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\begin{aligned} &\text{Alpha} = \frac{ \text{End Price} + \text{DPS} - \text{Start Price} }{ \text{Start Price} } \\ &\textbf{where:}\\ &\text{DPS} = \text{Distribution per share} \\ \end{aligned}
Alpha=Start PriceEnd Price+DPS−Start Pricewhere:DPS=Distribution per share
Here are the betas for three well-known stocks on November 2021:
We can see that Micron was 27% more volatile than the market as a whole, while Coca-Cola was 36% less volatile than the broader market. The SPDRs, or SPYs, have a beta of 1.00 because this ETF itself tracks the S&P 500 index.
\begin{aligned} &\text{Beta} = \frac{ \text{CR} }{ \text{Variance of Market's Return} } \\ &\textbf{where:}\\ &\text{CR} = \text{Covariance of asset's return with market's return} \\ \end{aligned}
Beta=Variance of Market’s ReturnCRwhere:CR=Covariance of asset’s return with market’s return
Fidelity. "All About Alpha, Beta, and Smart Beta."
Yahoo Finance. "Micron Technology, Inc. (MU)."
Yahoo Finance. "The Coca-Cola Company (KO)." |
15 July 2005 Cycles representing the top Chern class of the Hodge bundle on the moduli space of abelian varieties
Torsten Ekedahl, Gerard van der Geer
We give a generalization to higher genera of the famous formula
12 \lambda =\delta
for genus 1
Torsten Ekedahl. Gerard van der Geer. "Cycles representing the top Chern class of the Hodge bundle on the moduli space of abelian varieties." Duke Math. J. 129 (1) 187 - 199, 15 July 2005. https://doi.org/10.1215/S0012-7094-04-12917-3
Secondary: 14G35 , 14K10
Torsten Ekedahl, Gerard van der Geer "Cycles representing the top Chern class of the Hodge bundle on the moduli space of abelian varieties," Duke Mathematical Journal, Duke Math. J. 129(1), 187-199, (15 July 2005) |
Johansen constraint test - MATLAB jcontest - MathWorks í•œêµ
\mathit{p}
{\mathit{y}}_{1,\mathit{t}}
\left(1-L\right){y}_{1,t}=c+{\mathrm{ε}}_{t}.
\mathit{A}
{\mathit{a}}_{1}=0
\mathit{p}
\mathit{p}
{\mathit{y}}_{1,\mathit{t}}
{\mathit{y}}_{2,\mathit{t}}
\begin{array}{l}\left(1-L\right){y}_{1,t}={c}_{1,1}+{\mathrm{ε}}_{1,t}\\ \left(1-L\right){y}_{2,t}={c}_{2,1}+{\mathrm{ε}}_{2,t}.\end{array}
\mathit{A}
{\mathit{a}}_{1}=0
{\mathit{a}}_{2}=0
\mathit{j}
\left(1-L\right){y}_{t}=\left[\begin{array}{c}{a}_{1,j}\\ {a}_{2,j}\\ {a}_{3,j}\end{array}\right]\left({y}_{t-1}+{c}_{0}\right)+{c}_{1}+{\mathrm{ε}}_{t}.
Cons=1×3 cell array
StatTbl0=3×8 table
\left[\begin{array}{ccc}1& -1& -1\end{array}\right]â²
\mathit{A}
\mathit{B}
\stackrel{Ë}{\text{â}\mathit{C}}=\stackrel{Ë}{\mathit{A}}\stackrel{Ë}{\text{â}\mathit{B}}â²
AACon = 3×1
BACon = 3×1
CACon = 3×3
AAVec = 3×1
BAVec = 3×1
CAVec = 3×3
ABCon = 3×1
BBCon = 3×1
CBCon = 3×3
ABVec = 3×1
BBVec = 3×1
CBVec = 3×3
positive integer in [1,numDims − 1]
Common rank of A and B, specified by a positive integer in the interval [1,numDims − 1].
"ACon" R, a numDims-by-numCons numeric matrix Specifies numCons constraints on A given by R'A = 0, where numCons ≤ numDims − r.
"AVec" numDims1-by-numCons numeric matrix Specifies numCons equality constraints imposed on error-correction speed vectors in A, where numCons ≤ r.
"BCon" R, numDims-by-numCons numeric matrix Specifies numCons constraints on B given by R'B = 0, where numCons ≤ numDims − r.
"BVec" numDims1-by-numCons numeric matrix Specifies numCons equality constraints imposed on numCons of the cointegrating vectors in B, where numCons ≤ r.
A(B´yt−1+c0)+c1
"H*" A(B´yt−1+c0+d0t)+c1
"H" A(B´yt−1+c0+d0t)+c1+d1t
Example: Lags=1 includes Δyt – 1 in the model for all tests.
Example: Lags=[0 1] includes no lags in the model for the first test, and then includes Δyt – 1 in the model for the second test.
{A, B, B1, … Bq, c0, d0, c1, d1}.
EstCov Estimated covariance Q of the innovations process εt.
\begin{array}{c}\mathrm{Φ}\left(L\right)\left(1âL\right){y}_{t}=A\left(Bâ²{y}_{tâ1}+{c}_{0}+{d}_{0}t\right)+{c}_{1}+{d}_{1}t+{\mathrm{ε}}_{t}\\ =c+dt+C{y}_{tâ1}+{\mathrm{ε}}_{t},\end{array}
\mathrm{Φ}\left(L\right)=Iâ{\mathrm{Φ}}_{1}â{\mathrm{Φ}}_{2}â...â{\mathrm{Φ}}_{q}
r is the number of cointegrating relations and, in general, 0 ≤ r ≤ m.
C = AB′ is an m-by-m impact matrix with a rank of r.
Φj is an m-by-m matrix of short-run coefficients, where j = 1,...,q and Φq is not a matrix containing only zeros.
Constraint matrices R satisfying R′A = 0 or R′B = 0 are equivalent to A = Hφ or B = Hφ, where H is the orthogonal complement of R (null(R')) and φ is a vector of free parameters.
[2] Haug, A. “Testing Linear Restrictions on Cointegrating Vectors: Sizes and Powers of Wald Tests in Finite Samples.†Econometric Theory. v. 18, 2002, pp. 505–524. |
Interior_Schwarzschild_metric Knowpia
In Einstein's theory of general relativity, the interior Schwarzschild metric (also interior Schwarzschild solution or Schwarzschild fluid solution) is an exact solution for the gravitational field in the interior of a non-rotating spherical body which consists of an incompressible fluid (implying that density is constant throughout the body) and has zero pressure at the surface. This is a static solution, meaning that it does not change over time. It was discovered by Karl Schwarzschild in 1916, who earlier had found the exterior Schwarzschild metric.[1]
The interior Schwarzschild metric is framed in a spherical coordinate system with the body's centre located at the origin, plus the time coordinate. Its line element is[2][3]
{\displaystyle c^{2}{d\tau }^{2}={\frac {1}{4}}\left(3{\sqrt {1-{\frac {r_{s}}{r_{g}}}}}-{\sqrt {1-{\frac {r^{2}r_{s}}{r_{g}^{3}}}}}\right)^{2}c^{2}dt^{2}-\left(1-{\frac {r^{2}r_{s}}{r_{g}^{3}}}\right)^{-1}dr^{2}-r^{2}\left(d\theta ^{2}+\sin ^{2}\theta \,d\varphi ^{2}\right),}
{\displaystyle \tau }
is the proper time (time measured by a clock moving along the same world line with the test particle).
{\displaystyle c}
{\displaystyle t}
is the time coordinate (measured by a stationary clock located infinitely far from the spherical body).
{\displaystyle r}
is the Schwarzschild radial coordinate. Each surface of constan{\displaystyle t}
{\displaystyle r}
has the geometry of a sphere with measurable (proper) circumference
{\displaystyle 2\pi r}
{\displaystyle 4\pi r^{2}}
(as by the usual formulas), but the warping of space means the proper distance from each shell to the center of the body is greater than
{\displaystyle r}
{\displaystyle \theta }
is the colatitude (angle from north, in units of radians).
{\displaystyle \varphi }
is the longitude (also in radians).
{\displaystyle r_{s}}
is the Schwarzschild radius of the body, which is related to its mass
{\displaystyle M}
{\displaystyle r_{s}=2GM/c^{2}}
{\displaystyle G}
is the gravitational constant. (For ordinary stars and planets, this is much less than their proper radius.)
{\displaystyle r_{g}}
{\displaystyle r}
-coordinate at the body's surface. (This is less than its proper (measurable interior) radius, although for the Earth the difference is only about 1.4 millimetres.)
This solution is valid for
{\displaystyle r\leq r_{g}}
. For a complete metric of the sphere's gravitational field, the interior Schwarzschild metric has to be matched with the exterior one,
{\displaystyle c^{2}{d\tau }^{2}=\left(1-{\frac {r_{s}}{r}}\right)c^{2}dt^{2}-\left(1-{\frac {r_{s}}{r}}\right)^{-1}dr^{2}-r^{2}\left(d\theta ^{2}+\sin ^{2}\theta \,d\varphi ^{2}\right),}
at the surface. It can easily be seen that the two have the same value at the surface, i.e., at
{\displaystyle r=r_{g}}
Defining a parameter
{\displaystyle {\mathcal {R}}^{2}=r_{g}^{3}/r_{s}}
{\displaystyle c^{2}{d\tau }^{2}={\frac {1}{4}}\left(3{\sqrt {1-{\frac {r_{g}^{2}}{{\mathcal {R}}^{2}}}}}-{\sqrt {1-{\frac {r^{2}}{{\mathcal {R}}^{2}}}}}\right)^{2}c^{2}dt^{2}-\left(1-{\frac {r^{2}}{{\mathcal {R}}^{2}}}\right)^{-1}dr^{2}-r^{2}\left(d\theta ^{2}+\sin ^{2}\theta \,d\varphi ^{2}\right).}
We can also define an alternative radial coordinate
{\displaystyle \eta =\arcsin {\frac {r}{\mathcal {R}}}}
and a corresponding parameter
{\displaystyle \eta _{g}=\arcsin {\frac {r_{g}}{\mathcal {R}}}=\arcsin {\sqrt {\frac {r_{s}}{r_{g}}}}}
, yielding[4]
{\displaystyle c^{2}{d\tau }^{2}=\left({\frac {3\cos \eta _{g}-\cos \eta }{2}}\right)^{2}c^{2}dt^{2}-{\frac {dr^{2}}{\cos ^{2}\eta }}-r^{2}\left(d\theta ^{2}+\sin ^{2}\theta \,d\varphi ^{2}\right).}
{\displaystyle g_{rr}=(1-r_{s}r^{2}/r_{g}^{3})^{-1}}
{\displaystyle A=4\pi r^{2}}
the integral for the proper volume is
{\displaystyle V=\int _{0}^{r_{g}}A{\sqrt {g_{rr}}}\,{\rm {d}}r=2\pi \left({\frac {r_{g}^{9/2}\arcsin {\sqrt {\frac {r_{s}}{r_{g}}}}}{r_{s}^{3/2}}}-{\frac {r_{g}^{4}{\sqrt {1-{\frac {r_{s}}{r_{g}}}}}}{r_{s}}}\right)}
which is larger than the volume of a euclidean reference shell.
The fluid has a constant density by definition. It is given by
{\displaystyle \rho ={\frac {M}{{\frac {4\pi }{3}}r_{g}^{3}}}={\frac {3}{\kappa {\mathcal {R}}^{2}}},}
{\displaystyle \kappa =8\pi G/c^{2}}
is the Einstein gravitational constant.[3][5] It may be counterintuitive that the density is the mass divided by the volume of a sphere with radius
{\displaystyle r_{g}}
, which seems to disregard that this is less than the proper radius, and that space inside the body is curved so that the volume formula for a "flat" sphere shouldn't hold at all. However,
{\displaystyle M}
is the mass measured from the outside, for example by observing a test particle orbiting the gravitating body (the "Kepler mass"), which in general relativity is not necessarily equal to the proper mass. This mass difference exactly cancels out the difference of the volumes.
Pressure and stabilityEdit
The pressure of the incompressible fluid can be found by calculating the Einstein tensor
{\displaystyle G_{\mu \nu }}
from the metric. The Einstein tensor is diagonal (i.e., all off-diagonal elements are zero), meaning there are no shear stresses, and has equal values for the three spatial diagonal components, meaning pressure is isotropic. Its value is
{\displaystyle p=\rho c^{2}{\frac {\cos \eta -\cos \eta _{g}}{3\cos \eta _{g}-\cos \eta }}.}
As expected, the pressure is zero at the surface of the sphere and increases towards the centre. It becomes infinite at the centre if
{\displaystyle \cos \eta _{g}=1/3}
{\displaystyle r_{s}={\frac {8}{9}}r_{g}}
{\displaystyle \eta _{g}\approx 70.5^{\circ }}
, which is true for a body that is extremely dense or large. Such a body suffers gravitational collapse into a black hole. As this is a time dependent process, the Schwarzschild solution does not hold any longer.[2][3]
RedshiftEdit
Gravitational redshift for radiation from the sphere's surface (for example, light from a star) is
{\displaystyle z={\frac {1}{\cos \eta _{g}}}-1.}
From the stability condition
{\displaystyle \cos \eta _{g}>1/3}
{\displaystyle z<2}
Embedding of a Schwarzschild metric's slice in three-dimensional Euclidean space. The interior solution is the darker cap at the bottom.
This embedding should not be confused with the unrelated concept of a gravity well.
The spatial curvature of the interior Schwarzschild metric can be visualized by taking a slice (1) with constant time and (2) through the sphere's equator, i.e.
{\displaystyle t=const.,\theta =\pi /2}
. This two-dimensional slice can be embedded in a three-dimensional Euclidean space and then takes the shape of a spherical cap with radius
{\displaystyle {\mathcal {R}}}
and half opening angle
{\displaystyle \eta _{g}}
. Its Gaussian curvature
{\displaystyle K}
is proportional to the fluid's density and equals
{\displaystyle {\mathcal {R}}^{-2}=r_{s}/r_{g}^{3}=\rho \kappa /3}
. As the exterior metric can be embedded in the same way (yielding Flamm's paraboloid), a slice of the complete solution can be drawn like this:[5][6]
In this graphic, the blue circular arc represents the interior metric, and the black parabolic arcs with the equation
{\displaystyle w=2{\sqrt {r_{s}(r-r_{s})}}}
represent the exterior metric, or Flamm's paraboloid. The
{\displaystyle \eta }
-coordinate is the angle measured from the centre of the cap, that is, from "above" the slice. The proper radius of the sphere – intuitively, the length of a measuring rod spanning from its centre to a point on its surface – is half the length of the circular arc, or
{\displaystyle \eta _{g}{\mathcal {R}}}
This is a purely geometric visualization and does not imply a physical "fourth spatial dimension" into which space would be curved. (Intrinsic curvature does not imply extrinsic curvature.)
Here are the relevant parameters for some astronomical objects, disregarding rotation and inhomogeneities such as deviation from the spherical shape and variation in density.
{\displaystyle r_{g}}
{\displaystyle r_{s}}
{\displaystyle {\mathcal {R}}}
{\displaystyle \eta _{g}}
{\displaystyle z}
Earth 6,370 km 8.87 mm 170,000,000 km
9.5 light-minutes 7.7″ 7×10−10
Sun 696,000 km 2.95 km 338,000,000 km
19 light-minutes 7.0′ 2×10−6
White dwarf with 1 solar mass 5000 km 2.95 km 200,000 km 1.4° 3×10−4
Neutron star with 2 solar masses 20 km 6 km 37 km 30° 0.15
The interior Schwarzschild solution was the first static spherically symmetric perfect fluid solution that was found. It was published on 24 February 1916, only three months after Einstein's field equations and one month after Schwarzschild's exterior solution.[1][2]
^ a b Karl Schwarzschild (1916). "Über das Gravitationsfeld eines Massenpunktes nach der Einsteinschen Theorie" [On the gravitational field of a point mass following Einstein's theory]. Sitzungsberichte der Königlich-Preussischen Akademie der Wissenschaften (in German). Berlin: 189–196.
^ a b c Karl Schwarzschild (1916). "Über das Gravitationsfeld einer Kugel aus inkompressibler Flüssigkeit nach der Einsteinschen Theorie" [On the gravitational field of a ball of incompressible fluid following Einstein's theory]. Sitzungsberichte der Königlich-Preussischen Akademie der Wissenschaften (in German). Berlin: 424–434.
^ a b c d Torsten Fließbach (2003). Allgemeine Relativitätstheorie [General Theory of Relativity] (in German) (4th ed.). Spektrum Akademischer Verlag. pp. 231–241. ISBN 3-8274-1356-7.
^ R. Burghardt (2009). "Interior Schwarzschild Solution and Free Fall" (PDF). Austrian Reports on Gravitation.
^ a b P. S. Florides (1974). "A New Interior Schwarzschild Solution". Proceedings of the Royal Society of London. Series A, Mathematical and Physical Sciences. 337 (1611): 529–535. Bibcode:1974RSPSA.337..529F. doi:10.1098/rspa.1974.0065. JSTOR 78530.
^ R. Burghardt (2009). "New Embedding of Schwarzschild Geometry. II. Interior Solution" (PDF). Austrian Reports on Gravitation. |
Biochemistry - Vocabulary - Course Hero
General Chemistry/Biochemistry/Vocabulary
helical structure of proteins that adopts a right-handed-spiral conformation, where chain backbone
{\rm{N}{-}\rm{H}}
{\rm{C}{=}\rm{O}}
groups of nearby amino acids along the protein chain
protein-building molecule that contains an amino group (
{\rm{{-}NH}}_2
\rm{{-}COOH}
), and between them a carbon atom attached to a side chain
structure of amino acids that takes the shape of a pleated sheet folded at regular intervals
carbohydrate made of two monosaccharides
monosaccharide (simple sugar) used for energy in biological systems
hydrophobic macromolecule that may provide storage, structure, or nutrients in organisms
single (simple) sugar molecule
large molecule made of nucleotides
organic compound that forms the basis of a genetic sequence, consisting of a sugar, a phosphate, and a nitrogenous base
covalent amide (
\rm{{-}CONH{-}}
) bond that forms between two adjacent amino acids along the same peptide chain. The bond forms between the amine group of one amino acid and the carboxyl (
\rm{{-}COOH}
) group of another.
lipid that usually consists of two fatty acid tails covalently linked to a common phosphate group
carbohydrate polymer consisting of many monosaccharides that form a long chain
sequence of amino acids linked by peptide (amide) bonds to form a polypeptide chain
two or more chains of proteins that interact with one another
structure of a protein that is the initial folding of the amino acid polymer that arises from hydrogen bonding between the amine and carboxyl groups of amino acids in neighboring parts of the chain
lipid that contains a core with four fused ring structures (three 6- and one 5-membered ring) that serves as a precursor to hormones, cholesterol, or other molecules
structure of a protein produced by interactions between the R groups of the amino acids in the chain and with the environment around them
<Overview>Amino Acids, Proteins, and Enzymes |
torch.range — PyTorch 1.11.0 documentation
torch.range
torch.range¶
torch.range(start=0, end, step=1, *, out=None, dtype=None, layout=torch.strided, device=None, requires_grad=False) → Tensor¶
\left\lfloor \frac{\text{end} - \text{start}}{\text{step}} \right\rfloor + 1
with values from start to end with step step. Step is the gap between two values in the tensor.
\text{out}_{i+1} = \text{out}_i + \text{step}.
This function is deprecated and will be removed in a future release because its behavior is inconsistent with Python’s range builtin. Instead, use torch.arange(), which produces values in [start, end).
start (float) – the starting value for the set of points. Default: 0.
step (float) – the gap between each pair of adjacent points. Default: 1.
dtype (torch.dtype, optional) – the desired data type of returned tensor. Default: if None, uses a global default (see torch.set_default_tensor_type()). If dtype is not given, infer the data type from the other input arguments. If any of start , end , or stop are floating-point, the dtype is inferred to be the default dtype, see get_default_dtype(). Otherwise, the dtype is inferred to be torch.int64 .
>>> torch.range(1, 4)
>>> torch.range(1, 4, 0.5) |
Can you eat 50%of the cake Can you eat 100%of the cake Can you eat 150 % of - Maths - Comparing Quantities - 10033343 | Meritnation.com
Can you eat 50%of the cake?
Can you eat 100%of the cake?
Can you eat 150 % of the cake?
1 ) Yes , you eat 50 % of the cake .
As : 50 % of a cake =
\frac{50 × \mathrm{Cake}}{100 } = \left(\frac{1}{2}\right)\left( \mathrm{Cake} \right)
And we can eat half of a cake .
2 ) Yes , you eat 100 % of the cake .
As : 100 % of a cake =
\frac{100 × \mathrm{Cake}}{100 } = \mathrm{Cake}
And we can eat a full cake .
3 ) Now , you eat 150 % of the cake .
\frac{150 × \mathrm{Cake}}{100 } = \left(\frac{3}{2}\right)\left( \mathrm{Cake} \right) = 1 \mathrm{Cake} + \left(\frac{1}{2}\right)\left( \mathrm{Cake} \right)
And we can't eat a cake which is more than itself.
Hope this information will clear your doubts about Comparing Quantities.
If the total cake is 50% or 100% we can consume .
If the cake is 100% or more we can consume .
If the cake is 150% or 200% or more we can consume it .
We can eat 50% of a cake ,
even 100%. of cake,
but can't eat 150% of a cake
Nimish Gupta answered this
we can eat 50%
" " " 100%
but can't eat 150% |
Solve this: Q) Is the following in a standard form -4-15 - Maths - Rational Numbers - 11838205 | Meritnation.com
\frac{-4}{-15}
https://www.meritnation.com/ask-answer/question/what-is-standard-form-in-rational-numbers-explain-briefly/rational-numbers/4042564
\mathrm{The} \mathrm{standard} \mathrm{form} \mathrm{of} \frac{-4}{-15} \mathrm{is} \frac{4}{15}
no the minus signs should be cut down it should be 4 upon 15
Shriyansh Ranjan answered this
Its standard form is 4/15.
Soma Sekhar Sabat answered this
Yes it is in its standard form.
Krish Invincible answered this |
A borrows RS8000 at 12% per annum simple intrest and B borrows RS 9100 at 10% per annum simple intrest - Maths - Comparing Quantities - 6882124 | Meritnation.com
A borrows RS8000 at 12% per annum simple intrest and B borrows RS 9100 at 10% per annum simple intrest.In how many years will their amount be equal
Let the amount of A and B be equal after t years.
Now simple interest for A after t years =
\frac{8000\times 12\times t}{100}\quad =\quad 960t
Then, the total amount of A = 8000 + 960t......(i)
Similarly the simple interest for B after t years =
\frac{9100\times 10\times t}{100}\quad =\quad 910t
Then, the total amount of B = 9100 + 910t.....(ii)
Now, equating (i) and (ii) we have;
8000+960t\quad =\quad 9100+910t\phantom{\rule{0ex}{0ex}}\Rightarrow 960t-910t\quad =\quad 9100-8000\phantom{\rule{0ex}{0ex}}\Rightarrow 50t\quad =\quad 1100\phantom{\rule{0ex}{0ex}}\Rightarrow t\quad =\quad \frac{1100}{50}\quad =\quad 22
So the amount of A and B will be equal after 22 years. |
A number of methodological papers published during the last years testify that a need for a thorough revision of the research methodology is felt by the operations research community - see, for example, [Barr et al., J. Heuristics 1 (1995) 9-32; Eiben and Jelasity, Proceedings of the 2002 Congress on Evolutionary Computation (CEC'2002) 582-587; Hooker, J. Heuristics 1 (1995) 33-42; Rardin and Uzsoy, J. Heuristics 7 (2001) 261-304]. In particular, the performance evaluation of nondeterministic methods, including widely studied metaheuristics such as evolutionary computation and ant colony optimization, requires the definition of new experimental protocols. A careful and thorough analysis of the problem of evaluating metaheuristics reveals strong similarities between this problem and the problem of evaluating learning methods in the machine learning field. In this paper, we show that several conceptual tools commonly used in machine learning - such as, for example, the probabilistic notion of class of instances and the separation between the training and the testing datasets - fit naturally in the context of metaheuristics evaluation. Accordingly, we propose and discuss some principles inspired by the experimental practice in machine learning for guiding the performance evaluation of optimization algorithms. Among these principles, a clear separation between the instances that are used for tuning algorithms and those that are used in the actual evaluation is particularly important for a proper assessment.
Classification : 68T05, 68T20, 68W20, 68W40, 90C27
author = {Birattari, Mauro and Zlochin, Mark and Dorigo, Marco},
title = {Towards a theory of practice in metaheuristics design : a machine learning perspective},
AU - Birattari, Mauro
AU - Zlochin, Mark
AU - Dorigo, Marco
TI - Towards a theory of practice in metaheuristics design : a machine learning perspective
Birattari, Mauro; Zlochin, Mark; Dorigo, Marco. Towards a theory of practice in metaheuristics design : a machine learning perspective. RAIRO - Theoretical Informatics and Applications - Informatique Théorique et Applications, Tome 40 (2006) no. 2, pp. 353-369. doi : 10.1051/ita:2006009. http://www.numdam.org/articles/10.1051/ita:2006009/
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[2] M. Birattari, The Problem of Tuning Metaheuristics, as Seen from a Machine Learning Perspective. Ph.D. thesis, Université Libre de Bruxelles, Brussels, Belgium (2004). | Zbl 1101.68748
[3] M. Birattari, T. Stützle, L. Paquete and K. Varrentrapp, A racing algorithm for configuring metaheuristics, in Proceedings of the Genetic and Evolutionary Computation Conference (GECCO-2002), edited by W.B. Langdon, et al. Morgan Kaufmann Publishers, San Francisco, CA (2002) 11-18.
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𝒩P
-Completeness. Freeman, San Francisco, CA (1979). | MR 519066 | Zbl 0411.68039
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[22] K. Liang, X. Yao and C. Newton, Adapting self-adaptive parameters in evolutionary algorithms. Appl. Intell. 15 (2001) 171-180. | Zbl 0992.68231
[23] H.R. Lourenço, O. Martin and T. Stützle, Iterated local search, in Handbook of Metaheuristics. International Series in Operations Research & Management Science, edited by F. Glover and G. Kochenberger. Kluwer Academic Publishers, Norwell, MA 57 (2002) 321-353. | Zbl 1116.90412
[24] D.G. Luenberger, Introduction to Linear and Nonlinear Programming. Addison-Wesley Publishing Company, Reading, MA (1973). | Zbl 0297.90044
[25] O. Maron and A.W. Moore, Hoeffding races: Accelerating model selection search for classification and function approximation, in Advances in Neural Information Processing Systems, edited by J.D. Cowan, G. Tesauro and J. Alspector. Morgan Kaufmann Publishers, San Francisco, CA 6 (1994) 59-66.
[26] C.C. Mcgeogh, Toward an experimental method for algorithm simulation. INFORMS J. Comput. 2 (1996) 1-15. | Zbl 0854.68038
[27] Z. Michalewicz and D.B. Fogel, How to Solve it: Modern Heuristics. Springer-Verlag, Berlin, Germany (2000). | MR 1730907 | Zbl 0943.90002
[28] A.W. Moore and M.S. Lee, Efficient algorithms for minimizing cross validation error, in International Conference on Machine Learning. Morgan Kaufmann Publishers, San Francisco, CA (1994) 190-198.
[29] B.L. Nelson, J. Swann, D. Goldsman and W. Song, Simple procedures for selecting the best simulated system when the number of alternatives is large. Oper. Res. 49 (2001) 950-963.
[30] R.R. Rardin and R. Uzsoy, Experimental evaluation of heuristic optimization algorithms: A tutorial. J. Heuristics 7 (2001) 261-304. | Zbl 0972.68634
[31] I. Rechenberg, Evolutionsstrategie - Optimierung technischer Systeme nach Prinzipien der biologischen Information. Fromman Verlag, Freiburg, Germany (1973).
[32] H.-P. Schwefel, Numerical Optimization of Computer Models. John Wiley & Sons, Chichester, UK (1981). | Zbl 0451.65043
[33] I. Sommerville, Software Engineering. Addison Wesley, Harlow, UK, sixth edition (2001). | Zbl 0557.68006
[34] M. Toussaint, Self-adaptive exploration in evolutionary search. Technical Report IRINI-2001-05, Institut für Neuroinformatik, Ruhr-Universität Bochum, Bochum, Germany (2001).
[35] D.H. Wolpert and W.G. Macready, No free lunch theorems for optimization. IEEE Transactions on Evolutionary Computation 1 (1997) 67-82.
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Inverse discrete cosine transform (IDCT) of input - Simulink - MathWorks Deutschland
Inverse discrete cosine transform (IDCT) of input
The IDCT block computes the inverse discrete cosine transform (IDCT) of each channel in the M-by-N input matrix, u.
y = idct(u) % Equivalent MATLAB code
For all N-D input arrays, the block computes the IDCT across the first dimension. The size of the first dimension (frame size), must be a power of two. To work with other frame sizes, use the Pad block to pad or truncate the frame size to a power-of-two length.
When the input is an M-by-N matrix, the block treats each input column as an independent channel containing M consecutive samples. The block outputs an M-by-N matrix whose lth column contains the length-M IDCT of the corresponding input column.
y\left(m,l\right)=\sum _{k=1}^{M}w\left(k\right)u\left(k,l\right)\mathrm{cos}\frac{\pi \left(2m-1\right)\left(k-1\right)}{2M},\begin{array}{ccc}& & m=1,...,M\end{array}
w\left(k\right)=\left\{\begin{array}{l}\frac{1}{\sqrt{M}}\\ \sqrt{\frac{2}{M}}\end{array}\begin{array}{c}\begin{array}{l}\\ ,\\ \\ ,\end{array}\\ \end{array}\begin{array}{ll}\hfill & \hfill \\ \hfill & k=1\hfill \\ \hfill & \hfill \\ \hfill & 2\le k\le M\hfill \\ \hfill & \hfill \end{array}
The following diagrams show the data types used within the IDCT block for fixed-point signals. You can set the sine table, accumulator, product output, and output data types displayed in the diagrams in the IDCT block dialog as discussed in Parameters.
Inputs to the IDCT block are first cast to the output data type and stored in the output buffer. Each butterfly stage processes signals in the accumulator data type, with the final output of the butterfly being cast back into the output data type.
W{L}_{idealoutput}=W{L}_{input}+floor\left({\mathrm{log}}_{2}\left(DCTlength-1\right)\right)+1
F{L}_{idealoutput}=F{L}_{input}
DCT | IFFT |
Null controllability of the heat equation with boundary Fourier conditions : the linear case
In this paper, we prove the global null controllability of the linear heat equation completed with linear Fourier boundary conditions of the form
\frac{\partial y}{\partial n}+\beta \phantom{\rule{0.166667em}{0ex}}y=0
. We consider distributed controls with support in a small set and nonregular coefficients
\beta =\beta \left(x,t\right)
. For the proof of null controllability, a crucial tool will be a new Carleman estimate for the weak solutions of the classical heat equation with nonhomogeneous Neumann boundary conditions.
Mots clés : controllability, heat equation, Fourier conditions
title = {Null controllability of the heat equation with boundary {Fourier} conditions : the linear case},
TI - Null controllability of the heat equation with boundary Fourier conditions : the linear case
Fernández-Cara, Enrique; González-Burgos, Manuel; Guerrero, Sergio; Puel, Jean-Pierre. Null controllability of the heat equation with boundary Fourier conditions : the linear case. ESAIM: Control, Optimisation and Calculus of Variations, Tome 12 (2006) no. 3, pp. 442-465. doi : 10.1051/cocv:2006010. http://www.numdam.org/articles/10.1051/cocv:2006010/
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[6] O.Yu. Imanuvilov and M. Yamamoto, Carleman estimate for a parabolic equation in a Sobolev space of negative order and its applications, Dekker, New York. Lect. Notes Pure Appl. Math. 218 (2001). | MR 1817179 | Zbl 0977.93041
[7] G. Lebeau and L. Robbiano, Contrôle exacte de l'equation de la chaleur (French). Comm. Partial Differ. Equat. 20 (1995) 335-356. | Zbl 0819.35071
[8] D.L. Russell, A unified boundary controllability theory for hyperbolic and parabolic partial differential equations. Studies Appl. Math. 52 (1973) 189-211. | Zbl 0274.35041 |
Characterizations of Semihyperrings by Their (∈γ,∈γ∨qδ)-Fuzzy Hyperideals
2013 Characterizations of Semihyperrings by Their (
{\in }_{\gamma },{\in }_{\gamma }\vee {q}_{\delta }
)-Fuzzy Hyperideals
Xiaokun Huang, Yunqiang Yin, Jianming Zhan
The concepts of (
{\in }_{\gamma },{\in }_{\gamma }\vee {q}_{\delta }
)-fuzzy bi-hyperideals and (
{\in }_{\gamma },{\in }_{\gamma }\vee {q}_{\delta }
)-fuzzy quasi-hyperideals of a semihyperring are introduced, and some related properties of such (
{\in }_{\gamma },{\in }_{\gamma }\vee {q}_{\delta }
)-fuzzy hyperideals are investigated. In particular, the notions of hyperregular semihyperrings and left duo semihyperrings are given, and their characterizations in terms of hyperideals and (
{\in }_{\gamma },{\in }_{\gamma }\vee {q}_{\delta }
)-fuzzy hyperideals are studied.
Xiaokun Huang. Yunqiang Yin. Jianming Zhan. "Characterizations of Semihyperrings by Their (
{\in }_{\gamma },{\in }_{\gamma }\vee {q}_{\delta }
)-Fuzzy Hyperideals." J. Appl. Math. 2013 1 - 13, 2013. https://doi.org/10.1155/2013/348203
Xiaokun Huang, Yunqiang Yin, Jianming Zhan "Characterizations of Semihyperrings by Their (
{\in }_{\gamma },{\in }_{\gamma }\vee {q}_{\delta }
)-Fuzzy Hyperideals," Journal of Applied Mathematics, J. Appl. Math. 2013(none), 1-13, (2013) |
Write the underlying principle of a moving coil galvanometer. VIEW SOLUTION
Why are microwaves considered suitable for radar systems used in aircraft navigation? VIEW SOLUTION
Define 'quality factor' of resonance in a series LCR circuit. What is its SI unit? VIEW SOLUTION
A point charge +Q is placed at point O, as shown in the figure. Is the potential difference VA – VB positive, negative or zero?
How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased? VIEW SOLUTION
A nucleus with mass number A = 240 and BE/A = 7.6 MeV breaks into two fragments, each of A = 120 with BE/A = 8.5 MeV. Calculate the released energy.
Calculate the energy in fusion reaction:
{}_{1}{}^{2}\mathrm{H}+{}_{1}{}^{2}\mathrm{H}\to {}_{2}{}^{3}\mathrm{He}+\mathrm{n},
where BE of
{}_{1}{}^{2}\mathrm{H}=2.23
MeV and of
{}_{2}{}^{3}\mathrm{He}=7.73
MeV VIEW SOLUTION
Two cells of emf 1.5 V and 2.0 V, respectively, having internal resistances 0.2 Ω and 0.3 Ω, respectively, are connected in parallel. Calculate the emf and internal resistance of the equivalent cell. VIEW SOLUTION
State Brewster's law.
The value of Brewster's angle for a transparent medium is different for light of different colours. Give reason. VIEW SOLUTION
Explain the terms (i) Attenuation and (ii) Demodulation used in Communication System. VIEW SOLUTION
Plot a graph showing variation of de Broglie wavelength λ versus
\frac{1}{\sqrt{\mathrm{V}}},
where V is accelerating potential for two particles A and B, carrying the same charge but different masses m1, m2 (m1 > m2). Which one of the two represents a particle of smaller mass and why? VIEW SOLUTION
(i) Define mutual inductance.
(ii) A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil? VIEW SOLUTION
Two parallel-plate capacitors X and Y have the same area of plates and same separation between them. X has air between the plates, while Y contains a dielectric medium of εr = 4.
(i) Calculate capacitance of each capacitor if the equivalent capacitance of the combination is 4 μF.
(ii) Calculate the potential difference between the plates X and Y.
(iii) Estimate the ratio of electrostatic energies stored in X and Y. VIEW SOLUTION
Two long, straight, parallel conductors carry steady currents, I1 and I2, separated by a distance d. If the currents are flowing in the same direction, show how the magnetic field set up in one produces an attractive force on the other? Obtain the expression for this force. Hence, define one ampere. VIEW SOLUTION
A charge is distributed uniformly over a ring of radius 'a'. Obtain an expression for the electric intensity E at a point on the axis of the ring. Hence, show that for points at large distance from the ring, it behaves like a point charge. VIEW SOLUTION
Write three characteristic features in photoelectric effect that cannot be explained on the basis of wave theory of light, but can be explained only using Einstein's equation. VIEW SOLUTION
(a) Write the expression for the magnetic force acting on a charged particle moving with velocity v in the presence of magnetic field B.
(b) A neutron, an electron and an alpha particle, moving with equal velocities, enter a uniform magnetic field going into the plane of the paper, as shown. Trace their paths in the field and justify your answer.
(a) Calculate the distance of an object of height h from a concave mirror of radius of curvature 20 cm, so as to obtain a real image of magnification 2. Find the location of image also.
(b) Using mirror formula, explain why does a convex mirror always produce a virtual image. VIEW SOLUTION
(i) State Bohr's quantization condition for defining stationary orbits. How does the de Broglie hypothesis explain the stationary orbits?
(ii) Find the relation between three wavelengths λ1, λ2 and λ3 from the energy-level diagram shown below.
Draw a schematic ray diagram of a reflecting telescope showing how rays coming from a distant object are received at the eyepiece. Write its two important advantages over a refracting telescope. VIEW SOLUTION
How are em waves produced by oscillating charges?
Draw a sketch of linearly polarized em waves propagating in the Z-direction. Indicate the directions of the oscillating electric and magnetic fields.
Write Maxwell's generalization of Ampere's circuital law. Show that in the process of charging a capacitor, the current produced within the plates of the capacitor is
I={\epsilon }_{\mathit{0}}\frac{\mathit{d}{\mathit{\Phi }}_{\mathit{E}}}{\mathit{d}\mathit{t}}
, where ΦE is the electric flux produced during charging of the capacitor plates. VIEW SOLUTION
(a) Explain any two factors that justify the need of modulating a low-frequency signal.
(b) Write two advantages of frequency modulation over amplitude modulation. VIEW SOLUTION
(i) Write the functions of three segments of a transistor.
(ii) Draw a circuit diagram for studying the input and output characteristics of a n-p-n transistor in common emitter configuration. Using the circuit, explain how input, output characteristics are obtained. VIEW SOLUTION
Meeta's father was driving her to school. At the traffic signal, she noticed that each traffic light was made of many tiny lights instead of a single bulb. When Meeta asked this question to her father, he explained the reason for this.
Answer the following questions based on above information:
(i) What were the values displayed by Meeta and her father?
(ii) What answer did Meeta's father give?
(iii) What are the tiny lights in traffic signals called and how do these operate? VIEW SOLUTION
(i) Define the term drift velocity.
(ii) On the basis of electron drift, derive an expression for resistivity of a conductor in terms of number density of free electrons and relaxation time. On what factors does resistivity of a conductor depend?
(iii) Why alloys like constantan and manganin are used for making standard resistors?
(i) State the principle of working of a potentiometer.
(ii) In the following potentiometer circuit, AB is a uniform wire of length 1 m and resistance 10 Ω. Calculate the potential gradient along the wire and balance length AO (= l).
(i) An a.c. source of voltage V = V0 sin ωt is connected to a series combination of L, C and R. Use the phasor diagram to obtain expression for impedance of a circuit and the phase angle between voltage and current. Find the condition when current will be in phase with the voltage. What is the circuit in this condition called?
(ii) In a series LR circuit, XL = R and power factor of the circuit is P1. When capacitor with capacitance C such that XL = XC is put in series, the power factor becomes P2. Calculate P1/P2.
(i) Write the function of a transformer. State its principle of working with the help of a diagram. Mention various energy losses in this device.
(ii) The primary coil of an ideal step-up transformer has 100 turns and the transformation ratio is also 100. The input voltage and power are 220 V and 1100 W, respectively. Calculate the
(a) number of turns in secondary
(b) current in primary
(c) voltage across secondary
(d) current in secondary
(e) power in secondary VIEW SOLUTION
(i) In Young's double-slit experiment, deduce the condition for (a) constructive and (b) destructive interferences at a point on the screen. Draw a graph showing variation of intensity in the interference pattern against position 'x' on the screen.
(b) Compare the interference pattern observed in Young's double-slit experiment with single-slit diffraction pattern, pointing out three distinguishing features.
(i) Plot a graph to show variation of the angle of deviation as a function of angle of incidence for light passing through a prism. Derive an expression for refractive index of the prism in terms of angle of minimum deviation and angle of prism.
(ii) What is dispersion of light? What is its cause?
(iii) A ray of light incident normally on one face of a right isosceles prism is totally reflected, as shown in fig. What must be the minimum value of refractive index of glass? Give relevant calculations. |
Mathematics | Special Issue : Mathematical Analysis and Boundary Value Problems II
Mathematical Analysis and Boundary Value Problems II
issue 1 2 3 4 5 6 7 8 9 10 11 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
Special Issue "Mathematical Analysis and Boundary Value Problems II"
Facultade de Matemáticas,Campus Vida, 15782 Santiago de Compostela, Galicia, Spain
Interests: ordinary differential equations; boundary value problems; Green's functions; comparison results; nonlinear analysis
Special Issue in Mathematics: Mathematical Analysis and Boundary Value Problems
Special Issue in Axioms: Mathematical Control and Applications
Special Issue in Mathematics: Reviews in Mathematics and Applications
Prof. Dr. José Ángel Cid
Departamento de Matemáticas, Universidade de Vigo, Campus de Ourense, 32004 Ourense, Spain
Interests: differential equations; ordinary differential equations; mathematical analysis
Dr. Lucía López-Somoza
Departamento de Estatística, Análise Matemática e Optimización Instituto de Matemáticas, Facultade de Matemáticas, Universidade de Santiago de Compostela, Santiago de Compostela, 15782 Galicia, Spain
Interests: differential equations; boundary value problems; mathematical analysis
The study of the existence, nonexistence, and the uniqueness of solutions of boundary value problems, coupled to its stability, plays a fundamental role in the research of different kinds of differential equations (ordinary, fractional, and partial). One of the main tools developed in this area consists of fixed point theory and critical point theory.
The aim of this Special Issue is to study this type of problem in a broad sense. The development of theories that ensure the existence of solutions via topological or variational methods will contribute to the enrichment of this topic and will broaden the knowledge of this area.
This issue is a continuation of the previous successful Special Issue “Mathematical Analysis and Boundary Value Problems”.
topological methods in differential equations
Monotone Iterative Technique for a New Class of Nonlinear Sequential Fractional Differential Equations with Nonlinear Boundary Conditions under the ψ-Caputo Operator
Zidane Baitiche
Choukri Derbazi
The main crux of this work is to study the existence of extremal solutions for a new class of nonlinear sequential fractional differential equations (NSFDEs) with nonlinear boundary conditions (NBCs) under the
\psi
-Caputo operator. The obtained outcomes of the proposed problem are [...] Read more.
\psi
-Caputo operator. The obtained outcomes of the proposed problem are derived by means of the monotone iterative technique (MIT) associated with the method of upper and lower solutions. Lastly, the desired findings are well illustrated by an example. Full article
(This article belongs to the Special Issue Mathematical Analysis and Boundary Value Problems II)
Madalina Osiceanu
Two contact models are considered, with the behavior of the materials being described by a constitutive law governed by the subdifferential of a convex map. We deliver variational formulations based on the theory of bipotentials. In this approach, the unknowns are pairs consisting [...] Read more.
Two contact models are considered, with the behavior of the materials being described by a constitutive law governed by the subdifferential of a convex map. We deliver variational formulations based on the theory of bipotentials. In this approach, the unknowns are pairs consisting of the displacement field and the Cauchy stress tensor. The two-field weak solutions are sought into product spaces involving variable convex sets. Both models lead to variational systems which can be cast in an abstract setting. After delivering some abstract results, we apply them in order to study the weak solvability of the mechanical models as well as the data dependence of the weak solutions. Full article |
A Parameterized Splitting Preconditioner for Generalized Saddle Point Problems
2013 A Parameterized Splitting Preconditioner for Generalized Saddle Point Problems
Wei-Hua Luo, Ting-Zhu Huang
By using Sherman-Morrison-Woodbury formula, we introduce a preconditioner based on parameterized splitting idea for generalized saddle point problems which may be singular and nonsymmetric. By analyzing the eigenvalues of the preconditioned matrix, we find that when α is big enough, it has an eigenvalue at 1 with multiplicity at least
n
, and the remaining eigenvalues are all located in a unit circle centered at 1. Particularly, when the preconditioner is used in general saddle point problems, it guarantees eigenvalue at 1 with the same multiplicity, and the remaining eigenvalues will tend to 1 as the parameter
\alpha \to 0
. Consequently, this can lead to a good convergence when some GMRES iterative methods are used in Krylov subspace. Numerical results of Stokes problems and Oseen problems are presented to illustrate the behavior of the preconditioner.
Wei-Hua Luo. Ting-Zhu Huang. "A Parameterized Splitting Preconditioner for Generalized Saddle Point Problems." J. Appl. Math. 2013 (SI03) 1 - 6, 2013. https://doi.org/10.1155/2013/489295
Wei-Hua Luo, Ting-Zhu Huang "A Parameterized Splitting Preconditioner for Generalized Saddle Point Problems," Journal of Applied Mathematics, J. Appl. Math. 2013(SI03), 1-6, (2013) |
2019 Certain Subclasses of Bi-Close-to-Convex Functions Associated with Quasi-Subordination
U=\left\{z:\left|z\right|<\mathrm{1}\right\}
\left|{a}_{\mathrm{2}}\right|
\left|{a}_{\mathrm{3}}\right|
Gurmeet Singh. Gurcharanjit Singh. Gagandeep Singh. "Certain Subclasses of Bi-Close-to-Convex Functions Associated with Quasi-Subordination." Abstr. Appl. Anal. 2019 1 - 6, 2019. https://doi.org/10.1155/2019/6947020
Received: 13 January 2019; Accepted: 7 March 2019; Published: 2019
Gurmeet Singh, Gurcharanjit Singh, Gagandeep Singh "Certain Subclasses of Bi-Close-to-Convex Functions Associated with Quasi-Subordination," Abstract and Applied Analysis, Abstr. Appl. Anal. 2019(none), 1-6, (2019) |
Median Sets of Isometries in CAT(0) Cube Complexes and Some Applications
2021 Median Sets of Isometries in
\mathrm{CAT}\left(0\right)
Cube Complexes and Some Applications
In this paper, we associate with isometries of
\mathrm{CAT}\left(0\right)
cube complexes specific subspaces, referred to as median sets, which play a similar role as minimizing sets of semisimple isometries in
\mathrm{CAT}\left(0\right)
spaces. Various applications are deduced, including a cubulation of centralizers, a splitting theorem, a proof that Dehn twists in mapping class groups must be elliptic for every action on a
\mathrm{CAT}\left(0\right)
cube complex, a cubical version of the flat torus theorem, and a structural theorem about polycyclic groups acting on
\mathrm{CAT}\left(0\right)
cube complexes.
Anthony Genevois. "Median Sets of Isometries in
\mathrm{CAT}\left(0\right)
Cube Complexes and Some Applications." Michigan Math. J. Advance Publication 1 - 46, 2021. https://doi.org/10.1307/mmj/20195823
Received: 28 October 2019; Revised: 2 December 2020; Published: 2021
Anthony Genevois "Median Sets of Isometries in
\mathrm{CAT}\left(0\right)
Cube Complexes and Some Applications," Michigan Mathematical Journal, Michigan Math. J. Advance Publication, 1-46, (2021) |
Genetics and Evolutionary Psychology - Course Hero
Introduction to Psychology/Brains, Biology, and Behavior/Genetics and Evolutionary Psychology
Learn all about genetics and evolutionary psychology in just a few minutes! Brooke Miller, Ph.D., instructor of psychology at the University of Texas at Austin, details how genetic and environmental factors influence physical and psychological traits.
Genes, made up of DNA, are the basic units of heredity. Genes carry instructions for manufacturing proteins, which guide development and influence physical traits and behavior.
A gene is a unit of heritable material that codes for a particular trait. Genes carry instructions for manufacturing proteins that guide development. These instructions can influence an organism's behavior and psychological traits, just as they influence physical traits. Deoxyribonucleic acid (DNA) is an organic molecule containing coded instructions for the life processes of all living things. This genetic code is found inside the cells of every organism, providing the blueprint for the traits the organism expresses. When an organism grows, DNA replicates itself so each new cell created has an identical copy of the genetic information. This DNA, the genetic material that is passed from one generation to the next, is contained within a threadlike strand called a chromosome. Chromosomes are in the cell nucleus, which controls cell activity.
Humans have 23 pairs of chromosomes. One set of chromosomes for each pair comes from a person's mother, and the other set comes from the father. Genes are sequences of DNA, and each chromosome contains many genes. A useful metaphor is that DNA constitutes a sequence of letters that spell out the genetic code. Genes are like meaningful words and sentences made of letters. Humans have about 20,000 genes, and each one influences development or day-to-day functions.
Location of Gene within DNA Strand
Genes are segments of DNA (the genetic code) contained within chromosomes. Chromosomes are located in the cell nucleus, which regulates cell activity.
The genetic code in an organism's cells is called the genotype. Humans and other living organisms carry two copies of each gene, although the copies may be different. A version of a gene is known as an allele. When one allele masks the expression of another, it is a dominant allele. The allele whose expression is masked is a recessive allele. When a gene has two identical alleles in the same location, it is homozygous. When a gene has two different alleles in the same location, it is heterozygous. An individual's phenotype involves all of the observable characteristics that result from genetic and environmental influences. For example, a person may have the genetic potential to grow very tall (genotype). However, if that person is malnourished in childhood, they may end up with below-average height (phenotype).
A heterozygous gene will express a phenotype related to the dominant allele. However, an organism with a heterozygous gene can pass the recessive allele to offspring. For example, the genotype of a brown-eyed individual may include two genes for the color brown. Alternatively, a brown-eyed individual may have one dominant gene for the color brown and one recessive gene for the color blue. If a brown-eyed person with the recessive gene for blue eyes has children with someone who also has a gene for blue eyes, their child can have blue eyes.
Heritability is the proportion of phenotypic variance attributable to genetic variance. For example, if individuals in a population differ in weight, these differences may be due to differences in their genetic makeup or their diet. The heritability of weight can be calculated as
\mathrm{Heritability}=\frac{\mathrm{Phenotypic}\;\mathrm{variability}\;\mathrm{due}\;\mathrm{to}\;\mathrm{differences}\;\mathrm{in}\;\mathrm{genes}}{\mathrm{Phenotypic}\;\mathrm{variability}\;\mathrm{due}\;\mathrm{to}\;\mathrm{differences}\;\mathrm{in}\;\mathrm{genes}\;\mathrm{and}\;\mathrm{environment}}
To calculate heritability, there must be genetic variability. If a gene occurs in every member of a population and is passed on to every offspring in that population, then the heritability of that gene is zero. Some traits are more heritable than others, in part because they are less strongly shaped by environmental influences. For example, heritability of height is about 0.90, of personality about 0.50, and of religious beliefs about 0.30–0.40. Most traits are shaped by multiple genes. There is no one gene that determines height or personality.
Genes that code for beneficial traits are likely to be passed on to future generations because they help an organism survive and reproduce.
Evolution is the change in relative frequency of genes in a population over time. There are four mechanisms of evolutionary change: mutation, genetic drift, gene flow, and natural selection. Mutation refers to errors that occur during DNA replication that can result in new heritable traits. In genetic drift, changes in gene frequency occur due to chance. For example in 1775 a typhoon reduced the population of the Micronesian island of Pingelap to only 20. One of the survivors had a genetic variation that causes complete color blindness. The prevalence of color blindness in Pingelap is now 5%. In most populations, it is less than 1%. Gene flow refers to organisms migrating to new geographic locations, thereby changing the frequency distribution in the old and new locations.
Natural selection is a mechanism of evolution in which individuals that are better adapted to their environment survive and reproduce more successfully than less well adapted individuals do. Natural selection was proposed by English naturalist Charles Darwin in his book On the Origin of Species, published in 1859. Variation exists in the traits of members of most species, some of which is heritable. If these heritable traits confer survival advantages, individuals with these traits will be healthier or survive longer. This allows them to produce more offspring than members of their species with less successful traits. Individual reproductive success is referred to as fitness, defined as the number of copies of one's genes passed to subsequent generations. This is the type of fitness referred to in the common phrase survival of the fittest. The phrase does not mean that the physically strongest survive. Instead it means that those best suited to their environment survive. The outcome of natural selection is adaptation to specific environments. An adaptation is a process or trait that improves an organism's ability to survive or reproduce in a given environment.
Structural adaptations are the heritable physical features of an organism. An example is sickle cell anemia, an inherited blood disorder in which the body makes an abnormal form of hemoglobin (a protein that transports oxygen in blood). The disease is caused by inheriting two defective copies of a particular gene. The gene has spread because inheriting only one defective copy offers protection against the mosquito-borne disease malaria. The highest proportion of the gene variant is found among people from regions where malaria is prevalent, such as West Africa, Southeast Asia, the Middle East, and China.
Behavioral adaptations are the things organisms do to survive in their physical and social environments. These adaptations can be inherited, learned, or a combination of both. For example, birds migrating south in the winter to find food is an inherited behavioral adaptation. The tendency of humans and other animals to cooperate with others is an adaptation that is inherited but also learned.
Genes and the environment interact to shape physical and psychological traits. Genetic factors can influence environmental responses, and environmental factors can alter gene expression.
Heredity and the environment work together to shape physical and psychological traits. Genetic factors can influence environmental responses. In turn, the environment can shape genetic activity.
Psychologists use twin and adoption studies to understand genetic and environmental influences. In these studies, identical and fraternal twins raised in the same family are compared to twins who were separated at a young age and raised in separate families. These studies show that genes contribute to many physical and psychological traits, including intelligence, personality, careers, hobbies, weight, political beliefs, religious beliefs, and mental health. Twin studies also reveal the importance of the environment. Identical twins have very similar gene expression early in life. As they age, twins raised in different environments develop more epigenetic differences. Twins raised separately differ more in psychological traits and health histories than those raised together.
Genetic factors influence risk for mental illness. For instance, depression occurs when levels of a neurotransmitter called serotonin drop too low. Variation exists in a serotonin transporter gene (5-HTTLPR) that influences risk of depression following stressful life events. People with a short variant of the gene are more vulnerable to depression than those with a long variant. Individuals with the short variant experience increased activity in the amygdala (part of the brain involved with emotions such as fear and aggression) following a stressful event and are more likely to dwell on negative experiences. Researchers estimate that genes account for 50% of depression risk, environmental events account for 10%, and intentional activities (e.g., rumination) account for 40%.
Environmental factors can also lead to changes in the expression of the genetic code. Epigenetics is the study of these heritable changes in the expression of genes that are caused by the environment. Many genes require specific environmental circumstances to be expressed and may therefore never be expressed.
Gene expression is regulated through the process of methylation. Each gene has a "tail" called a histone. The addition of a methyl (CH3) molecule to this tail deactivates the gene. Removing the tag allows the gene to become active. Environmental factors such as diet and stress levels can influence gene methylation. Epigenetic change can be passed down across generations because children can inherit methyl tags from parents.
For example, all mammals have a gene called agouti. Mice with methylated agouti genes are slim, healthy, and covered in brown fur. Mice with unmethylated agouti genes are obese, prone to diabetes and cancer, and covered in yellow fur. Pregnant yellow mice who are fed a methyl-rich diet give birth to pups with brown fur who remain healthy for life.
Epigenetic Change in Gene Expression
Genes can be turned off by adding a small molecule called a methyl group to DNA. Genes can be turned on by adding a different molecule, called an acetyl group, to a part of the DNA strand called the histone tail. These changes to genes can endure over time and even be passed on to offspring.
Epigenetics can affect psychological traits as well. Children who grow up in poverty have more methylation in an area of the brain that responds to fear stimuli (the amygdala) than children not raised in poverty. Chronic stress also leads to decreases in the size of telomeres, a protective cap at the end of chromosomes. Shorter telomeres predict more rapid aging and age-related health problems.
<Vocabulary>Neural Communication |
Gravity - Vixrapedia
Gravity is a basic concept of modern physics.
2 Relativistic gravity
3 Measurement of gravitational constant
4 Changing gravitational constant
Newtonian gravity belongs to the classical theory of gravity. Newtonian theory of gravity believes that there is mutual attraction between two masses. This kind of gravitation is also called universal gravitation. The interaction between gravity is expressed by the following formula.
{\displaystyle F=G{\frac {Mm}{r^{2}}}}
Relativistic gravity[edit]
Relativity theory of gravity is a theory that deals with gravity in a geometric way. Study the strength of gravitational interaction through the change of space-time structure. In general relativity, when a mass appears in space-time, it will cause the space-time to bend, thereby producing gravitational interaction.
Measurement of gravitational constant[edit]
The measurement of the gravitational constant is a process requiring very high accuracy. Need to combine the most advanced science and technology to complete. Due to the rapid development of science and technology, especially the development of materials science, the accuracy of the gravitational constant has been improved. Nevertheless, the accuracy of the gravitational constant is still very low compared to other physical constants, such as electronic charge and Planck's constant.
Changing gravitational constant[edit]
Because the gravitational constants measured by different laboratories vary greatly at different times, and the accuracy is not high, there are many theories that the gravitational constants may be constantly changing.
Retrieved from "https://www.vixrapedia.org/w/index.php?title=Gravity&oldid=4235" |
Two-ray propagation channel - MATLAB - MathWorks Switzerland
10\mu s
\begin{array}{l}\stackrel{\to }{R}={\stackrel{\to }{x}}_{s}-{\stackrel{\to }{x}}_{r}\\ {R}_{los}=|\stackrel{\to }{R}|=\sqrt{{\left({z}_{r}-{z}_{s}\right)}^{2}+{L}^{2}}\\ {R}_{1}=\frac{{z}_{r}}{{z}_{r}+{z}_{z}}\sqrt{{\left({z}_{r}+{z}_{s}\right)}^{2}+{L}^{2}}\\ {R}_{2}=\frac{{z}_{s}}{{z}_{s}+{z}_{r}}\sqrt{{\left({z}_{r}+{z}_{s}\right)}^{2}+{L}^{2}}\\ {R}_{rp}={R}_{1}+{R}_{2}=\sqrt{{\left({z}_{r}+{z}_{s}\right)}^{2}+{L}^{2}}\\ \mathrm{tan}{\theta }_{los}=\frac{\left({z}_{s}-{z}_{r}\right)}{L}\\ \mathrm{tan}{\theta }_{rp}=-\frac{\left({z}_{s}+{z}_{r}\right)}{L}\\ {{\theta }^{\prime }}_{los}=-{\theta }_{los}\\ {{\theta }^{\prime }}_{rp}={\theta }_{rp}\end{array}
{E}_{los}={E}_{0}\left(\frac{{R}_{0}}{{R}_{los}}\right){e}^{i\omega \left(t-{R}_{los}/c\right)}
{E}_{rp}={L}_{G}{E}_{0}\left(\frac{{R}_{0}}{{R}_{rp}}\right){e}^{i\omega \left(t-{R}_{rp}/c\right)}
\begin{array}{l}{G}_{p}=\frac{{Z}_{1}\mathrm{cos}{\theta }_{1}-{Z}_{2}\mathrm{cos}{\theta }_{2}}{{Z}_{1}\mathrm{cos}{\theta }_{1}+{Z}_{2}\mathrm{cos}{\theta }_{2}}=\frac{\mathrm{cos}{\theta }_{1}-\frac{{Z}_{2}}{{Z}_{1}}\mathrm{cos}{\theta }_{2}}{\mathrm{cos}{\theta }_{1}+\frac{{Z}_{2}}{{Z}_{1}}\mathrm{cos}{\theta }_{2}}\\ {G}_{s}=\frac{{Z}_{2}\mathrm{cos}{\theta }_{1}-{Z}_{1}\mathrm{cos}{\theta }_{2}}{{Z}_{2}\mathrm{cos}{\theta }_{1}+{Z}_{1}\mathrm{cos}{\theta }_{2}}=\frac{\mathrm{cos}{\theta }_{2}-\frac{{Z}_{2}}{{Z}_{1}}\mathrm{cos}{\theta }_{1}}{\mathrm{cos}{\theta }_{2}+\frac{{Z}_{2}}{{Z}_{1}}\mathrm{cos}{\theta }_{1}}\\ {Z}_{1}=\sqrt{\frac{{\mu }_{1}}{{\epsilon }_{1}}}\\ {Z}_{2}=\sqrt{\frac{{\mu }_{2}}{{\epsilon }_{2}}}\end{array}
\begin{array}{l}{G}_{p}=\frac{\sqrt{\rho }\mathrm{cos}{\theta }_{1}-\mathrm{cos}{\theta }_{2}}{\sqrt{\rho }\mathrm{cos}{\theta }_{1}+\mathrm{cos}{\theta }_{2}}\\ {G}_{s}=\frac{\sqrt{\rho }\mathrm{cos}{\theta }_{2}-\mathrm{cos}{\theta }_{1}}{\sqrt{\rho }\mathrm{cos}{\theta }_{2}+\mathrm{cos}{\theta }_{1}}\end{array}
\gamma ={\gamma }_{o}\left(f\right)+{\gamma }_{w}\left(f\right)=0.1820f{N}^{″}\left(f\right).
{N}^{″}\left(f\right)=\sum _{i}{S}_{i}{F}_{i}+{{N}^{″}}_{D}^{}\left(f\right)
{S}_{i}={a}_{1}×{10}^{-7}{\left(\frac{300}{T}\right)}^{3}\mathrm{exp}\left[{a}_{2}\left(1-\left(\frac{300}{T}\right)\right]P.
{S}_{i}={b}_{1}×{10}^{-1}{\left(\frac{300}{T}\right)}^{3.5}\mathrm{exp}\left[{b}_{2}\left(1-\left(\frac{300}{T}\right)\right]W.
W=\frac{\rho T}{216.7}.
{\gamma }_{c}={K}_{l}\left(f\right)M,
{\gamma }_{R}=k{R}^{\alpha },
r=\frac{1}{0.477{d}^{0.633}{R}_{0.01}^{0.073\alpha }{f}^{0.123}-10.579\left(1-\mathrm{exp}\left(-0.024d\right)\right)} |
\underset{2}{\overset{3}{\int }}{3}^{x} dx.
\mathrm{f}\left(x\right)=\left\{\begin{array}{ll}\frac{kx}{\left|x\right|},& \mathrm{if} x < 0\\ 3,& \mathrm{if} x \ge 0\end{array}\right\
\int \frac{\mathrm{d}x}{{x}^{2}+4x+8}
f\left(x\right)=4{x}^{3}-18{x}^{2}+27x-7
\mathrm{ℝ}
y={\mathrm{sin}}^{-1}\left(6x\sqrt{1-9{x}^{2}}\right), -\frac{1}{3\sqrt{2}}<x<\frac{1}{3\sqrt{2}}
\frac{dy}{dx}
\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=\stackrel{^}{\mathrm{i}}
\stackrel{\to }{\mathrm{c}}={\mathrm{c}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{c}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{c}}_{3}\stackrel{^}{\mathrm{k}},
\stackrel{\to }{\mathrm{a}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}
\stackrel{\to }{\mathrm{c}}
\stackrel{\to }{\mathrm{a}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}
\stackrel{\to }{\mathrm{c}}
\stackrel{\to }{a}, \stackrel{\to }{b}, \stackrel{\to }{c}
\stackrel{\to }{a}+ \stackrel{\to }{b}+ \stackrel{\to }{c}
\stackrel{\to }{a}, \stackrel{\to }{b} \mathrm{and} \stackrel{\to }{c}
\stackrel{\to }{a}+ \stackrel{\to }{b}+ \stackrel{\to }{c}
\stackrel{\to }{a} \mathrm{or} \stackrel{\to }{b} \mathrm{or} \stackrel{\to }{c}
\mathrm{\Sigma }{p}_{\mathit{i}}{x}_{i}^{2}=2\mathrm{\Sigma }{p}_{\mathit{i}}{x}_{\mathit{i}}
\left|\begin{array}{ccc}x& x+y& x+2y\\ x+2y& x& x+y\\ x+y& x+2y& x\end{array}\right|=9{y}^{2}\left(x+y\right).
A=\left(\begin{array}{cc}2& -1\\ 3& 4\end{array}\right), B=\left(\begin{array}{cc}5& 2\\ 7& 4\end{array}\right), C=\left(\begin{array}{cc}2& 5\\ 3& 8\end{array}\right)
{\left(\mathrm{sin} x\right)}^{x}+{\mathrm{sin}}^{-1}\sqrt{x}
{x}^{m}{y}^{n}={\left(x+y\right)}^{m+n}
\frac{{d}^{2}y}{d{x}^{2}}=0
\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{x \mathrm{sin} x}{1+{\mathrm{cos}}^{2}x}\mathrm{d}x
\underset{0}{\overset{3/2}{\int }}\left|x \mathrm{sin} \pi x\right|\mathrm{d}x
Solve the following L.P.P graphically:
Show that the family of curves for which
\frac{dy}{dx}=\frac{{x}^{2}+{y}^{2}}{2xy}
, is given by x2 – y2 = cx. VIEW SOLUTION
\int \frac{\left(3 \mathrm{sin} x-2\right) \mathrm{cos} x}{13-{\mathrm{cos}}^{2} x-7 \mathrm{sin} x}dx
\mathrm{cos} \left({\mathrm{tan}}^{-1} x\right)=\mathrm{sin} \left({\mathrm{cot}}^{-1}\frac{3}{4}\right)
\sqrt{3}\mathrm{y}=x
\underset{r}{\to }·\left(2\stackrel{\mathit{^}}{i}-3\stackrel{\mathit{^}}{j}+4\stackrel{\mathit{^}}{k}\right)=1
\underset{r}{\to }·\left(\stackrel{^}{i}-\stackrel{^}{j}\right)+4=0
\underset{r}{\to }·\left(2\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}\right)+8=0
\frac{x-8}{3}=\frac{\mathrm{y}+19}{-16}=\frac{\mathrm{z}-10}{7}
\frac{x-15}{3}=\frac{\mathrm{y}-29}{8}=\frac{\mathrm{z}-5}{-5}
{\mathrm{f}}^{-1}\left(y\right)\left(\frac{\sqrt{y+6}-1}{3}\right)
{\mathrm{f}}^{-1}\left(y\right)=\frac{4}{3},
\frac{\mathrm{\pi }}{3}.
\mathrm{A}=\left(\begin{array}{rrr}2& 3& 1\\ 1& 2& 2\\ –3& 1& -1\end{array}\right)
, find A–1 and hence solve the system of equations 2x + y – 3z = 13, 3x + 2y + z = 4, x + 2y – z = 8. VIEW SOLUTION
\mathrm{tan} x·\frac{dy}{dx}=2x \mathrm{tan} x+{x}^{2}-y;\left(\mathrm{tan} x\ne 0\right)
x=\frac{\pi }{2} |
Bit-Wise Operations - MATLAB & Simulink - MathWorks Nordic
Bit Masking with Logical Operators
Reading Consecutive Bits
Reading Nonconsecutive Bits
This topic shows how to use bit-wise operations in MATLAB® to manipulate the bits of numbers. Operating on bits is directly supported by most modern CPUs. In many cases, manipulating the bits of a number in this way is quicker than performing arithmetic operations like division or multiplication.
Any number can be represented with bits (also known as binary digits). The binary, or base 2, form of a number contains 1s and 0s to indicate which powers of 2 are present in the number. For example, the 8-bit binary form of 7 is
A collection of 8 bits is also called 1 byte. In binary representations, the bits are counted from the right to the left, so the first bit in this representation is a 1. This number represents 7 because
{2}^{2}+{2}^{1}+{2}^{0}=7.
When you type numbers into MATLAB, it assumes the numbers are double precision (a 64-bit binary representation). However, you can also specify single-precision numbers (32-bit binary representation) and integers (signed or unsigned, from 8 to 64 bits). For example, the most memory efficient way to store the number 7 is with an 8-bit unsigned integer:
a = uint8(7)
You can even specify the binary form directly using the prefix 0b followed by the binary digits (for more information, see Hexadecimal and Binary Values). MATLAB stores the number in an integer format with the fewest number of bits. Instead of specifying all the bits, you need to specify only the left-most 1 and all the digits to the right of it. The bits to the left of that bit are trivially zero. So the number 7 is:
MATLAB stores negative integers using two's complement. For example, consider the 8-bit signed integer -8. To find the two's complement bit pattern for this number:
Start with the bit pattern of the positive version of the number, 8: 00001000.
Next, flip all of the bits: 11110111.
Finally, add 1 to the result: 11111000.
The result, 11111000, is the bit pattern for -8:
n = 0b11111000s8
n = int8
MATLAB does not natively display the binary format of numbers. For that, you can use the dec2bin function, which returns a character vector of binary digits for positive integers. Again, this function returns only the digits that are not trivially zero.
dec2bin(b)
You can use bin2dec to switch between the two formats. For example, you can convert the binary digits 10110101 to decimal format with the commands
data = [1 0 1 1 0 1 0 1];
dec = bin2dec(num2str(data))
The cast and typecast functions are also useful to switch among different data types. These functions are similar, but they differ in how they treat the underlying storage of the number:
cast — Changes the underlying data type of a variable.
typecast — Converts data types without changing the underlying bits.
Because MATLAB does not display the digits of a binary number directly, you must pay attention to data types when you work with bit-wise operations. Some functions return binary digits as a character vector (dec2bin), some return the decimal number (bitand), and others return a vector of the bits themselves (bitget).
MATLAB has several functions that enable you to perform logical operations on the bits of two equal-length binary representations of numbers, known as bit masking:
bitand — If both digits are 1, then the resulting digit is also a 1. Otherwise, the resulting digit is 0.
bitor — If either digit is 1, then the resulting digit is also a 1. Otherwise, the resulting digit is 0.
bitxor — If the digits are different, then the resulting digit is a 1. Otherwise, the resulting digit is 0.
In addition to these functions, the bit-wise complement is available with bitcmp, but this is a unary operation that flips the bits in only one number at a time.
One use of bit masking is to query the status of a particular bit. For example, if you use a bit-wise AND operation with the binary number 00001000, you can query the status of the fourth bit. You can then shift that bit to the first position so that MATLAB returns a 0 or 1 (the next section describes bit shifting in more detail).
n = 0b10111001;
n4 = bitand(n,0b1000);
n4 = bitshift(n4,-3)
n4 = uint8
Bit-wise operations can have surprising applications. For example, consider the 8-bit binary representation of the number
\mathit{n}=8
8 is a power of 2, so its binary representation contains a single 1. Now consider the number
\left(\mathit{n}-1\right)=7
By subtracting 1, all of the bits starting at the right-most 1 are flipped. As a result, when
\mathit{n}
is a power of 2, corresponding digits of
\mathit{n}
\left(\mathit{n}-1\right)
are always different, and the bit-wise AND returns zero.
n = 0b1000;
bitand(n,n-1)
\mathit{n}
is not a power of 2, then the right-most 1 is for the
{2}^{0}
bit, so
\mathit{n}
\left(\mathit{n}-1\right)
have all the same bits except for the
{2}^{0}
bit. For this case, the bit-wise AND returns a nonzero number.
n = 0b101;
This operation suggests a simple function that operates on the bits of a given input number to check whether the number is a power of 2:
function tf = isPowerOfTwo(n)
tf = n && ~bitand(n,n-1);
The use of the short-circuit AND operator && checks to make sure that n is not zero. If it is, then the function does not need to calculate bitand(n,n-1) to know that the correct answer is false.
Because bit-wise logical operations compare corresponding bits in two numbers, it is useful to be able to move the bits around to change which bits are compared. You can use bitshift to perform this operation:
bitshift(A,N) shifts the bits of A to the left by N digits. This is equivalent to multiplying A by
{2}^{\mathit{N}}
bitshift(A,-N) shifts the bits of A to the right by N digits. This is equivalent to dividing A by
{2}^{\mathit{N}}
These operations are sometimes written A<<N (left shift) and A>>N (right shift), but MATLAB does not use << and >> operators for this purpose.
When the bits of a number are shifted, some bits fall off the end of the number, and 0s or 1s are introduced to fill in the newly created space. When you shift bits to the left, the bits are filled in on the right; when you shift bits to the right, the bits are filled in on the left.
For example, if you shift the bits of the number 8 (binary: 1000) to the right by one digit, you get 4 (binary: 100).
bitshift(n,-1)
Similarly, if you shift the number 15 (binary: 1111) to the left by two digits, you get 60 (binary: 111100).
bitshift(15,2)
When you shift the bits of a negative number, bitshift preserves the signed bit. For example, if you shift the signed integer -3 (binary: 11111101) to the right by 2 digits, you get -1 (binary: 11111111). In these cases, bitshift fills in on the left with 1s rather than 0s.
n = 0b11111101s8;
You can use the bitset function to change the bits in a number. For example, change the first bit of the number 8 to a 1 (which adds 1 to the number):
bitset(8,1)
By default, bitset flips bits to on or 1. You can optionally use the third input argument to specify the bit value.
bitset does not change multiple bits at once, so you need to use a for loop to change multiple bits. Therefore, the bits you change can be either consecutive or nonconsecutive. For example, change the first two bits of the binary number 1000:
bits = [1 2];
c = 0b1000;
for k = 1:numel(bits)
c = bitset(c,bits(k));
dec2bin(c)
Another common use of bitset is to convert a vector of binary digits into decimal format. For example, use a loop to set the individual bits of the integer 11001101.
dec = 0b0u8;
dec = bitset(dec,n+1-k,data(k));
dec = uint8
dec2bin(dec)
Another use of bit shifting is to isolate consecutive sections of bits. For example, read the last four bits in the 16-bit number 0110000010100000. Recall that the last four bits are on the left of the binary representation.
n = 0b0110000010100000;
dec2bin(bitshift(n,-12))
To isolate consecutive bits in the middle of the number, you can combine the use of bit shifting with logical masking. For example, to extract the 13th and 14th bits, you can shift the bits to the right by 12 and then mask the resulting four bits with 0011. Because the inputs to bitand must be the same integer data type, you can specify 0011 as an unsigned 16-bit integer with 0b11u16. Without the -u16 suffix, MATLAB stores the number as an unsigned 8-bit integer.
m = 0b11u16;
dec2bin(bitand(bitshift(n,-12),m))
Another way to read consecutive bits is with bitget, which reads specified bits from a number. You can use colon notation to specify several consecutive bits to read. For example, read the last 8 bits of n.
bitget(n,16:-1:8)
You can also use bitget to read bits from a number when the bits are not next to each other. For example, read the 5th, 8th, and 14th bits from n.
bits = [14 8 5];
bitget(n,bits)
bitand | bitor | bitxor | bitget | bitset | bitshift | bitcmp |
Hulls and kernels from topological dynamical systems and their applications to homeomorphism $C^{*}$ -algebras
April, 2004 Hulls and kernels from topological dynamical systems and their applications to homeomorphism $C^{*}$ -algebras
For a topological dynamical system
\Sigma
=\left(X,\sigma \right)
\sigma
is a homeomorphism in an arbitrary compact Hausdorff space
X
, we consider the noncommutative hulls and kernels with respect to the action
\sigma
in the associated
C\text{*}-
A\left(\Sigma \right)
. We show that several ideals important for the structure of
A\left(\Sigma \right)
have the form of such kernels and give topological characterizations of their hulls from the behavior of orbits in the dynamical system.
Jun TOMIYAMA. "Hulls and kernels from topological dynamical systems and their applications to homeomorphism $C^{*}$ -algebras." J. Math. Soc. Japan 56 (2) 349 - 364, April, 2004. https://doi.org/10.2969/jmsj/1191418634
Primary: 37A55 , 46L55 , 54H15
Keywords: crossed product , homeomorphism , noncommutative hull-kernel , periodic point , recurrent point
Jun TOMIYAMA "Hulls and kernels from topological dynamical systems and their applications to homeomorphism $C^{*}$ -algebras," Journal of the Mathematical Society of Japan, J. Math. Soc. Japan 56(2), 349-364, (April, 2004) |
Constrained equiripple FIR filter - MATLAB firceqrip - MathWorks í•œêµ
firceqrip
Filter design using firceqrip
Inverse-Dirichlet-Sinc-Shaped Passband
Constrained equiripple FIR filter
B = firceqrip(n,Fo,DEV)
B = firceqrip(...,'slope',r)
B = firceqrip('minorder',[Fp Fst],DEV)
B = firceqrip(...,'passedge')
B = firceqrip(...,'stopedge')
B = firceqrip(...,'high')
B = firceqrip(...,'min')
B = firceqrip(...,'invsinc',C)
B = firceqrip(...,'invdiric',C)
B = firceqrip(n,Fo,DEV) designs an order n filter (filter length equal n + 1) lowpass FIR filter with linear phase.
firceqrip produces the same equiripple lowpass filters that firpm produces using the Parks-McClellan algorithm. The difference is how you specify the filter characteristics for the function.
The input argument Fo specifies the frequency at the upper edge of the passband in normalized frequency (0<Fo<1). The two-element vector dev specifies the peak or maximum error allowed in the passband and stopbands. Enter [d1 d2] for dev where d1 sets the passband error and d2 sets the stopband error.
B = firceqrip(...,'slope',r) uses the input keyword 'slope' and input argument r to design a filter with a nonequiripple stopband. r is specified as a positive constant and determines the slope of the stopband attenuation in dB/normalized frequency. Greater values of r result in increased stopband attenuation in dB/normalized frequency.
B = firceqrip('minorder',[Fp Fst],DEV) designs filter with the minimum number of coefficients required to meet the deviations in DEV = [d1 d2] while having a transition width no greater than Fst – Fp, the difference between the stopband and passband edge frequencies. You can specify 'mineven' or 'minodd' instead of 'minorder' to design minimum even order (odd length) or minimum odd order (even length) filters, respectively. The 'minorder' option does not apply when you specify the 'min' (minimum-phase), 'invsinc', or the 'invdiric' options.
B = firceqrip(...,'passedge') designs a filter where Fo specifies the frequency at which the passband starts to rolloff.
B = firceqrip(...,'stopedge') designs a filter where Fo specifies the frequency at which the stopband begins.
B = firceqrip(...,'high') designs a high pass FIR filter instead of a lowpass filter.
B = firceqrip(...,'min') designs a minimum-phase filter.
B = firceqrip(...,'invsinc',C) designs a lowpass filter whose magnitude response has the shape of an inverse sinc function. This may be used to compensate for sinc-like responses in the frequency domain such as the effect of the zero-order hold in a D/A converter. The amount of compensation in the passband is controlled by C, which is specified as a scalar or two-element vector. The elements of C are specified as follows:
If C is supplied as a real-valued scalar or the first element of a two-element vector, firceqrip constructs a filter with a magnitude response of 1/sinc(C*pi*F) where F is the normalized frequency.
If C is supplied as a two-element vector, the inverse-sinc shaped magnitude response is raised to the positive power C(2). If we set P=C(2), firceqrip constructs a filter with a magnitude response 1/sinc(C*pi*F)P.
If this FIR filter is used with a cascaded integrator-comb (CIC) filter, setting C(2) equal to the number of stages compensates for the multiplicative effect of the successive sinc-like responses of the CIC filters.
Since the value of the inverse sinc function becomes unbounded at C=1/F, the value of C should be greater the reciprocal of the passband edge frequency. This can be expressed as Fo<1/C. For users familiar with CIC decimators, C is equal to 1/2 the product of the differential delay and decimation factor.
B = firceqrip(...,'invdiric',C) designs a lowpass filter with a passband that has the shape of an inverse Dirichlet sinc function. The frequency response of the inverse Dirichlet sinc function is given by
\left\{rC{\left(\frac{\mathrm{sin}\left(f/2r\right)}{\mathrm{sin}\left(Cf/2\right)}\right\}}^{p}
where C, r, and p are scalars. The input C can be a scalar or vector containing 2 or 3 elements. If C is a scalar, p and r equal 1. If C is a two-element vector, the first element is C and the second element is p, [C p]. If C is a three-element vector, the third element is r, [C p r].
To introduce a few of the variations on FIR filters that you design with firceqrip, these five examples cover both the default syntax b = firceqrip(n,wo,del) and some of the optional input arguments. For each example, the input arguments n, wo, and del remain the same.
Design a 30th order FIR filter using firceqrip.
b = firceqrip(30,0.4,[0.05 0.03]); fvtool(b)
Design a minimum order FIR filter using firceqrip. The passband edge and stopband edge frequencies are 0.35
\mathrm{Ï}
\mathrm{Ï}
rad/sample. The allowed deviations are 0.02 and 1e-4.
b = firceqrip('minorder',[0.35 0.45],[0.02 1e-4]); fvtool(b)
Design a 30th order FIR filter with the stopedge keyword to define the response at the edge of the filter stopband.
b = firceqrip(30,0.4,[0.05 0.03],'stopedge'); fvtool(b)
Design a 30th order FIR filter with the slope keyword and r = 20.
b = firceqrip(30,0.4,[0.05 0.03],'slope',20,'stopedge'); fvtool(b)
Design a 30th order FIR filter defining the stopband and specifying that the resulting filter is minimum phase with the min keyword.
b = firceqrip(30,0.4,[0.05 0.03],'stopedge','min'); fvtool(b)
Comparing this filter to the filter in Figure 1. The cutoff frequency wo = 0.4 now applies to the edge of the stopband rather than the point at which the frequency response magnitude is 0.5.
Viewing the zero-pole plot shown here reveals this is a minimum phase FIR filter - the zeros lie on or inside the unit circle, z = 1
fvtool(b,'polezero')
Design a 30th order FIR filter with the invsinc keyword to shape the filter passband with an inverse sinc function.
b = firceqrip(30,0.4,[0.05 0.03],'invsinc',[2 1.5]); fvtool(b)
The inverse sinc function being applied is defined as 1/sinc(2*w)^1.5.
Design two order 30 constrained equiripple FIR filters with inverse-Dirichlet-sinc-shaped passbands. The cutoff frequency in both designs is pi/4 radians/sample. Set C=1 in one design C=2 in the second design. The maximum passband and stopband ripple is 0.05. Set p=1 in one design and p=2 in the second design.
Design the filters.
b1 = firceqrip(30,0.25,[0.05 0.05],'invdiric',[1 1]);
Obtain the filter frequency responses using freqz. Plot the magnitude responses.
[h1,~] = freqz(b1,1);
plot(w,abs(h1)); hold on;
plot(w,abs(h2),'r');
axis([0 pi 0 1.5]);
xlabel('Radians/sample');
legend('C=1 p=1','C=2 p=2');
Inspect the stopband ripple in the design with C=1 and p=1. The constrained design sets the maximum ripple to be 0.05. Zoom in on the stopband from the cutoff frequency of pi/4 radians/sample to 3pi/4 radians/sample.
plot(w,abs(h1));
set(gca,'xlim',[pi/4 3*pi/4]);
diric | fdesign.decimator | firhalfband | firnyquist | firgr | ifir | iirgrpdelay | iirlpnorm | iirlpnormc | fircls | firls | firpm | sinc |
If A is a 3 × 3 invertible matrix, then what will be the value of k if det(A–1) = (det A)k. VIEW SOLUTION
Determine the value of the constant 'k' so that function
\mathrm{f}\left(x\right)=\left\{\begin{array}{ll}\frac{kx}{\left|x\right|},& \mathrm{if} x < 0\\ 3,& \mathrm{if} x \ge 0\end{array}\right\
is continuous at x = 0. VIEW SOLUTION
\underset{2}{\overset{3}{\int }}{3}^{x} dx.
If a line makes angles 90° and 60° respectively with the positive directions of x and y axes, find the angle which it makes with the positive direction of z-axis. VIEW SOLUTION
Show that all the diagonal elements of a skew symmetric matrix are zero. VIEW SOLUTION
\frac{dy}{dx} \mathrm{at} x=1, y=\frac{\pi }{4} \mathrm{if} {\mathrm{sin}}^{2}y+\mathrm{cos} xy = K
The volume of a sphere is increasing at the rate of 3 cubic centimeter per second. Find the rate of increase of its surface area, when the radius is 2 cm. VIEW SOLUTION
f\left(x\right)=4{x}^{3}-18{x}^{2}+27x-7
is always increasing on
\mathrm{ℝ}
Find the vector equation of the line passing through the point A(1, 2, –1) and parallel to the line 5x – 25 = 14 – 7y = 35z. VIEW SOLUTION
Prove that if E and F are independent events, then the events E and F' are also independent. VIEW SOLUTION
It is being given that at least one of each must be produced. VIEW SOLUTION
\int \frac{\mathrm{d}x}{{x}^{2}+4x+8}
\mathrm{tan} \left\{\frac{\pi }{4}+\frac{1}{2}{\mathrm{cos}}^{-1}\frac{a}{b}\right\}+\mathrm{tan}\left\{\frac{\pi }{4}-\frac{1}{2}{\mathrm{cos}}^{-1}\frac{a}{b}\right\}=\frac{2b}{a}
\left|\begin{array}{ccc}x& x+y& x+2y\\ x+2y& x& x+y\\ x+y& x+2y& x\end{array}\right|=9{y}^{2}\left(x+y\right).
A=\left(\begin{array}{cc}2& -1\\ 3& 4\end{array}\right), B=\left(\begin{array}{cc}5& 2\\ 7& 4\end{array}\right), C=\left(\begin{array}{cc}2& 5\\ 3& 8\end{array}\right)
, find a matrix D such that CD − AB = O. VIEW SOLUTION
Differentiate the function
{\left(\mathrm{sin} x\right)}^{x}+{\mathrm{sin}}^{-1}\sqrt{x}
{x}^{m}{y}^{n}={\left(x+y\right)}^{m+n}
\frac{{d}^{2}y}{d{x}^{2}}=0
\int \frac{2x}{\left({x}^{2}+1\right){\left({x}^{2}+2\right)}^{2}}dx
\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{x \mathrm{sin} x}{1+{\mathrm{cos}}^{2}x}\mathrm{d}x
\underset{0}{\overset{3/2}{\int }}\left|x \mathrm{sin} \pi x\right|\mathrm{d}x
Prove that x2 – y2 = c(x2 + y2)2 is the general solution of the differential equation (x3 – 3xy2)dx = (y3 – 3x2y)dy, where C is parameter. VIEW SOLUTION
\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=\stackrel{^}{\mathrm{i}}
\stackrel{\to }{\mathrm{c}}={\mathrm{c}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{c}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{c}}_{3}\stackrel{^}{\mathrm{k}},
(a) Let c1 = 1 and c2 = 2, find c3 which makes
\stackrel{\to }{\mathrm{a}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}
\stackrel{\to }{\mathrm{c}}
coplanar.
(b) If c2 = –1 and c3 = 1, show that no value of c1 can make
\stackrel{\to }{\mathrm{a}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}
\stackrel{\to }{\mathrm{c}}
coplanar. VIEW SOLUTION
\stackrel{\to }{a}, \stackrel{\to }{b}, \stackrel{\to }{c}
are mutually perpendicular vectors of equal magnitudes, show that the vector
\stackrel{\to }{a}+ \stackrel{\to }{b}+ \stackrel{\to }{c}
is equally inclined to
\stackrel{\to }{a}, \stackrel{\to }{b} \mathrm{and} \stackrel{\to }{c}
. Also, find the angle which
\stackrel{\to }{a}+ \stackrel{\to }{b}+ \stackrel{\to }{c}
makes with
\stackrel{\to }{a} \mathrm{or} \stackrel{\to }{b} \mathrm{or} \stackrel{\to }{c}
The random variable X can take only the values 0, 1, 2, 3. Give that P(X = 0) = P(X = 1) = p and P(X = 2) = P(X = 3) such that
\mathrm{\Sigma }{p}_{\mathit{i}}{x}_{i}^{2}=2\mathrm{\Sigma }{p}_{\mathit{i}}{x}_{\mathit{i}}
, find the value of p. VIEW SOLUTION
Often it is taken that a truthful person commands, more respect in the society. A man is known to speak the truth 4 out of 5 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.
Do you also agree that the value of truthfulness leads to more respect in the society? VIEW SOLUTION
Solve the following L.P.P. graphically:
Use product
\left[\begin{array}{ccc}1& -1& 2\\ 0& 2& -3\\ 3& -2& 4\end{array}\right]\left[\begin{array}{ccc}-2& 0& 1\\ 9& 2& -3\\ 6& 1& -2\end{array}\right]
to solve the system of equations x + 3z = 9, −x + 2y − 2z = 4, 2x − 3y + 4z = −3. VIEW SOLUTION
Consider f : R+ → [−5, ∞), given by f(x) = 9x2 + 6x − 5. Show that f is invertible with
{\mathrm{f}}^{-1}\left(y\right)\left(\frac{\sqrt{y+6}-1}{3}\right)
(i) f−1(10)
(ii) y if
{\mathrm{f}}^{-1}\left(y\right)=\frac{4}{3},
where R+ is the set of all non-negative real numbers.
Discuss the commutativity and associativity of binary operation '*' defined on A = Q − {1} by the rule a * b = a − b + ab for all, a, b ∊ A. Also find the identity element of * in A and hence find the invertible elements of A. VIEW SOLUTION
If the sum of lengths of the hypotenuse and a side of a right angled triangle is given, show that the area of the triangle is maximum, when the angle between them is
\frac{\mathrm{\pi }}{3}.
Using integration, find the area of region bounded by the triangle whose vertices are (–2, 1), (0, 4) and (2, 3).
Find the area bounded by the circle x2 + y2 = 16 and the line
\sqrt{3}\mathrm{y}=x
in the first quadrant, using integration. VIEW SOLUTION
x\frac{\mathrm{dy}}{\mathrm{d}x}+y=x \mathrm{cos} x + \mathrm{sin} x,
x=\frac{\mathrm{\pi }}{2}
Find the equation of the plane through the line of intersection of
\underset{r}{\to }·\left(2\stackrel{\mathit{^}}{i}-3\stackrel{\mathit{^}}{j}+4\stackrel{\mathit{^}}{k}\right)=1
\underset{r}{\to }·\left(\stackrel{^}{i}-\stackrel{^}{j}\right)+4=0
\underset{r}{\to }·\left(2\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}\right)+8=0
. Hence find whether the plane thus obtained contains the line x − 1 = 2y − 4 = 3z − 12.
Find the vector and Cartesian equations of a line passing through (1, 2, –4) and perpendicular to the two lines
\frac{x-8}{3}=\frac{\mathrm{y}+19}{-16}=\frac{\mathrm{z}-10}{7}
\frac{x-15}{3}=\frac{\mathrm{y}-29}{8}=\frac{\mathrm{z}-5}{-5} |
By: A. Alexander Beaujean
Keywords:correlation; IQ; motivation; partial correlation
A partial correlation is a measure of linear association between two variables after variability in at least one other variable is removed from both variables. The traditional formula for the partial correlation between, say, variables X and Y after controlling for Z, denoted rXY·Z, is
{r}_{X}{}_{Y\cdot Z}=\frac{{r}_{XY}-{r}_{XZ}×{r}_{YZ}}{\sqrt{1-{r}_{YZ}^{2}}×\sqrt{1-{r}_{YZ}^{2}}},
where r is the Pearson correlation between two variables.
rX(Y·Z) is sometimes referred to as a first-order partial correlation to note that the correlation only controls for one other variable. If controlling for two variables, it would be a second-order, and so on. Zero-order correlations do not control for any other variables.
While not obvious from Equation 1, the partial correlation is really the correlation between the residuals of X and Y after regressing both variables on Z. A path model representing this ...
Beaujean, A. (2018). Partial correlations. In B. Frey (Ed.), The SAGE encyclopedia of educational research, measurement, and evaluation (Vol. 1, pp. 1213-1214). SAGE Publications, Inc., https://dx.doi.org/10.4135/9781506326139.n504
Beaujean, A. Alexander. "Partial Correlations." In The SAGE Encyclopedia of Educational Research, Measurement, and Evaluation, edited by Frey, Bruce B., 1213-14. Thousand Oaks,, CA: SAGE Publications, Inc., 2018. https://dx.doi.org/10.4135/9781506326139.n504.
Beaujean, A. 2018. Partial Correlations. In: Bruce B. Frey Editor, 2018. The SAGE Encyclopedia of Educational Research, Measurement, and Evaluation, Thousand Oaks,, CA: SAGE Publications, Inc. pp. 1213-1214 Available at: <https://dx.doi.org/10.4135/9781506326139.n504> [Accessed 24 May 2022].
Beaujean, A. Alexander. "Partial Correlations." The SAGE Encyclopedia of Educational Research, Measurement, and Evaluation. Edited by Bruce B. Frey. Vol. 1. Thousand Oaks,: SAGE Publications, Inc., 2018, pp. 1213-14. SAGE Knowledge. 24 May 2022, doi: https://dx.doi.org/10.4135/9781506326139.n504. |
If a * b denotes the larger of 'a' and 'b' and if
a\circ b
= (a * b) + 3, then write the value of
\left(5\right)\circ \left(10\right)
, where * and
\circ
are binary operations. VIEW SOLUTION
Find the magnitude of each of two vectors
\stackrel{\to }{a}
\stackrel{\to }{b}
\frac{9}{2}
A=\left[\begin{array}{ccc}0& a& -3\\ 2& 0& -1\\ b& 1& 0\end{array}\right]
is skew symmetric, find the values of 'a' and 'b'. VIEW SOLUTION
{\mathrm{tan}}^{-1}\sqrt{3}-{\mathrm{cot}}^{-1}\left(-\sqrt{3}\right).
The total cost C(x) associated with the production of x units of an item is given by C(x) = 0.005x3 – 0.02x2 + 30x + 5000. Find the marginal cost when 3 units are produced, where by marginal cost we mean the instantaneous rate of change of total cost at any level of output. VIEW SOLUTION
{\mathrm{tan}}^{-1}\left(\frac{1+\mathrm{cos} x}{\mathrm{sin} x}\right)
with respect to x. VIEW SOLUTION
A=\left[\begin{array}{cc}2& -3\\ -4& 7\end{array}\right]
, compute A–1 and show that 2A–1 = 9I – A. VIEW SOLUTION
3 {\mathrm{sin}}^{-1}x={\mathrm{sin}}^{-1} \left(3x-4{x}^{3}\right), x\in \left[-\frac{1}{2}, \frac{1}{2}\right]
A black and a red die are rolled together. Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4. VIEW SOLUTION
\stackrel{^}{i}-2\stackrel{^}{j}+3\stackrel{^}{k} \mathrm{and} 3\stackrel{^}{i}-2\stackrel{^}{j}+\stackrel{^}{k}
, find sin θ. VIEW SOLUTION
Find the differential equation representing the family of curves y = aebx+5, where a and b are arbitrary constants. VIEW SOLUTION
\int \frac{\mathrm{cos} 2x+2{\mathrm{sin}}^{2} x}{{\mathrm{cos}}^{2} x}dx
If y = sin (sin x), prove that
\frac{{d}^{2}y}{d{x}^{2}}+\mathrm{tan} x \frac{dy}{dx}+y {\mathrm{cos}}^{2} x= 0.
Find the particular solution of the differential equation ex tan y dx + (2 – ex) sec2 y dy = 0, give that
y=\frac{\pi }{4}
when x = 0.
\frac{dy}{dx}+2y \mathrm{tan} x=\mathrm{sin} x,
given that y = 0 when
x=\frac{\pi }{3}
\stackrel{\to }{r}=\left(4\stackrel{^}{i}-\stackrel{^}{j}\right)+\lambda \left(\stackrel{^}{i}+2\stackrel{^}{j}-3\stackrel{^}{k}\right)
\stackrel{\to }{r}=\left(\stackrel{^}{i}-\stackrel{^}{j}+2\stackrel{^}{k}\right)+µ\left(2\stackrel{^}{i}+4\stackrel{^}{j}-5\stackrel{^}{k}\right).
Two numbers are selected at random (without replacement) from the first five positive integers. Let X denote the larger of the two numbers obtained. Find the mean and variance of X. VIEW SOLUTION
\left|\begin{array}{ccc}1& 1& 1+3x\\ 1+3y& 1& 1\\ 1& 1+3z& 1\end{array}\right|=9\left(3xyz+xy+yz+zx\right)
Find the equations of the tangent and the normal, to the curve 16x2 + 9y2 = 145 at the point (x1, y1), where x1 = 2 and y1 > 0.
Find the intervals in which the function
f\left(x\right)=\frac{{x}^{4}}{4}-{x}^{3}-5{x}^{2}+24x+12
is (a) strictly increasing, (b) strictly decreasing.
\int \frac{2 \mathrm{cos} x}{\left(1-\mathrm{sin} x\right) \left(1+{\mathrm{sin}}^{2} x\right)}dx
Suppose a girl throws a die. If she gets 1 or 2, she tosses a coin three times and notes the number of tails. If she gets 3, 4, 5 or 6, she tosses a coin once and notes whether a 'head' or 'tail' is obtained. If she obtained exactly one 'tail', what is the probability that she threw 3, 4, 5 or 6 with the die?
\stackrel{\to }{a}=4\stackrel{^}{i}+5\stackrel{^}{j}-\stackrel{^}{k}, \stackrel{\to }{b}=\stackrel{^}{i}-4\stackrel{^}{j}+5\stackrel{^}{k} \mathrm{and} \stackrel{\to }{c}=3\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}.
\stackrel{\to }{d}
\stackrel{\to }{c} \mathrm{and} \stackrel{\to }{b} \mathrm{and} \stackrel{\to }{d}·\stackrel{\to }{a}=21
An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of material will be least when depth of the tank is half of its width. If the cost is to be borne by nearby settled lower income families, for whom water will be provided, what kind of value is hidden in this question? VIEW SOLUTION
{\left({x}^{2}+{y}^{2}\right)}^{2}=xy, \mathrm{find} \frac{dy}{dx}
If x = a (2θ – sin 2θ) and y = a (1 – cos 2θ), find
\frac{dy}{dx}
\mathrm{\theta }=\frac{\pi }{3}
\underset{0}{\overset{\pi /4}{\int }}\frac{\mathrm{sin} x+\mathrm{cos} x}{16+9 \mathrm{sin} 2x}dx
\underset{1}{\overset{3}{\int }}\left({x}^{2}+3x+{e}^{x}\right)dx,
as the limit of the sum. VIEW SOLUTION
A factory manufactures two types of screws A and B, each type requiring the use of two machines, an automatic and a hand-operated. It takes 4 minutes on the automatic and 6 minutes on the hand-operated machines to manufacture a packet of screws 'A' while it takes 6 minutes on the automatic and 3 minutes on the hand-operated machine to manufacture a packet of screws 'B'. Each machine is available for at most 4 hours on any day. The manufacturer can sell a packet of screws 'A' at a profit of 70 paise and screws 'B' at a profit of Rs 1. Assuming that he can sell all the screws he manufactures, how many packets of each type should the factory owner produce in a day in order to maximize his profit? Formulate the above LPP and solve it graphically and find the maximum profit. VIEW SOLUTION
Let A = {x ∈ Z : 0 ≤ x ≤ 12}. Show that
R = {(a, b) : a, b ∈ A, |a – b| is divisible by 4}is an equivalence relation. Find the set of all elements related to 1. Also write the equivalence class [2].
Show that the function f : ℝ → ℝ defined by
f\left(x\right)=\frac{x}{{x}^{2}+1},\forall x\in \mathrm{ℝ}
is neither one-one nor onto. Also, if g : ℝ → ℝ is defined as g(x) = 2x – 1, find fog(x). VIEW SOLUTION
Using integration, find the area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x2 + y2 = 32. VIEW SOLUTION
A=\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right]
, find A–1. Use it to solve the system of equations
x + y – 2z = –3.
Using elementary row transformations, find the inverse of the matrix
A=\left[\begin{array}{ccc} 1& 2& 3\\ 2& 5& 7\\ -2& -4& -5\end{array}\right]
Find the distance of the point (–1, –5, –10) from the point of intersection of the line
\stackrel{\to }{r}=2\stackrel{^}{i}-\stackrel{^}{j}+2k+\lambda \left(3\stackrel{^}{i}+4\stackrel{^}{j}+2\stackrel{^}{k}\right)
\stackrel{\to }{r}·\left(\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}\right)=5 |
Antenna Toolbox Coordinate System - MATLAB & Simulink - MathWorks í•œêµ
Phi (Φ) and Theta (θ) Angles
Antenna Toolbox™ uses two types of coordinate system: rectangular coordinate system and spherical coordinate system .
Antenna Toolbox uses the rectangular coordinate system to visualize antenna or array geometry. The toolbox uses the spherical coordinate system to visualize antenna radiation patterns.
Visualize the geometry of a default monopoleTopHat antenna from the antenna library.
m = monopoleTopHat;
The toolbox displays the top-hat monopole antenna in the rectangular or Cartesian coordinate system.
The rectangular coordinate system also called Cartesian coordinate system specifies a position in space as an ordered 3-tuple of real numbers, (x,y,z), with respect to the origin (0,0,0).
You can view the 3-tuple as a point in space, or equivalently as a vector in three-dimensional Euclidean space. When viewed as a vector in space, the coordinate axes are basis vectors and the vector gives the direction to a point in space from the origin. Every vector in space is uniquely determined by a linear combination of the basis vectors. The most common set of basis vectors for three-dimensional Euclidean space are the standard unit basis vectors:
\left\{\left[1\text{â}\text{â}0\text{â}\text{â}0\right],\left[0\text{â}\text{â}1\text{â}\text{â}0\right],\left[0\text{â}\text{â}0\text{â}\text{â}1\right]\right\}
Any three linearly independent vectors define a basis for three-dimensional space. However, the Antenna Toolbox assumes that the basis vectors you use are orthogonal.
\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}
\sqrt{{\left({x}_{0}â{x}_{1}\right)}^{2}+{\left({y}_{0}â{y}_{1}\right)}^{2}+{\left({z}_{0}â{z}_{1}\right)}^{2}}
Visualize the radiation pattern of the default monopoleTopHat antenna.
pattern(m,75e6);
The toolbox displays the radiation pattern of the top-hat monopole using spherical coordinate system represented by azimuth and elevation angles.
The spherical coordinate system defines a vector or point in space with a distance R and two angles. You can represent the angles in this coordinate system:
The azimuth angle is the angle from the positive x-axis to the vector's orthogonal projection onto the xy plane, moving in the direction towards the y-axis. The azimuth angle is in the range –180 and 180 degrees.
The elevation angle is the angle from the vector's orthogonal projection on the xy plane toward the positive z-axis, to the vector. The elevation angle is in the –90 and 90 degrees.
The φ angle is the angle from the positive x-axis to the vector's orthogonal projection onto the xy plane, moving in the direction towards the y-axis. The azimuth angle is between –180 and 180 degrees.
The θ angle is the angle from the positive z-axis to the vector itself. The θ angle is in the range 0 degrees and 180 degrees.
These angles are an alternative to using azimuth and elevation angles to express the location of point in a unit sphere.
You can define u and v in terms of φ and θ:
\begin{array}{l}u=\mathrm{sin}\mathrm{θ}\mathrm{cos}\mathrm{Ï}\\ v=\mathrm{sin}\mathrm{θ}\mathrm{sin}\mathrm{Ï}\end{array}
In terms of azimuth and elevation angles, the u and v coordinates are:
\begin{array}{l}u=\mathrm{cos}el\mathrm{sin}az\\ v=\mathrm{sin}el\end{array}
The values of u and v satisfy the inequalities:
\begin{array}{l}â1â¤uâ¤1\\ â1â¤vâ¤1\\ {u}^{2}+{v}^{2}â¤1\end{array}
The φ and θ angles in terms of u and v are:
\begin{array}{l}\mathrm{tan}\mathrm{Ï}=u/v\\ \mathrm{sin}\mathrm{θ}=\sqrt{{u}^{2}+{v}^{2}}\end{array}
The azimuth and elevation angles in terms of u and v are:
\begin{array}{l}\mathrm{sin}el=v\\ \mathrm{tan}az=\frac{u}{\sqrt{1â{u}^{2}â{v}^{2}}}\end{array}
Convert rectangular coordinates to spherical coordinates (az, el, R) using:
\begin{array}{l}R=\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}\\ az={\mathrm{tan}}^{â1}\left(\frac{y}{x}\right)\\ el=\mathrm{tan}{}^{â1}\left(\frac{z}{\sqrt{{x}^{2}+{y}^{2}}}\right)\end{array}
Convert spherical coordinates (az, el, R) to rectangular coordinates using:
\begin{array}{l}x=R\mathrm{cos}\left(el\right)\mathrm{cos}\left(az\right)\\ y=R\mathrm{cos}\left(el\right)\mathrm{sin}\left(az\right)\\ z=R\mathrm{sin}\left(el\right)\end{array}
R is the distance from the antenna
el and az are the azimuth and elevation angles |
General Instructions: i. All questions are compulsory. ii. The question paper consists of 29 questions divided into three sections A, B and C. Section A comprises of 10 questions of one mark each, Section B comprises of 12 questions of four marks each, and Section C comprises of 7 questions of six marks each. iii. All questions in section A are to be answered in one word, one sentence or as per the exact requirements of the question. iv. There is no overall choice. However, internal choice has been provided in 4 questions of four marks each and 2 questions of six marks each. You have to attempt only one of the alternatives in all such questions. v. Use of calculators is not permitted. You may ask for logarithmic tables, if required.
Let R = {(a, a3) : a is a prime number less than 5} be a relation. Find the range of R. VIEW SOLUTION
{\mathrm{cos}}^{-1}\left(-\frac{1}{2}\right) + 2 {\mathrm{sin}}^{-1} \left(\frac{1}{2}\right)
Use elementary column operations
{\mathrm{C}}_{2} \to {\mathrm{C}}_{2} - 2{\mathrm{C}}_{1}
in the matrix equation
\left(\begin{array}{cc}4& 2\\ 3& 3\end{array}\right) = \left(\begin{array}{cc}1& 2\\ 0& 3\end{array}\right)\left(\begin{array}{cc}2& 0\\ 1& 1\end{array}\right)
\left(\begin{array}{cc}\mathrm{a} + 4& 3\mathrm{b}\\ 8& -6\end{array}\right) = \left(\begin{array}{cc}2\mathrm{a} + 2& \mathrm{b} + 2\\ 8& \mathrm{a} - 8\mathrm{b}\end{array}\right),
write the value of a − 2b. VIEW SOLUTION
If A is a 3 × 3 matrix,
\left|\mathrm{A}\right|\ne 0 \mathrm{and} \left|3\mathrm{A}\right|=\mathrm{k}\left|\mathrm{A}\right|
, then write the value of k. VIEW SOLUTION
\int \frac{\mathrm{dx}}{{\mathrm{sin}}^{2} \mathrm{x} {\mathrm{cos}}^{2} \mathrm{x}}
\underset{0}{\overset{\frac{\pi }{4}}{\int }}\mathrm{tan} \mathrm{x} \mathrm{dx}
Write the projection of vector
\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}
\stackrel{^}{\mathrm{j}}
2 \stackrel{^}{\mathrm{i}}-3 \stackrel{^}{\mathrm{j}}+6 \stackrel{^}{\mathrm{k}}
which has magnitude 21 units. VIEW SOLUTION
\stackrel{\to }{r}=2\stackrel{^}{i}-5\stackrel{^}{j}+\stackrel{^}{k}+\lambda \left(3\stackrel{^}{i}+2\stackrel{^}{j}+6\stackrel{^}{k}\right) \mathrm{and} \stackrel{\to }{r}=7\stackrel{^}{i}-6\stackrel{^}{k}+\mu \left(\stackrel{^}{i}+2\stackrel{^}{j}+2\stackrel{^}{k}\right)
Let f : W → W be defined as f(x) = x − 1 if x is odd and f(x) = x + 1 if x is even. Show that f is invertible. Find the inverse of f, where W is the set of all whole numbers. VIEW SOLUTION
\mathrm{cos} \left({\mathrm{tan}}^{-1}x\right)=\mathrm{sin} \left({\mathrm{cot}}^{-1}\frac{3}{4}\right)
cot−1 7 + cot−1 8 + cot−1 18 = cot−1 3 VIEW SOLUTION
\left|\begin{array}{ccc}\mathrm{a}+\mathrm{x}& \mathrm{y}& \mathrm{z}\\ \mathrm{x}& \mathrm{a}+\mathrm{y}& \mathrm{z}\\ \mathrm{x}& \mathrm{y}& \mathrm{a}+\mathrm{z}\end{array}\right|={\mathrm{a}}^{2}\left(\mathrm{a}+\mathrm{x}+\mathrm{y}+\mathrm{z}\right)
If x = a cos θ + b sin θ and y = a sin θ − b cos θ, show that
{\mathrm{y}}^{2}\frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}-\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=0.
If xmyn = (x + y)m+n, prove that
\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}}{\mathrm{x}}.
Find the approximate value of f(3.02), up to 2 places of decimal, where f(x) = 3x2 + 5x + 3.
\mathrm{f}\left(\mathrm{x}\right)=\frac{3}{2}{\mathrm{x}}^{4}-4{\mathrm{x}}^{3}-45{\mathrm{x}}^{2}+51
(b) strictly decreasing VIEW SOLUTION
\int \frac{\mathrm{x} {\mathrm{cos}}^{-1} \mathrm{x}}{\sqrt{1-{\mathrm{x}}^{2}}}\mathrm{dx}
\int \left(3\mathrm{x}-2\right) \sqrt{{\mathrm{x}}^{2}+\mathrm{x}+1 }\mathrm{dx}
Solve the differential equation (x2 − yx2) dy + (y2 + x2y2) dx = 0, given that y = 1, when x = 1. VIEW SOLUTION
\frac{\mathrm{dy}}{\mathrm{dx}}
+ y cot x = 2 cos x, given that y = 0 when x =
\frac{\mathrm{\pi }}{2}
\stackrel{\to }{\mathrm{a},} \stackrel{\to }{\mathrm{b},} \stackrel{\to }{\mathrm{c} }
are coplanar if and only if
\stackrel{\to }{\mathrm{a} }+ \stackrel{\to }{\mathrm{b} }
\stackrel{\to }{\mathrm{b} }+ \stackrel{\to }{\mathrm{c} }
\stackrel{\to }{\mathrm{c} }+ \stackrel{\to }{\mathrm{a} }
Find a unit vector perpendicular to both the vectors
\stackrel{\to }{\mathrm{a} }+ \stackrel{\to }{\mathrm{b} }\mathrm{and} \stackrel{\to }{\mathrm{a} }- \stackrel{\to }{\mathrm{b} }
\stackrel{\to }{\mathrm{a} }= \stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}, \stackrel{\to }{\mathrm{b} }= \stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}
\stackrel{\to }{\mathrm{r}}=\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\mathrm{\lambda }\left(2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)
\stackrel{\to }{\mathrm{r}}=2\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}+\mathrm{\mu }\left(3\stackrel{^}{\mathrm{i}}-5\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right).
Three cards are drawn at random (without replacement) from a well-shuffled pack of 52 playing cards. Find the probability distribution of the number of red cards. Hence, find the mean of the distribution. VIEW SOLUTION
Two schools P and Q want to award their selected students on the values of tolerance, kindness and leadership. School P wants to award Rs x each, Rs y each and Rs z each for the three respective values to 3, 2 and 1 students, respectively, with a total award money of Rs 2,200. School Q wants to spend Rs 3,100 to award 4, 1 and 3 students on the respective values (by giving the same award money to the three values as school P). If the total amount of award for one prize on each value is Rs 1,200, using matrices, find the award money for each value. VIEW SOLUTION
Show that a cylinder of a given volume, which is open at the top, has minimum total surface area when its height is equal to the radius of its base. VIEW SOLUTION
\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{\mathrm{x} \mathrm{tan} \mathrm{x}}{\mathrm{sec} \mathrm{x} + \mathrm{tan} \mathrm{x}}\mathrm{dx}
Find the area of the smaller region bounded by the ellipse
\frac{{x}^{2}}{9}+\frac{{y}^{2}}{4}=1
\frac{x}{3}+\frac{y}{2}=1.
Find the equation of the plane that contains the point (1, –1, 2) and is perpendicular to both the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8. Hence, find the distance of point P (–2, 5, 5) from the plane obtained
Find the distance of the point P (–1, –5, –10) from the point of intersection of the line joining the points A (2, –1, 2) and B (5, 3, 4) with the plane x – y + z = 5. VIEW SOLUTION
A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes 2 hours on the grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 20 hours and the grinding/cutting machine for at most 12 hours. The profit from the sale of a lamp is Rs 25 and that from a shade is Rs 15. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit? Formulate an LPP and solve it graphically. VIEW SOLUTION
An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probabilities of an accident for them are 0.01, 0.03 and 0.15, respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver or a car driver?
Five cards are drawn one by one, with replacement, from a well-shuffled deck of 52 cards. Find the probability that
(i) all the five cards diamonds
(ii) only 3 cards are diamonds
(iii) none is a diamond VIEW SOLUTION |
GIVbacks | Giveth Docs
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Stationary preserving ideals over Pκλ
July, 2003 Stationary preserving ideals over
{\mathcal{P}}_{\kappa }\lambda
Yo MATSUBARA
We investigate the properties of ideals such that their corresponding partial orders preserve stationarity. We show that these ideals exhibit many large cardinal-like consequences. We also prove the existence of a certain non-reflecting stationary subset of
{\mathcal{P}}_{\kappa }\lambda
under some hypotheses.
Yo MATSUBARA. "Stationary preserving ideals over
{\mathcal{P}}_{\kappa }\lambda
." J. Math. Soc. Japan 55 (3) 827 - 835, July, 2003. https://doi.org/10.2969/jmsj/1191419004
Keywords: Forcing , large cardinals
Yo MATSUBARA "Stationary preserving ideals over
{\mathcal{P}}_{\kappa }\lambda
," Journal of the Mathematical Society of Japan, J. Math. Soc. Japan 55(3), 827-835, (July, 2003) |
01 November 05 Generic Singular Spectrum For Ergodic Schrödinger Operators
Artur Avila, David Damanik
Artur Avila,1 David Damanik2
1Laboratoire de Probabilités et Modèles aléatoires, Université Pierre et Marie Curie–Boîte courier 188
2Department of Mathematics, Mathematics 253-37, California Institute of Technology
Duke Math. J. 130(2): 393-400 (01 November 05). DOI: 10.1215/S0012-7094-05-13035-6
We consider Schrödinger operators with ergodic potential
{V}_{\omega }\left(n\right)=f\left({T}^{n}\left(\omega \right)\right)
n\in ℤ
\omega \in \Omega
T:\Omega \to \Omega
is a nonperiodic homeomorphism. We show that for generic
f\in C\left(\Omega \right)
, the spectrum has no absolutely continuous component. The proof is based on approximation by discontinuous potentials which can be treated via Kotani theory
Artur Avila. David Damanik. "Generic Singular Spectrum For Ergodic Schrödinger Operators." Duke Math. J. 130 (2) 393 - 400, 01 November 05. https://doi.org/10.1215/S0012-7094-05-13035-6
Artur Avila, David Damanik "Generic Singular Spectrum For Ergodic Schrödinger Operators," Duke Mathematical Journal, Duke Math. J. 130(2), 393-400, (01 November 05) |
From S. E. (Elizabeth) Wedgwood to Hensleigh Wedgwood [16] November [1836]1
My dear Hensleigh
Charles’s visit has been exceedingly pleasant. Caroline went in the phaeton to meet him on Saturday, and as he did not come by the mail she waited till several more coaches came in & then came back—and we quite gave him, when in a quarter of an hour after her, he appeared. He looks very much thinner, but it has improved his looks, and his countenance is so pleasant that his plainness does not signify— his manners are uncommonly pleasant it is impossible not to be quite fond of him— it is very pleasant to see him so ready to enjoy himself. We had to distribute proper shares of his company to everybody. Eliza came here on Saturday to stay over Sunday & Harry came both those days & went home in the evening—& on Monday again he appeared directly after breakfast but diverged from Madeley (where they all walked to see my father’s Yew-tree, that he has never been able to make anybody curious about before, but now Charles is going actually to do it the honour to describe it to Mr Henslow) & we saw him no more. Frank & Fanny came on Sunday, & we had Aunt Sarah—
We were particularly anxious that Aunt Sarah’s day should go off well, and by great good luck he got upon just the subjects that would suit her—& was remarkably pleasant all the evening—(we had begun to be low at dinner for he was shy, & we could not get on) and the next morning after breakfast we fell again into such a pleasant fit of talk that she actually ran to the door when the phaeton was there, and ordered it away again for an hour! Charles gave us Capt Fitzroy’s letter to read, which Sarah was very much pleased with—indeed we are all duly impressed with a proper profundity of admiration for the Captain— Harry enjoyed his cross questioning of Charles thoroughly— We got several little geological lectures— Harry was rather anxious to know whether we are going up or going down— I learn that the learned world are coming fast round to Mr Lyell, only that earth-quakes were bigger in those days— It is very lucky that Charles’s bones are turning out so valuable—it was so great a chance, for of course before he went he could not have had much opportunity of knowing what was already known. I perceive that he rather skips over his residence at Cambridge and looks forward to settling in London with the most pleasure— It will be very pleasing for his sisters if he does— There is a certain German pamphlet on the corals of the red sea,2 that he amused us with his account of Eras translating for him, & how pale he looked after his labours— I offered to finish translating it for him but he refused. I dont know whether from not caring about it or not to give me the trouble— If Eras thinks he would really like to have it & that I could do it (though I dont know how he is to know that—but I have read a great deal of Kozebue)3 ask him to send it me—it is a job I should like. Charles feels he must learn German, but I dont see how he can get time— we gave him a vocab—and looked for Robinson der Junger—4 but we could not find it— has Fanny taken it for Snow?
They were to have gone yesterday morning—but by great good ⟨lu⟩ck Caroline wrote for a chaise by post to Tern-hill & the chaise did not arrive here till two o’clock—and we persuad⟨ed⟩ them it was quite too late to go that day, and would only disturb the Dr when once he had given them up— so instead we set out for a good walk with Caroline on the pony & had the pleasure to see the empty chaise rattle away past us back again— They are gone in good earnest today—& Emma to Betley court—5 I have luckily had Ettore Fieramosca6 to console me, and I have done no duty & not gone out, but sat in an arm-chair by the fire reading it, till I heroically tore myself from it to write you this letter—but it was in a dull part— It is a pleasing & interesting book—on the model of P. Sposi7 & Walter Scott—but not so charming as that—but shewing as that does good & nice feelings in the author—everywhere but about some asses that he throws head over heels in a cruel manner. I was very glad indeed to get Caroline to stay here for Charles— she enjoyed his visit so much— I have not seen her a long time in such good spirits—only the longer she stay⟨s⟩ the more sorry she is to part with him. Charles told us his boat story,8 & very striking it was—from his own feeling about it—and a dreadful story of Miss Martineau’s about two young women, that made Sarah look quite as if she was suffering acutely for some time after— it was an inconceivable thing to happen amongst civilized Christians— In the last Emancipa⟨tor⟩9 there is the most barefaced thing one ever heard of yet—a pamphlet advertised with a full account of the trial & execution of 10 men, 5 black & 5 white by Lynch law! & it appeared as if the negroes were examined by flogging before they were hanged— The names date and all the particulars are given⟨.⟩
Clem Broughton10 has offered Charles the curacy of Norbury—but I have no doubt Frank has employed some of his leis⟨ure⟩ which is likely to last for the next 3 months in writing to you & will have told you all about it. Robert is delighted with it, and Eliza has taken hold of the occasion to write to Mrs R. a friendly letter which is a very kind thing of her. Jessie is going on quite well. We are expecting Uncle Allen & Emma tomorrow for a fortnight & Mrs Holland & Louisa ought to arrive tomorrow too, as they have never written to tell us they have altered the day but I dont expect them. Charlotte has been giving her first dinner party & suffering most severely—but I have no room to give you Charles’s account of her troubles
Dont you think one of our large family of lamps would do for Fanny? & then perhaps she would let you get your books.
We have been keeping an eye on the sky but all in vain— I was up from
\frac{1}{2}
past 2 to 4 on the night—but nothing was to be seen—not that I got up for that but my mother was not comfortable & was sitting up in bed.
CD arrived from London on 12 November, left on the 16th (see Correspondence vol. 1, Appendix I).
Ehrenberg 1834, an important source for CD’s Coral reefs. A heavily annotated copy is in ‘Philosophical tracts’, Darwin Library–CUL.
Otto von Kotzebue 1821, 1830.
[J. C. L.] H[aken] 1802, Robinsons des Jüngern, an elementary German reader. Two German dictionaries are in the Darwin Library–Down: Rabenhorst’s pocket dictionary (1829), inscribed ‘Erasmus Darwin’, and volume one of a two-volume set edited by Johann Gottfried Flügel, Leipzig (1838).
The home of the Tollet family.
Tapparelli 1833.
Manzoni 1827.
CD may have told the story of a narrow escape in Tierra del Fuego, when an exploring party from the Beagle was nearly marooned. Robert FitzRoy, in Narrative 2: 217, states ‘had not Mr. Darwin, and two or three of the men, run to them [the boats] instantly, they would have been swept away from us irrecoverably.’
Abolitionist weeklies of that name were published in Boston and New York.
Clement Francis Broughton. It is not clear to whom the offer was made. Possibly Charles Langton is meant, but he was rector of Onibury at the time, and the next sentences suggest that Elizabeth intended to write ‘Robert’. In any case, the offer was not accepted. In the next Clergy list (London 1841) no curate is listed for Norbury.
Ehrenberg, Christian Gottfried. 1834. Über die Natur und Bildung der Coralleninseln und Corallenbänke im rothen Meere. Berlin.
Kotzebue, Otto von. 1821. Entdeckungs-Reise in die SüdSee und nach der Berings- Strasse zur Erforschung einer nordöstlichen Durchfahrt. Unternommen in den Jahren 1815 … 1818. 3 vols. Weimar.
Manzoni, Alessandro F. T. A. 1827. I promessi sposi. Storia milanese del secolo XVII. 3 vols. Milan.
Tapparelli, Massimo (Marchese d’Azeglio). 1833. Ettore Fieramosca, o la disfida di Barletta. Florence.
Describes CD’s visit to Maer on his return from Beagle voyage.
Sarah Elizabeth (Elizabeth) Wedgwood
inc AL incomplete? |
The essence of Dirac's problem was simply that:
\sqrt{p^2c^2 + m_0^2c^4} \neq pc + m_0c^2
and, in general:
\sqrt{a^2 + b^2} \neq a + b
(for non-zero
and
b
), as everyone learns (or should learn) at school.
At which point mere mortals give up. Geniuses try harder.
(Norman McCubbin, "Beauty in physics: the legacy of Paul Dirac" , in Contemporary Physics, Volume 45, Number 4, July-August 2004, pp. 319–333) |
class torch.nn.Upsample(size=None, scale_factor=None, mode='nearest', align_corners=None, recompute_scale_factor=None)[source]¶
align_corners (bool, optional) – if True, the corner pixels of the input and output tensors are aligned, and thus preserving the values at those pixels. This only has effect when mode is 'linear', 'bilinear', 'bicubic', or 'trilinear'. Default: False
recompute_scale_factor (bool, optional) – recompute the scale_factor for use in the interpolation calculation. If recompute_scale_factor is True, then scale_factor must be passed in and scale_factor is used to compute the output size . The computed output size will be used to infer new scales for the interpolation. Note that when scale_factor is floating-point, it may differ from the recomputed scale_factor due to rounding and precision issues. If recompute_scale_factor is False, then size or scale_factor will be used directly for interpolation.
(N, C, W_{in})
(N, C, H_{in}, W_{in})
(N, C, D_{in}, H_{in}, W_{in})
(N, C, W_{out})
(N, C, H_{out}, W_{out})
(N, C, D_{out}, H_{out}, W_{out})
D_{out} = \left\lfloor D_{in} \times \text{scale\_factor} \right\rfloor
H_{out} = \left\lfloor H_{in} \times \text{scale\_factor} \right\rfloor
W_{out} = \left\lfloor W_{in} \times \text{scale\_factor} \right\rfloor |
\left[\begin{array}{ccc}0& 2\mathrm{b}& -2\\ 3& 1& 3\\ 3\mathrm{a}& 3& -1\end{array}\right]
is given to be symmetric, find values of a and b. VIEW SOLUTION
Find the position vector of a point which divides the join of points with position vectors
\stackrel{\to }{a}-2\stackrel{\to }{b} \mathrm{and} 2\stackrel{\to }{a} + \stackrel{\to }{b}
externally in the ratio 2 : 1. VIEW SOLUTION
\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}} \mathrm{and} 3\stackrel{^}{\mathrm{i}}-\mathrm{j}+4\stackrel{^}{\mathrm{k}}
represent the two sides AB and AC, respectively of a ∆ABC. Find the length of the median through A. VIEW SOLUTION
Find the vector equation of a plane which is at a distance of 5 units from the origin and its normal vector is
2\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}
\left|\begin{array}{ccc}1& 1& 1\\ 1& 1+\mathrm{sin} \mathrm{\theta }& 1\\ 1& 1& 1+\mathrm{cos} \mathrm{\theta } \end{array}\right|
If A is a square matrix such that A2 = I, then find the simplified value of (A – I)3 + (A + I)3 – 7A. VIEW SOLUTION
Show that the equation of normal at any point t on the curve x = 3 cos t – cos3t and y = 3 sin t – sin3t is
4 (y cos3t – sin3t) = 3 sin 4t. VIEW SOLUTION
\int \frac{\left(3 \mathrm{sin} \mathrm{\theta }-2\right)\mathrm{cos} \mathrm{\theta }}{5-{\mathrm{cos}}^{2} \mathrm{\theta }-4 \mathrm{sin} \mathrm{\theta }}\mathrm{d\theta }
{\int }_{0}^{\mathrm{\pi }}{e}^{2x}· \mathrm{sin}\left(\frac{\mathrm{\pi }}{4}+x\right) dx
\int \frac{\sqrt{x}}{\sqrt{{a}^{3}-{x}^{3}}}dx
\underset{-1}{\overset{2}{\int }}\left|{x}^{3}-x\right| dx.
(1 – y2) (1 + log x) dx + 2xy dy = 0, given that y = 0 when x = 1. VIEW SOLUTION
Find the general solution of the following differential equation :
\left(1+{y}^{2}\right)+\left(x-{e}^{{\mathrm{tan}}^{-1}y}\right) \frac{dy}{dx}=0
\stackrel{\to }{a}, \stackrel{\to }{b} \mathrm{and} \stackrel{\to }{c}
\stackrel{\to }{a}+\stackrel{\to }{b}, \stackrel{\to }{b}+\stackrel{\to }{c} \mathrm{and} \stackrel{\to }{c}+\stackrel{\to }{a}
Find the vector and Cartesian equations of the line through the point (1, 2, −4) and perpendicular to the two lines.
\stackrel{\to }{\mathrm{r}}=\left(8\stackrel{^}{\mathrm{i}}-19\stackrel{^}{\mathrm{j}}+10\stackrel{^}{\mathrm{k}}\right)+\lambda \left(3\stackrel{^}{\mathrm{i}}-16\stackrel{^}{\mathrm{j}}+7\stackrel{^}{\mathrm{k}}\right)
\stackrel{\to }{\mathrm{r}}=\left(15\stackrel{^}{\mathrm{i}}+29\stackrel{^}{\mathrm{j}}+5\stackrel{^}{\mathrm{k}}\right)+\mathrm{\mu }\left(3\stackrel{^}{\mathrm{i}}+8\stackrel{^}{\mathrm{j}}-5\stackrel{^}{\mathrm{k}}\right)
A and B throw a pair of dice alternately. A wins the game if he gets a total of 7 and B wins the game if he gets a total of 10. If A starts the game, then find the probability that B wins. VIEW SOLUTION
{\mathrm{tan}}^{-1}\frac{1}{5}+{\mathrm{tan}}^{-1}\frac{1}{7}+{\mathrm{tan}}^{-1}\frac{1}{3}+{\mathrm{tan}}^{-1}\frac{1}{8}=\frac{\mathrm{\pi }}{4}
2 {\mathrm{tan}}^{-1}\left(\mathrm{cos} x\right)={\mathrm{tan}}^{-1}\left(2\mathrm{cosec} x\right)
The monthly incomes of Aryan and Babban are in the ratio 3 : 4 and their monthly expenditures are in the ratio 5 : 7. If each saves Rs 15,000 per month, find their monthly incomes using matrix method. This problem reflects which value? VIEW SOLUTION
If x = a sin 2t (1 + cos 2t) and y = b cos 2t (1 – cos 2t), find the values of
\frac{dy}{dx}
\mathrm{t}=\frac{\mathrm{\pi }}{4}
\mathrm{t}=\frac{\mathrm{\pi }}{3}.
\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^{2}}-\frac{1}{y}{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}^{2}-\frac{y}{x}=0.
Find the values of p and q for which
\mathrm{f}\left(x\right)=\left\{\begin{array}{ll}\frac{1-{\mathrm{sin}}^{3}x}{3 {\mathrm{cos}}^{2}x}& , \mathrm{if} x<\frac{x}{2}\\ \mathrm{p}& , \mathrm{if} x=\pi /2\\ \frac{\mathrm{q}\left(1-\mathrm{sin} x\right)}{{\left(\pi -2x\right)}^{2}}& , \mathrm{if} x>\pi /2\end{array}\right\
is continuous at x = π/2. VIEW SOLUTION
Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is
\frac{4\mathrm{r}}{3}.
Also find maximum volume in terms of volume of the sphere.
Find the intervals in which f(x) = sin 3x – cos 3x, 0 < x < π, is strictly increasing or strictly decreasing. VIEW SOLUTION
Using integration find the area of the region
\left\{\left(x, y\right) : {x}^{2}+{y}^{2}⩽ 2ax, {y}^{2}⩾ ax, x, y ⩾ 0\right\}
Find the coordinate of the point P where the line through A(3, –4, –5) and B(2, –3, 1) crosses the plane passing through three points L(2, 2, 1), M(3, 0, 1) and N(4, –1, 0).
Also, find the ratio in which P divides the line segment AB. VIEW SOLUTION
An urn contains 3 white and 6 red balls. Four balls are drawn one by one with replacement from the urn. Find the probability distribution of the number of red balls drawn. Also find mean and variance of the distribution. VIEW SOLUTION
A manufacturer produces two products A and B. Both the products are processed on two different machines. The available capacity of first machine is 12 hours and that of second machine is 9 hours per day. Each unit of product A requires 3 hours on both machines and each unit of product B requires 2 hours on first machine and 1 hour on second machine. Each unit of product A is sold at Rs 7 profit and B at a profit of Rs 4. Find the production level per day for maximum profit graphically. VIEW SOLUTION
\mathrm{f} : \mathrm{N}\to \mathrm{N}
f\left(x\right)=9{x}^{2}+6x-5
\mathrm{f} : \mathrm{N}\to \mathrm{S}
, where S is the range of f, is invertible. Find the inverse of f and hence find
{\mathrm{f}}^{-1}\left(43\right) \mathrm{and} {\mathrm{f}}^{-1}\left(163\right)
\left|\begin{array}{ccc}yz-{x}^{2}& zx-{y}^{2}& xy-{z}^{2}\\ zx-{y}^{2}& xy-{z}^{2}& yz-{x}^{2}\\ xy-{z}^{2}& yz-{x}^{2}& zx-{y}^{2}\end{array}\right|
is divisible by (x + y + z) and hence find the quotient.
Using elementary transformations, find the inverse of the matrix
\mathrm{A}=\left(\begin{array}{ccc}8& 4& 3\\ 2& 1& 1\\ 1& 2& 2\end{array}\right)
and use it to solve the following system of linear equations :
x + 2y + 2z = 7 VIEW SOLUTION |
Create box collision geometry - MATLAB - MathWorks Italia
Create and Visualize Box Collision Geometry
Create box collision geometry
Use collisionBox to create a box collision geometry centered at the origin.
BOX = collisionBox(X,Y,Z)
BOX = collisionBox(X,Y,Z) creates an axis-aligned box collision geometry centered at the origin with X, Y, and Z as its side lengths along the corresponding axes in the geometry-fixed frame. By default, the geometry-fixed frame collocates with the world frame.
X — Side length of box geometry
Side length of box geometry along the x-axis, specified as a positive scalar. Units are in meters.
Y — Side length of box geometry
Side length of box geometry along the y-axis, specified as a positive scalar. Units are in meters.
Z — Side length of box geometry
Side length of box geometry along the z-axis, specified as a positive scalar. Units are in meters.
Create a box collision geometry centered at the origin. The side lengths in the x-, y-, and z-directions are 3, 1, and 2 meters, respectively.
box = collisionBox(3,1,2)
collisionBox with properties:
Visualize the box.
show(box)
title('Box')
Create two homogeneous transformation matrices. The first matrix is a rotation about the z-axis by
\pi /2
radians, and the second matrix is a rotation about the x-axis of
\pi /8
matZ = axang2tform([0 0 1 pi/2]);
matX = axang2tform([1 0 0 pi/8]);
Create a second box collision geometry with the same dimensions as the first. Change its pose to the product of the two matrices. The product corresponds to first rotation about the z-axis followed by rotation about the x-axis. Visualize the result.
box2 = collisionBox(3,1,2);
box2.Pose = matZ*matX;
show(box2)
title('Box2')
collisionCylinder | collisionMesh | collisionSphere | checkCollision |
Sick Universe | Wikipedia Mobile | Encyclopedia | What is / means Almost
For other uses, see Almost (disambiguation).
Find sources: "Almost" – news · newspapers · books · scholar · JSTOR (January 2021) (Learn how and when to remove this template message)
In set theory, when dealing with sets of infinite size, the term almost or nearly is used to refer to all but a negligible amount of elements in the set. The notion of "negligible" depends on the context, and may mean "of measure zero" (in a measure space), "finite" (when infinite sets are involved), or "countable" (when uncountably infinite sets are involved).
{\displaystyle S=\{n\in \mathbb {N} \,|\,n\geq k\}}
{\displaystyle \mathbb {N} }
{\displaystyle k}
{\displaystyle \mathbb {N} }
, because only finitely many natural numbers are less than
{\displaystyle k}
The set of prime numbers is not almost
{\displaystyle \mathbb {N} }
, because there are infinitely many natural numbers that are not prime numbers.
The set of transcendental numbers are almost
{\displaystyle \mathbb {R} }
, because the algebraic real numbers form a countable subset of the set of real numbers (which is uncountable).[1]
The Cantor set is uncountably infinite, but has Lebesgue measure zero.[2] So almost all real numbers in (0, 1) are members of the complement of the Cantor set.
Look up almost in Wiktionary, the free dictionary.
^ "Almost All Real Numbers are Transcendental - ProofWiki". proofwiki.org. Retrieved 2019-11-16.
^ "Theorem 36: the Cantor set is an uncountable set with zero measure". Theorem of the week. 2010-09-30. Retrieved 2019-11-16.
Subset · Superset
Zermelo–Fraenkel
von Neumann–Bernays–Gödel
Morse–Kelley
Kripke–Platek
Tarski–Grothendieck
Kurt Gödel |
What are the direction cosines of a line that makes equal angles with the co-ordinate axes? VIEW SOLUTION
\stackrel{\to }{\mathrm{a}}·\stackrel{\to }{\mathrm{a}}=0 \mathrm{and} \stackrel{\to }{\mathrm{a}}·\stackrel{\to }{\mathrm{b}}=0
, then what can be concluded about the vector
\stackrel{\to }{\mathrm{b}}
Write the position vector of the mid-point of the vector joining the points P(2, 3, 4) and Q(4, 1, −2). VIEW SOLUTION
\underset{1}{\overset{\sqrt{3}}{\int }}\frac{\mathrm{dx}}{1+{\mathrm{x}}^{2}}
\left|\begin{array}{cc}\mathrm{x}& \mathrm{x}\\ 1& \mathrm{x}\end{array}\right|=\left|\begin{array}{cc}3& 4\\ 1& 2\end{array}\right|
, write the positive value of x. VIEW SOLUTION
Write the order of the product matrix:
\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right] \left[\begin{array}{ccc}2& 3& 4\end{array}\right]
Write the values of x − y + z from the following equation:
\left[\begin{array}{ccccc}x& +& y& +& z\\ & x& +& z& \\ & y& +& z& \end{array}\right]=\left[\begin{array}{c}9\\ 5\\ 7\end{array}\right]
Write the principal value of tan−1(−1). VIEW SOLUTION
Write fog if f : R → R and g : R → R are given by
f(x) = |x| and g(x) = |5x − 2|. VIEW SOLUTION
\int \frac{{e}^{2x}-{e}^{-2x}}{{e}^{2x}+{e}^{-2x}}dx
Find the mean number of heads in three tosses of a fair coin. VIEW SOLUTION
\stackrel{\to }{a}=2\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}, \stackrel{\to }{b}=-\stackrel{^}{i}+2\stackrel{^}{j}+\stackrel{^}{k} \mathrm{and} \stackrel{\to }{c}=3\stackrel{^}{i}+\stackrel{^}{j}
\stackrel{\to }{a}+\mathrm{\lambda }\stackrel{\to }{b}
\stackrel{\to }{c}
, find the value of λ. VIEW SOLUTION
(1 + e2x) dy + (1 + y2) ex dx = 0, given that y = 1 when x = 0. VIEW SOLUTION
∫ e2x sin x dx
\int \frac{3x+5}{\sqrt{{x}^{2}-8x+7}}dx
\frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{\mathrm{x}}{2}\sqrt{{\mathrm{a}}^{2} -{\mathrm{x}}^{2}} + \frac{{a}^{2}}{2} \mathrm{sin}{ }^{-1} \left(\frac{\mathrm{x}}{\mathrm{a}}\right)\right] = \sqrt{{\mathrm{a}}^{2} - {\mathrm{x}}^{2}}\phantom{\rule{0ex}{0ex}}
\mathrm{If} \mathrm{y} = \mathrm{log} \left[\mathrm{x} +\sqrt{{\mathrm{x}}^{2} +1}\right], \mathrm{prove} \mathrm{that} \left({\mathrm{x}}^{2} + 1\right) \frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}} + \mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}} = 0.
f(x) = sin x + cos x, 0 ≤ x ≤ 2π
is strictly increasing or strictly decreasing.
Find the points on the curve y = x3 at which the slope of the tangent is equal to the y-coordinate of the point. VIEW SOLUTION
\frac{9\pi }{8} -\frac{9}{4}{\mathrm{sin}}^{-1} \left(\frac{1}{3}\right) = \frac{9}{4} \mathrm{sin}{ }^{-1}\left(\frac{2\sqrt{2}}{3}\right)
Solve the following equation for x :
{\mathrm{tan}}^{-1} \left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right) = \frac{1}{2} {\mathrm{tan}}^{-1} \left(\mathrm{x}\right), \mathrm{x}>0
\mathrm{f} : {\mathrm{R}}_{+} \to \left[4, \infty \right]
given by f(x) = x2 + 4. Show that f is invertible with the inverse (f−1) of f
{\mathrm{f}}^{-1} \left(\mathrm{y}\right) = \sqrt{\mathrm{y} -4,}
where R+ is the set of all non-negative real numbers. VIEW SOLUTION
Prove using properties of determinants :
\left|\begin{array}{ccc}\mathrm{a}-\mathrm{b}-\mathrm{c}& 2\mathrm{a}& 2\mathrm{a}\\ 2\mathrm{b}& \mathrm{b}-\mathrm{c}-\mathrm{a}& 2\mathrm{b}\\ 2\mathrm{c}& 2\mathrm{c}& \mathrm{c}-\mathrm{a}-\mathrm{b}\end{array}\right| = {\left(\mathrm{a} + \mathrm{b} + \mathrm{c}\right)}^{3}
Find the value of k, so that the function f defined by
\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{ll}\mathrm{kx}+1,& \mathrm{if} \mathrm{x}\le \mathrm{\pi }\\ \mathrm{cos} \mathrm{x},& \mathrm{if} \mathrm{x}>\mathrm{\pi }\end{array}\right\
is continuous at x = π. VIEW SOLUTION
\frac{\mathrm{dy}}{\mathrm{dx}}+2\mathrm{y} \mathrm{tan} \mathrm{x}= \mathrm{sin} \mathrm{x},
x=\frac{\mathrm{\pi }}{3}
Find the shortest distance between the given lines:
\stackrel{\to }{\mathrm{r}}=\left(\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\right)+\mathrm{\lambda }\left(\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right) \mathrm{and}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{\mathrm{r}}=\left(4\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}\right)+\mathrm{\mu }\left(2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right).
A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of grinding/cutting machine and a sprayer. It takes 2 hours on the grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes one hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at most 20 hours and the grinding/cutting machine for at most 12 hours. The profit from the sale of a lamp is Rs 5 and that from a shade is Rs 3. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit? Make an L.P.P. and solve it graphically. VIEW SOLUTION
\underset{1}{\overset{4}{\int }} \left({\mathrm{x}}^{2}-\mathrm{x}\right) \mathrm{dx}
as a limit of sums.
\underset{0}{\overset{\frac{\pi }{4}}{\int }} \frac{\mathrm{sin} \mathrm{x}+\mathrm{cos} \mathrm{x}}{9+16 \mathrm{sin} 2\mathrm{x}}\mathrm{dx}
Using the method of integration, find the area of the region bounded by the following lines :
x − 3y + 5 = 0 VIEW SOLUTION
A window is in the form of a rectangle surmounted by a semi-circular opening. The total perimeter of the window is 10 metres. Find the dimensions of the rectangle so as to admit maximum light through the whole opening. VIEW SOLUTION
\left[\begin{array}{ccc}1& -1& 2\\ 0& 2& -3\\ 3& -2& 4\end{array}\right]\left[\begin{array}{ccc}-2& 0& 1\\ 9& 2& -3\\ 6& 1& -2\end{array}\right]
to solve the system of equation:
3x − 2y + 4z = 2.
Using elementary transformations, find the inverse of the matrix :
\left(\begin{array}{ccc}2& 0& -1\\ 5& 1& 0\\ 0& 1& 3\end{array}\right)
Find the vector equation of the plane passing through the points A(2, 2, −1), B (3, 4, 2) and C (7, 0, 6) Also, find the Cartesian equation of the plane. VIEW SOLUTION
Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from bag I to bag II and then a ball is drawn from bag II at random. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black. VIEW SOLUTION |
Find the projection of the vector
\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+7\stackrel{^}{\mathrm{k}}
2\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}
Write the vector equation of the plane, passing through the point (a, b, c) and parallel to the plane
\stackrel{\to }{\mathrm{r}}·\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)=2
Write the antiderivative of
\left(3\sqrt{x}+\frac{1}{\sqrt{x}}\right).
2\left[\begin{array}{cc}3& 4\\ 5& x\end{array}\right] + \left|\begin{array}{cc}1& y\\ 0& 1\end{array}\right| =\left[\begin{array}{cc}7& 0\\ 10& 5\end{array}\right], \mathrm{find} \left(x-y\right).
Solve the following matrix equation for x:
\left[x 1\right] \left[\begin{array}{cc}1& 0\\ -2& 0\end{array}\right]=\mathrm{O}.
\left|\begin{array}{cc}2x& 5\\ 8& x\end{array}\right|=\left|\begin{array}{cc}6& -2\\ 7& 3\end{array}\right|
, write the value of x. VIEW SOLUTION
\mathrm{sin} \left({\mathrm{sin}}^{-1}\frac{1}{5}+{\mathrm{cos}}^{-1} x\right)=1,
then find the value of x. VIEW SOLUTION
Let * be a binary operation, on the set of all non-zero real numbers, given by
a*b=\frac{ab}{5}
for all a, b ∈ R – {0}. Find the value of x, given that 2 * (x * 5) = 10. VIEW SOLUTION
\int {\mathrm{cos}}^{-1}\left(\mathrm{sin} x\right) \mathrm{d}x
\stackrel{\to }{\mathrm{a}} \mathrm{and} \stackrel{\to }{\mathrm{b}}
\left|\stackrel{\to }{\mathrm{a}}\right|=3, \left|\stackrel{\to }{\mathrm{b}}\right|=\frac{2}{3} \mathrm{and} \stackrel{\to }{\mathrm{a}} ×\stackrel{\to }{\mathrm{b}}
\stackrel{\to }{\mathrm{a}} \mathrm{and} \stackrel{\to }{\mathrm{b}}
Find the intervals in which the function f(x) = 3x4 − 4x3 − 12x2 + 5 is
Find the equations of the tangent and normal to the curve x = a sin3θ and y = a cos3θ at
\mathrm{\theta }=\frac{\mathrm{\pi }}{4}.
\int \frac{{\mathrm{sin}}^{6}x+{\mathrm{cos}}^{6}x}{{\mathrm{sin}}^{2}x . {\mathrm{cos}}^{2}x}dx
\int \left(x-3\right)\sqrt{{x}^{2}+3x-18}dx
\left({x}^{2}-1\right)\frac{dy}{dx}+2xy=\frac{2}{{x}^{2}-1}
If y = xx, prove that
\frac{{\mathrm{d}}^{2}\mathrm{y}}{\mathrm{d}{x}^{2}}-\frac{1}{\mathrm{y}}{\left(\frac{\mathrm{dy}}{\mathrm{d}x}\right)}^{2}-\frac{\mathrm{y}}{x}=0.
Prove that, for any three vectors
\stackrel{\to }{\mathrm{a}}, \stackrel{\to }{\mathrm{b}}, \stackrel{\to }{\mathrm{c}}
\left[\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}, \stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}, \stackrel{\to }{\mathrm{c}}+\stackrel{\to }{\mathrm{a}}\right]=2 \left[\stackrel{\to }{\mathrm{a}}, \stackrel{\to }{\mathrm{b}}, \stackrel{\to }{\mathrm{c}}\right]
\stackrel{\to }{\mathrm{a}}, \stackrel{\to }{\mathrm{b}}
\stackrel{\to }{\mathrm{c}}
\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}=\stackrel{\to }{0 }\mathrm{and} \left|\stackrel{\to }{\mathrm{a}}\right|=3, \left|\stackrel{\to }{\mathrm{b}}\right|=5 \mathrm{and} \left|\stackrel{\to }{\mathrm{c}}\right|=7
\stackrel{\to }{\mathrm{a}} \mathrm{and} \stackrel{\to }{\mathrm{b}}
{\mathrm{cot}}^{-1}\left(\frac{\sqrt{1+\mathrm{sin} x}+\sqrt{1-\mathrm{sin} x}}{\sqrt{1+\mathrm{sin} }x-\sqrt{1-\mathrm{sin} x}}\right)=\frac{x}{2}; x \in \left(0, \frac{\mathrm{\pi }}{4}\right)
2 {\mathrm{tan}}^{-1} \left(\frac{1}{5}\right)+ {\mathrm{sec}}^{-1}\left(\frac{5\sqrt{2}}{7}\right)+2 {\mathrm{tan}}^{-1} \left(\frac{1}{8}\right)=\frac{\mathrm{\pi }}{4}
Let A = {1, 2, 3,......, 9} and R be the relation in A × A defined by (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) in A × A. Prove that R is an equivalence relation. Also, obtain the equivalence class [(2, 5)]. VIEW SOLUTION
Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls? Give that
(i) the youngest is a girl.
(ii) at least one is a girl. VIEW SOLUTION
Prove the following using properties of determinants :
\left|\begin{array}{ccc}\mathrm{a}+\mathrm{b}+2\mathrm{c}& \mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{b}+\mathrm{c}+2\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{a}& \mathrm{c}+\mathrm{a}+2\mathrm{b}\end{array}\right|=2\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right){}^{3}
{\mathrm{tan}}^{-1}\left(\frac{x}{\sqrt{1-{x}^{2}}}\right)
{\mathrm{sin}}^{-1}\left(2x\sqrt{1-{x}^{2}}\right)
\mathrm{cosec} x \mathrm{log} \mathrm{y}\frac{\mathrm{dy}}{\mathrm{d}x}+{x}^{2}{\mathrm{y}}^{2}=0
\frac{5-x}{-4}=\frac{\mathrm{y}-7}{4}=\frac{\mathrm{z}+3}{-5} \mathrm{and} \frac{x-8}{7}=\frac{2\mathrm{y}-8}{2}=\frac{\mathrm{z}-5}{3}
A dealer in rural area wishes to purchase a number of sewing machines. He has only Rs 5,760 to invest and has space for at most 20 items for storage. An electronic sewing machine cost him Rs 360 and a manually operated sewing machine Rs 240. He can sell an electronic sewing machine at a profit of Rs 22 and a manually operated sewing machine at a profit of Rs 18. Assuming that he can sell all the items that he can buy, how should he invest his money in order to maximize his profit? Make it as a LPP and solve it graphically. VIEW SOLUTION
A card from a pack of 52 playing cards is lost. From the remaining cards of the pack three cards are drawn at random (without replacement) and are found to be all spades. Find the probability of the lost card being a spade.
From a lot of 15 bulbs which include 5 defectives, a sample of 4 bulbs is drawn one by one with replacement. Find the probability distribution of number of defective bulbs. Hence find the mean of the distribution. VIEW SOLUTION
Two schools P and Q want to award their selected students on the values of discipline, politeness and punctuality. The school P wants to award Rs x each, Rs y each and Rs z each for the three respective values to its 3, 2 and 1 students with a total award money of Rs 1,000. School Q wants to spend Rs 1,500 to award its 4, 1 and 3 students on the respective values (by giving the same award money for the three values as before). If the total amount of awards for one prize on each value is Rs 600, using matrices, find the award money for each value.
Apart from the above three values, suggest one more value for awards. VIEW SOLUTION
Find the distance between the point (7, 2, 4) and the plane determined by the points A(2, 5, −3), B(−2, −3, 5) and C(5, 3, −3).
\stackrel{\to }{\mathrm{r}}=2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}+\lambda \left(3\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)
\stackrel{\to }{\mathrm{r}}.\left(\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right) = 5.
Find the area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x2 + y2 = 32. VIEW SOLUTION
\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{x \mathrm{tan} x}{\mathrm{sec} x · \mathrm{cosec} x}\mathrm{d}x
Prove that the semi-vertical angle of the Right circular cone of given volume and least curved surface area is
{\mathrm{cot}}^{-1} \sqrt{2} |
The Elegant Geometry Behind Rwanda’s COVID-19 Pooled Testing Strategy | CDJ
The Elegant Geometry Behind Rwanda’s COVID-19 Pooled Testing Strategy
December 18, 2020By Eric Chen, Medha Bulumulla, Megan Rochlin
Due to a strict lockdown and some tech savvy solutions, Rwanda has remarkably been able to keep its COVID-19 prevalence low. Up until Dec. 16, the Rwanda Biomedical Center has reported 7,032 cases in the country. Pennsylvania, which has a similar population of 12.5 million, has reported 481,819 cases.
Part of the reason Rwanda has been so successful in preventing the spread of COVID-19 is because of severe action from the government. As reported by the Wall Street Journal, over 70,000 people have been arrested for social distancing infractions.
But the country has also developed some innovative methods and technologies for dealing with COVID-19. Researchers at the University of Rwanda, for example, devised a unique geometric method to reduce the number of COVID-19 tests needed. This powerful geometric method, also called pooled testing, is surprisingly elegant, and I’ll explain why.
The way pooled testing works is simple. First, each patient is swabbed for COVID-19. These samples are combined into a single batch. If the batch tests negative, then each sample in the batch is COVID-19 negative. If the batch tests positive, each sample is then tested individually for COVID-19.
Cornell, for example, employs pooled testing to screen 93 samples at a time. With the help of robots to automate the pooling process, the College of Veterinary Medicine tests about 6,000 people a day. Despite this impressive result, we can do better. By creatively splitting and recombining batch samples, thousands of dollars can be saved when testing large populations.
To demonstrate how efficient the University of Rwanda’s geometric pooled testing strategy is, let’s start with a simple example. Suppose we want to test 27 people. If we wanted to test each person individually, we would need 27 tests.
However, with the geometric pooling method, we can reduce the number of tests needed to 9. First, let’s organize the samples onto something that resembles a Rubik’s cube. The 3-dimensional cube has a side length of 3. We cut the cube into three slices in three different directions. In total, we have nine slices to test.
X-Axis Cuts
Y-Axis Cuts
Z-Axis Cuts
Next, we perform a test on each sample. Three slices happen to test positive.
X-Axis Results
Z-Axis Results
Finally, with these nine tests, we can tell which sample is positive by seeing where the positive slices overlap.
Using this geometric slicing strategy, we were able to test 27 people with only 9 tests. To curb the spread of COVID-19, a reduction in tests is critical. In Rwanda, where a test can cost a laboratory $50, stepping down from 27 tests to 9 can save $900.
Utilizing higher dimensions
Although the rubik’s cube approach is efficient, a limitation of it is that it can only house 27 samples at a time. To increase its efficiency, something creative we can do is place the samples on a 4-dimensional cube — a hypercube. Constructing a hypercube is quite hard to wrap your head around, but I’ll try to keep it simple.
In the same way a 2-dimensional object is a series of 1-dimensional objects and a 3-dimensional object is a series of 2-dimensional objects, a 4-dimensional object is a series of 3-dimensional objects.
For example, a plane, which is 2-dimensional, is a combination of 3 lines, which are 1-dimensional objects.
A cube, which is 3-dimensional, is a combination of 3 planes.
Therefore, a 4D hypercube, which is 4-dimensional, should be a combination of 3 cubes, which are 3-dimensional. Here, the w
term denotes which subcube a sample is in.
W-Axis Cuts
w=0
w=1
w=2
Because each cube individually contains 27 samples, the 4D cube will hold 3 times 27 samples, which is equal to 81 samples. Now that we have our 4D hypercube, we can slice it in each principle direction just like how we would slice the 3D cube. To cut the cube in the x-axis, we will cut each cube individually in the x-axis and group together the slices to form the new rearranged cubes.
x=0
x=1
x=2
The same thing can be done in the y-axis and z-axis.
y=0
y=1
y=2
z=0
z=1
z=2
Let’s now say that we now want to identify two positives in a group of 81 samples. We place the samples on a 4D hypercube and perform the first round of slice tests. We see that in each axis, two of the three slices test positive.
x=0
x=1
x=2
y=0
y=1
y=2
z=0
z=1
z=2
w=0
w=1
w=2
Notice that in each principal direction, two out of the three slices test positive. This is different from our 3D example because in that example, only one of three slices tested positive. To determine which samples test positive, another round of testing is needed. The other positive is found by elimination. To find the first positive sample, we take one of the slices and slice it again. For instance, let’s take the cube w=1
Slices of the cube w=1
We see that sample which corresponds to the slices x=1
, y=2
, z=2
and w=1
tests positive. With this information, we can update our previous slices by removing the known positive.
x=0
x=1
x=2
y=0
y=1
y=2
z=0
z=1
z=2
w=0
w=1
w=2
By seeing where the slices intersect, we see that the remaining positive is at slices x=0
, y=0
, z=1
, and w=2
The positive sample on cube w=2
In total, we screened 81 samples with 21 tests, a significant improvement!
Pooled testing has the potential to reduce COVID-19 caseload significantly. Cornell Professor Peter Frazier estimates that if the entire U.S. population was tested in a four-week period, then the COVID-19 prevalence rate would drop to 0.3%, allowing 90% of adults to return to work.
To demonstrate how efficient geometric pooled testing, let’s use Cornell as an example. On Nov. 13, Cornell saw a spike in COVID-19 cases due to several social gatherings. Out of 5,990 people tested that day, 9 tests came back positive. With Cornell’s naive pooling method, the 5,990 samples would first be split up into 65 pools of 93 samples each. Each pool would then be tested for COVID-19. If the pool tests positive, each of the 93 samples is then tested individually to identify the infected individuals. Based on the prevalence rate, about 8.39 out of the 65 pools would test positive. On average,
65 + 8.39(93) = 845
tests would be performed.
Instead of naive pooling, let’s now try the geometric pooling method. First, we split the 5,990 samples onto 74 4D hypercubes with length 3. Each hypercube holds 81 samples.
In the first round of testing, each hypercube is tested as a whole. On average, 8.35 hypercubes test positive.
99.1% of the time, each positive hypercube only contains 1 positive sample. In this case, 12 tests are performed. The other 0.9% of the time, the hypercube contains 2 positive samples. Here, 21 tests would be needed. There is also the extreme case that each hypercube contains 3 or more positive samples, but the probability of that occurring is negligible. Taking a weighted average of these probabilities, the expected number of testes required per positive cube is
(99.1\%)(12) + (0.09\%)(21) = 12.081
(99.1\%)(12) + (0.09\%)(21) = 12.081
tests. Therefore, our total number of tests required is on average,
74 + (8.35)(12.081) = 175
tests. We see that geometric pooling requires 4.83\times
4.83\times
fewer tests than naive pooling!
If pooled testing is so efficient, why aren’t more countries adopting it? In reality, pooling is tough. It’s easy to imagine how mistakes can be made when sorting pooling by hand. Also, with how sensitive the COVID-19 tests are, pooling may increase the rate of false positives or negatives.
Nonetheless, pooled testing has profound implications in other areas of research such as computer science. For instance, this group testing method can be used to rapidly find a broken lightbulb in a series of Christmas lights. In practice, group testing is used to locate incorrect bits in computer programs.
In Rwanda, the researchers are now developing software to automate the geometric pooling process. Although the process is still in its infancy, field trials have begun in the country. In South Africa, a rugby team has also begun to use the testing method to screen its players and staff. With a dramatic expansion in testing, hopefully, we can see an end to this pandemic soon. |
Standard deviation of input or sequence of inputs - Simulink - MathWorks Switzerland
Find the standard deviation value over
Standard deviation of input or sequence of inputs
The Standard Deviation block computes the standard deviation of each row or column of the input, or along vectors of a specified dimension of the input. It can also compute the standard deviation of the entire input. You can specify the dimension using the Find the standard deviation value over parameter. The Standard Deviation block can also track the standard deviation in a sequence of inputs over a period of time. To track the standard deviation in a sequence of inputs, select the Running standard deviation parameter.
The Running mode in the Standard Deviation block will be removed in a future release. To compute the running standard deviation in Simulink®, use the Moving Standard Deviation block instead.
This port is unnamed until you select the Running standard deviation parameter and set the Reset port parameter to any option other than None.
Specify the reset event that causes the block to reset the running standard deviation. The sample time of the Rst input must be a positive integer multiple of the input sample time.
To enable this port, select the Running standard deviation parameter and set the Reset port parameter to any option other than None.
Port_1 — Standard deviation along the specified dimension
When you do not select the Running standard deviation parameter, the block computes the standard deviation in each row or column of the input, or along vectors of a specified dimension of the input. It can also compute the standard deviation of the entire input at each individual sample time. Each element in the output array y is the standard deviation of the corresponding column, row, or entire input. The output array y depends on the setting of the Find the standard deviation value over parameter. Consider a three-dimensional input signal of size M-by-N-by-P. When you set Find the standard deviation value over to:
Entire input — The output at each sample time is a scalar that contains the standard deviation of the M-by-N-by-P input matrix.
Each row — The output at each sample time consists of an M-by-1-by-P array, where each element contains the standard deviation of each vector over the second dimension of the input. For an M-by-N matrix input, the output at each sample time is an M-by-1 column vector.
Each column — The output at each sample time consists of a 1-by-N-by-P array, where each element contains the standard deviation of each vector over the first dimension of the input. For an M-by-N matrix input, the output at each sample time is a 1-by-N row vector.
Specified dimension — The output at each sample time depends on the value of the Dimension parameter. If you set the Dimension to 1, the output is the same as when you select Each column. If you set the Dimension to 2, the output is the same as when you select Each row. If you set the Dimension to 3, the output at each sample time is an M-by-N matrix containing the standard deviation of each vector over the third dimension of the input.
When you select Running standard deviation, the block tracks the standard deviation of each channel in a time sequence of inputs. In this mode, you must also specify a value for the Input processing parameter.
Elements as channels (sample based) — The block treats each element of the input as a separate channel. For a three-dimensional input signal of size M-by-N-by-P, the block outputs an M-by-N-by-P array. Each element yijk of the output contains the standard deviation of the element uijk for all inputs since the last reset.
When a reset event occurs, the running standard deviation yijk in the current frame is reset to the element uijk.
Columns as channels (frame based) — The block treats each column of the input as a separate channel. This option does not support input signals with more than two dimensions. For a two-dimensional input signal of size M-by-N, the block outputs an M-by-N matrix. Each element yij of the output contains the standard deviation of the elements in the jth column of all inputs since the last reset, up to and including the element uij of the current input.
When a reset event occurs, the running standard deviation for each channel becomes the standard deviation of all the samples in the current input frame, up to and including the current input sample.
Running standard deviation — Option to select running standard deviation
When you select the Running standard deviation parameter, the block tracks the standard deviation value of each channel in a time sequence of inputs.
Find the standard deviation value over — Dimension over which the block computes the standard deviation
Each column — The block outputs the standard deviation over each column.
Each row — The block outputs the standard deviation over each row.
Entire input — The block outputs the standard deviation over the entire input.
Specified dimension — The block outputs the standard deviation over the dimension, specified in the Dimension parameter.
To enable this parameter, clear the Running standard deviation parameter.
Specify the dimension (one-based value) of the input signal over which the standard deviation is computed. The value of this parameter must be greater than 0 and less than the number of dimensions in the input signal.
To enable this parameter, set Find the standard deviation value over to Specified dimension.
When your inputs are of variable size, and you select the Running standard deviation parameter, then:
When the input size difference is in the length of channels (number of rows), no reset occurs and the running operation is carried out as usual.
To enable this parameter, select the Running standard deviation parameter.
The block resets the running standard deviation whenever a reset event is detected at the optional Rst port. The reset sample time must be a positive integer multiple of the input sample time.
When a reset event occurs while the Input processing parameter is set to Elements as channels (sample based), the running standard deviation for each channel is initialized to the value in the corresponding channel of the current input. Similarly, when the Input processing parameter is set to Columns as channels (frame based), the running standard deviation for each channel becomes the standard deviation of all the samples in the current input frame, up to and including the current input sample.
The standard deviation of a discrete-time signal is the square root of the variance of the signal.
Standard deviation gives a measure of deviation of the signal from its mean value.
For purely real or imaginary input, u, of size M-by-N, the standard deviation is given by the following equation:
y=\sigma =\sqrt{\frac{\sum _{i=1}^{M}\sum _{j=1}^{N}{|{u}_{ij}|}^{2}-\frac{{|\sum _{i=1}^{M}\sum _{j=1}^{N}{u}_{ij}|}^{2}}{M*N}}{M*N-1}}
For complex inputs, the standard deviation is given by the following equation:
\sigma =\sqrt{{\sigma }_{\mathrm{Re}}{}^{2}+{\sigma }_{\mathrm{Im}}{}^{2}}
When you clear the Running standard deviation parameter in the block and specify a dimension, the block produces results identical to the MATLAB® std function, when it is called as y = std(u,0,D).
y is the standard deviation along the specified dimension.
The standard deviation along the entire input is identical to calling the std function as y = std(u(:)).
For a complex input signal, the standard deviation is the square root of the sum of the variances of the real and imaginary parts.
\sigma =\sqrt{{\sigma }_{\mathrm{Re}}{}^{2}+{\sigma }_{\mathrm{Im}}{}^{2}} |
h = fwind1(Hd,win1,win2)
h = fwind1(f1,f2,Hd,___)
The fwind1 function designs 2-D FIR filters using the window method. fwind1 uses a 1-D window specification to design a 2-D FIR filter based on the desired frequency response. fwind1 works with 1-D windows only. Use fwind2 to work with 2-D windows.
h = fwind1(Hd,win) creates a 2-D FIR filter h based on the desired frequency response Hd. The fwind1 function uses the 1-D window win to form an approximately circularly symmetric 2-D window using Huang's method.
h = fwind1(Hd,win1,win2) uses two 1-D windows, win1 and win2, to create a separable 2-D window.
h = fwind1(f1,f2,Hd,___) enables you to specify the desired frequency response Hd at arbitrary frequencies f1 and f2 along the x- and y-axes.
\pi
Design the 1-D window. This example uses a Hamming window of length 21.
win = 0.54 - 0.46*cos(2*pi*(0:20)/20);
Plot the 1-D window.
plot(linspace(-1,1,21),win);
Using the 1-D window, design the filter that best produces this frequency response
Desired frequency response, specified as a numeric matrix. Hd is sampled at equally spaced points between -1.0 and 1.0 (in normalized frequency, where 1.0 corresponds to half the sampling frequency, or π radians) along the x and y frequency axes. For accurate results, create Hd by using the freqspace function.
1-D window, specified as a numeric matrix. If you have Signal Processing Toolbox™ software, then you can specify win using windows such as hamming (Signal Processing Toolbox), hann (Signal Processing Toolbox), bartlett (Signal Processing Toolbox), blackman (Signal Processing Toolbox), kaiser (Signal Processing Toolbox), or chebwin (Signal Processing Toolbox).
win1 — 1-D window
2-D FIR filter, returned as a numeric matrix. The length of the window controls the size of the resulting filter.
If you specify a single window win of length n, then the size of h is n-by-n.
If you specify two windows win1 and win2 of length n and m respectively, then the size of h is m-by-n.
If Hd is of data type single, then h is of data type single. Otherwise, h is of data type double.
The fwind1 function takes a one-dimensional window specification and forms an approximately circularly symmetric two-dimensional window using Huang's method,
w\left({n}_{1},{n}_{2}\right)={w\left(t\right)|}_{t=\sqrt{{n}_{{}_{1}}^{2}+{n}_{2}^{2}}},
where w(t) is the one-dimensional window and w(n1,n2) is the resulting two-dimensional window.
Given two windows, the fwind1 function forms a separable two-dimensional window:
w\left({n}_{1},{n}_{2}\right)={w}_{1}\left({n}_{1}\right){w}_{2}\left({n}_{2}\right).
The fwind1 function calls the fwind2 with the desired frequency response Hd and the two-dimensional window. The fwind2 function calculates h using an inverse Fourier transform and multiplication by the two-dimensional window:
{h}_{d}\left({n}_{1},{n}_{2}\right)=\frac{1}{{\left(2\pi \right)}^{2}}{\int }_{-\pi }^{\pi }{\int }_{-\pi }^{\pi }{H}_{d}\left({\omega }_{1},{\omega }_{2}\right){e}^{j{\omega }_{1}{n}_{1}}{e}^{j{\omega }_{2}{n}_{2}}d{\omega }_{1}d{\omega }_{2}
h\left({n}_{1},{n}_{2}\right)={h}_{d}\left({n}_{1},{n}_{2}\right)w\left({n}_{1},{n}_{2}\right)
[1] Lim, Jae S., Two-Dimensional Signal and Image Processing, Englewood Cliffs, NJ, Prentice Hall, 1990. |
\underset{2}{\overset{3}{\int }}{3}^{x} dx.
\mathrm{f}\left(x\right)=\left\{\begin{array}{ll}\frac{kx}{\left|x\right|},& \mathrm{if} x < 0\\ 3,& \mathrm{if} x \ge 0\end{array}\right\
\int \frac{\mathrm{d}x}{{x}^{2}+4x+8}
\frac{dy}{dx} \mathrm{at} x=1, y=\frac{\pi }{4} \mathrm{if} {\mathrm{sin}}^{2}y+\mathrm{cos} xy = K
f\left(x\right)=4{x}^{3}-18{x}^{2}+27x-7
\mathrm{ℝ}
For the curve y = 5x – 2x3, if x increases at the rate of 2 units/sec, then fine the rate of change of the slope of the curve when x = 3. VIEW SOLUTION
\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{x \mathrm{sin} x}{1+{\mathrm{cos}}^{2}x}\mathrm{d}x
\underset{0}{\overset{3/2}{\int }}\left|x \mathrm{sin} \pi x\right|\mathrm{d}x
\stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}=\stackrel{^}{\mathrm{i}}
\stackrel{\to }{\mathrm{c}}={\mathrm{c}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{c}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{c}}_{3}\stackrel{^}{\mathrm{k}},
\stackrel{\to }{\mathrm{a}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}
\stackrel{\to }{\mathrm{c}}
\stackrel{\to }{\mathrm{a}},\text{\hspace{0.17em}}\stackrel{\to }{\mathrm{b}}
\stackrel{\to }{\mathrm{c}}
\mathrm{tan} \left\{\frac{\pi }{4}+\frac{1}{2}{\mathrm{cos}}^{-1}\frac{a}{b}\right\}+\mathrm{tan}\left\{\frac{\pi }{4}-\frac{1}{2}{\mathrm{cos}}^{-1}\frac{a}{b}\right\}=\frac{2b}{a}
\left|\begin{array}{ccc}x& x+y& x+2y\\ x+2y& x& x+y\\ x+y& x+2y& x\end{array}\right|=9{y}^{2}\left(x+y\right).
A=\left(\begin{array}{cc}2& -1\\ 3& 4\end{array}\right), B=\left(\begin{array}{cc}5& 2\\ 7& 4\end{array}\right), C=\left(\begin{array}{cc}2& 5\\ 3& 8\end{array}\right)
{\left(\mathrm{sin} x\right)}^{x}+{\mathrm{sin}}^{-1}\sqrt{x}
{x}^{m}{y}^{n}={\left(x+y\right)}^{m+n}
\frac{{d}^{2}y}{d{x}^{2}}=0
The random variable X can take only the values 0, 1, 2, 3. Given that P(2) = P(3) = p and P(0) = 2P(1). If
\mathrm{\Sigma }{p}_{\mathit{i}}{x}_{i}^{2}=2\mathrm{\Sigma }{p}_{\mathit{i}}{x}_{\mathit{i}}
Using vectors find the area of triangle ABC with vertices A(1, 2, 3), B(2, −1, 4) and C(4, 5, −1). VIEW SOLUTION
Solve the following L.P.P. graphically
Maximise Z = 4x + y
Subject to following constraints
\int \frac{2x}{\left({x}^{2}+1\right) \left({x}^{4}+4\right)}dx
\sqrt{3}\mathrm{y}=x
x\frac{\mathrm{dy}}{\mathrm{d}x}+y=x \mathrm{cos} x + \mathrm{sin} x,
x=\frac{\mathrm{\pi }}{2}
\underset{r}{\to }·\left(2\stackrel{\mathit{^}}{i}-3\stackrel{\mathit{^}}{j}+4\stackrel{\mathit{^}}{k}\right)=1
\underset{r}{\to }·\left(\stackrel{^}{i}-\stackrel{^}{j}\right)+4=0
\underset{r}{\to }·\left(2\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}\right)+8=0
\frac{x-8}{3}=\frac{\mathrm{y}+19}{-16}=\frac{\mathrm{z}-10}{7}
\frac{x-15}{3}=\frac{\mathrm{y}-29}{8}=\frac{\mathrm{z}-5}{-5}
{\mathrm{f}}^{-1}\left(y\right)\left(\frac{\sqrt{y+6}-1}{3}\right)
{\mathrm{f}}^{-1}\left(y\right)=\frac{4}{3},
A metal box with a square base and vertical sides is to contain 1024 cm3. The material for the top and bottom costs Rs 5 per cm2 and the material for the sides costs Rs 2.50 per cm2. Find the least cost of the box. VIEW SOLUTION
\mathrm{A}=\left(\begin{array}{crr}2& 3& 10\\ 4& -6& 5\\ 6& 9& -20\end{array}\right)
, find A–1. Using A–1 solve the system of equations
\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=2; \frac{4}{x}-\frac{6}{y}+\frac{5}{z}=5;\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=-4 |
Loan Knowpia
Personal loanEdit
SecuredEdit
A mortgage loan is a very common type of loan, used by many individuals to purchase residential or commercial property. The lender, usually a financial institution, is given security – a lien on the title to the property – until the mortgage is paid off in full. In the case of home loans, if the borrower defaults on the loan, the bank would have the legal right to repossess the house and sell it, to recover sums owing to it.
Other forms of secured loans include loans against securities – such as shares, mutual funds, bonds, etc. This particular instrument issues customers a line of credit based on the quality of the securities pledged. Gold loans are issued to customers after evaluating the quantity and quality of gold in the items pledged. Corporate entities can also take out secured lending by pledging the company's assets, including the company itself. The interest rates for secured loans are usually lower than those of unsecured loans. Usually, the lending institution employs people (on a roll or on a contract basis) to evaluate the quality of pledged collateral before sanctioning the loan.
UnsecuredEdit
Demand loans are short-term loans[1] that typically do not have fixed dates for repayment. Instead, demand loans carry a floating interest rate, which varies according to the prime lending rate or other defined contract terms. Demand loans can be "called" for repayment by the lending institution at any time.[2] Demand loans may be unsecured or secured.
SubsidizedEdit
ConcessionalEdit
A concessional loan, sometimes called a "soft loan", is granted on terms substantially more generous than market loans either through below-market interest rates, by grace periods, or a combination of both.[4] Such loans may be made by foreign governments to developing countries or may be offered to employees of lending institutions as an employee benefit (sometimes called a perk).
Target marketsEdit
Loans can also be categorized according to whether the debtor is an individual person (consumer) or a business.
Common personal loans include mortgage loans, car loans, home equity lines of credit, credit cards, installment loans, and payday loans. The credit score of the borrower is a major component in and underwriting and interest rates (APR) of these loans. The monthly payments of personal loans can be decreased by selecting longer payment terms, but overall interest paid increases as well.[5] A personal loan can be obtained from banks, alternative (non-bank) lenders, online loan providers and private lenders.
Loans to businesses are similar to the above but also include commercial mortgages and corporate bonds and government guaranteed loans Underwriting is not based upon credit score but rather credit rating.
Loan paymentEdit
{\displaystyle P=L\cdot {\frac {c\,(1+c)^{n}}{(1+c)^{n}-1}}}
For more information, see monthly amortized loan or mortgage payments.
Abuses in lendingEdit
Predatory lending is one form of abuse in the granting of loans. It usually involves granting a loan in order to put the borrower in a position that one can gain advantage over them; subprime mortgage-lending[7] and payday-lending[8] are two examples, where the moneylender is not authorized or regulated, the lender could be considered a loan shark.
Usury is a different form of abuse, where the lender charges excessive interest. In different time periods and cultures, the acceptable interest rate has varied, from no interest at all to unlimited interest rates. Credit card companies in some countries have been accused by consumer organizations of lending at usurious interest rates and making money out of frivolous "extra charges".[9]
Abuses can also take place in the form of the customer defrauding the lender by borrowing without intending to repay the loan.
United States taxesEdit
Most of the basic rules governing how loans are handled for tax purposes in the United States are codified by both Congress (the Internal Revenue Code) and the Treasury Department (Treasury Regulations – another set of rules that interpret the Internal Revenue Code).[10]: 111
1. A loan is not gross income to the borrower.[10]: 111 Since the borrower has the obligation to repay the loan, the borrower has no accession to wealth.[10]: 111 [11]
2. The lender may not deduct (from own gross income) the amount of the loan.[10]: 111 The rationale here is that one asset (the cash) has been converted into a different asset (a promise of repayment).[10]: 111 Deductions are not typically available when an outlay serves to create a new or different asset.[10]: 111
3. The amount paid to satisfy the loan obligation is not deductible (from own gross income) by the borrower.[10]: 111
4. Repayment of the loan is not gross income to the lender.[10]: 111 In effect, the promise of repayment is converted back to cash, with no accession to wealth by the lender.[10]: 111
5. Interest paid to the lender is included in the lender's gross income.[10]: 111 [12] Interest paid represents compensation for the use of the lender's money or property and thus represents profit or an accession to wealth to the lender.[10]: 111 Interest income can be attributed to lenders even if the lender doesn't charge a minimum amount of interest.[10]: 112
6. Interest paid to the lender may be deductible by the borrower.[10]: 111 In general, interest paid in connection with the borrower's business activity is deductible, while interest paid on personal loans are not deductible.[10]: 111 The major exception here is interest paid on a home mortgage.[10]: 111
Income from discharge of indebtednessEdit
Although a loan does not start out as income to the borrower, it becomes income to the borrower if the borrower is discharged of indebtedness.[10]: 111 [13] Thus, if a debt is discharged, then the borrower essentially has received income equal to the amount of the indebtedness. The Internal Revenue Code lists "Income from Discharge of Indebtedness" in Section 61(a)(12) as a source of gross income.
Annual percentage rate (a.k.a. Effective annual rate)
Bank, Fractional-reserve banking, Building society
Debt, Consumer debt, Debt consolidation, Government debt
Finance, Personal finance, Settlement (finance)
Interest-only loan, Negative amortization, PIK loan
George D. Sax and the Exchange National Bank of Chicago - Innovation of instant loans
^ Signoriello, Vincent J. (1991), Commercial Loan Practices and Operations, ISBN 978-1-55520-134-0
^ CCH Incorporated; CCH Tax Law Editors (April 2008). Federal Estate & Gift Taxes: Code & Regulations (Including Related Income Tax Provisions), As of March 2008. CCH. pp. 631–. ISBN 978-0-8080-1853-7. {{cite book}}: |author2= has generic name (help)
^ Subsidized Loan - Definition and Overview at About.com. Retrieved 2011-12-21.
^ Concessional Loans, Glossary of Statistical Terms, oecd.org, Retrieved on 5/5/2013
^ "Average new-car loan a record 65 months in fourth quarter". Reuters. August 6, 2017. Retrieved 2017-08-06.
^ Guttentag, Jack (October 6, 2007). "The Math Behind Your Home Loan". The Washington Post. Retrieved May 11, 2010.
^ "Predators try to steal home". money.cnn.com. [CNN]. 18 Apr 2000. Retrieved 7 Mar 2018.
^ Horsley, Scott; Arnold, Chris (2 Jun 2016). "New Rules To Ban Payday Lending 'Debt Traps'". National Public Radio. Retrieved 7 Mar 2018.
^ "Credit cardholders pay Rs 6,000 cr 'extra'". The Financial Express (India). Chennai, India]. 3 May 2007. Archived from the original on January 20, 2019. Alt URL
^ a b c d e f g h i j k l m n o p Samuel A. Donaldson, Federal Income Taxation of Individuals: Cases, Problems and Materials, 2nd Ed. (2007).
^ See Commissioner v. Glenshaw Glass Co., 348 U.S. 426 (1955) (giving the three-prong standard for what is "income" for tax purposes: (1) accession to wealth, (2) clearly realized, (3) over which the taxpayer has complete dominion).
^ 26 U.S.C. 61(a)(4)(2007).
^ 26 U.S.C. 61(a)(12)(2007).
^ 26 U.S.C. 108(2007).
^ EUGENE A. LUDWIG AND PAUL A. VOLCKER, 16 November 2012 Banks Need Long-Term Rainy Day Funds |
Roshani reads 1/6 hours on the first day 1/4 hours on the second day and 1/3 hours on the third - Maths - Rational Numbers - 6933678 | Meritnation.com
On the first day Roshani reads 1/6 hours
On the second day she reads 1/4 hours
On the third day she reads 1/3 hours
So the pattern will be as
\frac{1}{6},\quad \frac{1}{4},\quad \frac{1}{3},...
S, the given pattern shows an arithmetic progression in which common difference is given by;
d\quad =\quad \frac{1}{3}-\frac{1}{4}\quad =\quad \frac{1}{4}-\frac{1}{6}\quad =\quad \frac{1}{12}
First term; a =
\frac{1}{6}
and for 5th day, n = 5
So, nth term =
{t}_{n}\quad =\quad a+\left(n-1\right)d
Then, 5th term =
{t}_{5}\quad =\frac{1}{6}+\quad \left(5-1\right)\times \frac{1}{12}\quad =\quad \frac{1}{6}+\frac{1}{3}\quad =\quad \frac{3}{6}\quad =\quad \frac{1}{2}
Therefore, on 5th day Roshni reads
\frac{1}{2} |
PID Controller Design in the Live Editor - MATLAB & Simulink - MathWorks Switzerland
Refine Controller Design
Examine Disturbance Rejection Performance
Compare Two Controller Designs
This example shows how to use the Tune PID Controller task in the Live Editor to generate code for designing a PID controller for a linear plant model. The Tune PID Controller task lets you interactively refine the performance of the controller to adjust loop bandwidth and phase margin, or to favor setpoint tracking or disturbance. The task generates a response plot that lets you monitor controller performance while you adjust the tuning parameters.
Open this example to see a preconfigured script containing the Tune PID Controller task. For more information about Live Editor tasks generally, see Add Interactive Tasks to a Live Script.
In the Live Editor, create an LTI model for your plant.
G = zpk(-5,[-1 -2 -3 -4],6);
To design a PID controller for this plant, open the Tune PID Controller Live Editor task. On the Live Editor tab, select Task > Tune PID Controller. This action inserts the task into your script.
To generate an initial PID controller design, in the Plant menu, select the plant you created, G. Tune PID Controller automatically generates a PI controller that balances performance and robustness, assuming the standard unit-feedback control configuration of the following diagram.
The task also generates a step response plot showing the closed-loop step response from r to y using the initial controller design.
Select System response characteristics to display numeric values of some time-domain characteristics of this response.
The initial controller design has a rise time of about 1.5 seconds, with about 8% overshoot. Experiment with the Response Time and Transient Behavior sliders to change the design targets and see their effect on the step response.
With a PI controller and this plant, it is difficult to decrease the response time without introducing instability or otherwise degrading system response. Try switching to a PID controller to see if you can achieve a better response time. In the Controller Type drop-down menu, select PID.
You can now reduce the response time of the controller. Experiment with the sliders again, observing the effect on the step response. For an example that shows in more detail how the Response Time and Transient Behavior sliders affect controller performance, see Tune PID Controller to Favor Reference Tracking or Disturbance Rejection (PID Tuner). That example uses the PID Tuner app instead of the Tune PID Controller task in the Live Editor, but the behavior and effect of the sliders is the same in both tools.
The task automatically generates code to to tune a PID controller for the plant with the specified design goals. To see the generated code, click at the bottom of the task. The task expands to show the generated code.
As you change parameters such as controller structure, performance goals, and response-plot type, the generate code updates automatically to reflect the new settings.
Suppose that you are interested in the closed-loop system response to a disturbance at the plant input. To generate a plot of the step response from
{\mathit{d}}_{1}
to y, in the Output Plot drop-down menu, select Step Plot: Input disturbance rejection. The plot updates to show the new response. Depending on how you set the performance goals when you change the response plot, you might see a response that looks like the following.
You can now experiment again with controller parameters and observe their effect on disturbance rejection. For an example that shows in more detail how you can use the sliders and other design parameters to improve disturbance rejection performance, see Tune PID Controller to Favor Reference Tracking or Disturbance Rejection (PID Tuner). That example uses the PID Tuner app instead of the Tune PID Controller task in the Live Editor, but the behavior and effect of design parameters is the same in both tools.
Tune PID Controller automatically writes the tuned controller to the MATLAB® workspace as a pid, pidstd, pid2, or pidstd2 model object, whichever is appropriate for your controller settings. The task stores the controller using the variable name specified in the task summary line. By default, that variable name is C. When you change controller settings, performance goals, or other tuning parameters, by default the task writes over the variable C.
You can save a controller design to use as a baseline for comparison while you experiment further with controller types, performance goals, and other settings. To do so, type a new variable name in the task summary line. For instance, change the output controller name to Cnew.
Now, the current design is stored in the MATLAB workspace as C. Any further changes to the design are stored as Cnew.
To use C as a baseline for comparison, in the Baseline controller menu, choose Select from workspace. Then, select C in the menu that appears.
Now, as you experiment further with the controller design, the plot displays both the system response with controller C (dotted line) and with controller Cnew (solid line).
Because the Tune PID Controller task saves the controller to the MATLAB workspace, you can use the controller as you would use any other PID model object for control design and analysis. For instance, examine the controller performance against a slightly different plant model, to get a sense of the robustness of the closed-loop system against parameter variation.
G1 = zpk(-5,[-0.75 -2 -3 -4],8);
CL1 = getPIDLoopResponse(C,G1,'closed-loop');
CL = getPIDLoopResponse(C,G,'closed-loop');
step(CL,CL1) |
Global Existence of Solutions for a Nonstrictly Hyperbolic System
De-yin Zheng, Yun-guang Lu, Guo-qiang Song, Xue-zhou Lu, "Global Existence of Solutions for a Nonstrictly Hyperbolic System", Abstract and Applied Analysis, vol. 2014, Article ID 691429, 7 pages, 2014. https://doi.org/10.1155/2014/691429
De-yin Zheng,1 Yun-guang Lu,1 Guo-qiang Song,2 and Xue-zhou Lu3
2College of Health Administration, Anhui Medical University, Hefei 230032, China
3Laboratoire Ondes and Milieux Complexes UMR 6294, CNRS-Universite du Havre, 76058 Le Havre, France
We obtain the global existence of weak solutions for the Cauchy problem of the nonhomogeneous, resonant system. First, by using the technique given in Tsuge (2006), we obtain the uniformly bounded estimates and when is increasing (similarly, and when is decreasing) for the -viscosity and -flux approximation solutions of nonhomogeneous, resonant system without the restriction or as given in Klingenberg and Lu (1997), where and are Riemann invariants of nonhomogeneous, resonant system; is a uniformly bounded function of depending only on the function given in nonhomogeneous, resonant system, and is the bound of . Second, we use the compensated compactness theory, Murat (1978) and Tartar (1979), to prove the convergence of the approximation solutions.
The following system describes the evolution of an isothermal fluid in a nozzle with discontinuous cross-sectional area , where and stand for the density and the particle velocity of the fluid under consideration, respectively, and denotes the pressure function (See [1]). The existence of global weak solutions for the Cauchy problem or the initial boundary value problem of system (1) has been studied in [1–3]. In [4–6], the authors showed the global existence of BV entropy solutions to a more general class of nonhomogeneous, resonant system by the generalized Glimm scheme.
The Riemann problem for a more general resonant system of equations, was resolved in [7], where and is a smooth function.
To study the existence of entropy solutions of the Cauchy problem (1), the main difficulty is to establish boundedness of solutions because the equations are not in conservative form and the Conley-Chuey-Smoller principle of invariant regions does not apply (See [1] for the details about the physical background of system (1) and its difficulty in analysis). For the polytropic gas and the adiabatic exponent , the definition of a finite energy solution (unbounded) is given and its existence is obtained by using the compensated compactness method in [1].
For smooth solution, system (1) is equivalent to the following conservation laws of three equations: or the system of two equations where . When is a constant, system (3) or system (4) itself has many different physical backgrounds. For instance, it is a scaling limit system of Newtonian dynamics with long-range interaction for a continuous distribution of mass in (cf. [8, 9]) and also a hydrodynamic limit for the Vlasov equation (cf. [10]). Its global weak solution was obtained by using the random choice method [2] in [11] and by using the compensated compactness theory in [12, 13].
By simple calculations, two eigenvalues of system (4) are with corresponding Riemann invariants where is a constant.
The existence of weak solutions for the Cauchy problem (4) with bounded initial data was first studied in [12], where is a smooth function and the technical condition or on the initial data is imposed for obtaining the a-priori, uniform estimate of or .
Without the condition or , the reasonable estimate, depending on the variable , was first obtained in [2] for system (1) when , by using a modified Godunov scheme, and in [14] for general pressure function and smooth function by using the compensated compactness.
In this paper, using the vanishing viscosity method and the maximum principle coupled with the flux approximation proposed in [15] for the homogeneous system of isentropic gas dynamics, we extend the results in [2, 12] to the Cauchy problem (4)–(7) for any bounded initial data and for the function satisfying the conditions , .
We first construct the sequence of hyperbolic systems to approximate system (4), where denotes the flux approximation constant and the approximation pressure , and is the smooth approximation of , being a mollifier. If is a monotonic function, as required in Theorem 2, and and converge to zero much faster than , then it is easy to prove that and satisfy Second, we add the viscosity terms to the right-hand side of (8) to obtain the following parabolic system: with initial data where are given in (7).
Lemma 1. Let or equivalently for two constants and . If is increasing or equivalently , then we can choose a function satisfying , , and where the positive constants , and depend on , and , but are independent of .
The proof of Lemma 1 is trivial.
By applying the maximum principle to the Cauchy problem (11)-(12), we first obtain the estimates and when is increasing (similarly and when is decreasing) for a suitable positive, bounded function given in Lemma 1; then by using the compensated compactness theory and the already existed compact frameworks given in [12, 13], we give the following global existence theorem of weak solutions.
Theorem 2. Let and for two positive constants and .
(A) Let be increasing and we choose satisfying all conditions in Lemma 1 and, moreover, or , where and for a sufficiently small constant . Then the Riemann invariants and of system (4) with respect to the approximated solutions of the Cauchy problem (11)-(12) satisfy the estimate if and if .
(B) For such function and the initial data satisfying the conditions in Part , if either , , or , where is a positive constant, then there exists a subsequence of , which converges pointwise to a pair of bounded functions as , and tend to a zero, and the limit is a weak entropy solution of the Cauchy problem (4)–(7).
Definition 3. For integrable function , a pair of bounded measurable functions is called a weak entropy solution of the Cauchy problem (4)–(7), if hold for all test function and holds for any nonnegative test function , where is a pair of convex entropy-entropy flux of system (4).
We can easily construct many functions , and satisfying the conditions in Theorem 2.
Example 4. Let and choose Then (13) is satisfied since and satisfies all the conditions in Theorem 2. In fact, we may choose and then and for a positive constant .
We are going to prove Theorem 2 in the next section.
By simple calculations, two eigenvalues of system (8) are with corresponding right eigenvectors and Riemann invariants which are similar to the Riemann invariants of system (4) given by (6).
We multiply (11) by and , respectively, to obtain Let . Then or or where and .
Using the first equation of (11), we have the a-priori estimate . Since the conditions on in Theorem 2, the following two terms on the left-hand side of (28): Now, we consider the other terms on the left-hand side of (28).
First, we have from that Second, we have from , that and so since .
Thus, (28) is reduced to the following inequality about : and we can prove that or if applying for the maximum principle to (34).
To prove the estimate of , we have from (25) that where .
Let where is the upper bound of and , and are the bounds of and obtained from the local solution. Then
We have from (35)–(37) that We argue by assuming that (38) is violated for at a point in . Let be the least upper bound of values of at which . Then, by the continuity we see that at some points . So , , and at ; that is, But from (35) and (36), Since on ; then Thus, at from the relation of , and given by (23). So the right-hand side of (40) is negative, which yields a conclusion contradicting (39). So (38) is proved. Therefore for any point in , which yields the desired estimate if we let in (42), and, hence, complete the proof of Part (A) in Theorem 2.
For the homogeneous case (), the convergence of as , and tend to zero in the Part (B) was given in [13] when , , and given in [12] when by using the compensated compactness theory [16, 17] coupled with some basic ideas of the kinetic formulation [18, 19].
Now, we are going to prove the convergence of as , and tend to zero for the inhomogeneous system (11).
Any entropy-entropy flux pair of the original hyperbolic system (4) satisfies the additional system. Consider Eliminating the from (44), we have Similarly, any entropy-entropy flux pair of the approximated hyperbolic system (8) satisfies By eliminating the from (46), we have also the same entropy equation (45). Therefore, system (4) and system (8) have the same entropies.
For any entropy-entropy flux pair of system (4), by multiplying to system (11), we have where is the entropy flux of the approximated system (8) corresponding to entropy . Since is uniformly integrable, then the last term on the right-hand side of system (47) is compact in , for some , by the Sobolev embedding theorems. It is obvious that the term on the left-hand side is compact in . Therefore, using the same techniques given in [12, 13] for the homogeneous system, we may prove that is compact in and so the convergence of as , and tend to zero. Furthermore, the limit satisfies (16).
If precisely using (10), we can prove that the limit satisfies the following conservation form: In fact, we multiply the first equation in (11) by to obtain which yields (48) when goes to zero.
Since both systems (4) and (8) have the same entropies, we can easily prove that the limit satisfies the entropy condition (17). So we complete the proof of Theorem 2.
Yun-guang Lu and De-yin Zheng’s work was partially supported by the Natural Science Foundation of Zhejiang Province of China (Grant no. LY12A01030 and no. LZ13A010002) and the National Natural Science Foundation of China (Grant no. 11271105) and Guo-qiang Song’s work was partially supported by the Natural Science Foundation of Anhui Education Committee (Grant no. KJ2012A171).
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Copyright © 2014 De-yin Zheng 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. |
The Mechanics of Rate of Return Regulation | EME 801: Energy Markets, Policy, and Regulation
The determination of utility rates (the prices that are paid for electricity by you and me, or by other types of electricity customers such as businesses) occurs through an administrative process known as a "rate case." Rate cases are typically held on a periodic basis (say, every few years) or when special circumstances dictate. Examples of such special circumstances might include the utility wanting to introduce a new type of electric rate, or the utility wanting to raise rates due to an increase in fuel prices.
Please have a look at Rate of Return Regulation by Mark Jamison (A PDF file is available to registered students in the Lesson 05 module in Canvas.) that describes the process of utility ratemaking in some detail.
I won't repeat all of the details from the Jamison reading here, but there are a few important points and implications of utility ratemaking to hit on.
First, the ratemaking process involves determining which utility costs can be passed through to customers directly and which activities would qualify for inclusion in the "rate base," which determines how profitable the utility is. The sum of these two types of costs is called the "revenue requirement," which is how much money the utility would need to bring in during a given time period in order to cover its costs and retain its profit. Jamison's way of writing the revenue requirement is common; here is another version of that same equation:
RR=O+\left(V-D\right)×r
Where RR is the revenue requirement, O indicates total operating costs (fuel, labor, maintenance, taxes - these are pass-through costs on which the utility is not allowed to earn any profit); V is the accounting value of the utility's assets; D is the accounting depreciation on those assets; and r is the rate of return that the utility is allowed to earn. The term V-D is often referred to as the "rate base." We will discuss some of these accounting issues later in the course but there are two important aspects to the ratemaking equation. Operating expenses are generally passed through to customers with few challenges, and utilities are not allowed to profit on operational decision-making. On the other hand, the larger the rate base, the more profit that the utility earns. So the structure of utility ratemaking encourages capital investment.
Second, the ratemaking process involves decisions regarding how to allocate costs to different types of customers (e.g., residential, commercial, industrial). It can be hard to identify exactly how much of a utility's costs are attributable to serving specific customers or even types of customers. This is because the power grid is effectively a common resource used by all customers to facilitate reliable delivery of electric power. The process of allocating utility costs among the various customer types is thus done by negotiation (as part of the rate case) rather than by some magical calculation. Historically, large commercial and industrial customers have had rates that are lower than average while small commercial and residential customers have had higher rates. This has led to some accusations of "cross-subsidization" and the politicization of electric rate setting.
Third, the ratemaking process must also determine the rate of return for the utility. Generally, utilities are entitled to earn a "fair" rate of return. While the term "fair" is not all that meaningful in most economics discussions, it has a very specific definition in the public utility world. A "fair" rate of return is one that allows the utility to raise whatever capital it needs to make needed investments in infrastructure. Basically, the utility has to be profitable enough (and sufficiently low risk) that investors will be willing to lend it money. We will explore this concept in a future lesson, but the requirement for a "fair" rate of return implies that there is some connection between the cost of capital for the utility (here, "cost of capital" refers to the interest rate rather than the actual cost of any capital investment) and the rate of return that its regulator needs to allow. The rates of return allowed by public utility commissions varies, but a return on the rate base of 8% to 10% per year is a good representative figure.
‹ Electricity Industry Structure and Regulation up Economic Dispatch and Operations of Electric Utilities › |
a*b=\frac{ab}{5}
\mathrm{sin} \left({\mathrm{sin}}^{-1}\frac{1}{5}+{\mathrm{cos}}^{-1} x\right)=1,
2\left[\begin{array}{cc}3& 4\\ 5& x\end{array}\right] + \left|\begin{array}{cc}1& y\\ 0& 1\end{array}\right| =\left[\begin{array}{cc}7& 0\\ 10& 5\end{array}\right], \mathrm{find} \left(x-y\right).
\left[x 1\right] \left[\begin{array}{cc}1& 0\\ -2& 0\end{array}\right]=\mathrm{O}.
\left|\begin{array}{cc}2x& 5\\ 8& x\end{array}\right|=\left|\begin{array}{cc}6& -2\\ 7& 3\end{array}\right|
\left(3\sqrt{x}+\frac{1}{\sqrt{x}}\right).
{\int }_{0}^{3}\frac{\mathrm{d}x}{9+{x}^{2}}
\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+7\stackrel{^}{\mathrm{k}}
2\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}
\stackrel{\to }{\mathrm{a}} \mathrm{and} \stackrel{\to }{\mathrm{b}}
\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}
\stackrel{\to }{\mathrm{a}} \mathrm{and} \stackrel{\to }{\mathrm{b}}
\stackrel{\to }{\mathrm{r}}·\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)=2
{\mathrm{cot}}^{-1}\left(\frac{\sqrt{1+\mathrm{sin} x}+\sqrt{1-\mathrm{sin} x}}{\sqrt{1+\mathrm{sin} }x-\sqrt{1-\mathrm{sin} x}}\right)=\frac{x}{2}; x \in \left(0, \frac{\mathrm{\pi }}{4}\right)
2 {\mathrm{tan}}^{-1} \left(\frac{1}{5}\right)+ {\mathrm{sec}}^{-1}\left(\frac{5\sqrt{2}}{7}\right)+2 {\mathrm{tan}}^{-1} \left(\frac{1}{8}\right)=\frac{\mathrm{\pi }}{4}
\left|\begin{array}{ccc}2y& y-z-x& 2y\\ 2z& 2z& z-x-y\\ x-y-z& 2x& 2x\end{array}\right|={\left(x+y+z\right)}^{3}
Differentiate tan−1
\left(\frac{\sqrt{1-{x}^{2}}}{x}\right)
with respect to cos−1
\left(2x\sqrt{1-{x}^{2}}\right)
, when x ≠ 0. VIEW SOLUTION
\frac{{\mathrm{d}}^{2}\mathrm{y}}{\mathrm{d}{x}^{2}}-\frac{1}{\mathrm{y}}{\left(\frac{\mathrm{dy}}{\mathrm{d}x}\right)}^{2}-\frac{\mathrm{y}}{x}=0.
\mathrm{\theta }=\frac{\mathrm{\pi }}{4}.
\int \frac{{\mathrm{sin}}^{6}x+{\mathrm{cos}}^{6}x}{{\mathrm{sin}}^{2}x . {\mathrm{cos}}^{2}x}dx
\int \left(x-3\right)\sqrt{{x}^{2}+3x-18}dx
{\mathrm{e}}^{x}\sqrt{1-{\mathrm{y}}^{2}} \mathrm{d}x+\frac{\mathrm{y}}{x} \mathrm{dy}=0,
given that y = 1 when x = 0. VIEW SOLUTION
\left({x}^{2}-1\right)\frac{dy}{dx}+2xy=\frac{2}{{x}^{2}-1}
\stackrel{\to }{\mathrm{a}}, \stackrel{\to }{\mathrm{b}}, \stackrel{\to }{\mathrm{c}}
\left[\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}, \stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}, \stackrel{\to }{\mathrm{c}}+\stackrel{\to }{\mathrm{a}}\right]=2 \left[\stackrel{\to }{\mathrm{a}}, \stackrel{\to }{\mathrm{b}}, \stackrel{\to }{\mathrm{c}}\right]
\stackrel{\to }{\mathrm{a}}, \stackrel{\to }{\mathrm{b}}
\stackrel{\to }{\mathrm{c}}
\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}=\stackrel{\to }{0 }\mathrm{and} \left|\stackrel{\to }{\mathrm{a}}\right|=3, \left|\stackrel{\to }{\mathrm{b}}\right|=5 \mathrm{and} \left|\stackrel{\to }{\mathrm{c}}\right|=7
\stackrel{\to }{\mathrm{a}} \mathrm{and} \stackrel{\to }{\mathrm{b}}
\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7} \mathrm{and} \frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}
intersect. Also find their point of intersection. VIEW SOLUTION
Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is
{\mathrm{cos}}^{-1}\frac{1}{\sqrt{3}}.
\underset{\mathrm{\pi }/6}{\overset{\mathrm{\pi }/3}{\int }}\frac{dx}{1+\sqrt{\mathrm{cot} x}}
\stackrel{\to }{\mathrm{r}}=2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}+\lambda \left(3\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)
\stackrel{\to }{\mathrm{r}}.\left(\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right) = 5. |
Integer and Fractional General T-System and Its Application to Control Chaos and Synchronization
2015 Integer and Fractional General
T
-System and Its Application to Control Chaos and Synchronization
Mihaela Neamţu, Anamaria Liţoiu, Petru C. Strain
We propose a three-dimensional autonomous nonlinear system, called the general
T
system, which has potential applications in secure communications and the electronic circuit. For the general
T
system with delayed feedback, regarding the delay as bifurcation parameter, we investigate the effect of the time delay on its dynamics. We determine conditions for the existence of the Hopf bifurcations and analyze their direction and stability. Also, the fractional order general
T
-system is proposed and analyzed. We provide some numerical simulations, where the chaos attractor and the dynamics of the Lyapunov coefficients are taken into consideration. The effectiveness of the chaotic control and synchronization on schemes for the new fractional order chaotic system are verified by numerical simulations.
Mihaela Neamţu. Anamaria Liţoiu. Petru C. Strain. "Integer and Fractional General
T
-System and Its Application to Control Chaos and Synchronization." Abstr. Appl. Anal. 2015 1 - 14, 2015. https://doi.org/10.1155/2015/413540
Mihaela Neamţu, Anamaria Liţoiu, Petru C. Strain "Integer and Fractional General
T
-System and Its Application to Control Chaos and Synchronization," Abstract and Applied Analysis, Abstr. Appl. Anal. 2015(none), 1-14, (2015) |
Euler characteristics and elliptic curves II
January, 2001 Euler characteristics and elliptic curves II
John COATES, Susan HOWSON
This paper describes a generalisation of the methods of Iwasawa Theory to the field
{F}_{\infty }
obtained by adjoining the field of definition of all the
p
-power torsion points on an elliptic curve,
E
, to a number field,
F
. Everything considered is essentially well-known in the case
E
has complex multiplication, thus it is assumed throughout that
E
has no complex multiplication. Let
{G}_{\infty }
denote the Galois group of
{F}_{\infty }
F
. Then the main focus of this paper is on the study of the
{G}_{\infty }
-cohomology of the
{p}^{\infty }\text{-}
Selmer group of
E
{F}_{\infty }
, and the calculation of its Euler characteristic, where possible. The paper also describes proposed natural analogues to this situation of the classical Iwasawa
\lambda
-invariant and the condition of having
\mu
-invariant equal to 0.
The final section illustrates the general theory by a detailed discussion of the three elliptic curves of conductor 11, at the prime
p=5
John COATES. Susan HOWSON. "Euler characteristics and elliptic curves II." J. Math. Soc. Japan 53 (1) 175 - 235, January, 2001. https://doi.org/10.2969/jmsj/05310175
John COATES, Susan HOWSON "Euler characteristics and elliptic curves II," Journal of the Mathematical Society of Japan, J. Math. Soc. Japan 53(1), 175-235, (January, 2001) |
Evaluate clustering solutions - MATLAB evalclusters - MathWorks Deutschland
Evaluate a Matrix of Clustering Solutions
Specify Clustering Algorithm with a Function Handle
For Silhouette and Gap
For Silhouette Only
For Gap Only
Evaluate clustering solutions
eva = evalclusters(x,clust,criterion)
eva = evalclusters(x,clust,criterion,Name,Value)
eva = evalclusters(x,clust,criterion) creates a clustering evaluation object containing data used to evaluate the optimal number of data clusters.
eva = evalclusters(x,clust,criterion,Name,Value) creates a clustering evaluation object using additional options specified by one or more name-value pair arguments.
The data contains length and width measurements from the sepals and petals of three species of iris flowers.
eva = evalclusters(meas,'kmeans','CalinskiHarabasz','KList',1:6)
Use an input matrix of proposed clustering solutions to evaluate the optimal number of clusters.
Use kmeans to create an input matrix of proposed clustering solutions for the sepal length measurements, using 1, 2, 3, 4, 5, and 6 clusters.
clust = zeros(size(meas,1),6);
clust(:,i) = kmeans(meas,i,'emptyaction','singleton',...
'replicate',5);
Each row of clust corresponds to one sepal length measurement. Each of the six columns corresponds to a clustering solution containing 1 to 6 clusters.
Evaluate the optimal number of clusters using the Calinski-Harabasz criterion.
eva = evalclusters(meas,clust,'CalinskiHarabasz')
Use a function handle to specify the clustering algorithm, then evaluate the optimal number of clusters.
Use a function handle to specify the clustering algorithm.
myfunc = @(X,K)(kmeans(X, K, 'emptyaction','singleton',...
'replicate',5));
Evaluate the optimal number of clusters for the sepal length data using the Calinski-Harabasz criterion.
eva = evalclusters(meas,myfunc,'CalinskiHarabasz',...
'klist',[1:6])
Input data, specified as an N-by-P matrix. N is the number of observations, and P is the number of variables.
clust — Clustering algorithm
'kmeans' | 'linkage' | 'gmdistribution' | matrix of clustering solutions | function handle
Clustering algorithm, specified as one of the following.
'kmeans' Cluster the data in x using the kmeans clustering algorithm, with 'EmptyAction' set to 'singleton' and 'Replicates' set to 5.
'linkage' Cluster the data in x using the clusterdata agglomerative clustering algorithm, with 'Linkage' set to 'ward'.
'gmdistribution' Cluster the data in x using the gmdistribution Gaussian mixture distribution algorithm, with 'SharedCov' set to true and 'Replicates' set to 5.
If criterion is 'CalinskiHarabasz', 'DaviesBouldin', or 'silhouette', you can specify a clustering algorithm using a function handle. The function must be of the form C = clustfun(DATA,K), where DATA is the data to be clustered, and K is the number of clusters. The output of clustfun must be one of the following:
A vector of integers representing the cluster index for each observation in DATA. There must be K unique values in this vector.
A numeric n-by-K matrix of score for n observations and K classes. In this case, the cluster index for each observation is determined by taking the largest score value in each row.
If criterion is 'CalinskiHarabasz', 'DaviesBouldin', or 'silhouette', you can also specify clust as a n-by-K matrix containing the proposed clustering solutions. n is the number of observations in the sample data, and K is the number of proposed clustering solutions. Column j contains the cluster indices for each of the N points in the jth clustering solution.
Data Types: single | double | char | string | function_handle
criterion — Clustering evaluation criterion
'CalinskiHarabasz' | 'DaviesBouldin' | 'gap' | 'silhouette'
Clustering evaluation criterion, specified as one of the following.
'CalinskiHarabasz' Create a CalinskiHarabaszEvaluation clustering evaluation object containing Calinski-Harabasz index values. For more information, see Calinski-Harabasz Criterion.
'DaviesBouldin' Create a DaviesBouldinEvaluation cluster evaluation object containing Davies-Bouldin index values. For more information, see Davies-Bouldin Criterion.
'gap' Create a GapEvaluation cluster evaluation object containing gap criterion values. For more information, see Gap Value.
'silhouette' Create a SilhouetteEvaluation cluster evaluation object containing silhouette values. For more information, see Silhouette Value and Criterion.
Example: 'KList',[1:5],'Distance','cityblock' specifies to test 1, 2, 3, 4, and 5 clusters using the city block distance metric.
KList — List of number of clusters to evaluate
List of number of clusters to evaluate, specified as the comma-separated pair consisting of 'KList' and a vector of positive integer values. You must specify KList when clust is a clustering algorithm name or a function handle. When criterion is 'gap', clust must be a character vector, a string scalar, or a function handle, and you must specify KList.
Example: 'KList',[1:6]
'sqEuclidean' (default) | 'Euclidean' | 'cityblock' | vector | function | ...
Distance metric used for computing the criterion values, specified as the comma-separated pair consisting of 'Distance' and one of the following.
'Euclidean' Euclidean distance. This option is not valid for the kmeans clustering algorithm.
'Hamming' Percentage of coordinates that differ. This option is only valid for the Silhouette criterion.
'Jaccard' Percentage of nonzero coordinates that differ. This option is only valid for the Silhouette criterion.
For detailed information about each distance metric, see pdist.
You can also specify a function for the distance metric using a function handle. The distance function must be of the form d2 = distfun(XI,XJ), where XI is a 1-by-n vector corresponding to a single row of the input matrix X, and XJ is an m2-by-n matrix corresponding to multiple rows of X. distfun must return an m2-by-1 vector of distances d2, whose kth element is the distance between XI and XJ(k,:).
Distance only accepts a function handle if the clustering algorithm clust accepts a function handle as the distance metric. For example, the kmeans clustering algorithm does not accept a function handle as the distance metric. Therefore, if you use the kmeans algorithm and then specify a function handle for Distance, the software errors.
If criterion is 'silhouette', you can also specify Distance as the output vector created by the function pdist.
When clust is 'kmeans' or 'gmdistribution', evalclusters uses the distance metric specified for Distance to cluster the data.
If clust is 'linkage', and Distance is either 'sqEuclidean' or 'Euclidean', then the clustering algorithm uses the Euclidean distance and Ward linkage.
If clust is 'linkage' and Distance is any other metric, then the clustering algorithm uses the specified distance metric and average linkage.
In all other cases, the distance metric specified for Distance must match the distance metric used in the clustering algorithm to obtain meaningful results.
Example: 'Distance','Euclidean'
'empirical' (default) | 'equal'
Prior probabilities for each cluster, specified as the comma-separated pair consisting of 'ClusterPriors' and one of the following.
'empirical' Compute the overall silhouette value for the clustering solution by averaging the silhouette values for all points. Each cluster contributes to the overall silhouette value proportionally to its size.
'equal' Compute the overall silhouette value for the clustering solution by averaging the silhouette values for all points within each cluster, and then averaging those values across all clusters. Each cluster contributes equally to the overall silhouette value, regardless of its size.
Example: 'ClusterPriors','empirical'
B — Number of reference data sets
100 (default) | positive integer value
Number of reference data sets generated from the reference distribution ReferenceDistribution, specified as the comma-separated pair consisting of 'B' and a positive integer value.
Example: 'B',150
ReferenceDistribution — Reference data generation method
'PCA' (default) | 'uniform'
Reference data generation method, specified as the comma-separated pair consisting of 'ReferenceDistributions' and one of the following.
'PCA' Generate reference data from a uniform distribution over a box aligned with the principal components of the data matrix x.
'uniform' Generate reference data uniformly over the range of each feature in the data matrix x.
Example: 'ReferenceDistribution','uniform'
SearchMethod — Method for selecting optimal number of clusters
'globalMaxSE' (default) | 'firstMaxSE'
Method for selecting the optimal number of clusters, specified as the comma-separated pair consisting of 'SearchMethod' and one of the following.
'globalMaxSE'
Evaluate each proposed number of clusters in KList and select the smallest number of clusters satisfying
\text{Gap}\left(K\right)\ge GAPMAX-\text{SE}\left(GAPMAX\right),
where K is the number of clusters, Gap(K) is the gap value for the clustering solution with K clusters, GAPMAX is the largest gap value, and SE(GAPMAX) is the standard error corresponding to the largest gap value.
'firstMaxSE'
\text{Gap}\left(K\right)\ge \text{Gap}\left(K+1\right)-\text{SE}\left(K+1\right),
where K is the number of clusters, Gap(K) is the gap value for the clustering solution with K clusters, and SE(K + 1) is the standard error of the clustering solution with K + 1 clusters.
Example: 'SearchMethod','globalMaxSE'
eva — Clustering evaluation data
clustering evaluation object
Clustering evaluation data, returned as a clustering evaluation object.
CalinskiHarabaszEvaluation | SilhouetteEvaluation | GapEvaluation | DaviesBouldinEvaluation |
Other Games - Course Hero
Microeconomics/Game Theory/Other Games
The Prisoner's Dilemma, in which both players have a dominant strategy that leads to a worse outcome than if they had chosen the opposite strategy, is one type of game. For a different example, consider Quik Mart and Shop & Save, two convenience stores located down the street from one another in a small city. Both are considering adding an ice cream shop to their stores to draw in more customers. Adding the section may increase sales, but it will also lead to higher costs of running the store. The payoffs that each store would receive can be shown in a payoff matrix displaying their daily profits if they build or do not build the additions to their stores.
In this game, only Quik Mart has a dominant strategy-the best response for a player regardless of what the other player chooses, shown by the green check marks. Shop & Save does not have dominant strategy because its best option depends on what Quik Mart decides to do.
The best response analysis (in which a payoff maximizes a player's well-being, given the other player's choice) can be used to solve this game and look for a Nash equilibrium, the outcome in which no participant can gain by a change of strategy if the strategies of others remain unchanged.
If Quik Mart believes that Shop & Save is not going to build an addition, it is better off building an addition, because the payoff for building ($9,000) is greater than the payoff for not building ($7,000). If Quik Mart believes that Shop & Save is going to build the addition, it is still better off building an addition (
\$10\text{,}000 \gt \$9\text{,}000
). Here, building the addition is a dominant strategy for Quik Mart, because it is the best strategy regardless of what Shop & Save chooses to do.
If Shop & Save believes that Quik Mart will decide not to build the addition, Shop & Save would be better off building its own addition (
\$7\text{,}000 \gt \$5\text{,}000
). However, if Shop & Save believes that Quik Mart will build the addition, its best option is to not build, because its payoff for building the addition ($6,000) is less than the payoff for not building ($7,000). Shop & Save does not have a dominant strategy because its best option depends on what Quik Mart chooses.
There is a Nash equilibrium in this case. We know that Quik Mart will follow its dominant strategy and build the ice cream shop addition. Given that decision, Shop & Save will choose its best option and not build an addition to its store. This is a Nash equilibrium because each player is choosing its best option, given the choice made by the other. However, this game is different from the Prisoner's Dilemma because only one player has a dominant strategy.
Some games will not have a Nash equilibrium, while others may have more than one. Consider the following example. Rachel and Monica are roommates. Rachel loves spending time with Monica, but Monica would rather be alone. Each is deciding what she is going to do this evening, and the payoff matrix represents the level of happiness the roommates receive from going to dinner or a movie, conditional on what the other chooses.
In this game, what each player does depends on what the other chooses. There is no Nash equilibrium, in which players have the best outcome even if the other player changes their mind. Neither player has a dominant strategy, in which they have a best choice regardless of the other player's choice. There is no outcome that leaves both players happy.
If Rachel believes that Monica is going to dinner, she should also choose to go to dinner because her payoff is larger (
8 \gt 6
). If Rachel believes that Monica is going to a movie, she should also choose to go to a movie (
3 \gt 2
). Rachel does not have a dominant strategy because her best option changes when Monica changes her decision.
If Monica believes that Rachel will go to dinner, her best option is to go to a movie (
5 \gt -1
). If she thinks that Rachel will opt for a movie, Monica should choose to go out to dinner (
4 \gt -8)
. Monica has no dominant strategy either. Her best choice depends on the option that Rachel chooses.
In this game, there is no Nash equilibrium. No matter what Monica and Rachel choose, one of them will always want to alter her choice. For example, if both choose dinner, Monica will wish she had opted to go to a movie. However, if Monica opted for a movie when Rachel opted for dinner, Rachel will wish she had chosen to go to dinner. There is no outcome in which both players are happy, so there is no Nash equilibrium.
<The Prisoner's Dilemma>Repeated Prisoner's Dilemma |
\mathrm{A}=\left(\begin{array}{cc}3& 5\\ 7& 9\end{array}\right)
is written as A = P + Q, where P is a symmetric matrix and Q is skew symmetric matrix, then write the matrix P. VIEW SOLUTION
\stackrel{\to }{\mathrm{a}}, \stackrel{\to }{\mathrm{b}}, \stackrel{\to }{\mathrm{c}}
\stackrel{\to }{\mathrm{a}} + \stackrel{\to }{\mathrm{b}} + \stackrel{\to }{\mathrm{c}}=\stackrel{\to }{\mathrm{0}},
\stackrel{\to }{\mathrm{a}}·\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{b}}·\stackrel{\to }{\mathrm{c}}+\stackrel{\to }{\mathrm{c}}·\stackrel{\to }{\mathrm{a}}.
{\left|\stackrel{\to }{\mathrm{a}}×\stackrel{\to }{\mathrm{b}}\right|}^{2}+{\left|\stackrel{\to }{\mathrm{a}}·\stackrel{\to }{\mathrm{b}}\right|}^{2}=400
\left|\stackrel{\to }{\mathrm{a}}\right|=5,
\left|\stackrel{\to }{\mathrm{b}}\right|.
Write the equation of a plane which is at a distance of
5\sqrt{3}
units from origin and the normal to which is equally inclined to coordinate axes. VIEW SOLUTION
\left(2 1 3\right) \left(\begin{array}{ccc}-1& 0& -1\\ -1& 1& 0\\ 0& 1& 1\end{array}\right)\left(\begin{array}{c}1\\ 0\\ -1\end{array}\right)=\mathrm{A}
, then write the order of matrix A. VIEW SOLUTION
\left|\begin{array}{ccc}x& \mathrm{sin} \mathrm{\theta }& \mathrm{cos} \mathrm{\theta }\\ -\mathrm{sin} \mathrm{\theta }& -x& 1\\ \mathrm{cos} \mathrm{\theta }& 1& x\end{array}\right|=8
f\left(x\right)=\left\{\begin{array}{ccc}{x}^{2}+3x+a& ,& x⩽1\\ bx+2& ,& x>1\end{array}\right\
is differentiable at x = 1. VIEW SOLUTION
{\mathrm{tan}}^{-1}\left(\frac{\sqrt{1+{x}^{2}}-1}{x}\right) \mathrm{w}.\mathrm{r}.\mathrm{t}. {\mathrm{sin}}^{-1}\frac{2x}{1+{x}^{2}},
if x ∈ (–1, 1)
If x = sin t and y = sin pt, prove that
\left(1-{x}^{2}\right)\frac{{d}^{2}y}{d{x}^{2}}-x\frac{dy}{dx}+{p}^{2}y=0
Find the angle of intersection of the curves
{y}^{2}=4ax \mathrm{and} {x}^{2}=4by
\underset{0}{\overset{\pi }{\int }}\frac{x}{1+\mathrm{sin} \alpha \mathrm{sin} x}dx
\int \left(2x+5\right)\sqrt{10-4x-3{x}^{2}}dx
\int \frac{\left({x}^{2}+1\right)\left({x}^{2}+4\right)}{\left({x}^{2}+3\right)\left({x}^{2}-5\right)}dx
\int \frac{x {\mathrm{sin}}^{-1}x}{\sqrt{1-{x}^{2}}}dx
{y}^{2}dx+\left({x}^{2}-xy+{y}^{2}\right)dy=0
\left({\mathrm{cot}}^{-1}y+x\right) dy=\left(1+{y}^{2}\right) dx
\stackrel{\to }{a}×\stackrel{\to }{b}=\stackrel{\to }{c}×\stackrel{\to }{d} \mathrm{and} \stackrel{\to }{a}×\stackrel{\to }{c}=\stackrel{\to }{b}×\stackrel{\to }{d}
\stackrel{\to }{a}-\stackrel{\to }{d}
\stackrel{\to }{b}-\stackrel{\to }{c}
\stackrel{\to }{a}\ne \stackrel{\to }{d} \mathrm{and} \stackrel{\to }{b}\ne \stackrel{\to }{c}
Prove that the line through A(0, –1, –1) and B(4, 5, 1) intersects the line through C(3, 9, 4) and D(–4, 4, 4). VIEW SOLUTION
A box has 20 pens of which 2 are defective. Calculate the probability that out of 5 pens drawn one by one with replacement, at most 2 are defective.
\mathrm{P}\left(\mathrm{X} = x\right) = \left\{\begin{array}{lll}kx& ,& \mathrm{if} x = 0 \mathrm{or} 1\\ 2 kx& ,& \mathrm{if} x = 2\\ k\left(5-x\right)& ,& \mathrm{if} x = 3 \mathrm{or} 4\\ 0& ,& \mathrm{if} x > 4\end{array}\right\
where k is a positive constant. Find the value of k. Also find the probability that you will get admission in (i) exactly one college (ii) at most 2 colleges (iii) at least 2 colleges. VIEW SOLUTION
{\mathrm{cot}}^{-1}\frac{\sqrt{1+\mathrm{sin}x}+\sqrt{1-\mathrm{sin}x}}{\sqrt{1+\mathrm{sin}x}-\sqrt{1-\mathrm{sin}x}}=\frac{\mathrm{x}}{2}, 0<x<\frac{\mathrm{\pi }}{2}
{\mathrm{tan}}^{-1} \left(\frac{x-2}{x-1}\right)+{\mathrm{tan}}^{-1} \left(\frac{x+2}{x+1}\right)=\frac{\mathrm{\pi }}{4}
A coaching institute of English (subject) conducts classes in two batches I and II and fees for rich and poor children are different. In batch I, it has 20 poor and 5 rich children and total monthly collection is Rs 9,000, whereas in batch II, it has 5 poor and 25 rich children and total monthly collection is Rs 26,000. Using matrix method, find monthly fees paid by each child of two types. What values the coaching institute is inculcating in the society? VIEW SOLUTION
Using integration find the area of the region bounded by the curves
y=\sqrt{4-{x}^{2}}, {x}^{2}+{y}^{2}-4x=0
and the x-axis. VIEW SOLUTION
Find the equation of the plane which contains the line of intersection of the planes
x+2y+3z-4=0 \mathrm{and} 2x+y-z+5=0
and whose x-intercept is twice its z-intercept. VIEW SOLUTION
Bag A contains 3 red and 5 black balls, while bag B contains 4 red and 4 black balls. Two balls are transferred at random from bag A to bag B and then a ball is drawn from bag B at random. If the ball drawn from bag B is found to be red find the probability that two red balls were transferred from A to B. VIEW SOLUTION
In order to supplement daily diet, a person wishes to take X and Y tablets. The contents (in milligrams per tablet) of iron, calcium and vitamins in X and Y are given as below :
Tablets Iron Calcium Vitamin
The person needs to supplement at least 18 milligrams of iron, 21 milligrams of calcium and 16 milligrams of vitamins. The price of each tablet of X and Y is Rs 2 and Rs 1 respectively. How many tablets of each type should the person take in order to satisfy the above requirement at the minimum cost? Make an LPP and solve graphically. VIEW SOLUTION
If f, g : R → R be two functions defined as f(x) = |x| + x and g(x) = |x| –
x,\forall x\in \mathrm{R}
. Then find fog and gof. Hence find fog(–3), fog(5) and gof (–2). VIEW SOLUTION
If a, b and c are all non-zero and
\left|\begin{array}{ccc}1+\mathrm{a}& 1& 1\\ 1& 1+\mathrm{b}& 1\\ 1& 1& 1+\mathrm{c}\end{array}\right|=0,
\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{b}}+\frac{1}{\mathrm{c}}+1=0
\mathrm{A}=\left(\begin{array}{ccc}\mathrm{cos}\alpha & -\mathrm{sin}\alpha & 0\\ \mathrm{sin}\alpha & \mathrm{cos}\alpha & 0\\ 0& 0& 1\end{array}\right),
find adj·A and verify that A(adj·A) = (adj·A)A = |A| I3. VIEW SOLUTION
The sum of the surface areas of a cuboid with sides x, 2x and
\frac{x}{3}
and a sphere is given to be constant. Prove that the sum of their volumes is minimum, if x is equal to three times the radius of sphere. Also find the minimum value of the sum of their volumes.
Find the equation of tangents to the curve y = cos(x + y), –2π ≤ x ≤ 2π that are parallel to the line x + 2y = 0. VIEW SOLUTION |
statistical approach for modeling the relationship between a scalar dependent variable and one or more explanatory variables
Linear regression is a way to look at how something changes when other things change using math. A linear regression uses a dependent variable and one or more explanatory variables to create a straight line. This straight line is known as a "line of regression".
Linear regression was the first of many ways of performing regression analysis. This is because models which depend linearly on their unknown parameters are easier to fit than models which are non-linearly related to their parameters. Another advantage of linear regression is that the statistical properties of the resulting estimators are easier to determine.
{\displaystyle {\hat {y}}} |
Speeded up robust features - Wikipedia
Robust local feature detector
For other uses, see SURF (disambiguation).
In computer vision, speeded up robust features (SURF) is a patented local feature detector and descriptor. It can be used for tasks such as object recognition, image registration, classification, or 3D reconstruction. It is partly inspired by the scale-invariant feature transform (SIFT) descriptor. The standard version of SURF is several times faster than SIFT and claimed by its authors to be more robust against different image transformations than SIFT.
SURF descriptors have been used to locate and recognize objects, people or faces, to reconstruct 3D scenes, to track objects and to extract points of interest.
SURF was first published by Herbert Bay, Tinne Tuytelaars, and Luc Van Gool, and presented at the 2006 European Conference on Computer Vision. An application of the algorithm is patented in the United States.[1] An "upright" version of SURF (called U-SURF) is not invariant to image rotation and therefore faster to compute and better suited for application where the camera remains more or less horizontal.
The image is transformed into coordinates, using the multi-resolution pyramid technique, to copy the original image with Pyramidal Gaussian or Laplacian Pyramid shape to obtain an image with the same size but with reduced bandwidth. This achieves a special blurring effect on the original image, called Scale-Space and ensures that the points of interest are scale invariant.
1 Algorithm and features
1.1.1 Scale-space representation and location of points of interest
1.2.1 Orientation assignment
1.2.2 Descriptor based on the sum of Haar wavelet responses
Algorithm and features[edit]
The SURF algorithm is based on the same principles and steps as SIFT; but details in each step are different. The algorithm has three main parts: interest point detection, local neighborhood description, and matching.
SURF uses square-shaped filters as an approximation of Gaussian smoothing. (The SIFT approach uses cascaded filters to detect scale-invariant characteristic points, where the difference of Gaussians (DoG) is calculated on rescaled images progressively.) Filtering the image with a square is much faster if the integral image is used:
{\displaystyle S(x,y)=\sum _{i=0}^{x}\sum _{j=0}^{y}I(i,j)}
The sum of the original image within a rectangle can be evaluated quickly using the integral image, requiring evaluations at the rectangle's four corners.
SURF uses a blob detector based on the Hessian matrix to find points of interest. The determinant of the Hessian matrix is used as a measure of local change around the point and points are chosen where this determinant is maximal. In contrast to the Hessian-Laplacian detector by Mikolajczyk and Schmid, SURF also uses the determinant of the Hessian for selecting the scale, as is also done by Lindeberg. Given a point p=(x, y) in an image I, the Hessian matrix H(p, σ) at point p and scale σ, is:
{\displaystyle H(p,\sigma )={\begin{pmatrix}L_{xx}(p,\sigma )&L_{xy}(p,\sigma )\\L_{yx}(p,\sigma )&L_{yy}(p,\sigma )\end{pmatrix}}}
{\displaystyle L_{xx}(p,\sigma )}
etc. is the convolution of the second-order derivative of gaussian with the image
{\displaystyle I(x,y)}
{\displaystyle p}
The box filter of size 9×9 is an approximation of a Gaussian with σ=1.2 and represents the lowest level (highest spatial resolution) for blob-response maps.
Scale-space representation and location of points of interest[edit]
Interest points can be found at different scales, partly because the search for correspondences often requires comparison images where they are seen at different scales. In other feature detection algorithms, the scale space is usually realized as an image pyramid. Images are repeatedly smoothed with a Gaussian filter, then they are subsampled to get the next higher level of the pyramid. Therefore, several floors or stairs with various measures of the masks are calculated:
{\displaystyle \sigma _{\text{approx}}={\text{current filter size}}\times \left({\frac {\text{base filter scale}}{\text{base filter size}}}\right)}
The scale space is divided into a number of octaves, where an octave refers to a series of response maps of covering a doubling of scale. In SURF, the lowest level of the scale space is obtained from the output of the 9×9 filters.
Hence, unlike previous methods, scale spaces in SURF are implemented by applying box filters of different sizes. Accordingly, the scale space is analyzed by up-scaling the filter size rather than iteratively reducing the image size. The output of the above 9×9 filter is considered as the initial scale layer at scale s =1.2 (corresponding to Gaussian derivatives with σ = 1.2). The following layers are obtained by filtering the image with gradually bigger masks, taking into account the discrete nature of integral images and the specific filter structure. This results in filters of size 9×9, 15×15, 21×21, 27×27,.... Non-maximum suppression in a 3×3×3 neighborhood is applied to localize interest points in the image and over scales. The maxima of the determinant of the Hessian matrix are then interpolated in scale and image space with the method proposed by Brown, et al. Scale space interpolation is especially important in this case, as the difference in scale between the first layers of every octave is relatively large.
The goal of a descriptor is to provide a unique and robust description of an image feature, e.g., by describing the intensity distribution of the pixels within the neighbourhood of the point of interest. Most descriptors are thus computed in a local manner, hence a description is obtained for every point of interest identified previously.
The dimensionality of the descriptor has direct impact on both its computational complexity and point-matching robustness/accuracy. A short descriptor may be more robust against appearance variations, but may not offer sufficient discrimination and thus give too many false positives.
The first step consists of fixing a reproducible orientation based on information from a circular region around the interest point. Then we construct a square region aligned to the selected orientation, and extract the SURF descriptor from it.
Orientation assignment[edit]
In order to achieve rotational invariance, the orientation of the point of interest needs to be found. The Haar wavelet responses in both x- and y-directions within a circular neighbourhood of radius
{\displaystyle 6s}
around the point of interest are computed, where
{\displaystyle s}
is the scale at which the point of interest was detected. The obtained responses are weighted by a Gaussian function centered at the point of interest, then plotted as points in a two-dimensional space, with the horizontal response in the abscissa and the vertical response in the ordinate. The dominant orientation is estimated by calculating the sum of all responses within a sliding orientation window of size π/3. The horizontal and vertical responses within the window are summed. The two summed responses then yield a local orientation vector. The longest such vector overall defines the orientation of the point of interest. The size of the sliding window is a parameter that has to be chosen carefully to achieve a desired balance between robustness and angular resolution.
Descriptor based on the sum of Haar wavelet responses[edit]
To describe the region around the point, a square region is extracted, centered on the interest point and oriented along the orientation as selected above. The size of this window is 20s.
The interest region is split into smaller 4x4 square sub-regions, and for each one, the Haar wavelet responses are extracted at 5x5 regularly spaced sample points. The responses are weighted with a Gaussian (to offer more robustness for deformations, noise and translation).
By comparing the descriptors obtained from different images, matching pairs can be found.
Local energy-based shape histogram (LESH)
^ US 2009238460, Ryuji Funayama, Hiromichi Yanagihara, Luc Van Gool, Tinne Tuytelaars, Herbert Bay, "ROBUST INTEREST POINT DETECTOR AND DESCRIPTOR", published 2009-09-24
Herbert Bay, Andreas Ess, Tinne Tuytelaars, and Luc Van Gool, "Speeded Up Robust Features", ETH Zurich, Katholieke Universiteit Leuven
Andrea Maricela Plaza Cordero,Jorge Luis Zambrano Martínez, " Estudio y Selección de las Técnicas SIFT, SURF y ASIFT de Reconocimiento de Imágenes para el Diseño de un Prototipo en Dispositivos Móviles" , 15º Concurso de Trabajos Estudiantiles, EST 2012
Herbert Bay, Andreas Ess, Tinne Tuytelaars, Luc Van Gool "SURF: Speeded Up Robust Features", Computer Vision and Image Understanding (CVIU), Vol. 110, No. 3, pp. 346–359, 2008
Christopher Evans "Notes on the OpenSURF Library", MSc Computer Science, University of Bristol; source code and documentation archived here
J an Knopp, Mukta Prasad, Gert Willems, Radu Timofte, and Luc Van Gool, "Hough Transform and 3D SURF for Robust Three Dimensional Classification", European Conference on Computer Vision (ECCV), 2010
Website of SURF: Speeded Up Robust Features
First publication of Speeded Up Robust Features (2006)
Revised publication of SURF (2008)
Retrieved from "https://en.wikipedia.org/w/index.php?title=Speeded_up_robust_features&oldid=1053072748" |
On a Dual Model with Barrier Strategy
2012 On a Dual Model with Barrier Strategy
We consider the dual of the generalized Erlang
\left(n\right)
risk model with a barrier dividend strategy. We derive integro-differential equations with boundary conditions satisfied by the expectation of the sum of discounted dividends until ruin and the moment-generating function of the discounted dividend payments until ruin, respectively. The results are illustrated by several examples.
Yuzhen Wen. Chuancun Yin. "On a Dual Model with Barrier Strategy." J. Appl. Math. 2012 1 - 13, 2012. https://doi.org/10.1155/2012/343794
Yuzhen Wen, Chuancun Yin "On a Dual Model with Barrier Strategy," Journal of Applied Mathematics, J. Appl. Math. 2012(none), 1-13, (2012) |
Solve optimization problem or equation problem - MATLAB solve - MathWorks India
{x}^{2}+{y}^{2}\le 4
-5\le x,y\le 5
\underset{x}{\mathrm{min}}\left(-3{x}_{1}-2{x}_{2}-{x}_{3}\right)\phantom{\rule{0.2777777777777778em}{0ex}}subject\phantom{\rule{0.2777777777777778em}{0ex}}to\left\{\begin{array}{l}{x}_{3}\phantom{\rule{0.2777777777777778em}{0ex}}binary\\ {x}_{1},{x}_{2}\ge 0\\ {x}_{1}+{x}_{2}+{x}_{3}\le 7\\ 4{x}_{1}+2{x}_{2}+{x}_{3}=12\end{array}
\begin{array}{l}\mathrm{exp}\left(-\mathrm{exp}\left(-\left({x}_{1}+{x}_{2}\right)\right)\right)={x}_{2}\left(1+{x}_{1}^{2}\right)\\ {x}_{1}\mathrm{cos}\left({x}_{2}\right)+{x}_{2}\mathrm{sin}\left({x}_{1}\right)=\frac{1}{2}\end{array} |
Measurement of Energy | EGEE 102: Energy Conservation and Environmental Protection
How is energy measured? It is measured in various units by various industries or countries in much the same way as the value of goods is expressed in Dollars in the U.S. and Yen in Japan and Pounds in Britain.
The table below identifies different units for measuring energy. A lot of it also has some historical context. Our early studies of energy involved heating things up, so we names units based on how hard it was to heat things. Makes sense, right? Now we pass electrical energy to operate many devices, so now we use units that "better" capture this process.
Different Units for Measuring Energy
A unit of energy equal to the amount of energy needed to raise the temperature of one pound of water by one degree Fahrenheit. Equivalent to energy found in the tip of a match stick. Heating and Cooling industries 1 BTU = 1055 Joules (J)
Calorie or small calorie (calorie)
The amount of energy needed to raise the temperature of one gram of water by one degree Celsius. Science and Engineering 1 calorie = 0.003969 BTUs
Food Calorie, Kilocalorie or large calorie (Cal, kcal, Calorie)
The amount of energy needed to raise the temperature of one kilogram of water one degree Celsius. The food calorie is often used when measuring the energy content of food. Nutrition 1 Cal = 1000 cal, 4,187 J or 3.969 BTUs
It is a smaller quantity of energy than calorie and much smaller than a BTU. Science and Engineering 1 Joule = 0.2388 calories and 0.0009481 BTUs
An amount of energy from the steady production or consumption of one kilowatt of power for a period of one hour. Electrical fields 1 kWh = 3,413 BTUs or 3,600,000 J
A unit describing the energy contained in natural gas. Home heating appliances 1 therm = 100,000 BTUs
When writing BTUs, one uses a base of “10” raised to a particular exponent.
10,000\text{ }BTUs\text{ }=\text{ }{10}^{4}\quad BTUs
100,000\text{ }BTUs\text{ }=\text{ }{10}^{5}\quad BTUs
1,000,000\text{ }BTUs\text{ }=\text{ }{10}^{6}\quad BTUs
More specific notation involves the following:
10,000\text{ }BTUs\text{ }=\text{ }1\text{ }x\text{ }{10}^{4}\quad BTUs
100,000\text{ }BTUs\text{ }=\text{ }1\text{ }x\text{ }{10}^{5}\quad BTUs
1,000,000\text{ }BTUs\text{ }=1\text{ }x\text{ }{10}^{6\quad }BTUs
To express measurements greater than those with a base of 10, you would do the following:
50,000\text{ }BTUs\text{ }=\text{ }5\text{ }x\text{ }{10}^{4}\quad BTUs
700,000\text{ }BTUs\text{ }=\text{ }7\text{ }x\text{ }{10}^{5}\quad BTUs
9,000,000\text{ }BTUs\text{ }=\text{ }9\text{ }x\text{ }{10}^{6}\quad BTUs
Here is a fun way to understand your energy use
Prof. Bruce Logan of Penn State published a fascinating way to view your energy and climate impact. Using what you learned in this section you can start to piece together just how much energy each of us uses to maintain our busy lifestyles.
The premise of this approach is to define (another!) unit of energy, but one with a bit more meaning. The daily energy unit, D. We are all supposed to eat about 2000 food Calories a day to survive. So, let’s set this amount of energy to equal 1 D. Now, how many Ds does the typical U.S. home each day (normally in KWh) or operate a car (normally joules or BTUs )? This method of comparing energy consumption allows us to better understand the scale of our energy habits (which might be shocking!) and tell you how many big mac-powered humans it would take to take to do what your car does…
Here are a few examples he shows to give you an idea.
Food for 1 day = 1 D
Running a single 100 W light bulb all day = 1.03 D
Average daily electricity use for a US house = 13 D
1 gallon of gasoline used in an average car (goes 18 miles) = 15.2 D
Natural gas for daily heating a US house = 31 D
Once we tally up all the energy it takes to fuel our lifestyle (professional + personal uses), each person consumed about 101 D of energy! (remember this is daily) For comparison, a Swiss citizen consumes about 54 D. Check out his website for more comparisons.
‹ Activity: Day-to-Day Conversion Devices up Sources of Energy › |
String harmonic - Wikipedia
(Redirected from Artificial harmonic)
Find sources: "String harmonic" – news · newspapers · books · scholar · JSTOR (August 2016) (Learn how and when to remove this template message)
Playing a harmonic on a string. Here, "+7" indicates that the string is held down at the position for raising the pitch by 7 semitones.
Playing a string harmonic (a flageolet) is a string instrument technique that uses the nodes of natural harmonics of a musical string to isolate overtones. Playing string harmonics produces high pitched tones, often compared in timbre to a whistle or flute.[1][2] Overtones can be isolated "by lightly touching the string with the finger instead of pressing it down" against the fingerboard (without stopping).[3]
3 Artificial harmonics
4.1 Pinch harmonics
4.2 Tapped harmonics
4.3 String harmonics driven by a magnetic field
Overtones[edit]
Main article: Harmonic
When a string is plucked or bowed normally, the ear hears the fundamental frequency most prominently, but the overall sound is also colored by the presence of various overtones (frequencies greater than the fundamental frequency). The fundamental frequency and its overtones are perceived by the listener as a single note; however, different combinations of overtones give rise to noticeably different overall tones (see timbre).[4] A harmonic overtone has evenly spaced nodes along the string, where the string does not move from its resting position.
Table of harmonics, indicating in colors on which positions the same overtones occur
The nodes of natural harmonics are located at the following points along the string:
2 octave octave (P8) 1,200.0 0.0 1⁄2 Play (help·info)
3 just perfect fifth P8 + just perfect fifth (P5) 1,902.0 702.0 1⁄3, 2⁄3 Play (help·info)
4 just perfect fourth 2P8 2,400.0 0.0 1⁄4, 3⁄4 Play (help·info)
5 just major third 2P8 + just major third (M3) 2,786.3 386.3 1⁄5 to 4⁄5 Play (help·info)
6 just minor third 2P8 + P5 3,102.0 702.0 1⁄6, 5⁄6
7 septimal minor third 2P8 + septimal minor seventh (m7) 3,368.8 968.8 1⁄7 to 6⁄7 Play (help·info)
8 septimal major second 3P8 3,600.0 0.0 1⁄8, 3⁄8, 5⁄8, 7⁄8
9 Pythagorean major second 3P8 + Pythagorean major second (M2) 3,803.9 203.9 1⁄9, 2⁄9, 4⁄9, 5⁄9, 7⁄9, 8⁄9 Play (help·info)
10 just minor whole tone 3P8 + just M3 3,986.3 386.3 1⁄10, 3⁄10, 7⁄10, 9⁄10
11 greater undecimal neutral second 3P8 + lesser undecimal tritone 4,151.3 551.3 1⁄11 to 10⁄11 Play (help·info)
12 lesser undecimal neutral second 3P8 + P5 4,302.0 702.0 1⁄12, 5⁄12, 7⁄12, 11⁄12
13 tridecimal 2/3-tone 3P8 + tridecimal neutral sixth (n6) 4,440.5 840.5 1⁄13 to 12⁄13 Play (help·info)
14 2/3-tone 3P8 + P5 + septimal minor third (m3) 4,568.8 968.8 1⁄14, 3⁄14, 5⁄14, 9⁄14, 11⁄14, 13⁄14
15 septimal (or major) diatonic semitone 3P8 + just major seventh (M7) 4,688.3 1,088.3 1⁄15, 2⁄15, 4⁄15, 7⁄15, 8⁄15, 11⁄15, 13⁄15, 14⁄15 Play (help·info)
16 just (or minor) diatonic semitone 4P8 4,800.0 0.0 1⁄16, 3⁄16, 5⁄16, 7⁄16, 9⁄16, 11⁄16, 13⁄16, 15⁄16
Above, the length fraction is the point, with respect to the length of the whole string, the string is lightly touched. It is expressed as a fraction n/m, where m is the mode (2 through 16 are given above), and n the node number. The node number for a given mode can be any integer from 1 to m − 1. However, certain nodes of higher harmonics are coincident with nodes of lower harmonics, and the lower sounds overpower the higher ones. For example, mode number 4 can be fingered at nodes 1 and 3; it will occur at node 2 but will not be heard over the stronger first harmonic. Ineffective nodes to finger are not listed above.
The fret number, which shows the position of the node in terms of half tones (or frets on a fretted instrument) then is given by:
{\displaystyle F=\log _{s}{\frac {m}{m-n}}}
With s equal to the twelfth root of two, notated s because it's the first letter of the word "semitone".
Artificial harmonics on a G fundamental, as written (below) and as sounding (top). The round note (below) is pressed with one finger, and the square note is lightly touched with another one. Play (help·info)
Natural versus artificial harmonic
When a string is only lightly pressed by one finger (that is, isolating overtones of the open string), the resulting harmonics are called natural harmonics. However, when a string is held down on the neck in addition to being lightly pressed on a node, the resulting harmonics are called artificial harmonics. In this case, as the total length of the string is shortened, the fundamental frequency is raised, and the positions of the nodes shift accordingly (that is, by the same number of frets), thereby raising the frequency of the overtone by the same interval as the fundamental frequency.
Artificial harmonics are produced by stopping the string with the first or second finger, and thus making an artificial 'nut,' and then slightly pressing the node with the fourth finger. By this means harmonics in perfect intonation can be produced in all scales.
— Grove's Dictionary of Music and Musicians (1879)[5]
Artificial harmonics are more difficult to play than natural harmonics, but they are not limited to the overtone series of the open strings, meaning they have much greater flexibility to play chromatic passages. Unlike natural harmonics, they can be played with vibrato.[6]
This technique, like natural harmonics, works by canceling out the fundamental tone and one or more partial tones by deadening their modes of vibration. It is traditionally notated using two or three simultaneous noteheads in one staff: a normal notehead for the position of the firmly held finger, a square notehead for the position of the lightly pressed finger, and sometimes, a small notehead for the resulting pitch.[7]
The most commonly used artificial harmonic, due to its relatively easy and natural fingering, is that in which, "the fourth finger lightly touches the nodal point a perfect fourth above the first finger. (Resulting harmonic sound: two octaves above the first finger or new fundamental.),"[8] followed by the artificial harmonic produced when, "the fourth finger lightly touches the nodal point a perfect fifth above the first finger (Resulting harmonic sound: a twelfth above the first finger or new fundamental.),"[8] and, "the third finger lightly touches the nodal point a major third above the first finger. (Resulting harmonic sound: two octaves and a major third above the first finger or new fundamental.)"[8][9]
The fundamental and the double- and triple-frequency overtones of a guitar string.
There are a few harmonic techniques unique to guitar.
Pinch harmonics[edit]
Pinch harmonics performed on an acoustic guitar
Example of pinch harmonic (0:02)
Pinch harmonic example on the 3rd fret of the G string
A pinch harmonic (also known as squelch picking, pick harmonic or squealy) is a guitar technique to achieve artificial harmonics in which the player's thumb or index finger on the picking hand slightly catches the string after it is picked,[10] canceling (silencing) the fundamental frequency of the string, and letting one of the overtones dominate. This results in a high-pitched sound which is particularly discernible on an electrically amplified guitar as a "squeal".
Tapped harmonics[edit]
Tapped harmonics were popularized by Eddie Van Halen. This technique is an extension of the tapping technique. The note is fretted as usual, but instead of striking the string the excitation energy required to sound the note is achieved by tapping at a harmonic nodal point. The tapping finger bounces lightly on and off the fret. The open string technique can be extended to artificial harmonics. For instance, for an octave harmonic (12-fret nodal point) press at the third fret, and tap the fifteenth fret, as 12 + 3 = 15.
String harmonics driven by a magnetic field[edit]
This technique is used by effect devices producing a magnetic field that can agitate fundamentals and harmonics of steel strings. There are harmonic mode switches as provided by newer versions of the EBow and by guitar build in sustainers like the Fernandes Sustainer and the Moog Guitar. Harmonics control by harmonic mode switching and by the playing technique is applied by the Guitar Resonator where harmonics can be alternated by changing the string driver position at the fretboard while playing.
^ Kamien, Roger (2008). Music: An Appreciation, p.13. Sixth "brief" edition. McGraw Hill. ISBN 978-0-07-340134-8.
^ Palisca, Claude V.; ed. (1996). Norton Anthology of Western Music, Volume 1: Ancient to Baroque, glossary, p.601. Third edition. W. W. Norton. ISBN 0-393-96906-1.
^ "The Physics of Everyday Stuff - The Guitar". www.bsharp.org. Retrieved 2015-08-03.
^ Grove, George (1879). A Dictionary of Music and Musicians (A.D. 1450-1889), Vol. 2, p.665. Macmillan. [ISBN unspecified].
^ Sevsay, Ertuğrul (2013). The Cambridge Guide to Orchestration, unpaginated. Cambridge. ISBN 9781107067486.
^ Potter, Louis (1995). The Art of Cello Playing, p.194. Alfred Music. ISBN 9781457400353.
^ a b c Grimson, Samuel B. and Forsyth, Cecil (1920). Modern Violin-Playing, p.79-80. New York: H. W. Gray (Novello). [ISBN unspecified].
^ Hurwitz (2006), p.87.
^ "Guitar Techniques: Pick Harmonic", XtremeMusician.com. Accessed: December 19, 2017.
Guitar techniques and strums
Ska stroke
Retrieved from "https://en.wikipedia.org/w/index.php?title=String_harmonic&oldid=1064744679#Artificial_harmonics" |
A boat traveling due north sees another boat traveling due east cross
800
meters directly in front of it. The first boat is traveling at
4
meters per second and the second is traveling at
3
Find the angle (from due north) that a person on the first boat needs to angle his telescope to watch the second boat. Express your answer in terms of
t
, the number of seconds since the second boat was due north of the first.
Diagram (a):
Write an equation with
θ
t
. Use sine, cosine, or tangent.
If the boat that is traveling north wants to catch the boat that is traveling east, what heading will the boat need to take and how long will it take to catch up to the eastbound boat?
Draw a new diagram. The first boat will now travel on the diagonal.
Diagram (b):
Write equations with
θ
t
. Use sine, cosine, tangent and/or the Pythagorean theorem. |
I am ashamed to see the date on this & to compare it with that upon yours1 most especially as my excuse for delay is insufficient when I come to look fairly at it although when hazily regarded in the way one sometimes looks at such things it seemed to me to assume larger proportions than it does now when I bring it into focus. I don’t know indeed how to present it, so insignificant does it look without the hazy medium which I can’t give but the whole is that finding one of my reagent bottles empty I wanted to replenish it from my stock at the Hospital2 & regularly forgot to replenish while there till so many days had passed that I thought I must needs give you the whole answer to your other questions as well as that regarding the milk in my reply but unfortunately find I can’t do so till some day next week & so I write now as I am thoroughly ashamed of my delay.3 The part of the slide nearest the label & which seemed to you the best contained small refracting bodies which were either oil or resin. They were soluble in ether but so was the resin also with which the glass was secured I suppose however that they presented much the same appearance before the balsam was put on & therefore think that one may say with confidence that they were oil. In the other part of the slide were several bacteria. They seemed to refract light more strongly than usual but Dr. Ferrier who worked at the subject of low organisms for several months felt almost certain that they were bacteria.4
It is very odd that adding a minute drop of hydrochloric acid of 1 per cent should stop the digestion of albumen while a drop of
\frac{1}{2}
per cent should not. Too strong an acid certainly stops the digestive action of pepsin but the strength of 1 per cent (of ordinary acid = ·318 of real acid) is not very far from that which Brucke found to digest albumen most actively.5 For fibrin the best strength was 2·7 & 2·76 per cent of commercial acid when raised to 4· per cent digestion became much slower. When the acid was diluted digestion became gradually slower to 1·38 & 1·41 & when the dilution had reached ·69 digestion had become considerably slower. The best strength for neutralized white of egg was between 3·77 & 5. For not neutralized albumen 5·46. These numbers are taken from the abstract of Brücke’s paper in Canstatts Jahresbericht & I have converted the amount of absolute ClH in a litre which he gives into commercial acid per cent.6
Commercial pepsin consists of the cells lining the stomach scraped off with a blunt knife & dried & contains a good deal of albuminous material. I don’t think I have ever got it quite dissolved in any of my experiments but I have not yet ascertained even approximately what proportion remains undissolved I hope to be able in a few days to give you the results of the digestion of bones & of urea & shortly after some facts about digestion by the papaw7
I remain my dear Sir | Yours very truly | T Lauder Brunton
1.1 I … delay. 1.11] crossed blue crayon
1.11 The] opening square bracket blue crayon
1.11 The … resin. 1.13] ‘Milk’ added blue crayon
1.14 I … oil. 1.16] ‘(Yes)’ added blue crayon
1.16 In … were bacteria. 1.19] crossed blue crayon
2.1 It is … should not. 2.2] ‘[Probably] leaf injured’ added blue crayon
2.3 (of … actively. 2.5] double scored blue crayon
2.5 For … cent. 2.12] crossed blue crayon
3.6 papaw] underl blue crayon
Top of letter: ‘Casein of Milk | Acid not too strong for digestion’ ink
See Correspondence vol. 22, letter to T. L. Brunton, 11 May 1874.
Brunton was a physician at St Bartholomew’s Hospital, London.
CD had asked Brunton to analyse a microscope slide on which he had smeared the remnants of skimmed milk that had been placed on the glands of Drosera (sundew) for several hours, and to answer other questions concerning the digestive processes of insectivorous plants (see Correspondence vol. 22, letter to T. L. Brunton, 11 May 1874). CD had made the experiment to ascertain whether the digestive secretion would dissolve casein in milk (Insectivorous plants, p. 114).
David Ferrier had published an article on the bacterial species Sarcina ventriculi in 1872 (Ferrier 1872).
Ernst Wilhelm von Brücke. The concept of a ‘real acid’, that is, a hypothetical acid stripped of its water content, was developed in the late eighteenth century by Richard Kirwan (see Taylor 2008 for more on Kirwan’s theory and its reception). Kirwan produced tables for determining the amount of real acid in any concentration of acid, given a known specific gravity (Kirwan 1797).
Brunton obtained his information from the abstract of Brücke 1861 that appeared in Canstatt’s Jahresbericht über die Fortschritte der Gesammten Medicin in allen Ländern 5 (1862): 74–5. ‘Absolute ClH’ probably refers to the amount of chloride of hydrogen in aqueous solution.
Brunton described the digestive action of the ferment papain, derived from the pawpaw fruit (Carica papaya) in Brunton 1885, pp. 777–8. |
Numerically Gorenstein surface singularities are homeomorphic to Gorenstein ones
15 September 2011 Numerically Gorenstein surface singularities are homeomorphic to Gorenstein ones
Patrick Popescu-Pampu1
1Université Lille 1
Duke Math. J. 159(3): 539-559 (15 September 2011). DOI: 10.1215/00127094-1433412
Consider a normal complex analytic surface singularity. It is called Gorenstein if the canonical line bundle is holomorphically trivial in some punctured neighborhood of the singular point and is called numerically Gorenstein if this line bundle is topologically trivial. The second notion depends only on the topological type of the singularity. Laufer proved in 1977 that, given a numerically Gorenstein topological type of singularity, every analytical realization of it is Gorenstein if and only if one has either a Kleinian or a minimally elliptic topological type. The question to know if any numerically Gorenstein topology was realizable by some Gorenstein singularity was left open. We prove that this is indeed the case. Our method is to plumb holomorphically meromorphic
2
-forms obtained by adequate pullbacks of the natural holomorphic symplectic forms on the total spaces of the canonical line bundles of complex curves. More generally, we show that any normal surface singularity is homeomorphic to a
\mathbb{Q}
-Gorenstein singularity whose index is equal to the smallest common denominator of the coefficients of the canonical cycle of the starting singularity.
Patrick Popescu-Pampu. "Numerically Gorenstein surface singularities are homeomorphic to Gorenstein ones." Duke Math. J. 159 (3) 539 - 559, 15 September 2011. https://doi.org/10.1215/00127094-1433412
Secondary: 32S25 , 32S50
Patrick Popescu-Pampu "Numerically Gorenstein surface singularities are homeomorphic to Gorenstein ones," Duke Mathematical Journal, Duke Math. J. 159(3), 539-559, (15 September 2011) |
Backscatter radar target - MATLAB - MathWorks í•œêµ
ShhPattern
SvvPattern
ShvPattern
Backscatter Nonpolarized Signal
Backscatter Polarized Signal
Backscatter radar target
The phased.BackscatterRadarTarget System object™ models the backscattering of a signal from a target. Backscattering is a special case of radar target scattering when the incident and reflected angles are the same. This type of scattering applies to monostatic radar configurations. The radar cross-section determines the backscattering response of a target to an incoming signal. This System object lets you specify an angle-dependent radar cross-section model that covers a range of incident angles.
The phased.BackscatterRadarTarget System object creates a backscattered signal for polarized and nonpolarized signals. While electromagnetic radar signals are polarized, you can often ignore polarization in your simulation and process the signals as scalar signals. To ignore polarization, specify the EnablePolarization property as false. To employ polarization, specify the EnablePolarization property as true.
For nonpolarized signals, you specify the radar cross section as an array of radar cross-section (RCS) values at discrete azimuth and elevation points. The System object interpolates values for incident angles between array points. For polarized signals, you specify the radar scattering matrix using three arrays defined at discrete azimuth and elevation points. These three arrays correspond to the HH, HV, and VV polarization components. The VH component is computed from the conjugate symmetry of the HV component.
For both nonpolarized and polarized signal cases, you can employ one of four Swerling models to generate random fluctuations in the RCS or radar scattering matrix. Choose the model using the Model property. Then, use the SeedSource and Seed properties to control the fluctuations.
true ShhPattern, SvvPattern, and ShvPattern
To model a backscattered radar signal:
Define and set up your radar target. You can set phased.BackscatterRadarTarget System object properties at construction time or leave them to their default values. See Construction. Some properties that you set at construction time can be changed later. These properties are tunable.
To compute the reflected signal, call the step method of phased.BackscatterRadarTarget. The output of the method depends on the properties of the phased.BackscatterRadarTarget System object. You can change tunable properties at any time.
target = phased.BackscatterRadarTarget creates a backscatter radar target System object, target.
target = phased.BackscatterRadarTarget(Name,Value) creates a backscatter radar target System object, target, with each specified property Name set to the specified Value. You can specify additional name and value pair arguments in any order as (Name1,Value1,...,NameN,ValueN).
EnablePolarization — Enable polarized signals
Option to enable processing of polarized signals, specified as false or true. Set this property to true to allow the target to simulate the reflection of polarized radiation. Set this property to false to ignore polarization.
Azimuth angles used to define the angular coordinates of each column of the matrices specified by the RCSPattern, ShhPattern, ShvPattern, or SvvPattern properties. Specify the azimuth angles as a length P vector. P must be greater than two. Angle units are in degrees.
Elevation angles used to define the angular coordinates of each row of the matrices specified by the RCSPattern, ShhPattern, ShvPattern, or SvvPattern properties. Specify the elevation angles as a length Q vector. Q must be greater than two. Angle units are in degrees.
ones(181,361) (default) | Q-by-P real-valued matrix | Q-by-P-by-M real-valued array | 1-by-P real-valued vector | M-by-P real-valued matrix
Radar cross-section (RCS) pattern, specified as a Q-by-P real-valued matrix or a Q-by-P-by-M real-valued array. Q is the length of the vector in the ElevationAngles property. P is the length of the vector in the AzimuthAngles property. M is the number of target patterns. The number of patterns corresponds to the number of signals passed into the step method. You can, however, use a single pattern to model multiple signals reflecting from a single target. Pattern units are square-meters.
You can also specify the pattern as a function only of azimuth for a single elevation. In this case, specify the pattern as either a 1-by-P vector or an M-by-P matrix. Each row is a separate pattern.
This property applies when the EnablePolarization property is false.
ShhPattern — Radar-scattering matrix HH polarization component
ones(181,361) (default) | Q-by-P complex-valued matrix | Q-by-P-by-M complex-valued array | 1-by-P complex-valued vector | M-by-P complex-valued matrix
Radar scattering matrix HH polarization component, specified as a Q-by-P complex-valued matrix or a Q-by-P-by-M complex-valued array. Q is the length of the vector in the ElevationAngles property. P is the length of the vector in the AzimuthAngles property. M is the number of target patterns. The number of patterns corresponds to the number of signals passed into the step method. You can, however, use a single pattern to model multiple signals reflecting from a single target. Scattering matrix units are meters.
You can also specify the pattern as a function only of azimuth for a single elevation. Then, specify the pattern as either a 1-by-P vector or an M-by-P matrix. Each row is a separate pattern.
This property applies when the EnablePolarization property is true.
Example: [1,1;1i,1i]
SvvPattern — Radar scattering matrix VV polarization component
Radar scattering matrix VV polarization component, specified as a Q-by-P complex-valued matrix or a Q-by-P-by-M complex-valued array. Q is the length of the vector in the ElevationAngles property. P is the length of the vector in the AzimuthAngles property. M is the number of target patterns. The number of patterns corresponds to the number of signals passed into the step method. You can, however, use a single pattern to model multiple signals reflecting from a single target. Scattering matrix units are meters.
ShvPattern — Radar scattering matrix HV polarization component
Radar scattering matrix HV polarization component, specified as a Q-by-P complex-valued matrix or a Q-by-P-by-M complex-valued array. Q is the length of the vector in the ElevationAngles property. P is the length of the vector in the AzimuthAngles property. M is the number of target patterns. The number of patterns corresponds to the number of signals passed into the step method. You can, however, use a single pattern to model multiple signals reflecting from a single target. Scattering matrix units are meters.
Model — Target fluctuation model
'Nonfluctuating' (default) | 'Swerling1' | 'Swerling2' | 'Swerling3' | 'Swerling4'
Target fluctuation model, specified as 'Nonfluctuating', 'Swerling1', 'Swerling2', 'Swerling3', or 'Swerling4'. If you set this property to a value other than 'Nonfluctuating', use the update input argument when calling step.
Example: 'Swerling3'
SeedSource — Seed source of random number generator for RCS fluctuation model
Seed source of random number generator for RCS fluctuation model, specified as 'Auto' or 'Property'. When you set this property to 'Auto', the System object generates random numbers using the default MATLAB® random number generator. When you set this property to 'Property', you specify the random number generator seed using the Seed property. This property applies when you set the Model property to'Swerling1', 'Swerling2', 'Swerling3', or 'Swerling4'. When you use this object with Parallel Computing Toolbox™ software, you set this property to 'Auto'.
Random number generator seed, specified as a nonnegative integer less than 232. This property applies when the SeedSource property is set to 'Property'.
step Backscatter incoming signal
Calculate the reflected radar signal from a nonfluctuating point target with a peak RCS of 10.0
{m}^{2}
. Use a simplified expression of an RCS pattern of a target for illustrative purposes. Real RCS patterns are more complicated. The RCS pattern covers a range of angles from 10°–30° in azimuth and 5°–15° in elevation. The RCS peaks at 20° azimuth and 10° elevation. Assume that the radar operating frequency is 1 GHz and that the signal is a sinusoid at 1 MHz.
Create and plot the RCS pattern.
azmax = 20.0;
elmax = 10.0;
azpatangs = [10.0:0.1:30.0];
elpatangs = [5.0:0.1:15.0];
rcspattern = 10.0*cosd(4*(elpatangs - elmax))'*cosd(4*(azpatangs - azmax));
imagesc(azpatangs,elpatangs,rcspattern)
title('RCS')
Generate and plot 50 samples of the radar signal.
foper = 1.0e9;
fs = 10*freq;
t = [0:(nsamp-1)]'/fs;
sig = sin(2*pi*freq*t);
plot(t*1e6,sig)
xlabel('Time (\mu seconds)')
Create the phased.BackscatterRadarTarget System object™.
target = phased.BackscatterRadarTarget('Model','Nonfluctuating',...
'AzimuthAngles',azpatangs,'ElevationAngles',elpatangs,...
'RCSPattern',rcspattern,'OperatingFrequency',foper);
For a sequence of incident angles at constant elevation angle, find and plot the scattered signal amplitude.
az0 = 13.0;
naz = 20;
az = az0 + [0:2:20];
naz = length(az);
ss = zeros(1,naz);
for k = 1:naz
y = target(sig,[az(k);el]);
ss(k) = max(abs(y));
plot(az,ss,'.')
ylabel('Scattered Signal Amplitude')
Calculate the polarized radar signal scattered from a Swerling1 fluctuating point target. Assume the target axis is rotated from the global coordinate system. Use simple expressions for the scattering patterns for illustration. Real scattering patterns are more complicated. For polarized signals, you need to specify the HH, HV, and VV components of the scattering matrix for a range of incident angles. In this example, the patterns cover the range 10°–30° in azimuth and 5°–15° in elevation. Angles are with respect to the target local coordinate system. Assume that the radar operating frequency is 1 GHz and that the signal is a sinusoid with a frequency of 1 MHz. The incident angle is 13.0° azimuth and 14.0° elevation with respect to the target orientation.
Create and plot the scattering matrix patterns.
shhpat = cosd(4*(elpatangs - elmax))'*cosd(4*(azpatangs - azmax));
shvpat = 1i*cosd(4*(elpatangs - elmax))'*sind(4*(azpatangs - azmax));
svvpat = sind(4*(elpatangs - elmax))'*sind(4*(azpatangs - azmax));
imagesc(azpatangs,elpatangs,abs(shhpat))
title('HH')
imagesc(azpatangs,elpatangs,abs(shvpat))
title('HV')
imagesc(azpatangs,elpatangs,abs(svvpat))
title('VV')
'Model','Swerling1','AzimuthAngles',azpatangs,...
'SvvPattern',svvpat);
Generate 50 samples of a polarized radar signal.
signal.X = exp(1i*2*pi*freq*t);
signal.Y = exp(1i*2*pi*freq*t + pi/3);
signal.Z = zeros(size(signal.X));
tgtaxes = azelaxes(60,10);
ang = [13.0;14.0];
Reflect the signal from the target and plot its components.
refl_signal = target(signal,ang,tgtaxes,true);
plot(t*1e6,real(refl_signal.X))
plot(t*1e6,real(refl_signal.Y))
plot(t*1e6,real(refl_signal.Z))
xlabel('Time \mu seconds')
Y=\sqrt{G}â
X,
G=\frac{4\mathrm{Ï}\mathrm{Ï}}{{\mathrm{λ}}^{2}}.
\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 [1] or [2].
[1] Mott, H. Antennas for Radar and Communications. New York: John Wiley & Sons, 1992.
phased.RadarTarget | phased.WidebandBackscatterRadarTarget | phased.BackscatterSonarTarget | backscatterPedestrian (Radar Toolbox) | backscatterBicyclist (Radar Toolbox)
Modeling Target Radar Cross Section (Radar Toolbox) |
Estimate Orientation Through Inertial Sensor Fusion - MATLAB & Simulink - MathWorks Benelux
Note that the ecompass algorithm correctly finds the location of north. However, because the function is memoryless, the estimated motion is not smooth. The algorithm could be used as an initialization step in an orientation filter or some of the techniques presented in the Lowpass Filter Orientation Using Quaternion SLERP could be used to smooth the motion.
Consider how an imufilter with a LinearAccelerationNoise of 9e-3
\left(m/{s}^{2}{\right)}^{2}
responds to a shaking trajectory, compared to one with a LinearAccelerationNoise of 9e-4
\left(m/{s}^{2}{\right)}^{2} |
torch.nn.functional.cosine_similarity — PyTorch 1.11.0 documentation
torch.nn.functional.cosine_similarity
torch.nn.functional.cosine_similarity¶
torch.nn.functional.cosine_similarity(x1, x2, dim=1, eps=1e-08) → Tensor¶
Returns cosine similarity between x1 and x2, computed along dim. x1 and x2 must be broadcastable to a common shape. dim refers to the dimension in this common shape. Dimension dim of the output is squeezed (see torch.squeeze()), resulting in the output tensor having 1 fewer dimension.
\text{similarity} = \dfrac{x_1 \cdot x_2}{\max(\Vert x_1 \Vert _2 \cdot \Vert x_2 \Vert _2, \epsilon)}
Supports type promotion.
x1 (Tensor) – First input.
x2 (Tensor) – Second input.
dim (int, optional) – Dimension along which cosine similarity is computed. Default: 1 |
2\sqrt{2}
\frac{1}{\sqrt{2}}
\sqrt{2}
The rms current in a circuit connected to a 50 Hz ac source is 15 A. The value of the current in the circuit
\left(\frac{1}{600}\right)
s after the instant the current is zero, is–
\frac{15}{\sqrt{2}} \mathrm{A}
15\sqrt{2} \mathrm{A}
\frac{\sqrt{2}}{15} \mathrm{A}
In a circuit the phase difference between the alternating current and the source voltage is
\frac{\mathrm{\pi }}{2}
. Which of the following cannot be the element(s) of the circuit?
\sqrt{\frac{1}{{\mathrm{X}}_{\mathrm{C}}^{2}}+\frac{1}{{\mathrm{X}}_{\mathrm{L}}^{2}}+{\mathrm{R}}^{2}}
\sqrt{{\mathrm{X}}_{\mathrm{L}}^{2}-{\mathrm{X}}_{\mathrm{C}}^{2}+{\mathrm{R}}^{2}}
\sqrt{{\mathrm{R}}^{2}+{\left({\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}\right)}^{2}}
When an alternating voltage
\mathrm{E}={\mathrm{E}}_{0}\mathrm{sin} \mathrm{\omega t}
is applied to a circuit, a current
\mathrm{I}={\mathrm{I}}_{0}\mathrm{sin}\left(\mathrm{\omega t}+\frac{\mathrm{\pi }}{2}\right)
flows through it. The average power dissipated in the circuit is
{\mathrm{E}}_{\mathrm{rms}}·{\mathrm{I}}_{\mathrm{rms}}
{\mathrm{E}}_{0}{\mathrm{I}}_{0}
\frac{{\mathrm{E}}_{0}{\mathrm{I}}_{0}}{\sqrt{2}}
The magnitude of electric field due to a point charge 2q, at distance r is E. Then the magnitude of electric field due to a uniformly charged thin spherical shell of radius R with total charge q at a distance
\frac{r}{2}\left(r>>R\right)
\frac{E}{4}
A square sheet of side ‘a' is lying parallel to XY plane at z = a. The electric field in the region is
\stackrel{\to }{E}=c{z}^{2 }\stackrel{^}{k}
. The electric flux through the sheet is
\frac{1}{3}{a}^{3}c
\frac{1}{3}{a}^{4}c
Three charges q, –q and q0, are placed as shown in figure. The magnitude of the net force on the charge q0 at point O is
\left[k=\frac{1}{\left(4\mathrm{\pi }{\in }_{0}\right)}\right]
\frac{2kq{q}_{0}}{{a}^{2}}
\frac{\sqrt{2}kq{q}_{0}}{{a}^{2}}
\frac{1}{\sqrt{2}}\frac{kq{q}_{0}}{{a}^{2}}
\frac{mgB}{l}
\frac{mgl}{B}
\frac{mg}{lB}
\frac{lB}{mg}
If n, e,
\tau
and m have their usual meanings, then the resistance of a wire of length l and cross-sectional area A is given by –
\frac{{\mathrm{ne}}^{2}\mathrm{A}}{2\mathrm{m\tau }l}
\frac{\mathrm{m}l}{{\mathrm{ne}}^{2}\mathrm{\tau A}}
\frac{\mathrm{m\tau A}}{{\mathrm{ne}}^{2}l}
\frac{{\mathrm{ne}}^{2}\mathrm{\tau A}}{2\mathrm{m}l}
A proton and an alpha particle move in circular orbits in a uniform magnetic field. Their speeds are in the ratio of 9 : 4. The ratio of radii of their circular orbits
\left(\frac{{\mathrm{r}}_{\mathrm{p}}}{{\mathrm{r}}_{\mathrm{alpha}}}\right)
\frac{3}{4}
\frac{4}{3}
\frac{8}{9}
\frac{9}{8}
5\sqrt{3} \mathrm{V}
50\sqrt{3} \mathrm{V}
5.0 \mathrm{V}
50.0 \mathrm{V}
\mathrm{\Omega }
resistor, an 80 mH inductor and a capacitor of capacitance C are connected in series with a 50 Hz ac source. If the source voltage and current in the circuit are in phase, then the value of capacitance is
\mathrm{\mu }
\mathrm{\mu }
\mathrm{\mu }
\mathrm{\mu }
F VIEW SOLUTION
\frac{F}{7}
\frac{F}{5}
\frac{F}{3}
\frac{F}{2}
\frac{{r}_{1}+{r}_{2}}{{r}_{2}-{r}_{1}}
\frac{{r}_{1}{r}_{2}}{{r}_{2}-{r}_{1}}
\frac{{r}_{1} + {r}_{2}}{{r}_{1}{r}_{2}}
\mathrm{\Omega }
\left(\frac{3}{2}\right) \mathrm{R\Omega }
(c) 2R
\mathrm{\Omega }
\left(\frac{2}{3}\right) \mathrm{R} \mathrm{\Omega }
A bar magnet has magnetic dipole moment
\stackrel{\to }{\mathrm{M}}
. Its initial position is parallel to the direction of uniform magnetic field
\stackrel{\to }{\mathrm{B}}
. In this position, the magnitudes of torque and force acting on it respectively are –
\left|\stackrel{\to }{\mathrm{M}}×\stackrel{\to }{\mathrm{B}}\right|
and 0 VIEW SOLUTION
Two charges 14 μC and –4 μC are placed at (–12 cm, 0, 0) and (12 cm, 0, 0) in an external electric field E =
\left(\frac{B}{{r}^{2}}\right)
, where B = 1.2 × 106 N/(cm2) and r is in metres. The electrostatic potential energy of the configuration is:-
A 300 Ω resistor and a capacitor of
\left(\frac{25}{\mathrm{\pi }}\right)
μF are connected in series to a 200 V – 50 Hz ac source. The current in the circuit is -
\frac{1}{4\mathrm{\pi }{\in }_{0}}\frac{2q}{\mathrm{L}}\left(1-\frac{1}{\sqrt{5}}\right)
\frac{1}{4\mathrm{\pi }{\in }_{0}}\frac{2q}{\mathrm{L}}\left(1+\frac{1}{\sqrt{5}}\right)
\frac{1}{4\mathrm{\pi }{\in }_{0}}\frac{q}{2 \mathrm{L}}\left(1-\frac{1}{\sqrt{5}}\right)
\frac{1}{3}\mathrm{A}
\frac{1}{2}\mathrm{A}
\frac{2}{3}\mathrm{A} |
Solar thermal energy - 3D CAD Models & 2D Drawings
Solar thermal energy (19976 views - Mechanical Engineering)
Licensed under Creative Commons Attribution-Share Alike 4.0 (Cachogaray).
3 Low-temperature solar heating and cooling systems
3.2 Heat storage in low-temperature solar thermal systems
8 Heat storage to stabilize solar-electric power generation
10 Conversion rates from solar energy to electrical energy
Solar thermal collectors are classified by the United States Energy Information Administration as low-, medium-, or high-temperature collectors. Low-temperature collectors are generally unglazed and used to heat swimming pools or to heat ventilation air. Medium-temperature collectors are also usually flat plates but are used for heating water or air for residential and commercial use. High-temperature collectors concentrate sunlight using mirrors or lenses and are generally used for fulfilling heat requirements up to 300 deg C / 20 bar pressure in industries, and for electric power production. Two categories include Concentrated Solar Thermal (CST) for fulfilling heat requirements in industries, and Concentrated Solar Power (CSP) when the heat collected is used for power generation. CST and CSP are not replaceable in terms of application. The largest facilities are located in the American Mojave Desert of California and Nevada. These plants employ a variety of different technologies. The largest examples include, Ivanpah Solar Power Facility (377 MW), Solar Energy Generating Systems installation (354 MW), and Crescent Dunes (110 MW). Spain is the other major developer of solar thermal power plant. The largest examples include, Solnova Solar Power Station (150 MW), the Andasol solar power station (150 MW), and Extresol Solar Power Station (100 MW).
Systems for utilizing low-temperature solar thermal energy include means for heat collection; usually heat storage, either short-term or interseasonal; and distribution within a structure or a district heating network. In some cases more than one of these functions is inherent to a single feature of the system (e.g. some kinds of solar collectors also store heat). Some systems are passive, others are active (requiring other external energy to function).[2]
Heating is the most obvious application, but solar cooling can be achieved for a building or district cooling network by using a heat-driven absorption or adsorption chiller (heat pump). There is a productive coincidence that the greater the driving heat from insulation, the greater the cooling output. In 1878, Auguste Mouchout pioneered solar cooling by making ice using a solar steam engine attached to a refrigeration device.[3]
Solar roof ponds are unique solar heating and cooling systems developed by Harold Hay in the 1960s. A basic system consists of a roof-mounted water bladder with a movable insulating cover. This system can control heat exchange between interior and exterior environments by covering and uncovering the bladder between night and day. When heating is a concern the bladder is uncovered during the day allowing sunlight to warm the water bladder and store heat for evening use. When cooling is a concern the covered bladder draws heat from the building's interior during the day and is uncovered at night to radiate heat to the cooler atmosphere. The Skytherm house in Atascadero, California uses a prototype roof pond for heating and cooling.[9]
Heat storage in low-temperature solar thermal systems
CSP facilities utilize high electrical conductivity materials, such as copper, in field power cables, grounding networks, and motors for tracking and pumping fluids, as well as in the main generator and high voltage transformers. (See: Copper in concentrating solar thermal power facilities.)
The federal government has dedicated nearly 2,000 times more acreage to oil and gas leases than to solar development. In 2010 the Bureau of Land Management approved nine large-scale solar projects, with a total generating capacity of 3,682 megawatts, representing approximately 40,000 acres. In contrast, in 2010, the Bureau of Land Management processed more than 5,200 applications gas and oil leases, and issued 1,308 leases, for a total of 3.2 million acres. Currently, 38.2 million acres of onshore public lands and an additional 36.9 million acres of offshore exploration in the Gulf of Mexico are under lease for oil and gas development, exploration and production.[27]
During the day the sun has different positions. For low concentration systems (and low temperatures) tracking can be avoided (or limited to a few positions per year) if nonimaging optics are used.[28][29] For higher concentrations, however, if the mirrors or lenses do not move, then the focus of the mirrors or lenses changes (but also in these cases nonimaging optics provides the widest acceptance angles for a given concentration). Therefore, it seems unavoidable that there needs to be a tracking system that follows the position of the sun (for solar photovoltaic a solar tracker is only optional). The tracking system increases the cost and complexity. With this in mind, different designs can be distinguished in how they concentrate the light and track the position of the sun.
The first commercial tower power plant was PS10 in Spain with a capacity of 11 MW, completed in 2007. Since then a number of plants have been proposed, several have been built in a number of countries (Spain, Germany, U.S., Turkey, China, India) but several proposed plants were cancelled as photovoltaic solar prices plummeted. A solar power tower is expected to come online in South Africa in 2014.[37] Ivanpah Solar Power Facility in California generates 392 MW of electricity from three towers, making it the largest solar power tower plant when it came online in late 2013.
CSP-Stirling is known to have the highest efficiency of all solar technologies (around 30%, compared to solar photovoltaic's approximately 15%), and is predicted to be able to produce the cheapest energy among all renewable energy sources in high-scale production and hot areas, semi-deserts, etc. A dish Stirling system uses a large, reflective, parabolic dish (similar in shape to a satellite television dish). It focuses all the sunlight that strikes the dish up onto a single point above the dish, where a receiver captures the heat and transforms it into a useful form. Typically the dish is coupled with a Stirling engine in a Dish-Stirling System, but also sometimes a steam engine is used.[38] These create rotational kinetic energy that can be converted to electricity using an electric generator.[39]
More energy is contained in higher frequency light based upon the formula of
{\displaystyle E=h\nu }
, where h is the Planck constant and
{\displaystyle \nu }
is frequency. Metal collectors down convert higher frequency light by producing a series of Compton shifts into an abundance of lower frequency light. Glass or ceramic coatings with high transmission in the visible and UV and effective absorption in the IR (heat blocking) trap metal absorbed low frequency light from radiation loss. Convection insulation prevents mechanical losses transferred through gas. Once collected as heat, thermos containment efficiency improves significantly with increased size. Unlike Photovoltaic technologies that often degrade under concentrated light, Solar Thermal depends upon light concentration that requires a clear sky to reach suitable temperatures.
Heat storage to stabilize solar-electric power generation
A variety of fluids have been tested to transport the sun's heat, including water, air, oil, and sodium, but Rockwell International[58] selected molten salt as best.[59] Molten salt is used in solar power tower systems because it is liquid at atmospheric pressure, provides a low-cost medium to store thermal energy, its operating temperatures are compatible with today's steam turbines, and it is non-flammable and nontoxic. Molten salt is used in the chemical and metals industries to transport heat, so industry has experience with it.
In 2016 SolarReserve proposed a 2 GW, $5 billion concentrated solar plant with storage in Nevada.
Solar parabolic trough plants have been built with efficiencies of about 20%.[citation needed] Fresnel reflectors have an efficiency that is slightly lower (but this is compensated by the denser packing).
Furthermore, efficiency does not directly relate to cost: on calculating total cost, both efficiency and the cost of construction and maintenance should be taken into account.
Electrical energyRotational energySolar cellSolar SystemRadiator (heating)Water heatingParabolic troughRenewable energySolar chimneySolar panelSunlightPhotovoltaicsPhotovoltaic systemSolar powerSolar cell efficiencyEnergy transformationEnergyThermal conduction
This article uses material from the Wikipedia article "Solar thermal energy", which is released under the Creative Commons Attribution-Share-Alike License 3.0. There is a list of all authors in Wikipedia |
Physics - Taking Temperature in 2D
Taking Temperature in 2D
February 2, 2018 • Physics 11, 12
Electron microscopy can produce nanometer-scale maps of the thermal expansion of 2D materials, which may be important for the development of nanoelectronic devices.
X. Hu et al., Phys. Rev. Lett. (2018)
A heated response. An electron microscopy technique records the thermal expansion in 2D materials (here, molybdenum diselenide). The colored pixels (each about 3 nanometers wide) represent the local thermal expansion coefficient, with red corresponding to high values and blue corresponding to low values. The background image, in which the shading depends on the number of atomic layers, shows that the region with two layers has higher values of the expansion coefficient than the region with four layers.A heated response. An electron microscopy technique records the thermal expansion in 2D materials (here, molybdenum diselenide). The colored pixels (each about 3 nanometers wide) represent the local thermal expansion coefficient, with red correspondi... Show more
To build robust electronic devices, it is crucial to know how the components change size in response to heat generated by electric currents. A new experiment demonstrates electron microscopy as a tool for creating a map—with nanometer resolution—of thermal expansion in 2D materials. Observations of 2D carbon and 2D semiconductors showed that the thermal response strongly depends on the number of atomic layers in the material, with single layered materials exhibiting surprisingly large expansion rates. This behavior, which the researchers have yet to fully explain, will need to be accounted for as engineers develop next-generation nanoelectronic devices relying on 2D materials.
Probing the temperature and thermal properties of very small objects has long been a challenge. Conventional thermometers are simply too large: they would alter the temperature of the target object if they were placed in contact with it. Researchers have developed less invasive techniques that use a sharp probe with sensors that can measure the heat flow between the object and the probe. Other methods record temperature by detecting light emitted by (or reflected from) the object. But such techniques cannot provide sufficient spatial resolution for studies of transistors and other components in future nanoscale devices.
Recent work has shown that scanning transmission electron microscopy (STEM) can measure temperature with nanometer spatial resolution [1]. The technique involves measuring the amount of kinetic energy lost by electrons passing through a material, a quantity that is temperature dependent. Robert Klie and his colleagues from the University of Illinois at Chicago have adapted this STEM method to the case of measuring thermal expansion in 2D materials.
The team investigated graphene (2D carbon) and several transition metal dichalcogenides (TMDs), which are 2D semiconductors with unusual electronic properties. They fabricated small flakes of various thicknesses and placed them on a heated mesh, which allowed the temperature of the samples to be controlled.
For the STEM measurements, the researchers scanned an electron beam across each sample and recorded the energy of transmitted electrons. They produced, at each location, a spectrum of energy lost through interactions with the sample. One of the main loss mechanisms is the excitation of plasmons, which are “collective oscillations of electrons in a material,” Klie explains. The energy of the plasmons depends on the wavelength of these collective oscillations, which in turn depends on the volume occupied by the electrons. So you can measure changes in the volume of the material by measuring the energy of the plasmon peak.
Klie and his colleagues recorded the position of the plasmon peak at various temperatures between and . As temperature increased, the peak shifted, reflecting a change in volume. The researchers used the shift to calculate the thermal expansion coefficient, which is the fractional volume change per degree of temperature increase. They found that this coefficient strongly depended on the number of atomic layers in the sample. For the TMDs, the expansion rate increased—in some cases by a factor of 45—when the number of layers decreased from four to one. For graphene, which experiences contraction rather than expansion when heated, the coefficient for a monolayer was 30 times that of samples with four or more layers. This behavior suggests that single layers respond more strongly to temperature changes because they are not constrained by interlayer interactions. However, these interactions are normally weak in 2D materials, so a more sophisticated model may be needed to explain why the expansion of multiple layers is so different from that of single layers, says Klie.
The team produced maps of the thermal expansion coefficient with spatial resolution of around 2 nanometers. Klie says that such maps could help in designing nanoelectromechanical systems with connections between different 2D materials. By observing how the thermal expansion varies across boundaries between different materials, engineers can avoid structural failure by adjusting the geometry or the type of materials used.
“Measuring the thermal expansion coefficient in 2D materials is useful for materials engineering,” says electron microscopy expert Matthew Mecklenburg from the University of Southern California. He is surprised by the dramatic increase observed in the thermal expansion coefficient as the number of layers decreases. For single layers, the expansion data suggests that 2D materials behave more like plastics than metals—an unexpected result that Mecklenburg would like to see verified.
M. Mecklenburg, W. A. Hubbard, E. R. White, R. Dhall, S. B. Cronin, S. Aloni, and B. C. Regan, “Nanoscale Temperature Mapping in Operating Microelectronic Devices,” Science 347, 629 (2015).
Xuan Hu, Poya Yasaei, Jacob Jokisaari, Serdar Öğüt, Amin Salehi-Khojin, and Robert F. Klie
Condensed Matter PhysicsGraphene
{\text{CaH}}_{6} |
About Projections of Solutions for Fuzzy Differential Equations
2013 About Projections of Solutions for Fuzzy Differential Equations
Moiseis S. Cecconello, Jefferson Leite, Rodney C. Bassanezi, Joao de Deus M. Silva
In this paper we propose the concept of fuzzy projections on subspaces of
ℱ\left({ℝ}^{n}\right)
, obtained from Zadeh's extension of canonical projections in
{ℝ}^{n}
, and we study some of the main properties of such projections. Furthermore, we will review some properties of fuzzy projection solution of fuzzy differential equations. As we will see, the concept of fuzzy projection can be interesting for the graphical representation of fuzzy solutions.
Moiseis S. Cecconello. Jefferson Leite. Rodney C. Bassanezi. Joao de Deus M. Silva. "About Projections of Solutions for Fuzzy Differential Equations." J. Appl. Math. 2013 1 - 9, 2013. https://doi.org/10.1155/2013/184950
Moiseis S. Cecconello, Jefferson Leite, Rodney C. Bassanezi, Joao de Deus M. Silva "About Projections of Solutions for Fuzzy Differential Equations," Journal of Applied Mathematics, J. Appl. Math. 2013(none), 1-9, (2013) |
In problem 11-40 you found that the polar equation
\frac { \operatorname { sin } ( \theta + \frac { \pi } { 4 } ) } { \operatorname { cos } ^ { 2 } ( \theta + \frac { \pi } { 4 } ) }
will generate a parabola that is rotated
45°
clockwise. Convert the equation into rectangular form.
\cos^2\left(θ + \frac{π}{4}\right)
Note: Steps can be done in various orders.
\sin\left(θ + \frac{π}{4}\right)
\cos\left(θ + \frac{π}{4}\right)
using the angle sum identities.
\sin\left( \frac{π}{4}\right)
\cos\left(\frac{π}{4}\right)
to their numeric values.
r
x
r \cos\left(θ\right) |
Numerically evaluate integral — Gauss-Kronrod quadrature - MATLAB quadgk - MathWorks India
Integral with Singularity at Endpoint
Complex Contour Integration
Examine Absolute Error
errbnd
Numerically evaluate integral — Gauss-Kronrod quadrature
q = quadgk(fun,a,b)
[q,errbnd] = quadgk(fun,a,b)
[___] = quadgk(fun,a,b,Name,Value)
q = quadgk(fun,a,b) integrates the function handle fun from a to b using high-order global adaptive quadrature and default error tolerances.
[q,errbnd] = quadgk(fun,a,b) additionally returns an approximate upper bound on the absolute error |q - I|, where I is the exact value of the integral.
[___] = quadgk(fun,a,b,Name,Value) specifies additional options with one or more name-value pair arguments using either of the previous output argument combinations. For example, specify 'Waypoints' followed by a vector of real or complex numbers to indicate specific points for the integrator to use.
\mathit{q}={\int }_{0}^{1}{\mathit{e}}^{\mathit{x}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{ln}\left(\mathit{x}\right)\text{\hspace{0.17em}}\mathrm{dx}.
This integral has a singularity at the point
\mathit{x}=0
\mathrm{ln}\left(0\right)
-\infty
Create an anonymous function for the integrand. The log function calculates
\mathrm{ln}\left(\mathit{x}\right)
f = @(x) exp(x).*log(x);
Integrate f from 0 to 1.
q = quadgk(f,0,1)
Integrate a complex function around a pole by specifying a contour.
\mathit{q}=\oint \frac{\mathrm{dz}}{2\mathit{z}-1}.
The integrand has a simple pole at
\mathit{z}=1/2
, so use a rectangular contour that encloses that point. The contour starts and ends at
\mathit{x}=1
on the real number line. Use the 'Waypoints' name-value pair to specify the piecewise segments in the contour.
f = @(z) 1./(2.*z-1);
contour_segments = [1+1i 0+1i 0-1i 1-1i];
q = quadgk(f,1,1,'Waypoints',contour_segments)
q = -0.0000 + 3.1416i
Use quadgk to evaluate an oscillatory integrand that is difficult to evaluate.
\mathit{Q}={\int }_{0}^{\pi }\mathrm{sin}\left(20000\pi \mathit{x}\right)\mathrm{dx}.
The integrand oscillates very quickly, so it is difficult to evaluate. Use quadgk to evaluate the integral, and specify two outputs to examine how close the error tolerances are to being met.
fun = @(x) sin(2e4*pi*x);
[Q,errbnd] = quadgk(fun,0,pi)
Warning: Reached the limit on the maximum number of intervals in use. Approximate bound on error is 5.7e-01. The integral may not exist, or it may be difficult to approximate numerically. Increase MaxIntervalCount to 1272 to enable QUADGK to continue for another iteration.
errbnd = 0.5723
The warning message indicates how to adjust MaxIntervalCount to allow for another iteration in the solution process.
Solve the integral again, but specify MaxIntervalCount as 1e5. With many more intervals, quadgk is able to meet the absolute error tolerance for the problem (1e-10 for double precision).
[Q,errbnd] = quadgk(fun,0,pi,'MaxIntervalCount',1e5)
Q = 1.6656e-06
Example: q = quadgk(@(x) exp(1-x.^2),a,b) integrates an anonymous function handle.
Example: q = quadgk(@myFun,a,b) integrates the function myFun, which is saved as a file.
Integration limits, specified as separate arguments of real or complex scalars. The limits a and b can be -Inf or Inf. If both are finite, they can be complex. If at least one is complex, the integral is approximated over a straight line path from a to b in the complex plane.
Example: quadgk(fun,0,1) integrates fun from 0 to 1.
Example: q = quadgk(fun,a,b,'Waypoints',[0.1 1.1 2.1]) uses the 'Waypoints' option to specify a few points of interest where the integrand should be evaluated.
1e-10 (double), 1e-5 (single) (default) | nonnegative real number
Absolute error tolerance, specified as the comma-separated pair consisting of 'AbsTol' and a nonnegative real number. quadgk uses the absolute error tolerance to limit an estimate of the absolute error, |q – I|, where q is the computed value of the integral and I is the (unknown) exact value. quadgk might provide more decimal places of precision if you decrease the absolute error tolerance.
quadgk attempts to satisfy
errbnd <= max(AbsTol,RelTol*abs(q))
This relation is absolute error control when |q| is sufficiently small and relative error control when |q| is larger. For pure absolute error control, use 'AbsTol' > 0 and 'RelTol'= 0. For pure relative error control use 'RelTol' > 0 and 'AbsTol' = 0. Except when using pure absolute error control, the minimum relative tolerance is 'RelTol' >= 100*eps(class(q)).
Example: quadgk(fun,a,b,'AbsTol',1e-12) sets the absolute error tolerance to approximately 12 decimal places of accuracy.
Example: quadgk(fun,a,b,'AbsTol',tol,'RelTol',0) uses a purely absolute error control, requiring that errbnd <= tol.
1e-6 (double), 1e-4 (single) (default) | nonnegative real number
Relative error tolerance, specified as the comma-separated pair consisting of 'RelTol' and a nonnegative real number. quadgk uses the relative error tolerance to limit an estimate of the relative error, |q - I|/|I|, where q is the computed value of the integral and I is the (unknown) exact value. quadgk might provide more significant digits of precision if you decrease the relative error tolerance.
Example: quadgk(fun,a,b,'RelTol',1e-9) sets the relative error tolerance to approximately 9 significant digits.
Example: quadgk(fun,a,b,'AbsTol',0,'RelTol',tol) uses a purely relative error tolerance, requiring that errbnd <= |I|*tol.
Example: 'Waypoints',[1+1i,1-1i] specifies two complex waypoints along the interval of integration.
MaxIntervalCount — Maximum number of intervals allowed
Maximum number of intervals allowed, specified as a scalar. This option limits the number of intervals that quadgk uses at any one time after the first iteration. A warning is issued if quadgk returns early because of this limit. Routinely increasing this value is not recommended, but it may be appropriate when errbnd is small enough that the desired accuracy has nearly been achieved.
Example: quadgk(fun,a,b,'MaxIntervalCount',700)
errbnd — Approximate upper bound on absolute error
Approximate upper bound on absolute error, returned as a scalar. The approximate upper bound on absolute error in the integration is errbnd = |q – I|, where q is the computed value of the integral and I is the (unknown) exact value. quadgk attempts to satisfy
Specify this output argument to see how well the integration meets the AbsTol and RelTol error tolerances. In cases where errbnd is close to the desired value, you might be able to reach the desired value by increasing the value of MaxIntervalCount.
quadgk and integral use essentially the same integration method. You should generally use integral rather than quadgk. However, you can use quadgk to:
Monitor solution accuracy with the errbnd output argument.
Specify a large value for MaxIntervalCount when integral warns about reaching the maximum number of intervals.
quadgk can integrate functions that are singular at finite endpoints if the singularities are not too strong. For example, it can integrate functions that behave at an endpoint c like log|x-c| or |x-c|p for p >= -1/2. If the function is singular at points inside the integration limits [a b], then write the integral as a sum of integrals over subintervals with the singular points as endpoints, compute them with quadgk, and add the results.
If the interval is infinite,
\left[a,\infty \right)
, then for the integral of fun(x) to exist, fun(x) must decay as x approaches infinity, and quadgk requires it to decay rapidly.
[1] Shampine, L.F. "Vectorized Adaptive Quadrature in MATLAB®." Journal of Computational and Applied Mathematics. Vol. 211, 2008, pp.131–140.
quad2d | integral | integral2 | integral3 |
Electrostatic Potential And Capacitance, Popular Questions: CBSE Class 12-science PHYSICS, Physics Part I - Meritnation
+ 15 \mu C and + 9 \mu C
\lambda
\overline{)E}={\overline{)E}}_{0}\stackrel{^}{j}
△\mathrm{\theta }
\left(\mathrm{\lambda }=1 \mathrm{kg}/\mathrm{m}, \mathrm{a}=2\mathrm{m}, \mathrm{p}=1/3 \mathrm{C}/\mathrm{m}, {\mathrm{E}}_{0}={\mathrm{\pi }}^{2} \mathrm{N}/\mathrm{C}\right)
4×{10}^{15} Hz to 8×{10}^{15 }Hz?\phantom{\rule{0ex}{0ex}}Given h= 6.4×{10}^{-34} J -s,e=1.6×{10}^{-19} C and c=3×{10}^{8} m{s}^{-1}
{R}_{1}
\frac{8}{5} A
64. \mathrm{Figure} \mathrm{shows} \mathrm{two} \mathrm{shells} \mathrm{of} \mathrm{radii} \mathrm{R} \mathrm{and} 2\mathrm{R}. \mathrm{The} \mathrm{inner} \mathrm{shell} \left(\mathrm{centre} \mathrm{at} \mathrm{A}\right) \mathrm{is} \mathrm{nonconducting} \mathrm{and} \mathrm{uniformly} \mathrm{charged} \mathrm{with} \mathrm{charge} \mathrm{Q} \mathrm{while} \mathrm{the} \mathrm{outer} \mathrm{shell} \left(\mathrm{centre} \mathrm{at} \mathrm{B}\right) \mathrm{is} \mathrm{conducting} \mathrm{and} \mathrm{uncharged}. \mathrm{The} \mathrm{potential} \mathrm{at} \mathrm{the} \mathrm{point} \mathrm{B} \mathrm{is} :\phantom{\rule{0ex}{0ex}}\left(\mathrm{a}\right) \mathrm{zero} \left(\mathrm{b}\right) \frac{\mathrm{KQ}}{\mathrm{R}} \left(\mathrm{c}\right) \frac{\mathrm{KQ}}{\mathrm{x}} \left(\mathrm{d}\right) \mathrm{None} \mathrm{of} \mathrm{these} |
Extension of the Drasin-Shea-Jordan theorem
July, 2000 Extension of the Drasin-Shea-Jordan theorem
Nicholas H. BINGHAM, Akihiko INOUE
Passing from regular variation of a function
f
to regular variation of its integral transform
k*f
of Mellin-convolution form with kernel
k
is an Abelian problem; its converse, under suitable Tauberian conditions, is a Tauberian one. In either case, one has a comparison statement that the ratio of
k*f
tends to a constant at infinity. Passing from a comparison statement to a regular-variation statement is a Mercerian problem. The prototype results here are the Drasin-Shea theorem (for non-negative
k
) and Jordan's theorem (for
k
which may change sign). We free Jordan's theorem from its non-essential technical conditions which reduce its applicability. Our proof is simpler than the counter-parts of the previous results and does not even use the Pólya Peak Theorem which has been so essential before. The usefulness of the extension is highlighted by an application to Hankel transforms.
Nicholas H. BINGHAM. Akihiko INOUE. "Extension of the Drasin-Shea-Jordan theorem." J. Math. Soc. Japan 52 (3) 545 - 559, July, 2000. https://doi.org/10.2969/jmsj/05230545
Keywords: Hankel transform , Mercerian theorem , regular variation
Nicholas H. BINGHAM, Akihiko INOUE "Extension of the Drasin-Shea-Jordan theorem," Journal of the Mathematical Society of Japan, J. Math. Soc. Japan 52(3), 545-559, (July, 2000) |
May 2021 Estimation of convex supports from noisy measurements
Victor-Emmanuel Brunel, Jason M. Klusowski, Dana Yang
Victor-Emmanuel Brunel,1 Jason M. Klusowski,2 Dana Yang3
1Department of Statistics, CREST UMR 9194, ENSAE, 5 avenue Henry Le Chatelier, TSA 96642, 91764 Palaiseau cedex, France
2Department of Statistics, Rutgers University, 110 Frelinghuysen Road, Piscataway, NJ 08854-8019, USA
3The Fuqua School of Business, Duke University, 100 Fuqua Drive, Durham, NC 27708, USA
A popular class of problems in statistics deals with estimating the support of a density from n observations drawn at random from a d-dimensional distribution. In the one-dimensional case, if the support is an interval, the problem reduces to estimating its end points. In practice, an experimenter may only have access to a noisy version of the original data. Therefore, a more realistic model allows for the observations to be contaminated with additive noise.
In this paper, we consider estimation of convex bodies when the additive noise is distributed according to a multivariate Gaussian (or nearly Gaussian) distribution, even though our techniques could easily be adapted to other noise distributions. Unlike standard methods in deconvolution that are implemented by thresholding a kernel density estimate, our method avoids tuning parameters and Fourier transforms altogether. We show that our estimator, computable in
{\left(\mathit{O}\left(log\mathit{n}\right)\right)}^{\left(\mathit{d}-1\right)/2}
time, converges at a rate of
{\mathit{O}}_{\mathit{d}}\left(loglog\mathit{n}/\sqrt{log\mathit{n}}\right)
in Hausdorff distance, in accordance with the polylogarithmic rates encountered in Gaussian deconvolution problems. Part of our analysis also involves the optimality of the proposed estimator. We provide a lower bound for the minimax rate of estimation in Hausdorff distance that is
{\mathrm{\Omega }}_{\mathit{d}}\left(1/{log}^{2}\mathit{n}\right)
Victor-Emmanuel Brunel. Jason M. Klusowski. Dana Yang. "Estimation of convex supports from noisy measurements." Bernoulli 27 (2) 772 - 793, May 2021. https://doi.org/10.3150/20-BEJ1229
Received: 1 April 2018; Revised: 1 April 2020; Published: May 2021
Keywords: Convex bodies , order statistics , support estimation , support function
Victor-Emmanuel Brunel, Jason M. Klusowski, Dana Yang "Estimation of convex supports from noisy measurements," Bernoulli, Bernoulli 27(2), 772-793, (May 2021) |
If A and B are square matrices of the same order 3, such that ∣A∣ = 2 and AB = 2I, write the value of ∣B∣. VIEW SOLUTION
If f(x) = x + 1, find
\frac{d}{dx}\left(fof\right) \left(x\right)
Find the order and the degree of the differential equation
{x}^{2}\frac{{d}^{2}y}{d{x}^{2}}={\left\{1+{\left(\frac{dy}{dx}\right)}^{2}\right\}}^{4}
If a line makes angles 90°, 135°, 45° with the x, y and z axes respectively, find its direction cosines.
Find the vector equation of the line which passes through the point (3, 4, 5) and is parallel to the vector
2\stackrel{^}{i}+2\stackrel{^}{j}-3\stackrel{^}{k}
Examine whether the operation *defined on R by a * b = ab + 1 is (i) a binary or not. (ii) if a binary operation, is it associative or not ? VIEW SOLUTION
Find a matrix A such that 2A − 3B + 5C = O, where
B =\left[\begin{array}{ccc}-2& 2& 0\\ 3& 1& 4\end{array}\right] \mathrm{and} C=\left[\begin{array}{ccc}2& 0& -2\\ 7& 1& 6\end{array}\right].
\mathrm{Find} : \int \frac{{\mathrm{sec}}^{2}x}{\sqrt{{\mathrm{tan}}^{2}x+4}}dx.
\mathrm{Find} : \int \sqrt{1-\mathrm{sin} 2x }dx, \frac{\mathrm{\pi }}{4}<x<\frac{\mathrm{\pi }}{2}
Find : ∫ sin−1 (2x) dx. VIEW SOLUTION
Form the differential equation representing the family of curves y = e2x (a + bx), where 'a' and 'b' are arbitrary constants. VIEW SOLUTION
If the sum of two unit vectors is a unit vector, prove that the magnitude of their difference is
\sqrt{3}
\stackrel{\to }{a}=2\stackrel{^}{i}+3\stackrel{^}{j}+\stackrel{^}{k}, \stackrel{\to }{b}=\stackrel{^}{i}-2\stackrel{^}{j}+\stackrel{^}{k} \mathrm{and} \stackrel{\to }{c}=-3\stackrel{^}{i}+\stackrel{^}{j}+2\stackrel{^}{k}, \mathrm{find} \left[\stackrel{\to }{a}\stackrel{\to }{b}\stackrel{\to }{c}\right]
A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event "number is even" and B be the event "number is marked red". Find whether the events A and B are independent or not. VIEW SOLUTION
A die is thrown 6 times. If "getting an odd number" is a "success", what is the probability of (i) 5 successes? (ii) atmost 5 successes?
The random variable X has a probability distribution P(X) of the following form, where 'k' is some number.
\mathrm{P}\left(\mathrm{X}=x\right)=\left\{\begin{array}{lll}k& ,& \mathrm{if} x=0\\ 2k& ,& \mathrm{if} x=1\\ 3k& ,& \mathrm{if} x=2\\ 0& ,& \mathrm{otherwise}\end{array}\right\
Determine the value of 'k'. VIEW SOLUTION
Show that the relation R on ℝ defined as R = {(a, b) : a ≤ b}, is reflexive, and transitive but not symmetric.
Prove that the function f : N → N, defined by f(x) = x2 + x + 1 is one-one but not onto. Find inverse of f : N → S, where S is range of f. VIEW SOLUTION
Solve: tan−1 4x + tan−1 6x =
\frac{\mathrm{\pi }}{4}
\left|\begin{array}{ccc}{a}^{2}+2a& 2a+1& 1\\ 2a+1& a+2& 1\\ 3& 3& 1\end{array}\right|={\left(a-1\right)}^{3}.
If log (x2 + y2) =
2{\mathrm{tan}}^{-1}\left(\frac{y}{x}\right)
\frac{dy}{dx}=\frac{x+y}{x-y}
If xy − yx = ab, find
\frac{dy}{dx}
If y = (sin−1x)2, prove that
\left(1-{x}^{2}\right)\frac{{d}^{2}y}{d{x}^{2}}-x\frac{dy}{dx}-2=0
Find the equation of tangent to the curve
y=\sqrt{3x-2}
which is parallel to the line 4x − 2y + 5 = 0. Also, write the equation of normal to the curve at the point of contact. VIEW SOLUTION
\int \frac{3x+5}{{x}^{2}+3x-18}dx
{\int }_{0}^{a}f\left(x\right)dx={\int }_{0}^{a}f\left(a-x\right)dx
, hence evaluate
{\int }_{0}^{\mathrm{\pi }}\frac{x\mathrm{sin}x}{1+{\mathrm{cos}}^{2}x}dx
Solve the differential equation: x dy − y dx =
\sqrt{{x}^{2}+{y}^{2}}dx
, given that y = 0 when x = 1.
\left(1+{x}^{2}\right)\frac{dy}{dx}+2xy-4{x}^{2}=0
, subject to the initial condition y(0) = 0. VIEW SOLUTION
\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k},2\stackrel{^}{i}+5\stackrel{^}{j},3\stackrel{^}{i}+2\stackrel{^}{j}-3\stackrel{^}{k}
\stackrel{^}{i}-6\stackrel{^}{j}-\stackrel{^}{k}
respectively are the position vectors A, B, C and D, then find the angle between the straight lines AB and CD. Find whether
\stackrel{\to }{\mathrm{AB}}
\stackrel{\to }{\mathrm{CD}}
are collinear or not. VIEW SOLUTION
Find the value of λ, so that the lines
\frac{1-x}{3}=\frac{7y-14}{\lambda }=\frac{z-3}{2}
\frac{7-7x}{3\lambda }=\frac{y-5}{1}=\frac{6-z}{5}
are at right angles. Also, find whether the lines are intersecting or not. VIEW SOLUTION
\mathrm{A}=\left[\begin{array}{ccc}1& 1& 1\\ 1& 0& 2\\ 3& 1& 1\end{array}\right],
find A–1. Hence, solve the system of equations x + y + z = 6, x + 2z = 7, 3x + y + z = 12.
Find the inverse of the following matrix using elementary operations.
\mathrm{A}=\left[\begin{array}{ccc}1& 2& -2\\ -1& 3& 0\\ 0& -2& 1\end{array}\right]
A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m3. If building of tank costs ₹ 70 per square metre for the base and ₹ 45 per square metre for the sides, what is the cost of least expensive tank? VIEW SOLUTION
Using integration, find the area of triangle ABC, whose vertices are A(2, 5), B(4, 7) and C(6, 2).
Find the area of the region lying above x-axis and included between the circle x2 + y2 = 8x and inside of the parabola
{y}^{2}=4x
Find the vector and Cartesian equations of the plane passing through the points (2, 2 –1), (3, 4, 2) and (7, 0, 6). Also find the vector equation of a plane passing through (4, 3, 1) and parallel to the plane obtained above.
\stackrel{\to }{r}=\left(\stackrel{^}{i}+\stackrel{^}{j}\right)+\lambda \left(\stackrel{^}{i}+2\stackrel{^}{j}-\stackrel{^}{k}\right)
and the point (–1, 3, –4). Also, find the length of the perpendicular drawn from the point (2, 1, 4) to the plane thus obtained. VIEW SOLUTION
A manufacture has three machine operators A, B and C. The first operator A produces 1% of defective items, whereas the other two operators B and C produces 5% and 7% defective items respectively. A is on the job for 50% of the time, B on the job 30% of the time and C on the job for 20% of the time. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by A? VIEW SOLUTION
A manufacturer has employed 5 skilled men and 10 semi-skilled men and makes two models A and B of an article. The making of one item of model A requires 2 hours work by a skilled man and 2 hours work by a semi-skilled man. One item of model B requires 1 hour by a skilled man and 3 hours by a semi-skilled man. No man is expected to work more than 8 hours per day. The manufacturer's profit on an item of model A is ₹ 15 and on an item of model B is ₹ 10. How many of items of each model should be made per day in order to maximize daily profit? Formulate the above LPP and solve it graphically and find the maximum profit. VIEW SOLUTION |
Exact Solutions to the Sharma-Tasso-Olver Equation by Using Improved G′/G-Expansion Method
2013 Exact Solutions to the Sharma-Tasso-Olver Equation by Using Improved
{G}^{\prime }/G
-Expansion Method
Yinghui He, Shaolin Li, Yao Long
This paper is concerned with a double nonlinear dispersive equation: the Sharma-Tasso-Olver equation. We propose an improved
{G}^{\prime }/G
-expansion method which is employed to investigate the solitary and periodic traveling waves of this equation. As a result, some new traveling wave solutions involving hyperbolic functions, the trigonometric functions, are obtained. When the parameters are taken as special values, the solitary wave solutions are derived from the hyperbolic function solutions, and the periodic wave solutions are derived from the trigonometric function solutions. The improved
{G}^{\prime }/G
-expansion method is straightforward, concise and effective and can be applied to other nonlinear evolution equations in mathematical physics.
Yinghui He. Shaolin Li. Yao Long. "Exact Solutions to the Sharma-Tasso-Olver Equation by Using Improved
{G}^{\prime }/G
-Expansion Method." J. Appl. Math. 2013 1 - 6, 2013. https://doi.org/10.1155/2013/247234
Yinghui He, Shaolin Li, Yao Long "Exact Solutions to the Sharma-Tasso-Olver Equation by Using Improved
{G}^{\prime }/G
-Expansion Method," Journal of Applied Mathematics, J. Appl. Math. 2013(none), 1-6, (2013) |
The argument diagonal controls which diagonal to consider. If diagonal = 0, all elements on and below the main diagonal are retained. A positive value includes just as many diagonals above the main diagonal, and similarly a negative value excludes just as many diagonals below the main diagonal. The main diagonal are the set of indices
\lbrace (i, i) \rbrace
i \in [0, \min\{d_{1}, d_{2}\} - 1]
d_{1}, d_{2} |
Fermat's_Last_Theorem_in_fiction Knowpia
Find sources: "Fermat's Last Theorem in fiction" – news · newspapers · books · scholar · JSTOR%5B%5BWikipedia%3AArticles+for+deletion%2FFermat%26%2339%3Bs+Last+Theorem+in+fiction+%283rd+nomination%29%5D%5DAFD
The problem in number theory known as "Fermat's Last Theorem" has repeatedly received attention in fiction and popular culture. It was proved by Andrew Wiles in 1994.
Jorge Luis Borges's 1939 essay titled "The Total Library", a precursor to the short story "The Library of Babel", mentions "the proof of Pierre Fermat's theorem" in an imaginative list of texts that would exist in a total library. Borges also alluded to Fermat's Last Theorem in his detective story "Ibn-Khakam al-Bokhari, Murdered in His Labyrinth".[1]
The theorem plays a key role in the 1948 mystery novel Murder by Mathematics by Hector Hawton.[2][3]
Arthur Porges' short story "The Devil and Simon Flagg" features a mathematician who bargains with the Devil that the latter cannot produce a proof of Fermat's Last Theorem within twenty-four hours.[4] The devil is not successful and is last seen beginning a collaboration with the hero. The story was first published in 1954 in The Magazine of Fantasy and Science Fiction.[5]
In Douglas Hofstadter's 1979 book Gödel, Escher, Bach, the statement, "I have discovered a truly remarkable proof of this theorem which this margin is too small to contain" is repeatedly rephrased and satirized, including a pun on "fermata".[6]
In Robert Forward's 1984/1985 science fiction novel Rocheworld, Fermat's Last Theorem is unproved far enough into the future for interstellar explorers to describe it to one of the mathematically inclined natives of another star system, who finds a proof.[7]
In the 2003 book The Oxford Murders by Guillermo Martinez, Wiles's announcement in Cambridge of his proof of Fermat's Last Theorem forms a peripheral part of the action.[8]
In Stieg Larsson's 2006 book The Girl Who Played With Fire, the main character Lisbeth Salander is mesmerized by the theorem, and claims (to herself) to having produced an unorthodox solution based on the premise that "a logician would have a greater chance at solving it than a mathematician", but an untimely confrontation suspends indefinitely her revealing it to the readers.[9][10] Fields medalist Timothy Gowers criticized Larsson's portrayal of the theorem as muddled and confused.[11]
In Jasper Fforde's 2007 book First Among Sequels, 9 year-old Tuesday Next, seeing the equation on the sixth-form's math classroom's chalkboard, and thinking it homework, finds a simple counterexample.[12]
Arthur C. Clarke and Frederik Pohl's 2008 novel The Last Theorem tells of the rise to fame and world prominence of a young Sri Lankan mathematician who devises an elegant proof of the theorem.[13]
"The Royale", an episode (first aired 27 March 1989) of Star Trek: The Next Generation, begins with Picard attempting to solve the puzzle in his ready room; he remarks to Riker that the theorem had remained unproven for 800 years.[14] The captain ends the episode with the line "Like Fermat's theorem, it is a puzzle we may never solve." Wiles' proof was released five years after the episode aired.[15] The theorem was again mentioned in a subsequent Star Trek: Deep Space Nine episode called "Facets" in June 1995,[16] in which Jadzia Dax comments that one of her previous hosts, Tobin Dax, had "the most original approach to the proof since Wiles over 300 years ago."
A sum, proved impossible by the theorem, appears in the 1995 episode of The Simpsons, "Treehouse of Horror VI". In the three-dimensional world in "Homer3", the equation
{\displaystyle 1782^{12}+1841^{12}=1922^{12}}
is visible, just as the dimension begins to collapse. The joke is that the twelfth root of the sum does evaluate to 1922 due to rounding errors when entered into most handheld calculators.[17] A second "counterexample" appeared in the 1998 episode, "The Wizard of Evergreen Terrace":
{\displaystyle 3987^{12}+4365^{12}=4472^{12}}
, again forming a near-miss that appears true when evaluated on a handheld calculator.[18]
In the Doctor Who 2010 episode "The Eleventh Hour", the Doctor transmits a proof of Fermat's Last Theorem by typing it in just a few seconds on a laptop, to prove his genius to a collection of world leaders discussing the latest threat to the human race.[19]
Fermat's equation appears in the 2000 film Bedazzled with Elizabeth Hurley and Brendan Fraser. Hurley plays the devil who, in one of her many forms, appears as a school teacher who assigns Fermat's Last Theorem as a homework problem.[16]
In the 2008 adaptation of The Oxford Murders, Fermat's Last Theorem is described or parodied as "Bormat's Last Theorem", which the protagonist's roommate, a Russian mathematician (played by Burn Gorman) obsessively wants to solve.[20]
In Tom Stoppard's 1993 play Arcadia, Septimus Hodge poses the problem of proving Fermat's Last Theorem to the precocious Thomasina Coverly (who is perhaps a mathematical prodigy), in an attempt to keep her busy. Thomasina responds that Fermat had no proof and claimed otherwise in order to torment later generations.[21] Shortly after Arcadia opened in London, Andrew Wiles announced his proof of Fermat's Last Theorem, a coincidence of timing that resulted in news stories about the proof quoting Stoppard.[22]
Fermat's Last Tango is a 2000 stage musical by Joanne Sydney Lessner and Joshua Rosenblum.[23] Protagonist "Daniel Keane" is a fictionalized Andrew Wiles.[24] The characters include Fermat, Pythagoras, Euclid, Newton, and Gauss, the singing, dancing mathematicians of "the aftermath".
^ Irwin, John T. (1994). The Mystery to a Solution: Poe, Borges, and the Analytic Detective Story. Johns Hopkins University Press. ISBN 9780801854668. OCLC 27895797.
^ Schaaf, William L. (1963). Recreational Mathematics: A Guide to the Literature (third ed.). National Council of Teachers of Mathematics.
^ "Review of Murder by Mathematics". Scripta Mathematica: 294. 1948. Reprinted in Sharp, John (September 1996). "Mathematics and murder". Newsletter of the British Society for the History of Mathematics. 11 (2): 27. doi:10.1080/09629419608000021.
^ Kasman, Alex (January 2003). "Mathematics in Fiction: An Interdisciplinary Course". PRIMUS. 13 (1): 1–16. doi:10.1080/10511970308984042. ISSN 1051-1970.
^ "Devilish Short Story | Simon Singh". simonsingh.net. Retrieved 2018-09-11.
^ Lask, Thomas (October 5, 1979). "Publishing: A Heavy Price for a Heavy Book". The New York Times. Retrieved August 2, 2021.
^ Kasman, Alex. "MathFiction: The Flight of the Dragonfly (aka Rocheworld) (Robert L. Forward)". College of Charleston. Retrieved 2018-09-11.
^ Gray, Mary W. (June 2007). "The Oxford Murders". The Mathematical Intelligencer. 29 (3): 77–78. doi:10.1007/bf02985700. ISSN 0343-6993.
^ Kasman, Alex. "MathFiction: The Girl Who Played With Fire (Stieg Larsson)". College of Charleston. Retrieved 2018-09-10.
^ Gray, Mary W. (2010-02-17). "A Person of Interest: A Novel by Susan Choi and Fermat's Room (La Habitación de Fermat) directed by Luis Piedrahita and Rodrigo Opeña and No One You Know by Michelle Richmond and Pythagoras' Revenge: A Mathematical Mystery by Arturo Sangalli and Pythagorean Crimes by Tefcros Michaelides and The Girl Who Played with Fire by Stieg Larsson". The Mathematical Intelligencer. 32 (3): 67–71. doi:10.1007/s00283-009-9129-8. ISSN 0343-6993.
^ Gowers, Timothy (2009-12-20). "Wiles Meets his Match". Gowers's Weblog. Retrieved 2018-09-10.
^ Fraser, Anne. "LibGuides: Mathematics: Maths Fiction". Assumption College. Retrieved 2018-09-11.
^ Berry, Michael (August 10, 2008). "Clarke and Pohl's 'The Last Theorem'". San Francisco Chronicle. Retrieved 2018-09-10.
^ Moseman, Andrew (2017-09-01). "Here's a Fun Math Goof in 'Star Trek: The Next Generation'". Popular Mechanics. Retrieved 2018-09-10.
^ Knudson, Kevin (2015-08-20). "The Math Of Star Trek: How Trying To Solve Fermat's Last Theorem Revolutionized Mathematics". Forbes. Retrieved 2019-09-03.
^ a b Garmon, Jay. "Geek Trivia: The math behind the myth". TechRepublic. Retrieved 2018-09-11.
^ Greenwald, Sarah J.; Nestler, Andrew (January 2004). "Engaging students with significant mathematical content from The Simpsons". PRIMUS. 14 (1): 29–39. doi:10.1080/10511970408984074.
^ Singh, Simon (2013). The Simpsons and Their Mathematical Secrets. A&C Black. pp. 35–36. ISBN 978-1-4088-3530-2.
^ Singh, Simon (2014-10-17). "Homer's Last Theorem". Boing Boing. Retrieved 2018-09-10.
^ Kasman, Alex. "The Oxford Murders (2004)". MathFiction. Retrieved December 4, 2020.
^ Guaspari, David (1996). "Stoppard's Arcadia". The Antioch Review. 54 (2): 222–238. doi:10.2307/4613314. JSTOR 4613314.
^ Jackson, Allyn (1995). "Love and the Second Law of Thermodynamics: Tom Stoppard's Arcadia" (PDF). Notices of the AMS. 42 (11): 1284–1287.
^ "Math Plus Music Equals Fermat's Last Tango, a World Preem, Opening Dec. 6". Playbill. 2000-12-06. Retrieved 2018-09-10.
^ Emmer, Michele (December 2003). "Fermat's last tango, a musical". The Mathematical Intelligencer. 25 (1): 77–78. doi:10.1007/bf02985645. ISSN 0343-6993. |
Coincidence - Vixrapedia
Coincidence is the effect of quantifying the degree to which two events are said to be uncorrelated in causality. Many real effects are often dismissed as just a coincidence, when really there is a possibly hidden or unrealized link between the two.
Coincidence can be expressed as a raw probability
{\displaystyle p:[0,1]}
or as an adjusted quantity, which is measured in Stans (
{\displaystyle \varsigma }
{\displaystyle C=\tan \left({\frac {\pi }{2}}(1-p)\right)}
The probability of rolling a number between 1 and 6 inclusive is
{\displaystyle p=1}
{\displaystyle 0\,\varsigma }
(0 Stans). The probability of 10 people rolling the same number is
{\displaystyle p=(1/6)^{10}}
{\displaystyle 38.5M\varsigma }
(38.5 million Stans).
Retrieved from "https://www.vixrapedia.org/w/index.php?title=Coincidence&oldid=4252" |
Economic Dispatch and Operations of Electric Utilities | EME 801: Energy Markets, Policy, and Regulation
Electricity is a unique commodity in that it cannot generally be stored at a large scale at reasonable cost, so the entities that operate the transmission grid need to make plans and take actions to keep supply and demand matched in "real-time" - from minute to minute and second to second. Except in circumstances where regulations have prohibited such actions, the electric utility industry has historically broken the process of meeting customer demands into three stages. The sections below discuss each of the three stages under the assumption that all relevant plans and actions are undertaken by a regulated vertically-integrated electric utility.
Days to years in advance, utilities have acquired stores of electric generation capacity through several mechanisms including the construction of new generation plants (and supporting transmission lines), long-term forward contracting, and smaller quantities of spot-market purchases. Construction of new generation plants typically needs to happen years in advance, while forward and spot market contracts may be signed for periods varying from years in advance to a day or hour in advance. Utilities operating in states that have by and large retained the regulated and vertically-integrated structure submit resource plans to their respective state regulatory commissions through a process known as "integrated resource planning."
Generation unit commitment
Days to hours in advance, utilities make "commitment" decisions to have certain quantities of existing generation plant available to produce electricity when the need arises. In some cases (such as nuclear plants or hydroelectric facilities), the commitment decisions are made many weeks or months in advance. The decision to commit a generating unit to be able to produce electricity means that the utility is willing to incur fixed costs related to unit startup in order to have that generating plant ready and available to produce electricity in real time.
Generators with large start-up costs or long minimum-run times (such as large combustion turbines and nuclear power plants) cannot run optimally if their output is determined using a single-period analysis (a "period" in the electric power industry usually refers to a length of time of about an hour). Instead, their operation must be scheduled over a longer time horizon of days or even weeks. A utility would need a forecast of demand weeks in advance before turning on a generator with a long minimum run time. They would need to look at the demand forecast over that period of weeks and decide the lowest-cost mix of generation plants that would meet the demand. This decision is referred to as the unit-commitment decision, and it is aptly named. The utility would, weeks in advance of consumption, have to commit to using a given mix of generation resources. If demand were to deviate wildly (and unexpectedly) from the utility's forecast, it would be faced with the decision of curtailing output from some of its generators (if it over-estimated demand) or utilizing generators with fast start times but high costs (if it under-estimated demand).
Close to the real-time point of consumption (i.e., minutes ahead of real time), LSEs issue dispatch orders to existing generation facilities, setting specific output levels for each. LSEs also utilize some generation facilities to provide "ancillary services," which represent very fast (sub-second to minutes) automatic adjustments to keep supply and demand in balance. The process of dispatching generation plants to meet customer demands within a specified control footprint is variously known as "economic dispatch" or "optimal power flow." The terminology here is suggestive that generation plants are dispatched in such a way as to minimize the total cost of delivered electricity.
The economic dispatch algorithm actually implemented is conceptually pretty simple. Begin by turning on (or "dispatching") your generation source with the lowest "marginal" or operational costs. Have that generation source increase output until all load is met or the generation source hits its capacity constraint, whichever comes first. If the cheapest generation source hits its capacity constraint before all the demand is met, the second-cheapest generation source is turned on, and increases power until either it reaches its capacity constraint or all the demand is met. The process continues by successively turning on more expensive generation sources until the entire load is served. It is important to realize that the economic dispatch algorithm does not consider any fixed costs of power plants - only those costs directly associated with plant operation (for fossil-fired power plants, this would be primarily fuel).
Economic dispatch is best illustrated using an example. Suppose that you were an electric utility that had three generators that could be used to meet electricity demand, as shown in Table 5.1. The fixed costs would represent land leases, fuel/transmission interconnections and any other costs that do not depend on the level of output. The marginal cost measures the amount of money that it costs each plant to produce one megawatt-hour of electric energy.
Table 5.3: Fleet of power plants for economic dispatch example.
Marginal Cost ($/MWh)
Colchester 100 50 10
Burke 20 10 60
Suppose that demand during some time period was 150 MWh. The process of determining economic dispatch would follow three steps.
First, order the plants from lowest to highest marginal cost, which will tell you which plants would be utilized to produce electricity given some level of demand. This picture is called the "dispatch curve," shown in Figure 5.6. Each step in the dispatch curve represents one power plant. The height of the step represents the marginal cost of electricity production for each plant. The width of each step represents the capacity of each plant. Note that the x-axis represents total capacity for the entire utility, not the capacity for any individual power plant.
Figure 5.6: Dispatch curve for the three-generator example.
Dispatch Curve By the Numbers
Colchester 0-100 100 10
Warren 100-175 75 30
Burke 175-195 20 60
To determine which plants would be operating if demand was 150 MWh, you would draw a straight vertical line through the amount 150 MWh. Any capacity to the left of this line would be dispatched, while any capacity to the right of the line would not be dispatched. This is shown in Figure 5.7.
Figure 5.7: Economic dispatch for the three generator example.
In this case, to meet the demand of 150 MWh, the utility would dispatch Colchester at its maximum output (100 MWh) and would meet the remaining 50 MWh of demand using the Warren plant. The Burke plant (the most expensive of the lot) is not dispatched.
We can use the dispatch information to calculate several cost measures for our hypothetical electric utility.
Total Cost is the sum of all costs incurred to meet electricity demand. To obtain total cost, we would multiply quantity produced times marginal cost for each plant, and then add in the fixed costs for all plants. Note that we need to add fixed costs even for plants that do not produce anything.
Average Cost is Total Cost divided by the total amount of electricity produced.
System Marginal Cost is the marginal cost of the generating plant that meets the last MWh of electricity demanded. The system marginal cost is also referred to as the "system lambda."
We can now calculate these cost measures for our example. The economic dispatch of all power plants in the system is:
Colchester produces 100 MWh
Warren produces 50 MWh
Burke produces 0 MWh.
\text{Total cost}=(\$50+\$25+\$10)+\$10/MWh\times 100MWh+\$30/MWh\times 50MWh+\$60/MWh\times 0MWh=\$2,585
\text{Average cost}=\$2,585/150MWh=\$17.23/MWh
The "marginal generator" (the plant used to meet the last MWh of demand) is the Warren plant. So, the system lambda would just be the marginal cost of the Warren plant, which is $30/MWh.
Note the different units for total cost ($) versus average cost and marginal cost ($/MWh).
Here is an exercise that you can try on your own. Take the same three-generator example and suppose that demand was 190 MWh. Verify that:
Average cost = $22.29/MWh
System lambda = $60/MWh
‹ The Mechanics of Rate of Return Regulation up Summary and Final Tasks › |
2\left[\begin{array}{cc}3& 4\\ 5& x\end{array}\right] + \left|\begin{array}{cc}1& y\\ 0& 1\end{array}\right| =\left[\begin{array}{cc}7& 0\\ 10& 5\end{array}\right], \mathrm{find} \left(x-y\right).
\left[x 1\right] \left[\begin{array}{cc}1& 0\\ -2& 0\end{array}\right]=\mathrm{O}.
\left|\begin{array}{cc}2x& 5\\ 8& x\end{array}\right|=\left|\begin{array}{cc}6& -2\\ 7& 3\end{array}\right|
\left(3\sqrt{x}+\frac{1}{\sqrt{x}}\right).
\mathrm{sin} \left({\mathrm{sin}}^{-1}\frac{1}{5}+{\mathrm{cos}}^{-1} x\right)=1,
a*b=\frac{ab}{5}
\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+7\stackrel{^}{\mathrm{k}}
2\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}
\stackrel{\to }{\mathrm{r}}·\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)=2
\underset{0}{\overset{\pi /2}{\int }}{e}^{x}\left(\mathrm{sin} x-\mathrm{cos} x\right) dx.
\stackrel{\to }{\mathrm{a}}=2\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-5\stackrel{^}{\mathrm{k}}
\stackrel{\to }{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-7\stackrel{^}{\mathrm{k}}.
\stackrel{\to }{\mathrm{a}}, \stackrel{\to }{\mathrm{b}}, \stackrel{\to }{\mathrm{c}}
\left[\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}, \stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}, \stackrel{\to }{\mathrm{c}}+\stackrel{\to }{\mathrm{a}}\right]=2 \left[\stackrel{\to }{\mathrm{a}}, \stackrel{\to }{\mathrm{b}}, \stackrel{\to }{\mathrm{c}}\right]
\stackrel{\to }{\mathrm{a}}, \stackrel{\to }{\mathrm{b}}
\stackrel{\to }{\mathrm{c}}
\stackrel{\to }{\mathrm{a}}+\stackrel{\to }{\mathrm{b}}+\stackrel{\to }{\mathrm{c}}=\stackrel{\to }{0 }\mathrm{and} \left|\stackrel{\to }{\mathrm{a}}\right|=3, \left|\stackrel{\to }{\mathrm{b}}\right|=5 \mathrm{and} \left|\stackrel{\to }{\mathrm{c}}\right|=7
\stackrel{\to }{\mathrm{a}} \mathrm{and} \stackrel{\to }{\mathrm{b}}
\left({x}^{2}-1\right)\frac{dy}{dx}+2xy=\frac{2}{{x}^{2}-1}
\int \frac{{\mathrm{sin}}^{6}x+{\mathrm{cos}}^{6}x}{{\mathrm{sin}}^{2}x . {\mathrm{cos}}^{2}x}dx
\int \left(x-3\right)\sqrt{{x}^{2}+3x-18}dx
\mathrm{\theta }=\frac{\mathrm{\pi }}{4}.
{\mathrm{cot}}^{-1}\left(\frac{\sqrt{1+\mathrm{sin} x}+\sqrt{1-\mathrm{sin} x}}{\sqrt{1+\mathrm{sin} }x-\sqrt{1-\mathrm{sin} x}}\right)=\frac{x}{2}; x \in \left(0, \frac{\mathrm{\pi }}{4}\right)
2 {\mathrm{tan}}^{-1} \left(\frac{1}{5}\right)+ {\mathrm{sec}}^{-1}\left(\frac{5\sqrt{2}}{7}\right)+2 {\mathrm{tan}}^{-1} \left(\frac{1}{8}\right)=\frac{\mathrm{\pi }}{4}
\frac{{\mathrm{d}}^{2}\mathrm{y}}{\mathrm{d}{x}^{2}}-\frac{1}{\mathrm{y}}{\left(\frac{\mathrm{dy}}{\mathrm{d}x}\right)}^{2}-\frac{\mathrm{y}}{x}=0.
\left|\begin{array}{ccc}{x}^{2}+1& x\mathrm{y}& x\mathrm{z}\\ x\mathrm{y}& {\mathrm{y}}^{2}+1& yz\\ x\mathrm{z}& y\mathrm{z}& {\mathrm{z}}^{2}+1\end{array}\right|=1+{x}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}
{\mathrm{tan}}^{-1}\left(\frac{\sqrt{1+{x}^{2}}-1}{x}\right)
{\mathrm{sin}}^{-1} \left(\frac{2x}{1+{x}^{2}}\right),
when x ≠ 0. VIEW SOLUTION
\frac{\mathrm{dy}}{\mathrm{d}x}=\frac{x\left(2 \mathrm{log} x+1\right)}{\mathrm{sin} \mathrm{y}+\mathrm{y} \mathrm{cos} \mathrm{y}}
\mathrm{y}=\frac{\pi }{2}
when x = 1. VIEW SOLUTION
Show that lines
\stackrel{\to }{r}=\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\right)+\mathrm{\lambda }\left(3\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}\right)
\stackrel{\to }{r}=\left(4\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{k}}\right)+\mathrm{\mu }\left(2\stackrel{^}{\mathrm{i}}+3\stackrel{^}{k}\right)
intersect. Also, find their point of intersection. VIEW SOLUTION
\stackrel{\to }{\mathrm{r}}=2\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}+\lambda \left(3\stackrel{^}{\mathrm{i}}+4\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)
\stackrel{\to }{\mathrm{r}}.\left(\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right) = 5.
\underset{0}{\overset{\pi /2}{\int }}\frac{x \mathrm{sin} x \mathrm{cos} x}{{\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x}\mathrm{d}x
Of all the closed right circular cylindrical cans of volume 128π cm3, find the dimensions of the can which has minimum surface area. VIEW SOLUTION |
Double-Sided Transferred Carbon Nanotube Arrays for Improved Thermal Interface Materials | J. Electron. Packag. | ASME Digital Collection
Double-Sided Transferred Carbon Nanotube Arrays for Improved Thermal Interface Materials
Andrew J. McNamara,
Zhuomin Zhang,
Kyoung-sik Moon,
Kyoung-sik Moon
Ziyin Lin,
Ziyin Lin
Yagang Yao,
Ching-Ping Wong,
Albany Nanotechnology Center,
Contributed by the Electronic and Photonic Packaging Division of ASME for publication in the JOURNAL OF ELECTRONIC PACKAGING. Manuscript received December 4, 2014; final manuscript received April 22, 2015; published online July 14, 2015. Assoc. Editor: Gongnan Xie.
McNamara, A. J., Joshi, Y., Zhang, Z., Moon, K., Lin, Z., Yao, Y., Wong, C., and Lin, W. (September 1, 2015). "Double-Sided Transferred Carbon Nanotube Arrays for Improved Thermal Interface Materials." ASME. J. Electron. Packag. September 2015; 137(3): 031014. https://doi.org/10.1115/1.4030802
Recently, much attention has been given to reducing the thermal resistance attributed to thermal interface materials (TIMs) in electronic devices, which contribute significantly to the overall package thermal resistance. Thermal transport measured experimentally through several vertically aligned carbon nanotube (VACNT) array TIMs anchored to copper and silicon substrates is considered. A steady-state infrared (IR) microscopy experimental setup was designed and utilized to measure the cross-plane total thermal resistance of VACNT TIMs. Overall thermal resistance for the anchored arrays ranged from
4 to 50 mm2 KW-1
. These values are comparable to the best current TIMs used for microelectronic packaging. Furthermore, thermal stability after prolonged exposure to a high-temperature environment and thermal cycling tests shows limited deterioration for an array anchored using a silver-loaded thermal conductive adhesive (TCA).
Nanotechnolgy, Thermal analysis
Carbon nanotubes, Steady state, Temperature, Thermal resistance, Solders, Thermal conductivity
Characterization of Nanostructured Thermal Interface Materials—A Review
Thermal and Structural Characterizations of Individual Single-, Double-, and Multi-Walled Carbon Nanotubes
Contact Mechanics and Thermal Conductance of Carbon Nanotube Array Interfaces
Water Vapor Treatment for Decreasing the Adhesion Between Vertically Aligned Carbon Nanotubes and the Growth Substrate
.10.1002/cvde.201304319
Infrared Imaging Microscope as an Effective Tool for Measuring Thermal Resistance of Emerging Interface Materials
Paper No. AJTEC2011-44421.10.1115/AJTEC2011-44421
Robust Vertically Aligned Carbon Nanotube–Carbon Fiber Paper Hybrid as Versatile Electrodes for Supercapacitors and Capacitive Deionization
Interface Effect on Thermal Conductivity of Carbon Nanotube Composites
Model for the Effective Thermal Conductivity of Carbon Nanotube Composites
A Combined Process of In Situ Functionalization and Microwave Treatment to Achieve Ultrasmall Thermal Expansion of Aligned Carbon Nanotube–Polymer Nanocomposites: Toward Applications as Thermal Interface Materials
Effect of Bending Buckling of Carbon Nanotubes on Thermal Conductivity of Carbon Nanotube Materials |
The Special Cubic Formula - Wikiversity
The Special Cubic Formula
1 Part I: The Special Cubic Formula
2 Part II: Derivation of the Special Cubic Formula
3 Part III: Limitations of the Formula
4 Part IV: Examples
Part I: The Special Cubic Formula[edit | edit source]
This article discusses a way to solve special cubic equations in the form of
{\displaystyle ax^{3}+bx^{2}+cx+d=0,\quad {\text{where }}c={\frac {b^{2}}{3a}}}
If the cubic equation satisfies that condition, then you can use the special cubic formula to find the value of
{\displaystyle x}
{\displaystyle x={\frac {-b+{\sqrt[{3}]{b^{3}-27a^{2}d}}}{3a}}}
Part II: Derivation of the Special Cubic Formula[edit | edit source]
{\displaystyle ax^{3}+bx^{2}+cx+d=0}
1.) subtract
{\displaystyle d}
from both sides of the equation and divide both sides by
{\displaystyle a}
{\displaystyle x^{3}+{\frac {b}{a}}x^{2}+{\frac {c}{a}}x=-{\frac {d}{a}}}
2.) find the value of
{\displaystyle k}
{\displaystyle x^{3}+3kx^{2}+3k^{2}x+k^{3}=(x+k)^{3}}
There’s a problem with this that puts a limitation on the values of
{\displaystyle b}
{\displaystyle c}
{\displaystyle {\frac {b}{3a}}}
{\displaystyle {\sqrt {\frac {c}{3a}}}}
{\displaystyle \quad c={\frac {b^{2}}{3a}}}
for the formula to work.
If this condition is true, then the value of
{\displaystyle k}
{\displaystyle {\frac {b}{3a}}}
3.) add
{\displaystyle \left({\frac {b}{3a}}\right)^{3}}
(which is
{\displaystyle k^{3}}
) to both sides of the equation
{\displaystyle x^{3}+{\frac {b}{a}}x^{2}+{\frac {c}{a}}x+{\frac {b^{3}}{27a^{3}}}=-{\frac {d}{a}}+{\frac {b^{3}}{27a^{3}}}}
4.) factor the left side of the equation
{\displaystyle \left(x+{\frac {b}{3a}}\right)^{3}=-{\frac {d}{a}}+{\frac {b^{3}}{27a^{3}}}}
5.) rearrange the right side of the equation
{\displaystyle \left(x+{\frac {b}{3a}}\right)^{3}={\frac {b^{3}-27a^{2}d}{27a^{3}}}}
6.) take the cubic root of both sides of the equation
{\displaystyle \left(x+{\frac {b}{3a}}\right)={\frac {\sqrt[{3}]{b^{3}-27a^{2}d}}{3a}}}
{\displaystyle {\frac {b}{3a}}}
from both sides of the equation
{\displaystyle x={\frac {\sqrt[{3}]{b^{3}-27a^{2}d}}{3a}}-{\frac {b}{3a}}}
8.) simplify the equation
{\displaystyle x={\frac {-b+{\sqrt[{3}]{b^{3}-27a^{2}d}}}{3a}}}
Part III: Limitations of the Formula[edit | edit source]
As stated above, this formula can only be used in special cases where
{\displaystyle c}
{\displaystyle b}
are dependent on each other. The equations that display this are:
{\displaystyle c={\frac {b^{2}}{3a}}}
{\displaystyle b=\pm {\sqrt {3ac}}}
If the cubic equation in question does not obey these equations, then a much longer formula must be used to find the solution. These two equations also restrict the cubic formula to cubic equations that only have one solution.
Part IV: Examples[edit | edit source]
{\displaystyle 3x^{3}+6x^{2}+4x+9=0}
Step 1: Check if the equation obeys the limitations
{\displaystyle c=4\quad =\quad {\frac {b^{2}}{3a}}={\frac {6^{2}}{3\cdot 3}}=4}
Step 2: Since the equation obeys the criteria of a special cubic equation, the special cubic formula may be applied
{\displaystyle x={\frac {-6+{\sqrt[{3}]{6^{3}-27\cdot 3^{2}\cdot 9}}}{3\cdot 3}}={\frac {-6-{\sqrt[{3}]{1971}}}{9}}\approx -2.05978}
{\displaystyle 3(-2.05978)^{3}+6(-2.05978)^{2}+4(-2.05978)+9\approx 0}
{\displaystyle 3x^{3}+21x^{2}+2x+3=0}
{\displaystyle c=2\quad \neq \quad {\frac {b^{2}}{3a}}={\frac {21^{2}}{3\cdot 3}}=49}
This equation doesn’t obey the limitations, so it is not a special cubic equation.
{\displaystyle 3x^{3}-6x^{2}+4x-5=0}
{\displaystyle c=4\quad =\quad {\frac {b^{2}}{3a}}={\frac {(-6)^{2}}{3\cdot 3}}=4}
{\displaystyle x={\frac {6+{\sqrt[{3}]{-6^{3}-27\cdot 3^{2}\cdot -5}}}{3\cdot 3}}={\frac {6+{\sqrt[{3}]{999}}}{9}}\approx 1.7774}
{\displaystyle 3(1.7774)^{3}-6(1.7774)^{2}+4(1.7774)-5\approx 0}
Retrieved from "https://en.wikiversity.org/w/index.php?title=The_Special_Cubic_Formula&oldid=2189972" |
torch.kron — PyTorch 1.11.0 documentation
torch.kron
torch.kron¶
torch.kron(input, other, *, out=None) → Tensor¶
\otimes
(a_0 \times a_1 \times \dots \times a_n)
tensor and other is a
(b_0 \times b_1 \times \dots \times b_n)
tensor, the result will be a
(a_0*b_0 \times a_1*b_1 \times \dots \times a_n*b_n)
tensor with the following entries:
(\text{input} \otimes \text{other})_{k_0, k_1, \dots, k_n} = \text{input}_{i_0, i_1, \dots, i_n} * \text{other}_{j_0, j_1, \dots, j_n},
k_t = i_t * b_t + j_t
0 \leq t \leq n
. If one tensor has fewer dimensions than the other it is unsqueezed until it has the same number of dimensions.
This function generalizes the typical definition of the Kronecker product for two matrices to two tensors, as described above. When input is a
(m \times n)
matrix and other is a
(p \times q)
matrix, the result will be a
(p*m \times q*n)
block matrix:
\mathbf{A} \otimes \mathbf{B}=\begin{bmatrix} a_{11} \mathbf{B} & \cdots & a_{1 n} \mathbf{B} \\ \vdots & \ddots & \vdots \\ a_{m 1} \mathbf{B} & \cdots & a_{m n} \mathbf{B} \end{bmatrix}
where input is
\mathbf{A}
and other is
\mathbf{B}
other (Tensor) –
out (Tensor, optional) – The output tensor. Ignored if None. Default: None
>>> mat1 = torch.eye(2)
>>> mat2 = torch.ones(2, 2)
>>> torch.kron(mat1, mat2)
>>> mat2 = torch.arange(1, 5).reshape(2, 2) |
3\mathrm{A}-\mathrm{B}=\left[\begin{array}{cc}5& 0\\ 1& 1\end{array}\right] \mathrm{and} \mathrm{B}=\left[\begin{array}{cc}4& 3\\ 2& 5\end{array}\right]
, then find the matrix A. VIEW SOLUTION
Write the order and the degree of the following differential equation:
{x}^{3}{\left(\frac{{d}^{2}y}{d{x}^{2}}\right)}^{2}+x{\left(\frac{dy}{dx}\right)}^{4}=0
\frac{d}{dx}\left(fof\right) \left(x\right)
2\stackrel{^}{i}+2\stackrel{^}{j}-3\stackrel{^}{k}
Find : ∫ sin x ⋅log cos x dx VIEW SOLUTION
\underset{-\mathrm{\pi }}{\overset{\mathrm{\pi }}{\int }} \left(1-{x}^{2}\right)\mathrm{sin}x {\mathrm{cos}}^{2}x dx.
\underset{-1}{\overset{2}{\int }}\frac{\left|x\right|}{x}dx.
B =\left[\begin{array}{ccc}-2& 2& 0\\ 3& 1& 4\end{array}\right] \mathrm{and} C=\left[\begin{array}{ccc}2& 0& -2\\ 7& 1& 6\end{array}\right].
\mathrm{P}\left(\mathrm{X}=x\right)=\left\{\begin{array}{lll}k& ,& \mathrm{if} x=0\\ 2k& ,& \mathrm{if} x=1\\ 3k& ,& \mathrm{if} x=2\\ 0& ,& \mathrm{otherwise}\end{array}\right\
\sqrt{3}
\stackrel{\to }{a}=2\stackrel{^}{i}+3\stackrel{^}{j}+\stackrel{^}{k}, \stackrel{\to }{b}=\stackrel{^}{i}-2\stackrel{^}{j}+\stackrel{^}{k} \mathrm{and} \stackrel{\to }{c}=-3\stackrel{^}{i}+\stackrel{^}{j}+2\stackrel{^}{k}, \mathrm{find} \left[\stackrel{\to }{a}\stackrel{\to }{b}\stackrel{\to }{c}\right]
\left|\begin{array}{ccc}a& b& c\\ a-b& b-c& c-a\\ b+c& c+a& a+b\end{array}\right|={a}^{3}+{b}^{3}+{c}^{3}-3abc.
\frac{\mathrm{\pi }}{4}
y=\sqrt{3x-2}
2{\mathrm{tan}}^{-1}\left(\frac{y}{x}\right)
\frac{dy}{dx}=\frac{x+y}{x-y}
\frac{dy}{dx}
\left(1-{x}^{2}\right)\frac{{d}^{2}y}{d{x}^{2}}-x\frac{dy}{dx}-2=0
{\int }_{0}^{a}f\left(x\right)dx={\int }_{0}^{a}f\left(a-x\right)dx
{\int }_{0}^{\mathrm{\pi }}\frac{x\mathrm{sin}x}{1+{\mathrm{cos}}^{2}x}dx
\int \frac{\mathrm{cos}x}{\left(1+\mathrm{sin}x\right)\left(2+\mathrm{sin}x\right)}dx
\frac{dy}{dx}-\frac{2x}{1+{x}^{2}}y={x}^{2}+2
\left(x+1\right)\frac{dy}{dx}=2{e}^{-y}-1;y\left(0\right)=0
\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k},2\stackrel{^}{i}+5\stackrel{^}{j},3\stackrel{^}{i}+2\stackrel{^}{j}-3\stackrel{^}{k}
\stackrel{^}{i}-6\stackrel{^}{j}-\stackrel{^}{k}
\stackrel{\to }{\mathrm{AB}}
\stackrel{\to }{\mathrm{CD}}
\frac{1-x}{3}=\frac{7y-14}{\lambda }=\frac{z-3}{2}
\frac{7-7x}{3\lambda }=\frac{y-5}{1}=\frac{6-z}{5}
\mathrm{A}=\left[\begin{array}{ccc}1& 1& 1\\ 1& 0& 2\\ 3& 1& 1\end{array}\right],
\mathrm{A}=\left[\begin{array}{ccc}1& 2& -2\\ -1& 3& 0\\ 0& -2& 1\end{array}\right]
Prove that the curves y2 = 4x and x2 = 4y divide the area of the square bounded by sides x = 0, x = 4, y = 4 and y = 0 into three equal parts.
Using integration, find the area of the triangle whose vertices are (2, 3), (3, 5) and (4, 4). VIEW SOLUTION
\stackrel{\to }{r}=\left(\stackrel{^}{i}+\stackrel{^}{j}\right)+\lambda \left(\stackrel{^}{i}+2\stackrel{^}{j}-\stackrel{^}{k}\right)
Two cards are drawn simultaneously (or successively without replacement) from a well shuffled pack of 52 cards. Find the mean and variance of the number of kings. VIEW SOLUTION |
Exploring Learner Predictions with Partial Dependence and Functional ANOVA | R-bloggers
Exploring Learner Predictions with Partial Dependence and Functional ANOVA
Learners use features to make predictions but how those features are used is often not apparent. mlr can estimate the dependence of a learned function on a subset of the feature space using generatePartialDependenceData.
Partial dependence plots reduce the potentially high dimensional function estimated by the learner, and display a marginalized version of this function in a lower dimensional space. For example suppose $\mathbb{E}[Y \ | \ X = x] = f(x)$. With $(x, y)$ pairs drawn independently, a learner may estimate $\hat{f}$, which, if $X$ is high dimensional can be uninterpretable. Suppose we want to approximate the relationship between some column-wise subset of $X$. We partition $X$ into two sets, $X_s$ and $X_c$ such that $X = X_s \cup X_c$, where $X_s$ is a subset of $X$ of interest.
The partial dependence of $f$ on $X_c$ is
We can use the following estimator:
This is described by Friedman (2001) and in Hastie, Tibsharani, and Friedman (2009).
The individual conditional expectation of an observation can also be estimated using the above algorithm absent the averaging, giving $\hat{f}^{(i)}_{x_s}$ as described in Goldstein, Kapelner, Bleich, and Pitkin (2014). This allows the discovery of features of $\hat{f}$ that may be obscured by an aggregated summary of $\hat{f}$.
The partial derivative of the partial dependence function, $\frac{\partial \hat f_{x_s}}{\partial x_s}$, and the individual conditional expectation function, $\frac{\partial \hat{f}^{(i)}_{x_s}}{\partial x_s}$, can also be computed. For regression and survival tasks the partial derivative of a single feature $x_s$ is the gradient of the partial dependence function, and for classification tasks where the learner can output class probabilities the Jacobian. Note that if the learner produces discontinuous partial dependence (e.g., piecewise constant functions such as decision trees, ensembles of decision trees, etc.) the derivative will be 0 (where the function is not changing) or trending towards positive or negative infinity (at the discontinuities where the derivative is undefined). Plotting the partial dependence function of such learners may give the impression that the function is not discontinuous because the prediction grid is not composed of all discontinuous points in the predictor space. This results in a line interpolating that makes the function appear to be piecewise linear (where the derivative would be defined except at the boundaries of each piece).
The partial derivative can be informative regarding the additivity of the learned function in certain features. If $\hat{f}^{(i)}_{x_s}$ is an additive function in a feature $x_s$, then its partial derivative will not depend on any other features ($x_c$) that may have been used by the learner. Variation in the estimated partial derivative indicates that there is a region of interaction between $x_s$ and $x_c$ in $\hat{f}$. Similarly, instead of using the mean to estimate the expected value of the function at different values of $x_s$, instead computing the variance can highlight regions of interaction between $x_s$ and $x_c$.
Again, see Goldstein, Kapelner, Bleich, and Pitkin (2014) for more details and their package ICEbox for the original implementation. The algorithm works for any supervised learner with classification, regression, and survival tasks.
Our implementation, following mlr’s visualization pattern, consists of the above mentioned function generatePartialDependenceData and plotPartialDependence. The former generates input (objects of class PartialDependenceData) for the latter.
The first step executed by generatePartialDependenceData is to generate a feature grid for every element of the character vector features passed. The data are given by the input argument, which can be a Task or a data.frame. The feature grid can be generated in several ways. A uniformly spaced grid of length gridsize (default 10) from the empirical minimum to the empirical maximum is created by default, but arguments fmin and fmax may be used to override the empirical default (the lengths of fmin and fmax must match the length of features). Alternatively the feature data can be resampled, either by using a bootstrap or by subsampling.
Results from generatePartialDependenceData can be visualized with plotPartialDependence.
## Loading required package: ParamHelpers
lrn.classif = makeLearner("classif.ksvm", predict.type = "prob")
fit.classif = train(lrn.classif, iris.task)
pd = generatePartialDependenceData(fit.classif, iris.task, "Petal.Width")
## PartialDependenceData
## Task: iris-example
## Features: Petal.Width
## Target: setosa, versicolor, virginica
## Derivative: FALSE
## Interaction: FALSE
## Individual: FALSE
## Class Probability Petal.Width
## 1 setosa 0.4872634 0.1000000
## ... (30 rows, 3 cols)
plotPartialDependence(pd, data = iris)
As noted above, $x_s$ does not have to be unidimensional. If it is not, the interaction flag must be set to TRUE. Then the individual feature grids are combined using the Cartesian product, and the estimator above is applied, producing the partial dependence for every combination of unique feature values. If the interaction flag is FALSE (the default) then by default $x_s$ is assumed unidimensional, and partial dependencies are generated for each feature separately. The resulting output when interaction = FALSE has a column for each feature, and NA where the feature was not used. With one feature and a regression task the output is a line plot, with a point for each point in the corresponding feature’s grid. For classification tasks there is a line for each class (except for binary classification tasks, where the negative class is automatically dropped). The data argument to plotPartialPrediction allows the training data to be input to show the empirical marginal distribution of the data.
pd.lst = generatePartialDependenceData(fit.classif, iris.task, c("Petal.Width", "Petal.Length"))
head(pd.lst$data)
## Class Probability Petal.Width Petal.Length
## 1 setosa 0.4872634 0.1000000 NA
tail(pd.lst$data)
## Class Probability Petal.Width Petal.Length
## 55 virginica 0.2622712 NA 3.622222
plotPartialDependence(pd.lst, data = iris)
pd.int = generatePartialDependenceData(fit.classif, iris.task, c("Petal.Width", "Petal.Length"), interaction = TRUE)
pd.int
## Features: Petal.Width, Petal.Length
## Interaction: TRUE
## 1 setosa 0.6043750 0.1000000 1
## ... (300 rows, 4 cols)
plotPartialDependence(pd.int, facet = "Petal.Length")
When interaction = TRUE, plotPartialDependence can either facet over one feature, showing the conditional relationship between the other feature and $\hat{f}$ in each panel, or a tile plot. The latter is, however, not possible with multiclass classification (an example of a tile plot will be shown later).
At each step in the estimation of $\hat{f}_{x_s}$ a set of predictions of length $N$ is generated. By default the mean prediction is used. For classification where predict.type = "prob" this entails the mean class probabilities. However, other summaries of the predictions may be used. For regression and survival tasks the function used here must either return one number or three, and, if the latter, the numbers must be sorted lowest to highest. For classification tasks the function must return a number for each level of the target feature.
As noted, the fun argument can be a function which returns three numbers (sorted low to high) for a regression task. This allows further exploration of relative feature importance. If a feature is relatively important, the bounds are necessarily tighter because the feature accounts for more of the variance of the predictions, i.e., it is “used” more by the learner. More directly setting fun = var identifies regions of interaction between $x_s$ and $x_c$. This can also be accomplished by computing quantiles. The wider the quantile bounds, the more variation in $\hat{f}$ is due to features other than $x_s$ that is shown in the plot.
lrn.regr = makeLearner("regr.ksvm")
fit.regr = train(lrn.regr, bh.task)
pd.ci = generatePartialDependenceData(fit.regr, bh.task, "lstat",
fun = function(x) quantile(x, c(.25, .5, .75)))
pd.ci
## Task: BostonHousing-example
## Features: lstat
## Target: medv
## medv lstat lower upper
## 1 24.92828 1.730000 21.33593 29.56459
## 4 20.70007 13.810000 18.70272 23.63418
plotPartialDependence(pd.ci)
In addition to bounds based on a summary of the distribution of the conditional expectation of each observation, learners which can estimate the variance of their predictions can also be used. The argument bounds is a numeric vector of length two which is added (so the first number should be negative) to the point prediction to produce a confidence interval for the partial dependence. The default is the .025 and .975 quantiles of the Gaussian distribution.
fit.se = train(makeLearner("regr.randomForest", predict.type = "se"), bh.task)
pd.se = generatePartialDependenceData(fit.se, bh.task, c("lstat", "crim"))
head(pd.se$data)
## medv lstat crim lower upper
## 1 31.16866 1.730000 NA 27.33761 34.99971
## 4 22.04595 13.810000 NA 20.29177 23.80014
tail(pd.se$data)
## medv lstat crim lower upper
## 15 21.80198 NA 39.54849 19.50150 24.10246
plotPartialDependence(pd.se)
As previously mentioned if the aggregation function is not used, i.e., it is the identity, then the conditional expectation of $\hat{f}^{(i)}_{x_s}$ is estimated. If individual = TRUE then generatePartialDependenceData returns $N$ partial dependence estimates made at each point in the prediction grid constructed from the features.
pd.ind.regr = generatePartialDependenceData(fit.regr, bh.task, "lstat", individual = TRUE)
pd.ind.regr
## Individual: TRUE
## Predictions centered: FALSE
## medv lstat idx
## 1 27.34360 1.730000 1
## 4 23.63205 13.810000 1
## ... (5060 rows, 3 cols)
plotPartialDependence(pd.ind.regr)
The resulting output, particularly the element data in the returned object, has an additional column idx which gives the index of the observation to which the row pertains.
For classification tasks this index references both the class and the observation index.
pd.ind.classif = generatePartialDependenceData(fit.classif, iris.task, "Petal.Length", individual = TRUE)
pd.ind.classif
## Features: Petal.Length
## Class Probability Petal.Length idx
## 1 setosa 0.9824128 1 1.setosa
plotPartialDependence(pd.ind.classif)
The plots, at least in these forms, are difficult to interpet. Individual estimates of partial dependence can also be centered by predictions made at all $N$ observations for a particular point in the prediction grid created by the features. This is controlled by the argument center which is a list of the same length as the length of the features argument and contains the values of the features desired.
pd.ind.classif = generatePartialDependenceData(fit.classif, iris.task, "Petal.Length", individual = TRUE, center = list("Petal.Length" = min(iris$Petal.Length)))
Partial derivatives can also be computed for individual partial dependence estimates and aggregate partial dependence. This is restricted to a single feature at a time. The derivatives of individual partial dependence estimates can be useful in finding regions of interaction between the feature for which the derivative is estimated and the features excluded. Applied to the aggregated partial dependence function they are not very informative, but when applied to the individual conditional expectations, they can be used to find regions of interaction.
pd.regr.der.ind = generatePartialDependenceData(fit.regr, bh.task, "lstat", derivative = TRUE, individual = TRUE)
head(pd.regr.der.ind$data)
## medv lstat idx
## 1 -0.2446031 1.730000 1
## 4 -0.3059314 13.810000 1
plotPartialDependence(pd.regr.der.ind)
pd.classif.der.ind = generatePartialDependenceData(fit.classif, iris.task, "Petal.Width", derivative = TRUE, individual = TRUE)
head(pd.classif.der.ind$data)
## Class Probability Petal.Width idx
## 1 setosa 0.04088195 0.1 1.setosa
plotPartialDependence(pd.classif.der.ind)
This suggests that Petal.Width interacts with some other feature in the neighborhood of $(1.5, 2)$ for classes “virginica” and “versicolor”.
Hooker (2004) proposed the decomposition of a learned function $\hat{f}$ as a sum of lower dimensional functions
f(\mathbf{x}) = g_0 + \sum_{i = 1}^p g_{i}(x_i) + \sum_{i \neq j} g_{ij}(x_{ij}) + \ldots
where $p$ is the number of features. generateFunctionalANOVAData estimates the individual $g$ functions using partial dependence. When functions depend only on one feature, they are equivalent to partial dependence, but a $g$ function which depends on more than one feature is the “effect” of only those features: lower dimensional “effects” are removed.
Here $u$ is a subset of ${1, \ldots, p}$. When $|v| = 1$ $g_v$ can be directly computed by computing the bivariate partial dependence of $\hat{f}$ on $x_u$ and then subtracting off the univariate partial dependences of the features contained in $v$.
Although this decomposition is generalizable to classification it is currently only available for regression tasks.
fa = generateFunctionalANOVAData(fit.regr, bh.task, "lstat", depth = 1, fun = median)
## FunctionalANOVAData
## effect medv lstat
## 1 lstat 24.88912 1.730000
## 4 lstat 20.67813 13.810000
pd.regr = generatePartialDependenceData(fit.regr, bh.task, "lstat", fun = median)
pd.regr
## medv lstat
## 1 24.88912 1.730000
## 4 20.67813 13.810000
The depth argument is similar to the interaction argument in generatePartialDependenceData but instead of specifying whether all of joint “effect” of all the features is computed, it determines whether “effects” of all subsets of the features given the specified depth are computed. So, for example, with $p$ features and depth 1, the univariate partial dependence is returned. If, instead, depth = 2, then all possible bivariate functional ANOVA effects are returned. This is done by computing the univariate partial dependence for each feature and subtracting it from the bivariate partial dependence for each possible pair.
fa.bv = generateFunctionalANOVAData(fit.regr, bh.task, c("crim", "lstat", "age"), depth = 2)
fa.bv
## Features: crim, lstat, age
## effect medv crim lstat age
## 1 crim:lstat -22.68839 0.006320 1.73 NA
## 3 crim:lstat -24.78243 19.777404 1.73 NA
names(table(fa.bv$data$effect)) ## interaction effects estimated
## [1] "crim:age" "crim:lstat" "lstat:age"
Plotting univariate and bivariate functional ANOVA components works the same as for partial dependence.
fa.bv = generateFunctionalANOVAData(fit.regr, bh.task, c("crim", "lstat"), depth = 2)
plotPartialDependence(fa.bv, geom = "tile", data = getTaskData(bh.task))
When overplotting the training data on the plot it is easy to see that much of the variation of the effect is due to extrapolation. Although it hasn’t been implemented yet, weighting the functional ANOVA appropriately can ensure that the estimated effects do not depend (or depend less) on regions of the feature space which are sparse.
I also plan on implementing the faster estimation algorith for expanding the functionality of the functional ANOVA function include faster computation using the algorithm in Hooker (2007) and weighting (in order to avoid excessive reliance on points of extrapolation) using outlier detection or joint density estimation. |
Electrochemical spectral induced polarization modeling of artificial sulfide-sand mixturesElectrochemical SIP modeling | Geophysics | GeoScienceWorld
Edmundo Placencia-Gómez;
Edmundo Placencia-Gómez
, Department of Civil and Environmental Engineering, Aalto,
. E-mail: edmundo.placencia@aalto.fi.
Lee D. Slater
, Department of Earth & Environmental Sciences, Newark, New Jersey,
. E-mail: lslater@andromeda.rutgers.edu.
Edmundo Placencia-Gómez, Lee D. Slater; Electrochemical spectral induced polarization modeling of artificial sulfide-sand mixtures. Geophysics 2014;; 79 (6): EN91–EN106. doi: https://doi.org/10.1190/geo2014-0034.1
We examined the sensitivity of the electrochemical spectral induced polarization (SIP) model developed by Wong to the oxidation extent of pyrite and pyrrhotite minerals disseminated in silica sand. The sensitivity of this model to the oxidation of sulfide minerals was mainly related to the model parameters defining the ratio of the active to the inactive passive ions
(c2/co)
dissolved in the pore water, and the variation of the current reaction parameters
α
β
. The increase in these parameters as well as in the associated exchange current densities,
io(α)
io(β)
was consistent with an increase in the activation of the charge transfer at the metal-electrolyte interface, resulting in the decrease in polarization of such an interface, which was reflected by a decrease in the SIP phase response as previously argued by Wong. Under this premise, the model described fairly well measurements below 500 Hz from a laboratory experiment, being consistent with the depletion of the SIP phase response associated with the oxidation degree promoted on the disseminate sulfides analyzed here. This suggested that electrochemical modeling of SIP measurements can provide information to assess the oxidation state of sulfides and also to infer the formation of passivating layers coating the metal minerals during oxidation-dissolution processes. Our results suggested a possible alternative for the monitoring of mine waste deposits producing acid mine drainage and the stability of sequestered harmful metals during remedial treatments by means of the SIP method.
Combining spectral induced polarization with X-ray tomography to investigate the importance of DNAPL geometry in sand samples |
Region growing - Wikipedia
Find sources: "Region growing" – news · newspapers · books · scholar · JSTOR (February 2017) (Learn how and when to remove this template message)
Region growing is a simple region-based image segmentation method. It is also classified as a pixel-based image segmentation method since it involves the selection of initial seed points.
This approach to segmentation examines neighboring pixels of initial seed points and determines whether the pixel neighbors should be added to the region. The process is iterated on, in the same manner as general data clustering algorithms. A general discussion of the region growing algorithm is described below.
1 Region-based segmentation
2 Basic concept of seed points
3.1 Suitable selection of seed points
3.2 More information of the image is better
3.3 Minimum area threshold
3.4 Similarity threshold value
Region-based segmentation[edit]
The main goal of segmentation is to partition an image into regions. Some segmentation methods such as thresholding achieve this goal by looking for the boundaries between regions based on discontinuities in grayscale or color properties. Region-based segmentation is a technique for determining the region directly. The basic formulation is:[1]
{\displaystyle (a){\text{ }}\bigcup _{i=1}^{n}{R_{i}=R.}}
{\displaystyle (b){\text{ }}R_{i}{\text{ is a connected region}},{\text{ i}}={\text{1}},{\text{ 2}},{\text{ }}...,{\text{n}}}
{\displaystyle (c){\text{ }}R_{i}\bigcap R_{j}=\varnothing ,i\neq j}
{\displaystyle (d){\text{ }}P(R_{i})=\mathrm {TRUE} {\text{ for }}i=1,2,...,n.}
{\displaystyle (e){\text{ }}P(R_{i}\bigcup R_{j})=\mathrm {FALSE} {\text{ for any adjacent region }}R_{i}{\text{ and }}R_{j}.}
{\displaystyle P(R_{i})}
is a logical predicate defined over the points in set
{\displaystyle R_{i}}
{\displaystyle \varnothing }
is the null set.
(a) means that the segmentation must be complete; that is, every pixel must be in a region.
(b) requires that points in a region must be connected in some predefined sense.
(c) indicates that the regions must be disjoint.
(d) deals with the properties that must be satisfied by the pixels in a segmented region. For example,
{\displaystyle P(R_{i})={\text{TRUE}}}
if all pixels in
{\displaystyle R_{i}}
have the same grayscale.
(e) indicates that region
{\displaystyle R_{i}}
{\displaystyle R_{j}}
are different in the sense of predicate
{\displaystyle P}
Basic concept of seed points[edit]
The first step in region growing is to select a set of seed points. Seed point selection is based on some user criterion (for example, pixels in a certain grayscale range, pixels evenly spaced on a grid, etc.). The initial region begins as the exact location of these seeds.
The regions are then grown from these seed points to adjacent points depending on a region membership criterion. The criterion could be, for example, pixel intensity, grayscale texture, or colour.
Since the regions are grown on the basis of the criterion, the image information itself is important. For example, if the criterion were a pixel intensity threshold value, knowledge of the histogram of the image would be of use, as one could use it to determine a suitable threshold value for the region membership criterion.
One can use 4-connected neighborhood to grow from the seed points. An alternative for pixels adjacent relationship is the 8-connected neighborhood. Pixels adjacent to the seed points are examined and classified into the seed points if they have the same intensity value. It is an iterated process until there are no change in two successive iterative stages. Other criteria can be chosen; the main goal is to classify the similarity of the image into regions.
Suitable selection of seed points[edit]
The selection of seed points is depending on the users. For example, in a grayscale lightning image, we may want to segment the lightning from the background. Then probably, we can examine the histogram and choose the seed points from the highest range of it.
More information of the image is better[edit]
Obviously, the connectivity or pixel adjacent information is helpful for us to determine the threshold and seed points.
Minimum area threshold[edit]
No region in region growing method result will be smaller than this threshold in the segmented image.
Similarity threshold value[edit]
If the difference of pixel-value or the difference value of average grayscale of a set of pixels less than “Similarity threshold value”, the regions will be considered as a same region.
The criteria of similarities or so called homogeneity we choose are also important. It usually depends on the original image and the segmentation result we want.[2]
Some criteria often used are grayscale (average intensity or variance), color, and texture or shape.
Can correctly separate the regions that have the same properties we define.
Can provide the original images which have clear edges with good segmentation results.
Simple concept: only need a small number of seed points to represent the property we want, then grow the region.
Can determine the seed points and the criteria we want to make.
Can choose the multiple criteria at the same time.
Theoretical very efficient due to visiting each pixel by a limited bound of times.
Unless image has had a threshold function applied, a continuous path of points related to color may exist, which connects any two points in the image.
Practically random memory access slows down the algorithm, so adaption might be needed
Watershed (image processing)
^ Pal, Nikhil R; Pal, Sankar K (1993). "A review on image segmentation techniques". Pattern Recognition. 26 (9): 1277–1278. doi:10.1016/0031-3203(93)90135-J.
^ Adoui, Mohammed El; Drisis, Stylianos; Benjelloun, Mohammed (2017-07-21). Analyzing breast tumor heterogeneity to predict the response to chemotherapy using 3D MR images registration. ACM. pp. 56–63. doi:10.1145/3128128.3128137. ISBN 9781450352819.
Jian-Jiun Ding, The class of "Time-Frequency Analysis and Wavelet Transform", the Department of Electrical Engineering, National Taiwan University (NTU), Taipei, Taiwan, 2007.
Jian-Jiun Ding, The class of "Advanced Digital Signal Processing", the Department of Electrical Engineering, National Taiwan University (NTU), Taipei, Taiwan, 2008.
W. K. Pratt, Digital Image Processing 4th Edition, John Wiley & Sons, Inc., Los Altos, California, 2007
M. Petrou and P. Bosdogianni, Image Processing the Fundamentals, Wiley, UK, 2004.
R. C. Gonzalez and R.E. Woods, Digital Image Processing 2nd Edition, Prentice Hall, New Jersey, 2002.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Region_growing&oldid=1050260545" |
Complete blow up and global behaviour of solutions of $u_t - \Delta u = g (u)$
Complete blow up and global behaviour of solutions of
{u}_{t}-\Delta u=g\left(u\right)
title = {Complete blow up and global behaviour of solutions of $u_t - \Delta u = g (u)$},
AU - Martel, Yvan
TI - Complete blow up and global behaviour of solutions of $u_t - \Delta u = g (u)$
Martel, Yvan. Complete blow up and global behaviour of solutions of $u_t - \Delta u = g (u)$. Annales de l'I.H.P. Analyse non linéaire, Tome 15 (1998) no. 6, pp. 687-723. http://www.numdam.org/item/AIHPC_1998__15_6_687_0/
[1] P. Baras and L. Cohen, Complete blow after Tmax for the solution of a semilinear heat equation, J. Funct. Anal., Vol. 71, 1987, pp. 142-174. | MR 879705 | Zbl 0653.35037
[2] P. Baras and M. Pierre, Critère d'existence de solutions positives pour des equations semi-linéaires non monotones, Ann. Inst. Henri Poincaré, Vol. 2, 1985, pp. 185-212. | Numdam | MR 797270 | Zbl 0599.35073
[3] H. Brezis, T. Cazenave, Y. Martel and A. Ramiandrisoa, Blow up for ut - Δu = g(u) revisited, Adv. Diff. Eq., Vol. 1, 1996, pp. 73-90. | MR 1357955 | Zbl 0855.35063
[4] H. Fujita, On the nonlinear equations Δu + eu = 0 and ∂u/ ∂t = Δu + eu, Bull. Amer. Math. Soc., Vol. 75, 1969, pp. 132-135. | MR 239258 | Zbl 0216.12101
[5] V.A. Galaktionov and J.L. Vazquez, Continuation of blow up solutions of nonlinear heat equations in several space dimensions, to appear.
[6] A.A. Lacey and D.E. Tzanetis, Global, unbounded solutions to a parabolic equation, J. Diff. Eq., Vol. 101, 1993, pp. 80-102. | MR 1199484 | Zbl 0799.35123
[7] Y. Martel, Uniqueness of weak extremal solutions of nonlinear elliptic problems, to appear in Houston Journal of Math., 1997. | MR 1688823 | Zbl 0884.35037
[8] P. Mironescu and V.D. Radulescu, The study of a bifurcation problem associated to an asymptotically linear function, Nonlinear Analysis TMA, Vol. 26, 1996, pp. 857-875. | MR 1362758 | Zbl 0842.35008
[9] W.-M. Ni, P.E. Sacks and J. Tavantzis, On the asymptotic behavior of solutions of certain quasilinear parabolic equations, J. Diff. Eq., Vol. 54, 1984, pp. 97-120. | MR 756548 | Zbl 0565.35053 |
Kohn–Sham equations - Wikipedia
Schrödinger equation of a fictitious system of non-interacting particles
In physics and quantum chemistry, specifically density functional theory, the Kohn–Sham equation is the non-interacting Schrödinger equation (more clearly, Schrödinger-like equation) of a fictitious system (the "Kohn–Sham system") of non-interacting particles (typically electrons) that generate the same density as any given system of interacting particles.[1][2] The Kohn–Sham equation is defined by a local effective (fictitious) external potential in which the non-interacting particles move, typically denoted as vs(r) or veff(r), called the Kohn–Sham potential. If the particles in the Kohn–Sham system are non-interacting fermions (non-fermion DFT has been researched[3][4]), the Kohn–Sham wavefunction is a single Slater determinant constructed from a set of orbitals that are the lowest-energy solutions to
{\displaystyle \left(-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}+v_{\text{eff}}(\mathbf {r} )\right)\varphi _{i}(\mathbf {r} )=\varepsilon _{i}\varphi _{i}(\mathbf {r} ).}
This eigenvalue equation is the typical representation of the Kohn–Sham equations. Here εi is the orbital energy of the corresponding Kohn–Sham orbital
{\displaystyle \varphi _{i}}
, and the density for an N-particle system is
{\displaystyle \rho (\mathbf {r} )=\sum _{i}^{N}|\varphi _{i}(\mathbf {r} )|^{2}.}
The Kohn–Sham equations are named after Walter Kohn and Lu Jeu Sham, who introduced the concept at the University of California, San Diego, in 1965.
Kohn–Sham potential[edit]
In Kohn–Sham density functional theory, the total energy of a system is expressed as a functional of the charge density as
{\displaystyle E[\rho ]=T_{s}[\rho ]+\int d\mathbf {r} \,v_{\text{ext}}(\mathbf {r} )\rho (\mathbf {r} )+E_{\text{H}}[\rho ]+E_{\text{xc}}[\rho ],}
where Ts is the Kohn–Sham kinetic energy, which is expressed in terms of the Kohn–Sham orbitals as
{\displaystyle T_{s}[\rho ]=\sum _{i=1}^{N}\int d\mathbf {r} \,\varphi _{i}^{*}(\mathbf {r} )\left(-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}\right)\varphi _{i}(\mathbf {r} ),}
vext is the external potential acting on the interacting system (at minimum, for a molecular system, the electron–nuclei interaction), EH is the Hartree (or Coulomb) energy
{\displaystyle E_{\text{H}}[\rho ]={\frac {e^{2}}{2}}\int d\mathbf {r} \int d\mathbf {r} '\,{\frac {\rho (\mathbf {r} )\rho (\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|}},}
and Exc is the exchange–correlation energy. The Kohn–Sham equations are found by varying the total energy expression with respect to a set of orbitals, subject to constraints on those orbitals,[5] to yield the Kohn–Sham potential as
{\displaystyle v_{\text{eff}}(\mathbf {r} )=v_{\text{ext}}(\mathbf {r} )+e^{2}\int {\frac {\rho (\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|}}\,d\mathbf {r} '+{\frac {\delta E_{\text{xc}}[\rho ]}{\delta \rho (\mathbf {r} )}},}
where the last term
{\displaystyle v_{\text{xc}}(\mathbf {r} )\equiv {\frac {\delta E_{\text{xc}}[\rho ]}{\delta \rho (\mathbf {r} )}}}
is the exchange–correlation potential. This term, and the corresponding energy expression, are the only unknowns in the Kohn–Sham approach to density functional theory. An approximation that does not vary the orbitals is Harris functional theory.
The Kohn–Sham orbital energies εi, in general, have little physical meaning (see Koopmans' theorem). The sum of the orbital energies is related to the total energy as
{\displaystyle E=\sum _{i}^{N}\varepsilon _{i}-E_{\text{H}}[\rho ]+E_{\text{xc}}[\rho ]-\int {\frac {\delta E_{\text{xc}}[\rho ]}{\delta \rho (\mathbf {r} )}}\rho (\mathbf {r} )\,d\mathbf {r} .}
Because the orbital energies are non-unique in the more general restricted open-shell case, this equation only holds true for specific choices of orbital energies (see Koopmans' theorem).
^ Kohn, Walter; Sham, Lu Jeu (1965). "Self-Consistent Equations Including Exchange and Correlation Effects". Physical Review. 140 (4A): A1133–A1138. Bibcode:1965PhRv..140.1133K. doi:10.1103/PhysRev.140.A1133.
^ Parr, Robert G.; Yang, Weitao (1994). Density-Functional Theory of Atoms and Molecules. Oxford University Press. ISBN 978-0-19-509276-9. OCLC 476006840. OL 7387548M.
^ Wang, Hongmei; Zhang, Yunbo (2013). "Density-functional theory for the spin-1 bosons in a one-dimensional harmonic trap". Physical Review A. 88 (2). arXiv:1304.1328. doi:10.1103/PhysRevA.88.023626. S2CID 119280339.
^ Hu, Yayun; Murthy, G.; Rao, Sumathi; Jain, J. K. (2021). "Kohn-Sham density functional theory of Abelian anyons". Physical Review B. 103 (3). arXiv:2010.09872. doi:10.1103/PhysRevB.103.035124. S2CID 224802789.
^ Tomas Arias (2004). "Kohn–Sham Equations". P480 notes. Cornell University. Archived from the original on 2020-02-18. Retrieved 2021-01-14.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Kohn–Sham_equations&oldid=1070831947" |
The binomial terms you found in parts (c) and (d) of the previous problem are called the complex conjugates of the complex number. You can use the complex conjugate to simplify division problems involving complex numbers. Simplify the following division problems by multiplying the numerator and the denominator by the complex conjugate of the denominator.
\frac { 5 } { 3 + 2 i }
\frac{5}{3+2\textit{i}}\left( \frac{3-2\textit{i}}{3-2\textit{i}} \right)
\frac{15-10\textit{i}}{9-6\textit{i}+6\textit{i}-4\textit{i}^2}
\frac{15-10\textit{i}}{9+4}
\frac{15}{13}-\frac{10}{13}\textit{i}
\frac { 2 - i } { 2 + i }
\frac{3}{5}-\frac{4}{5}\textit{i}
\frac { 2 i } { \sqrt { 3 } + 7 i }
\frac{2\textit{i}}{\sqrt{3}+7\textit{i}}\left( \frac{\sqrt{3}-7\textit{i}}{\sqrt{3}-7\textit{i}} \right)
\frac { \sqrt { 3 } + i \sqrt { 2 } } { \sqrt { 3 } - i \sqrt { 2 } }
\frac{\sqrt{3}+\textit{i}\sqrt{2}}{\sqrt{3}-\textit{i}\sqrt{2}}\left( \frac{\sqrt{3}+\textit{i}\sqrt{2}}{\sqrt{3}+\textit{i}\sqrt{2}} \right) |
Discrete-time or continuous-time lead-lag compensator - Simulink - MathWorks í•œêµ
General Control​
Lead-Lag (Discrete or Continuous)
Bypass Compensator Dynamics
Lead time constant, T1
Lag time constant, T2
Discrete-time or continuous-time lead-lag compensator
The Lead-Lag (Discrete or Continuous) block implements a lead-lag compensator in conformance with IEEE 421.5-2016[1].
You can switch between continuous and discrete implementations of the block using the Sample time parameter.
To configure the compensator for continuous time, set the Sample time property to 0. This representation is equivalent to the continuous transfer function:
G\left(s\right)=\frac{{T}_{1}s+1}{{T}_{2}s+1},
T1 is the lead time constant.
T2 is the lag time constant.
From the preceeding transfer function, the compensator defining equations are:
\left\{\begin{array}{c}\stackrel{Ë}{x}\left(t\right)=\frac{1}{{T}_{2}}\left(u\left(t\right)âx\left(t\right)\right)\\ y\left(t\right)=\frac{{T}_{1}}{{T}_{2}}u\left(t\right)+\left(1â\frac{{T}_{1}}{{T}_{2}}\right)x\left(t\right)\end{array}\text{â}\text{â}\text{â}y\left(0\right)=x\left(0\right)={u}_{0},
To configure the compensator for discrete time, set the Sample time property to a positive, nonzero value, or to -1 to inherit the sample time from an upstream block. The discrete representation is equivalent to the transfer function:
\frac{{T}_{1}z+\left({T}_{s}â{T}_{1}\right)}{{T}_{2}z+\left({T}_{s}â{T}_{2}\right)},
Ts is the compensator sample time.
From the discrete transfer function, the compensator equations are defined using the forward Euler method:
\left\{\begin{array}{c}x\left(n+1\right)=\left(1â\frac{{T}_{s}}{{T}_{2}}\right)x\left(n\right)+\left(\frac{{T}_{s}}{{T}_{2}}\right)u\left(n\right)\\ y\left(n\right)=\left(1â\frac{{T}_{1}}{{T}_{2}}\right)x\left(n\right)+\left(\frac{{T}_{1}}{{T}_{2}}\right)u\left(n\right)\end{array}\text{â}\text{â}\text{â}y\left(0\right)=x\left(0\right)={u}_{0},
Set the Upper saturation limit and Lower saturation limit parameters to use the anti-windup saturation method.
The anti-windup method limits the compensator state between the lower saturation limit A and upper saturation limit B:
A<=x<=B\text{â}.
Because the state is limited, the output can respond immediately to a reversal of the input sign when the integral is saturated.
This block does not provide a windup saturation method. To use the windup saturation method, set the Upper saturation limit parameter to inf, the Lower saturation limit parameter to -inf, and attach a Saturation block to the output.
Set the lag time constant to zero or to a value equal to that of the lead time constant to ignore the dynamics of the compensator. When bypassed, the block feeds the input directly to the output:
\begin{array}{c}{T}_{1}=0\\ {T}_{2}=0\\ {T}_{1}={T}_{2}\end{array}\right\}\text{â}\text{â}y=u\text{â}.
In the continuous case, both the sample time and at least one time constant must be zero.
u — Compensator input
Lead-lag compensator input signal. The block uses the input initial value to determine the state initial value.
y — Compensator output
Lead-lag compensator output.
Lead time constant, T1 — Lead time constant
Compensator lead time constant. To bypass the dynamics of the compensator. set this value to 0 or to the value of the Lag time constant, T2 parameter.
Lag time constant, T2 — Lag time constant
Compensator lag time constant. To bypass the dynamics of the compensator. set this value to 0 or to the value of the Lead time constant, T1 parameter.
Upper saturation limit — State upper limit
inf (default) | real number
Compensator upper state limit. Set this to inf for an unsaturated upper limit, or to a finite value to prevent upper windup of the system's integrator.
Lower saturation limit — State lower limit
-inf (default) | real number
Compensator lower state limit. Set this to -inf for an unsaturated lower limit, or to a finite value to prevent lower windup of the system's integrator.
Filtered Derivative (Discrete or Continuous) | Washout (Discrete or Continuous) | Integrator (Discrete or Continuous) | Integrator with Wrapped State (Discrete or Continuous) | Low-Pass Filter (Discrete or Continuous) |
Is T completely arbitrary in the definition of reachability? - Murray Wiki
Q Is T completely arbitrary in the definition of reachability, or is it limited by the system? For instance, if the system is a car, x0 = LA, xf = NYC, does the fact that I can't get there from here in 1 second mean that it isn't reachable?
A If a system is continuous time and linear, then the property has to be veryfied for any time T. Indeed, when you check using the reachability matrix
{\displaystyle [B\quad AB\quad A^{2}B\quad ...\quad A^{n-1}B]}
there is no time dependence in this.
On the other hand, the notion of reachability for discrete time systems (you have not seen this, and won't probably see it) requires that a desired state
{\displaystyle {\bar {x}}}
be reached within
{\displaystyle n-1}
{\displaystyle n}
is always the order (dimension) of the system.
For a nonlinear system, things are more complicated: one has to talk about reachable regions in a certain finite or desired time. So unfortunately your car is a highly constrained nonlinear system, therefore NY is in your reachability region, but you takes a while to get there...
Retrieved from "https://murray.cds.caltech.edu/index.php?title=Is_T_completely_arbitrary_in_the_definition_of_reachability%3F&oldid=5110" |
Stoichiometry of Chemical Reactions - Course Hero
General Chemistry/Stoichiometry of Chemical Reactions
In chemistry, knowledge of reactants and products, chemical reactions and types, and solutions and precipitations is vital. These concepts must be studied deeply, qualitatively, and quantitatively. To study a reaction quantitatively, the masses and volumes of reactants and products involved and the percent yields from a reaction are important factors in conducting laboratory investigations and further study of substances. The relationship between the amounts, or quantities, of substances involved in a reaction or compound is described using stoichiometry. The quantitative measure of reactants required to produce useful and practical quantities of products is the foundation for several mass industrial productions of medicines, dyes, fertilizers, pesticides, and other important chemicals. The percent yield of a useful element from extracted raw ore decides whether an extraction of ore through a certain process is beneficial. Understanding how the amounts of substances involved in reactions change, the concepts of molar mass and molar volume, calculations of percent yields, and the reactants that control the yields as limiting reactants is essential to quantitative analysis of chemicals and substances.
A mole is the amount of substance that has
6.023\times10^{23}
particles, and the molar mass is the mass of one mole of a substance.
Mass can be converted to moles by dividing the mass by the molar mass of the substance.
Moles can be converted to mass by multiplying the number of moles of the substance by the molar mass of the substance.
To find the mass of a product, first convert the mass of the given reactant to moles. Then, use the mole ratio from the balanced chemical equation to convert moles of the reactant to moles of the product. Finally, change the mole of the product to grams.
Moles of a substance are converted to moles of another substance by comparing the mole ratio of a balanced chemical equation.
The mole ratio from a balanced chemical equation must be used when it is necessary to convert the mass of one substance to the moles of another substance within a chemical reaction. |
Explain the feature 'interdependence of firms' in an oligopoly market. VIEW SOLUTION
Giving reason comment on the shape of the Production Possibilities curve based on the following schedule :
A consumer spends Rs. 100 on a good priced at Rs. 4 per unit. When price falls by 50 per cent, the consumer continues to spend Rs. 100 on the good. Calculate price elasticity of demand by percentage method. VIEW SOLUTION
Market for a good is in equilibrium. Supply of the good 'increases'. Explain the chain of effects of this change. VIEW SOLUTION
\frac{1}{MPC}
\frac{1}{MPS}
\frac{1}{1-MPS}
\frac{1}{MPC-1}
If the Real GDP is Rs. 300 and Nominal GDP is Rs. 330, calculate Price Index (base = 100). VIEW SOLUTION
An economy is in equilibrium. Find Marginal Propensity to Consume from the following :
Calculate 'Net Domestic Product at Factor Cost' and 'Gross National Disposable Income':
(iii) Net imports (−)20
(iv) Net domestic capital formation 100
(vi) Depreciation 50
(vii) Change in stocks 17
(viii) Net indirect tax 120
(ix) Government final consumption expenditure 200
(x) Exports 30 |
Classification of normal quartic surfaces with irrational singularities
July, 2004 Classification of normal quartic surfaces with irrational singularities
Yuji ISHII, Noboru NAKAYAMA
If a normal quartic surface admits a singular point that is not a rational double point, then the surface is determined by the triplet
\left(M,D,E\right)
consisting of the minimal desingularization
M
D
of a general hyperplane section, and a nonzero effective anti-canonical divisor
E
M
. Geometric constructions of all the possible triplets
\left(M,D,E\right)
Yuji ISHII. Noboru NAKAYAMA. "Classification of normal quartic surfaces with irrational singularities." J. Math. Soc. Japan 56 (3) 941 - 965, July, 2004. https://doi.org/10.2969/jmsj/1191334093
Primary: 14E30 , 14J25 , 14J26 , 14J70
Keywords: extremal ray , Quartic surface , ruled surface
Yuji ISHII, Noboru NAKAYAMA "Classification of normal quartic surfaces with irrational singularities," Journal of the Mathematical Society of Japan, J. Math. Soc. Japan 56(3), 941-965, (July, 2004) |
Sick Universe | Wikipedia Mobile | Encyclopedia | What is / means Addition
{\displaystyle \scriptstyle \left.{\begin{matrix}\scriptstyle {\text{term}}\,+\,{\text{term}}\\\scriptstyle {\text{summand}}\,+\,{\text{summand}}\\\scriptstyle {\text{addend}}\,+\,{\text{addend}}\\\scriptstyle {\text{augend}}\,+\,{\text{addend}}\end{matrix}}\right\}\,=\,}
{\displaystyle \scriptstyle {\text{sum}}}
Subtraction (−)
{\displaystyle \scriptstyle \left.{\begin{matrix}\scriptstyle {\text{term}}\,-\,{\text{term}}\\\scriptstyle {\text{minuend}}\,-\,{\text{subtrahend}}\end{matrix}}\right\}\,=\,}
{\displaystyle \scriptstyle {\text{difference}}}
Multiplication (×)
{\displaystyle \scriptstyle \left.{\begin{matrix}\scriptstyle {\text{factor}}\,\times \,{\text{factor}}\\\scriptstyle {\text{multiplier}}\,\times \,{\text{multiplicand}}\end{matrix}}\right\}\,=\,}
{\displaystyle \scriptstyle {\text{product}}}
Division (÷)
{\displaystyle \scriptstyle \left.{\begin{matrix}\scriptstyle {\frac {\scriptstyle {\text{dividend}}}{\scriptstyle {\text{divisor}}}}\\\scriptstyle {\text{ }}\\\scriptstyle {\frac {\scriptstyle {\text{numerator}}}{\scriptstyle {\text{denominator}}}}\end{matrix}}\right\}\,=\,}
{\displaystyle {\begin{matrix}\scriptstyle {\text{fraction}}\\\scriptstyle {\text{quotient}}\\\scriptstyle {\text{ratio}}\end{matrix}}}
{\displaystyle \scriptstyle {\text{base}}^{\text{exponent}}\,=\,}
{\displaystyle \scriptstyle {\text{power}}}
nth root (√)
{\displaystyle \scriptstyle {\sqrt[{\text{degree}}]{\scriptstyle {\text{radicand}}}}\,=\,}
{\displaystyle \scriptstyle {\text{root}}}
{\displaystyle \scriptstyle \log _{\text{base}}({\text{anti-logarithm}})\,=\,}
{\displaystyle \scriptstyle {\text{logarithm}}}
{\displaystyle 1+1=2}
{\displaystyle 2+2=4}
{\displaystyle 1+2=3}
{\displaystyle 5+4+2=11}
{\displaystyle 3+3+3+3=12}
Columnar addition – the numbers in the column are to be added, with the sum written below the underlined number.
{\displaystyle 3{\frac {1}{2}}=3+{\frac {1}{2}}=3.5.}
{\displaystyle \sum _{k=1}^{5}k^{2}=1^{2}+2^{2}+3^{2}+4^{2}+5^{2}=55.}
All of the above terminology derives from Latin. "Addition" and "add" are English words derived from the Latin verb addere, which is in turn a compound of ad "to" and dare "to give", from the Proto-Indo-European root *deh₃- "to give"; thus to add is to give to.[10] Using the gerundive suffix -nd results in "addend", "thing to be added".[a] Likewise from augere "to increase", one gets "augend", "thing to be increased".
Combining sets[edit]
This interpretation is easy to visualize, with little danger of ambiguity. It is also useful in higher mathematics (for the rigorous definition it inspires, see § Natural numbers below). However, it is not obvious how one should extend this version of addition to include fractional numbers or negative numbers.[16]
Extending a length[edit]
Commutativity[edit]
Identity element[edit]
Performing addition[edit]
Childhood learning[edit]
¹
————
Since the end of the XXth century, some US programs, including TERC, decided to remove the traditional transfer method from their curriculum.[36] This decision was criticized[37] that is why some states and counties didn’t support this experiment.
Decimal fractions[edit]
————————————
Scientific notation[edit]
Main article: Scientific notation § Basic operations
{\displaystyle x=a\times 10^{b}}
{\displaystyle a}
{\displaystyle 10^{b}}
{\displaystyle 2.34\times 10^{-5}+5.67\times 10^{-6}=2.34\times 10^{-5}+0.567\times 10^{-5}=2.907\times 10^{-5}}
Non-decimal[edit]
0 + 0 → 0
1 + 1 → 0, carry 1 (since 1 + 1 = 2 = 0 + (1 × 21))
5 + 5 → 0, carry 1 (since 5 + 5 = 10 = 0 + (1 × 101))
—————————————
The abacus, also called a counting frame, is a calculating tool that was in use centuries before the adoption of the written modern numeral system and is still widely used by merchants, traders and clerks in Asia, Africa, and elsewhere; it dates back to at least 2700–2300 BC, when it was used in Sumer.[42]
Addition of numbers[edit]
Natural numbers[edit]
{\displaystyle N(A\cup B)}
Here, A ∪ B is the union of A and B. An alternate version of this definition allows A and B to possibly overlap and then takes their disjoint union, a mechanism that allows common elements to be separated out and therefore counted twice.
For an integer n, let |n| be its absolute value. Let a and b be integers. If either a or b is zero, treat it as an identity. If a and b are both positive, define a + b = |a| + |b|. If a and b are both negative, define a + b = −(|a| + |b|). If a and b have different signs, define a + b to be the difference between |a| and |b|, with the sign of the term whose absolute value is larger.[60] As an example, −6 + 4 = −2; because −6 and 4 have different signs, their absolute values are subtracted, and since the absolute value of the negative term is larger, the answer is negative.
Although this definition can be useful for concrete problems, the number of cases to consider complicates proofs unnecessarily. So the following method is commonly used for defining integers. It is based on the remark that every integer is the difference of two natural integers and that two such differences, a – b and c – d are equal if and only if a + d = b + c. So, one can define formally the integers as the equivalence classes of ordered pairs of natural numbers under the equivalence relation
The equivalence class of (a, b) contains either (a – b, 0) if a ≥ b, or (0, b – a) otherwise. If n is a natural number, one can denote +n the equivalence class of (n, 0), and by –n the equivalence class of (0, n). This allows identifying the natural number n with the equivalence class +n.
{\displaystyle (a,b)+(c,d)=(a+c,b+d).}
Rational numbers (fractions)[edit]
{\displaystyle {\frac {a}{b}}+{\frac {c}{d}}={\frac {ad+bc}{bd}}.}
{\displaystyle {\frac {3}{4}}+{\frac {1}{8}}={\frac {3\times 8+4\times 1}{4\times 8}}={\frac {24+4}{32}}={\frac {28}{32}}={\frac {7}{8}}}
{\displaystyle {\frac {a}{c}}+{\frac {b}{c}}={\frac {a+b}{c}}}
{\displaystyle {\frac {1}{4}}+{\frac {2}{4}}={\frac {1+2}{4}}={\frac {3}{4}}}
Adding π2/6 and e using Dedekind cuts of rationals.
{\displaystyle a+b=\{q+r\mid q\in a,r\in b\}.}
Adding π2/6 and e using Cauchy sequences of rationals.
{\displaystyle \lim _{n}a_{n}+\lim _{n}b_{n}=\lim _{n}(a_{n}+b_{n}).}
{\displaystyle (a+bi)+(c+di)=(a+c)+(b+d)i.}
{\displaystyle (a,b)+(c,d)=(a+c,b+d).}
Matrix addition is defined for two matrices of the same dimensions. The sum of two m × n (pronounced "m by n") matrices A and B, denoted by A + B, is again an m × n matrix computed by adding corresponding elements:[74][75]
{\displaystyle {\begin{aligned}\mathbf {A} +\mathbf {B} &={\begin{bmatrix}a_{11}&a_{12}&\cdots &a_{1n}\\a_{21}&a_{22}&\cdots &a_{2n}\\\vdots &\vdots &\ddots &\vdots \\a_{m1}&a_{m2}&\cdots &a_{mn}\\\end{bmatrix}}+{\begin{bmatrix}b_{11}&b_{12}&\cdots &b_{1n}\\b_{21}&b_{22}&\cdots &b_{2n}\\\vdots &\vdots &\ddots &\vdots \\b_{m1}&b_{m2}&\cdots &b_{mn}\\\end{bmatrix}}\\&={\begin{bmatrix}a_{11}+b_{11}&a_{12}+b_{12}&\cdots &a_{1n}+b_{1n}\\a_{21}+b_{21}&a_{22}+b_{22}&\cdots &a_{2n}+b_{2n}\\\vdots &\vdots &\ddots &\vdots \\a_{m1}+b_{m1}&a_{m2}+b_{m2}&\cdots &a_{mn}+b_{mn}\\\end{bmatrix}}\\\end{aligned}}}
{\displaystyle {\begin{bmatrix}1&3\\1&0\\1&2\end{bmatrix}}+{\begin{bmatrix}0&0\\7&5\\2&1\end{bmatrix}}={\begin{bmatrix}1+0&3+0\\1+7&0+5\\1+2&2+1\end{bmatrix}}={\begin{bmatrix}1&3\\8&5\\3&3\end{bmatrix}}}
In modular arithmetic, the set of available numbers is restricted to a finite subset of the integers, and addition "wraps around" when reaching a certain value, called the modulus. For example, the set of integers modulo 12 has twelve elements; it inherits an addition operation from the integers that is central to musical set theory. The set of integers modulo 2 has just two elements; the addition operation it inherits is known in Boolean logic as the "exclusive or" function. A similar "wrap around" operation arises in geometry, where the sum of two angle measures is often taken to be their sum as real numbers modulo 2Ï€. This amounts to an addition operation on the circle, which in turn generalizes to addition operations on many-dimensional tori.
General theory[edit]
Set theory and category theory[edit]
Related operations[edit]
Subtraction can be thought of as a kind of addition—that is, the addition of an additive inverse. Subtraction is itself a sort of inverse to addition, in that adding x and subtracting x are inverse functions.
Multiplication can be thought of as repeated addition. If a single term x appears in a sum n times, then the sum is the product of n and x. If n is not a natural number, the product may still make sense; for example, multiplication by −1 yields the additive inverse of a number.
{\displaystyle e^{a+b}=e^{a}e^{b}.}
Division is an arithmetic operation remotely related to addition. Since a/b = a(b−1), division is right distributive over addition: (a + b) / c = a/c + b/c.[83] However, division is not left distributive over addition; 1 / (2 + 2) is not the same as 1/2 + 1/2.
The maximum operation "max (a, b)" is a binary operation similar to addition. In fact, if two nonnegative numbers a and b are of different orders of magnitude, then their sum is approximately equal to their maximum. This approximation is extremely useful in the applications of mathematics, for example in truncating Taylor series. However, it presents a perpetual difficulty in numerical analysis, essentially since "max" is not invertible. If b is much greater than a, then a straightforward calculation of (a + b) − b can accumulate an unacceptable round-off error, perhaps even returning zero. See also Loss of significance.
{\displaystyle a+\max(b,c)=\max(a+b,a+c).}
{\displaystyle \log(a+b)\approx \max(\log a,\log b),}
{\displaystyle \max(a,b)=\lim _{h\to 0}h\log(e^{a/h}+e^{b/h}).}
Other ways to add[edit]
Addition is also used in the musical set theory. George Perle gives the following example: “do-mi, re-fa♯, miâ™-sol are different requirements of one interval... or other type of equality... and are connected with the axis of symmetry. Do-mi belongs to the family of symmetrically connected dyads as it is shown further:â€
re re♯ mi fa fa♯ sol sol♯
re do♯ do si la♯ la sol♯
Thus, do-mi is a part of an interval family-4 and a part of sum family -2 (at G♯ = 0).
A tonal range of Alban Berg’s Lyric Suite {0,11,7,4,2,9,3,8,10,1,5,6} is a series of six dyads with their total number being 11.[96] If the line is turned and inverted, then it is {0,6,5,1,...} with all dyads in total being 6.[97]
do sol re re♯ la♯ mi♯
si mi la sol♯ do♯ fa♯
^ a b Shmerko, V.P.; Yanushkevich [ÂnuÅ¡keviÄ], Svetlana N. [Svitlana N.]; Lyshevski, S.E. (2009). Computer arithmetics for nanoelectronics. CRC Press. p. 80.
^ Karpinski pp. 56–57, reproduced on p. 104
^ Karpinski pp. 150–153
^ Mosley, F. (2001). Using number lines with 5–8 year olds. Nelson Thornes. p. 8
^ Bronstein, Ilja NikolaeviÄ; Semendjajew, Konstantin AdolfoviÄ (1987) [1945]. "2.4.1.1.". In Grosche, Günter; Ziegler, Viktor; Ziegler, Dorothea (eds.). Taschenbuch der Mathematik (in German). Vol. 1. Translated by Ziegler, Viktor. Weiß, Jürgen (23 ed.). Thun and Frankfurt am Main: Verlag Harri Deutsch (and B.G. Teubner Verlagsgesellschaft, Leipzig). pp. 115–120. ISBN 978-3-87144-492-0.
^ Kaplan pp. 69–71
^ a b Henry, Valerie J.; Brown, Richard S. (2008). "First-grade basic facts: An investigation into teaching and learning of an accelerated, high-demand memorization standard". Journal for Research in Mathematics Education. 39 (2): 153–183. doi:10.2307/30034895. JSTOR 30034895.
^ Schmidt, W., Houang, R., & Cogan, L. (2002). "A coherent curriculum". American Educator, 26(2), 1–18.
^ Truitt and Rogers pp. 1;44–49 and pp. 2;77–78
^ Jean Marguin, p. 48 (1994) ; Quoting René Taton (1963)
^ Flynn and Overman pp. 1–9
^ Yeo, Sang-Soo, et al., eds. Algorithms and Architectures for Parallel Processing: 10th International Conference, ICA3PP 2010, Busan, Korea, May 21–23, 2010. Proceedings. Vol. 1. Springer, 2010. p. 194
^ Joshua Bloch, "Extra, Extra – Read All About It: Nearly All Binary Searches and Mergesorts are Broken" Archived 2016-04-01 at the Wayback Machine. Official Google Research Blog, June 2, 2006.
^ Ferreirós p. 223
^ Schyrlet Cameron, and Carolyn Craig (2013)Adding and Subtracting Fractions, Grades 5–8 Mark Twain, Inc.
^ Ferreirós p. 135; see section 6 of Stetigkeit und irrationale Zahlen Archived 2005-10-31 at the Wayback Machine.
^ Cheng, pp. 124–132
Ferreirós, José (1999). Labyrinth of Thought: A History of Set Theory and Its Role in Modern Mathematics. Birkhäuser. ISBN 978-0-8176-5749-9.
Baez, J.; Dolan, J. (2001). Mathematics Unlimited – 2001 and Beyond. From Finite Sets to Feynman Diagrams. p. 29. arXiv:math.QA/0004133. ISBN 3-540-66913-2.
Mikhalkin, Grigory (2006). Sanz-Solé, Marta (ed.). Proceedings of the International Congress of Mathematicians (ICM), Madrid, Spain, August 22–30, 2006. Volume II: Invited lectures. Tropical Geometry and its Applications. Zürich: European Mathematical Society. pp. 827–852. arXiv:math.AG/0601041. ISBN 978-3-03719-022-7. Zbl 1103.14034.
Viro, Oleg (2001). Cascuberta, Carles; Miró-Roig, Rosa Maria; Verdera, Joan; Xambó-Descamps, Sebastià (eds.). European Congress of Mathematics: Barcelona, July 10–14, 2000, Volume I. Dequantization of Real Algebraic Geometry on Logarithmic Paper. Progress in Mathematics. Vol. 201. Basel: Birkhäuser. pp. 135–146. arXiv:math/0005163. Bibcode:2000math......5163V. ISBN 978-3-7643-6417-5. Zbl 1024.14026.
Marguin, Jean (1994). Histoire des Instruments et Machines à Calculer, Trois Siècles de Mécanique Pensante 1642–1942 (in French). Hermann. ISBN 978-2-7056-6166-3.
Taton, René (1963). Le Calcul Mécanique. Que Sais-Je ? n° 367 (in French). Presses universitaires de France. pp. 20–28.
Poonen, Bjorn (2010). "Addition". Girls' Angle Bulletin. 3 (3–5). ISSN 2151-5743.
Steinhaus–Moser notation |
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