass
m
fly by with velocity
\vec{v}
, they'll measure the momentum
p=m\vec{v}
, whereas somebody driving in a car that has velocity
\vec{u}
in the direction of the bullet will measure the momentum
p^\prime = m \left(\vec{v} - \vec{u}\right)
Cite as: Conservation of Momentum. Brilliant.org. Retrieved from https://brilliant.org/wiki/identifying-when-momentum-conserved-isolated-syste/ |
Hyperbolic cosecant - MATLAB csch - MathWorks España
Hyperbolic Cosecant of Vector
Graph of Hyperbolic Cosecant
Y = csch(X)
Y = csch(X) returns the hyperbolic cosecant of the elements of X. The csch function operates element-wise on arrays. The function accepts both real and complex inputs. All angles are in radians.
Create a vector and calculate the hyperbolic cosecant of each value.
Plot the hyperbolic cosecant over the domain
-\pi <x<0
0<x<\pi .
y1 = csch(x1);
The hyperbolic cosecant of x is equal to the inverse of the hyperbolic sine
\text{csch}\left(x\right)=\frac{1}{\mathrm{sinh}\left(x\right)}=\frac{2}{{e}^{x}-{e}^{-x}}.
In terms of the traditional cosecant function with a complex argument, the identity is
\text{csch}\left(x\right)=i\mathrm{csc}\left(ix\right)\text{\hspace{0.17em}}.
acsch | csc | sinh | cosh |
Calculate output reflection coefficient of two-port network - MATLAB gammaout - MathWorks France
Output Reflection Coefficient Calculation using Network Data
Output Reflection Coefficient Calculation of S-Parameters Object
Calculate output reflection coefficient of two-port network
coefficient = gammaout(s_params,z0,zs)
coefficient = gammaout(hs,zs)
coefficient = gammaout(s_params,z0,zs) calculates the output reflection coefficient of a two-port network. z0 is the reference impedance Z0; its default value is 50 ohms. zs is the source impedance Zs; its default value is also 50 ohms. coefficient is an M-element complex vector.
coefficient = gammaout(hs,zs) calculates the output reflection coefficient of the two-port network represented by the S-parameter object hs.
Calculate the output reflection coefficient using network data from a file.
zs = 100;
Specify the source impedance.
Calculate the output reflection coefficient using the gammaout function. .
coefficient = gammaout(s_params,zs)
Source impedance, specified as a positive scalar.
coefficient — Output reflection coefficient
Output reflection coefficient, returned as a M element complex vector.
{\Gamma }_{out}={S}_{22}+\frac{{S}_{12}{S}_{21}{\Gamma }_{S}}{1-{S}_{11}{\Gamma }_{S}}
{\Gamma }_{S}=\frac{{Z}_{s}-{Z}_{0}}{{Z}_{s}+{Z}_{0}} |
Factoring Polynomials With Large Coefficients: Factoring by Extraction | Brilliant Math & Science Wiki
Trevor Arashiro, Mahindra Jain, Samir Khan, and
Dr. Bela Palnitkar
Factoring polynomials, in general, is quite difficult, but some special ones can be factored using certain tricks.
Factorizing Quadratics with Large Numbers
Factorizing High-degree Polynomials with Large Numbers
Today, I will discuss how to factor polynomials with large coefficients such as
3x^2+10x-1000
with ease. I know that this will be a long note, but I feel that it is worth reading everything including the generalized form at the bottom except for the proof (unless you want to).
While sitting in my math class today, I discovered a trick to factoring second-degree polynomials with large or irrational second and third coefficients. For example, try factoring
3x^2+10x-1000
. It's relatively simple to factor it to
(3x-50)(x+20),
but that would take a little while or at least longer than the way that I'm about to discuss.
We begin with the expression
3x^2+10x-1000
. Then we divide the second coefficient by 10 and the third by 100, and we are left with the expression
3x^2+x-10,
which we can easily factor to
(3x-5)(x+2)
. Finally, we multiply the second term in each factor by 10 and have
(3x-50)(x+20)
. Looks familiar, doesn't it?
Basically, what I have done is that I divided the second coefficient by any one of its factors (in this case 10) and then divided the third coefficient by the square of that factor while leaving the first untouched.
This method applies to irrational and imaginary coefficients as well:
4x^2+8\sqrt2x+8
2\sqrt2
from the second coefficient and 8 from the third, and then we are left with
4x^2+4x+1=(2x+1)^2.
By multiplying back
2\sqrt2
to the second term in each factor, which in this case is
(2x+1),
\left(2x+2\sqrt2\right)^2. \ _\square
This also works in reverse from the first coefficient:
25x^2-60x+36
We can factor out 5 from the middle coefficient and 25 from the first, and then we are left with
x^2-12x+36
. Next, we can factor out 6 from the second coefficient and 36 from the final coefficient, being left with
x^2-2x+1=(x-1)^2
. Finally, we factor back in 5 to the first coefficient and 6 to the last one to obtain
(5x-6)^2
_\square
This also works for unfactorable expressions. This method will not make unfactorable equations factorable; however, it will make the quadratic formula much easier to use. This is a little tougher to do because, depending on which way you factor a number out, the formula changes.
ax^2+bx+c\Rightarrow ax^2+\frac{bx}{d}+\frac{c}{d^2}
This changes the quadratic equation to
\dfrac{-\frac{b}{d}\pm \sqrt{\frac{b^2}{d^2}-\frac{4ac}{d^2}}}{2a}=\dfrac{-\frac{b}{d}\pm \frac{\sqrt{b^2-4ac}}{d}}{2a}=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a\color{#3D99F6}{\textbf{d}}}.
Thus, once our answer is achieved, we must multiply the answer by the number
\color{#3D99F6}{\textbf{d}}
that we extracted at the start.
x^2+60x+2025
By factoring out
15
from the second coefficient and
15^2=225
from the final coefficient, we have
\begin{aligned} x^2+\frac{60}{15}x+\frac{2025}{225} &=x^2+4x+9 \\ &=\left(x-\frac{-4+\sqrt{16-36}}{2}\right)\left(x-\frac{-4+\sqrt{16-36}}{2}\right)\\ &=\left(x-\big(-2+\sqrt{5}i\big)\right)\left(x-\big(-2-\sqrt{5}i\big)\right). \end{aligned}
Now, by multiplying back
15,
\begin{aligned} x^2+60x+2025 &=\left(x-\big(-(15)2+(15)\sqrt{5}i\big)\right)\left(x-\big(-(15)2-(15)\sqrt{5}i\big)\right)\\ &=\left(x-\big(-30+15\sqrt{5}i\big)\right)\left(x-\big(-30-15\sqrt{5}i\big)\right).\ _\square \end{aligned}
ax^2+bx+c\Rightarrow \frac{ax^2}{d^2}+\frac{bx}{d}+c
This changes the quadratic to
\dfrac{-\frac{b}{d}\pm \sqrt{\frac{b^2}{d^2}-\frac{4ac}{d^2}}}{\frac{2a}{d^2}}=\dfrac{d^2\left(-\frac{b}{d}\pm \frac{\sqrt{b^2-4ac}}{d}\right)}{2a}=\frac{{\color{#3D99F6}{\textbf{d}}}\big(-b\pm \sqrt{b^2-4ac}\big)}{2a}.
Thus, once our answer is achieved, we must divide the answer by the number
\color{#3D99F6}{\textbf{d}}
Now, we shall prove why this works:
Make the general expression
ax^2+bx+c,
which can be factored into
(dx+e)(fx+g).
a=df, b=dg+ef,
c=eg.
The last step of our method requires us to multiply both of the second coefficients in our binomials by
n
(n
being the number that we factored out of
b).
So our expression becomes
\big(dx+\frac{e}{n}\big)\big(fx+\frac{g}{n}\big),
a=df, \frac{b}{n}=\frac{dg+ef}{n}, \frac{c}{n^2}=\frac{eg}{n^2}
. This is why we factor out
n
n^2
from the second and third coefficients, respectively.
_\square
This rule can actually apply to polynomials of any degree. Unfortunately, the higher the degree of the polynomial, the less convenient this becomes.
But, say, we have a polynomial with degree
n
, which can be factored into
(a_1x+k_1)(a_2x+k_2)(a_3x+k_3)\cdots (a_nx+k_n)
. This does not necessarily have integer or real coefficients. Next, we assume that all coefficients after the second (inclusive) are divisible by
r^t,
t
represents the location of
relative to the second coefficient.
This means that we have
\displaystyle\prod_{t=0}^n (a_tx+b_tr),
\displaystyle\sum_{t=0}^n \left(r^{n-t}\big(p_tx^t\big)\right).
Upon expansion, we get
p_nx^n+r\big(p_{n-1}x^{n-1}\big)+r^2\big(p_{n-2}x^{n-2}\big)+r^3\big(p_{n-3}x^{n-3}\big)+\cdots+r^{n-1}\big(p_1x^1\big)+r^n(p_0).
As we can see, this method works whenever every coefficient after the first is divisible by
r^t.
A useful method for seeing if this works is by prime-factoring every coefficient after the second and looking for a common term. That has multiple powers throughout. Say we have
x^4-3x^3-63x^2+27x+486
. First, we prime-factor the numbers to get
x^4-3x^3-3^2\cdot7x^2+3^3x+2\cdot3^5.
As we can see, they share 3 in increasing powers. Therefore, we can eliminate 3 in increasing powers from each coefficient and are left with
\begin{aligned} x^4-x^3-7x^2+x+6&=x^3(x-1)-(7x+6)(x-1)\\ &=\left(x\big(x^2-1\big)-6(x+1)\right)(x-1)\\ &= \big(x^2-x-6\big)(x+1)(x-1)\\ &=(x-3)(x+2)(x+1)(x-1). \end{aligned}
Finally, we factor back in the three that we took out at the start and are left with
(x-9)(x+6)(x+3)(x-3).
Cite as: Factoring Polynomials With Large Coefficients: Factoring by Extraction. Brilliant.org. Retrieved from https://brilliant.org/wiki/factoring-polynomials-with-large-coefficients/ |
Limitations of Linear Regression Practice Problems Online | Brilliant
Linear regression is clearly a very useful tool. Whether you are analyzing crop yields or estimating next year’s GDP, it is always a powerful machine learning technique.
However, it does have limitations. Possibly the most obvious is that it will not be effective on data which isn’t linear. Using linear regression means assuming that the response variable changes linearly with the predictor variables.
Limitations of Linear Regression
Alfred’s done some thinking, and he wants to account for fertilizer in his tree growing efforts. Assume that for every ton of fertilizer he uses each seed is about 1.5 times more likely to sprout.
Over the past few years, he has compiled a large data set in which he records fertilizer use, seeds planted, and trees sprouted. Is ordinary linear regression likely to give good predictions for the number of sprouting trees given the amount of fertilizer used and number of seeds planted?
Outliers are another confounding factor when using linear regression. These are elements of a data set that are far removed from the rest of the data.
Outliers are problematic because they are often far enough from the rest of the data that the best-fit line will be strongly skewed by them, even when they are present because of a mistake in recording or an unlikely fluke.
Commonly, outliers are dealt with simply by excluding elements which are too distant from the mean of the data. A slightly more complicated method is to model the data and then exclude whichever elements contribute disproportionately to the error.
It is not impossible for outliers to contain meaningful information though. One should be careful removing test data.
Another major setback to linear regression is that there may be multicollinearity between predictor variables. This is the term for when several of the input variables appear to be strongly related.
Multicollinearity has a wide range of effects, some of which are outside the scope of this lesson. However, the major concern is that multicollinearity allows many different best-fit equations to appear almost equivalent to a regression algorithm.
As a result, tools such as least squares regression tend to produce unstable results when multicollinearity is involved. There are generally many coefficient values which produce almost equivalent results. This is often problematic, especially if the best-fit equation is intended to extrapolate to future situations where multicollinearity is no longer present.
Another issue is that it becomes difficult to see the impact of single predictor variables on the response variable. For instance, say that two predictor variables
x_1
x_2
are always exactly equal to each other and therefore perfectly correlated. We can immediately see that multiple weightings, such as
m \cdot x_1 + m\cdot x_2
2m\cdot x_1 + 0\cdot x_2
, will lead to the exact same result. Now it’s impossible to meaningfully predict how much the response variable will change with an increase in
x_1
because we have no idea which of the possible weightings best fits reality. This both decreases the utility of our results and makes it more likely that our best-fit line won’t fit future situations.
We can see the effects of multicollinearity clearly when we take the problem to its extreme. Say that we have two predictor variables,
x_1
x_2
, and one response variable
y
. Using the test data given in the table below, determine which candidate best-fit equation has the lowest SSE:
\begin{array}{c|c|c} x_1 & x_2 & y \\ \hline 5&10&3 \\ \hline 2 & 4 & 1\\ \hline 7 & 14 & 6 \\ \hline 2.5 & 5 & 2 \\ \end{array}
y = 2x_1 + x_2
y = x_1 + 1.5x_2
The sums of square errors are equivalent
The property of heteroscedasticity has also been known to create issues in linear regression problems. Heteroscedastic data sets have widely different standard deviations in different areas of the data set, which can cause problems when some points end up with a disproportionate amount of weight in regression calculations.
A data set is displayed on the scatterplot below. Which section of the graph will have the greatest weight in linear regression?
Another classic pitfall in linear regression is overfitting, a phenomenon which takes place when there are enough variables in the best-fit equation for it to mold itself to the data points almost exactly.
Although this sounds useful, in practice it means that errors in measurement, outliers, and other deviations in the data have a large effect on the best-fit equation. An overfitted function might perform well on the data used to train it, but it will often do very badly at approximating new data. Useless variables may become overvalued in order to more exactly match data points, and the function may behave unpredictably after leaving the space of the training data set. |
Algebra Warmups - Cryptograms | Brilliant Math & Science Wiki
Algebra Warmups - Cryptograms
Zandra Vinegar, Jimin Khim, and Calvin Lin contributed
This page is an introduction to basic cryptogram problem-solving. For more advanced cryptogram problem-solving strategies, please check out the main cryptograms page.
A cryptogram is a mathematical puzzle, where various symbols are used to represent digits, and a given system has to be true. The most common form is a mathematical equation (such as the example below), but sometimes there can be multiple equations or statements.
This example is from the cryptograms warmup quiz.
\begin{array} {c c c } & 1 & E \\ \times & & E \\ \hline & 9 & E \\ \end{array}
E
would make the above multiplication true?
Let's start by testing a random digit in this equation, say 8. Check the long multiplication for yourself:
\begin{array} {c c c } & 1 & 8 \\ \times & & 8 \\ \hline 1 & 4 & 4 \\ \end{array}
Nope - 8 didn't work, but look at why not -- we could have predicted that 8 would fail because we know that
8 \times 8 = 64
, and the 1's place of the desired solution,
9 E
, isn't 8. So let's consider all of the possible digits and find one where the 1's place of
X^2 = X \times X
ends with the digit
X
\begin{array}{rrl} 1^2 &= 1 &\rightarrow \text{ so 1 works and could be the answer.}\\ 2^2 &= 4 &\rightarrow \text{ so 2 does not work.} \\ 3^2 &= 9 &\rightarrow \text{ so 3 does not work.} \\ 4^2 &= 16 &\rightarrow \text{ so 4 does not work.} \\ 5^2 &= 25 &\rightarrow \text{ so 5 works and could be the answer.} \\ 6^2 &= 36 &\rightarrow \text{ so 6 works and could be the answer.} \\ 7^2 &= 49 &\rightarrow \text{ so 7 does not work.} \\ 8^2 &= 64 &\rightarrow \text{ so 8 does not work.}\\ 9^2 &= 81 &\rightarrow \text{ so 9 does not work.} \end{array}
So we only have to check 3 possible values that might be
E: 1, 5,
6
. And...
\begin{array} {c c c } & 1 & 6 \\ \times & & 6 \\ \hline & 9 & 6\\ \end{array}
6 is the one that works! So
E = 6
Guessing and checking
E=8
initially wasn't a waste of time, but it's what led us to realize that instead of wasting time with 9 long-multiplication checks, we can just look at the last digits of the perfect squares, and thereby limit our test-cases to 3.
Mathematicians need to be ready (and are usually eager) to get their hands dirty. Even though we are working with numbers instead of physical tools, chemicals, and dangerously high voltage power sources, there's an element of risk to committing to any experiment. Will 5 work? 17? Answering each small question is a bit of work, and it can require a lot of grit to be wrong 1000 times before you're finally right. But, don't let yourself be afraid to guess and be wrong.
\begin{array} { l l l } & & & X & Y & Z \\ & & & X & Y & Z \\ &+ & & X & Y & Z \\ \hline & & & Z & Z & Z \\ \end{array}
X, Y,
Z
are distinct digits, then what is the value of
X \times Y \times Z ?
Want to solve more cryptogram puzzles? Try the Cryptograms Warmup Quiz.
Quick tips to become a master at solving cryptograms:
Don't be afraid to start by simply making a guess or two and checking to see what happens. Even if the numbers you guess don't work, you can learn a lot from watching for why they don't work.
Ask yourself, "What are the properties of the numbers involved in this question and the relevance of the positions that they're in?"
If a question is too big to tackle all at once, break it into parts. The answers to small questions are frequently necessary stepping stones to discovering the answers to huge, beautiful questions.
Read the solution if you get stumped, and take careful note of the strategies used so that you can improve your own skills!
Looking for an extra challenge? There are also harder forms for these cryptogram puzzles and more advanced strategies for working on them. If you're intrigued, check out the main, Brilliant Cryptogram Problem Solving page.
Additional articles on Brilliant that are related to solving cryptograms include:
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Article Practice Quiz
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\hspace{10mm}
\hspace{20mm}
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(Base 10) Simplifying Algebraic Expressions
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Divisibility Rules Alternative Base Systems
Connecting Algebra And Number Theory
The really deep mathematical patterns that lead to a lot of the logic relevant to solving cryptograms derives from the same reasoning as the divisibility rules, which are also based on digit-by-digit analysis, carried out in base 10. Ever wonder why 1, 5, and 6 are the only numbers whose units digits, when squared, are equal to the original number? Number Theory, with a focus on understanding the base-10 number system, can explain this. But if you're looking to kick it up another few notches, consider that the patterns used to solve cryptograms in other bases would work differently.
Cite as: Algebra Warmups - Cryptograms. Brilliant.org. Retrieved from https://brilliant.org/wiki/cryptic-cryptograms/ |
Euler's number | Brilliant Math & Science Wiki
展豪 張, Pi Han Goh, Kishore S. Shenoy, and
This wiki is quite incomplete. Please do help in improving it.
Euler's number (also known as Napier's constant),
e
, is a mathematical constant, which is approximately equal to
2.7182818284590452353602874713526624977572470936999595749669676277240766303535475945713829178...
It can be expressed as the limit,
\displaystyle e = \lim_{n\to\infty} \left( 1 + \dfrac{1}{n} \right)^n
, and it can also be expressed as the infinite sum
\displaystyle \sum_{j=0}^\infty \dfrac{1}{j!} = \dfrac{1}{0!} + \dfrac{1}{1!} + \dfrac{1}{2!} + \cdots
It is not be confused with
\gamma
(Euler-Mascheroni constant).
Proof of Equivalence of Definitions
\displaystyle\lim_{n\to\infty} \left(1+\frac 1n\right)^n=\sum_{j=0}^\infty \frac 1{j!}
Apply binomial expansion on the left hand side, we have:
\displaystyle\;\;\;\;\lim_{n\to\infty}(1+\frac 1n)^n\\ \displaystyle=\lim_{n\to\infty}(1+\frac n{1!}\cdot\frac 1n+\frac{n(n-1)}{2!} \frac 1{n^2}+\frac{n(n-1)(n-2)}{3!}\frac 1{n^3}+\cdots)\\ \displaystyle=\lim_{n\to\infty}(1+\frac 1{1!}\frac nn+\frac 1{2!}\frac{n(n-1)}{n^2}+\frac 1{3!}\frac{n(n-1)(n-2)}{n^3}+\cdots)\\ \displaystyle=\lim_{n\to\infty}(1+\frac 1{1!}(1)+\frac 1{2!}(1)(1-\frac 1n)+\frac 1{3!}(1)(1-\frac 1n)(1-\frac 2n)+\cdots)\\ \displaystyle=1+\frac 1{1!}+\frac 1{2!}+\frac 1{3!}+\cdots\\ \displaystyle=\frac 1{0!}+\frac 1{1!}+\frac 1{2!}+\frac 1{3!}+\cdots\\ \displaystyle=\sum_{j=0}^\infty \frac 1{j!}
e^{i\theta} = \cos\theta + i\sin \theta
Cite as: Euler's number. Brilliant.org. Retrieved from https://brilliant.org/wiki/eulers-number/ |
PSSCH decoding - MATLAB ltePSSCHDecode - MathWorks í•œêµ
{n}_{\text{ssf}}^{\text{PSSCH}}
NRE is NPRB × 144 for D2D normal cyclic prefix or NPRB × 120 for D2D extended cyclic prefix and V2X. NPRB is the number of physical resource blocks (PRB) used for transmission.
Log-likelihood ratio (LLR) soft bits, returned as a vector with Nbps × NRE softbits. Nbps is the number of bits per symbol. PSSCH modulation is either QPSK (2 bits per symbol) or 16 QAM (4 bits per symbol). NRE is the number of PSSCH resource elements in the subframe. The LLR of the punctured soft bits associated with the last SC-FDMA symbol are set to 0.
For PSSCH, the input codeword length is Mbits = NRE × Nbps, where Nbps is the number of bits per symbol. PSSCH modulation is either QPSK (2 bits per symbol) or 16 QAM (4 bits per symbol).
The number of PSSCH resource elements (NRE) in a subframe is NRE = NPRB × NREperPRB × NSYM and includes symbols associated with the sidelink SC-FDMA guard symbol.
{c}_{\text{init}}={n}_{\text{ID}}^{\text{X}}Ã{2}^{14}+{n}_{\text{ssf}}^{\text{PSSCH}}Ã{2}^{9}+510
{n}_{\text{ID}}^{\text{SA}}
{n}_{\text{ID}}^{\text{SA}}
{n}_{\text{ssf}}^{\text{PSSCH}} |
This almost class-less CSS library turns your HTML document into a website that looks like a LaTeX document.
To install this, run
npm install latex-tailwind
Then in your tailwind.config.js, add it as a plugin like this
plugins: [require("latex-tailwind")],
Then add latex-style to the outermost div you want to have LaTeX style
There are two options for this plugin, footnotes and syntax, they can be enabled like so
latex: {
Footnotes will only work with some markdown processors, only tested with remark and the remark-footnotes plugin. This will style the footnotes in a similar way to the rest of the document. It is optional as the footnotes class is external to the div it is created in, meaning that when enabled, it will apply for all footnotes classes in the document.
Syntax provides syntax highlighting in the style of minted, designed to work with prism.js.
Excluding components from the layers to purge is important as Tailwind doesn't see the classes created by Prism or the footnote generator
This has been inspired by Tailwind CSS Typography and LaTeX.css, with many of the styles being taken directly from the latter, and a few of the list styles being taken from the former.
The source code of this project can be found at samrobbins85/latex-tailwind
theorem class:
The real numbers are uncountable
\mathbb{R}
proof class:
\mathbb{R}
is countable, then
[0,1]
is countable
lemma class:
An even number plus an even number is an even number
A definition is a statement
Also, as a nice little extra, you can style the LaTeX logo by typing
<span class="latex">L<span>a</span>T<span>e</span>X</span>
General Markdown
The text below is taken directly from the tailwind Typography plugin, and works well to show how things work, along with ensuring that nothing looks weird.
A longer example including footnotes for good measure
A description of the non-local means denoising algorithm
1 Non local means denoising uses samples from all around the image, instead of conventional denoising which will just look at the area around the given pixel to increase the accuracy of the colour. The reason it does this is due to the fact that patterns and shapes will be repeated in images, meaning that there will likely be an area somewhere else in the image that looks very similar to the patch around the pixel looking to be corrected. By finding these areas and taking averages of the pixels in similar areas, the noise will reduce as the random noise will converge around the true value.
So the method by which non-local means runs is to look at many patches throughout the image, and compare the similarities of those patches with the patch around the pixel looking to be denoised. This comparison then allows for assigning a weight to each patch looked at, which are then used (along with the colour of the pixel in the centre of the patch) in the calculation of the colour of the pixel to be denoised.
Various implementations of the algorithm and their efficiency
2Taking an image u and a pixel in it you want to denoise, p, you first need to decide a patch size, given by r, as the dimensions of the patch (blue) are
(2r+1)\times(2r+1)
. You then look at all the other pixels,
q\in Q
, but as it is intensive to do the calculations, specifying a research zone (red) allows you to make the processing faster as fewer comparisons have to be done. When looking at the other pixels, calculate their patch of the same size as the patch of p, then compare each pixel in the patch of q with the corresponding pixel in the patch of p. This similarity is then used to compute the similarity between the patch around p and the patch around q, and a weighting is given to q to describe this. These weightings are then averaged with the colours of the pixels to provide a more accurate representation of the pixel.
Patchwise
2 The main way in which patchwise differs from pixelwise is in the formulation of the weighting, as you can see below
C(p)=\sum_{q \in B(p, r)} w(p, q)
C=\sum_{Q=Q(q, f) \in B(p, r)} w(B, Q)
By calculating weights for pixels instead of patches we can make one calculation per patch, therefore not needing to do
(2f+1)^2
calculations per pixel, providing a large increase in performance. The overall quality of the two methods are the same, and so the patchwise method is preferred as it has no drawbacks for an improvement in speed.
The strengths and limitations of non-local means compared to other denoising algorithms
3Definition (method noise). Let u be a (not necessarily noisy) image and
D_h
a denoising operator depending on h. Then we define the method noise of u as the image difference
n(D_h,u)=u-D_h(u)
This method noise should be as similar to white noise as possible. The image below is sourced from Buades, A., Coll, B., and Morel, J. 2005 3
From left to right and from top to bottom: original image,Gaussian convolution, mean curvature motion, total variation, Tadmor–Nezzar–Vese iterated total variation, Osher et al. total variation, neighborhood filter, soft TIWT, hard TIWT, DCT empirical Wiener filter, and the NL-means algorithm.
You can see that the NL means algorithm is closest to white noise, as it is very difficult to make out the original image from the method noise, and so is the best in this area
4The mean square error measures the average squared difference between the estimated values and what is estimated. In images this acts as a measure of how far from the true image the denoised image is. These results are taken from Buades, A., Coll, B., and Morel, J. 2005 3
Here it can be seen that the NL-means algorithm gives images that are closest to the true image, and so performs best for image denoising under this measurement.
The influence of the algorithmic parameters on the output
In the following images I am changing the values of h, the template window size and the search window size, from a standard set at h=5, template window size=7 and search window size =21. I will adjust each one in turn to show the differences yielded by changing them.
By adjusting the value of h you get a large change in the amount of smoothing, although a large amount of noise is still present. Increasing the value of h does increase the PSNR from 28.60 to 29.66
The effects from adjusting the template width are much more subtle than that of adjusting h, it can be noticed in the wires overhead that a larger template width has reduced the detail. An increase in the template width yields a small reduction in the PSNR from 28.68 to 28.51.
The effects for the value of the search window are also very subtle, and again can only be noticed fully in the overhead wires. An increase in the search window yields a marginal increase in the PSNR from 28.51 to 28.52.
Modifications and extensions of the algorithm that have been proposed in the literature
Testing stationarity
3 One proposed modification is one to test stationarity. The original algorithm works under the conditional expectation process:
Theorem - Conditional Expectation Theorem
Z_j=\{X_j,Y_j\}
j=1,2,...
be a strictly stationary and mixing process. For
i\in I
X
Y
be distributed as
X_i
Y_i
. Let J be a compact subset
J\subset \mathbb{R}^p
\inf\{f_X(x);x\in J\}>0
However this is not true everywhere, as each image may contain exceptional, non-repeated structures, these would be blurred out by the algorithm, so the algorithm should have a detection phase and special treatment of nonstationary points. In order to use this strategy a good estimate of the mean and variance at every pixel is needed, fortunately the non-local means algorithm converges to the conditional mean, and the variance can just be calculated using
EX^2-(EX)^2
Multiscale version
Another improvement to make is one to speed up the algorithm, this is proposed using a multiscale algorithm.
Zoom out the image
u_0
by a factor of 2. This gives the new image
u_1
Apply the NL means algorithm to
u_1
, so that with each pixel of
u_1
, a list of windows centered in
(i_1,j_1)...(i_k,j_k)
is associated
For each pixel of
u_0
(2i+r,2j+s)
r,s\in \{0,1\}
, we apply the NL means algorithm. However instead of comparing with all the windows in the search zone, we just compare with the 9 neighbouring windows of each pixel
This procedure can be applied in a pyramid fashion
Applications of the original algorithm and its extensions
5 It has been proposed that Non-Local means can be used in X-Ray imaging, allowing for a reduction of noise in the scans, making them easier to interpret. In CT scans a higher dose can be given to give a clearer image, but with that is more dangerous, however by applying the NL means algorithm a lower dose can be given for the same clarity. It benefits from the improvement stated above to test stationarity as the the noise and streak artifacts are non stationary. The original algorithm was also not good at removing the streak artifacts in low-flux CT images resulting from photon starvation. However by applying one-dimensional nonlinear diffusion in the stationary wavelet domain before applying the non-local means algorithm these could be reduced.
6 NLM can also be applied in video denoising, it has an adaptation as the denoising can be improved by using the data from sequential frames. In the implementation proposed in the paper, the current input frame and prior output frame are used to form the current output frame. In the paper the measurements they make fail to show that this algorithm is an improvement from current algorithms, however the algorithm does have much better subjective visual performance.
Buades, A., Coll, B., and Morel, J.M. 2005. A Non-Local Algorithm for Image Denoising. In 2005 IEEE Computer Society Conference on Computer Vision and Pattern Recognition (CVPR'05) (pp. 60–65). IEEE.↩
Buades, A., Coll, B., and Morel, J.M. 2011. Non-Local Means Denoising. Image Processing On Line, 1.↩
Buades, A., Coll, B., and Morel, J. 2005. A Review of Image Denoising Algorithms, with a New One. Multiscale Modeling & Simulation, 4(2), p.490–530.↩
Machine learning: an introduction to mean squared error and regression lines. URL https://www.freecodecamp.org/news/machine-learning-mean-squared-error-regression-line-c7dde9a26b93/.↩
Zhang, H., Zeng, D., Zhang, H., Wang, J., Liang, Z., and Ma, J. 2017. Applications of nonlocal means algorithm in low-dose X-ray CT image processing and reconstruction: A review. Medical Physics, 44(3), p.1168–1185.↩
Ali, R., and Hardie, R. 2017. Recursive non-local means filter for video denoising. EURASIP Journal on Image and Video Processing, 2017(1), p.29.↩ |
Add polynomials over Galois field - MATLAB gfadd - MathWorks América Latina
Add Two GF Arrays
Add polynomials over Galois field
This function performs computations in GF(pm) where p is prime. To work in GF(2m), apply the + operator to Galois arrays of equal size. For details, see Example: Addition and Subtraction.
c = gfadd(a,b) adds two GF(2) polynomials, a and b, which can be either polynomial character vectors or numeric vectors. If a and b are vectors of the same orientation but different lengths, then the shorter vector is zero-padded. If a and b are matrices they must be of the same size.
c = gfadd(a,b,p) adds two GF(p) polynomials, where p is a prime number. a, b, and c are row vectors that give the coefficients of the corresponding polynomials in order of ascending powers. Each coefficient is between 0 and p-1. If a and b are matrices of the same size, the function treats each row independently.
c = gfadd(a,b,p,len) adds row vectors a and b as in the previous syntax, except that it returns a row vector of length len. The output c is a truncated or extended representation of the sum. If the row vector corresponding to the sum has fewer than len entries (including zeros), extra zeros are added at the end; if it has more than len entries, entries from the end are removed.
c = gfadd(a,b,field) adds two GF(pm) elements, where m is a positive integer. a and b are the exponential format of the two elements, relative to some primitive element of GF(pm). field is the matrix listing all elements of GF(pm), arranged relative to the same primitive element. c is the exponential format of the sum, relative to the same primitive element. See Representing Elements of Galois Fields for an explanation of these formats. If a and b are matrices of the same size, the function treats each element independently.
2+3x+{x}^{2}
4+2x+3{x}^{2}
over GF(5).
Add the two polynomials and display the first two elements.
For prime number p and exponent m, create a matrix listing all elements of GF(p^m) given primitive polynomial
2+2x+{x}^{2}
{A}^{2}
{A}^{4}
A |
Solve and check the following equation. \frac{x-7}{5}+\frac{1}{5}=-\frac
Solve and check the following equation. \frac{x-7}{5}+\frac{1}{5}
Solve and check the following equation.
\frac{x-7}{5}+\frac{1}{5}=-\frac{x}{10}
A. The solution set is (_).
B. The solution is the empty set.
An equation can be defined as a mathematical statement consisting of an equal symbol between two algebraic expressions that have the same value and the process of finding the value of the variable is called solving the equation.
The value of the variable for which the equations are equal is known as the solution of the equations and if there are more than one such solution, it is known as the solution set of the equations. Solution is basically the value of the variable for which the equality holds true.
\frac{x-7}{5}+\frac{1}{5}=-\frac{x}{10}
\frac{a}{b}+\frac{c}{b}=\frac{a+c}{b}
⇒\frac{x-7+1}{5}=-\frac{x}{10}
⇒\frac{x-6}{5}=-\frac{x}{10}
⇒x-6=5\left(-\frac{x}{10}\right)
⇒x-6=-\frac{x}{2}
⇒2\left(x-6\right)=-x
⇒2x-12=-x
⇒2x+x=12
⇒3x=12
⇒x=\frac{12}{3}
x=4
Hence, the answer is option A with the answer 4.
Write an algebraic expression of the verbal expression. 1 minus the quotient of r and 7
\frac{2{y}^{2}}{-5y}
x=3
y=5
Write an algebraic expression for: 6 more than a number c.
9B+C=A
B=?
How do you add
2x±5x?
-4\left(n-7\right)-4\left(-8n+6\right)
3.2\left(x+y\right)+2.3\left(x+y\right)+4x |
Moran's I - Wikipedia
The white and black squares are perfectly dispersed so Moran's I would be −1 using a Rook neighbors definition. If the white squares were stacked to one half of the board and the black squares to the other, Moran's I approaches +1 as N increases. A random arrangement of square colors would give Moran's I a value that is close to 0.
In statistics, Moran's I is a measure of spatial autocorrelation developed by Patrick Alfred Pierce Moran.[1][2] Spatial autocorrelation is characterized by a correlation in a signal among nearby locations in space. Spatial autocorrelation is more complex than one-dimensional autocorrelation because spatial correlation is multi-dimensional (i.e. 2 or 3 dimensions of space) and multi-directional.
1.1 Defining weights matrices
Moran's I is defined as
{\displaystyle I={\frac {N}{W}}{\frac {\sum _{i=1}^{N}\sum _{j=1}^{N}w_{ij}(x_{i}-{\bar {x}})(x_{j}-{\bar {x}})}{\sum _{i=1}^{N}(x_{i}-{\bar {x}})^{2}}}}
{\displaystyle N}
is the number of spatial units indexed by
{\displaystyle i}
{\displaystyle j}
{\displaystyle x}
is the variable of interest;
{\displaystyle {\bar {x}}}
{\displaystyle x}
{\displaystyle w_{ij}}
is a matrix of spatial weights with zeroes on the diagonal (i.e.,
{\displaystyle w_{ii}=0}
); and
{\displaystyle W}
{\displaystyle w_{ij}}
Defining weights matrices[edit]
{\displaystyle I}
can depend quite a bit on the assumptions built into the spatial weights matrix
{\displaystyle w_{ij}}
. The idea is to construct a matrix that accurately reflects your assumptions about the particular spatial phenomenon in question. A common approach is to give a weight of 1 if two zones are neighbors, and 0 otherwise, though the definition of 'neighbors' can vary. Another common approach might be to give a weight of 1 to
{\displaystyle k}
nearest neighbors, 0 otherwise. An alternative is to use a distance decay function for assigning weights. Sometimes the length of a shared edge is used for assigning different weights to neighbors. The selection of spatial weights matrix should be guided by theory about the phenomenon in question. The value of
{\displaystyle I}
is quite sensitive to the weights and can influence the conclusions you make about a phenomenon, especially when using distances.
Expected value[edit]
The expected value of Moran's I under the null hypothesis of no spatial autocorrelation is
{\displaystyle E(I)={\frac {-1}{N-1}}}
The null distribution used for this expectation is that the
{\displaystyle x}
input is permuted by a permutation
{\displaystyle \pi }
picked uniformly at random (and the expectation is over picking the permutation).
At large sample sizes (i.e., as N approaches infinity), the expected value approaches zero.
Its variance equals
{\displaystyle \operatorname {Var} (I)={\frac {NS_{4}-S_{3}S_{5}}{(N-1)(N-2)(N-3)W^{2}}}-(E(I))^{2}}
{\displaystyle S_{1}={\frac {1}{2}}\sum _{i}\sum _{j}(w_{ij}+w_{ji})^{2}}
{\displaystyle S_{2}=\sum _{i}\left(\sum _{j}w_{ij}+\sum _{j}w_{ji}\right)^{2}}
{\displaystyle S_{3}={\frac {N^{-1}\sum _{i}(x_{i}-{\bar {x}})^{4}}{(N^{-1}\sum _{i}(x_{i}-{\bar {x}})^{2})^{2}}}}
{\displaystyle S_{4}=(N^{2}-3N+3)S_{1}-NS_{2}+3W^{2}}
{\displaystyle S_{5}=(N^{2}-N)S_{1}-2NS_{2}+6W^{2}}
Values of I usually range from −1 to +1. Values significantly below -1/(N-1) indicate negative spatial autocorrelation and values significantly above -1/(N-1) indicate positive spatial autocorrelation. For statistical hypothesis testing, Moran's I values can be transformed to z-scores.
Moran's I is inversely related to Geary's C, but it is not identical. Moran's I is a measure of global spatial autocorrelation, while Geary's C is more sensitive to local spatial autocorrelation.
Moran's I is widely used in the fields of geography and geographic information science. Some examples include:
The analysis of geographic differences in health variables.[4]
Characterising the impact of lithium concentrations in public water on mental health.[5]
In dialectology to measure the significance of regional language variation.[6]
Defining an objective function for meaningful terrain segmentation for geomorphological studies[7]
^ Moran, P. A. P. (1950). "Notes on Continuous Stochastic Phenomena". Biometrika. 37 (1): 17–23. doi:10.2307/2332142. JSTOR 2332142. PMID 15420245.
^ Li, Hongfei; Calder, Catherine A.; Cressie, Noel (2007). "Beyond Moran's I: Testing for Spatial Dependence Based on the Spatial Autoregressive Model". Geographical Analysis. 39 (4): 357–375. doi:10.1111/j.1538-4632.2007.00708.x.
^ Cliff and Ord (1981), Spatial Processes, London
^ Getis, Arthur (3 Sep 2010). "The Analysis of Spatial Association by Use of Distance Statistics". Geographical Analysis. 24 (3): 189–206. doi:10.1111/j.1538-4632.1992.tb00261.x.
^ Helbich, M; Leitner, M; Kapusta, ND (2012). "Geospatial examination of lithium in drinking water and suicide mortality". Int J Health Geogr. 11 (1): 19. doi:10.1186/1476-072X-11-19. PMC 3441892. PMID 22695110.
^ Grieve, Jack (2011). "A regional analysis of contraction rate in written Standard American English". International Journal of Corpus Linguistics. 16 (4): 514–546. doi:10.1075/ijcl.16.4.04gri.
^ Alvioli, M.; Marchesini, I.; Reichenbach, P.; Rossi, M.; Ardizzone, F.; Fiorucci, F.; Guzzetti, F. (2016). "Automatic delineation of geomorphological slope units with r.slopeunits v1.0 and their optimization for landslide susceptibility modeling". Geoscientific Model Development. 9: 3975–3991. doi:10.5194/gmd-9-3975-2016.
Indicators of spatial association
Retrieved from "https://en.wikipedia.org/w/index.php?title=Moran%27s_I&oldid=1081457830" |
Grid Puzzles | Brilliant Math & Science Wiki
Ivan Koswara and Jimin Khim contributed
\hspace{2.3cm}
(The Art of Sudoku puzzle 3, by Thomas Snyder)
Puzzles come in many forms. A common way to make puzzles to be culture-neutral (does not depend on any knowledge of some culture or language) is to make them grid puzzles. These puzzles tend to have two parts, the rules and the actual puzzle in a grid. Sudoku is perhaps the most familiar example of grid puzzles.
There is no clear-cut definition of grid puzzles. Even the most natural definition of puzzles involving grids isn't very accurate, either. Grid puzzles usually have two parts: the rules part, which outlines the goal and the rules that the solver must follow, and the actual puzzle, which is presented in a grid, giving the name. The rules part is generally constant, which gives grid puzzles certain names (known as genres) depending on the rules; for example, a Sudoku puzzle is a puzzle where the rules part says "put 1-9 to the cells such that each row/column/box has one of each number," while a crossword puzzle is a puzzle where the rules part says "put a letter to each white cell such that each clue, reading across/down, is satisfied."
Since a grid puzzle requires both parts to form a complete puzzle, it's possible to give the rules in advance without spoiling the puzzle. In particular, it's possible to avoid any cultural requirements (knowledge of culture or language) in the puzzle itself, and since the rules can be given out, they can also be translated; this makes most grid puzzles culture-neutral. Sudoku is culture-neutral, since after you know the rules, you don't need any knowledge of culture or language to be able to finish a Sudoku, but crossword isn't. Word search is culture-neutral, since you don't need to know the meanings of the words. Elimination grid isn't, since you need to understand the language the facts are given in.
It's impossible to list every grid puzzle known to mankind, since there are so many, and it's very easy to construct a new one. Instead, this list tries to cover major genres.
Sudoku is a grid puzzle where the solver is presented by a
9 \times 9
grid divided into nine
3 \times 3
blocks, where some of the small cells have been filled with a digit between 1-9 inclusive. The solver needs to complete the grid, filling all remaining white cells with a single digit between 1-9 each, such that every row, column, and outlined
3 \times 3
block has exactly one of each digit between 1-9.
There are endless variations of Sudoku that add to, modify, or remove from the rules. For example, diagonal Sudoku also requires each diagonal to have exactly one of each 1-9. Irregular Sudoku replaces the blocks with irregular shapes; this time, instead of each
3 \times 3
block, each of these shapes needs to have exactly one of each 1-9. Non-consecutive Sudoku disallows two consecutive digits to share a side.
Crossword is a grid puzzle where the solver is presented by a grid where some cells are blackened. A contiguous line of two or more white cells running in the same row or column makes a word. Each word has an identifier (usually a small number at the top-left of the leftmost/topmost cell), and each identifier has a clue associated with it. The solver needs to fill a letter from the target alphabet (for example, in English crosswords, it's the standard A-Z) to each white cell, such that the letters making up a word satisfy the given clue. For example, a word of length nine that has an associated clue "smart; best website" can be filled by "BRILLIANT."
Crossword grids normally are just white and black cells. As a variation, the grid is entirely white instead, but there are thicker borders that separate words in the same line. Sometimes the grid can also be hexagonal.
A plain crossword isn't culture-neutral, since it requires knowledge of at least the language used. A way to make it more culture-neutral is math crossword, where the alphabet is the digits 0-9, and the clues are mathematical questions. To spice up math crosswords, one can use self-referential clues, where clues refer to the answers of other clues instead of standing on its own. Another attempt at making culture-neutral crosswords is a regular expression crossword.
Another variation is cryptic crossword, where the clues are cryptic. Generally cryptic clues are composed of two parts, a normal definition clue and a word puzzle clue. For example, "lap dancing friend" with three letters clues "PAL"; "friend" is its definition, while "lap dancing" is the word puzzle clue: "dancing" indicates that "lap" is to be anagrammed (its letters reordered).
Word search is a grid puzzle where the solver is presented by a grid full of letters, together with a word list. The solver has to find all words in the list in the grid. Words are always in a straight line, and generally can go in all eight compass directions (although some word searches only limit to the four cardinal directions or even just right/down).
Word search, on its own, isn't a particularly interesting puzzle. Ways to spice it up include adding words that aren't on the grid (the solver needs to identify these extraneous words), removing letters from the grid (the solver needs to identify what letters go in the missing spaces), and introducing new ways on how the words are located, for example by hiding them in a fractal grid.
Cite as: Grid Puzzles. Brilliant.org. Retrieved from https://brilliant.org/wiki/miscellaneous-grid-puzzle/ |
Find a basis for the eigenspace corresponding to each listed
Find a basis for the eigenspace corresponding to each listed eigenvalue.
dictetzqh 2021-11-21 Answered
A=\left[\begin{array}{cc}5& 0\\ 2& 1\end{array}\right],\lambda =1,5
To get the basis for the eigenspace, we first solve the system
\left(A-\lambda l\right)x=0
\lambda =1,A-l=
\left[\begin{array}{cc}5-1& 0\\ 2& 1-1\end{array}\right]
\left[\begin{array}{cc}4& 0\\ 2& 0\end{array}\right]
The augment matrix of
\left(A-l\right)x=0
\left[\begin{array}{ccc}4& 0& 0\\ 2& 0& 0\end{array}\right]
{x}_{1}=0
{x}_{2}
is free, the solution can be written in the form:
x={x}_{2}
\left[\begin{array}{c}0\\ 1\end{array}\right]
\left[\begin{array}{c}0\\ 1\end{array}\right]
Is a basis for the eigenspace.
\lambda =5,A-5l=
\left[\begin{array}{cc}5-5& 0\\ 2& 1-5\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 2& -4\end{array}\right]
The augmented matrix of (A-5l)x=0 is
\left[\begin{array}{ccc}0& 0& 0\\ 2& -4& 0\end{array}\right]
That leads to
2{x}_{1}-4{x}_{2}=0\to {x}_{1}=2{x}_{2}
. The solution can be writtem in the form:
x={x}_{2}
\left[\begin{array}{c}2\\ 1\end{array}\right]
\left[\begin{array}{c}2\\ 1\end{array}\right]
Jones figures that the total number of thousands of miles that a used auto can be driven before it would need to be junked is an exponential random variable with parameter
\frac{1}{20}
Smith has a used car that he claims has been driven only 10,000 miles.
If Jones purchases the car, what is the probability that she would get at least 20,000 additional miles out of it?
Repeat under the assumption that the lifetime mileage of the car is not exponentially distributed but rather is (in thousands of miles) uniformly distributed over (0, 40).
A pair of forces with equal magnitudes, opposite directions,and different lines of action is called a "couple". When acouple acts on a rigid object, the couple produces a torque thatdoes not depend on the location of the axis. The drawing shows acouple acting on a tire wrench, each force being perpendicular tothe wrench. Determine an expression for the torque produced by thecouple when the axis is perpendicular to the tired and passesthrough (a) point A, (b) point B, and (c) point C. Express youranswers in terms of the magnitude F of the force and the length Lof the wrench
A 2.0-kg projectile is fired with initial velocity components
{v}_{0x}=30
{v}_{0y}=40
m/s from a point on the earth's surface. Neglect any effects due to air resistance. What is the kinetic energy of the projectile when it reaches the highest point in its trajectory? How much work was done in firing the projectile?
In a scene in an action movie, a stuntman jumps from the top of one building to the top of another building 4.0m away. After a running start, he leaps at a velocity of 5.0 m/s at an angle of 15degrees with respect to the flat roof. Will he make it to the other roof, which is 2.5m shorter than the building he jumps from?
A.Which figure shows the loop that the must beused as the Ampèrean loop for finding for inside the solenoid?
B.Find , the z component of the magnetic field insidethe solenoid where Ampères
You hear a sound at 65 dB. What is the sound intensity level if the intensity of the sound is doubled?
Consider the following data and corresponding weights.
\begin{array}{|cc|}\hline {x}_{i}& Weight\left(Wi\right)\\ 3.2& 6\\ 2.0& 3\\ 2.5& 2\\ 5.0& 8\\ \hline\end{array}
Compute the weighted mean. b. Compute the sample mean of the four data values without weighting. Note the dif-ference in the results provided by the two computations. |
The first experiment (Exp1) deals with conventional average values such as oil flow patterns, static pressure measured on large times. Thus no information about unsteadiness will be included in the data. In Exp1, conventional wake surveys are performed with 10 hole probes. Drag is measured by strain gauge balance. The contribution of the drag is estimated for each part of the model. (front, slant rear, vertical rear base). The Reynolds number based on the model total length is 4.29 x 106.
The second experiment (Exp2) uses a two-components LDV system. Averages are performed on a high number of samples (40000) for long time durations, typically 5 minutes.
External turbulence level
EXP 1 Ahmed original (1984) 4.29x106 60ms-1 0.5% 5°, 12.5°, 25°, 30° Pw, Ui Cx, Flow structure
EXP2 Lienhart et al. (2000) 2.78x106 40ms-1 0.25% 25°, 35° First, second and third moments Pw, Flow structure
The test section is a ¾ open test section. Only the floor is a solid boundary. The homogeneity and far field conditions are not given.
The incoming turbulence intensity is less than 0.5% for 60 ms-1. No details on the incoming turbulence are available.
The size of the nozzle at the entrance of the test section is 3x3 m2.
The model is supposed to be smooth. No info is available on the turbulent/laminar nature of the boundary layers on the model. The influence of possible transition can be tested by CFD;
No information is available on the precision of the alignment of the model in the flow, although the symmetry on the visualizations give some confidence on this point. This sensitivity can also be checked by CFD.
No details on the typical time scales are provided. The influence of the unsteadiness of the wake on the averaging process can be estimated through URANS or LES.
Flow angle precision ± 0.4°.
Free stream dynamic pressure 1%.
Forces and moments are measured with balances with uncertainty of
± 0.2 N and ± 0.1Nm.
The test data include measurements of:
- Wall pressure
- Visualization of flow patterns on rear (slant) surface
- Wake survey (velocity vector plots, average values) :
- Mean velocity distribution in wake central plane
- Cross flow velocity for several downstream locations
- Drag coefficient: contributions of the pressure and friction drags to the total drag are estimated, as well as the repartition of the pressure drag among the front, slant part and vertical base.
All these data are provided for slant angles j = 5°, 12.5°, 25° and 30°.
An additional test is performed by fixing a splitter plate vertically in the wake of the body, in the plane of symmetry.
Some salient features of the time-averaged ground vehicle wake, S.R. Ahmed, G. Ramm and G. Faltin, SAE paper series Technical paper 840300, Detroit, 1984
The test section is a ¾ open test section. Only the floor is a solid boundary. The homogeneity and far field conditions are not given. However the blockage is assumed to be less than 4%.
The incoming turbulence intensity is less than 0.25% for 40 ms-1 measured by hot wire anemometry 400 mm upstream of the model. The viscosity ratio is about 10.
The models are supposed to be smooth. Transition to turbulence of the boundary layer on the front part is triggered.
No information is available on the accuracy of the alignment of the model in the flow, although the symmetry on the visualizations give some confidence on this point. This sensitivity can also be checked by CFD.
No detail on the typical time scales are provided. The influence of the unsteadiness of the wake on the averaging process can be estimated through URANS or LES.
Error on mean velocities is less than 0.005% of local mean in the outer flow. In the wake region the accuracy is assumed to be 1% for mean values and 1.5% for RMS.
LDA measurements of mean velocities: U, V, W, Reynolds stresses
{\displaystyle {\overline {u'u'}},{\overline {v'v'}},{\overline {w'w'}},{\overline {u'v'}},{\overline {u'w'}}}
and third order moments
{\displaystyle {\overline {u'u'u'}},{\overline {v'v'v'}},{\overline {w'w'w'}},{\overline {u'u'v'}},{\overline {u'u'w'}},{\overline {u'v'v'}},{\overline {u'w'w'}}}
in some planes for 2 slant angles:
25° slant angle:
planes: Ahmed_25_y=0_global.dat (whole flow); Ahmed_25_y=0_focus.dat (focus on the slant part); y=100; y=180; y=195; y=-195.dat x=-178; x=-138;
x=-88; x=-38; x=0; x=80; x=200; x=500
planes: Ahmed_35_y=0_global.dat (whole flow); Ahmed_35_y=0_focus.dat (focus on the slant part); y=100; y=180
x=-88; x=0; x=80; x=200; x=500
Hot wire measurements in the boundary layer in the symmetry plane at different x-location: mean velocities, Reynolds stresses and third moments (only u-w components):
25° slant angle:x= -243, -223, -203, -183, -163, -143, -123, -103, -83, -63, -43, -23, -3
Pressure coefficients on the rear of the body:
25° slant angle
Inlet.dat
Flow and Turbulence Structures in the Wake of a Simplified Car Model (Ahmed model),
H. Lienhart, C. Stoots and S. Becker, DGLR Fach Symp. Der AG STAB, Stuttgart University, 15-17 nov. 2000
H. Lienhart and S. Becker, Flow and turbulence structures in the wake of a simplified car model, SAE Paper 2003-01-0656, 2003. |
Solve the equation: \tan^2x-3\tan x-4=0
{\mathrm{tan}}^{2}x-3\mathrm{tan}x-4=0
{\mathrm{tan}}^{2}x-3\mathrm{tan}x-4=0
by using inverse trigonometry
{\mathrm{tan}}^{2}x-3\mathrm{tan}x-4=0
\left(\mathrm{tan}x-4\right)\left(\mathrm{tan}x+1\right)=0
\mathrm{tan}x-4=0
\mathrm{tan}x+1=0
\mathrm{tan}x=4
\mathrm{tan}x=-1
x=\mathrm{arctan}\left(4\right)
x=\mathrm{arctan}\left(-1\right)
x=\eta \pi +\mathrm{arctan}\left(4\right)
x=\eta \pi -\frac{\pi }{4}
\mathrm{sin}t=,\mathrm{cos}t=,\text{ }\text{and}\text{ }\mathrm{tan}t=
Evaluate the integral using trigonometric substitutions.
\int \frac{x}{\sqrt{3-2x-{x}^{2}}}dx
\int \frac{{x}^{2}}{\mathrm{tan}\left\{x\right\}-x}dx
x\in \left(0,\frac{\pi }{2}\right)
\mathrm{sin}2\theta +\mathrm{cos}2\theta =\mathrm{sin}\theta +\mathrm{cos}\theta
I tried to do like this
=2\mathrm{sin}\theta \mathrm{cos}\theta +{\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta
=\mathrm{sin}\theta \mathrm{cos}\theta +{\mathrm{cos}}^{2}\theta +\mathrm{sin}\theta \mathrm{cos}\theta -{\mathrm{sin}}^{2}\theta
=\mathrm{cos}\theta \left(\mathrm{sin}\theta +\mathrm{cos}\theta \right)+\mathrm{sin}\theta \left(\mathrm{cos}\theta -\mathrm{sin}\theta \right)
\frac{6\mathrm{tan}x}{1-{\mathrm{tan}}^{2}x}
3\mathrm{tan}2x
16{\mathrm{cos}}^{6}\left(t\right)+16{\mathrm{sin}}^{6}\left(t\right)+48{\mathrm{sin}}^{2}\left(t\right){\mathrm{cos}}^{2}\left(t\right)
\underset{x\to 0}{lim}\frac{\mathrm{sin}\left(\pi {\mathrm{cos}}^{2}\left(\frac{x}{2}\right)\right)}{\mathrm{sin}\left(\mathrm{sin}\left(x\right)\right)}
I have the following solution, only the first equality of which bothers me:
\underset{x\to 0}{lim}\frac{\mathrm{sin}\left(\pi {\mathrm{cos}}^{2}\left(\frac{x}{2}\right)\right)}{\mathrm{sin}\left(\mathrm{sin}\left(x\right)\right)}=\underset{x\to 0}{lim}\frac{\mathrm{sin}\left(\pi {\mathrm{sin}}^{2}\left(\frac{x}{2}\right)\right)}{\mathrm{sin}\left(\mathrm{sin}\left(x\right)\right)}=\underset{x\to 0}{lim}\pi \frac{\mathrm{sin}\left(\frac{x}{2}\right)}{2\mathrm{cos}\left(\frac{x}{2}\right)}.\frac{2\mathrm{sin}\left(\frac{x}{2}\right)\mathrm{cos}\left(\frac{x}{2}\right)}{\mathrm{sin}\left(2\mathrm{sin}\left(\frac{x}{2}\right)\mathrm{cos}\left(\frac{x}{2}\right)\right)}.\frac{\mathrm{sin}\left(\pi {\mathrm{sin}}^{2}\left(\frac{x}{2}\right)\right)}{\pi {\mathrm{sin}}^{2}\left(\frac{x}{2}\right)}=0
What is the justification for replacing
{\mathrm{cos}}^{2}
{\mathrm{sin}}^{2} |
Batch Compute Steady-State Operating Points for Parameter Variation - MATLAB & Simulink - MathWorks Switzerland
Vary Single Parameter
Multidimensional Parameter Grids
Vary Multiple Parameters
Batch Trim Model for Parameter Variations
Batch Trim Model at Known States Derived from Parameter Values
Block parameters configure a Simulink® model in several ways. For example, you can use block parameters to specify various coefficients or controller sample times. You can also use a discrete parameter, like the control input to a Multiport Switch block, to control the data path within a model. Varying the value of a parameter helps you understand its impact on the model behavior. Also, you can vary the parameters of a plant model in a control system to study the robustness of the controller to plant variations.
When trimming a model using findop, you can specify a set of parameter values for which to trim the model. The full set of values is called a parameter grid or parameter samples. findop computes an operating point for each value combination in the parameter grid. You can vary multiple parameters, thus extending the parameter grid dimension.
You can vary any model parameter with a value given by a variable in the model workspace, the MATLAB® workspace, or a data dictionary. In cases where the varying parameters are all tunable, findop requires only one model compilation to find operating points for varying parameter values. This efficiency is especially advantageous for models that are expensive to compile repeatedly.
To vary the value of a single parameter for batch trimming with findop, specify the parameter grid as a structure having two fields. The Name field contains the name of the workspace variable that specifies the parameter. The Value field contains a vector of values for that parameter to take during trimming.
After you create the structure param, pass it to findop as the param input argument.
When you vary more than one parameter at a time, you generate parameter grids of higher dimension. For example, varying two parameters yields a parameter matrix, and varying three parameters yields a 3-D parameter grid. Consider the following parameter grid used for batch trimming:
Here, you vary the values of three parameters, a, b, and c. The samples form a 3-by-4-by-5 grid. op is an array with same dimensions that contains corresponding trimmed operating point objects.
To vary the value of multiple parameters for batch trimming with findop, specify parameter samples as a structure array. The structure has an entry for each parameter whose value you vary. The structure for each parameter is the same as described in Vary Single Parameter. You can specify the Value field for a parameter as an array of any dimension. However, the size of the Value field must match for all parameters. Corresponding array entries for all the parameters, also referred to as a parameter grid points, must map to a specified parameter combination. When the software trims the model, it computes an operating point for each grid point.
\begin{array}{l}a=\left\{a1,a2\right\}\\ b=\left\{b1,b2\right\}\end{array}
You want to trim the model for every combination of a and b, also referred to as a full grid:
\left\{\begin{array}{cc}\left({a}_{1},{b}_{1}\right),& \left({a}_{1},{b}_{2}\right)\\ \left({a}_{2},{b}_{1}\right),& \left({a}_{2},{b}_{2}\right)\end{array}\right\}
If your model is complex or you vary the value of many parameters, trimming the model for the full grid can become expensive. In this case, you can specify a subset of the full grid using a table-like approach. Using the example in Specify Full Grid, suppose that you want to trim the model for the following combinations of a and b:
\left\{\left({a}_{1},{b}_{1}\right),\left({a}_{1},{b}_{2}\right)\right\}
This example shows how to obtain multiple operating points for a model by varying parameter values. You can study the controller robustness to plant variations by batch linearizing the model using the trimmed operating points.
Create a default operating point specification for the model, which specifies that both model states are unknown and must be at steady state in the trimmed operating point.
By default, findop displays an operating point search report in the Command Window for each trimming operation. To suppress the report display, create a trimming option set and turn off the operating point search report display.
Trim the model using the specified operating point specification, parameter grid, and option set.
findop trims the model for each parameter combination. The software uses only one model compilation. op is a 3-by-4 array of operating point objects that correspond to the specified parameter grid points.
View the operating point in the first row and first column of op.
op(1,1)
This example shows how to batch trim a model when the specified parameter variations affect the known states for trimming.
In the Batch Trim Model for Parameter Variations example, the model is trimmed to meet a single operating point specification that contains unknown states. In other cases, the model states are known for trimming, but depend on the values of the varying parameters. In this case, you cannot batch trim the model using a single operating point specification. You must create a separate specification for each parameter value grid point.
In this model, the aerodynamic forces and moments depend on the speed, , and incidence, .
Vary the and parameters, and create a 6-by-4 parameter grid.
nA = 6; % number of alpha values
nV = 4; % number of V values
[alphaGrid,vGrid] = ndgrid(alphaRange,vRange);
Since some known state values for trimming depend on the values of and , you must create a separate operating point specification object for each parameter combination.
for j = 1:nV
% Set parameter values in model.
alpha_ini = alphaGrid(i,j);
v_ini = vGrid(i,j);
% Create default specifications based on the specified parameters.
opspec(i,j) = operspec(sys);
% Specify which states are known and which states are at steady state.
opspec(i,j).States(1).Known = [1;1];
opspec(i,j).States(1).SteadyState = [0;0];
opspec(i,j).States(2).Known = 1;
opspec(i,j).States(2).SteadyState = 0;
Create a parameter structure for batch trimming. Specify a name and value grid for each parameter.
params(1).Name = 'alpha_ini';
params(1).Value = alphaGrid;
params(2).Name = 'v_ini';
params(2).Value = vGrid;
Trim the model using the specified parameter grid and operating point specifications. When you specify an array of operating point specifications and varying parameter values, the dimensions of the specification array must match the parameter grid dimensions.
op = findop(sys,opspec,params,opt);
findop trims the model for each parameter combination. op is a 6-by-4 array of operating point objects that correspond to the specified parameter grid points.
findop | operspec | linearize |
ChangeVariables - Maple Help
Home : Support : Online Help : Education : Student Packages : ODEs : ChangeVariables
transform an ODE using a change of variables
ChangeVariables(tr, ODE)
ChangeVariables(tr, ODE, y(x))
ChangeVariables(tr, ODE, y(x), u(t))
a set or list of two transformation equations
name; the existing dependent variable
name; the existing independent variable
name; the new dependent variable
name; the new independent variable
ChangeVariables(tr, ODE, y(x), u(t)) applies the change of variables tr to the given ODE.
The second and third arguments, y(x) and u(t), representing the existing and new dependent variables, are optional; however they must be given if the existing and dependent variables cannot be determined from the form of the transformation.
There must be two transformation equations, each of which specifies the new form of one of the following three entities:
x,y\left(x\right),\frac{ⅆ}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}y\left(x\right)
If the left-hand sides of the transformation equations are x and diff(y(x),x), then the ODE should not contain y(x) (independently of diff(y(x),x)). If the left-hand sides of the transformation equations are y(x) and diff(y(x),x), then the ODE should not contain x (independently of y(x) and diff(y(x),x)).
\mathrm{with}\left(\mathrm{Student}[\mathrm{ODEs}]\right):
\mathrm{ode1}≔\frac{{x}^{2}\left(y\left(x\right)+1\right)}{y\left(x\right)}+\left(x-1\right)\mathrm{diff}\left(y\left(x\right),x\right)={x}^{2}\left(y\left(x\right)+\frac{1}{y\left(x\right)}\right)
\textcolor[rgb]{0,0,1}{\mathrm{ode1}}\textcolor[rgb]{0,0,1}{≔}\frac{{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\right)}{\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\right)}\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\right)\textcolor[rgb]{0,0,1}{}\left(\frac{\textcolor[rgb]{0,0,1}{ⅆ}}{\textcolor[rgb]{0,0,1}{ⅆ}\textcolor[rgb]{0,0,1}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\right)\right)\textcolor[rgb]{0,0,1}{=}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\right)}\right)
\mathrm{ChangeVariables}\left({x=u\left(t\right),y\left(x\right)=t},\mathrm{ode1}\right)
\frac{{\textcolor[rgb]{0,0,1}{u}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\right)}{\textcolor[rgb]{0,0,1}{t}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{u}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}}{\frac{\textcolor[rgb]{0,0,1}{ⅆ}}{\textcolor[rgb]{0,0,1}{ⅆ}\textcolor[rgb]{0,0,1}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\textcolor[rgb]{0,0,1}{u}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)}\textcolor[rgb]{0,0,1}{=}{\textcolor[rgb]{0,0,1}{u}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{t}}\right)
\mathrm{ode2}≔\mathrm{diff}\left(y\left(x\right),x,x\right)=\frac{{x}^{2}}{x-1}\mathrm{diff}\left(y\left(x\right),x\right)-\frac{{x}^{2}}{x-1}
\textcolor[rgb]{0,0,1}{\mathrm{ode2}}\textcolor[rgb]{0,0,1}{≔}\frac{{\textcolor[rgb]{0,0,1}{ⅆ}}^{\textcolor[rgb]{0,0,1}{2}}}{\textcolor[rgb]{0,0,1}{ⅆ}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{=}\frac{{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\left(\frac{\textcolor[rgb]{0,0,1}{ⅆ}}{\textcolor[rgb]{0,0,1}{ⅆ}\textcolor[rgb]{0,0,1}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\right)\right)}{\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{-}\frac{{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}}{\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}}
\mathrm{ChangeVariables}\left({x=t,\mathrm{diff}\left(y\left(x\right),x\right)=u\left(t\right)},\mathrm{ode2}\right)
\frac{\textcolor[rgb]{0,0,1}{ⅆ}}{\textcolor[rgb]{0,0,1}{ⅆ}\textcolor[rgb]{0,0,1}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\textcolor[rgb]{0,0,1}{u}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)\textcolor[rgb]{0,0,1}{=}\frac{{\textcolor[rgb]{0,0,1}{t}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{u}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)}{\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{-}\frac{{\textcolor[rgb]{0,0,1}{t}}^{\textcolor[rgb]{0,0,1}{2}}}{\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}}
The Student[ODEs][ChangeVariables] command was introduced in Maple 2021. |
Clutch schedule for a four-speed carrier ring-carrier ring transmission - MATLAB - MathWorks Nordic
4-Speed CR-CR
Clutch schedule for a four-speed carrier ring-carrier ring transmission
The 4-Speed CR-CR block consists of two planetary gear sets and five disk friction clutches. The follower shaft connects to the planet gear carrier of the output planetary gear and the ring gear of the input planetary gear. The clutches are configured to implement four ratios, one reverse ratio, and neutral. The reverse clutch and two of the clutches determine which gears the base shaft connects to. The other two clutches act as brakes, grounding various gears of the planetary sets to the transmission housing.
This diagram shows a four-speed carrier ring-carrier ring transmission. The labels for the gear components are superimposed on the input and output gears. The table lists the gear and clutch components that are labeled in the diagram.
I.P.G. Input planetary gear
O.P.G. Output planetary gear
R Reverse clutch
A–B Forward clutches that control the power flow path
C–D Forward, braking clutches
{g}_{1}=\frac{{N}_{RI}}{{N}_{SI}},
{g}_{2}=\frac{{N}_{RO}}{{N}_{SO}},
NRI is the number of teeth in the planetary ring gear on the input shaft side.
NSI is the number of teeth in the planetary sun gear on the input shaft side.
NRO is the number of teeth in the planetary ring gear on the output shaft side.
NSO is the number of teeth in the planetary sun gear on the output shaft side.
The table shows the clutch schedule, drive-ratio expressions, drive-ratio default values, and the power-flow diagrams for each gear of the 4-Speed CR-CR block.
\frac{{g}_{1}}{1+{g}_{1}}
1
\frac{{g}_{1}+{g}_{2}+1}{1+{g}_{1}}
{g}_{2}+1
-{g}_{1}
4-Speed Ravigneaux | 6-Speed Lepelletier | 7-Speed Lepelletier | 8-Speed | 9-Speed | 10-Speed |
\frac{\left(a+b\right)}{\left(u+v\right)}
\frac{\left(a+b\right)}{\left(u-v\right)}
\left(\sqrt{b}:\sqrt{a}\right)
[66*(5/18)] m/sec = (55/3) m/sec
Time taken to pass the man = [110*(3/55)]m/sec = 6 sec.
A train travelling at a speed of 75 mph enters a tunnel miles long. The train is mile long. How long does it take for the train to pass through the tunnel from the moment the front enters to the moment the rear emerges?
A) 2.5 min B) 3 min
D) 3.5 min
Answer & Explanation Answer: B) 3 min
Total distance covered =miles =miles
Time taken = hrs = hrs = = 3 min
Relative speed = (60 + 40) km/hr =[ 100 x ( 5 / 18 ) ]m/sec = ( 250 /9 ) m/sec.
Required time = [ 300 x ( 9/250 ) ] sec = ( 54/ 5 )sec = 10.8 sec.
A) 50 m B) 150 m
C) 200 m D) data inadequate
Answer & Explanation Answer: B) 150 m
Then, (x/y)= 15 y =(x/15)
(x+100)/25 = x/15
=> 15(x + 100) = 25x
=> 15x + 1500 = 25x
=> 1500 = 10x
=> x = 150 m.
A train moves with a speed of 108 kmph. Its speed in metres per second is :
108 kmph = 108*[5/18] m/sec = 30 m / s.
Two trains are moving in opposite directions at 60 km/hr and 90 km/hr. Their lengths are 1.10 km and 0.9 km respectively. The time taken by the slower train to cross the faster train in seconds is ?
Relative speed = 60 + 90 = 150 km/hr.
Distance covered = 1.10 + 0.9 = 2 km = 2000 m.
Required time = 2000 x 3/125 = 48 sec.
Two cogged wheels of which one has 32 cogs and other 54 cogs, work into each other. If the latter turns 80 times in three quarters of a minute, how often does the other turn in 8 seconds?
Less Cogs more turns and less time less turns
Number of turns required=80 × 54/32 × 8/45 = 24 times
Two trains of equal length, running with the speeds of 60 and 40 kmph, take 50 seconds to cross each other while they are running in the same direction. What time will they take to cross each other if they are running in opposite directions ?
C) 12 sec D) 8 sec
Answer & Explanation Answer: A) 10 sec
Relative Speed = 60 -40 = 20 x 5/18 = 100/18
Distance = 50 x 100/18 = 2500/9
Relative Speed = 60 + 40 = 100 x 5/18
Time = 2500/9 x 18/500 = 10 sec. |
Classify whether this series converges sum_(k = 1)^oo ((-1)^k xx k)
Classify whether this series convergessum_(k = 1)^oo ((-1)^k xx k)
Classify whether this series converges
\sum _{k=1}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{k}×k}{6{k}^{4}+1}
Ayesha Gomez
\sum _{k=1}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{k}×k}{6{k}^{4}+1}
{a}_{k}=\frac{k}{6{k}^{4}+1}
{b}_{k}=\frac{k}{6{k}^{4}+1}
{b}_{k}\ge 0
\underset{k+0}{lim}\frac{k}{6{k}^{4}+1}=0
Then, by alternative series test we can converge.
The term of series are alternative and limit of absolute value is 0, so series converge by alternative series test.
How do you find the Maclaurin series of
f\left(x\right)=\mathrm{ln}\left(1+x\right)?
Tell whether the series converge. If it converges, find the sum.
\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{\pi }{2}\right)}^{n}
\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{3}{7}\right)}^{n}
\sum _{n=2}^{\mathrm{\infty }}\frac{1}{n\sqrt{\mathrm{ln}n}}
The next number in the series 3, 4, 6, 7, 9, 10 is: 12 11 15 17 13
Find the sim of each of the following series.
\sum _{n=1}^{\mathrm{\infty }}n{x}^{n},\text{ }|x|<1
\sum _{n=1}^{\mathrm{\infty }}\frac{n}{{8}^{n}}
In series A, the first term is 2, and the common ratio is 0.5. In series B, the first term is 3, and the two infinite series have the same sum. What is the ratio in series B?
Write out he first four terms of the Maclaurin series of f(x) if
f\left(0\right)=2,\text{ }{f}^{\prime }\left(0\right)=3,\text{ }f{}^{″}\left(0\right)=4,\text{ }f{}^{‴}\left(0\right)=12 |
Home : Support : Online Help : Mathematics : Calculus : Integration : Elliptic
Elliptic integrals are integrals of the form
{\int }_{a}^{b}R\left(x,\sqrt{y}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}ⅆx
with R a rational function and y a polynomial of degree 3 or 4. This is the algebraic form of an elliptic integral. There are also trig forms (rational functions of sin and cos and a square root of a quadratic polynomial in sin and cos) and hyperbolic trig forms.
Elliptic integrals are reduced to their Legendre normal form in terms of elementary functions and the Elliptic functions EllipticF, EllipticE, and EllipticPi (or their complete versions).
Elementary answer
\mathrm{int}\left(\frac{\mathrm{sqrt}\left(1+{x}^{4}\right)}{1-{x}^{4}},x=0..\frac{1}{2}\right)
\frac{\sqrt{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{ln}}\textcolor[rgb]{0,0,1}{}\left(\sqrt{\textcolor[rgb]{0,0,1}{17}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{2}}\right)}{\textcolor[rgb]{0,0,1}{8}}\textcolor[rgb]{0,0,1}{-}\frac{\sqrt{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{ln}}\textcolor[rgb]{0,0,1}{}\left(\sqrt{\textcolor[rgb]{0,0,1}{17}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{2}}\right)}{\textcolor[rgb]{0,0,1}{8}}\textcolor[rgb]{0,0,1}{-}\frac{\sqrt{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{arctan}}\textcolor[rgb]{0,0,1}{}\left(\frac{\sqrt{\textcolor[rgb]{0,0,1}{17}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{2}}}{\textcolor[rgb]{0,0,1}{4}}\right)}{\textcolor[rgb]{0,0,1}{4}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{\mathrm{\pi }}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{2}}}{\textcolor[rgb]{0,0,1}{8}}
\mathrm{assume}\left(0<k,k<1\right)
\mathrm{int}\left(\frac{{x}^{2}}{\mathrm{sqrt}\left(\left(1-{x}^{2}\right)\left(1-{k}^{2}{x}^{2}\right)\right)},x=0..k\right)
\frac{\textcolor[rgb]{0,0,1}{\mathrm{EllipticF}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{\mathrm{k~}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{k~}}\right)}{{\textcolor[rgb]{0,0,1}{\mathrm{k~}}}^{\textcolor[rgb]{0,0,1}{2}}}\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{\mathrm{EllipticE}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{\mathrm{k~}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{k~}}\right)}{{\textcolor[rgb]{0,0,1}{\mathrm{k~}}}^{\textcolor[rgb]{0,0,1}{2}}}
Answer as sum of roots
\mathrm{ans}≔\mathrm{int}\left(\frac{1}{\left({x}^{4}+2\right)\mathrm{sqrt}\left(4-5{x}^{2}+{x}^{4}\right)},x=0..\frac{1}{4}\right)
\textcolor[rgb]{0,0,1}{\mathrm{ans}}\textcolor[rgb]{0,0,1}{≔}\frac{\left(\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{\sum }_{\textcolor[rgb]{0,0,1}{\mathrm{_α}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{RootOf}}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{\mathrm{_Z}}}^{\textcolor[rgb]{0,0,1}{4}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{2}\right)}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{EllipticPi}}\textcolor[rgb]{0,0,1}{}\left(\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{4}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}\frac{{\textcolor[rgb]{0,0,1}{\mathrm{_α}}}^{\textcolor[rgb]{0,0,1}{2}}}{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{2}}\right)\right)}{\textcolor[rgb]{0,0,1}{16}}
Can evaluate to floating point:
\mathrm{evalf}\left(\mathrm{ans}\right)
\textcolor[rgb]{0,0,1}{0.06331207100}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{0.}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{I}
\mathrm{evalf}\left(\mathrm{ans},20\right)
\textcolor[rgb]{0,0,1}{0.063312071018173992738}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{0.}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{I}
Trig form
\mathrm{int}\left(\mathrm{sqrt}\left(1+2\mathrm{sin}\left(x\right)\right),x=0..\frac{\mathrm{\pi }}{2}\right)
\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{EllipticK}}\textcolor[rgb]{0,0,1}{}\left(\frac{\sqrt{\textcolor[rgb]{0,0,1}{3}}}{\textcolor[rgb]{0,0,1}{2}}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{EllipticF}}\textcolor[rgb]{0,0,1}{}\left(\frac{\sqrt{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{3}}}{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{,}\frac{\sqrt{\textcolor[rgb]{0,0,1}{3}}}{\textcolor[rgb]{0,0,1}{2}}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{EllipticE}}\textcolor[rgb]{0,0,1}{}\left(\frac{\sqrt{\textcolor[rgb]{0,0,1}{3}}}{\textcolor[rgb]{0,0,1}{2}}\right)\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{EllipticPi}}\textcolor[rgb]{0,0,1}{}\left(\frac{\sqrt{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{3}}}{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{3}}{\textcolor[rgb]{0,0,1}{4}}\textcolor[rgb]{0,0,1}{,}\frac{\sqrt{\textcolor[rgb]{0,0,1}{3}}}{\textcolor[rgb]{0,0,1}{2}}\right)
Indefinite trig form
\mathrm{Itrig}≔\mathrm{int}\left(\frac{1}{\mathrm{sqrt}\left(1+2\mathrm{cos}\left(x\right)\right)},x\right)
\textcolor[rgb]{0,0,1}{\mathrm{Itrig}}\textcolor[rgb]{0,0,1}{≔}\frac{\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{InverseJacobiAM}}\textcolor[rgb]{0,0,1}{}\left(\frac{\textcolor[rgb]{0,0,1}{x}}{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{3}}}{\textcolor[rgb]{0,0,1}{3}}\right)}{\textcolor[rgb]{0,0,1}{3}}
\mathrm{simplify}\left(\mathrm{combine}\left(\mathrm{diff}\left(\mathrm{Itrig},x\right)-\frac{1}{\mathrm{sqrt}\left(1+2\mathrm{cos}\left(x\right)\right)},\mathrm{trig}\right)\right)
\textcolor[rgb]{0,0,1}{0}
Labahn, G., and Mutrie, M. "Reduction of Elliptic Integrals to Legendre Normal Form." University of Waterloo Tech Report 97-21, Department of Computer Science, 1997. |
Dilation in Math | Definition & Examples (Video)
Dilation in Math (Definition & Examples)
Dilation Definition
Dilation is the enlarging or shrinking of a mathematical element (a point on a coordinate grid, polygon, line segment) using a specific scale factor.
Dilation is one of the five major transformations in geometry. Dilation does not change the shape of the object from preimage to image. The position and size of a figure can change, but not the shape.
Dilation Examples
Dilations on the Coordinate Plane
Dilations not on the Coordinate Plane
All dilations begin with a center. This can be a single point on a coordinate grid, the middle of a polygon, or any fixed point in space.
From that center of dilation, the preimage – the mathematical element before scaling – is enlarged, inverted, or shrunk to form the image. The preimage and image are similar figures.
You can think of the preimage as the original figure, and the image as the new figure.
Here is a square. Its center of dilation is its exact middle, so any dilation from the square will still be a square with all parts equidistant from the center point:
The center of the dilation does not need to be inside the shape. It could be one vertex of the polygon. Here we still have a square as the preimage, but the center of thedilation is the top-left vertex, so the dilated images (one smaller, one larger) all share that same vertex.
The scale factor of a dilation is the amount by which all original terms are enlarged or shrunk, usually on a coordinate plane. If you multiply the original coordinates:
By whole numbers other than
1
, you enlarge the preimage in producing the image.
1
, you produce an image congruent to the preimage.
By fractions or decimals, you shrink the preimage to produce the image.
By negative numbers, you will produce an image that is the inverse (upside down) of the preimage, equidistant from the center of dilation but on the opposite side.
Let's see the scale factor at work on a coordinate plane. Here is a trapezoid on a coordinate grid with the origin
\left(0, 0\right)
as the center of dilation:
If we choose a scale factor of
2
, every plotted point on the polygon will be multiplied by
2
to create the enlarged image.
Take the coordinates of
vertex A
in the preimage at
\left(-2, 1\right)
and multiply times
2
, producing
\left(-4, 2\right)
as the new
A\text{'}
for the image.
Vertex D
\left(-3, -1\right)
D\text{'}
\left(-6, -2\right)
Can you calculate the coordinates for vertices
B\text{'}
C\text{'}
Point B
\left(1, 1\right)
enlarges to
B\text{'}
\left(2, 2\right)
Point C
\left(3, -1\right)
C\text{'}
\left(6, -2\right)
The image or enlargement has sides twice the length of the original preimage trapezoid. The new dilation is also the same shape as the preimage. The two polygons are similar:
Notice that the vertices of our image points share almost the same designation as the preimage vertices, but with the prime indicator:
A\text{'}
B\text{'}
C\text{'}
D\text{'}
The trapezoid used our origin
\left(0, 0\right)
as the center of dilation. You can move the center to any place on the coordinate plane you wish. The "distance" from the center of the dilation to each point is calculated as the difference between the two sets of coordinate points. Here is a point
\left(4, 5\right)
with the center of dilation at
\left(1, 3\right)
. You are asked to plot the image point using a scale factor of
4
You are essentially calculating the slope of the line from the center of dilation to both coordinate pairs (preimage and image):
\left(1, 3\right)
Plotted point at
\left(4, 5\right)
4
The plotted point of our preimage is
3
horizontal units away from the center of dilation. It is
2
vertical units away. Multiplying these distances times the scale factor,
4
, means our new point must be
12
horizontal (x-axis) units from the center of dilation, and
8
vertical (y-axis) units from the same center of dilation, at
\left(13, 11\right)
You can check this by drawing a line from the center of dilation through the preimage point. The dilation (image) must lie on that line, which it does.
When you do not have coordinates for the points of a figure on a coordinate grid, you must calculate each line segment of the figure and multiply it times the absolute value of the scale factor:
Image = Preimage × |Scale Factor|
If the scale factor is negative, you will be going in the opposite direction from the point of dilation, but you must take the absolute value to get the actual distance. (Physical distances cannot be negative.)
Suppose you have a preimage of a polygon with a side,
AB
9 cm
long, and a scale factor of
-6
\left|-6\right|
6
, so the image will be larger by a factor of
6
. This makes side
A\text{'}B\text{'}
54 cm
This new side will be opposite the center of dilation from the preimage, as will all the other sides of the polygon since the scale factor was negative.
Suppose you have a preimage of a polygon with a side
CD
9 cm
long and a scale factor of
\frac{1}{3}
. In this case, the image side
C\text{'}D\text{'}
\frac{1}{3}
the preimage length, or
3 cm
The definition of dilation in math
What the center of dilation is
To understand the scale factor of a dilation
How to perform a dilation with and without a coordinate grid |
Recovering a Rotation Matrix From Three Direction Cosines | IDETC-CIE | ASME Digital Collection
Recovering a Rotation Matrix From Three Direction Cosines
Mengdi Xu,
Xu, M, & Chirikjian, GS. "Recovering a Rotation Matrix From Three Direction Cosines." Proceedings of the ASME 2018 International Design Engineering Technical Conferences and Computers and Information in Engineering Conference. Volume 5B: 42nd Mechanisms and Robotics Conference. Quebec City, Quebec, Canada. August 26–29, 2018. V05BT07A079. ASME. https://doi.org/10.1115/DETC2018-85825
In this paper, we propose a new parameterization method to represent rotation matrices using the angles
ϕ→
recovered from the three direction cosines that lie on the diagonal. The map from the possible configuration space of the new variable
ϕ→
to the solid ball model in axis-angle coordinates is constructed. We also introduce a bi-invariant metric and two left-invariant metrics for measuring the distance in configuration space which could be the foundation for path planning in
ϕ→
space. We further analyze the Jacobian matrix and singularities to better understand the manipulability.
Rotation, Jacobian matrices, Path planning
Singularity Analysis of a Novel 4-DOFs Parallel Robot H4 by Using Screw Theory
Type Synthesis of Uncoupled 2T2R Parallel Manipulators
The Shortest Path for a Point Passing Through Obstacles Represented by Quadratic Curves
The Research of Local Path Planning for Mobile Robots Based on Grid Method |
Write and solve a proportion for the following problem.
In a recent survey for the student council, Dominique found that
150
students out of a total of
800
students on campus did not like soda. If half of the student body was going to attend a dance, how many students could she expect would want soda?
150
students dislike soda, how many students like soda?
800-150=650
650
students out of
800
like soda. How many students are attending the school dance?
If half of the student body is attending the dance, then:
\frac{800}{2}=400
400
students are attending the school dance. How many of those
400
students are expected to like soda?
650
800
that like soda, we can set up a proportion to find the expected number of students at the dance that will like soda.
\frac{650}{800}=\frac{x}{400}
x
x=325 |
Graphene/hexagonal boron nitride (h-BN) is a type of hybrid material to regulate the electronic properties of pristine graphene and is recognized as having potential application in functional devices. Its fracture behavior is one of the most important parameters to affect the device performance. In this work, the fracture behaviors of hybrid graphene/h-BN sheets with cracks were studied using molecular dynamics method. Effects of the crack size, type and location on the failure behavior of the hybrid sheets were considered and analyzed. For most of the models, both Young’s modulus and the fracture strength reduced with the increasing crack size. A threshold of the crack size was found for the models: when the crack size was larger than 0.1
L
L
is the periodic length of the sheet), Young’s modulus dropped rapidly, while the reduction of the fracture strength slowed down. Crack location had no obvious effect on the fracture strength of the hybrid sheets with a crack of
c=0.05L
. However, the fracture strength exhibited more dependence on the crack location for a relatively large crack (
c=0.15L
0.3L
). The fracture process of the hybrid sheet with a crack usually started from the crack tips where stress concentration existed. If the crack was located in or close to the graphene/h-BN interface, the fracture usually happened in the h-BN domain. The work would provide useful mechanical property information for the applications of hybrid graphene/h-BN sheets in material devices.
hybrid graphene/hexagonal boron nitride sheet, interface, nanocrack, fracture, failure
Shandong Analysis and Test Center, Qilu University of Technology (Shandong Academy of Sciences), Jin
Civil Engineering and geo-Environmental Laboratory, Lille University, Lille 59000, France |
Linear Time Invariant Systems | Brilliant Math & Science Wiki
Alex Chumbley, João Areias, Christopher Williams, and
Linear time-invariant systems (LTI systems) are a class of systems used in signals and systems that are both linear and time-invariant. Linear systems are systems whose outputs for a linear combination of inputs are the same as a linear combination of individual responses to those inputs. Time-invariant systems are systems where the output does not depend on when an input was applied. These properties make LTI systems easy to represent and understand graphically.
LTI systems are superior to simple state machines for representation because they have more memory. LTI systems, unlike state machines, have a memory of past states and have the ability to predict the future. LTI systems are used to predict long-term behavior in a system. So, they are often used to model systems like power plants. Another important application of LTI systems is electrical circuits. These circuits, made up of inductors, transistors, and resistors, are the basis upon which modern technology is built.
Properties of LTI Systems
Discrete LTI System: Example
Continuous LTI System: Example
LTI systems are those that are both linear and time-invariant.
Linear systems have the property that the output is linearly related to the input. Changing the input in a linear way will change the output in the same linear way. So if the input
x_1(t)
produces the output
y_1(t)
x_2(t)
y_2(t)
, then linear combinations of those inputs will produce linear combinations of those outputs. The input
\big(x_1(t) + x_2(t)\big)
will produce the output
\big(y_1(t) + y_2(t)\big)
. Further, the input
\big(a_1 \cdot x_1(t) + a_2 \cdot x_2(t)\big)
(a_1 \cdot y_1(t) + a_2 \cdot y_2(t))
a_1
a_2
In other words, for a system
T
t
, composed of signals
x_1(t)
x_2(t)
with outputs
y_1(t)
y_2(t)
T\big[a_1x_1(t) + a_2x_2(t)\big] = a_1T\big[x_1(t)\big] + a_2T\big[x_2(t)\big] = a_1y_1(t) + a_2y_2(t),
a_1
a_2
Further, the output of a linear system for an input of 0 is also 0.
Time-invariant systems are systems where the output for a particular input does not change depending on when that input was applied. A time-invariant systems that takes in signal
x(t)
and produces output
y(t)
will also, when excited by signal
x(t + \sigma)
, produce the time-shifted output
y(t + \sigma)
Thus, the entirety of an LTI system can be described by a single function called its impulse response. This function exists in the time domain of the system. For an arbitrary input, the output of an LTI system is the convolution of the input signal with the system's impulse response.
Conversely, the LTI system can also be described by its transfer function. The transfer function is the Laplace transform of the impulse response. This transformation changes the function from the time domain to the frequency domain. This transformation is important because it turns differential equations into algebraic equations, and turns convolution into multiplication. In the frequency domain, the output is the product of the transfer function with the transformed input. The shift from time to frequency is illustrated in the following image:
Shifting from the time to the frequency domain[1]
In addition to linear and time-invariant, LTI systems are also memory systems, invertible, casual, real, and stable. That means they have memory, they can be inverted, they depend only on current and past events, they have fully real inputs and outputs, and they produce bounded output for bounded input.
Because of the properties of LTI systems, the general form of an LTI system with output
y[n]
and input
x[n]
n
, and constants
c_k
d_j
is defined a
y[n] = c_0y[n-1] + c_1y[n-2] + ... + c_{k-1}y[n-k] + d_0x[n] + d_1x[n-1] + ... + d_jx[n-j] .
The state of this system depends on the previous
k
output values and
j
input values. Because of the linearity property, the output at time
n
is just a linear combination of the previous outputs, previous inputs, and current input.
Further, if a string of LTI systems are cascaded together, the output of that new system does not depend on the order in which the systems were cascaded. This property follows from the associative property and the commutative property.
We can take the general form of the LTI system, and write it as an operator equation, and with some manipulation we can turn it into a useful formula:
\begin{aligned} Y &= c_0\mathcal{R}Y + c_1\mathcal{R}^2Y + \cdots + c_{k-1}\mathcal{R}^{k}Y + d_0X + d_1\mathcal{R}X + \cdots + d_j\mathcal{R}^jX \\ &= Y\big(c_0\mathcal{R} + c_1\mathcal{R}^2 + \cdots + c_{k-1}\mathcal{R}^{k}\big) + X\big(d_0 + d_1\mathcal{R} + \cdots + d_j\mathcal{R}^j\big). \end{aligned}
This is the same equation as
Y\big(1 - c_0\mathcal{R} - c_1\mathcal{R}^2 - ... - c_{k-1}\mathcal{R}^{k}\big) = X\big(d_0 + d_1\mathcal{R} + \cdots + d_j\mathcal{R}^j\big) .
We can then do some division to create an equation that describes the quotient of the output signal and the input signal:
\frac{Y}{X} = \frac{d_0 + d_1\mathcal{R} + \cdots + d_j\mathcal{R}^j}{1 - c_0\mathcal{R} - c_1\mathcal{R}^2 - ... - c_{k-1}\mathcal{R}^{k}} .
This is the system function of the LTI system, and it is typically written as the polynomial
\frac{Y}{X} = \frac{b_0 + b_1\mathcal{R} + b_2\mathcal{R}^2 + \cdots}{a_0 + a_1\mathcal{R} + a_2\mathcal{R}^2 + \cdots} .
Note that both the numerator and the denominator are polynomials in
\mathcal{R}
, the delay variable. Understanding the different roles that the numerator and denominator play is important.
1. In a feedforward system, what will be the value of the denominator in the system function?
A feedforward system has no dependence whatsoever on previous values of
Y
. So, the denominator will be equal to 1.
The impulse response is an especially important property of any LTI system. We can use it to describe an LTI system and predict its output for any input. To understand the impulse response, we need to use the unit impulse signal, one of the signals described in the Signals and Systems wiki. It has many important applications in sampling. The unit impulse signal is simply a signal that produces a signal of 1 at time = 0. It is zero everywhere else. With that in mind, an LTI system's impulse function is defined as follows:
The impulse response for an LTI system is the output,
y(t)
, when the input is the unit impulse signal,
\sigma(t)
\mbox{when}\ \ x(t) = \sigma(t) ,\ \ h(t) = y(t) .
Essentially, the impulse function for an LTI system basically asks this: If we introduce a unit impulse signal at a certain time, what will be the output of the system at a later time? Sometimes, we can even find the impulse response by doing just that: introducing an impulse signal and seeing what happens.
Convolution is a representation of signals as a linear combination of delayed input signals. In other words, we're just breaking down a signal into the inputs that were used to create it. However, it is used differently between discrete time signals and continuous time signals because of their underlying properties. Discrete time signals are simply linear combinations of discrete impulses, so they can be represented using the convolution sum. Continuous signals, on the other hand, are continuous. Much like calculating the area under the curve of a continuous function, these signals require the convolution integral.
y[n] = \sum_{k = -\infty}^{\infty}x[k]\, h[n - k]
y(t) = \int_{-\infty}^{\infty}h(\tau)x(t-\tau)\,d\tau = x(t) \ast h(t)
\ast
is the mathematical convolution symbol.
All LTI systems can be described using this integral or sum, for a suitable function
h()
h()
is the impulse function for the signal. The output of any LTI system can be calculated using the input and the impulse function for that system.
Convolution has many important properties:
x(t) \ast h(t) = h(t) \ast x(t)
\big[x(t) \ast h_1(t)\big] \ast h_2(t) = x(t) \ast \big[h_1(t) \ast h_2(t)\big]
Distributivity of Addition:
x(t) \ast \big[h_1(t) + h_2(t)\big] = x(t) \ast h_1(t) + x(t) \ast h_2(t)
Identity Element:
x(t) \ast h(t) = h(t)
The transfer function of an LTI system is given by the Laplace transform of the impulse response of the system and it gives valuable information of the system's behavior and can greatly simplify the computation of the output response.
If the impulse response of a system
y(t)
h(t)
then the transfer function of that system is given by
H(S) = \mathcal{L}(h(t))
The equation describing a causal LTI system is given by:
\ddot{y(t)} + \dot{y(t)} = x(t)
We can compute the impulse response by replacing
x(t)
\sigma(t)
and solve it using the Laplace transform which will give us:
y(t) = h(t) = \mathcal{L}^{-1}(\frac{1}{s^2 + s}) = u(t) - e^{-t}
For this differential equation, the transfer function is given by:
H(S) = \frac{1}{s^2 + s}
We've seen previously that an LTI system can be written as
\frac{Y}{X} = \frac{b_0 + b_1\mathcal{R} + b_2\mathcal{R}^2 + \cdots}{a_0 + a_1\mathcal{R} + a_2\mathcal{R}^2 + \cdots} .
The transfer function of any (causal) LTI can then be given by:
\frac{Y(S)}{X(S)} = \frac{a_0 + a_1\mathcal{S} + a_2\mathcal{S}^2 + \cdots}{b_0 + b_1\mathcal{S} + b_2\mathcal{S}^2 + \cdots} .
An LTI system can be represented by:
b_n\mathcal{R}^nY + b_{n-1}\mathcal{R}^{n-1}Y \cdots b_0Y = a_m\mathcal{R}^mX + a_{m-1}\mathcal{R}^{m-1}X a_m\mathcal{R}^mX \cdots a_0X
\mathcal{L}\big\{f^{(n)}\big\}=s^n\mathcal{L}\{f\}-\displaystyle\sum_{i=1}^{n}s^{n-i}f^{(i-1)}(0).
and that in a causal system
f^{(n)}(0) = 0
, taking the Laplace transform of the previous equation will yield:
b_n\mathcal{S}^nY(S) + b_{n-1}\mathcal{S}^{n-1}Y(S) \cdots b_0Y(S) = a_m\mathcal{S}^mX(S) + a_{m-1}\mathcal{S}^{m-1}X(S) a_m\mathcal{S)}^mX(S) \cdots a_0X(S)
Y(S)(b_n\mathcal{S}^n + b_{n-1}\mathcal{S}^{n-1} \cdots b_0) = X(S)(a_m\mathcal{S}^m + a_{m-1}\mathcal{S}^{m-1} a_m\mathcal{S)}^m \cdots a_0
\frac{Y(S)}{X(S)} = \frac{a_0 + a_1\mathcal{S} + a_2\mathcal{S}^2 + \cdots}{b_0 + b_1\mathcal{S} + b_2\mathcal{S}^2 + \cdots} .
x(t) = \sigma(t)
\mathcal{L}\{\sigma(t)\} = 1
H(S) = \frac{a_0 + a_1\mathcal{S} + a_2\mathcal{S}^2 + \cdots}{b_0 + b_1\mathcal{S} + b_2\mathcal{S}^2 + \cdots} .
Transfer function and the output
We know that the output of an LTI system will be given by the convolution of the signal with the impulse response. Since the convolution in the time domain is equivalent to a multiplication in the Laplace domain, the output
Y(S)
of a system with the transfer function
H(S)
to the input
X(S)
will be given by:
Y(S) = H(S)X(S)
One can easily calculate the output in the time domain by
y(t) = \mathcal{L}^{-1}(Y(S))
What is the output of the system described by
\ddot{y(t)} + \dot{y(t)} = x(t)
x(t) = e^t
We know from previously that:
H(S) = \frac{1}{s^2 + s}
X(S) = \mathcal{L}\{x(t)\} = \frac{1}{S-1}
Y(S) = X(S)H(S) = \frac{1}{S(S+1)(S-1)}
The output is then given by:
y(t) = \mathcal{L}^{-1}\{Y(S)\} = \mathcal{L}^{-1}\{\frac{1}{S(S+1)(S-1)}\} = 0.5e^{t} + 0.5e^{-t} - u(t)
Since the transfer function is described by the division of two polinomials, we can factor those polinomials into:
H(S) = \frac{(S + z_0)(S + z_1)(S + z_2)\cdots}{(S + p_0)(S + p_1)(S + p_2)\cdots}
z_0, z_1, z_2 \cdots
are the complex zeros of the system and
p_0, p_1, p_2 \cdots
are the complex poles. They give interesting information on the system's behaviour and can be seing in more detailed on the wiki Predicting System Behavior.
Discrete time signals are simply a collection of individual signals. These discrete signals can be a product of sampling a continuous time signal, or it can be a product of truly discrete phenomena. These discrete signals can be represented in a graph with individual points connected to the
x
-axis, as in the graphic below.
Discrete Time Signal[2]
Here, time is on the
x
-axis and the signal is on the
y
-axis. It is discretized, meaning the signal function is not continuous. So, as mentioned earlier, a sum is needed to calculate its output at any given time.
We have a discrete LTI system. Given the following input function and impulse response function, calculate the output of the system at time
n
u[n]
is the unit step function.
x[n] \ast h[n]
\begin{aligned} u[n] &= \begin{cases} 1 & \mbox{if } n \geq 0 \\ 0 & \mbox{if } n \lt 0\end{cases}\\ x[n] &= u[n] \\ h[n] &= 2^nu[n]. \end{aligned}
y[n] = \sum_{k=-\infty}^{\infty}x[k]\, h[n-k] = \sum_{k=-\infty}^{\infty}u[k]2^{n-k}u[n-k] = 2^n\sum_{k=-\infty}^{\infty}2^{-k}u[k]\, u[n-k] .
Now, we need to analyze the limits of this sum. When
k \lt 0
x[k]\, k[n-k] = 0
, so we can ignore any value of
k
that is less than 0. When
k \geq 0
, the two functions overlap only in the range
\{0, n\}.
So, those are the limits we need to use. Therefore, the equation reduces to
\begin{aligned} y[n] &= 2^n\sum_{k=0}^{n}2^{-k} \\ &= 2^n\frac{1 - \frac{1}{2}^{n+1}}{1-\frac{1}{2}} \\ &= \frac{1 - 2^{n+1}}{1 - 2} \\ &= 2^{n+1} - 1.\ _\square \end{aligned}
Continuous LTI systems have signals that are defined at all possible time values. So, we need to use integrals to properly understand this type of system.
Let's say that we have an LTI system with an impulse response function
h(t)
. We want to figure out how an input,
x(t)
, will affect this system. To do so, we need convolution! Assume for this problem that
is greater than zero (if it wasn't, the answer would always be zero!).
Try convolving the following two functions. Solve for
x(t) \ast h(t)
u(t)
is the unit step function:
\begin{aligned} u(t) &= \begin{cases} 1 & \mbox{if } t \geq 0 \\ 0 & \mbox{if } t \lt 0\end{cases}\\ x(t) &= u(t) \\ h(t) &= e^{-3t}u(t). \end{aligned}
x(t) \ast h(t)
\int_{-\infty}^{\infty}h(\tau)u(t - \tau)\, d\tau.
To understand these integral bounds, it's useful to think about the functions and where there are non-zero.
h(\tau)
is zero to the left of the
y
-axis, or for all negative numbers.
u(t - \tau)
is zero for all values greater than
\tau
. So, our bounds are
\{0, \tau\}.
The inside of our integral is the product of two signals, and we're really just calculating the area under the curve. So, now we have
\int_{0}^{t}e^{-3\tau}d\tau = \frac{1}{-3}\left. e^{-3\tau} \right|^{t}_{0} = \frac{1}{-3}\left[e^{-3t} - 1\right].\ _\square
Calvert, J. Time and Frequency Domains. Retrieved April 10, 2016, from http://mysite.du.edu/~etuttle/electron/elect6.htm
Ho Ahn, S. Feedforward. Retrieved June 16, 2016, from http://www.songho.ca/dsp/signal/signals.html
Cite as: Linear Time Invariant Systems. Brilliant.org. Retrieved from https://brilliant.org/wiki/linear-time-invariant-systems/ |
How do i show: \sin^2(x+y)-\sin^2(x-y)\equiv\sin(2x)\sin(2y)
Ingrid Senior 2022-03-01 Answered
How do i show:
{\mathrm{sin}}^{2}\left(x+y\right)-{\mathrm{sin}}^{2}\left(x-y\right)\equiv \mathrm{sin}\left(2x\right)\mathrm{sin}\left(2y\right)
Malaika Ridley
{\left(\mathrm{sin}\left(x+y\right)\right)}^{2}-{\left(\mathrm{sin}\left(x-y\right)\right)}^{2}={\left(\mathrm{sin}x\mathrm{cos}y+\mathrm{cos}x\mathrm{sin}y\right)}^{2}-{\left(\mathrm{sin}x\mathrm{cos}y+\mathrm{cos}x\mathrm{sin}y\right)}^{2}
=\left({\mathrm{sin}}^{2}{\mathrm{cos}}^{2}y+2\mathrm{sin}x\mathrm{cos}y\mathrm{cos}x\mathrm{sin}y+{\mathrm{cos}}^{2}x{\mathrm{sin}}^{2}y\right)-\left({\mathrm{sin}}^{2}{\mathrm{cos}}^{2}y-2\mathrm{sin}x\mathrm{cos}y\mathrm{cos}x\mathrm{sin}y+{\mathrm{cos}}^{2}x{\mathrm{sin}}^{2}y\right)
=4\mathrm{sin}x\mathrm{cos}y\mathrm{cos}x\mathrm{sin}y
=\left(2\mathrm{sin}x\mathrm{cos}x\right)\left(2\mathrm{sin}y\mathrm{cos}y\right)
=\mathrm{sin}\left(2x\right)\mathrm{sin}\left(2y\right)
meizhen85ulg
{\mathrm{sin}}^{2}a=\frac{1-\mathrm{cos}2a}{2}
\mathrm{cos}x-\mathrm{cos}y=-2\mathrm{sin}\left(\frac{x-y}{2}\right)\mathrm{sin}\left(\frac{x+y}{2}\right)
{\mathrm{sin}}^{2}\left(x+y\right)-{\mathrm{sin}}^{2}\left(x-y\right)
=\frac{1-\mathrm{cos}2\left(x+y\right)-\left(1-\mathrm{cos}2\left(x-y\right)\right)}{2}
=\frac{\mathrm{cos}2\left(x-y\right)-\mathrm{cos}2\left(x+y\right)}{2}
=\mathrm{sin}\left(2x\right)\mathrm{sin}\left(2y\right)
\mathrm{sin}t=,\mathrm{cos}t=,\text{ }\text{and}\text{ }\mathrm{tan}t=
1+m\mathrm{cos}2\theta +\left(\begin{array}{c}m\\ 2\end{array}\right)\mathrm{cos}4\theta +\dots +\left(\begin{array}{c}m\\ r\end{array}\right)\mathrm{cos}2r\theta +\dots +\mathrm{cos}2m\theta ={2}^{n}{\mathrm{cos}}^{m}\theta \mathrm{cos}m\theta
Prove the inequalities
1-\frac{{x}^{2}}{2}\le \mathrm{cos}\left(x\right)\le 1-\frac{{x}^{2}}{2}+\frac{{x}^{4}}{24}
\mathrm{sin}\left(5\theta \right)=16{\mathrm{sin}}^{5}\left(\theta \right)-20{\mathrm{sin}}^{3}\left(\theta \right)+5\mathrm{sin}\left(\theta \right)
3\mathrm{sin}2x+4\mathrm{cos}2x-2\mathrm{cos}x+6\mathrm{sin}x-6=0
6\mathrm{sin}x\mathrm{cos}x+4\left({\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x\right)-2\mathrm{cos}x+6\mathrm{sin}x-6=0
i\mathrm{sin}\left(x\right)
By Euler's formula, I can express i in the following way:
i=\mathrm{cos}\left(\frac{\pi }{2}\right)+i\mathrm{sin}\left(\frac{\pi }{2}\right)=\mathrm{exp}\left(i\frac{\pi }{2}\right)
I wonder if it is legitimate to write
i\mathrm{sin}x=\mathrm{exp}\left(i\frac{\pi }{2}\right)\cdot \frac{\mathrm{exp}\left(ix\right)-\mathrm{exp}\left(-ix\right)}{2i}
=\frac{\mathrm{exp}\left(ix\right)\mathrm{exp}\left(i\frac{\pi }{2}\right)-\mathrm{exp}\left(-ix\right)\mathrm{exp}\left(i\frac{\pi }{2}\right)}{2i}
=\frac{\mathrm{exp}\left(i\left(x+\frac{\pi }{2}\right)\right)-\mathrm{exp}\left(-i\left(x+\frac{\pi }{2}\right)\right)}{2i}
=\mathrm{sin}\left(x+\frac{\pi }{2}\right)
I don't feel like this is right, because it would imply a lot of weird things. So where is my mistake?
{\mathrm{sec}}^{2}x+{\mathrm{tan}}^{2}x=\left(1-{\mathrm{sin}}^{4}x\right){\mathrm{sec}}^{4}x |
Pick's Theorem | Brilliant Math & Science Wiki
Mei Li, Pi Han Goh, ShreyesRamanuja Balaji, and
Pick's theorem gives a way to find the area of polygons in a plane whose endpoints have integer vertices.
Lattice points are points whose coordinates are both integers, such as
(1,2), (-4, 11)
(0,5)
. The set of all lattice points forms a grid. A lattice polygon is a shape made of straight lines whose vertices are all lattice points and Pick's theorem gives a formula for the area of a lattice polygon.
First, observe that for any lattice polygon
P
, the polygon contains some lattice points on its boundary edges
(
including the lattice points that are the vertices of
P)
and may contain some lattice points in its interior (not counting the points on the boundary). Let
\begin{aligned} B(P) &= \mbox{ number of points on the boundary of the polygon}\\ I(P) &= \mbox{ number of points in the interior of the polygon}. \end{aligned}
P
be a lattice polygon, let
B(P)
be the points on the boundary of the polygon, and let
I(P)
be the number of points in the interior of the polygon. Then
(\mbox{Area of the polygon } P) = I(P) + \frac{1}{2} B(P) -1.
Notice that Pick's theorem applies to any polygon, not only convex polygons.
For a rectangle
R
l
h,
B(R) = 2l + 2h
points along the boundary of the rectangle, and
I(R) = (l-1)(h-1) = lh - l - h + 1
points in the interior of the rectangle. Applying Pick's theorem gives
\begin{aligned} \mbox{Area}(R) &= I(R) + \frac{1}{2} B(R) - 1\\ &= lh - l - h + 1 + \frac{1}{2} (2l + 2h) - 1\\ &= lh - l - h + 1 + l + h - 1\\ &= lh. \end{aligned}
Without Pick's theorem, we might calculate the area of a lattice polygon by decomposing the lattice polygon into triangles, computing the area of each triangle using the sine rule, and then summing the resulting triangle areas to obtain the area of the lattice polygon. As a powerful tool, the Shoelace theorem works side by side finding the area of any figure given the coordinates. Pick's theorem gives a way to find the area of a lattice polygon without performing all of these calculations. Pick's theorem also implies the following interesting corollaries:
The area of a lattice polygon is always an integer or half an integer.
Proof: By Pick's theorem, the area of a lattice polygon
P
\mbox{Area}(P) = I(P) + \frac{1}{2} B(P) - 1.
In a lattice polygon, the number of points in the interior of
P
and the number of points on the boundary of
P
are both integers. Then
\mbox{Area}(P) = I(P) + \frac{1}{2} B(P) - 1
is an integer if
B(P)
is even and is half an integer if
B(P)
_\square
It is impossible to draw an equilateral triangle as a lattice polygon.
Proof: Suppose we could draw an equilateral triangle as a lattice polygon with lattice vertices
A
B
C
s
. By distance formula and Pythagorean theorem,
s^2
represents the square of distance between any of these 2 vertices, and must be an integer. Also note that the area of an equilateral triangle can be expressed as
\text{Area} = \frac{s^2 \sqrt3}4
, thus the area must be irrational.
On the other hand, the area of
\triangle ABC
must be an integer or half an integer by the previous theorem, which means the area must be rational, a contradiction. Therefore, it is impossible to draw an equilateral triangle as a lattice polygon.
_\square
An integer lattice point is a point with coordinates
(n, m)
n
m
are integers. As
N
ranges from 1 to 905, what is the maximum number of integer lattice points in the interior of a triangle with vertices
(0,0),
(N, 907-N),
(N+1, 907-N-1)?
Note: The point
(0,0)
is not in the interior of any of the triangles described above.
We can prove this using the following sequence of steps:
It is true for rectangles whose sides are parallel to the axes: This is obvious. An
a \times b
rectangle has
2a+2b
points on the boundary,
(a-1)(b-1)
points in the interior, and an area of
(a-1)(b-1) + \frac{1}{2} (2a+2b) - 1 = ab
It is true for right triangles whose bases are parallel to the axes: We can double the triangle to form a rectangle and track the points that lie on the hypotenuse.
It is true for an arbitrary triangle: We take the rectangle whose sides are parallel to the axes that contains this triangle and then subtract off the right triangles whose bases are parallel to the axes.
It is true for any polygon: We triangulate the polygon.
The main idea here is to show that the sum and difference of polygons still obey Pick's theorem, which is what allows us to glue/detach triangles.
Cite as: Pick's Theorem. Brilliant.org. Retrieved from https://brilliant.org/wiki/picks-theorem/ |
Bayesian linear regression model with conjugate prior for data likelihood - MATLAB - MathWorks í•œêµ
The Bayesian linear regression model object conjugateblm specifies that the joint prior distribution of the regression coefficients and the disturbance variance, that is, (β, σ2) is the dependent, normal-inverse-gamma conjugate model. The conditional prior distribution of β|σ2 is multivariate Gaussian with mean μ and variance σ2V. The prior distribution of σ2 is inverse gamma with shape A and scale B.
\underset{t=1}{\overset{T}{â}}\mathrm{Ï}\left({y}_{t};{x}_{t}\mathrm{β},{\mathrm{Ï}}^{2}\right),
where ϕ(yt;xtβ,σ2) is the Gaussian probability density evaluated at yt with mean xtβ and variance σ2. The specified priors are conjugate for the likelihood, and the resulting marginal and conditional posterior distributions are analytically tractable. For details on the posterior distribution, see Analytically Tractable Posteriors.
PriorMdl = conjugateblm(NumPredictors) creates a Bayesian linear regression model object (PriorMdl) composed of NumPredictors predictors and an intercept, and sets the NumPredictors property. The joint prior distribution of (β, σ2) is the dependent normal-inverse-gamma conjugate model. PriorMdl is a template that defines the prior distributions and the dimensionality of β.
Prior covariance matrix of β (V)
Mean parameter of the Gaussian prior on β, specified as a numeric scalar or vector.
Conditional covariance matrix of Gaussian prior on β, specified as a c-by-c symmetric, positive definite matrix. c can be NumPredictors or NumPredictors + 1.
\left[\begin{array}{cccc}1e5& 0& \cdots & 0\\ 0& & & \\ ⋮& & V& \\ 0& & & \end{array}\right].
V is the prior covariance of β up to a factor of σ2.
{\text{GNPR}}_{t}={\mathrm{β}}_{0}+{\mathrm{β}}_{1}{\text{IPI}}_{t}+{\mathrm{β}}_{2}{\text{E}}_{t}+{\mathrm{β}}_{3}{\text{WR}}_{t}+{\mathrm{ε}}_{t}.
time points,
{\mathrm{ε}}_{t}
{\mathrm{Ï}}^{2}
\mathrm{β}|{\mathrm{Ï}}^{2}â¼{N}_{4}\left(M,{\mathrm{Ï}}^{2}V\right)
M
V
{\mathrm{Ï}}^{2}â¼IG\left(A,B\right)
A
B
\mathrm{β}
{\mathrm{Ï}}^{2}
\mathrm{β}
{\mathrm{Ï}}^{2}
\mathrm{β}
{\mathrm{Ï}}^{2}=2
\mathrm{β}
{\mathrm{Ï}}^{2}
\mathrm{β}
\mathrm{β}
\mathrm{β}
\mathrm{β}
{t}_{68}
\mathrm{â}\left(\mathrm{β},{\mathrm{Ï}}^{2}|y,x\right)=\underset{t=1}{\overset{T}{â}}\mathrm{Ï}\left({y}_{t};{x}_{t}\mathrm{β},{\mathrm{Ï}}^{2}\right). |
Supported Energy Derivative Functions - MATLAB & Simulink - MathWorks Benelux
\mathrm{max}\left(0,{S}_{av}-X\right)
\mathrm{max}\left(0,X-{S}_{av}\right)
\mathrm{max}\left(0,S-{S}_{av}\right)
\mathrm{max}\left(0,{S}_{av}-S\right)
{S}_{av}
S
X
{S}_{av}
Price an Asian option from a Cox-Ross-Rubinstein binomial tree.
Price an Asian option from an Equal Probabilities binomial tree.
Price an Asian option using an implied trinomial tree (ITT).
Price an Asian option using a standard trinomial tree.
A barrier option is similar to a vanilla put or call option, but its life either begins or ends when the price of the underlying asset passes a predetermined barrier value. There are four types of barrier options.
This option becomes effective when the price of the underlying asset passes above a barrier that is above the initial asset price. Once the barrier has knocked in, it will not knock out even if the price of the underlying instrument moves below the barrier again.
This option terminates when the price of the underlying asset passes above a barrier that is above the initial stock price. Once the barrier has knocked out, it will not knock in even if the price of the underlying instrument moves below the barrier again.
This option becomes effective when the price of the underlying asset passes below a barrier that is below the initial stock price. Once the barrier has knocked in, it will not knock out even if the price of the underlying instrument moves above the barrier again.
This option terminates when the price of the underlying asset passes below a barrier that is below the initial stock price. Once the barrier has knocked out, it will not knock in even if the price of the underlying instrument moves above the barrier again.
Price European or American barrier options using Monte Carlo simulations.
Price European barrier options using Black-Scholes option pricing model.
Price a barrier option from a Cox-Ross-Rubinstein binomial tree.
Price a barrier option from an Equal Probabilities binomial tree.
Price a barrier option using an implied trinomial tree (ITT).
Price a barrier options using a standard trinomial tree.
\mathrm{max}\left(St-K,0\right)
\mathrm{max}\left(K-St,0\right)
Price European, Bermudan, or American vanilla options using the Longstaff-Schwartz model.
Calculate European, Bermudan, or American vanilla option prices and sensitivities using the Longstaff-Schwartz model.
Calculate American options prices using Barone-Adesi and Whaley option pricing model.
Calculate American options prices and sensitivities using Barone-Adesi and Whaley option pricing model.
Calculate American call option prices using Roll-Geske-Whaley option pricing model.
Calculate American call option prices or sensitivities using Roll-Geske-Whaley option pricing model.
Price American options using Bjerksund-Stensland 2002 option pricing model.
Determine American option prices or sensitivities using Bjerksund-Stensland 2002 option pricing model.
Price an option from a Cox-Ross-Rubinstein binomial tree.
Price an option from an Equal Probabilities binomial tree.
Price an option using an implied trinomial tree (ITT).
Price an option using a standard trinomial tree.
\mathrm{max}\left(X1-X2-K,0\right)
Price European or American spread options using the Alternate Direction Implicit (ADI) and Crank-Nicolson finite difference methods.
Calculate price and sensitivities of European or American spread options using the Alternate Direction Implicit (ADI) and Crank-Nicolson finite difference methods.
For more information on using spread options, see Pricing European and American Spread Options.
A lookback option is a path-dependent option based on the maximum or minimum value the underlying asset (e.g. electricity, stock) achieves during the entire life of the option. Basically the holder of the option can ‘look back’ over time to determine the payoff. This type of option provides price protection over a selected period, reduces uncertainties with the timing of market entry, moderates the need for the ongoing management, and therefore, is usually more expensive than vanilla options.
Lookback call options give the holder the right to buy the underlying asset at the lowest price. Lookback put options give the right to sell the underlying asset at the highest price.
Financial Instruments Toolbox™ software supports two types of lookback options: fixed and floating. The difference is related to how the strike price is set in the contract. Fixed lookback options have a specified strike price and the option pays out the maximum of the difference between the highest (lowest) observed price of the underlying during the life of the option and the strike. Floating lookback options have a strike price determined at maturity, and it is set at the lowest (highest) price of the underlying reached during the life of the option. This means that for a floating strike lookback call (put), the holder has the right to buy (sell) the underlying asset at its lowest (highest) price observed during the life of the option. So, there are a total of four lookback option types, each with its own characteristic payoff formula:
\mathrm{max}\left(0,{S}_{\mathrm{max}}-X\right)
\mathrm{max}\left(0,X-{S}_{\mathrm{min}}\right)
\mathrm{max}\left(0,S-{S}_{\mathrm{min}}\right)
\mathrm{max}\left(0,{S}_{\mathrm{max}}-S\right)
{S}_{\mathrm{max}}
is the maximum price of underlying asset.
{S}_{\mathrm{min}}
is the minimum price of underlying asset.
S
is the price of the underlying asset at maturity.
X
Calculate prices of European lookback fixed and floating strike options using the Conze-Viswanathan and Goldman-Sosin-Gatto models.
Calculate prices and sensitivities of European fixed and floating strike lookback options using the Conze-Viswanathan and Goldman-Sosin-Gatto models.
Calculate prices of lookback fixed and floating strike options using the Longstaff-Schwartz model.
Calculate prices and sensitivities of lookback fixed and floating strike options using the Longstaff-Schwartz model.
Price a lookback option from a Cox-Ross-Rubinstein binomial tree.
Price a lookback option from an Equal Probabilities binomial tree.
Price a lookback option using an implied trinomial tree (ITT).
Price a lookback option using a standard trinomial tree.
Lookback options and Asian options are instruments used in the electricity market to manage purchase timing risk. Electricity purchasers cover part of their expected electricity consumption on the forward market to avoid the volatility and limited liquidity of the spot market. Using Asian options as a hedging tool is a passive approach to solving the purchase timing problem. An Asian option instrument diminishes the wrong timing risk but it also reduces any potential benefit to the buyer from falling prices. On the other hand, lookback options allow the purchasers to buy electricity at the lowest price, but as mentioned before, this instrument is more expensive than Asian and vanilla options.
A forward option is a non-standardized contract between two parties to buy or to sell an asset at a specified future time at a price agreed upon today. The buyer of a forward option contract has the right to hold a particular forward position at a specific price any time before the option expires. The forward option seller holds the opposite forward position when the buyer exercises the option. A call option is the right to enter into a long forward position and a put option is the right to enter into a short forward position. A closely related contract is a futures contract. A forward is like a futures in that it specifies the exchange of goods for a specified price at a specified future date. The following table displays some of the characteristics of forward and futures contracts.
{f}_{T}={S}_{T}-K
{f}_{T}=K-{S}_{T}
F\left(t,T\right)
F\left(s,T\right)-F\left(t,T\right)
F\left(T,T\right)
F\left(T,T\right)
spreadbyls | spreadsensbyls | asianbyls | asiansensbyls | lookbackbyls | lookbacksensbyls | optstockbyls | optstocksensbyls | optpricebysim | spreadbykirk | spreadsensbykirk | spreadbybjs | spreadsensbybjs | asianbykv | asiansensbykv | asianbylevy | asiansensbylevy | lookbackbycvgsg | lookbacksensbycvgsg | optstockbyblk | optstocksensbyblk | spreadbyfd | spreadsensbyfd |
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