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Imkeller, Peter ; Pavlyukevich, Ilya
We consider a dynamical system in
ℝ
driven by a vector field
-{U}^{\text{'}}
U
is a multi-well potential satisfying some regularity conditions. We perturb this dynamical system by a Lévy noise of small intensity and such that the heaviest tail of its Lévy measure is regularly varying. We show that the perturbed dynamical system exhibits metastable behaviour i.e. on a proper time scale it reminds of a Markov jump process taking values in the local minima of the potential
U
. Due to the heavy-tail nature of the random perturbation, the results differ strongly from the well studied purely gaussian case.
Classification : 60E07, 60F10
Mots clés : Lévy process, jump diffusion, heavy tail, regular variation, metastability, extreme events, first exit time, large deviations
author = {Imkeller, Peter and Pavlyukevich, Ilya},
title = {Metastable behaviour of small noise {L\'evy-driven} diffusions},
AU - Imkeller, Peter
AU - Pavlyukevich, Ilya
TI - Metastable behaviour of small noise Lévy-driven diffusions
Imkeller, Peter; Pavlyukevich, Ilya. Metastable behaviour of small noise Lévy-driven diffusions. ESAIM: Probability and Statistics, Tome 12 (2008), pp. 412-437. doi : 10.1051/ps:2007051. http://www.numdam.org/articles/10.1051/ps:2007051/
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[3] A. Bovier, V. Gayrard and M. Klein, Metastability in reversible diffusion processes II: Precise asymptotics for small eigenvalues. Eur. Math. Soc. 7 (2005) 69-99. | MR 2120991 | Zbl 1105.82025
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[5] S. Cerrai, Second order PDE's in finite and infinite dimension. A probabilistic approach. Lect. Notes Math. Springer, Berlin Heidelberg (2001). | MR 1840644 | Zbl 0983.60004
[6] A.V. Chechkin, V. Yu Gonchar, J. Klafter and R. Metzler, Barrier crossings of a Lévy flight. EPL 72 (2005) 348-354. | MR 2213557
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\alpha
-stable noise induced millenial climate changes from an ice record. Geophysical Research Letters 26 (1999) 1441-1444.
[10] M.I. Freidlin and A.D. Wentzell, Random perturbations of dynamical systems, Grundlehren der Mathematischen Wissenschaften 260. Springer, New York, NY, second edition (1998). | MR 1652127 | Zbl 0522.60055
[11] A. Galves, E. Olivieri and M.E. Vares, Metastability for a class of dynamical systems subject to small random perturbations. Ann. Probab. 15 (1987) 1288-1305. | MR 905332 | Zbl 0709.60058
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Simulate Bates, Heston, and CIR sample paths by quadratic-exponential discretization scheme - MATLAB simByQuadExp - MathWorks Australia
{X}_{t}=P\left(t,{X}_{t}\right)
\begin{array}{l}dS\left(t\right)=\gamma \left(t\right)S\left(t\right)dt+\sqrt{V\left(t\right)}S\left(t\right)d{W}_{S}\left(t\right)\\ dV\left(t\right)=\kappa \left(\theta -V\left(t\right)\right)dt+\sigma \sqrt{V\left(t\right)}d{W}_{V}\left(t\right)\end{array}
d{X}_{t}=S\left(t\right)\left[L\left(t\right)-{X}_{t}\right]dt+D\left(t,{X}_{t}^{\frac{1}{2}}\right)V\left(t\right)d{W}_{t}
d{X}_{1t}=B\left(t\right){X}_{1t}dt+\sqrt{{X}_{2t}}{X}_{1t}d{W}_{1t}+Y\left(t\right){X}_{1t}d{N}_{t}
d{X}_{2t}=S\left(t\right)\left[L\left(t\right)-{X}_{2t}\right]dt+V\left(t\right)\sqrt{{X}_{2t}}d{W}_{2t} |
Increase Image Resolution Using Deep Learning - MATLAB & Simulink Example - MathWorks América Latina
{\mathit{Y}}_{\mathrm{high}\mathrm{res}}
{\mathit{Y}}_{\mathrm{lowres}}
{\mathit{Y}}_{\mathrm{lowres}}
{\mathit{Y}}_{\mathrm{residual}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}{\mathit{Y}}_{\mathrm{highres}}-{\mathit{Y}}_{\mathrm{lowres}}
Create an imageDataAugmenter (Deep Learning Toolbox) that specifies the parameters of data augmentation. Use data augmentation during training to vary the training data, which effectively increases the amount of available training data. Here, the augmenter specifies random rotation by 90 degrees and random reflections in the x-direction.
Create a randomPatchExtractionDatastore that performs randomized patch extraction from the upsampled and residual image datastores. Patch extraction is the process of extracting a large set of small image patches, or tiles, from a single larger image. This type of data augmentation is frequently used in image-to-image regression problems, where many network architectures can be trained on very small input image sizes. This means that a large number of patches can be extracted from each full-sized image in the original training set, which greatly increases the size of the training set.
convolution2dLayer (Deep Learning Toolbox) - 2-D convolution layer for convolutional neural networks
Train the network using stochastic gradient descent with momentum (SGDM) optimization. Specify the hyperparameter settings for SGDM by using the trainingOptions (Deep Learning Toolbox) function. The learning rate is initially 0.1 and decreased by a factor of 10 every 10 epochs. Train for 100 epochs.
To train the VDSR network, set the doTraining variable in the following code to true. Train the network using the trainNetwork (Deep Learning Toolbox) function.
Convert the low-resolution image from the RGB color space to luminance (Iy) and chrominance (Icb and Icr) channels by using the rgb2ycbcr function.
Pass the upscaled luminance component, Iy_bicubic, through the trained VDSR network. Observe the activations (Deep Learning Toolbox) from the final layer (a regression layer). The output of the network is the desired residual image.
Concatenate the high-resolution VDSR luminance component with the upscaled color components. Convert the image to the RGB color space by using the ycbcr2rgb function. The result is the final high-resolution color image using VDSR.
randomPatchExtractionDatastore | rgb2ycbcr | ycbcr2rgb | trainingOptions (Deep Learning Toolbox) | trainNetwork (Deep Learning Toolbox) | transform | combine |
Metadata API - ImageKit.io Docs
Get image metadata for uploaded media files
Get image metadata from remote URL
You can programmatically get image exif, pHash, and other metadata using either of the API below:
Get image metadata from remote URL if you don't want to upload image files to ImageKit.io media library or,
Get image metadata for uploaded media files if you want to fetch metadata for already uploaded image files in your ImageitKit.io media library.
Metadata Object Structure
"hasColorProfile": true,
"hasTransparency": false,
"pHash": "f06830ca9f1e3e90",
"Software": "GIMP 2.4.5",
"ExifOffset": 214,
"GPSInfo": 978
"ThumbnailOffset": 1090,
"ThumbnailLength": 1378
"ExposureTime": 0.00625,
"FNumber": 7.1,
"ShutterSpeedValue": 7.375,
"ApertureValue": 5.625,
"ExposureCompensation": 0,
"FocalLength": 135,
"ExifImageWidth": 100,
"ExifImageHeight": 68,
"InteropOffset": 948,
"FocalPlaneXResolution": 4438.356164383562,
"FocalPlaneYResolution": 4445.969125214408,
"SceneCaptureType": 0
"GPSVersionID": [
"interoperability": {
"InteropIndex": "R98",
"InteropVersion": "0100"
"makernote": {}
For more information about the Exif standard, please refer to the specification found on http://www.exif.org. A comprehensive list of available Exif attributes and their meaning can be found on http://www.sno.phy.queensu.ca/~phil/exiftool/TagNames/.
Perceptual Hash (pHash)
Perceptual hashing allows you to construct a hash value that uniquely identifies an input image based on the image's contents. It is different from cryptographic hash functions like MD5 and SHA1. pHash provides similar hash value after minor distortions, like small rotations, blurring, and compression in the image.
ImageKit.io metadata API returns the pHash value of an image in the metadata response as a hexadecimal string. More information about pHash can be found on https://www.phash.org/.
Using pHash to find similar or duplicate images
The hamming distance between two pHash values determines how similar or different the images are.
The pHash value returned by ImageKit.io metadata API is a hexadecimal string of 64bit pHash. The distance between two hash can be between 0 and 64. A lower distance means similar images. If the distance is 0, that means two images are identical.
To calculate a similarity score between 0 and 1, we can do:
SimilarityScore = 1 - (phashDistance(phash1, phash2) / 64
For example, consider these two images. The first image with pHash value 63433b3ccf8e1ebe
pHash = 63433b3ccf8e1ebe
Second with pHash value f5d2226cd9d32b16
pHash = f5d2226cd9d32b16
The distance between two pHash values can be calculated using the utility function provided by ImageKit.io server-side SDKs.
The hamming distance between 63433b3ccf8e1ebe and f5d2226cd9d32b16 is 27.
publicKey : "your_public_api_key",
privateKey : "your_private_api_key",
urlEndpoint : "https://ik.imagekit.io/your_imagekit_id/"
imagekit.pHashDistance("63433b3ccf8e1ebe", "f5d2226cd9d32b16");
public_key='your_public_api_key',
private_key='your_private_api_key',
url_endpoint = 'https://ik.imagekit.io/your_imagekit_id/'
# distance is 27
distance = imagekit.phash_distance("63433b3ccf8e1ebe", "f5d2226cd9d32b16"),
$public_key = "your_public_api_key";
$your_private_key = "your_private_api_key";
$url_end_point = "https://ik.imagekit.io/your_imagekit_id";
$public_key,
$your_private_key,
$url_end_point
$distance = $imageKit->pHashDistance("63433b3ccf8e1ebe", "f5d2226cd9d32b16");
SimilarityScore = 1-27/64 = 0.578125
The similarity score is 57%. This means the two images are not similar.
Now let's consider a case of two similar images. The first image with pHash value 63433b3ccf8e1ebe
Let's resize & crop this image to 300x400 and reduce the quality using aggressive compression. The pHash value of the slightly modified image is 61433b3fcf8f9faf
pHash = 61433b3fcf8f9faf
The hamming distance between these two pHash values is 8
imagekit.pHashDistance("63433b3ccf8e1ebe", "61433b3fcf8f9faf");
SimilarityScore = 1-8/64 = 0.875
The similarity score is 87%, so it is safe to say that the two images are similar. |
Ask Answer - Triangles - Popular Questions for School Students
Q). In fig. Angle ACB is a right angle, AC = CD, CDEF is a rectangle and angle BAC = 50 degree. Calculate
(i) angle BDE;
(ii) the angle between the diagonals CE, DF of the rectangle.
In a triangle ABC ID,IE and IF are the perpendicular bisector of BC, AC and AB respectively. Prove that I is equidistant from the vertices of the triangle.
(ii) ∆AOD ≅ ∆COD
Prove this triangle as an isosceles triangle when ,there altitudes are equal .Please tell me by taking the triangles ABD and triangle ACE
Vijeta Mohanty
Q.3. In the following figure, AB = AC; BC = CD and DE is parallel to BC. Calculate :
\angle
\angle
in a triangle abc ab =ac,ba is produced to d and ae is drawn parallel to bc ,prove that ae bisects angle dac
please give me the answer of the question on the pic
Solve the following question ,along with the explanation
In the figure,O is the centre of the circular arc ABC.Find the angles of triangle ABC
Q. In the figure, O is the centre of the circular arc ABC. Find the angles of triangle ABC
A point O is taken inside a rhombus ABCD such that its distances from the vertices B and D are equal. Show that AOC is a straight line.
pqrs is a quadrilateral and o is a point inside it (other than the point of intersection of the diagonals),prove that( op+oq+or+os) greater than(pr+qs)
Ananyamadhur Rastogi
In the given figure:AB is parallel to FD , AC is parallel GE and BD = CE, PROVE THAT :
1- BG = DF
2- CF = EG
In ? PQR PQ=PR,A is point in PQ and B is a point in PR,so that QR=RA=AB=BP
1. show that angle P:angle R=1:3
2.find the value of angle q
Savikaran
Solve q 12 plz
two lines ab and cd are parallel to each other,the tranversal bd bisects the other transversal ac,prove that ac also bisect bd
Prove that the difference of any two sides of a triangle is less than the third side
Q. In the adjoining figure, QX and RX are bisectors of angles Q and R respectively of the triangle PQR. If XS is perpendicular to QR and XT is perpendicular to PQ, prove that PX bisects angle P.
please give me the answer of the question written on the pic
pqr is a triangle ans ps bisects angle qrp,show that a)pq greater than qs,b)pr greater than rs,c)pq+pr greater than qr
Soham Jhingran
Q24 and Q26 with proper equation
which is the longer QM or QR if LM> LR and angle LMQ = angle LRQ
ANS:10m,24m,26m
No links , please .... |
A manufacturer has 600 L of a 12% solution of acid How many litres of a 30% acid solution - Maths - Linear Inequalities - 7793747 | Meritnation.com
A manufacturer has 600 L of a 12% solution of acid. How many litres of a 30% acid solution must be added to it so that the acid content in the resulting mixture will be more than 15% but less than 18%?
Let x litres of 30% acid solution is required to be added. Then
Total mixture = (x + 600) litres
30% x + 12% of 600 > 15% of (x + 600)
and 30% x + 12% of 600 < 18% of (x + 600)
\mathrm{or} \frac{30\mathrm{x}}{100}+\frac{12}{100}\left(600\right)>\frac{15}{100}\left(\mathrm{x}+600\right)\phantom{\rule{0ex}{0ex}}\mathrm{and} \frac{30\mathrm{x}}{100}+\frac{12}{100}\left(600\right)<\frac{18}{100}\left(\mathrm{x}+600\right)\phantom{\rule{0ex}{0ex}}\mathrm{or} 30\mathrm{x}+7200>15\mathrm{x}+9000\phantom{\rule{0ex}{0ex}}\mathrm{and} 30 \mathrm{x} +7200<18\mathrm{x}+10800\phantom{\rule{0ex}{0ex}}\mathrm{or} 15\mathrm{x}>1800 \mathrm{and} 12\mathrm{x}<3600\phantom{\rule{0ex}{0ex}}\mathrm{or} \mathrm{x} >120 \mathrm{and} \mathrm{x}<300\phantom{\rule{0ex}{0ex}}\mathrm{So} 120<\mathrm{x}<300\phantom{\rule{0ex}{0ex}}\mathrm{Thus} , \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{liters} \mathrm{of} \mathrm{the} 30% \mathrm{solution} \mathrm{of} \mathrm{acid} \mathrm{will} \mathrm{have} \mathrm{to} \mathrm{be} \mathrm{more}\phantom{\rule{0ex}{0ex}}\mathrm{than} 120 \mathrm{liters} \mathrm{but} \mathrm{less} \mathrm{than} 300 \mathrm{litres}. |
Mixed Integer ga Optimization - MATLAB & Simulink - MathWorks España
Solving Mixed Integer Optimization Problems
Mixed Integer Optimization of Rastrigin's Function
Characteristics of the Integer ga Solver
Example: Integer Programming with a Nonlinear Equality Constraint
Effective Integer ga
Integer ga Algorithm
ga can solve problems when certain variables are integer-valued. Give intcon, a vector of the x components that are integers:
[x,fval,exitflag] = ga(fitnessfcn,nvars,A,b,[],[],...
lb,ub,nonlcon,intcon,options)
intcon is a vector of positive integers that contains the x components that are integer-valued. For example, if you want to restrict x(2) and x(10) to be integers, set intcon to [2,10].
The surrogateopt solver also accepts integer constraints.
Restrictions exist on the types of problems that ga can solve with integer variables. In particular, ga does not accept nonlinear equality constraints when there are integer variables. For details, see Characteristics of the Integer ga Solver.
ga solves integer problems best when you provide lower and upper bounds for every x component.
This example shows how to find the minimum of Rastrigin's function restricted so the first component of x is an integer. The components of x are further restricted to be in the region
5\pi \le x\left(1\right)\le 20\pi ,\phantom{\rule{0.5em}{0ex}}-20\pi \le x\left(2\right)\le -4\pi
Set up the bounds for your problem
lb = [5*pi,-20*pi];
ub = [20*pi,-4*pi];
Set a plot function so you can view the progress of ga
opts = optimoptions('ga','PlotFcn',@gaplotbestf);
Call the ga solver where x(1) has integer values
[x,fval,exitflag] = ga(@rastriginsfcn,2,[],[],[],[],...
lb,ub,[],intcon,opts)
ga converges quickly to the solution.
There are some restrictions on the types of problems that ga can solve when you include integer constraints:
No nonlinear equality constraints. Any nonlinear constraint function must return [] for the nonlinear equality constraint. For a possible workaround, see Example: Integer Programming with a Nonlinear Equality Constraint.
Only doubleVector population type.
No hybrid function. ga overrides any setting of the HybridFcn option.
ga ignores the ParetoFraction, DistanceMeasureFcn, InitialPenalty, and PenaltyFactor options.
The listed restrictions are mainly natural, not arbitrary. For example, no hybrid functions support integer constraints. So ga does not use hybrid functions when there are integer constraints.
This example attempts to locate the minimum of the Ackley function (included with your software) in five dimensions with these constraints:
x(1), x(3), and x(5) are integers.
norm(x) = 4.
The Ackley function is difficult to minimize. Adding integer and equality constraints increases the difficulty.
To include the nonlinear equality constraint, give a small tolerance tol that allows the norm of x to be within tol of 4. Without a tolerance, the nonlinear equality constraint is never satisfied, and the solver does not realize when it has a feasible solution.
Write the expression norm(x) = 4 as two “less than zero” inequalities:
norm(x) - 4 ≤ 0
-(norm(x) - 4) ≤ 0. (1)
Allow a small tolerance in the inequalities:
norm(x) - 4 - tol ≤ 0
-(norm(x) - 4) - tol ≤ 0. (2)
Write a nonlinear inequality constraint function that implements these inequalities:
function [c, ceq] = eqCon(x)
confcnval = norm(x) - rad;
c = [confcnval - tol;-confcnval - tol];
MaxStallGenerations = 50 — Allow the solver to try for a while.
FunctionTolerance = 1e-10 — Specify a stricter stopping criterion than usual.
MaxGenerations = 300 — Allow more generations than default.
PlotFcn = @gaplotbestfun — Observe the optimization.
opts = optimoptions('ga','MaxStallGenerations',50,'FunctionTolerance',1e-10,...
'MaxGenerations',300,'PlotFcn',@gaplotbestfun);
Set lower and upper bounds to help the solver:
[x,fval,exitflag] = ga(@ackleyfcn,nVar,[],[],[],[], ...
lb,ub,@eqCon,[1 3 5],opts);
Examine the solution:
x,fval,exitflag,norm(x)
The odd x components are integers, as specified. The norm of x is 4, to within the given relative tolerance of 1e-3.
Despite the positive exit flag, the solution is not the global optimum. Run the problem again and examine the solution:
opts = optimoptions('ga',opts,'Display','off');
[x2,fval2,exitflag2] = ga(@ackleyfcn,nVar,[],[],[],[], ...
Examine the second solution:
x2,fval2,exitflag2,norm(x2)
-2.0000 2.8930 0 -1.9095 0
The second run gives a better solution (lower fitness function value). Again, the odd x components are integers, and the norm of x2 is 4, to within the given relative tolerance of 1e-3.
Be aware that this procedure can fail; ga has difficulty with simultaneous integer and equality constraints.
To use ga most effectively on integer problems, follow these guidelines.
Bound each component as tightly as you can. This practice gives ga the smallest search space, enabling ga to search most effectively.
If you cannot bound a component, then specify an appropriate initial range. By default, ga creates an initial population with range [-1e4,1e4] for each component. A smaller or larger initial range can give better results when the default value is inappropriate. To change the initial range, use the InitialPopulationRange option.
If you have more than 10 variables, set a population size that is larger than default by using the PopulationSize option. The default value is 200 for six or more variables. For a large population size:
ga can take a long time to converge. If you reach the maximum number of generations (exit flag 0), increase the value of the MaxGenerations option.
For information on options, see the ga options input argument.
Integer programming with ga involves several modifications of the basic algorithm (see How the Genetic Algorithm Works). For integer programming:
By default, special creation, crossover, and mutation functions enforce variables to be integers. For details, see Deep et al. [2].
If you use nondefault creation, crossover, or mutation functions, ga enforces linear feasibility and feasibility with respect to integer constraints at each iteration.
The genetic algorithm attempts to minimize a penalty function, not the fitness function. The penalty function includes a term for infeasibility. This penalty function is combined with binary tournament selection by default to select individuals for subsequent generations. The penalty function value of a member of a population is:
If the member is feasible, the penalty function is the fitness function.
If the member is infeasible, the penalty function is the maximum fitness function among feasible members of the population, plus a sum of the constraint violations of the (infeasible) point.
[2] Deep, Kusum, Krishna Pratap Singh, M.L. Kansal, and C. Mohan. A real coded genetic algorithm for solving integer and mixed integer optimization problems. Applied Mathematics and Computation, 212(2), pp. 505–518, 2009. |
LowerPCentralSeries - Maple Help
Home : Support : Online Help : Mathematics : Group Theory : LowerPCentralSeries
construct the lower p-central series of a group
LowerPCentralSeries( p, G )
The lower p-central series of a group
G
p
, is the descending normal series of
G
{G}_{0}=G
0<k
{G}_{k}={G}^{{p}_{G,{G}_{k-1}}}
G={G}_{0}▹{G}_{1}▹\dots ▹{G}_{c}
is called the lower p-central series of
G
p
-residual
{G}_{c}
is the trivial group, then
G
p
-group. In this case, the number
c
is called the p- class of
G
The LowerPCentralSeries( G ) command constructs the lower
p
-central series of a group G.
p
-central series of G is represented by a series data structure which admits certain operations common to all series. See GroupTheory[Series].
\mathrm{with}\left(\mathrm{GroupTheory}\right):
G≔\mathrm{PermutationGroup}\left({[[1,2]],[[1,2,3],[4,5]]}\right)
\textcolor[rgb]{0,0,1}{G}\textcolor[rgb]{0,0,1}{≔}〈\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\right)\textcolor[rgb]{0,0,1}{,}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\right)\left(\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}\right)〉
\mathrm{LowerPCentralSeries}\left(2,G\right)
〈\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\right)\textcolor[rgb]{0,0,1}{,}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\right)\left(\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}\right)〉\textcolor[rgb]{0,0,1}{▹}〈\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\right)〉
\mathrm{LowerPCentralSeries}\left(3,G\right)
〈\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\right)\textcolor[rgb]{0,0,1}{,}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\right)\left(\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}\right)〉
\mathrm{LowerPCentralSeries}\left(2,\mathrm{QuaternionGroup}\left(\right)\right)
\textcolor[rgb]{0,0,1}{Q}\textcolor[rgb]{0,0,1}{▹}〈\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\right)\left(\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\right)\left(\textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{8}\right)\left(\textcolor[rgb]{0,0,1}{6}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{7}\right)〉\textcolor[rgb]{0,0,1}{▹}〈〉
\mathrm{LowerPCentralSeries}\left(2,\mathrm{DihedralGroup}\left(4\right)\right)
{\textcolor[rgb]{0,0,1}{\mathbf{D}}}_{\textcolor[rgb]{0,0,1}{4}}\textcolor[rgb]{0,0,1}{▹}〈\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\right)\left(\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\right)〉\textcolor[rgb]{0,0,1}{▹}〈〉
\mathrm{LowerPCentralSeries}\left(2,\mathrm{DihedralGroup}\left(5\right)\right)
{\textcolor[rgb]{0,0,1}{\mathbf{D}}}_{\textcolor[rgb]{0,0,1}{5}}\textcolor[rgb]{0,0,1}{▹}〈\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\right)\textcolor[rgb]{0,0,1}{,}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\right)〉
The GroupTheory[LowerPCentralSeries] command was introduced in Maple 2019. |
Compare Markov Chain Mixing Times - MATLAB & Simulink - MathWorks Switzerland
Fast-Mixing Chain
Slow-Mixing Chain
Dumbbell Chain Mixing Time
This example compares the estimated mixing times of several Markov chains with different structures. Convergence theorems typically require ergodic unichains. Therefore, before comparing mixing time estimates, this example ensures that the Markov chains are ergodic unichains.
Create a 23-state Markov chain from a random transition matrix containing 250 infeasible transitions of 529 total transitions. An infeasible transition is a transition whose probability of occurring is zero. Plot a digraph of the Markov chain and identify classes by using node colors and markers.
Zeros1 = 250;
mc1 = mcmix(numStates,'Zeros',Zeros1);
graphplot(mc1,'ColorNodes',true);
mc1 represents a unichain because it is a single, recurrent, aperiodic class.
tf1 = isergodic(mc1)
tf1 = 1 indicates that mc1 represents an ergodic unichain.
The pink disc in the plot shows the spectral gap (the difference between the two largest eigenvalue moduli). The spectral gap determines the mixing time of the Markov chain. Large gaps indicate faster mixing, whereas thin gaps indicate slower mixing. In this case, the gap is large, indicating a fast-mixing chain.
Estimate the mixing time of the chain.
On average, it takes 0.8357 steps for the total variation distance to decay by a factor of
{e}^{1}
The expected first hitting time for a target state is another way to view the mixing rate of a Markov chain. The hitting time computation does not require an ergodic Markov chain.
Plot a digraph of the Markov chain with node colors representing the expected first hitting times for regime 1.
hittime(mc1,1,'Graph',true);
The expected first hitting time for regime 1 beginning from regime 2 is approximately 16 time steps.
Create another 23-state Markov chain from a random transition matrix containing 475 infeasible transitions. With fewer feasible transitions, this chain should take longer to mix. Plot a digraph of the Markov chain and identify classes by using node colors and markers.
mc2 represents a unichain because it has a single, recurrent, aperiodic class and several transient classes.
tf2 = 0 indicates that mc2 is not ergodic.
Extract the recurrent subchain from mc2. Determine whether the subchain is ergodic.
[bins,~,ClassRecurrence] = classify(mc2);
sc2 = subchain(mc2,recurrentState);
tf2 = isergodic(sc2)
sc2 represents an ergodic unichain.
Plot the eigenvalues of the subchain on the complex plane.
eigplot(sc2);
The spectral gap in the subchain is much thinner than the gap in mc1, which indicates that the subchain mixes more slowly.
Estimate the mixing time of the subchain.
[~,tMix2] = asymptotics(sc2)
{e}^{1}
Plot a digraph of the Markov chain with node colors representing the expected first hitting times for the first regime in the recurrent subclass.
sc2.StateNames(1)
hittime(sc2,1,'Graph',true);
The expected first hitting time for regime 2 beginning from regime 8 is about 30 time steps.
w = 10; % Dumbbell weights
DBar = [0 1 0; 1 0 1; 0 1 0]; % Dumbbell bar
DB = blkdiag(rand(w),DBar,rand(w)); % Transition matrix
mc3 = dtmc(DB);
Plot a directed graph of the dumbbell chain and identify classes by using node colors and markers. Suppress node labels.
h = graphplot(mc3,'ColorNodes',true);
mc3 represents a unichain because it has a single, recurrent, aperiodic class.
tf3 = 1 indicates that mc3 is ergodic.
Plot the eigenvalues of the dumbbell on the complex plane.
The spectral gap in the subchain is very thin, which indicates that the dumbbell chain mixes very slowly.
Estimate the mixing time of the dumbbell chain.
tMix3 = 90.4334
On average, it takes 90.4334 steps for the total variation distance to decay by a factor of
{e}^{1}
The expected first hitting time for regime 1 beginning from regime 15 is about 300 time steps.
eigplot | asymptotics | hittime |
Consider the bases B = left(begin{array}{c}begin{bmatrix}2 3 end{bmatrix}, begi
Consider the bases B = left(begin{array}{c}begin{bmatrix}2
Consider the bases
B=\left(\begin{array}{c}\left[\begin{array}{c}2\\ 3\end{array}\right],\left[\begin{array}{c}3\\ 5\end{array}\right]\end{array}\right)of{R}^{2}\text{ }and\text{ }C=\left(\begin{array}{c}\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]\end{array},\left[\begin{array}{c}0\\ 1\\ 1\end{array}\right]\right)of{R}^{3}
and the linear maps
S\in L\left({R}^{2},{R}^{3}\right)\text{ }and\text{ }T\in L\left({R}^{3},{R}^{2}\right)
given given (with respect to the standard bases) by
\left[S{\right]}_{E,E}=\left[\begin{array}{cc}2& -1\\ 5& -3\\ -3& 2\end{array}\right]\text{ }and\text{ }\left[T{\right]}_{E,E}=\left[\begin{array}{ccc}1& -1& 1\\ 1& 1& -1\end{array}\right]
Find each of the following coordinate representations.
\left(b\right){\left[S\right]}_{E,C}
\left(c\right){\left[S\right]}_{B,C}
B=\left(\begin{array}{c}\left[\begin{array}{c}2\\ 3\end{array}\right],\left[\begin{array}{c}3\\ 5\end{array}\right]\end{array}\right)of{R}^{2},C=\left(\begin{array}{c}\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]\end{array},\left[\begin{array}{c}0\\ 1\\ 1\end{array}\right]\right)of{R}^{3}
S\in L\left({R}^{2},{R}^{3}\right),T\in L\left({R}^{3},{R}^{2}\right)
\left[S{\right]}_{E,E}=\left[\begin{array}{cc}2& -1\\ 5& -3\\ -3& 2\end{array}\right],\left[T{\right]}_{E,E}=\left[\begin{array}{ccc}1& -1& 1\\ 1& 1& -1\end{array}\right]
(b) To find
{\left[S\right]}_{E,C}:
S\left(\left[\begin{array}{c}1\\ 0\end{array}\right]\right)=\left[\begin{array}{c}2\\ 5\\ -3\end{array}\right]=a\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]+b\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]+c\left[\begin{array}{c}0\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}a+b\\ a+c\\ b+c\end{array}\right]
\therefore a+b=2,a+c=5,b+c=-3
b=2-a⇒b+c=2-a+c=-3⇒-a+c=-5
\begin{array}{c}\therefore a+c=5\\ +-a+c=-5\\ 2c=0\end{array}\begin{array}{c}⇒c=0,b=-3,a=5\\ ⇒a=5,b=-3,c=0\end{array}
S\left(\left[\begin{array}{c}0\\ 1\end{array}\right]\right)=\left[\begin{array}{c}-1\\ -3\\ 2\end{array}\right]=a\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]+b\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]+c\left[\begin{array}{c}0\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}a+b\\ a+c\\ b+c\end{array}\right]
\therefore a+b=-1,a+c=-3,b+c=2
\therefore b=-1-a⇒b+c=-1-a+c=2⇒-a+c=3
\begin{array}{c}\therefore a+c=-3\\ +-a+c=3\\ 2c=0\end{array}\begin{array}{c}⇒c=0,a=-3,b=2\\ ⇒a=-3,b=2,c=0\end{array}
\therefore \left[S{\right]}_{E,C}=\left[\begin{array}{cc}5& -3\\ -3& 2\\ 0& 0\end{array}\right]
c) To find
To plot: Thepoints which has polar coordinate
\left(2,\frac{7\pi }{4}\right)
also two alternaitve sets for the same.
To find: The equivalent polar equation for the given rectangular-coordinate equation.
{x}^{2}+{y}^{2}+8x=0
Solving Systems Graphically- One Solution
Find or create an example of a system of equations with one solution.
Graph and label the lines on a coordinate plane. Provide their equations.
State the accurate solution to the system.
The coordinates of the point in the \(\displaystyle{x}
Consider the following vectors in
{R}^{4}:
{v}_{1}=\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right],{v}_{2}=\left[\begin{array}{c}0\\ 1\\ 1\\ 1\end{array}\right]{v}_{3}=\left[\begin{array}{c}0\\ 0\\ 1\\ 1\end{array}\right],{v}_{4}=\left[\begin{array}{c}0\\ 0\\ 0\\ 1\end{array}\right]
d. If
x=\left[\begin{array}{c}23\\ 12\\ 10\\ 19\end{array}\right],find{\left\{x\right\}}_{B}e
{x}_{B}=\left[\begin{array}{c}3\\ 1\\ -4\\ -4\end{array}\right]
To fill: The blanck spaces in the statement " The origin in the rectangular coordinate system concedes with the ? in polar coordinates. The positive x-axis in rectangular coordinates coincides with the ? in polar coordinates".
Given the elow bases for
{R}^{2}
and the point at the specified coordinate in the standard basis as below, (40 points)
\left(B1=\left\{\left(1,0\right),\left(0,1\right)\right\}
B2=\left(1,2\right),\left(2,-1\right)\right\}
{3}^{\ast }\left(1,2\right)-\left(2,1\right)
B2=\left(1,1\right),\left(-1,1\right)\left(3,7={5}^{\ast }\left(1,1\right)+{2}^{\ast }\left(-1,1\right)
B2=\left(1,2\right),\left(2,1\right)\text{ }\text{ }\text{ }\left(0,3\right)={2}^{\ast }\left(1,2\right)-{2}^{\ast }\left(2,1\right)
\left(8,10\right)={4}^{\ast }\left(1,2\right)+{2}^{\ast }\left(2,1\right)
B2 = (1, 2), (-2, 1) (0, 5) =
a. Use graph technique to find the coordinate in the second basis. (10 points) b. Show that each basis is orthogonal. (5 points) c. Determine if each basis is normal. (5 points) d. Find the transition matrix from the standard basis to the alternate basis. (15 points) |
siddhartha-gadgil/LTS2019 - Gitter
siddhartha-gadgil/LTS2019
Bump addressable from 2.5.2 to … Merge pull request #285 from si… (compare)
siddhartha-gadgil closed #285
Bump rake from 12.0.0 to 12.3.3… Merge pull request #283 from si… (compare)
Questions about LTS2019 welome.
@adithyaupadhya
How do you write a type that shows that an implication implies its contrapositive? The type of such a thing looks like this, I suppose : impliesContrapositive : (A : Type) -> (B : Type) -> (A -> B) -> ((B -> Void) -> (A -> Void))
Is such a thing possible to write, or is it preferable to do a different thing to achieve the same result?
The law of excluded middle is not normally part of the foundations, and is needed for contrapositive being equivalent to the statement. The reason we don't have it is that it does not give a concrete value, and we want concrete values.
By the law of excluded middle for the type $A$ is $A \oplus (A \to Void)$
A \oplus (A \to 0)
Here, I meant the 0 type , so should have said
lem(A) = A \oplus (A \to \mathbb{0}
$$lem(A)= A \oplus (A \to \mathbb{0})
lem(A)= A \oplus (A \to \mathbb{0})
The contrapositive in the form you said it is true though,
(A \to B) \to (B \to 0) \to (A \to 0)
This can be proved, i.e., we get a term
(f : A \to B) \mapsto (c : B \to \mathbb{0}) \mapsto a : A \mapsto c(f(a))
bnag098
@bnag098
I am having some trouble in using the replace function.
I was trying to substitute $ 0 $ using the equality
0 = b \cdot 0
c = 0 + c
to hopefully get
c = b \cdot 0 + c
But in the function call below:
replace (sym (multZeroRightZero b)) (sym (plusZeroLeftNeutral c))
I am getting an error claiming that in the second variable of replace, idris expects a value of the type P 0. The documentation on this P function couldn't explain much. Any help here would be appreciated.
@SS-C4
Is it possible to case split on a dependent pair?
For example, if I want to use the pair (x **pf) in a function, where I want to case split on x being Z or (S k), and I need the proof when this happens.
case (fst (x**pf)) of typechecks, but the proof type does not change, and stays the same as the original.
Oh, I found it. with helps here.
To capture the notion of a total order, in which exactly one of three things can occur, a<=b and !(a>=b), !(a<=b) and (a>=b) or (a<=b) and (a>=b), would the best thing to do be to define a new type along the lines of Either, or would it be better to try to work with existing types to get something with the same behaviour?
Generally best to define a new type, though this one may be in the standard library or contrib (look for trichotomy).
Using combinations of pairs, dependent pairs and coproducts (Either) makes stuff hard to understand, parse by idris and also more error-prone (a wrong type may be accepted because of the same structure).
Would it be preferable to define the order on the naturals as arising from addition rather than the inductive one defined in the standard library?
Yes, it may work better. In any case if you prove the two are equivalent,
you can use both definitions
shafilmaheenn
@shafilmaheenn
Is there a way to check whether 2 functions have the same name in the entire code folder?
Atom has a search in folder, ctrl-shift-f, which can also be filtered, here using *.idr for filename
As a lot of theorems about gcd need the bezout's lemma, may I use assert_total in the bez and GCDCalc function which essentially follow Euclid's algorithm?. I have no clue on how to convince idris that Euclid's algorithm is total without that.
Sure. But eventually we will fix this.
I have written and posted a total gcd calculator as BoundedGCD.idr.
Could we possibly get a summary of what has been completed so far and what remains to be done? I don't know where to pick up from and the amount of code that is up makes it hard to tell.
you can read our reports to get a rough idea.
fundamental theorem of arithmetic remains to be done |
Interest Model - DeFiner.org
u is the capital utilization rate of a certain token
Compound Supply Rate: the real-time supply rate on the money market
Compound Borrow Rate: the real-time borrow rate on the money market
Compound Supply Rate Weight: the weight parameter of the Compound Supply Rate
Compound Borrow Rate Weight: the weight parameter of the Compound Borrow Rate
Compound Supply Ratio: the percentage of capital deployed on money market
Borrow Rate Model
Borrow APR= Compound Supply Rate Weights \times Compound Supply Rate + Compound Borrow Rate Weights \times Compound Borrow Rate () + RateCurve Constant\div(1-u)
u
Rate Curve Constant\div(1-u) = Rate Curve Constant \div(1-0.98)= RateCurveConstant\times50
For assets that are not available on Compound or other money markets, Compound Supply Rate Weights=0, Compound Borrow Rate Weights=0,
In summary, there are two factors that decided the Borrow APR, the prevailing market rate that is available in the market and the capital utilization rate in the DeFiner protocol. Also, it's a non-linear model. The borrowing interest can adapt quickly if the utilization of the pool approaches a relatively high level.
Based on different parameter sets, we have three different strategies: Conservative Mode, Moderate Model, and Aggressive Model.
_RateCurveConstant
Below is how the borrow interest rate curve varies at different capital utilization levels based on three strategies.
if(isSupportedOnCompound) {
if (u>0.999) {
BorrowAPR= compoundSupplyRateWeights*(compoundSupplyRate) + compoundBorrowRateWeights*(compoundBorrowRate) + RateCurveConstant*(1000);
BorrowAPR= compoundSupplyRateWeights*(compoundSupplyRate) + compoundBorrowRateWeights*(compoundBorrowRate) + RateCurveConstant/(1-u);
BorrowAPR = RateCurveConstant*(1000);
BorrowAPR = RateCurveConstant/(1-u);
uint256 capitalUtilizationRatio = getCapitalUtilizationRatio(_token);
// rateCurveConstant = <'3 * (10)^16'_rateCurveConstant_configurable>
uint256 rateCurveConstant = globalConfig.rateCurveConstant();
// compoundSupply = Compound Supply Rate * <'0.4'_supplyRateWeights_configurable>
uint256 compoundSupply = compoundPool[_token].depositRatePerBlock.mul(globalConfig.compoundSupplyRateWeights());
// compoundBorrow = Compound Borrow Rate * <'0.6'_borrowRateWeights_configurable>
uint256 compoundBorrow = compoundPool[_token].borrowRatePerBlock.mul(globalConfig.compoundBorrowRateWeights());
// nonUtilizedCapRatio = (1 - U) // Non utilized capital ratio
uint256 nonUtilizedCapRatio = INT_UNIT.sub(capitalUtilizationRatio);
bool isSupportedOnCompound = globalConfig.tokenInfoRegistry().isSupportedOnCompound(_token);
uint256 compoundSupplyPlusBorrow = compoundSupply.add(compoundBorrow).div(10);
uint256 rateConstant;
// if the token is supported in third party (like Compound), check if U = 1
if(capitalUtilizationRatio > ((10**18) - (10**15))) { // > 0.999
// if U = 1, borrowing rate = compoundSupply + compoundBorrow + ((rateCurveConstant * 100) / BLOCKS_PER_YEAR)
rateConstant = rateCurveConstant.mul(1000).div(BLOCKS_PER_YEAR);
return compoundSupplyPlusBorrow.add(rateConstant);
// if U != 1, borrowing rate = compoundSupply + compoundBorrow + ((rateCurveConstant / (1 - U)) / BLOCKS_PER_YEAR)
rateConstant = rateCurveConstant.mul(10**18).div(nonUtilizedCapRatio).div(BLOCKS_PER_YEAR);
// If the token is NOT supported by the third party, check if U = 1
// if U = 1, borrowing rate = rateCurveConstant * 100
return rateCurveConstant.mul(1000).div(BLOCKS_PER_YEAR);
// if 0 < U < 1, borrowing rate = 3% / (1 - U)
return rateCurveConstant.mul(10**18).div(nonUtilizedCapRatio).div(BLOCKS_PER_YEAR);
Deposit Rate Model
Deposit Rate= CompoundSupplyRatio\times CompoundSupplyRate +BorrowRate\times u
For assets that are not available on Compound or other money markets, Compound Supply Rate Weights=0, Compound Borrow Rate Weights=0
function getDepositRatePerBlock(address _token) public view returns(uint) {
uint borrowRatePerBlock = getBorrowRatePerBlock(_token);
uint capitalUtilRatio = getCapitalUtilizationRatio(_token);
if(!isSupportedOnCompound) {
DepositAPR = borrowRatePerBlock * capitalUtilRatio;
DepositAPR = borrowRatePerBlock * capitalUtilRatio + CompoundSupplyRate*(u);
uint256 borrowRatePerBlock = getBorrowRatePerBlock(_token);
uint256 capitalUtilRatio = getCapitalUtilizationRatio(_token);
return borrowRatePerBlock.mul(capitalUtilRatio).div(INT_UNIT);
return borrowRatePerBlock.mul(capitalUtilRatio).add(compoundPool[_token].depositRatePerBlock
.mul(compoundPool[_token].capitalRatio)).div(INT_UNIT);
Interest Accounting System
Deposit principle: the crypto assets that users deposited
Deposit interest: interest that the depositor earned
Deposit storage interest: the interest that depositor accrued
Deposit accrual interest: the deposit interest that has not accrued
Deposit Interest per block: interest that user earned for every block
BlocksPerYear: annual expected blocks of the blockchain
Deposit Interest Rate Per Block = BorrowAPR\div BlocksPerYear
Deposit Interest Per Block = (Deposit Principle+Deposit Storage Interest) \times Deposit Interest Rate Per Block
DepositInterest(block_t)=DepositInterest(block_t -_1))+DepositInterestPerBlock
BorrowAPR will be updated in the contract if there were any users who have deposits of the token performs a transaction
The interest earned between the last transaction block of the user and the latest transaction block will be accrued if the user performed a transaction and will be added to the Deposit Storage Interest. |
IN THE ADJOINING FIGURE,IF ANGLE 1=ANGLE 2,ANGLE 3 =ANGLE 4 ANGLE 2=ANGLE4, THEN FIND THE RELATION BETWEEN ANGLE1 AND ANGLE - Maths - Introduction to Euclid\s Geometry - 7999173 | Meritnation.com
IN THE ADJOINING FIGURE,IF ANGLE 1=ANGLE 2,ANGLE 3 =ANGLE 4.ANGLE 2=ANGLE4, THEN FIND THE RELATION BETWEEN ANGLE1 AND ANGLE 3, USING EUCLIDS AXIOM.
\angle
\angle
2 ------------------ ( 1 )
\angle
\angle
\angle
\angle
And we know by first axiom of Euclid's " Things which are equal to the same thing are also equal to one another. "
So , from equation 2 and 3 , we get
\angle
\angle
3 ----------- ( 4 ) ( As both angles equal to same angle 4 , so we apply given axiom )
form equation 1 and 4 , we get
As , we know " Things which are equal to the same thing are also equal to one another. "
\angle
\angle
3 ( Ans )
Sakthi Ramu.... answered this
The things tht r equal to the same thing r equal to each other... As
By the same axiom we can prove tht
Gargi Srinivasa answered this
L is angle sign .
L1 = L2 ---(1)
So we can substitute L2 with L4 because they are equal to each other in (1).
L1 = L3 [ becauseTHINGS WHICH ARE EQUAL TO THE SAME THING ARE EQUAL TO EACH OTHER ] |
Explain the answer of 9.
what is oxidation number of metal in IUPAC name of complex cation and complex anion
Experts can we say that lizard could be a suitable example for axesual reproduction?
Please tell what's the correct hybridisation, magnetic behaviour, and formula of geometrical or optical isomer for the complexes written in the page.
Specially tell about the complexes for which i have mention question mark.
I had already the mention things for the complexes in case if it is wrong( if any found for any complex) then correct me.
Note ;- I don’t want link for similar query.
I have marked x for those complexex whose particular isomeric form doesn’t exist and right for those whose exist.
\left[Ni\left({H}_{2}O{\right)}_{2}\left({C}_{2}{O}_{4}{\right)}_{2}{\right]}^{-2}\phantom{\rule{0ex}{0ex}}\left[Co\left(en{\right)}_{2}C{l}_{2}{\right]}^{+}\phantom{\rule{0ex}{0ex}}\left[Cr\left({C}_{2}{O}_{4}{\right)}_{3}{\right]}^{-3}\phantom{\rule{0ex}{0ex}}\left[Co\left(N{H}_{3}{\right)}_{3}C{l}_{3}\right]\phantom{\rule{0ex}{0ex}}\left[Fe\left(CN{\right)}_{6}{\right]}^{-4}
Please tell what's the correct hybridisation, magnetic behaviour, and formula of geometrical or optical isomer for the complexes written in the page. Specially tell about the complexes for which i have mention question mark. I had already the mention things for the complexes in case if it is wrong( if any found for any complex) then correct me. Note ;- I don’t want link for similar query. I have marked x for those complexex whose particular isomeric form doesn’t exist and right for those whose exist.
Please tell what's the correct hybridisation, magnetic behaviour, and formula of geometrical or optical isomer for the complexes
\left[Cr\left(N{H}_{3}{\right)}_{4}C{l}_{2}\right]Cl\phantom{\rule{0ex}{0ex}}\left[Co\left(N{H}_{3}{\right)}_{4}C{l}_{2}\right]Cl\phantom{\rule{0ex}{0ex}}\left[Co\left(N{H}_{3}\right)\left(en{\right)}_{2}O{\right]}^{+2}\phantom{\rule{0ex}{0ex}}\left[Ni\left({H}_{2}O{\right)}_{2}\left({C}_{2}{O}_{4}{\right)}_{2}{\right]}^{-2}\phantom{\rule{0ex}{0ex}}\left[Co\left(en{\right)}_{2}C{l}_{2}{\right]}^{+2}
\left[Cr\left(N{H}_{3}{\right)}_{4}C{l}_{2}\right]Cl\phantom{\rule{0ex}{0ex}}\left[Co\left(N{H}_{3}{\right)}_{4}C{l}_{2}\right]Cl\phantom{\rule{0ex}{0ex}}\left[Fe\left(N{H}_{3}{\right)}_{4}C{l}_{2}\right]\phantom{\rule{0ex}{0ex}}\left[Co\left(en{\right)}_{3}\right]
\left[Co\left(N{H}_{3}{\right)}_{5}Cl\right]{O}_{2}\phantom{\rule{0ex}{0ex}}\left[Co\left({H}_{2}O{\right)}_{2}\left({C}_{2}{O}_{4}{\right)}_{2}\right]\phantom{\rule{0ex}{0ex}}\left[Fe\left(N{H}_{3}{\right)}_{4}C{l}_{2}\right]\phantom{\rule{0ex}{0ex}}\left[Co\left(en{\right)}_{3}\right]
Help for 3 to 5
\left[Mn\left(CN{\right)}_{6}{\right]}^{-3}\phantom{\rule{0ex}{0ex}}\left[Co\left({C}_{2}{O}_{4}{\right)}_{3}{\right]}^{-3}\phantom{\rule{0ex}{0ex}}\left[Fe\left(CN{\right)}_{6}{\right]}^{-3}
I m confused .. I think it should be pent-3-nitride
Ch3cch3ch3-oh |
An object of mass 3 kg is traveling counterclockwise around the ellipse
4{x}^{2}+9{y}^{2}=36
. When it reaches the point
\left(0,-2\right)
its acceleration vector is
3 \mathbf{i}+5 \mathbf{j}
. What is its speed?
\left(0,-2\right)
is on the branch of the ellipse defined by
y\left(x\right)= -2\sqrt{1-{\left(x/3\right)}^{2}}
, which has curvature
\mathrm{κ}=\frac{|y″|}{{\left(1+{\left(y\prime \right)}^{2}\right)}^{3/2}}
\frac{162}{{\left(81-5 {x}^{2}\right)}^{3/2}}
x=0
\mathbf{T}\left(0\right)=\mathbf{i}
\mathbf{N}\left(0\right)=\mathbf{j}
\mathrm{κ}\left(0\right)=162/{81}^{3/2}=2/9
\mathbf{F}=m \mathbf{a}
\left[\begin{array}{c}3\\ 5\end{array}\right]
3(\stackrel{.}{v} \left[\begin{array}{c}1\\ 0\end{array}\right]+\frac{2}{9} {v}^{2} \left[\begin{array}{c}0\\ 1\end{array}\right])
from which the equation
5=2 {v}^{2}/3
can be extracted. Since the speed must be positive,
v=\sqrt{15/2} |
Home : Support : Online Help : Mathematics : Factorization and Solving Equations : RegularChains : Intersect
compute the common solutions of a polynomial and a regular chain
Intersect(f, rc, R)
The command Intersect(f, rc, R) computes the common solutions of the polynomial f and the regular chain rc in the following sense. Let V be the hypersurface defined by f, that is, the solutions of the equation f = 0 . Let W be the quasi-component of rc. Then Intersect(f, rc, R) returns regular chains such that the union of their quasi-components contains the intersection of V and W, and this union is contained in the intersection of V and the Zariski closure of W. See ConstructibleSetTools for a definition of a quasi-component.
When the regular chain rc has dimension zero, Intersect(f, rc, R) computes exactly the intersection of V and W. This is also the case when W is a variety (that is a closed set for Zariski topology) or when rc has dimension one and f is regular w.r.t. the saturated ideal of rc. In all other cases, Intersect(f, rc, R) computes a superset of the intersection of V and W. However this superset is very close to this intersection.
In summary and in broad terms, Intersect(f, rc, R) computes a sharp approximation of the intersection of V and W by means of regular chains.
You can use the function Intersect to solve systems of equations incrementally, that is, one equation after the other. The example below illustrates this strategy.
Another way of understanding the Intersect command is to observe that it specializes the solutions of rc with the constraint f = 0 .
\mathrm{with}\left(\mathrm{RegularChains}\right):
\mathrm{with}\left(\mathrm{ChainTools}\right):
\mathrm{vars}≔[x,y,z]:
R≔\mathrm{PolynomialRing}\left(\mathrm{vars}\right):
\mathrm{sys}≔[{x}^{2}+y+z-1,x+{y}^{2}+z-1,x+y+{z}^{2}-1]
\textcolor[rgb]{0,0,1}{\mathrm{sys}}\textcolor[rgb]{0,0,1}{≔}[{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{z}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}]
Define the empty regular chain.
\mathrm{rc}≔\mathrm{Empty}\left(R\right)
\textcolor[rgb]{0,0,1}{\mathrm{rc}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{regular_chain}}
\mathrm{dec}≔\mathrm{Intersect}\left(\mathrm{sys}[1],\mathrm{rc},R\right);
\mathrm{map}\left(\mathrm{Equations},\mathrm{dec},R\right)
\textcolor[rgb]{0,0,1}{\mathrm{dec}}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{\mathrm{regular_chain}}]
[[{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}]]
Solve the first and second equations.
\mathrm{dec}≔[\mathrm{seq}\left(\mathrm{op}\left(\mathrm{Intersect}\left(\mathrm{sys}[2],\mathrm{rc},R\right)\right),\mathrm{rc}=\mathrm{dec}\right)];
\mathrm{map}\left(\mathrm{Equations},\mathrm{dec},R\right)
\textcolor[rgb]{0,0,1}{\mathrm{dec}}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{\mathrm{regular_chain}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{regular_chain}}]
[[\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{z}]]
Solve the three equations together.
\mathrm{dec}≔[\mathrm{seq}\left(\mathrm{op}\left(\mathrm{Intersect}\left(\mathrm{sys}[3],\mathrm{rc},R\right)\right),\mathrm{rc}=\mathrm{dec}\right)];
\mathrm{map}\left(\mathrm{Equations},\mathrm{dec},R\right)
\textcolor[rgb]{0,0,1}{\mathrm{dec}}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{\mathrm{regular_chain}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{regular_chain}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{regular_chain}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{regular_chain}}]
[[\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{z}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{z}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{z}]] |
Model voltage controlled oscillator - Simulink - MathWorks Switzerland
VCO Subsystem
Model voltage controlled oscillator
VCO or voltage controlled oscillator is a voltage to frequency converter. It produces an output square wave signal whose frequency is controlled by the voltage at the input vctrl port. The frequency of the output signal, F is determined either by:
F=\left({K}_{VCO}·{V}_{ctrl}\right)+{F}_{o}
Kvco = voltage sensitivity (in Hz/V)
Vctrl = control voltage (in V)
Fo= free running frequency (in Hz)
or from linear interpolation using the mapping:
F=\text{interp}\left({F}_{out}\left({V}_{cntl}\right)\right)
Vctnl = vector of control voltages (in V)
Fout= vector of corresponding output frequencies (in Hz)
VCO control voltage used to control the output frequency of the VCO. In a phase-locked loop (PLL) system, vctrl is the output of the Loop Filter that contains the phase error information.
vco out — Output square wave signal determined by vctrl port
Output square wave signal of VCO. In a PLL system, vco out is the output clock generated by the PLL. It is also fed back to the PFD block through a clock divider to complete the control loop.
To enable this parameter, select Voltage sensitivity in Specify using in the Parameters tab.
To enable this parameter, select Output frequency vs. control voltage in Specify using in the Parameters tab.
Output amplitude (V) — Maximum amplitude of the VCO output voltage
Select to enable increased buffer size during simulation. This increases the buffer size of the Variable Pulse Delay block inside the VCO block. By default, this option is deselected.
Number of samples of the input buffering available during simulation, specified as a positive integer scalar. This sets the buffer size of the Variable Pulse Delay block inside the VCO block.
The frequency offsets of phase noise from the carrier frequency specified as a positive real valued vector in Hz.
The phase noise power in a 1 Hz bandwidth centered at the specified frequency offsets relative to the carrier specified as a negative real valued vector in dBc/Hz. The elements of Phase noise level corresponds to relative elements in the Phase noise frequency offset.
Find VCO metrics such as voltage sensitivity (Kvco) and quiescent frequency or free running frequency (Fo).
Validate the phase noise profile of a VCO device under test (DUT) using a VCO Testbench.
The VCO subsystem block consists of two subsystems, Ideal VCO and Real VCO encapsulated under one variant subsystem.
If Add phase noise impairment is disabled, then the Ideal VCO subsystem gets active. This produces the following two orthogonal output signals, without any phase noise impairment and hence the name Ideal VCO:
\begin{array}{l}{y}_{1}\left(t\right)=A\mathrm{cos}\int \left(2\pi {K}_{\text{vco}}*{V}_{\text{ctrl}}+2\pi {F}_{\text{out}}\right)dt\\ {y}_{2}\left(t\right)=-A\mathrm{sin}\int \left(2\pi {K}_{\text{vco}}*{V}_{\text{ctrl}}+2\pi {F}_{\text{out}}\right)dt\end{array}
Out of the two orthogonal outputs, only the real part of the signal, y1(t) is connected to the output port of VCO.
When the Add phase noise impairment is enabled, the Real VCO block becomes active which introduces phase noise as a function of frequency to the ideal VCO output. The subsystem consists of the Ideal VCO block with a phase noise generator block. The latter adds phase noise impairment to the input signal.
Loop Filter | PFD | VCO Testbench | Ring Oscillator VCO |
Fundamental theorem of linear programming - Wikipedia
Extremes of a linear function over a convex polygonal region occur at the region's corners
In mathematical optimization, the fundamental theorem of linear programming states, in a weak formulation, that the maxima and minima of a linear function over a convex polygonal region occur at the region's corners. Further, if an extreme value occurs at two corners, then it must also occur everywhere on the line segment between them.
{\displaystyle \min c^{T}x{\text{ subject to }}x\in P}
{\displaystyle P=\{x\in \mathbb {R} ^{n}:Ax\leq b\}}
{\displaystyle P}
is a bounded polyhedron (and thus a polytope) and
{\displaystyle x^{\ast }}
is an optimal solution to the problem, then
{\displaystyle x^{\ast }}
is either an extreme point (vertex) of
{\displaystyle P}
, or lies on a face
{\displaystyle F\subset P}
of optimal solutions.
Suppose, for the sake of contradiction, that
{\displaystyle x^{\ast }\in \mathrm {int} (P)}
{\displaystyle \epsilon >0}
such that the ball of radius
{\displaystyle \epsilon }
{\displaystyle x^{\ast }}
{\displaystyle P}
{\displaystyle B_{\epsilon }(x^{\ast })\subset P}
{\displaystyle x^{\ast }-{\frac {\epsilon }{2}}{\frac {c}{||c||}}\in P}
{\displaystyle c^{T}\left(x^{\ast }-{\frac {\epsilon }{2}}{\frac {c}{||c||}}\right)=c^{T}x^{\ast }-{\frac {\epsilon }{2}}{\frac {c^{T}c}{||c||}}=c^{T}x^{\ast }-{\frac {\epsilon }{2}}||c||<c^{T}x^{\ast }.}
{\displaystyle x^{\ast }}
is not an optimal solution, a contradiction. Therefore,
{\displaystyle x^{\ast }}
must live on the boundary of
{\displaystyle P}
{\displaystyle x^{\ast }}
is not a vertex itself, it must be the convex combination of vertices of
{\displaystyle P}
{\displaystyle x_{1},...,x_{t}}
{\displaystyle x^{\ast }=\sum _{i=1}^{t}\lambda _{i}x_{i}}
{\displaystyle \lambda _{i}\geq 0}
{\displaystyle \sum _{i=1}^{t}\lambda _{i}=1}
{\displaystyle 0=c^{T}\left(\left(\sum _{i=1}^{t}\lambda _{i}x_{i}\right)-x^{\ast }\right)=c^{T}\left(\sum _{i=1}^{t}\lambda _{i}(x_{i}-x^{\ast })\right)=\sum _{i=1}^{t}\lambda _{i}(c^{T}x_{i}-c^{T}x^{\ast }).}
{\displaystyle x^{\ast }}
is an optimal solution, all terms in the sum are nonnegative. Since the sum is equal to zero, we must have that each individual term is equal to zero. Hence,
{\displaystyle c^{T}x^{\ast }=c^{T}x_{i}}
{\displaystyle x_{i}}
, so every
{\displaystyle x_{i}}
is also optimal, and therefore all points on the face whose vertices are
{\displaystyle x_{1},...,x_{t}}
, are optimal solutions.
http://www.linearprogramming.info/fundamental-theorem-of-linear-programming-and-its-properties/
http://demonstrations.wolfram.com/TheFundamentalTheoremOfLinearProgramming/
Retrieved from "https://en.wikipedia.org/w/index.php?title=Fundamental_theorem_of_linear_programming&oldid=982852281" |
function bounds - Maple Help
Home : Support : Online Help : Programming : Logic : Boolean : verify : function bounds
verify/function_bounds
verify approximate equality between two function plots
verify(P, Q, function_bounds)
The verify(P, Q, function_bounds) calling sequence verifies the approximate equality between two function plots.
The parameters, P and Q, are assumed to be either PLOT data structures, sets or lists of CURVES data structures, or a CURVES data structure.
The verify(P, Q, function_bounds) function returns true for CURVES data-structures P and Q by checking that neither curve has extreme points or constant regions which do not appear in the other curve or in the union of the two curves.
If you are comparing two curves, false is returned by a list with false in the first operand and a plot data-structure showing the points where the curves differ.
a≔\mathrm{plot}\left(\mathrm{piecewise}\left(x<1,x,2-x\right),x=0..2\right):
b≔\mathrm{plot}\left(\mathrm{piecewise}\left(x<1,x,2-x\right),x=0..2,\mathrm{numpoints}=10,\mathrm{adaptive}=\mathrm{false}\right):
c≔\mathrm{plot}\left(1.001\left(\mathrm{piecewise}\left(x<1,x,2-x\right)\right),x=0..2,\mathrm{numpoints}=100\right):
\mathrm{verify}\left(a,b,\mathrm{function_bounds}\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{verify}\left(a,c,\mathrm{function_bounds}\right)
[\textcolor[rgb]{0,0,1}{\mathrm{false}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{PLOT}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{\mathrm{...}}\right)]
Note that the plot 'b' does not have the last maxima.
a≔\mathrm{plot}\left(\mathrm{sin}\left(x\right),x=0..40\right):
b≔\mathrm{plot}\left(\mathrm{sin}\left(x\right),x=0..40,\mathrm{numpoints}=20,\mathrm{adaptive}=\mathrm{false}\right):
c≔\mathrm{plot}\left(\mathrm{sin}\left(x\right),x=0..40,\mathrm{numpoints}=30,\mathrm{adaptive}=\mathrm{false}\right):
\mathrm{verify}\left(a,b,\mathrm{function_bounds}\right)
[\textcolor[rgb]{0,0,1}{\mathrm{false}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{PLOT}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{\mathrm{...}}\right)]
\mathrm{verify}\left(a,c,\mathrm{function_bounds}\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
verify/function_shells |
How can a median be greater than the mean?
The median of a set of numbers is the value that is in the middle (In a set with an odd number of values, it's the middle value. In a set with an even number of values, it's the mean of the two middle values).
The mean is the generally understood "average", where the sum of the values is divided by the number of values (sometimes referred to as the count of the values).
How can we set up a set of values so that the median is higher than the mean? We can do it by taking a set of numbers and skewing the values to be very low below the median and just above the median. For instance, if I take a set of five numbers and set the middle value as 10, I can place the two lower values at 1 and 2 and the higher values at 11:
=\frac{1+2+10+11+11}{5}=\frac{35}{5}=7
In fact, the mean will be lower than the median in any distribution where the values "fall off", or decrease from the middle value faster than they increase from the middle value.
Given the set: {9,0,6, -10, 2,x,3}, for what x would the mean of the set be -2?
Gender is on variable of interest in a study of the effectiveness of a new medication.For data entry purposes, the researcher conducting the study assigns 1 for Male and 2 for Female. Is the gender data quantitative or qualtitative?
Time and Work: Techniques and examples with solutionsToday I'm going to discuss a very important topic i.e. Time and Work of quantitative aptitude. In almost every exam at least 2-3 question are asked every time. In this chapter, I will tell you about a definite relationship between Time and work and easy method to solve the problems.
Find the mean for the data.
The data represents the costs of nine compact refrigerators rated very good or excellent by Consumer Reports on their website.
Given a set {-3, x,3,4,6,5, -3, 2}, for what x would the mean of the set be -1?
Given the set: {2, 1, -3, x, 1, 2, -4}, for what x would the mean of the set be 3? |
Mixtures | Brilliant Math & Science Wiki
Jordan Calmes, Rishabh Tripathi, Vishnuram Leonardodavinci, and
A mixture contains two or more distinct chemical substances. The components of a mixture interact physically, but no chemical reaction takes place. There is no rearrangement of valence in any of the substances involved in the mixture. Mixtures, unlike compounds, can be separated again using physical methods such as filtration or distillation.
Many mining and mineral extraction processes require separating mixtures to extract and purify the desired products. Crude oil is a more complex mixture, containing many different types of hydrocarbons. The components of the mixture are more industrially valuable once they have been separated.
Mixtures lack the drama and excitement of chemical reactions: there are no explosions and the color changes are subtle at best, but in their own way, mixtures are even more interesting than reactions because of that subtlety. Mixtures allow for stability. It would be bad for your health if every oxygen atom that entered your lungs reacted with the first molecule it came in contact with!
Solutions: One Type of Homogeneous Mixture
Suspensions: One Type of Heterogeneous Mixtures
Chemical mixtures can be as simple as a cube of sugar stirred into a mug of hot water. The sugar seems to disappear, but if the water is boiled or left sitting long enough to evaporate at room temperature, the mixture separates again, with solid sugar molecules left behind in the mug.
Waste disposal is another field where it is crucial to understand the properties of chemical mixtures and how to separate them. Hazardous waste separated from non-hazardous waste is cheaper to dispose of. This diagram shows how wastewater mixtures can be separated into their liquid and solid components, then treated appropriately such that the water can be recycled. [2]
All three phases of matter can participate in mixtures. Mixtures may contain only one phase (all gases or all liquids, for example), or they may contain multiple phases. Coca-cola and other carbonated sodas are a gas-liquid mixture.
Homogeneous mixtures are those which have a uniform composition throughout. The components of the mixture are interspersed at a molecular or atomic level. In the sugar example mentioned previously, the sugar molecules are separated from one another by the water molecules, and a sample from any part of the mixture would contain roughly the same proportion of sugar molecules and water molecules. Another way of phrasing that is to say homogeneous mixtures are defined as those in which the molar ratio of components remains constant.
By contrast, heterogeneous mixtures have distinguishable phases. If someone added plastic cubes resembling sugar cubes to a mug of hot water, the result would still be a mixture, but no amount of stirring would make it homogenous. A sample taken from the top of the mug might contain only water molecules.
We take 1L water(Density= 1 g/mL) in beaker and dissolve 10 g common salt (sodium chloride) in it. When the salt is completely dissolved, we took out 10 mL of the solution in another beaker and named it as sample A. Is A homogeneous? If yes then calculate the amount of salt present in it. (Assume that the salt has no effect on the total volume of the mixture.)
Yes, sample A is a homogeneous mixture because salt dissolves in water uniformly.
Molar ratio of salt and water
= \frac{\frac{1000}{18}}{\frac{10}{58.5}}
(You can learn how to calculate the number of moles here).
Number of moles of water in sample A
= \frac{10}{18}
Let number of moles of Sodium chloride in A
=x
Since the molar ratio remains constant in homogeneous solution,
\frac{5850}{18} = \frac{\frac{10}{18}}{x}
x= \frac{1}{585}
Mass of sodium chloride in A
= \frac{1}{585} \times 58.5 = 0.1 g
One of the components of the mixture is called the solvent. The solvent is the same phase as the resulting solution. Components that dissolve or dissociate in the solvent are called solutes. They are generally present at smaller molar concentrations than the solvent and may be the same phase or a different phase.
Identify the solute and solvent in each of the following examples
(1) A mixture of salt and water.
(2) A mixture of 10 moles of ethyl alcohol and 150 moles of water.
(3) A mixture of 1 L oxygen and 5 L steam at 373 K temperature.
(1) Salt is the solute and water is the solvent. The phase of the solution is liquid, which is same as the phase of water.
(2) Ethyl alcohol is the solute and water is the solvent. There are fewer moles of ethyl alcohol than of water in the solution.
(3) Oxygen is the solute and steam is the solvent. There are fewer moles of oxygen than of steam. (We can calculate the number of moles using the ideal gas law,
PV=nRT
If the number of moles of substance A and substance B are the same and they are both in the same phase, either of them could be called the solvent and the other would be called the solute, or both of them could be called solvents. However, both of them could not be called solutes. A solution cannot exist without a solvent.
Some properties of solutions are:
(1) The size of the particles involved is usually less than 1 nanometer.
(2) The solute and solvent cannot be separated by centrifugation, decantation, or filtration.
(3) They can usually be separated by distillation or fractional distillation.
Sulphur in water detergent in water Egg albumin in water Sodium chloride dissolved in water
The particles of X in water can not be seen even with microscope. It forms homogeneous transparent mixture and no residue is left on filter paper on filtration. Particles do not settle down due to gravity. X could be:
A new mother goes to the pharmacy to pick up an antibiotic for her child, who has an ear infection. The child is two years old, and cannot swallow tablets, he has been prescribed a liquid medication. The pharmacist tells the mother that this medicine, an amoxicillin suspension, must be shaken before every dose is given, but he doesn't tell her why. The mother is distracted and in a hurry. She quickly leaves with the medication, and dismisses what the pharmacist told her. Her child had an ear infection last year, and she didn't shake the bottle before giving him his medicine, and it worked fine. However, three days later the child is even sicker, even though he has been taking the antibiotic. What happened?
Some liquid antibiotics are homogenous solutions, and others are heterogenous suspensions. It's possible that the last time this child was given an antibiotic, it was a solution that had a uniform concentration of the drug. However, in a suspension, the drug separates and moves toward the bottom of the bottle. Without shaking the medication, the mixture at the top of the bottle does not contain the correct concentration of the drug, so the child is not actually getting enough medication to kill the infection.
Depending on the nature of a mixture, many physical processes can be used to separate it into its components. Some of them are listed below. Crude oil is separated by fractional distillation, because hydrocarbons with different masses boil at different temperatures [3].
Filtration, used commonly to separate solids from liquids in a heterogenous mixture.
Distillation, a process where a liquid is boiled and then the vapor is collected, leaving other components of the mixture behind.
Fractional Distillation, which is particularly useful for separating fluids that mix with each other but have different boiling points.
Decantation, where a less-dense liquid is carefully poured off the top of a heterogenous mixture into a new container.
Electromagnetic-Separation: which uses the electrical or magnetic properties of a material to separated it from a mixture, such as running a large magnet over a pile of scrap metal to find iron.
Chromatography can be used to separated gases, liquids, or dissolved substances. Many different types of chromatography exist. In each technique, the mixture is placed on or in a solid phase and a moving phase is allowed to pass through the system. The moving phase will separate the mixture based on physical properties, such as particle size or vapor pressure.
In thin layer chromatography (TLC), the mixture is dotted on a silica gel plate (the solid phase. The end of the plate is placed in a jar of solvent (the moving phase), and capillary action slowly carries the solvent up the plate. As the solvent moves, it encounters the mixture and moves particles from the mixture up the silica plate. The distance the particles moves depends on their polarity. This is a quick and cheap way to tell if a chemical is pure, or how many components are in a mixture. [4]
[1] Image from U.S. Energy Information Administration. Energy Kids http://www.eia.gov/kids/energy.cfm?page=oil_home-basics Accessed February 27, 2016.
[2] Image from New York State Department of Environmental Conservation. Recycling Biosolids from Wastewater Treatment Facilities http://www.dec.ny.gov/chemical/97463.html Accessed February 27, 2016.
[3] Image from U.S. Energy Information Administration. Crude oil distillation and the definition of refinery capacity http://www.eia.gov/todayinenergy/detail.cfm?id=6970 Accessed February 27, 2016.
[4] Image from https://commons.wikimedia.org/wiki/File:Paper.jpg under Creative Commons licensing for reuse and modification.
Cite as: Mixtures. Brilliant.org. Retrieved from https://brilliant.org/wiki/mixtures/ |
Stochastic frontier analysis - Wikipedia
Stochastic frontier analysis (SFA) is a method of economic modeling. It has its starting point in the stochastic production frontier models simultaneously introduced by Aigner, Lovell and Schmidt (1977) and Meeusen and Van den Broeck (1977).
The production frontier model without random component can be written as:
{\displaystyle y_{i}=f(x_{i};\beta )\cdot TE_{i}}
the best where yi is the observed scalar output of the producer i, i=1,..I, xi is a vector of N inputs used by the producer i, f(xi, β) is the production frontier, and
{\displaystyle \beta }
is a vector of technology parameters to be estimated.
TEi denotes the technical efficiency defined as the ratio of observed output to maximum feasible output. TEi = 1 shows that the i-th firm obtains the maximum feasible output, while TEi < 1 provides a measure of the shortfall of the observed output from maximum feasible output.
A stochastic component that describes random shocks affecting the production process is added. These shocks are not directly attributable to the producer or the underlying technology. These shocks may come from weather changes, economic adversities or plain luck. We denote these effects with
{\displaystyle \exp \left\{{v_{i}}\right\}}
. Each producer is facing a different shock, but we assume the shocks are random and they are described by a common distribution.
The stochastic production frontier will become:
{\displaystyle y_{i}=f(x_{i};\beta )\cdot TE_{i}\cdot \exp \left\{{v_{i}}\right\}}
We assume that TEi is also a stochastic variable, with a specific distribution function, common to all producers.
We can also write it as an exponential
{\displaystyle TE_{i}=\exp \left\{{-u_{i}}\right\}}
, where ui ≥ 0, since we required TEi ≤ 1. Thus, we obtain the following equation:
{\displaystyle y_{i}=f(x_{i};\beta )\cdot \exp \left\{{-u_{i}}\right\}\cdot \exp \left\{{v_{i}}\right\}}
Now, if we also assume that f(xi, β) takes the log-linear Cobb–Douglas form, the model can be written as:
{\displaystyle \ln y_{i}=\beta _{0}+\sum \limits _{n}{\beta _{n}\ln x_{ni}+v_{i}-u_{i}}}
where vi is the “noise” component, which we will almost always consider as a two-sided normally distributed variable, and ui is the non-negative technical inefficiency component. Together they constitute a compound error term, with a specific distribution to be determined, hence the name of “composed error model” as is often referred.
Stochastic frontier analysis has examined also "cost" and "profit" efficiency (see Kumbhakar & Lovell 2003). The "cost frontier" approach attempts to measure how far from full-cost minimization (i.e. cost-efficiency) is the firm. Modeling-wise, the non-negative cost-inefficiency component is added rather than subtracted in the stochastic specification. "Profit frontier analysis" examines the case where producers are treated as profit-maximizers (both output and inputs should be decided by the firm) and not as cost-minimizers, (where level of output is considered as exogenously given). The specification here is similar with the "production frontier" one.
Stochastic frontier analysis has also been applied in micro data of consumer demand in an attempt to benchmark consumption and segment consumers. In a two-stage approach, a stochastic frontier model is estimated and subsequently deviations from the frontier are regressed on consumer characteristics (Baltas 2005).
Extensions: The two-tier stochastic frontier modelEdit
Polacheck & Yoon (1987) have introduced a three-component error structure, where one non-negative error term is added to, while the other is subtracted from, the zero-mean symmetric random disturbance. This modeling approach attempts to measure the impact of informational inefficiencies (incomplete and imperfect information) on the prices of realized transactions, inefficiencies that in most cases characterize both parties in a transaction (hence the two inefficiency components, to disentangle the two effects).
Recently, various non-parametric and semi-parametric approaches were proposed in the literature, where no parametric assumption on the functional form of production relationship is made, see for example Parmeter and Kumbhakar (2014) and Park, Simar and Zelenyuk (2015) [1] and references cited therein.
^ Park, B., Simar, L. and V. Zelenyuk (2015) "Categorical data in local maximum likelihood: theory and applications to productivity analysis," Journal of Productivity Analysis 43:2, pp. 199-214.
Aigner, D.J.; Lovell, C.A.K.; Schmidt, P. (1977) Formulation and estimation of stochastic frontier production functions. Journal of Econometrics, 6:21–37.
Baltas, G., (2005). Exploring Consumer Differences in Food Demand: A Stochastic Frontier Approach. British Food Journal, 107(9): 685-692.
Coelli, T.J.; Rao, D.S.P.; O'Donnell, C.J.; Battese, G.E. (2005) An Introduction to Efficiency and Productivity Analysis, 2nd Edition. Springer, ISBN 978-0-387-24266-8.`
Greene, W. H. (2008) The Econometric Approach to Efficiency Analysis. In Fried, H. O., Knox Lovell, C. A., and Schmidt, P., editors, The Measurement of Productive Efficiency. Oxford University Press, New York and Oxford.
Parmeter, C.F., Kumbhakar, S.C., (2014) "Efficiency Analysis: A Primer on Recent Advances," Foundations and Trends in Econometrics, 7(3-4), 191-385.
Polachek, S. W. ; Yoon, B. J. (1987). A two-tiered earnings frontier estimation of employer and employee information in the labor market. Review of Economics and Statistics, 69(2), 296-302.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Stochastic_frontier_analysis&oldid=1020504896" |
Consider the following system of linear equations: 4x+2y=25 4x-y=-5 If the va
Consider the following system of linear equations: 4x+2y=25 4x-y=-5 If the value of y is 10 what is the value of X for this system: 1.1.25 2.11.25 3.1.45 4.5
If the value of y is 10 what is the value of X for this system:
Given the system of equations
If value of y is 10 then we need to find the value of x.
If value of y is 10 then it must satisfy to both equations
By putting the value of y in equation first then we get
4x+2×10=25
x=\frac{5}{4}
A=\left[\begin{array}{cc}3& 1\\ 1& 1\\ 1& 4\end{array}\right],b\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]
\stackrel{―}{x}=
Consider the the equations:
Equation 1 is 5x - 2y - 4z = 3
Equation 2 is 3x + 3y + 2z = -3.
Eliminate z by copying Equation 1, multiplying Equation 2 by 2, and then adding the equations.
24{n}^{2}-38n+15=0
\left\{\begin{array}{l}{x}^{2}+{y}^{2}=8\\ x+y=0\end{array}
Write a quadratic equation in standard form with the solution set {−3, 6}.
(Simplify your answer. Type an equation using x as the variable.)
{27}^{x-1}={9}^{2x-3} |
Solve frac{(sin theta+cos theta)}{cos theta}+frac{(sin theta-cos theta)}{cos theta}
\frac{\left(\mathrm{sin}\theta +\mathrm{cos}\theta \right)}{\mathrm{cos}\theta }+\frac{\left(\mathrm{sin}\theta -\mathrm{cos}\theta \right)}{\mathrm{cos}\theta }
\frac{\mathrm{sin}\left(\theta \right)+\mathrm{cos}\left(\theta \right)}{\mathrm{cos}\left(\theta \right)}+\frac{\mathrm{sin}\left(\theta \right)-\mathrm{cos}\left(\theta \right)}{\mathrm{cos}\left(\theta \right)}=
Apply rule
\frac{a}{c}±\frac{b}{c}=\frac{a±b}{c}
=\left(\mathrm{sin}\left(\theta \right)+\mathrm{cos}\left(\theta \right)+\mathrm{sin}\left(\theta \right)-\frac{\mathrm{cos}\left(\theta \right)}{\mathrm{cos}\left(\theta \right)}
=\frac{2\mathrm{sin}\left(\theta \right)}{\mathrm{cos}\left(\theta \right)}
Use the following identity:
\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)=\mathrm{tan}\left(x\right)
=2\mathrm{tan}\left(\theta \right)
\mathrm{sin}x+\mathrm{sin}y=a\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{cos}x+\mathrm{cos}y=b
\mathrm{tan}\left(x-\frac{y}{2}\right)
\mathrm{sin}\left(A+\pi \right)=-\mathrm{sin}A
\frac{\mathrm{sin}x+\mathrm{csc}y}{\mathrm{sin}y+\mathrm{csc}x}=\mathrm{csc}y\mathrm{sin}x
Find range of the function
f\left(x\right)=3|\mathrm{sin}x|-4|\mathrm{cos}x|
I tried to do by using the trigonometric identities
{\mathrm{sin}}^{2}x=\frac{1-\mathrm{cos}2x}{2};\text{ }\text{ }{\mathrm{cos}}^{2}x=\frac{1+\mathrm{cos}2x}{2}
f\left(x\right)=3\sqrt{\frac{1-\mathrm{cos}2x}{2}}-4\sqrt{\frac{1+\mathrm{cos}2x}{2}}
but don't know how to proceed from here
\mathrm{sin}\left(2x\right)-\mathrm{tan}\left(x\right)
1+\frac{\mathrm{cos}x}{1}-\mathrm{cos}x=\frac{{\mathrm{tan}}^{2}x}{{\left(\mathrm{sec}x-1\right)}^{2}} |
Redundancy (engineering) — Wikipedia Republished // WIKI 2
For other uses, see Redundancy.
Extensively redundant rear lighting installation on a Thai tour bus
In many safety-critical systems, such as fly-by-wire and hydraulic systems in aircraft, some parts of the control system may be triplicated,[1] which is formally termed triple modular redundancy (TMR). An error in one component may then be out-voted by the other two. In a triply redundant system, the system has three sub components, all three of which must fail before the system fails. Since each one rarely fails, and the sub components are expected to fail independently, the probability of all three failing is calculated to be extraordinarily small; often outweighed by other risk factors, such as human error. Redundancy may also be known by the terms "majority voting systems"[2] or "voting logic".[3]
A suspension bridge's numerous cables are a form of redundancy.
Q-SYS: System Redundancy (QSC Innovations)
VRRP configuration on L3 switch CCNP 300-115(v-17)
Example of a TMR system with Spare.© UPV
1 Forms of redundancy
1.1 Dissimilar redundancy
1.2 Geographic redundancy
2 Function of redundancy
4 Voting logic
5 Calculating the probability of system failure
Hardware redundancy, such as dual modular redundancy and triple modular redundancy
Information redundancy, such as error detection and correction methods
Software redundancy such as N-version programming
Structures are usually designed with redundant parts as well, ensuring that if one part fails, the entire structure will not collapse. A structure without redundancy is called fracture-critical, meaning that a single broken component can cause the collapse of the entire structure. Bridges that failed due to lack of redundancy include the Silver Bridge and the Interstate 5 bridge over the Skagit River.
The Distant Early Warning Line was an example of Geographic redundancy. Those radar sites were a minimum of 50 miles apart, but provided overlapping coverage.
The two functions of redundancy are passive redundancy and active redundancy. Both functions prevent performance decline from exceeding specification limits without human intervention using extra capacity.
Eyes and ears provide working examples of passive redundancy. Vision loss in one eye does not cause blindness but depth perception is impaired. Hearing loss in one ear does not cause deafness but directionality is lost. Performance decline is commonly associated with passive redundancy when a limited number of failures occur.
Electrical power distribution provides an example of active redundancy. Several power lines connect each generation facility with customers. Each power line includes monitors that detect overload. Each power line also includes circuit breakers. The combination of power lines provides excess capacity. Circuit breakers disconnect a power line when monitors detect an overload. Power is redistributed across the remaining lines.[citation needed] At the Toronto Airport, there are 4 redundant electrical lines. Each of the 4 lines supply enough power for the entire airport. A Spot network substation uses reverse current relays to open breakers to lines that fail, but lets power continue to flow the airport.
Charles Perrow, author of Normal Accidents, has said that sometimes redundancies backfire and produce less, not more reliability. This may happen in three ways: First, redundant safety devices result in a more complex system, more prone to errors and accidents. Second, redundancy may lead to shirking of responsibility among workers. Third, redundancy may lead to increased production pressures, resulting in a system that operates at higher speeds, but less safely.[4]
A more reliable form of voting logic involves an odd number of three devices or more. All perform identical functions and the outputs are compared by the voting logic. The voting logic establishes a majority when there is a disagreement, and the majority will act to deactivate the output from other device(s) that disagree. A single fault will not interrupt normal operation. This technique is used with avionics systems, such as those responsible for operation of the Space Shuttle.
{\displaystyle {p}=\prod _{i=1}^{n}p_{i}}
{\displaystyle n}
– number of components
{\displaystyle p_{i}}
– probability of component i failing
{\displaystyle p}
– the probability of all components failing (system failure)
This formula assumes independence of failure events. That means that the probability of a component B failing given that a component A has already failed is the same as that of B failing when A has not failed. There are situations where this is unreasonable, such as using two power supplies connected to the same socket in such a way that if one power supply failed, the other would too.
Air gap (networking) – Network security measure
Common cause and special cause (statistics)
Data redundancy
Double switching
Fault tolerance – Resilience of systems to component failures or errors
Radiation hardening – Processes and techniques used for making electronic devices resistant to ionizing radiation
Factor of safety – System strength beyond intended load
Reliability engineering – Sub-discipline of systems engineering that emphasizes dependability
Reliability theory of aging and longevity – Biophysics theory
Safety engineering – Engineering discipline which assures that engineered systems provide acceptable levels of safety
Reliability (computer networking)
Unidirectional network – Network device that permits data flow in only one direction
N+1 redundancy
fault-tolerant computer system
Byzantine fault – Fault in a computer system that presents different symptoms to different observers
Quantum Byzantine agreement
Two Generals' Problem – Thought experiment
^ Redundancy Management Technique for Space Shuttle Computers (PDF), IBM Research
^ R. Jayapal (2003-12-04). "Analog Voting Circuit Is More Flexible Than Its Digital Version". elecdesign.com. Archived from the original on 2007-03-03. Retrieved 2014-06-01.
^ "The Aerospace Corporation | Assuring Space Mission Success". Aero.org. 2014-05-20. Retrieved 2014-06-01.
^ a b Scott D. Sagan (March 2004). "Learning from Normal Accidents" (PDF). Organization & Environment. Archived from the original (PDF) on 2004-07-14.
^ [4] Protecting against the power of lightning | to protect against induced surges rather than direct lightning strikes. Feb 1st, 2005 Twisted pair
Secure Propulsion using Advanced Redundant Control
Using powerline as a redundant communication channel
Occupational safety and health
Occupational diseases
Byssinosis ("brown lung")
Chimney sweeps' carcinoma
Chronic solvent-induced encephalopathy
Coalworker's pneumoconiosis ("black lung")
Concussions in sport
De Quervain syndrome
Exposure to human nail dust
Farmer's lung
Fiddler's neck
Flock worker's lung
Glassblower's cataract
Golfer's elbow
Hearing loss
Hospital-acquired infection
Indium lung
Laboratory animal allergy
Lead poisoning
Metal fume fever
Mule spinners' cancer
Noise-induced hearing loss
Phossy jaw
Radium jaw
Silo-filler's disease
Tennis elbow
Writer's cramp
Prevention through design
Occupational epidemiology
Environmental health
Occupational health nursing
Safety engineering
Canadian Centre for Occupational Health and Safety
European Agency for Safety and Health at Work
UK Health and Safety Executive
International Labour Organization
US National Institute for Occupational Safety and Health
US Occupational Safety and Health Administration
National Institute for Safety and Health at Work (Spain)
Bangladesh Accord
Occupational Safety and Health Convention, 1981
Worker Protection Standard (US)
Working Environment Convention, 1977
Emergency evacuation
Redundancy (engineering)
Occupational Safety and Health Act (United States)
Environment, health and safety
Environmental toxicology
Indoor air quality
International Chemical Safety Card
National Day of Mourning (Canadian observance)
Process safety management
Toxic tort |
Price cap instrument from Heath-Jarrow-Morton interest-rate tree - MATLAB capbyhjm - MathWorks í•œêµ
Price a 3% Cap Instrument Using an HJM Forward-Rate Tree
Compute the Price of an Amortizing Cap Using the HJM Model
Price cap instrument from Heath-Jarrow-Morton interest-rate tree
[Price,PriceTree] = capbyhjm(HJMTree,Strike,Settle,Maturity)
[Price,PriceTree] = capbyhjm(___,CapReset,Basis,Principal,Options)
[Price,PriceTree] = capbyhjm(HJMTree,Strike,Settle,Maturity) computes the price of a cap instrument from a Heath-Jarrow-Morton interest-rate tree. capbyhjm computes prices of vanilla caps and amortizing caps.
[Price,PriceTree] = capbyhjm(___,CapReset,Basis,Principal,Options) adds optional arguments.
Load the file deriv.mat, which provides HJMTree. The HJMTree structure contains the time and forward-rate information needed to price the cap instrument.
Use capbyhjm to compute the price of the cap instrument.
Price = capbyhjm(HJMTree, Strike, Settle, Maturity)
Load deriv.mat to specify the HJMTree and then define the cap instrument.
Price = capbyhjm(HJMTree, Strike, Settle, Maturity, CapReset, Basis, Principal)
Settlement date for the cap, specified as a NINST-by-1 vector of serial date numbers or date character vectors. The Settle date for every cap is set to the ValuationDate of the HJM tree. The cap argument Settle is ignored.
\mathrm{max}\left(CurrentRateâCapRate,0\right)
cfbyhjm | floorbyhjm | hjmtree | swapbyhjm | capbynormal |
Fermat's Last Theorem/Leonhard Euler - Wikibooks, open books for an open world
Leonhard Euler[edit | edit source]
{\displaystyle \mathbb {T} }
he publication of Fermat’s writings had generated opposing opinions among mathematicians. The majority of them recognised the usefulness but the fact that the greater part of the theorems were without proof or with incomplete proofs obviously reduced the immediate usefulness of them even if some mathematicians took the theorems as challenges to face and win. Many were faced and resolved but that which subsequently would be called the last theorem resisted all attempted assaults. Leonhard Euler obtained the first results a century after Fermat. Euler was a Swiss mathematician born in 1707 in Basel and died in 1783 in St. Petersburg. Initially Euler was to have become a theologian but Johann Bernoulli became aware of the extraordinary ability of the young man and convinced his father to let Leonhard become a mathematician. This was an enormous good fortune for mathematics given that Euler’s contributions range over so many areas of mathematics and are so profound as to render Euler one of the greatest mathematicians of the XVIII century if not rightly the greatest. Euler analysing the notes written by Fermat found an outline proof of the case n=4. Fermat had written this proof within another proof. In order to prove that case Fermat made use of a technique called infinite descent, Euler sought to utilise this technique for the other cases in such a way as to find a proof for all values of n. Initially he confronted the case n=3. He succeeded in resolving this case but had to make use of complex numbers, in reality other mathematicians had sought to adapt infinite descent to the case n=3 but it took a creative person such as Euler in order to understand that complex numbers were necessary in order to obtain a valid proof. Euler also sought to resolve n=5 but without results.
Retrieved from "https://en.wikibooks.org/w/index.php?title=Fermat%27s_Last_Theorem/Leonhard_Euler&oldid=3651187" |
Home : Support : Online Help : Programming : Audio Processing : Duration
compute the duration of a recording, in seconds
Duration(audArray)
Array, Vector, or Matrix containing the audio data
The Duration command computes the duration of an audio recording, in seconds, based on the number of samples, and the samples per second as stored in the attributes of audArray.
The audArray parameter must be a dense, rectangular, one or two dimensional Array, Vector, or Matrix with datatype=float[8],
\mathrm{audiofile}≔\mathrm{cat}\left(\mathrm{kernelopts}\left(\mathrm{datadir}\right),"/audio/stereo.wav"\right):
\mathrm{with}\left(\mathrm{AudioTools}\right):
\mathrm{aud}≔\mathrm{Read}\left(\mathrm{audiofile}\right)
\textcolor[rgb]{0,0,1}{\mathrm{aud}}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{"Sample Rate"}& \textcolor[rgb]{0,0,1}{22050}\\ \textcolor[rgb]{0,0,1}{"File Format"}& \textcolor[rgb]{0,0,1}{\mathrm{PCM}}\\ \textcolor[rgb]{0,0,1}{"File Bit Depth"}& \textcolor[rgb]{0,0,1}{8}\\ \textcolor[rgb]{0,0,1}{"Channels"}& \textcolor[rgb]{0,0,1}{2}\\ \textcolor[rgb]{0,0,1}{"Samples/Channel"}& \textcolor[rgb]{0,0,1}{19962}\\ \textcolor[rgb]{0,0,1}{"Duration"}& \textcolor[rgb]{0,0,1}{0.90531}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{s}\end{array}]
\mathrm{Duration}\left(\mathrm{aud}\right)
\frac{\textcolor[rgb]{0,0,1}{1109}}{\textcolor[rgb]{0,0,1}{1225}}
\mathrm{evalf}\left(\right)
\textcolor[rgb]{0,0,1}{0.9053061224} |
Absolute Complement Practice Problems Online | Brilliant
Chef Connie Cookie prepared a meal that consisted of 5 different dishes. The dishes were supposed to be served to the customers in the order Appetizer, Soup, Salad, Entree, Dessert, however the waitstaff served them in the wrong order. How many different incorrect ways could the waitstaff have served the dishes?
A way is incorrect if at least one dish wasn't served in its intended place.
The city council of Springfield consists of
13
male members and
10
female members. How many ways are there to elect a president and vice president such that at least one of the two leaders is female?
A class of students takes a test of five true and false questions. If no student answers all questions correctly, no student answers all questions incorrectly, and no two students give the same sequence of answers, what is the maximum number of students in the class?
This problem is posed by Eeshan U.
S
has 6 elements. How many subsets of
S
are there with two or more elements?
There are two sets
X= \{ a, b, c, d \}
Y= \{ a, e, i, o, u \}
. How many of the functions whose domain is
X
and codomain is
Y
, are not one-to-one?
You may choose to read up on Function Terminology. |
Calculus/Infinite Limits - Wikibooks, open books for an open world
Calculus/Infinite Limits
← Finite Limits Calculus Continuity →
Informal Infinite LimitsEdit
Another kind of limit involves looking at what happens to
{\displaystyle f(x)}
{\displaystyle x}
gets very big. For example, consider the function
{\displaystyle f(x)={\frac {1}{x}}}
{\displaystyle x}
gets very big,
{\displaystyle {\frac {1}{x}}}
gets very small. In fact,
{\displaystyle {\frac {1}{x}}}
gets closer and closer to 0 the bigger
{\displaystyle x}
gets. Without limits it is very difficult to talk about this fact, because
{\displaystyle x}
can keep getting bigger and bigger and
{\displaystyle {\frac {1}{x}}}
never actually gets to 0; but the language of limits exists precisely to let us talk about the behavior of a function as it approaches something - without caring about the fact that it will never get there. In this case, however, we have the same problem as before: how big does
{\displaystyle x}
have to be to be sure that
{\displaystyle f(x)}
is really going towards 0?
In this case, we want to say that, however close we want
{\displaystyle f(x)}
to get to 0, for
{\displaystyle x}
big enough
{\displaystyle f(x)}
is guaranteed to get that close. So we have yet another definition.
Definition: (Definition of a limit at infinity)
{\displaystyle L}
{\displaystyle f(x)}
{\displaystyle x}
approaches infinity if
{\displaystyle f(x)}
becomes arbitrarily close to
{\displaystyle L}
{\displaystyle x}
When this holds we write
{\displaystyle \lim _{x\to \infty }f(x)=L}
{\displaystyle f(x)\to L\quad {\mbox{as}}\quad x\to \infty }
Similarly, we call
{\displaystyle L}
{\displaystyle f(x)}
{\displaystyle x}
approaches negative infinity if
{\displaystyle f(x)}
{\displaystyle L}
{\displaystyle x}
is sufficiently negative.
{\displaystyle \lim _{x\to -\infty }f(x)=L}
{\displaystyle f(x)\to L\quad {\mbox{as}}\quad x\to -\infty }
So, in this case, we write:
{\displaystyle \quad \lim _{x\to \infty }{\frac {1}{x}}=0}
and say "The limit, as
{\displaystyle x}
approaches infinity, equals
{\displaystyle 0}
," or "as
{\displaystyle x}
approaches infinity, the function approaches 0.
{\displaystyle \lim _{x\to -\infty }{\frac {1}{x}}=0}
because making
{\displaystyle x}
very negative also forces
{\displaystyle {\frac {1}{x}}}
{\displaystyle 0}
Notice, however, that infinity is not a number; it's just shorthand for saying "no matter how big." Thus, this is not the same as the regular limits we learned about in the last two chapters.
Limits at Infinity of Rational FunctionsEdit
One special case that comes up frequently is when we want to find the limit at
{\displaystyle \infty }
{\displaystyle -\infty }
) of a rational function. A rational function is just one made by dividing two polynomials by each other. For example,
{\displaystyle f(x)={\frac {x^{3}+x-6}{x^{2}-4x+3}}}
is a rational function. Also, any polynomial is a rational function, since
{\displaystyle 1}
is just a (very simple) polynomial, so we can write the function
{\displaystyle f(x)=x^{2}-3}
{\displaystyle f(x)={\frac {x^{2}-3}{1}}}
, the quotient of two polynomials.
Consider the numerator of a rational function as we allow the variable to grow very large (in either the positive or negative sense). The term with the highest exponent on the variable will dominate the numerator, and the other terms become more and more insignificant compared to the dominating term. The same applies to the denominator. In the limit, the other terms become negligible, and we only need to examine the dominating term in the numerator and denominator.
There is a simple rule for determining a limit of a rational function as the variable approaches infinity. Look for the term with the highest exponent on the variable in the numerator. Look for the same in the denominator. This rule is based on that information.
If the exponent of the highest term in the numerator matches the exponent of the highest term in the denominator, the limit (at both
{\displaystyle \infty }
{\displaystyle -\infty }
) is the ratio of the coefficients of the highest terms.
If the numerator has the highest term, then the fraction is called "top-heavy". If, when you divide the numerator by the denominator the resulting exponent on the variable is even, then the limit (at both
{\displaystyle \infty }
{\displaystyle -\infty }
{\displaystyle \infty }
. If it is odd, then the limit at
{\displaystyle \infty }
{\displaystyle \infty }
, and the limit at
{\displaystyle -\infty }
{\displaystyle -\infty }
If the denominator has the highest term, then the fraction is called "bottom-heavy" and the limit at both
{\displaystyle \pm \infty }
Note that, if the numerator or denominator is a constant (including 1, as above), then this is the same as
{\displaystyle x^{0}}
. Also, a straight power of
{\displaystyle x}
, like
{\displaystyle x^{3}}
, has coefficient 1, since it is the same as
{\displaystyle 1x^{3}}
{\displaystyle \lim _{x\to \infty }{\frac {x-5}{x-3}}}
{\displaystyle f(x)={\frac {x-5}{x-3}}}
is the quotient of two polynomials,
{\displaystyle x-5}
{\displaystyle x-3}
. By our rule we look for the term with highest exponent in the numerator; it's
{\displaystyle x}
. The term with highest exponent in the denominator is also
{\displaystyle x}
. So, the limit is the ratio of their coefficients. Since
{\displaystyle x=1x}
, both coefficients are 1,
{\displaystyle \lim _{x\to \infty }{\frac {x-5}{x-3}}={\frac {1}{1}}=1}
{\displaystyle \lim _{x\to \infty }{\frac {x^{3}+x-6}{x^{2}-4x+3}}}
Using L'Hôpital's rule
{\displaystyle d/dx(x^{3}+x-6)=3x^{2}+1}
{\displaystyle d/dx(x^{2}-4x+3)=2x-4}
We look at the terms with the highest exponents; for the numerator, it is
{\displaystyle x^{3}}
, while for the denominator it is
{\displaystyle x^{2}}
. Since the exponent on the numerator is higher, we know the limit at
{\displaystyle \infty }
{\displaystyle \infty }
{\displaystyle \lim _{x\to \infty }{\frac {x^{3}+x-6}{x^{2}-4x+3}}=\infty }
Retrieved from "https://en.wikibooks.org/w/index.php?title=Calculus/Infinite_Limits&oldid=3472632" |
Using Symbolic Mathematics with Optimization Toolbox Solvers - MATLAB & Simulink - MathWorks Benelux
First Example: Unconstrained Minimization with Hessian
Second Example: Constrained Minimization Using the fmincon Interior-Point Algorithm
Cleaning Up Symbolic Variables
This example shows how to use the Symbolic Math Toolbox™ functions jacobian and matlabFunction to provide analytical derivatives to optimization solvers. Optimization Toolbox™ solvers are usually more accurate and efficient when you supply gradients and Hessians of the objective and constraint functions.
Problem-based optimization can calculate and use gradients automatically; see Automatic Differentiation in Optimization Toolbox. For a problem-based example using automatic differentiation, see Constrained Electrostatic Nonlinear Optimization, Problem-Based.
There are several considerations in using symbolic calculations with optimization functions:
Optimization objective and constraint functions should be defined in terms of a vector, say x. However, symbolic variables are scalar or complex-valued, not vector-valued. This requires you to translate between vectors and scalars.
Optimization gradients, and sometimes Hessians, are supposed to be calculated within the body of the objective or constraint functions. This means that a symbolic gradient or Hessian has to be placed in the appropriate place in the objective or constraint function file or function handle.
Calculating gradients and Hessians symbolically can be time-consuming. Therefore you should perform this calculation only once, and generate code, via matlabFunction, to call during execution of the solver.
Evaluating symbolic expressions with the subs function is time-consuming. It is much more efficient to use matlabFunction.
matlabFunction generates code that depends on the orientation of input vectors. Since fmincon calls the objective function with column vectors, you must be careful to call matlabFunction with column vectors of symbolic variables.
The objective function to minimize is:
f\left({x}_{1},{x}_{2}\right)=\mathrm{log}\left(1+3{\left({x}_{2}-\left({x}_{1}^{3}-{x}_{1}\right)\right)}^{2}+\left({x}_{1}-4/3{\right)}^{2}\right).
This function is positive, with a unique minimum value of zero attained at x1 = 4/3, x2 =(4/3)^3 - 4/3 = 1.0370...
We write the independent variables as x1 and x2 because in this form they can be used as symbolic variables. As components of a vector x they would be written x(1) and x(2). The function has a twisty valley as depicted in the plot below.
syms x1 x2 real
x = [x1;x2]; % column vector of symbolic variables
f = log(1 + 3*(x2 - (x1^3 - x1))^2 + (x1 - 4/3)^2)
\mathrm{log}\left({\left({x}_{1}-\frac{4}{3}\right)}^{2}+3 {\left(-{{x}_{1}}^{3}+{x}_{1}+{x}_{2}\right)}^{2}+1\right)
Compute the gradient and Hessian of f:
gradf = jacobian(f,x).' % column gradf
gradf =
\begin{array}{l}\left(\begin{array}{c}-\frac{6 \left(3 {{x}_{1}}^{2}-1\right) \left(-{{x}_{1}}^{3}+{x}_{1}+{x}_{2}\right)-2 {x}_{1}+\frac{8}{3}}{{\sigma }_{1}}\\ \frac{-6 {{x}_{1}}^{3}+6 {x}_{1}+6 {x}_{2}}{{\sigma }_{1}}\end{array}\right)\\ \\ \mathrm{where}\\ \\ \mathrm{ }{\sigma }_{1}={\left({x}_{1}-\frac{4}{3}\right)}^{2}+3 {\left(-{{x}_{1}}^{3}+{x}_{1}+{x}_{2}\right)}^{2}+1\end{array}
hessf = jacobian(gradf,x)
hessf =
\begin{array}{l}\left(\begin{array}{cc}\frac{6 {\left(3 {{x}_{1}}^{2}-1\right)}^{2}-36 {x}_{1} \left(-{{x}_{1}}^{3}+{x}_{1}+{x}_{2}\right)+2}{{\sigma }_{2}}-\frac{{{\sigma }_{3}}^{2}}{{{\sigma }_{2}}^{2}}& {\sigma }_{1}\\ {\sigma }_{1}& \frac{6}{{\sigma }_{2}}-\frac{{\left(-6 {{x}_{1}}^{3}+6 {x}_{1}+6 {x}_{2}\right)}^{2}}{{{\sigma }_{2}}^{2}}\end{array}\right)\\ \\ \mathrm{where}\\ \\ \mathrm{ }{\sigma }_{1}=\frac{\left(-6 {{x}_{1}}^{3}+6 {x}_{1}+6 {x}_{2}\right) {\sigma }_{3}}{{{\sigma }_{2}}^{2}}-\frac{18 {{x}_{1}}^{2}-6}{{\sigma }_{2}}\\ \\ \mathrm{ }{\sigma }_{2}={\left({x}_{1}-\frac{4}{3}\right)}^{2}+3 {\left(-{{x}_{1}}^{3}+{x}_{1}+{x}_{2}\right)}^{2}+1\\ \\ \mathrm{ }{\sigma }_{3}=6 \left(3 {{x}_{1}}^{2}-1\right) \left(-{{x}_{1}}^{3}+{x}_{1}+{x}_{2}\right)-2 {x}_{1}+\frac{8}{3}\end{array}
The fminunc solver expects to pass in a vector x, and, with the SpecifyObjectiveGradient option set to true and HessianFcn option set to 'objective', expects a list of three outputs: [f(x),gradf(x),hessf(x)].
matlabFunction generates exactly this list of three outputs from a list of three inputs. Furthermore, using the vars option, matlabFunction accepts vector inputs.
fh = matlabFunction(f,gradf,hessf,'vars',{x});
Now solve the minimization problem starting at the point [-1,2]:
options = optimoptions('fminunc', ...
'SpecifyObjectiveGradient', true, ...
'HessianFcn', 'objective', ...
'Algorithm','trust-region', ...
[xfinal,fval,exitflag,output] = fminunc(fh,[-1;2],options)
fminunc stopped because the final change in function value relative to
Compare this with the number of iterations using no gradient or Hessian information. This requires the 'quasi-newton' algorithm.
options = optimoptions('fminunc','Display','final','Algorithm','quasi-newton');
fh2 = matlabFunction(f,'vars',{x});
% fh2 = objective with no gradient or Hessian
[xfinal,fval,exitflag,output2] = fminunc(fh2,[-1;2],options)
The number of iterations is lower when using gradients and Hessians, and there are dramatically fewer function evaluations:
sprintf(['There were %d iterations using gradient' ...
' and Hessian, but %d without them.'], ...
output.iterations,output2.iterations)
'There were 14 iterations using gradient and Hessian, but 18 without them.'
sprintf(['There were %d function evaluations using gradient' ...
'There were 15 function evaluations using gradient and Hessian, but 81 without them.'
We consider the same objective function and starting point, but now have two nonlinear constraints:
5\mathrm{sinh}\left({x}_{2}/5\right)\ge {x}_{1}^{4}
5\mathrm{tanh}\left({x}_{1}/5\right)\ge {x}_{2}^{2}-1.
The constraints keep the optimization away from the global minimum point [1.333,1.037]. Visualize the two constraints:
% Z=2 where the second is satisfied, Z=3 where both are
plot3(.4396, .0373, 4,'o','MarkerEdgeColor','r','MarkerSize',8);
% best point
We plotted a small red circle around the optimal point.
Here is a plot of the objective function over the feasible region, the region that satisfies both constraints, pictured above in dark red, along with a small red circle around the optimal point:
W = log(1 + 3*(Y - (X.^3 - X)).^2 + (X - 4/3).^2);
% W = the objective function
W(Z < 3) = nan; % plot only where the constraints are satisfied
surf(X,Y,W,'LineStyle','none');
plot3(.4396, .0373, .8152,'o','MarkerEdgeColor','r', ...
'MarkerSize',8); % best point
The nonlinear constraints must be written in the form c(x) <= 0. We compute all the symbolic constraints and their derivatives, and place them in a function handle using matlabFunction.
The gradients of the constraints should be column vectors; they must be placed in the objective function as a matrix, with each column of the matrix representing the gradient of one constraint function. This is the transpose of the form generated by jacobian, so we take the transpose below.
We place the nonlinear constraints into a function handle. fmincon expects the nonlinear constraints and gradients to be output in the order [c ceq gradc gradceq]. Since there are no nonlinear equality constraints, we output [] for ceq and gradceq.
c1 = x1^4 - 5*sinh(x2/5);
c2 = x2^2 - 5*tanh(x1/5) - 1;
gradc = jacobian(c,x).'; % transpose to put in correct form
constraint = matlabFunction(c,[],gradc,[],'vars',{x});
The interior-point algorithm requires its Hessian function to be written as a separate function, instead of being part of the objective function. This is because a nonlinearly constrained function needs to include those constraints in its Hessian. Its Hessian is the Hessian of the Lagrangian; see the User's Guide for more information.
The Hessian function takes two input arguments: the position vector x, and the Lagrange multiplier structure lambda. The parts of the lambda structure that you use for nonlinear constraints are lambda.ineqnonlin and lambda.eqnonlin. For the current constraint, there are no linear equalities, so we use the two multipliers lambda.ineqnonlin(1) and lambda.ineqnonlin(2).
We calculated the Hessian of the objective function in the first example. Now we calculate the Hessians of the two constraint functions, and make function handle versions with matlabFunction.
hessc1 = jacobian(gradc(:,1),x); % constraint = first c column
hessc2 = jacobian(gradc(:,2),x);
hessfh = matlabFunction(hessf,'vars',{x});
hessc1h = matlabFunction(hessc1,'vars',{x});
To make the final Hessian, we put the three Hessians together, adding the appropriate Lagrange multipliers to the constraint functions.
myhess = @(x,lambda)(hessfh(x) + ...
lambda.ineqnonlin(1)*hessc1h(x) + ...
lambda.ineqnonlin(2)*hessc2h(x));
Set the options to use the interior-point algorithm, the gradient, and the Hessian, have the objective function return both the objective and the gradient, and run the solver:
options = optimoptions('fmincon', ...
'Algorithm','interior-point', ...
'SpecifyConstraintGradient',true, ...
'HessianFcn',myhess, ...
% fh2 = objective without Hessian
fh2 = matlabFunction(f,gradf,'vars',{x});
[xfinal,fval,exitflag,output] = fmincon(fh2,[-1;2],...
[],[],[],[],[],[],constraint,options)
Again, the solver makes many fewer iterations and function evaluations with gradient and Hessian supplied than when they are not:
% fh3 = objective without gradient or Hessian
% constraint without gradient:
constraint = matlabFunction(c,[],'vars',{x});
[xfinal,fval,exitflag,output2] = fmincon(fh3,[-1;2],...
' and Hessian, but %d without them.'],...
The symbolic variables used in this example were assumed to be real. To clear this assumption from the symbolic engine workspace, it is not sufficient to delete the variables. You must clear the assumptions of variables using the syntax
assume([x1,x2],'clear')
All assumptions are cleared when the output of the following command is empty
assumptions([x1,x2]) |
Create an example of sequence of numbers with an exponential growth pattern, and
Create an example of sequence of numbers with an exponential growth pattern, and explain how you know that the growth is exponential.
In the exponential patterns, the successive numbers increase or decreases by the same percent.
Exponential growth patterns: A sequence of numbers has an exponential pattern when each successive number increase by the same percentage.
There is one simple example of a sequence of an exponential growth:
The population of a dogs doubled every year and the sequence is given as:
2, 4, 8, 16, 32, 64, 128, 256 etc.
In the exponential function, over the time t values of function increase very fast.
Now, the formula for exponential growth function is:
y\left(t\right)=a{e}^{kt}
a=
t=
k=
rate of growth
y\left(t\right)=
value at time t
At the beginning of an environmental study, ad forest covered an area of
1500k{m}^{2}
.Since then, this area has decreased by 3.75% each year.Let t be the number of years since the start of the study.Let y be the area that the forest covers in
k{m}^{2}
Write an exponential function showing relationship between y and t.
A fruit fly population of 24 flies is in a closed container. The number of flies grows exponentially, reaching 384 in 18 days. Find the doubling time (time for the population to double) and write an equation that models this scenario.
The population of a town increased by 2.54% per year from the beginning of 2000 to the beginning of 2010. The town's population at the beginning of 2000 was 74,860.
Exponential growth and decay problems follow the model given by the equation
A\left(t\right)=P{e}^{rt}
-The model is a function of time t
-A(t) is the amount we have ater time t
-PIs the initial amount, because for t=0, notice how
A\left(0\right)=P{e}^{0×t}=P{e}^{0}=P
-Tis the growth or decay rate. It is positive for growth and negative for decay
Growth and decay problems can deal with money (interest compounded continuously), bacteria growth, radioactive decay. population growth etc.
So A(t) can represent any of these depending on the problem.
The growth of a certain bactenia population can be modeled by the function
A\left(t\right)=900{e}^{0.0534}
where A(t) is the number of bacteria and t represents the time in minutes.
What is the number of bactenia ater 15 minutes? (round to the nearest whole number of bacteria.)
If f(x) is an exponential function where f(−1)=8 and f(8.5)=87, then find the value of f(3), to the nearest hundredth.
A=7.7{\left(0.92\right)}^{t}
represents an exponential growth or decay function.
Does the function P represent exponential growth or decay? B. What is the initial quantity? C. What is the growth or decay factor?
Exponential Growth. B. The initial quantity is 7.7. C. The growth or decay factor is 0.92
Exponential Decay. B. The initial quantity is 7.7. C. The growth or decay facotor is 1.08
Exponential Growth. B. The initial quantity is 0.92. C. The growth or decay factor is 1.08
. Exponential Decay. B. The initial quantity is 7.7. C. The growth or decay factor is 0.92
Define thew term Exponential Growth and Decay? |
ON Modified -Contractive Mappings
Marwan Amin Kutbi, Muhammad Arshad, Aftab Hussain, "ON Modified -Contractive Mappings", Abstract and Applied Analysis, vol. 2014, Article ID 657858, 7 pages, 2014. https://doi.org/10.1155/2014/657858
Marwan Amin Kutbi,1 Muhammad Arshad,2 and Aftab Hussain2
Hussain et al. (2013) established new fixed point results in complete metric space. In this paper, we prove fixed point results of α-admissible mappings with respect to η, for modified contractive condition in complete metric space. An example is given to show the validity of our work. Our results generalize/improve several recent and classical results existing in the literature.
1. Preliminaries and Scope
The study of fixed point problems in nonlinear analysis has emerged as a powerful and very important tool in the last 60 years. Particularly, the technique of fixed point theory has been applicable to many diverse fields of sciences such as engineering, chemistry, biology, physics, and game theory. Over the years, fixed point theory has been generlized in many directions by several mathematicians (see [1–36]).
In 1973, Geraghty [12] studied different contractive conditions and established some useful fixed point theorems.
In 2012, Samet et al. [33] introduced a concept of -contractive type mappings and established various fixed point theorems for mappings in complete metric spaces. Afterwards Karapinar and Samet [10] refined the notions and obtained various fixed point results. Hussain et al. [17] extended the concept of -admissible mappings and obtained useful fixed point theorems. Subsequently, Abdeljawad [4] introduced pairs of -admissible mappings satisfying new sufficient contractive conditions different from those in [17, 33] and proved fixed point and common fixed point theorems. Lately, Salimi et al. [32] modified the concept of -contractive mappings and established fixed point results.
We define the family of nondecreasing functions such that , and for each where is the th term of .
Lemma 1 (see [32]). If , then for all .
Definition 2 (see [33]). Let be a metric space and let be a given mapping. We say that is an -contractive mapping if there exist two functions and such that for all .
Definition 3 (see [33]). Let and . One says that is -admissible if , .
Example 4. Consider . Define and by , for all and Then is -admissible.
Definition 5 (see [32]). Let and let be two functions. One says that is -admissible mapping with respect to if , . Note that if one takes , then this definition reduces to definition [33]. Also if we take , then one says that is an -subadmissible mapping.
In this section, we prove fixed point theorems for -admissible mappings with respect to , satisfying modified ()-contractive condition in complete metric space.
Theorem 6. Let be a complete metric space and let is -admissible mappings with respect to . Assume that there exists a function such that, for any bounded sequence of positive reals, implies such that for all where ; then suppose that one of the following holds: (i)is continuous;(ii)if is a sequence in such that for all and as , then If there exists such that , then has a unique fixed point.
Proof. Let and define We will assume that for each . Otherwise, there exists an such that . Then and is a fixed point of . Since and is -admissible mapping with respect to , we have By continuing in this way, we have for all . From (7), we have Thus applying the inequality (3), with and , we obtain which implies that We suppose that Then we prove that . It is clear that is a decreasing sequence. Therefore, there exists some positive number such that . Now we will prove that . From (10), we have Now by taking limit , we have By using property of function, we have . Thus Now we prove that sequence is Cauchy sequence. Suppose on contrary that is not a Cauchy sequence. Then there exists and sequences and such that, for all positive integers , we have , By the triangle inequality, we have for all . Now taking limit as in (16) and using (14), we have Again using triangle inequality, we have Taking limit as and using (14) and (17), we obtain By using (3), (17), and (19), we have which implies that Therefore, we have Now taking limit as in (22), we get Hence , which is a contradiction. Hence is a Cauchy sequence. Since is complete so there exists such that . Now we prove that . Suppose (i) holds; that is, is continuous, so we get Thus . Now we suppose that (ii) holds. Since for all . By the hypotheses of (ii), we have Using the triangle inequality and (3), we have which implies that Letting then we have . Thus . Let there exists to be another fixed point of , s.t ; which implies that By the property of function, , implies ; then we have . Hence has a unique fixed point.
If in Theorem 6, we get the following corollary.
Corollary 7 (see [17]). Let be a complete metric space and let be -admissible mapping. Assume that there exists a function such that, for any bounded sequence of positive reals, implies such that for all , where . Suppose that either(i) is continuous, or(ii)if is a sequence in such that for all and as , then If there exists such that ; then has a fixed point.
If in Theorem 6, we get the following corollary.
Corollary 8. Let be a complete metric space and let be -subadmissible mapping. Assume that there exists a function such that, for any bounded sequence of positive reals, implies such that for all where ; then suppose that one of the following holds:(i) is continuous;(ii)if is a sequence in such that for all and as , then If there exists such that , then has a fixed point.
Example 9. Let with usual metric for all and , and for all be defined by We prove that Corollary 7 can be applied to . Let ; clearly and , then of -admissible mapping , and , , and imply that If , then we have Let and ; then
Theorem 10. Let be a complete metric space and let be -admissible mappings with respect to . Assume that there exists a function such that, for any bounded sequence of positive reals, implies such that for all ; then suppose that one of the following holds:(i) is continuous;(ii)if is a sequence in such that for all and as , then If there exists such that , then has a fixed point.
Proof. Let and define We will assume that for each . Otherwise, there exists an such that . Then and is a fixed point of . Since and is -admissible mapping with respect to , we have By continuing in this way, we have for all . From (43), we have Thus applying the inequality (39), with and , we obtain which implies that We suppose that Then we prove that . It is clear that is a decreasing sequence. Therefore, there exists some positive number such that . Now we will prove that . From (47), we have Now by taking limit , we have By using property of function, we have . Thus Now we prove that sequence is Cauchy sequence. Suppose on contrary that is not a Cauchy sequence. Then there exists and sequences and such that, for all positive integers , we have , By the triangle inequality, we have for all . Now taking limit as in (52) and using (50), we have Again using triangle inequality, we have Taking limit as and using (50) and (53), we obtain By using (39), (53), and (55), we have which implies that Therfore, we have
Now taking limit as in (58), we get Hence , which is a contradiction. Hence is a Cauchy sequence. Since is complete so there exists such that . Now we prove that . Suppose (i) holds; that is, is continuous, so we get Thus . Now we suppose that (ii) holds. Since for all . By the hypotheses of (ii), we have Using the triangle inequality and (39), we have which implies that Letting , we have . Thus . Let there exists to be another fixed point of , s.t ; implies By the property of function, implies ; then we have . Hence has a unique fixed point.
If in Theorem 10, we get the following corollary.
Corollary 11 (see [17]). Let be a complete metric space and let be -admissible mapping. Assume that there exists a function such that, for any bounded sequence of positive reals, implies such that for all . Suppose that either(i) is continuous, or(ii)if is a sequence in such that for all and as , then If there exists such that , then has a fixed point. Our results are more general than those in [17, 32, 33] and improve several results existing in the literature.
Marwan Amin Kutbi gratefully acknowledges the support from the Deanship of Scientific Research (DSR) at King Abdulaziz University (KAU) during this research. The authors thank the editor and the referees for their valuable comments and suggestions which improved greatly the quality of this paper.
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We need find: The real zeros of polynomial function form an arithmetic sequence
We need find: The real zeros of polynomial function form an arithmetic sequence f(x) = x^{4} - 4x^{3} - 4x^{2} + 16x.
f\left(x\right)={x}^{4}-4{x}^{3}-4{x}^{2}+16x.
A rational expression is a fraction that is quotient of two polynomials. A rational function is defined by the two polynomial functions.
A function f of the form,
f\left(x\right)=p\frac{x}{q}\left(x\right)
Where, p(x) and g(x) are polynomial functions, with
g\left(x\right)\ne 0.
The given polynomial unction form an arithmetic sequence is
f\left(x\right)={x}^{4}-4{x}^{3}-4{x}^{2}+16x.
Here, the constant is 0.
f\left(x\right)=x\left({x}^{3}-4{x}^{2}-4x+16\right)
The possibilities for
\frac{p}{q}are±1,±2,±4,and±8.
Factoring the term
\left({x}^{3}-4{x}^{2}-4x+16\right)
\left({x}^{3}-4{x}^{2}-4x+16\right)=\left(x+2\right)\left({x}^{2}-6x+8\right)
\left({x}^{2}-6x+8\right)
\left({x}^{2}-6x+8\right)=\left(x-2\right)\left(x-4\right)
Combining all the terms, we get
\left({x}^{3}-4{x}^{2}-4x+16\right)=\left(x+2\right)\left(x-2\right)\left(x-4\right)
f\left(x\right)=x\left({x}^{3}-4{x}^{2}-4x+16\right)=x\left(x+2\right)\left(x-2\right)\left(x-4\right)
Thus, the real zeros are -2, 0, 2, and 4
\frac{1}{3},\frac{2}{9},\frac{3}{27},\frac{4}{81},...
3,8,13,18,...,48
Polynomial equation with real coefficients that has roots
i,\text{ }1+ii,\text{ }1+i
Solve the Numerical Analysis Explain how Newton`s interpolation formula better than the Lagrange formula.
\left(14-10\right)+8-5×10÷15
8x^3−48x^2+96x−64
0.3,1.2,2.1,3,...
{a}_{1}=-4
d=-\frac{4}{3}
{6}^{th}
-2,-\frac{7}{2},-5,-\frac{13}{2}...
{22}^{nd}
15.6,15,14.4,13.8,...
{32}^{nd}
-2,-1,-\frac{1}{2},-\frac{1}{4},... |
Ratio and Proportion Word Problems Practice Problems Online | Brilliant
Calvin went cycling in San Francisco, which is extremely hilly. He can pedal up a hill at a speed of 12 mph, and down a hill at a speed of 36 mph. If he went up and down a hill (back to the same point), what is his average speed (in mph) for the entire journey?
In a park, the ratio of red roses to white roses is
3 : 6
. If there are 30 red roses, how many white roses are there?
If it takes 33 minutes for a water faucet to fill up a tub, how many minutes will it take the water faucet to fill up
\frac {1}{3}
of the tub?
A grocer has a stack of apples and a stack of oranges. After arranging them in neat rows, he realizes that he has 4 apples for every 3 oranges. Given that he has 168 apples, how many oranges does he have?
If a vehicle can travel at 68 kilometers per hour, how far (in kilometers) can it go in 225 minutes? |
Convert VAR model to VEC model - MATLAB var2vec - MathWorks 日本
{y}_{t}=\left[\begin{array}{c}0.5\\ 1\\ -2\end{array}\right]+\left[\begin{array}{ccc}0.54& 0.86& -0.43\\ 1.83& 0.32& 0.34\\ -2.26& -1.31& 3.58\end{array}\right]{y}_{t-1}+\left[\begin{array}{ccc}0.14& -0.12& 0.05\\ 0.14& 0.07& 0.10\\ 0.07& 0.16& 0.07\end{array}\right]{y}_{t-3}+{\mathrm{ε}}_{t}.
{A}_{1}
{A}_{2}
{A}_{3}
{y}_{t-1}
{y}_{t-2}
{y}_{t-3}
\mathrm{Î}{y}_{t-1}
\mathrm{Î}{y}_{t-2}
{y}_{t-1}
\mathrm{Î}{y}_{t-1}
B1 = 3×3
\begin{array}{rcl}\mathrm{Î}{y}_{t}& =& \left[\begin{array}{c}0.5\\ 1\\ -2\end{array}\right]+\left[\begin{array}{ccc}-0.14& 0.12& -0.05\\ -0.14& -0.07& -0.10\\ -0.07& -0.16& -0.07\end{array}\right]\mathrm{Î}{y}_{t-1}+\left[\begin{array}{ccc}-0.14& 0.12& -0.05\\ -0.14& -0.07& -0.10\\ -0.07& -0.16& -0.07\end{array}\right]\mathrm{Î}{y}_{t-2}\\ & +& \left[\begin{array}{ccc}-0.32& 0.74& -0.38\\ 1.97& -0.61& 0.44\\ -2.19& -1.15& 2.65\end{array}\right]{y}_{t-1}+{\mathrm{ε}}_{t}\end{array}.
\left[\begin{array}{cc}0.54& -2.26\\ 1.83& 0.86\end{array}\right]{y}_{t}=\left[\begin{array}{cc}0.32& -0.43\\ -1.31& 0.34\end{array}\right]{y}_{t-1}+\left[\begin{array}{cc}0.07& 0.07\\ -0.01& -0.02\end{array}\right]{y}_{t-2}+{\mathrm{ε}}_{t}.
{A}_{0}
{A}_{1}
{A}_{2}
\left({A}_{0}-{A}_{1}L-{A}_{2}{L}^{2}\right){y}_{t}={\mathrm{ε}}_{t}.
L
{y}_{t}
\mathrm{Î}{y}_{t}
\mathrm{Î}{y}_{t}
{A}_{0}
{y}_{t}
\mathrm{Î}{y}_{t}
\left[\begin{array}{cc}0.54& -2.26\\ 1.83& 0.86\end{array}\right]\mathrm{Î}{y}_{t}=\left[\begin{array}{cc}-0.07& -0.07\\ 0.01& 0.02\end{array}\right]\mathrm{Î}{y}_{t-1}+\left[\begin{array}{cc}-0.15& 1.9\\ -3.15& -0.54\end{array}\right]{y}_{t-1}+{\mathrm{ε}}_{t}.
\begin{array}{l}\left\{\left[\begin{array}{ccc}1& 0.2& -0.1\\ 0.03& 1& -0.15\\ 0.9& -0.25& 1\end{array}\right]+\left[\begin{array}{ccc}0.5& -0.2& -0.1\\ -0.3& -0.1& 0.1\\ 0.4& -0.2& -0.05\end{array}\right]{L}^{4}+\left[\begin{array}{ccc}0.05& -0.02& -0.01\\ -0.1& -0.01& -0.001\\ 0.04& -0.02& -0.005\end{array}\right]{L}^{8}\right\}{y}_{t}=\\ \left\{\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]+\left[\begin{array}{ccc}-0.02& 0.03& 0.3\\ 0.003& 0.001& 0.01\\ 0.3& 0.01& 0.01\end{array}\right]{L}^{4}\right\}{\mathrm{ε}}_{t}\end{array}
{y}_{t}={\left[{y}_{1t}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}{y}_{2t}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}{y}_{3t}\right]}^{â²}
{\mathrm{ε}}_{t}={\left[{\mathrm{ε}}_{1t}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}{\mathrm{ε}}_{2t}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}{\mathrm{ε}}_{3t}\right]}^{â²}
{y}_{t}
{\mathrm{ε}}_{t}
\mathrm{Î}{y}_{t}
\mathrm{Î}{y}_{t-1}
\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]{y}_{t}=\left[\begin{array}{cc}0.1& 0.2\\ 1& 0.1\end{array}\right]{y}_{tâ1}+\left[\begin{array}{cc}â0.1& 0.01\\ 0.2& â0.3\end{array}\right]{y}_{tâ2}+{\mathrm{ε}}_{t}
\left(\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]â\left[\begin{array}{cc}0.1& 0.2\\ 1& 0.1\end{array}\right]â\left[\begin{array}{cc}â0.1& 0.01\\ 0.2& â0.3\end{array}\right]L\right){y}_{t}={\mathrm{ε}}_{t}.
{A}_{0}{y}_{t}=a+{A}_{1}{y}_{tâ1}+{A}_{2}{y}_{tâ2}+...+{A}_{p}{y}_{tâp}+{\mathrm{ε}}_{t}.
{B}_{0}\mathrm{Î}{y}_{t}=b+{B}_{1}\mathrm{Î}{y}_{tâ1}+{B}_{2}\mathrm{Î}{y}_{tâ2}+...+{B}_{q}\mathrm{Î}{y}_{tâq}+C{y}_{tâ1}+{\mathrm{ε}}_{t}.
A\left(L\right){y}_{t}=a+{\mathrm{ε}}_{t}
A\left(L\right)={A}_{0}â{A}_{1}Lâ{A}_{2}{L}^{2}â...â{A}_{p}{L}^{p}
{L}^{j}{y}_{t}={y}_{tâj}
B\left(L\right)\mathrm{Î}{y}_{t}=b+C{y}_{tâ1}+{\mathrm{ε}}_{t}
B\left(L\right)={B}_{0}â{B}_{1}Lâ{B}_{2}{L}^{2}â...â{B}_{q}{L}^{q}
{A}_{0}{y}_{t}=a+{A}_{1}{y}_{tâ1}+{A}_{2}{y}_{tâ2}+...+{A}_{p}{y}_{tâp}+{\mathrm{ε}}_{t}.
εt is an n-dimensional innovations series. The innovations are serially uncorrelated, and have a multivariate normal distribution with mean 0 and n-by-n covariance matrix Σ.
{B}_{0}\mathrm{Î}{y}_{t}=b+{B}_{1}\mathrm{Î}{y}_{tâ1}+{B}_{2}\mathrm{Î}{y}_{tâ2}+...+{B}_{q}\mathrm{Î}{y}_{tâq}+C{y}_{tâ1}+{\mathrm{ε}}_{t}.
Δ is the first difference operator, that is, Δyt = yt – yt–1.
Bj is the n-by-n coefficient matrix of Δyt–j, j = 1,...,q.
VECDEN is a cell vector containing p coefficients corresponding to the differenced response terms in VEC.Lags in difference-equation notation. The first element is the coefficient of Δyt, the second element is the coefficient of Δyt–1, and so on.
Rank zero, then the converted VEC model is a stable VAR(p – 1) model in terms of Δyt. |
Solve for y: (7y+3)/2=((8y+1)/6)+23 Simplify your answer as much as possible.
\frac{7y+3}{2}=\left(\frac{8y+1}{6}\right)+23
Liyana Mansell
\frac{7y+3}{2}=\left(\frac{8y+1}{6}\right)+23
Multiply both sides by the LCD which is 6:
\frac{7y+3}{2}\left(6\right)=\left(\left(\frac{8y+1}{6}\right)+23\right)\left(6\right)
21y+9=(8y+1)+138
21y+9=8y+139
Subtract 9 from both sides: 21y=8y+130
Subtract 8y from both sides: 13y=130
Divie both sides by 13: y=10
Ten red cards and ten black cards are placed in a bag. You choose one card and then another without replacing the first card. What is the probability that the first card will be red and the second card will be black?
In one study, the correlation between the educational level of husbands and wives in a certain town was about 0.50, both averaged 12 years of schooling completed, with an SD of 3 years.
a) Predict the educational level of a woman whose husband has completed 18 years of schooling b) Predict the educational level of a man whose wife has completed 15 years of schooling. c) Apparently, well-educated men marry women who are less well educated than themselves. But the women marry men with even less education. How is this possible?
A professor writes 40 discrete mathematics true/false questions. Of the statements in these questions, 17 are true. If the questions can be positioned in any order, how many different answer keys are possible?
Police response time to an emergency call is the difference between the times the call is first received by the dispatcher and the time a patrol car radios that it has arrived at the scene. over a long period of time it has been determined that the police responce time has a normal distribution with a mean of 8.4 minutes and a standard deviation of 1.7 minutes. for a randomly recieved emegency call, what is the probability that the response time will be:
a) between 5 and 10 min
b) less than 5 min
c) more than 10 min
The probability that an event will happen is
P\left(E\right)=\frac{24}{29}
. Find the probability that the event will not happen |
Vehicle and tire distances to objects - Simulink - MathWorks 日本
Vehicle Terrain Sensor
Hit Event
VehCntr
TireRadii
VehHitDist
TireHitDist
Distance to vehicle center
Distance to tire center
Distance from vehicle center to front, VehCntrLngthVal
Distance from tire center to ground, TireRadiiVal
Trace Lengths
Vehicle body x-axis trace length, VehRayLngth
Left front wheel z-axis trace length, LfRayLngth
Right front wheel z-axis trace length, RfRayLngth
Left rear wheel z-axis trace length, LrRayLngth
Right rear wheel z-axis trace length, RrRayLngth
Starting Point Offsets
Vehicle body x-axis trace offset, VehRayOffset
Left front wheel z-axis trace offset, LfRayOffset
Right front wheel z-axis trace offset, RfRayOffset
Left rear wheel z-axis trace offset, LrRayOffset
Right rear wheel z-axis trace offset, RrRayOffset
Trace line visualization
Vehicle and tire distances to objects
The Vehicle Terrain Sensor block implements ray tracing to detect the terrain below the tires and objects in front of the vehicle. Specifically, for these actor components, the block returns the hit location (in the world coordinate system) and the distance to an object.
Verify that the Vehicle Terrain Sensor block executes before the Simulation 3D Fisheye Camera block. That way, the Unreal Engine® 3D visualization environment prepares the data before the Vehicle Terrain Sensor block receives it. To check the block execution order, right-click the blocks and select Properties. On the General tab, confirm these Priority settings:
Vehicle Terrain Sensor — 1
To calculate the hit distances shown in the illustration, the block implements these equations.
Front of vehicle to object, DistToHitVhAdjust
DistToHitVh = GetLength(CntrLocVh,HitLocVh)
DistToHitVhAdjust = DistToHitVh - VehCntrLngthVal
EndLocVh = CntrLocVh + VehRayLngth - VehRayOffset
VehRayOffset = CntrLocVh - StartLocVh
VehRayLngth = StartLocVh - EndLocVh
Tires to terrain, DistToHitTrAdjust
DistToHitTr = GetLength(CntrLocTr, HitLocTr)
DistToHitTrAdjust = DistToHitTr - TireRadiiVal
EndLocTr = CntrLocTr + LengthTr - OffsetTr
OffsetTr = CntrLocTr - StartLocTr
LengthTr = StartLocTr - EndLocTr
This illustration and equations use these variables.
CntrLocVh
Vehicle center location
DistToHitVh
Distance from vehicle center location to object
DistToHitVhAdjust
Distance from the front of the vehicle to object
EndLocVh
Vehicle ray trace end
HitLocVh
Vehicle hit location
OffsetVh
Vehicle trace offset
StartLocVh
Vehicle ray trace start
VehRayLngth
Vehicle trace length
VehCntrLngthVal
Distance from vehicle center to front
CntrLocTr
Tire center location
DistToHitTr
Distance from tire center location to terrain
DistToHitTrAdjust
Distance from tire to terrain
HitLocTr
Tire hit location
EndLocTr
Tire ray trace end
Tire trace offset
StartLocTr
Tire ray trace start
LengthTr
Tire trace length
TireRadiiVal
To determine a hit event, the block uses the ray trace. The block provides the hit location in the world coordinate system.
To determine a miss event, the block uses the ray trace.
VehCntr — Vehicle distance from center to front
Distance from the vehicle center to front, VehCntrLngthVal, in m.
Creates Port
Creates Parameter
External input VehCntr None
TireRadii — Tire radii
Tire radii, TireRadiiVal, in m.
Distance to tire center Setting
External input TireRadii
Bus signal containing block values. The signals are arrays that depend on the wheel location.
HitFlg
Vehicle and wheel hit flag:
Hit an object – 1
Miss an object – 0
\left[\begin{array}{c}Vehicle\\ FrontLeft\\ FrontRight\\ RearLeft\\ RearRight\end{array}\right]
Vehicle, HitLocVh, and tire, HitLocTr, hit locations, in world coordinate system X-, Y, and Z- axes, respectively
\left[\begin{array}{ccc}Vehicl{e}_{X}& Vehicl{e}_{Y}& Vehicl{e}_{Z}\\ FrontLef{t}_{X}& FrontLef{t}_{Y}& FrontLef{t}_{Z}\\ FrontRigh{t}_{X}& FrontRigh{t}_{Y}& FrontRigh{t}_{Z}\\ RearLef{t}_{X}& RearLef{t}_{Y}& RearLef{t}_{Z}\\ RearRea{r}_{X}& RearRea{r}_{Y}& RearRea{r}_{Z}\end{array}\right]
Vehicle, StartLocVh, and tire, StartLocTr, ray trace start locations, in world coordinate system X-, Y, and Z- axes, respectively
VehHitDist — Front of vehicle distance to object
Distance from the front of the vehicle to object, DistToHitVhAdjust, in m.
TireHitDist — Tire distance to terrain
Distance from tire to terrain, DistToHitTrAdjust, in m.
DistToHitTrAdjust =
\left[\begin{array}{cccc}FrontLeft& FrontRight& RearLef& RearRight\right]\end{array}
SimulinVehicle1 (default) | character vector
Distance to vehicle center — Selection
Constant (default) | External input
Configure how to provide the distance to the vehicle center.
Distance to tire center — Selection
Configure how to provide the distance to the tire center.
Distance from vehicle center to front, VehCntrLngthVal — Vehicle center
Distance from tire center to ground, TireRadiiVal — Tire radii
Tire radius, TireRadiiVal, in m.
Vehicle body x-axis trace length, VehRayLngth — Trace length
Vehicle body trace length, VehRayLngth, in m.
Left front wheel z-axis trace length, LfRayLngth — Trace length
Left front wheel trace length, LfRayLngth and LengthTr, in m.
Right front wheel z-axis trace length, RfRayLngth — Trace length
Right front wheel trace length, RfRayLngth and LengthTr, in m.
Left rear wheel z-axis trace length, LrRayLngth — Trace length
Left rear wheel trace length, LrRayLngth and LengthTr, in m.
Right rear wheel z-axis trace length, RrRayLngth — Trace length
Right rear wheel trace length, RrRayLngth and LengthTr, in m.
Vehicle body x-axis trace offset, VehRayOffset — Offset the vehicle ray trace
Vehicle body trace offset, OffsetVh, in m.
Left front wheel z-axis trace offset, LfRayOffset — Offset the left front wheel ray trace
Left front wheel trace offset, LfRayOffset and OffsetTr, in m.
Right front wheel z-axis trace offset, RfRayOffset — Offset the right front wheel ray trace
Right front wheel trace offset, RfRayOffset and OffsetTr, in m.
Left rear wheel z-axis trace offset, LrRayOffset — Offset the left rear wheel ray trace
Left rear wheel trace offset, LrRayOffset and OffsetTr, in m.
Right rear wheel z-axis trace offset, RrRayOffset — Offset the right rear wheel ray trace
Right rear wheel trace offset, RrRayOffset and OffsetTr, in m.
Vehicle body — Enable vehicle body ray tracing
Enable vehicle body ray tracing.
Left front tire — Enable left front tire ray tracing
Enable left front tire ray tracing.
Right front tire — Enable right front tire ray tracing
Enable right front tire ray tracing.
Left rear tire — Enable left rear tire ray tracing
Enable left rear tire ray tracing.
Right rear tire — Enable right rear tire ray tracing
Enable right rear tire ray tracing.
Trace line visualization — Visualize ray traces
Enable trace line visualization.
Simulation 3D Camera Get | Simulation 3D Scene Configuration | Simulation 3D Vehicle | Simulation 3D Vehicle with Ground Following |
In perfect competition, TR increases at a constant rate as MR=AR=constant
But can we say that slope of TR = 1 ?????
Suppose MR=AR=10
Then TR at various levels of output (1 unit, 2 units, 3 units, 4 units....) is 10,20,30,40....
In this case slope of TR = 10 then how can the angle it makes with the X axis = 45 degrees????
How does the SMC curve above the SAVC curve represent the supply curve of a firm???
If AR curve indicates the demand of a commodity (it is the demand curve) then which cost curve indicates the supply of a commodity???
Explain the terms Normal Profits, Super Normal Profits , accounting costs , accounting profits, economic profits and economic costs explaining the differences between them
TVC is the summation of MC
'The area under MC curve is equal to TVC'
Kindly give me a mathematical derivation of this formula
Is Depreciation an Implicit Cost or an Implicit Cost???
Please explain the following in layman terms.
Fixed factors initally utilised with ↓ efficiency ⇒↓ in MP postponed by ↑efficiency ⇒Law fails
Q How after 33 (in tp) it is increasing at decreasing rate .....mean how we know decreasing rate?
Why AR is more elastic in monopolistic competition than monopoly...???
q.How does the nature of a commodity influence the price elasticity of demand?
Ans Elasticity of demand of a commodity is influenced by its nature.A commodity for a person may be necessities,a comfort or a luxuries.
.when a commodity is a necessity like food grains,vegetables,medicines,etc,its demand is generally inelastic as it required for human survival and demand does not fluctuate much with change in price.
.when a commodity is a comfort like fan,refrigerator etc its demand is generally elastic as a consumer can postpone it consumption
.when a commodity is a luxury like AC,DVD player ,etc its demand generally more elastic as compared to demand for comforts
It is given in On the Meritnation Study Material it is given that when AP reaches its maximum value its is equal to MP.However this does not happen in this example given on meritnation???
what is the reason behind this contradiction???
how do managerial economies lead to more availability of scarce finance???
As given in the Meritnation Study Material???
Also do Internal Economies reduce cost of production/unit or overall (total) cost of production????
why can't supply curve start from origin????
Q. How does "Trading on Equity" affect the choice of capital structure or a company. Explain with the help of suitable example.
Total Capital - Rs. 50 lakhs
Equity - Rs. 50 lakhs (4 lakhs share of Rs. 10 each)
Debt - Nil
E BIT - 7 Lakhs
Debt - 10 lakhs
Tax rate - 30% p.a
Interest on debt - 10%
Equity - Rs. 30 lakhs
Debt - Rs. 20 lakhs
Q7.Define unitary elastic demand and draw a curve for it.what is the significance of the unitary elastic demand?
ans.Unitary elastic demand -when the percentage change in the quantity demanded is equal to the percentage change in price ,then the demand for such commodity is said to be unitary elastic.in this case Ed=1 |
What is Life? Practice Problems Online | Brilliant
"All science is either physics or stamp collecting. That which is not measurable is not science." - Ernest Rutherford, known as the father of Nuclear physics.
Ernest Rutherford on a stamp.
Biology as a field of study occasionally gets little respect compared to its more quantitative cousins under the umbrellas of mathematics and physics. This isn’t aided by the introduction to biology that many of us encounter early in our schooling: first comes the study of kingdoms, phyla, and species, and later the memorization of mitochondria, chloroplasts, and cellular nuclei. All in all; a pile of facts and details, some of them interesting, but disappointingly unconnected by unifying themes or quantitative principles.
But biology, the natural science that studies life and living organisms, does have a unifying principle that connects every organism that has ever existed on Earth, and even unknown organisms that may exist elsewhere in the universe.
Let's get to know the light that will guide the rest of this course.
The diversity of life is astounding at nearly every scale: Humans are just one of the nearly six thousand different species of mammal. Other members of our furry and big-brained group have sizes spread over three orders of magnitude, from the bumblebee bat in the forests of Myanmar to the blue whale in the Antarctic Oceans.
But mammals are an evolutionary newcomer compared to others; over a million species of insects roam every continent on Earth. Other invertebrates have spread even to the hydrothermal vents on the ocean floor, where some have teamed up with sulfur-breathing bacteria to grow iron plate armor. While some bacteria breathe toxic chemicals to extract energy from deep-sea vents, others get energy straight from the sun or by consuming materials from other organisms.
Every handful of soil contains many billions of bacteria from millions of different species; only a tiny fraction of which have ever been isolated and studied in a lab.
Lifetimes have been spent categorizing all these different organisms, and enumerating their divergent diets, anatomies, metabolisms, and reproductive cycles. These studies and many others have been lumped together in the field of biology. Biology’s incredible diversity is obvious and evident to an observer, and this is one of the reasons that much of the scientific community was cautious to accept the theory of evolution. It's hard to deny that at first impression, more about life seems to be different than the same.
Some of Charles Darwin’s first and most thorough investigations of evolution came from studying the beaks and other anatomical features of finches. He was in search of shared structures and features that could hint at the relatedness of different bird species. Many identified this as a great way to make a family tree of birds, but not a unifying principle that could be applied to all life.
Darwin's sketches of Galapagos finches. From On the Origin of the Species.
Why might Darwin's theories not have been particularly convincing to his contemporaries, outside of ornithology (the study of birds)?
Not all organisms have shared features to be studied. Shared structures can only be used to find related individuals of the same species. Shared structures cannot be used to find related individuals in similar species.
Not all organisms have shared features like beaks and wings. At the microscopic scale, there are thousands of known bacterial strains that look completely identical under a microscope. Using anatomical structure and shared features to construct a unifying principle connecting all forms of life was doomed to failure, even though Darwin was conceptually correct. So how can we connect the family tree of birds to that of bats, snakes, or bacteria?
The answer has come from looking very closely. In the last
50
years we've encountered the flip side of biological diversity by studying molecular biology: At a smaller scale, all life is the same. Every form of life from bacteria to dinosaurs have a DNA genome, and its genetic information programs all the other molecules that make up an organism: proteins, RNA, carbohydrates and fats.
Darwin inferred the relatedness of finches by studying common features. What can we infer from recent findings in molecular biology?
All life must share a common ancestor which had a DNA genome. We can't be sure about life's ancestry without detailed DNA analysis. DNA genomes must contain random information.
Life at the macroscopic scale has remarkable diversity, which can be studied by anatomy, genealogy, paleontology, and other fields of biology. But quantitation and unifying principles are hard to come by at that scale.
All the diversity of different organisms must be present in their DNA genomes. Not in the form of cellular, skeletal or morphological structures that must be studied with an X-ray or a microscope, but in strings of encoded information. We'll learn throughout this course that this genetic code is nearly universal in all forms of life, and indicates that there must be a common ancestor that had a genome just like life today. Genetic information can be quantified and processed at a large scale in a way that traditional biological results simply cannot, as we'll soon see.
If the unity of life at the molecular scale is explained by our shared features with an ancient common ancestor, what principle of life could explain all of life's differences today?
Genetic information changes over time and is selected by competition and its environment. Genetic information must be completely randomized. There must be many different common ancestors for each existing organism.
At human scales, evolution is slow and subtle: it took about two million years for Darwin's finches to develop such different beaks. Only careful anatomical comparison can reveal the anatomy of a common ancestor to birds, bats and humans may have had. But the changes in genetic information are discrete and quantifiable, even though their signal-to-noise ratio may be low.
By comparing genetic information, it is possible to provide a quantitative measure of relatedness from molecules to organisms that Darwin couldn't have ever dreamed of. Genome analysis of Darwin's finches tracked down singular events over the last million years that lead to each finch's unique beaks. Furthermore, studying how genetic information is translated to traits like beaks and other features can provide far more insight into biology than a thousand years of anatomical study.
For this reason much of modern biology has stepped away from the detailed study of a wide variety of different organisms and their traits and features, and into an information-driven study of the genetic information that makes up the DNA in every form of life.
This shift has been enabled by rapid development in the field of DNA sequencing and been helped in a large part by the Human Genome Project. Understanding genetic information and how it connects to the rest of biology is only possible because of the torrent of sequencing information made available in the last 20 years since this project finished. The complete genomes of hundreds of animals and plants each consisting of gigabytes of data have been collected.
Number of genomes sequenced per year. Source: Illumina
The rise of genetic sequencing has been a gold rush for information theorists; renewing the techniques that used to be in communications, signal processing and statistical physics for the new frontier of biological information. Now that the data is here, what problems should biology research focus on to make the most of it?
Processing and interpreting large quantities of genetic information. Compiling the existing body of anatomical, cellular and structural biology data. Correlating traditional biological data with genetic information.
This course will explore the field of computational biology. First through an introduction to molecular biology with a focus on the information flow from DNA to more familiar biological features. Then the connection between genetic information and the biological structures of proteins, RNA and cells will be explored using folding algorithms implemented in Python using dynamic programming techniques.
Once we've gained the biological context, we'll shift gears to analyzing genetic sequences to gain insight into forensics, human history, and disease. Finally we'll retrace the steps of Darwin and reconstruct the tree of life — this time with firmly quantitative principles that let us compare birds, beans, and bats. Throughout, we will focus less on facts and details, and more on unifying principles that will come up again and again in our study of computational biology.
This isn't your grandpappy's biology course. |
Cognitive Bias | Brilliant Math & Science Wiki
Christopher Williams and Eli Ross contributed
Cognitive bias refers to individuals consistently making irrational decisions, often intuitively or unknowingly. Many humans have cognitive biases that appear in certain logic, economic, or interpersonal situations. Researchers suspect that many of biases are adaptive, developed over time to aid in human decision making, especially in social situations.
Understanding these bias can help individuals make better decisions or recognize situations where they may be being manipulated.
Anchoring is the cognitive bias where a person is first shown a number in some form deliberate or subtle and then told to perform an action. Persons shown a low number typically anchor towards lower numbers and persons shown a high number tend to anchor up.
Participants are given the same sequence of numbers, but in different orders, then are given not enough time to estimate the product of these numbers.
One group of users is given the numbers
8\times 7\times 6\times 5\times 4\times 3\times 2\times 1
and the other group is given the same numbers reversed.
1\times 2\times 3\times 4\times 5\times 6\times 7\times 8
. They are given only a few seconds to then guess at the product of these numbers. Participants in the first group, which started, or was anchored, at a high number, guessed an average number of
2,250
. Participants in the second group guessed an average of
512
. The correct answer is 40,320.
Anchoring is a common tactic used in sales or marketing, for instance clothing prices and sales or discounts. In many Western countries, retailers use a pricing strategy known as high-low pricing. Clothing items are given a high retail price, a high anchor price, and then frequent and varied sales are thrown with large percentage discounts. This tactic anchors the shopper to a high number, and wows them with a large discount, even though the retailer was planning to sell the items for an average affective price well below retail.
The conjunction fallacy is a logical fallacy where individuals assume that a set of specific conditions is more probable than a single, more general, position. Amos Tversky and Daniel Kahneman were amongst the first to identify this phenomena through a problem known as The Linda Problem.[1]
In experiments, the majority of test takers got this question wrong. The description of Linda seems almost designed to encourage a test taker to think that she is a feminist. So, when presented with that choice, many take it, even though the conjunction of multiple specific facts is less probable of occurring than just one of those facts by themselves.
Mathematically speaking, this is equivalent to the fact from probability that
P(A) \ge P(A \cap B).
A deeper look at these ideas can be seen in Bayes' Theorem.
Prospect theory, also known as decision weighting refers to how people choose between risky alternatives, specifically when the probabilities of the alternatives are known or can be guessed. In general, people overestimate the probability of unlikely events (for instance: a third party candidate being elected for president) and overweight unlikely events in their decisions
Daniel Kahneman[1] conducted an experiment in which participants were asked to make two decisions concurrently: 1st. Choose between:
a) a sure gain of $240
b) a 25% chance at $1,000
2nd. Choose between:
c) a sure loss of $750
d) a 75% chance to lose $1,000
The majority of people chose A then D, indeed this is most people's quick gut reaction. However choosing A and D means the chooser has a combined 75% chance to lose $760, and a 25% chance to gain $240. The more rational choice would be to choose B and C, which results in a 75% chance to lose $750, losing $10 fewer dollars than A and C, and a 25% chance to make $250, making $10 more than choosing A and C.
In competitions of luck, results that are on the extreme will eventually regress back to the mean or average. In experimentation it is often the case that if an extreme result occurs the first time, an average result will occur the second. And if the second result was extreme, it is likely that the first result was average.
Consider the case where two people are standing on a line. Another line is drawn behind them. They are then each handed a coin and told to throw it as close to the other line as they can without looking. If they only attempt this once, one of the test takers will do better than the other, and if that is left as the only result, they may conclude that they are superior at this task than the other coin tosser.
However if the test is conducted a second time, the results often switch. The person with the extreme case of accuracy the first time will do worse, and the person who did poorly the first time will do better. They are both regressing to the mean.
This is one of the reason, in scientific settings, that researchers repeat their experiments. Over enough trials, the results will average out to the most accurate result.
The sunk cost fallacy represents a contrast between microeconomic theory and intuitive decision making. Microeconomics holds that once costs have been incurred and are unrecoverable, agents should ignore these cost in making future decisions. However behavior economics has shown that agents do take these costs into account.
A company is behind and over budget ($10 Million more on top of $10 Million already spent) on an IT project that is estimated to generate $100 Million in profits. To finish the project they're going to have to spend an another $10 Million dollars. However this is all they'll have to spend for the year, and a manager in another department is proposing they drop the existing project and invest in a new project that is estimated to generate higher returns.
In many cases, the company chooses to continue with the existing project, suffering from loss aversion and taking sunk costs into their decision making.
[1] Kahneman D. Thinking, Fast and Slow. (New York, Farrar, Straus and Giroux, 2011)
Cite as: Cognitive Bias. Brilliant.org. Retrieved from https://brilliant.org/wiki/cognitive-bias/ |
Some Weighted Norm Estimates for the Composition of the Homotopy and Green’s Operator
Huacan Li, Qunfang Li, "Some Weighted Norm Estimates for the Composition of the Homotopy and Green’s Operator", Abstract and Applied Analysis, vol. 2014, Article ID 941658, 7 pages, 2014. https://doi.org/10.1155/2014/941658
Huacan Li1 and Qunfang Li2
1School of Science, Jiangxi University of Science and Technology, Ganzhou 341000, China
2Department of Mathematics, Ganzhou Teachers College, Ganzhou 341000, China
Academic Editor: Yuming Xing
We establish the -weighted integral inequality for the composition of the Homotopy and Green’s operator on a bounded convex domain and also motivated it to the global domain by the Whitney cover. At the same time, we also obtain some -type norm inequalities. Finally, as applications of above results, we obtain the upper bound for the norms of or in terms of norms of or .
Our purpose is to study the theory of the composition of the Homotopy and Green’s operator acting on differential forms on a bounded convex domain. Both operators play an important role in many fields, including harmonic analysis, potential theory, and partial equations (see [1–6]). In the present paper, we will obtain some -type norm inequalities for the composition of the Homotopy and Green’s operator and also prove the -weighted integral inequality on a bounded convex domain. These results will provide effective tools for studying behavior of solutions of -harmonic equations and related differential systems on manifolds.
We start this paper by introducing some notations and definitions. Let be a Riemannian, compact, oriented, and -smooth manifold without boundary on and let be an open subset of . Also, we use to denote Green’s operator throughout this paper. Furthermore, we use to denote a ball and to denote the ball with the same center as and with . We do not distinguish balls from cubs in this paper.
We assume that is the linear space of all -forms with summation over all ordered -tuples , . If the coefficient of -form is differential on , then we call a differential -form on . A differential -form on is a de Rham current (see [7]) on with values in . Let be the th exterior power of the cotangent bundle and be the space of smooth -forms on . As usual, we use to denote the space of all differential -forms and to denote the -form with the norm on . Thus is a Banach space. As usual, we still use to denote the Hodge star operator. Also, we use to denote the differential operator and use to denote the Hodge codifferential operator which is defined by on . The -dimensional Lebesgue measure of a set is denoted by . We call a weight if and . For , we denote the weighted -norm of a measurable function over by where is a real number.
Let be a bounded, convex domain. Iwaniec and Lutoborski in [1] first introduced a linear operator satisfying that and the decomposition . Then by averaging over all points in , they constructed a Homotopy operator satisfying that , where is normalized by . The -form is defined by , if , and if , then
2. Boundedness of the Composition of the Homotopy and Green’s Operator in Space
In this section, we will prove the -weighted norm inequality for the composition of the Homotopy and Green’s operator on a bounded convex domain. Then using the Whitney cover, we develop the local result to the global domain. In [8], Gol’dshtein and Troyanov proved the following lemma.
Lemma 1. Let be a bounded convex domain. The operator maps continuously to in the following cases:
From [3], we have the following lemma about -estimates for Green’s operator.
Lemma 2. Let and . Then there exists a constant , independent of , such that
Definition 3. We say that a weight satisfies the condition for and write , if a.e. and
For weight, we also need the following result which appears in [9].
Lemma 4. If , then there exist constants and , independent of , such that for all balls .
Theorem 5. Let be a bounded convex domain, , and let be the Homotopy operator, . Then there exists a constant , independent of , such that for any ball , , and .
Proof. Since , by Lemma 4, there exist constants and , independent of , such that for any ball .
Choosing , then by Hölder inequality with , we have Thus, substituting (11) into (12), we obtain Taking , it is easy to see that and . Hence communicating Lemmas 1 and 2, we have Combining (13) and (14), we have Using Hölder inequality with , we have Note ; then, Thus, observing (15) and (16), we immediately obtain that Here is a constant independent of . Thus we complete the proof of Theorem 5.
Furthermore, if is an -harmonic tensor on , and , , then there exists a constant , independent of , such that for all balls or cubs with (for more details about -harmonic tensors, see [10]). By the property of -harmonic tensor, using the same method developed in the proof of Theorem 5, we can easily extend into the following -weighted version.
Corollary 6. Let be a bounded convex domain, , be an -harmonic tensor, and be the Homotopy operator, . Then there exists a constant , independent of , such that for any ball , , and , , .
In order to obtain the boundedness of the composition , we need the following modified Whitney cover in [10] and see [11] for more details about Whitney cover.
Lemma 7. Each open subset has a modified Whitney cover of cubs satisfying and , for all and some , where is the characteristic function for the set .
Theorem 8. Let be a bounded convex domain, . Then the composite operator is bounded, . Here and .
Proof. From Lemma 7, we know that there exists a sequence of cubs such that and for all , where is some constant. Hence, for , we have where and is independent of and each . Thus, we complete the proof of Theorem 8.
3. Norm Estimates with Power-Type Weights
Let be a bounded domain and be a nonempty of . If we use to denote the distance of the point from the set , then for is called power-type weight. In this section, we will establish some strong -type norm inequalities with power-type weights for the composition of the Homotopy and Green’s operator acting on differential form. In the following proof, we will use the following Lemma which appears in [8].
Lemma 9. The operator is bounded provided that
Theorem 10. Let be a bounded convex domain, , , , and let be the Homotopy operator, . Then there exists a constant , independent of , such that for any .
Proof. From (4), we have the following decomposition: for any differential form , .
Note that is an element of , . From (4) and Lemmas 1 and 9, we have Here is a constant independent of . Applying (24) and (5), we have Applying Lemma 2 into (26), we obtain Thus Here is independent of . Thus, we complete the proof of Theorem 10.
Next, we consider the following norm comparison equipped with power-type weights.
Theorem 11. Let be a bounded convex domain, , , , let be the Homotopy operator, , and that continuous functions and defined in satisfy ; . Then there exists a constant , independent of , such that for any , , .
Proof. From Theorem 10, we know that there exists a constant , independent of , such that Fixing , then there exists such that for all with . Let and . Then for all , we have Therefore, by the continuity of , we know that there exists , such that for all . Thus we have Here . Communicating (30) and (33), we have Note that . Then there exists such that for all with . Let and . Then for all , we have Therefore, by the continuity of , we know that there exists , such that for all . Therefore, we obtain Here . By (34) and (37), we have Here is independent of . Thus, we complete the proof of Theorem 11.
In Theorem 11, if we choose and , , , we can easily obtain the following corollary.
Corollary 12. Let be a bounded convex domain, , , , and let be the Homotopy operator, . Then there exists a constant , independent of , such that Here , .
Note that, in the proof of Theorem 11, if we let the composite operator act on the solution of nonhomogeneous -harmonic equation, then we can drop . Next, we state the result as follows.
Corollary 13. Let be a bounded convex domain, , , , let be the Homotopy operator, and is a solution of nonhomogeneous -harmonic equation, . If continuous functions and defined in satisfy that , and . Then there exists a constant , independent of , such that for all balls with . Here is some constant.
It is easy to find that the above corollary does not hold for balls with but holds for those balls with . Next, we introduce the following singular integral inequality.
Theorem 14. Let be a bounded convex domain, , , , let be the Homotopy operator, and is a solution of nonhomogeneous -harmonic equation, . If continuous functions and defined in and is an increasing function, then there exists a constant , independent of , such that for all balls with and . Here is some constant.
Proof. Let . From , it is easy to see that . Using the Hölder inequality, we have Note that . Therefore, there exists a positive number such that for all . Furthermore, by the continuity of function in , has a positive lower bound in . Thus, from Theorem 10 and (42), we have where is a constant. Let and . Since is the solution of nonhomogenous -harmonic equation. By (19), we know where is a constant. It is easy to find that . Using the Hölder inequality, we have The continuity and monotonicity of function imply that Hence, combining (41)–(47), we have Here is dependent of and but independent of . Thus, we complete the proof of Theorem 11.
In this section, we will use the estimates in Section 3 to obtain the upper bound for the norms of or in terms of norms of or .
Example 15. For , let be a -form defined in by It is easy to find that If we choose the usual -type norm inequality to estimate and take , where is a ball, then by Theorem 10, we have However, if we choose the -type norm inequality to estimate and take , , then , satisfy the condition . Hence by using Theorem 10, we obtain Compare (51) and (52), we can easily find that if we choose different -type norm inequality to estimate the oscillation , we also obtain the different upper bound.
Example 16. In , consider that It is easy to check that is harmonic in the upper half plane. Note that Therefore, we have which implies that is a closed form and hence is a solution of nonhomogenous -harmonic equation. It is easy to see that Let denote a bound convex domain in the upper half plane and let be a closed ball without the points and . If and satisfy that , then both and have the upper bounds in . Thus, for the term it is usually not easy to be estimated due to the complexity of the compositions and the function . However, by Theorem 14, (57) can be controlled by the term Thus, we obtain an upper bound of (57).
The first author was supported by the foundation at the Jiangxi University of Science and Technology (no. jxxj12073) and by the Youth Foundation of Jiangxi Provincial Education Department of China (no. GJJ13376).
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{L}^{p}
theory of differential forms on manifolds,” Transactions of the American Mathematical Society, vol. 347, no. 6, pp. 2075–2096, 1995. View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
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Copyright © 2014 Huacan Li and Qunfang Li. 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. |
Find an equation of the plane. The plane through the points (2,1,2), (3,-8,6), and (-2,-3
Find an equation of the plane. The plane through the points (2,1,2), (3,-8,6), and (-2,-3,1)
The plane through the points (2,1,2), (3,-8,6), and (-2,-3,1)
To find the equation of the plane we need to find its normal vector.
Let us call the given points as: P(2,1,2), Q(3,-8,6), R(-2,-3,1)
\stackrel{\to }{PQ}
\stackrel{\to }{PR}
are vectors in the plane
Hence, their cross product will give us normal vector to the plane
\stackrel{\to }{PQ}=\left(3,-8,6\right)-\left(2,1,2\right)=\left(3-2,-8-1,6-2\right)=\left(1,-9,4\right)
\stackrel{\to }{PR}=\left(-2,-3,1\right)-\left(2,1,2\right)=\left(-2-2,-3-1,1-2\right)=\left(-4,-4,-1\right)
n=\stackrel{\to }{PQ}×\stackrel{\to }{PR}=|\begin{array}{ccc}i& j& k\\ 1& -9& 4\\ -4& -4& -1\end{array}|=\left(25,-15,-40\right)
Equation of a plane passing through the point (a,b,c) and having normal vector (l,m,n) is
l\left(x-a\right)+m\left(y-b\right)+n\left(z-c\right)=0
We found the normal vector in the previous cell. For a point on the plane, we can choose any of the three given points, I will choose P(2,1,2)
25\left(x-2\right)-15\cdot \left(y-1\right)-40\cdot \left(z-2\right)\right]=0
5\left(x-2\right)-3\cdot \left(y-1\right)-8\cdot \left(z-2\right)=0
5x-3y-8z+9=0
5x-3y-8z+9=0
\left(x,\text{ }y\right)=
How do you find the polar coordinates of the point
\left(-2,\text{ }0\right)?
Find a vector equation and parametric equations for the line segment that joins P to Q . P(2, 0, 0), Q(6, 2, -2)
For problems 1 the area of the region below the parameric curve given by the set of parametric equations. For each problem you may assume that each curve traces out exactly once from right to left for the given range of t. For these problems you should only use the given parametric equations to determine the answer.
x={t}^{2}+5t-1
y=40-{t}^{2}
-2\le t\le 5
Find two different sets of parametric equations for a rectangular equation
y-{x}^{2}-3
Replace the Cartesian equation with equivalent polar equations. xy = 2
Sketch the curve represented by the vector-valued function
r\left(t\right)=\left(t+1\right)i+\left(3t-1\right)j+2tk
and give the orientation of the curve. |
Laws of Chemical Combination | Brilliant Math & Science Wiki
Sravanth C., Tim O'Brien, Abhiram Rao, and
The laws of chemical combination describe the basic principles obeyed by interacting atoms and molecules, interactions that can include many different combinations that happen in many different ways. This amazing diversity of interactions allows for an astounding variety of chemical reactions and compounds. Spontaneous chemical reactions happen constantly, shaping the world around us, while humans engineer specific reactions to our benefit and attempt to curb reactions that hurt us. Though chemical reactions can be as complex as they are numerous, they are all fundamentally governed by these same guiding laws of chemical combination, which lay the groundwork for analysis of chemical reactions. They give a mathematical formulation and allow predictability given initial conditions. They are the launch pad from which we jump off to creating all sorts of wild compounds and phenomena. And while chemistry is still difficult and intricate, with the laws of chemical combination on our side, we can begin to make some headway.
\ce{Ca(OH)2 + CO2 -> CaCO3 + H2O}
The masses of
\ce{Ca}
\ce O
\ce H
\ce C
40\text{u}, 16\text{u}, 1\text{u},
12\text{u},
Since they obey the law of conservation of mass, the answer is yes. Let's verify it. The molecular masses are
\begin{array}{rrlrl} \ce{Ca(OH)2}: &40+32+2 &= &74 \\ \ce{CO2}: &12+32&=&44 \\ \ce{CaCO3}: &40+12+48&=&100\\ \ce{H2O}: &2+16&=&18&. \end{array}
Substituting these values into the equation,
\begin{aligned} 74+44 &= 100+18\\ 118&=118.\ _\square \end{aligned}
This means each compound has the same elements in the same proportions, regardless of where the compound was obtained, who prepared it, or its mass.
100\text{ ml}
\ce{CaCO3}
200\text{ ml}
and compared it to his friend's sample. Which of the two compounds has a greater ratio of
\ce{Ca}:\ce{C}?
Both contain equal ratios of
\ce{Ca}
\ce C
_\square
Carbon combines with oxygen to form two different compounds (under different circumstances). One is the most common gas
\ce{CO2}
\ce{CO}
12\text{ u}
and the mass of oxygen is
16\text{ u}
12\text{ g}
32\text{ g}
\ce{CO2}
12\text{ g}
16\text{ g}
\ce{CO}
2:1=\frac{32}{16}= 2,
_\square
The law of reciprocal proportions states that when two different elements combine with the same quantity of the third element, the ratio in which they will do so will be the same or a multiple of the proportion in which they combine with each other.
Oxygen and sulfur react with copper to create copper oxide and copper sulfide, respectively. Sulfur and oxygen also react with each other to form
\ce{SO2}.
\begin{array}{rrcr} \text{in } \ce{CuS}, & \ce{Cu}:\ce{S} &=& 63.5:32 \\ \text{in } \ce{CuO}, & \ce{Cu}:\ce{O} &=& 63.5:16 \\\\ \Rightarrow &\ce{S}: \ce{O} &=& 32:16 \\ & &=&2:1. \end{array}
\ce{SO2},
\ce{S}:\ce{O} = 32:32= 1:1.
Thus the ratio between the two ratios is the following:
\frac{2}{1} :\frac{1}{1} = 2:1,
which is a simple multiple ratio.
Cite as: Laws of Chemical Combination. Brilliant.org. Retrieved from https://brilliant.org/wiki/laws-of-chemical-combination/ |
Engineering Acoustics/Mechanical Resistance - Wikibooks, open books for an open world
Engineering Acoustics/Mechanical Resistance
2 Dashpots
3 Modeling the Damped Oscillator
4 Mechanical Impedance for Damped Oscillator
Mechanical ResistanceEdit
For most systems, a simple oscillator is not a very accurate model. While a simple oscillator involves a continuous transfer of energy between kinetic and potential form, with the sum of the two remaining constant, real systems involve a loss, or dissipation, of some of this energy, which is never recovered into kinetic nor potential energy. The mechanisms that cause this dissipation are varied and depend on many factors. Some of these mechanisms include drag on bodies moving through the air, thermal losses, and friction, but there are many others. Often, these mechanisms are either difficult or impossible to model, and most are non-linear. However, a simple, linear model that attempts to account for all of these losses in a system has been developed.
DashpotsEdit
The most common way of representing mechanical resistance in a damped system is through the use of a dashpot. A dashpot acts like a shock absorber in a car. It produces resistance to the system's motion that is proportional to the system's velocity. The faster the motion of the system, the more mechanical resistance is produced.
As seen in the graph above, a linear relationship is assumed between the force of the dashpot and the velocity at which it is moving. The constant that relates these two quantities is
{\displaystyle R_{M}}
, the mechanical resistance of the dashpot. This relationship, known as the viscous damping law, can be written as:
{\displaystyle F=R\cdot u}
Also note that the force produced by the dashpot is always in phase with the velocity.
The power dissipated by the dashpot can be derived by looking at the work done as the dashpot resists the motion of the system:
{\displaystyle P_{D}={\frac {1}{2}}\Re \left[{\hat {F}}\cdot {\hat {u^{*}}}\right]={\frac {|{\hat {F}}|^{2}}{2R_{M}}}}
Modeling the Damped OscillatorEdit
In order to incorporate the mechanical resistance (or damping) into the forced oscillator model, a dashpot is placed next to the spring. It is connected to the mass (
{\displaystyle M_{M}}
) on one end and attached to the ground on the other end. A new equation describing the forces must be developed:
{\displaystyle F-S_{M}x-R_{M}u=M_{M}a\rightarrow F=S_{M}x+R_{M}{\dot {x}}+M_{M}{\ddot {x}}}
It's phasor form is given by the following:
{\displaystyle {\hat {F}}e^{j\omega t}={\hat {x}}e^{j\omega t}\left[S_{M}+j\omega R_{M}+\left(-\omega ^{2}\right)M_{M}\right]}
Mechanical Impedance for Damped OscillatorEdit
Previously, the impedance for a simple oscillator was defined as
{\displaystyle \mathbf {\frac {F}{u}} }
. Using the above equations, the impedance of a damped oscillator can be calculated:
{\displaystyle {\hat {Z_{M}}}={\frac {\hat {F}}{\hat {u}}}=R_{M}+j\left(\omega M_{M}-{\frac {S_{M}}{\omega }}\right)=|{\hat {Z_{M}}}|e^{j\Phi _{Z}}}
For very low frequencies, the spring term dominates because of the
{\displaystyle {\frac {1}{\omega }}}
relationship. Thus, the phase of the impedance approaches
{\displaystyle {\frac {-\pi }{2}}}
for very low frequencies. This phase causes the velocity to "lag" the force for low frequencies. As the frequency increases, the phase difference increases toward zero. At resonance, the imaginary part of the impedance vanishes, and the phase is zero. The impedance is purely resistive at this point. For very high frequencies, the mass term dominates. Thus, the phase of the impedance approaches
{\displaystyle {\frac {\pi }{2}}}
and the velocity "leads" the force for high frequencies.
Based on the previous equations for dissipated power, we can see that the real part of the impedance is indeed
{\displaystyle R_{M}}
. The real part of the impedance can also be defined as the cosine of the phase times its magnitude. Thus, the following equations for the power can be obtained.
{\displaystyle W_{R}={\frac {1}{2}}\Re \left[{\hat {F}}{\hat {u^{*}}}\right]={\frac {1}{2}}R_{M}|{\hat {u}}|^{2}={\frac {1}{2}}{\frac {|{\hat {F}}|^{2}}{|{\hat {Z_{M}}}|^{2}}}R_{M}={\frac {1}{2}}{\frac {|{\hat {F}}|^{2}}{|{\hat {Z_{M}}}|}}cos(\Phi _{Z})}
Retrieved from "https://en.wikibooks.org/w/index.php?title=Engineering_Acoustics/Mechanical_Resistance&oldid=3232730" |
U= the set of natural numbers between 10 and 20
U= the set of natural numbers between 10 and 20 A= Even numbers B= Multiples of 3 where the number is less than 18. Find (A'UC) - B
U=
the set of natural numbers between 10 and 20
A=
Even numbers
B=
Multiples of 3 where the number is less than 18.
C=
\left({A}^{\prime }UC\right)-B
cheekabooy
U=
the set of natural numbers between 10 and 20.
U=\left\{11,12,13,14,15,16,17,18,19\right\}
A=
⇒\left\{2,4,6,...18\right\}
B=
multiples of three where the number 15 less than 18.
B=\left\{3,6,9,12,15\right\}
C=
=\left\{4,6,8,9,10,12,14,15,16,18,...\right\}
\left({A}^{\prime }UC\right)-B
{A}^{\prime }=\left\{1,3,5,7,9,11,13,...\right\}
\left({A}^{\prime }UC\right)=\left\{1,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,...\right\}
\left({A}^{\prime }UC\right)-B=\left\{1,4,5,7,8,10,11,13,14,16,17,19\right\}
h\left(t\right)=-16{t}^{2}+24t
7{C}_{2}
Consider each set of numbers and determine if the set has an additive identity, additive inverse, multiplicative identity, or a multiplicative inverse. Explain your reasoning for each. a. the set of natural numbers.
How many bit strings of length 10 have more 0s than 1s?
To find other numbers which are factors of a number whose factor is 18.
Find the sum of alll the multiples of three between 1 and 200 |
Specific Requirements of Membrane Structures | Dlubal Software
Home Support & Learning Support Knowledge Base Specific Requirements of Membrane Structures
Membrane structures are one of the current trends in civil engineering. They are beautiful, light, statically effective, and challenging. Due to their zero bending stiffness, it is impossible to separate the shape of a membrane structure from the prestress. The shapes cannot be selected freely; they have to be found. These multi-dimensional structures are manufactured from fabric or foil rolls. The cutting patterns are formed from planar material strips, and you reach the intended structure by connecting and stretching them in the final position. The determination of cutting patterns is a sensitive step in the planning process, and its quality strongly influences the quality of the entire structure. This article deals in detail with the two main processes - the shape determination of membrane structures, and the determination of cutting patterns. Special attention is paid to practical and useful insights for planning.
In this chapter, the physical principles of shape determination for membrane structures are described first. Afterward, the feasibility of prestress required by the civil engineer is discussed. The text is then supplemented by practical examples to illustrate the considerations and theories.
The planning of membrane structures differs significantly from the usual practice. Since the materials used practically have only a tension resistance, the shape cannot be selected freely. It is impossible to separate the shape from the prestress. In this case, the aesthetic and physical aspects of buildings are basically connected.
The shape of a membrane structure is determined by the boundary conditions and the spatial equilibrium system. The form-finding process can be described by Equation (1) below. The equilibrium shape is found if the virtual work does not change (δW = 0); that is, if the sum of the virtual work performing the required prestress σ and the virtual work performing the external load p (positive pressure, self-weight) is equal to zero.
- \mathrm{\delta W} = {\mathrm{\delta W}}^{\mathrm{int}} - {\mathrm{\delta W}}^{\mathrm{ext}} = \mathrm{t} · {\int }_{\mathrm{\Omega }}\mathrm{\sigma }:\mathrm{\delta êd\Omega } - {\int }_{\mathrm{\Omega }}\stackrel{\to }{\mathrm{p}}·\mathrm{\delta ud\Omega } = 0 \left(1\right)
In the equation above, t represents the thickness of the material used, δê is the change in material deformation, and δu is the deformation over the surface of the structure Ω.
In addition to some theoretical problems to be solved, there is a fundamental problem. The main problem is that a preset prestress is assumed. However, it is generally excluded. The membrane structures have a double curvature (that is, the Gaussian curvature is not equal to zero), which is why they exclude a homogeneous orthotropic prestress. Theoretically, a state where there is a specific prestress value in the warp direction and a precise prestress value in the weft direction at each point of the membrane is almost impossible. The only exception is the isotropic prestress, which can be achieved if the shape is physically real under the given boundary conditions.
Thus, the prestress itself must be found. The aim of the process (form-finding) is not only to find an unknown shape for a given prestress, but also to search for an unknown shape for a generally unknown prestress. This prestress is approximated by a value specified by the civil engineer for the warp and weft directions. A number of methods have been developed for form-finding. If you use different programs for problem solving, you can obtain more or less different results for the same input data. Then, of course, the question arises: which solution is the optimal one? Some examples of different structures and required prestresses are shown in the following text.
Image 01 - Basic Shapes of Membrane Structures [1]
For the first example, we will use a hyperbolic paraboloid (Figure 2 and Figure 3). Both isotropic and orhotropic prestress are applied. For the isotropic prestress, two different results are seen from the form-finding process (Figure 4 and Figure 5), which are also commented on briefly. For the isotropic prestress, nwarp = nweft = 2.00 kN/m is set. A relative cable sag s = 8.00% is set for the edge cables. The results are illustrated as vectors of the principal internal forces using a color scale.
Image 02 - Membrane Structure in Form of Hyperbolic Paraboloid
Image 03 - FE Mesh and Alignment of Warp (Red) and Weft (Green) Threads
Image 04 - Vectors of Principal Internal Forces n1, n2
If two different results are obtained for the same input data, the question naturally arises: which solution is the right one? In theory, both solutions are correct because both have reached an equilibrium state and both are also feasible. However, the solution shown on the left shows a uniform prestress that is not concentrated on corner areas. Such local effects are considered as undesirable because they reduce the load-bearing capacity of the structure and result in uneven rheological effects. Therefore, the solution shown on the left is advantageous. Generally, it is considered as favorable to find a shape with a uniformly distributed and not locally concentrated prestress. The membrane structure is thus well prestressed, and its load-bearing capacity is not reduced in some areas by excessive prestress.
As already mentioned, an isotropic prestress is the only homogeneous prestress that can be achieved precisely. The achievable accuracy is practically limited only by the size of the FE mesh. In the case of a roughly set mesh, an equilibrium state cannot be approximated exactly, and thus the values may deviate from the entered prestresses. However, such deviations should be within a small range, and a coarser mesh does not necessarily lead to a clearly more concentrated prestress.
The same boundary conditions are applied for the other calculation. The structure is given an orthotropic prestress of nwarp = 4.00 kN/m and nweft = 2.00 kN/m. A relative cable sag s = 8.00% is set for the edge cables. As mentioned, an exact homogeneous orthotropic prestress cannot be achieved due to the double curvature of the membrane structures. However, it is possible to achieve a shape with a prestress that comes very close to the specified values (Figure 5). The result is a uniformly distributed prestress approximating the input values. In this case, there is no reason for significant stress concentrations.
For most shapes, including hyperbolic paraboloids and arc-supported or pneumatic membranes (Figure 1), the resulting prestress can be distributed evenly without the need for local prestress concentrations. For high conical shapes, it is impossible to avoid areas with concentrated prestress. Any concentrations occur at the apex of the cone, but they are neither necessary nor desired in the bottom corners (Figure 6).
Image 06 - Vectors of Principal Internal Forces n1, n2 and Axial Forces N
Whether or not a concentrated prestress is necessary can be deduced intuitively from the following formula (2). The equation represents an equilibrium of forces at a point where n1 and n2 are the principal internal forces, 1/R1 and 1/R2 are the curvatures in the direction of these principal internal forces, and p is any external load.
\frac{\mathrm{n}1}{\mathrm{R}1} \frac{\mathrm{n}2}{\mathrm{R}2} - \mathrm{p} = 0 \left(2\right)
In the case of an anticlastic structure of which the self-weight barely influences the found shape, the equilibrium of forces in a node is given by the prestress and the curvatures in the opposite direction. The issue now is whether the curvature of the structure has to change so rapidly. If so, the locally concentrated prestress is inherent in the structure; otherwise, the prestress concentration is not necessary for the structure at all. This method can be applied to our examples. Shapes without conical areas (Figure 4, Figure 5, Figure 8, and Figure 10, except for the conical areas) do not require rapid changes in the curvature, which is why they can be prestressed uniformly. Conical areas show rapid changes of the radial and tangential curvatures; therefore, a quick change of the prestress cannot be avoided (Figure 6 and the conical areas in Figure 10).
Image 07 - Arch-Supported Membrane
Image 08 - Vectors of Principal Internal Forces n
Image 09 - Membrane Structure
Two more complex structures (Figure 7 and Figure 9) and their prestresses (Figure 8 and Figure 10) are shown at the end of this chapter. In order to achieve the most accurate results possible in the form-finding process as well as in the structural analysis, the structure should be modeled as a whole and not separated into parts. Thus, the interaction of all parts of the structure and the force redistribution due to the deformations are considered.
Cutting Membrane Structures
The process of determining cutting patterns is explained in the following text. It describes the individual steps of the process, then presents a practical example to show how the material properties can affect the shapes of the cutting patterns.
As mentioned, the double curvature is one of the typical features of membrane structures, which is why its shape cannot be developed in one plane. However, the membranes are made of rolls of planar fabrics. For this, a cutting (that is, individual planar cutting patterns) must be generated that approximates their corresponding patterns in space. The process of creating a cutting pattern consists of two steps. First, the membrane structure is divided into individual spatial cutting patterns by means of cutting lines; then, the best possible approximation of planar cutting patterns to the spatial one is found.
Theoretically, a membrane structure can be divided into partial strips by any cutting line. For practical reasons, however, geodetic cutting lines are usually used (Figure 11, left), which are preferred because of the straight axis of the cutting patterns after flattening (Figure 12, left). Plane sections (Figure 11, right) that are not straight after flattening (Figure 12, right) are used less often, and this results in a higher material requirement.
Image 11 - Hyperbolic Paraboloid Divided by Geodesic Section Lines (Left) and Plane Sections (Right)
Image 12 - Cutting Patterns Created by Means of Geodetic Cutting Lines (Left) and Plane Sections (Right)
The second step of creating a cutting pattern is much more complex: finding the best possible approximation of a planar cutting pattern to the corresponding spatial cutting pattern. For this process, a number of methods were designed; the historically oldest used a simplified geometric method and the later methods, advanced mathematical mapping. The current methods are based on continuum mechanics, with a nonlinear analysis using the finite element method (FEM) for the determination of the cutting pattern.
The latter method is considered to be the most general solution for an approximation problem and allows you to consider the material properties of the fabric or film used. If you do not want to consider the orthotropic properties of the textile material or the transverse contraction, you can apply an isotropic material with Poisson's ratio v = 0. However, if the material properties are to be included in the process of flattening the cutting pattern, the optimal shape of the cutting pattern can be achieved.
When testing the textile materials used for membrane structures, you usually determine the stiffnesses in the warp and weft directions and Poisson's ratio. The shear stiffness is usually neglected. The following example shows how the shear stiffness affects the shape of the resulting cutting pattern. For the example, we have selected one of the middle cutting patterns of the hyperbolic paraboloid (Figure 11). Two different materials are used for the cutting pattern.
The following values are provided for the first surface-treated fabric:
Ewarp = 1600 kN/m
Eweft = 1200 kN/m
vWarp/weft = 0.05
G = 400 kN/m
The other material, a textile mesh without surface treatment, has the following values:
G = 10 kN/m
The following figure shows the resulting plane cutting patterns. By moving the centers of gravity of both cutting patterns into the same point and enlarging the right part of the cutting patterns in the cutout (Figure 14), the difference between both shapes becomes clear. If you consider the material properties, you can achieve better quality cutting patterns. After assembling the structure, the real prestress is closer to the intended prestress.
Image 13 - Cutting Patterns for Fabric with Surface Treatment (Top) and for Textile Mesh Without Surface Treatment (Bottom)
Image 14 - Different Shape of Cutting Patterns when Using Different Materials
For determining the cutting patterns, a compensation is also used, which is determined by biaxial tests and simulates the dissolution of the prestress in the fabric.
A nonlinear calculation according to the finite element method provides an energetically optimal planar cutting pattern in relation to the spatial one. Since it is based on physical principles, this calculation method is the most natural.
In the process of creating a cutting pattern, you can also consider other design requirements. Mainly, maintaining equal lengths of the adjacent edges of adjacent cutting patterns is required. Often, the application of different compensation is required for some edges of cutting patterns. This is often referred to as decompensation of the edges. In compliance with these design requirements and using the nonlinear analysis, an energetically optimized cutting pattern is found.
The aim of this article was to explain the main processes involved in planning membrane structures. The physical principles should be explained and the individual theses illustrated by examples. The examples were developed in the RFEM program by Dlubal Software GmbH [2].
This article was created with the support of the FAST-J-15-2803 project.
[1] Otto, F.; Rasch, B.: Finding Form: Towards an Architecture of the Minimal. Fellbach: Edition Axel Menges, 1996
[2] Forster, B.; Mollaert, M.: European Design Guide for Tensile Surface Structures. Brussels: TensiNet, 2004
[3] Veenendaal, D.; Block, P.: An Overview and Comparison of Structural Form Finding Methods for General Networks, International Journal of Solids and Space Structures 49, pp. 3741 - 3753. Amsterdam: Elsevier, 2012
[4] Architen Landrell: Basic Theories of Tensile Fabric Architecture
[5] Bletzinger, K.-U.; Ramm, E.: A General Finite Element Approach to the Form-Finding of Tensile Structures by the Updated Reference Strategy, International Journal of Solids and Space Structures 14, pp. 131 - 146. Amsterdam: Elsevier, 1999
[6] Wüchner, R.; Bletzinger, K.-U.: Stress‐Adapted Numerical Form Finding of Pre‐Stressed Surfaces by the Updated Reference Strategy, International Journal for Numerical Methods in Engineering 64, pp. 143 - 166. Amsterdam: Elsevier, 2005
[7] Němec, I. et al.: Finite Element Analysis of Structures: Principles and Praxis. Aachen: Shaker, 2010
[8] Moncrieff, E.; Topping, B.-H.-V.: Computer Methods for the Generation of Membrane Cutting Patterns, Computers and Structures 37, pp. 441 - 450. Amsterdam: Elsevier, 1990
[9] Bletzinger, K.-U.; Linhard, J.; Wüchner, R.: Advanced Numerical Methods for the Form Finding and Patterning of Membrane Structures, CISM International Centre for Mechanical Sciences 519, pp. 133 - 154. Berlin: Springer, 2010
Ing. Rostislav Lang
doc. Ing. Ivan Němec, CSc.
Ing. Hynek Štekbauser
Institute of Mechanical Engineering, FAST VUT v Brně (College of Civil Engineering, Technical University of Brno), FEM consulting Brno
Prof. Ing. Jiří Studnička, DrSc., ČVUT v Praze (Czech Technical University in Prague)
Form-finding Cutting pattern Prestress Membrane Cable structure Warp Weft Cutting Textile construction Textile Force density
Cutting Pattern of Membranes and Cable Elements Form-Finding in RFEM New Wind Load Application Using RWIND 2 New Saving Models as Blocks in RFEM 6 |
Use Cramer’s Rule to solve (if possible) the system of linear equations. 4x-y-z
Use Cramer’s Rule to solve (if possible) the system of linear equations. 4x-y-z=1 2x+2y+3z=10 5x-2y-2z=-1
Use Cramer’s Rule to solve (if possible) the system of linear equations.
5x-2y-2z=-1
Given system of equations are
Firstly, we will find determinant of coefficient matrix.
D=\left[\begin{array}{ccc}4& -1& -1\\ 2& 2& 3\\ 5& -2& -2\end{array}\right]=4|\begin{array}{cc}2& 3\\ -2& -2\end{array}|-\left(-1\right)|\begin{array}{cc}2& 3\\ 5& -2\end{array}|-1|\begin{array}{cc}2& 2\\ 5& -2\end{array}|
D=4(-4+6)+(-4-15)-(-4-10)=8-19+14=3
Now, we will find the following determinants.
{D}_{1}=\left[\begin{array}{ccc}1& -1& -1\\ 10& 2& 3\\ -1& -2& -2\end{array}\right]=1|\begin{array}{cc}2& 3\\ -2& -2\end{array}|-\left(-1\right)|\begin{array}{cc}10& 3\\ -1& -2\end{array}|-1|\begin{array}{cc}10& 2\\ -1& -2\end{array}|
{D}_{1}=\left(-4+6\right)+\left(-20+3\right)-\left(-20+2\right)=2-17+18=3
{D}_{2}=\left[\begin{array}{ccc}4& 1& -1\\ 2& 10& 3\\ 5& -1& -2\end{array}\right]=4|\begin{array}{cc}10& 3\\ -1& -2\end{array}|-1|\begin{array}{cc}2& 3\\ 5& -2\end{array}|-1|\begin{array}{cc}2& 10\\ 5& -1\end{array}|
{D}_{2}=4\left(-20+3\right)-\left(-4-15\right)-\left(-2-50\right)=-68+19+52=3
{D}_{3}=\left[\begin{array}{ccc}4& -1& 1\\ 2& 2& 10\\ 5& -2& -1\end{array}\right]=4|\begin{array}{cc}2& 10\\ -2& -1\end{array}|-\left(-1\right)|\begin{array}{cc}2& 10\\ 5& -1\end{array}|+1|\begin{array}{cc}2& 2\\ 5& -2\end{array}|
{D}_{3}=4\left(-2+20\right)+\left(-2-50\right)+\left(-4-10\right)=6
Now, solution of given system of equations is
x=\frac{{D}_{1}}{D},y=\frac{{D}_{2}}{D},z=\frac{{D}_{3}}{D}
Result: Solution of given system of equations is
A=\left[\begin{array}{cc}3& 1\\ 1& 1\\ 1& 4\end{array}\right],b\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]
\stackrel{―}{x}=
Eliminate the parameter to express the following parametric equations as a single equation in x and y.
x=2\mathrm{sin}8t,y=2\mathrm{cos}8t
Solve the system of equations using Gaussian elimination
-x+2y=-3
x\frac{dy}{dx}-4y=2{x}^{4}{e}^{x}
Write the equation of the line through each point. We need to use slope-intercept form. (1, -1), parallel to
y=\frac{2}{5x-3}
Find the discriminant of each equation and determine whether the equation has (1) two nonreal complex solutions, (2) one real solution with a multiplicity of 2, or (3) two real solutions. Do not solve the equations.
7{x}^{2}-2x-14=0
Replace the Cartesian equation with equivalent polar equations.
{x}^{2}+\left(y-2{\right)}^{2}=4 |
CreatePDF - Maple Help
Home : Support : Online Help : Programming : eBookTools Package : CreatePDF
convert Maple worksheets to PDF book
CreatePDF(book, settings)
The CreatePDF command transforms Maple worksheets into PDF format.
\mathrm{with}\left(\mathrm{eBookTools}\right):
\mathrm{book}≔\mathrm{NewBook}\left("eBookSample","eBook Sample Book","Maplesoft, a division of Waterloo Maple Inc.","2012"\right):
\mathrm{AddChapter}\left(\mathrm{book},"legal",\mathrm{cat}\left(\mathrm{kernelopts}\left('\mathrm{datadir}'\right),"/eBookTools/Legal.mw"\right)\right):
\mathrm{AddChapter}\left(\mathrm{book},"preface",\mathrm{cat}\left(\mathrm{kernelopts}\left('\mathrm{datadir}'\right),"/eBookTools/Preface.mw"\right)\right):
\mathrm{AddChapter}\left(\mathrm{book},1,\mathrm{cat}\left(\mathrm{kernelopts}\left('\mathrm{datadir}'\right),"/eBookTools/GettingStartedWithMaple.mw"\right)\right):
\mathrm{CreatePDF}\left(\mathrm{book}\right)
The eBookTools[CreatePDF] command was introduced in Maple 16. |
Insert one of the symbols << of >> to make the following statements true. a) If x >>, the
Insert one of the symbols << of >> to make the following statements true. a) If x >>, then x + 8 ___ 3. b) If x >> -3, then -5x + 18 ___ 3.
Insert one of the symbols << of >> to make the following statements true.
x⟩
, then x + 8 ___ 3.
b) If
x⟩-3
, then -5x + 18 ___ 3.
We rewrite the inequality so that it matches the left side of the "then" part.
a)By adding 8 to each side, we have: x+8>5+8
So, insert>.
b)Multiply both sides by -5. Recall that when we multyply of divide both sides of an inequality by a negative number, we change the direction of the inequality: -5x<-15
Add 18 to both sides: -5x+18<3
So, insert <.
A bag contains 2 red checkers and 6 black checkers. A checker is selected, kept out of the bag, and then another checker is selected. What is P(black, then red)?
Sleep: Mean 7.35 and Range
5.5-8.8
Light physical activity: Mean 8.71 and Range
3.0-16.0
Moderate physical activity: Mean 3.36 and Range
0.71-8.3
Hard physical activity: Mean 0.78 and Range
0-4.1
Very hard physical activity: Mean 0.14 and Range
0-1.05
On the average. how many hours per day did the women sleep?
Let's say that I have two ordered sets of numbers
\left\{1,2\right\}
\left\{3,4\right\}
. I'm trying to figure out the number of possible ways to combine these two sets into one without breaking the ordering of the two sets.
So for instance,
\left\{1,2,3,4\right\}
\left\{3,4,1,2\right\}
\left\{1,3,2,4\right\}
are valid combinations, but
\left\{2,1,4,3\right\}
isn't. How do I figure out the number of valid combinations? This feels like something I should remember from college, but I'm drawing a blank. It feels somewhere in between a combination and a permutation.
There are 30 students in a class of Mr. Obniarillo. 18 are boys and 12 are girls. What is the probability of
choosing a random student?
p(x|y)? |
Which of the following is not sp2 hybridised?
Solid CO2 is dry ice in which carbon atomundergoes sp-hybridisation.
Which is correct regarding size of atom?
B < Ne
V > Ti
Na > K
The atomic radii of noble gases are by far the largest in their respective periods. This is due to the reason that noble gases have only van der Waals radii.
Choose the correctly paired gaseous cation and its magnetic (spin only) moment (in B.M.)
Ti2+,3.87 B.M.
Cr2+, 4.90 B.M.
Co3+,3.87 B.M.
Mn2+,4.90 B.M.
Using expression, µ =
\sqrt{n\left(n+2}
B.M. (where, n = no. of unpaired electrons)
Ion Outer configuration n
\mu
{}_{22}\mathrm{Ti}^{2+}
3d2 2 2.84
{}_{24}C{r}^{2+}
{}_{27}C{o}^{3+}
{}_{22}M{n}^{2+}
Which ofthe following statements is incorrect?
Li+ has minimum degree of hydration.
The oxidation state of K in KO2 is + 1.
Na is used to make a Na/Pb alloy.
MgSO4 is readily soluble in water.
The hydration enthalpies of alkali metal ions decreases with increase in ionic sizes Hence, the order is Li+ > Na+> K+ > Rb+ > Cs+.Therefore, Li+ has maximum degree of hydration.
Which of the following represents the correct bond order?
{O}_{2}^{+}<{O}_{2}^{-}>{O}_{2}^{2-}
{O}_{2}^{-}>{O}_{2}^{2-}>{O}_{2}^{+}
{O}_{2}^{2-}>{O}_{2}^{+}>{O}_{2}^{-}
{O}_{2}^{+}>{O}_{2}^{-}>{O}_{2}^{2-}
{O}_{2}^{+}>{O}_{2}^{-}>{O}_{2}^{2-}
Hence, the coorect Bond order is
{\text{O}}_{2}^{+}>{O}_{2}^{-}>{O}_{2}^{2-}
Total no of electron
MO configuration Bond order
{o}_{2}^{+}
{\text{KK σ2s}}^{2}\quad {\sigma }^{*}2{s}^{2\quad }\quad \sigma 2{p}_{x}^{2}\quad \pi 2{p}_{x}^{2}=\pi 2{p}_{y}^{2}{\pi }^{*}2{p}_{x}^{1}={\pi }^{*}2{p}_{y}^{0}
{o}_{2}^{-}
{\text{KK α2s}}^{2}{\sigma }^{*}2{s}^{2}\sigma 2{p}_{z}^{2}\pi 2{p}_{x}^{2}=\pi 2{p}_{y}^{2}{\pi }^{*}2{p}_{x}^{2}={\pi }^{*}2{p}_{y}^{1}
{o}_{2}^{2-}
{\text{KKσ2s}}^{2}\quad {\sigma }^{*}2{s}^{2}\quad \sigma 2{p}_{z}^{2}\quad \pi 2{p}_{x}^{2}=\pi 2{p}_{y}^{2}\quad {\pi }^{*}2{p}_{x}^{2}={\pi }^{*}2{p}_{y}^{2}
In O3 molecule, the formal charge on the central O-atom is
Lewis gave the structure of O, molecule as
Formal charge = [Total no. of valence electrons in the free atom] - [Total no. of non-bonding (lone pair) electrons]-
\frac{1}{2}
[Total no. of bonding (shared) electrons]
The formal charge on central O -atom i.e., no. 1
=6-2- -(6)=+1
A diatomic gas at pressure P, compressed adiabatically to half of its volume, what is the final pressure?
(2)1.4P
P/(2)1.4
(2)5/3P
P/(2)5/3
For adiabatic condition,
{\mathrm{PV}}^{\mathrm{\gamma }}
{\mathrm{P}}_{1}^{}
{\mathrm{V}}_{1}^{\mathrm{\gamma }}
{\mathrm{P}}_{2}
{\mathrm{V}}_{2}^{\mathrm{\gamma }}
; V2=
\frac{1}{2}
{\mathrm{P}}_{2}
{\mathrm{P}}_{1}
{\left[\frac{{\mathrm{V}}_{1}}{{\mathrm{V}}_{2}}\right]}^{\mathrm{\gamma }}
[For diantomic gas
\mathrm{\gamma }
= 1.4]
{\mathrm{P}}_{2}
{\mathrm{P}}_{1}
{\left(\frac{{\mathrm{V}}_{1\quad }\times 2}{{\mathrm{V}}_{1}}\right)}^{1.4}
{\text{P}}_{2}
{\text{P}}_{1}(2{)}^{1.4}\quad
(2{)}^{1.4\quad }\mathrm{P}
\frac{1}{2}
{{\text{H}}_{2}}_{\left(g\right)}
+
\frac{1}{2}
{\text{I}}_{{2}_{\left(g\right)}}
⇌
HI(g) is Kc
Equilibrium constant for the reaction 2HI(g)
⇌
H2(g) + I2(g) will be
1/Kc
1/(Kc)2
\frac{1}{2}{\mathrm{H}}_{{2}_{\left(\mathrm{g}\right)}}\quad +\frac{1}{2}{\mathrm{I}}_{{2}_{\left(\mathrm{g}\right)}}\quad \leftrightharpoons {\mathrm{HI}}_{\left(\mathrm{g}\right)}
{\mathrm{K}}_{\mathrm{C}}=\frac{\left[\mathrm{HI}\right]}{\left[{\mathrm{H}}_{2}{\right]}^{1/2}\left[{\mathrm{I}}_{2}{\right]}^{1/2}}
Now, reverse the equaion (i) and multiple by 2, we get
2\mathrm{HI}\quad \rightleftharpoons {\mathrm{H}}_{{2}_{\left(\mathrm{g}\right)}}+{\mathrm{I}}_{{2}_{\left(\mathrm{g}\right)}}
{}^{\quad \quad \quad \mathrm{Hence},\quad {\mathrm{K}}_{\mathrm{C}}^{\text{'}}=\frac{\left[{\mathrm{H}}_{2}\right]\quad \left[{\mathrm{I}}_{2}\right]}{[\mathrm{HI}{]}^{2}}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad }\quad
Equating equations (ii) and (iii), we get
{\text{K}}_{C}^{\text{'}}=\frac{1}{\left({K}_{C}{\right)}^{2}}
Which of the following pairs represent isotones?
{}_{33}{}^{77}\mathrm{As},{}_{34}{}^{\quad 78}\mathrm{Se}
{}_{78}{}^{195}\mathrm{Pt},{}_{76}{}^{190}\mathrm{Os}
{}_{47}{}^{108}\mathrm{Ag},{}_{48}{}^{112}\mathrm{Cd}
{}_{72}{}^{178}\mathrm{Hf},{}_{56}{}^{137}\mathrm{Ba}
{}_{33}{}^{77}\mathrm{As},{}_{34}{}^{\quad 78}\mathrm{Se}
Isotones have the same number of neutrons.
As = 77 - 33 = 44 ; Se = 78 - 34 = 44
What is the oxidation number of Br in KBrO4 ?
Let the oxidation no. of Br be x.
In KBrO4, + 1 + x + 4 (-2) =0,- 7 + x = 0, x=+7 |
You want to know whether people in different regions of the country are equally
You want to know whether people in different regions of the country are equally likely to vote Sarah Duterte, Peter Cayetano, Mar Roxas, or any candid
You want to know whether people in different regions of the country are equally likely to vote Sarah Duterte, Peter Cayetano, Mar Roxas, or any candidate other than the three in the next election. You would use
A. chi-square test of independence.
B. either chi-square test (goodness-of-fit or test of independence), depending on how you set up the problem.
C. chi-square goodness-of-fit test.
D. both chi-square tests, in order to check the results of one with the other.
Tuthornt
-Chi-Square Test of Association.
This test utilizes a contingency table to analyze the data. A contingency table (also known as a cross-tabulation, crosstab, or two-way table) is an arrangement in which data is classified according to two categorical variables.
The categories for one variable appear in the rows, and the categories for the other variable appear in columns. Each variable must have two or more categories. Each cell reflects the total count of cases for a specific pair of categories.
Chi-Square goodness of fit test is a non-parametric test that is used to find out how the observed value of a given phenomena is significantly different from the expected value. In Chi-Square goodness of fit test, the term goodness of fit is used to compare the observed sample distribution with the expected probability distribution. Chi-Square goodness of fit test determines how well theoretical distribution (such as normal, binomial, or Poisson) fits the empirical distribution. In Chi-Square goodness of fit test, sample data is divided into intervals. Then the numbers of points that fall into the interval are compared, with the expected numbers of points in each interval.
Since, we are to test whether people in different regions of the country are equally likely to vote Sarah Duterte, Peter Cayetano, Mar Roxas, or any candidate other than the three in the next election we will use Chi-Square goodness of fit test.
A Chi-Square for goodness of fit test is a test used to assess whether the observed data can be claimed to reasonably fit the expected data. Sometimes, a Chi-Square test for goodness of fit is referred as a test for multinomial experiments, because there is a fixed number of N categories, and each of the outcomes of the experiment falls in exactly one of those categories. Then, based on sample information, the test uses a Chi-Square statistic to assess if the expected proportions for all categories reasonably fit the sample data. The main properties of a one sample Chi-Square test for goodness of fit are:
- The distribution of the test statistic is the Chi-Square distribution, with n-1 degrees of freedom, where n is the number of categories
- The Chi-Square distribution is one of the most important distributions in statistics, together with the normal distribution and the F-distribution
{x}^{2}=\sum _{i=1}^{n}\frac{{\left({O}_{i}-{E}_{i}\right)}^{2}}{{E}_{i}}
One of the most common uses for this test is to assess whether a sample come from a population with a specific population(this is, for example, using this test we can assess if a sample comes from a normally distributed popelation or not).
What happens to the critical value for a chi-square test if the number of categories is increased?
a. The critical value increases.
b. The critical value decreases.
c. The critical value depends on the sample size, not the number of categories.
d. The critical value is determined entirely by the alpha level
You intend to conduct a test of homogeneity for a contingency table with 8 categories in the column variable and 2 categories in the row variable. You collect data from 352 subjects.
What are the degrees of freedom for the
{\chi }^{2}
distribution for this test?
Suppose I gave you a bag of M&M’s, but I didn’t let you see the original packaging so you can’t determine which plant made the M& M’s. Your job is to count the number of candies of each color in your bag and figure out which plant made your bag. What test should you do to determine this?
Suppose we wanted to see if there was a correlation between suspension and ethnicity. We have the following summary of data below. What statistical test would we use?
\begin{array}{|ccc|}\hline & \text{ Suspended }& \text{ Note Suspended }\\ \text{ Black }& 17& 17\\ \text{ White }& 65& 878\\ \hline\end{array}
For this study, why we would want to use Chi-Squared?
1.A study investigating the effects of second-hand smoke in working environments asked the following question: “How often do you experience second-hand smoke in a work environment/function? Never, Occasionally, Fairly Often, Very Often, Almost Always.” The question was asked of managers and employees to determine whether there was an association between position and the amount of second-hand smoke exposure.
A study compared women who viewed high levels of television violence as children with those who did not in order to study the differences with regard to physical abuse of their partners as adults. Use the table shown below to calculate the observed value of the chi-square statistic. The table shows both actual counts and expected counts (in parentheses).
\begin{array}{|ccc|}\hline & \text{ High TV Violence }& \text{ Low TV Violence }\\ \text{ Yes, Physical Abuse }& 14\left(8.42\right)& 26\left(31.58\right)\\ \text{ No Physical Abuse }& 22\left(27.58\right)& 109\left(103.42\right)\\ \hline\end{array}
{\chi }^{2}=\square
What is stated by the null hypothesis for the chi-square test for independence?
a.There is no relationship between the two populations regarding the two variables.
b.Both variables have the same frequency distribution.
c.The two variables have different frequency distributions.
d.There is a relationship between the two populations regarding the two variables. |
Cargo Tank Filling Limits on LNG and LPG tankers
Filling Limits for Cargo Tanks on Liquefied Gas Tankers
This article talks about Cargo Tank Filling Limits (higher filling limit – FL, and maximum loading limit – LL) on different types of gas carriers.
Information to be provided to the master
No cargo tanks shall have a higher tank filling limit (FL) than 98 % at the reference temperature, except as permitted beyond.
The maximum loading limit (LL) to which a cargo tank may be loaded shall be determined by the following formula:
LL=FL\left(\frac{{\rho }_{R}}{{\rho }_{L}}\right),
LL – loading limit expressed in percent which means the maximum allowable liquid volume relative to the tank volume to which the tank may be loaded;
FL – filling limit as specified above;
ρR – relative density of cargo at the reference temperature;
ρL – relative density of cargo at the loading temperature and pressure.
Inside in the tank on LNG tanker
The society may allow a greater filling limit (FL) than 98 % in beginning of this article at the reference temperature, taking into account the shape of the tank, arrangements of pressure relief valves, accuracy of level and temperature gauging and the difference between the loading temperature and the temperature corresponding to the vapour pressure of the cargo at the set pressure of the pressure relief valves, provided the conditions of “Cargo Temperature Control and Cargo Vent Systems”Cargo Vent Systems are maintained.
For the purpose of this article only, “reference temperature” means:
The temperature corresponding to the vapour pressure of the cargo at the set pressure relief valves when no cargo vapour pressure/temperature control as referred to in “Cargo Temperature Control and Cargo Vent Systems”Cargo temperature control and Vent system is provided.
The temperature of the cargo upon termination of loading, during transport, or at unloading, whichever is the greater, when a cargo vapour pressure/temperature control as referred to in “Cargo Temperature Control and Cargo Vent Systems”Cargo temperature control and Vent system is provided. If this reference temperature would result in the cargo tank becoming liquid full before the cargo reaches a temperature corresponding to the vapour pressure of the cargo at the set pressure of the relief valves required in “Cargo Temperature Control and Cargo Vent Systems”Pressure and Temperature Control, an additional pressure relieving system complying with “Cargo Temperature Control and Cargo Vent Systems”Additional pressure relieving system for liquid level contro is to be fitted.
The Society may allow type C tanks to be loaded according to the following formula provided that the tank vent system has been approved in accordance with “Cargo Temperature Control and Cargo Vent Systems”Cargo Vent Systems:
LL=FL\left(\frac{{\rho }_{R}}{{\rho }_{L}}\right),
ρR – relative density of cargo at the highest temperature which the cargo may reach upon termination of loading, during transport, or at unloading, under the ambient design temperature conditions described in “Cargo Temperature Control and Cargo Vent Systems”Cargo temperature control and Vent system;
This paragraph does not apply to products requiring a type 1G ship.
The maximum allowable loading limits for each cargo tank shall be indicated for each product which may be carried, for each loading temperature which may be applied and for the applicable maximum reference temperature, on a list to be approved by the Society. Pressures at which the pressure relief valves, including those valves required by “Cargo Temperature Control and Cargo Vent Systems”Additional pressure relieving system for liquid level contro, have been set shall be stated on the list. A copy of the list shall be permanently kept on board by the Master.
This article applies to all ships regardless of the date of construction.
Поскольку в программу подготовки судового персонала по использованию системы регулируемого выпуска паров должны быть включены вопросы устройства и при...
Природный газ, это смесь углеводородов, которая после сжижения образует чистую без цвета и запаха жидкость. Такой СПГ обычно транспортируется и хранит...
Горючие газы имеют высокую теплотворную способность. Теплотворная способность метана, например, в 1,6 раза больше, чем каменного угля, и в 1,2 раза бо... |
Simplifying Expressions Involving Radicals Simplify the expression and express t
Simplifying Expressions Involving Radicals Simplify the expression and express the answer using rational exponents. Assume that all letters denote positive numbers.NKSsqrt{x^{5}}x5
Simplifying Expressions Involving Radicals Simplify the expression and express the answer using rational exponents. Assume that all letters denote positive numbers.
\sqrt{{x}^{5}}{x}^{5}
"For any rational exponent
\frac{m}{n}\text{ }\frac{m}{n}
in lowest terms, where mm and nn are integers and
n>0,\text{ }n>0
{a}^{\frac{m}{n}}={\left(\sqrt{n}\left\{a\right\}\right)}^{m}=\sqrt{n}\left\{{a}^{m}\right\}a\frac{m}{n}\left(an\right)m=amn
If nn is even, then we require that
a\ge 0a\ge 0
\sqrt{{x}^{5}}x5
\sqrt{{x}^{5}}x5={\left({x}^{5}\right)}^{\frac{1}{2}}\left(x5\right)\frac{1}{2}
Apply Law of exponents
{a}^{\frac{m}{n}}={\left(\sqrt{n}\left\{a\right\}\right)}^{m}=\sqrt{n}\left\{{a}^{m}\right\}a\frac{m}{n}=\left(an\right)m=amn
{\left({x}^{5}\right)}^{\frac{1}{2}}={x}^{\frac{5}{2}}\left({x}^{5}\right)\frac{1}{2}={x}^{52}
Therefore the expression
\sqrt{{x}^{5}}x5
{x}^{\frac{5}{2}}{x}^{52}
The simplified value of the radical expression
\sqrt[3]{\frac{1}{2}}
To calculate: The restriction on the variable for rational expression
\frac{6c}{7{a}^{3}{b}^{2}}
The expression to triple the amount.
Simplify radical expression
\sqrt{\frac{2}{3a}}-2\sqrt{\frac{2}{3a}}
We need to calculate: the simplified form of
\left(3\sqrt[3]{{a}^{3}}\left(-5\sqrt[4]{{a}^{3}}\right)\right)
The expression with rational exponents as a radical expression.
5×x
to the one fourth power |
Descartes rule of sign can be used to isolate the intervals containing the real roots of a real poly
Unfortunately the answer is: subdivide and go on. The rule-of-signs-predicate is not able to tell whether there are any roots, and if you have no additional means to do so, you have no other option.
However, there are theorems which tell you that this won't happen too often. Short summary of the "easy" cases: If you count the sign variations v of the polynomial which "describes" the interval
\left[a,b\right]
v
will be 0 if no (complex) root of the input polynomial is contained in the disc in the complex plane with diameter
ab
v
will be 1 if only one (complex) root of the input is contained un the union of the circumcircles of the two equilateral triangles that have
ab
as one side, assuming that this root is simple.
There are more detailed versions of these theorems, but in a nutshell it boils down to: you will count the right thing unless there is a cluster of complex roots close to the interval (w.r.t. the scale/precision you are currently considering), and you have to "zoom in" to deblur and resolve the situation.
However, if you have reasons to believe that there is a wide range without any roots, note that "subdivide" is not necessarily the same as "bisect". You are free to choose other subdivision methods; this eventually leads to fancier algorithms like Continued Fraction solver (VAS; praised in practice, at least for some benchmarks, but their actual merit is disputed) or combinations of Newton iteration and Descartes
True or false: if
{a}_{n}
is any decreasing sequence of positive real numbers and
{b}_{n}
is any sequence of real numbers converges to
0
\frac{{a}_{n}}{{b}_{n}}
TP\text{ }\left(true\text{ }positive\right)\text{ }=\text{ }2739
TN\text{ }\left(true\text{ }negative\right)\text{ }=\text{ }103217
FP\text{ }\left(false\text{ }positive\right)\text{ }=\text{ }43423
FN\text{ }\left(false\text{ }negative\right)\text{ }=\text{ }5022
accuracy=\frac{TP+TN}{TP+TN+FP+FN}
In this case the accuracy is
0.68
. Can I say that I have low accuracy because the value false positive is high? There is any relathion between false positive and the parameters true positive or true negative? |
For the function f whose graph is given, state the following. (a) \lim_{
For the function f whose graph is given, state the following. 12110601381.jpg (a) \lim_{x \rightarrow \infty} f(x) b) \lim_{x \rightarrow -\infty} f(x) (c) \lim_{x \rightarrow 1} f(x) (d) \lim_{x \rightarrow 3} f(x) (e) the equations of the asymptotes Vertical:- ? Horizontal:-?
For the function f whose graph is given, state the following.
\underset{x\to \mathrm{\infty }}{lim}f\left(x\right)
\underset{x\to -\mathrm{\infty }}{lim}f\left(x\right)
\underset{x\to 1}{lim}f\left(x\right)
\underset{x\to 3}{lim}f\left(x\right)
(e) the equations of the asymptotes
Vertical:- ?
Horizontal:-?
Differentiate the following function respect to x
p={e}^{{x}^{2}\mathrm{sin}x}
g is related to one of the parent functions described in Section 1.6. Describe the sequence of transformations from f to g.
g\left(x\right)=\sqrt{\frac{1}{4}}x
Begin by graphing the standard quadratic functions.
f\left(x\right)={x}^{2}
Then use transformations of this graph to the given function.
r\left(x\right)=-{\left(x+1\right)}^{2}
f\left(x\right)={2}^{x}
Then use transformations of this graph to graph the given function. Be sure to graph and give equations of the asymptotes. Use the graphs to determine each functions
Graph the functions
y={5}^{x},y=-{5}^{x},y={5}^{x}+2,y={5}^{x}-2
y={10}^{x}
on the same screen. Compare
y=-{5}^{x},y={5}^{-x}
to the parent graph
y={5}^{x}
. Describe the transformations of the functions.
Describe the transformations that were applied to
y={x}^{4}
to get each of the following functions.
a\right)y=-25{\left(3\left(x+4\right)\right)}^{4}-60\text{ }\phantom{\rule{0ex}{0ex}}
b\right)y=8{\left(\frac{3}{4}x\right)}^{4}+43\text{ }
c\right)y={\left(-13x+26\right)}^{4}+13\text{ }
d\right)y=\frac{8}{11}{\left(-x\right)}^{4}-1
f\left(x\right)={2}^{x} |
The eccentricities of conic sections with one focus at the origin and the direct
The eccentricities of conic sections with one focus at the origin and the directrix corresponding and sketch a graph:displaystyle{e}={2}, directrix displaystyle{r}=-{2} sec{theta}.
e=2,
r=-2\mathrm{sec}\theta .
e=2,r=-2\mathrm{sec}\theta
e>1
, therefore the conic is a hyperbola.
Now consider the directrix,
r=-2\mathrm{sec}\theta
⇒r=-\frac{2}{\mathrm{cos}\theta }
⇒r\mathrm{cos}\theta =-2
⇒x=-2
Now comparing whith
x=-p,
p=2
Therefore equation is,
r=\frac{ep}{1-e\mathrm{cos}\theta }
⇒r=\frac{2\cdot 2}{1-2\mathrm{cos}\theta }
⇒r=\frac{4}{1-2\mathrm{cos}\theta }
Now the graph is,
How to find the position on ellipse (or hyperbola) arc if we know it's euclidean distance from given point and direction of movement?
Finding equation of a path in the plane
y=z
What is the easiest way to see that the path
\underset{―}{r}:\mathbb{R}\to {\mathbb{R}}^{3}:t↦\left(\mathrm{sin},t,\mathrm{cos},t,\mathrm{cos},t\right)
traces out an ellipse in the plane
y=z
I think first rotating
{\mathbb{R}}^{3}
\frac{\pi }{4}
about the x-axis will help but I am not sure how to proceed.
Graph the lines and conic sections
r=\frac{8}{4+\mathrm{cos}\theta }
Angle between normal vector of ellipse and the major-axis.
I am trying to derive the angle made between the major or x-axis and the normal vector of an ellipse of general shape
x=a\mathrm{cos}\left(t\right),y=b\mathrm{sin}\left(t\right)
with the parameter t reffering to Ellipse in polar coordinates. I need to solve it for any angle t. From standard reasoning I find the normal vector by its definition and checked it with the page on mathworld from wolfram and works well. Then since I know 2 points, namely a point ON the shape and a point on the normal vector I derive the angle of interest to be
\mathrm{tan}\left(\varphi \right)=\frac{a}{b}\mathrm{tan}\left(t\right)
Derivation. However this is very similar to the polar angle namely its simply the term a and b flipped. But when thinking about it I keep getting confused, am I correct or do I need the polar angle? If so where did I go wrong?
I also found Normal to Ellipse and Angle at Major Axis but this page confused me a bit, one idea I had was they use the polar angle vs the angle I am in need of
\left(\varphi \right)
then I would indeed get by combing
t={\mathrm{tan}}^{-1}\left(\frac{a}{b}\mathrm{tan}\left(\theta \right)\right)
\varphi ={\mathrm{tan}}^{-1}\left(\frac{a}{b}\mathrm{tan}\left(t\right)\right)
\varphi ={\mathrm{tan}}^{-1}\left(\frac{a}{b}\mathrm{tan}\left({\mathrm{tan}}^{-1}big\left(\frac{a}{b}\mathrm{tan}\theta big\right)\right)\right)={\mathrm{tan}}^{-1}\left(\frac{{a}^{2}}{{b}^{2}}\mathrm{tan}\theta \right)
My excuse for my rambling, I find these angles confusing...
r=\frac{4}{1+\mathrm{cos}\theta }
{x}^{2}\text{ }-\text{ }2xy\text{ }+\text{ }{y}^{2}\text{ }+\text{ }24x\text{ }-\text{ }8=0
Graph the conic section and make sure to label the coordinates in the graph. Give the standard form (SF) and the general form (GF) of the conic sections.
Center is at
\left(2,\text{ }-4\right).
the diameter's length is 6. The endpoints of the diameter is at
\left(-1,\text{ }-4\right)
\left(3,\text{ }6\right). |
Matrix Representation of Geometric Transformations - MATLAB & Simulink - MathWorks Switzerland
2-D Affine Transformations
2-D Projective Transformations
Create Composite 2-D Affine Transformations
Rotation Followed by Translation
You can use a geometric transformation matrix to perform a global transformation of an image. First, define a transformation matrix and use it to create a geometric transformation object. Then, apply a global transformation to an image by calling imwarp with the geometric transformation object. For an example, see Perform Simple 2-D Translation Transformation.
The table lists 2-D affine transformations with the transformation matrix used to define them. For 2-D affine transformations, the last column must contain [0 0 1] homogeneous coordinates.
Use any combination of 2-D transformation matrices to create an affine2d geometric transformation object. Use combinations of 2-D translation and rotation matrices to create a rigid2d geometric transformation object.
2-D Affine Transformation
Example (Original and Transformed Image)
tx specifies the displacement along the x axis
ty specifies the displacement along the y axis.
For more information about pixel coordinates, see Image Coordinate Systems.
sx specifies the scale factor along the x axis
sy specifies the scale factor along the y axis.
shx specifies the shear factor along the x axis
shy specifies the shear factor along the y axis.
q specifies the angle of rotation about the origin.
Projective transformation enables the plane of the image to tilt. Parallel lines can converge towards a vanishing point, creating the appearance of depth.
The transformation is a 3-by-3 matrix. Unlike affine transformations, there are no restrictions on the last column of the transformation matrix.
2-D Projective Transformation
\left[\begin{array}{ccc}1& 0& E\\ 0& 1& F\\ 0& 0& 1\end{array}\right]
E and F influence the vanishing point.
When E and F are large, the vanishing point comes closer to the origin and thus parallel lines appear to converge more quickly.
Note that when E and F are equal to 0, the transformation becomes an affine transformation.
Projective transformations are frequently used to register images that are out of alignment. If you have two images that you would like to align, first select control point pairs using cpselect. Then, fit a projective transformation matrix to control point pairs using fitgeotrans and setting the transformationType to 'projective'. This automatically creates a projective2d geometric transformation object. The transformation matrix is stored as a property in the projective2d object. The transformation can then be applied to other images using imwarp.
You can combine multiple transformations into a single matrix using matrix multiplication. The order of the matrix multiplication matters.
This example shows how to create a composite of 2-D translation and rotation transformations.
Create a checkerboard image that will undergo transformation. Also create a spatial reference object for the image.
cb = checkerboard(4,2);
cb_ref = imref2d(size(cb));
To illustrate the spatial position of the image, create a flat background image. Overlay the checkerboard over the background, highlighting the position of the checkerboard in green.
background = zeros(150);
imshowpair(cb,cb_ref,background,imref2d(size(background)))
Create a translation matrix, and store it as an affine2d geometric transformation object. This translation will shift the image horizontally by 100 pixels.
T = [1 0 0;0 1 0;100 0 1];
tform_t = affine2d(T);
Create a rotation matrix, and store it as an affine2d geometric transformation object. This translation will rotate the image 30 degrees clockwise about the origin.
R = [cosd(30) sind(30) 0;-sind(30) cosd(30) 0;0 0 1];
tform_r = affine2d(R);
Perform translation first and rotation second. In the multiplication of the transformation matrices, the translation matrix T is on the left, and the rotation matrix R is on the right.
TR = T*R;
tform_tr = affine2d(TR);
[out,out_ref] = imwarp(cb,cb_ref,tform_tr);
imshowpair(out,out_ref,background,imref2d(size(background)))
Reverse the order of the transformations: perform rotation first and translation second. In the multiplication of the transformation matrices, the rotation matrix R is on the left, and the translation matrix T is on the right.
RT = R*T;
tform_rt = affine2d(RT);
[out,out_ref] = imwarp(cb,cb_ref,tform_rt);
Notice how the spatial position of the transformed image is different than when translation was followed by rotation.
The following table lists the 3-D affine transformations with the transformation matrix used to define them. Note that in the 3-D case, there are multiple matrices, depending on how you want to rotate or shear the image. The last column must contain [0 0 0 1].
\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ {t}_{x}& {t}_{y}& {t}_{z}& 1\end{array}\right]
\left[\begin{array}{cccc}{s}_{x}& 0& 0& 0\\ 0& {s}_{y}& 0& 0\\ 0& 0& {s}_{z}& 0\\ 0& 0& 0& 1\end{array}\right]
x,y shear:
\begin{array}{l}x\text{'}=x+az\\ y\text{'}=y+bz\\ z\text{'}=z\end{array}
\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ a& b& 1& 0\\ 0& 0& 0& 1\end{array}\right]
x,z shear:
\begin{array}{l}x\text{'}=x+ay\\ y\text{'}=y\\ z\text{'}=z+cy\end{array}
\left[\begin{array}{cccc}1& 0& 0& 0\\ a& 1& c& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]
y, z shear:
\begin{array}{l}x\text{'}=x\\ y\text{'}=y+bx\\ z\text{'}=z+cx\end{array}
\left[\begin{array}{cccc}1& b& c& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]
About x axis:
\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& \mathrm{cos}\left(a\right)& \mathrm{sin}\left(a\right)& 0\\ 0& -\mathrm{sin}\left(a\right)& \mathrm{cos}\left(a\right)& 0\\ 0& 0& 0& 1\end{array}\right]
About y axis:
\left[\begin{array}{cccc}\mathrm{cos}\left(a\right)& 0& -\mathrm{sin}\left(a\right)& 0\\ 0& 1& 0& 0\\ \mathrm{sin}\left(a\right)& 0& \mathrm{cos}\left(a\right)& 0\\ 0& 0& 0& 1\end{array}\right]
About z axis:
\left[\begin{array}{cccc}\mathrm{cos}\left(a\right)& \mathrm{sin}\left(a\right)& 0& 0\\ -sin\left(a\right)& \mathrm{cos}\left(a\right)& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]
For N-D affine transformations, the last column must contain [zeros(N,1); 1]. imwarp does not support transformations of more than three dimensions.
imwarp | fitgeotrans | affine2d | affine3d | rigid2d | rigid3d | projective2d |
f(x) = \displaystyle{2+\frac{3}{1+(x+1)^2}}
Determine all critical values of [math]
f
. If there is more than one, enter the values as a comma-separated list.
Critical value(s) =
Construct a first derivative sign chart for [math]
f
and thus determine all intervals on which [math]
f
is increasing or decreasing. If there is more than one, enter the intervals as a comma-separated list. Use interval notation: for example, (-17,20) is the interval [math]
-17 < x < 20
, and (-inf, 40) is the interval [math]
x<40
Interval(s) where [math]
f
is increasing:
f
is decreasing:
Does [math]
f
have a global maximum? If so, enter its value. If not, enter DNE.
Global maximum =
\displaystyle{\lim_{x\rightarrow \infty}} f(x) =
\displaystyle{\lim_{x\rightarrow -\infty}} f(x) =
Explain why [math]
f(x) > 2
for every value of [math]
x
If you were logged into a WeBWorK course and this problem were assigned to you, you would be able to submit an essay answer that would be graded later by a human being.
f
have a global minimum? If so, enter its value. If not, enter DNE.
Global minimum = |
Classify Videos Using Deep Learning with Custom Training Loop - MATLAB & Simulink - MathWorks Deutschland
Create Sequence Classification Network
Train Sequence Classification Network
You can perform video classification without using a custom training loop by using the trainNetwork function. For an example, see Classify Videos Using Deep Learning. However, If trainingOptions does not provide the options you need (for example, a custom learning rate schedule), then you can define your own custom training loop as shown in this example.
Train a sequence classification network on the sequences to predict the video labels.
The following diagram illustrates the network architecture:
To extract features from the image sequences, use convolutional layers from the pretrained GoogLeNet network.
To classify the resulting vector sequences, include the sequence classification layers.
When training this type of network with the trainNetwork function (not done in this example), you must use sequence folding and unfolding layers to process the video frames independently. When you train this type of network with a dlnetwork object and a custom training loop (as in this example), sequence folding and unfolding layers are not required because the network uses dimension information given by the dlarray dimension labels.
After extracting the RAR file, make sure that the folder hmdb51_org contains subfolders named after the body motions. If it contains RAR files, you need to extract them as well. Use the supporting function hmdb51Files to get the file names and the labels of the videos. To speed up training at the cost of accuracy, specify a fraction in the range [0 1] to read only a random subset of files from the database. If the fraction input argument is not specified, the function hmdb51Files reads the full dataset without changing the order of the files.
[files,labels] = hmdb51Files(dataFolder,fraction);
Read the first video using the readVideo helper function, defined at the end of this example, and view the size of the video. The video is an H-by-W-by-C-by-T array, where H, W, C, and T are the height, width, number of channels, and number of frames of the video, respectively.
shoot_ball
To view the video, loop over the individual frames and use the image function. Alternatively, you can use the implay function (requires Image Processing Toolbox).
Use the convolutional network as a feature extractor: input video frames to the network and extract the activations. Convert the videos to sequences of feature vectors, where the feature vectors are the output of the activations function on the last pooling layer of the GoogLeNet network ("pool5-7x7_s1").
Read the video data using the readVideo function, defined at the end of this example, and resize it to match the input size of the GoogLeNet network. Note that this step can take a long time to run. After converting the videos to sequences, save the sequences and corresponding labels in a MAT file in the tempdir folder. If the MAT file already exists, then load the sequences and labels from the MAT file directly. In case a MAT file already exists but you want to overwrite it, set the variable overwriteSequences to true.
overwriteSequences = false;
if exist(tempFile,'file') && ~overwriteSequences
load(tempFile)
video = imresize(video,inputSize);
% Save the sequences and the labels associated with them.
save(tempFile,"sequences","labels","-v7.3");
View the sizes of the first few sequences. Each sequence is a D-by-T array, where D is the number of features (the output size of the pooling layer) and T is the number of frames of the video.
{1024×40 single}
Create Datastore for Data
Create an arrayDatastore object for the sequences and the labels, and then combine them into a single datastore.
dsXTrain = arrayDatastore(sequencesTrain,'OutputType','same');
dsYTrain = arrayDatastore(labelsTrain,'OutputType','cell');
Determine the classes in the training data.
Next, create a sequence classification network that can classify the sequences of feature vectors representing the videos.
Define the sequence classification network architecture. Specify the following network layers:
A sequence input layer with an input size corresponding to the feature dimension of the feature vectors.
A BiLSTM layer with 2000 hidden units with a dropout layer afterwards. To output only one label for each sequence, set the 'OutputMode' option of the BiLSTM layer to 'last'.
A dropout layer with a probability of 0.5.
A fully connected layer with an output size corresponding to the number of classes and a softmax layer.
Convert the layers to a layerGraph object.
Train for 15 epochs and specify a mini-batch size of 16.
Specify the options for Adam optimization. Specify an initial learning rate of 1e-4 with a decay of 0.001, a gradient decay of 0.9, and a squared gradient decay of 0.999.
decay = 0.001;
Create a minibatchqueue object that processes and manages mini-batches of sequences during training. For each mini-batch:
Use the custom mini-batch preprocessing function preprocessLabeledSequences (defined at the end of this example) to convert the labels to dummy variables.
Format the vector sequence data with the dimension labels 'CTB' (channel, time, batch). By default, the minibatchqueue object converts the data to dlarray objects with underlying type single. Do not add a format to the class labels.
Train on a GPU if one is available. By default, the minibatchqueue object converts each output to a gpuArray object if a GPU is available. Using a GPU requires Parallel Computing Toolbox™ and a supported GPU device. For information on supported devices, see GPU Support by Release (Parallel Computing Toolbox).
'MiniBatchFcn', @preprocessLabeledSequences,...
'MiniBatchFormat',{'CTB',''});
Initialize the average gradient and average squared gradient parameters for the Adam solver.
Train the model using a custom training loop. For each epoch, shuffle the data and loop over mini-batches of data. For each mini-batch:
Evaluate the model gradients, state, and loss using dlfeval and the modelGradients function and update the network state.
Determine the learning rate for the time-based decay learning rate schedule: for each iteration, the solver uses the learning rate given by
{\rho }_{\mathit{t}}=\frac{{\rho }_{0}}{1+\mathit{k}\text{\hspace{0.17em}}\mathit{t}}
{\rho }_{0}
Note that training can take a long time to run.
[dlX, dlY] = next(mbq);
[gradients,state,loss] = dlfeval(@modelGradients,dlnet,dlX,dlY);
[dlnet,averageGrad,averageSqGrad] = adamupdate(dlnet,gradients,averageGrad,averageSqGrad, ...
iteration,learnRate,gradDecay,sqGradDecay);
title("Epoch: " + epoch + " of " + numEpochs + ", Elapsed: " + string(D))
After training is complete, making predictions on new data does not require the labels.
To create a minibatchqueue object for testing:
Create an array datastore containing only the predictors of the test data.
Preprocess the predictors using the preprocessUnlabeledSequences helper function, listed at the end of the example.
For the single output of the datastore, specify the mini-batch format 'CTB' (channel, time, batch).
dsXValidation = arrayDatastore(sequencesValidation,'OutputType','same');
mbqTest = minibatchqueue(dsXValidation, ...
'MiniBatchFcn',@preprocessUnlabeledSequences, ...
'MiniBatchFormat','CTB');
Loop over the mini-batches and classify the images using the modelPredictions helper function, listed at the end of the example.
predictions = modelPredictions(dlnet,mbqTest,classes);
Evaluate the classification accuracy by comparing the predicted labels to the true validation labels.
accuracy = mean(predictions == labelsValidation)
To create a network that classifies videos directly, assemble a network using layers from both of the created networks. Use the layers from the convolutional network to transform the videos into vector sequences and the layers from the sequence classification network to classify the vector sequences.
To use convolutional layers to extract features, that is, to apply the convolutional operations to each frame of the videos independently, use the GoogLeNet convolutional layers.
When training this type of network with the trainNetwork function (not done in this example), you have to use sequence folding and unfolding layers to process the video frames independently. When training this type of network with a dlnetwork object and a custom training loop (as in this example), sequence folding and unfolding layers are not required because the network uses dimension information given by the dlarray dimension labels.
Add the sequence input layer to the layer graph. Connect the output of the input layer to the input of the first convolutional layer ("conv1-7x7_s2").
lgraph = addLayers(cnnLayers,inputLayer);
lgraph = connectLayers(lgraph,"input","conv1-7x7_s2");
Add Sequence Classification Layers
Add the previously trained sequence classification network layers to the layer graph and connect them.
Take the layers from the sequence classification network and remove the sequence input layer.
lstmLayers = dlnet.Layers;
Add the sequence classification layers to the layer graph. Connect the last convolutional layer pool5-7x7_s1 to the bilstm layer.
lgraph = addLayers(lgraph,lstmLayers);
lgraph = connectLayers(lgraph,"pool5-7x7_s1","bilstm");
Convert to dlnetwork
To be able to do predictions, convert the layer graph to a dlnetwork object.
dlnetAssembled = dlnetwork(lgraph)
dlnetAssembled =
Unzip the file pushup_mathworker.zip.
unzip("pushup_mathworker.zip")
The extracted pushup_mathworker folder contains a video of a push-up. Create a file datastore for this folder. Use a custom read function to read the videos.
ds = fileDatastore("pushup_mathworker", ...
'ReadFcn',@readVideo);
Read the first video from the datastore. To be able to read the video again, reset the datastore.
video = read(ds);
To preprocess the videos to have the input size expected by the network, use the transform function and apply the imresize function to each image in the datastore.
dsXTest = transform(ds,@(x) imresize(x,inputSize));
To manage and process the unlabeled videos, create a minibatchqueue:
Specify a mini-batch size of 1.
Preprocess the videos using the preprocessUnlabeledVideos helper function, listed at the end of the example.
For the single output of the datastore, specify the mini-batch format 'SSCTB' (spatial, spatial, channel, time, batch).
mbqTest = minibatchqueue(dsXTest,...
'MiniBatchFcn', @preprocessUnlabeledVideos,...
'MiniBatchFormat',{'SSCTB'});
Classify the videos using the modelPredictions helper function, defined at the end of this example. The function expects three inputs: a dlnetwork object, a minibatchqueue object, and a cell array containing the network classes.
[predictions] = modelPredictions(dlnetAssembled,mbqTest,classes)
predictions = categorical
Video Reading Function
The readVideo function reads the video in filename and returns an H-by-W-by-C-by-T array, where H, W, C, and T are the height, width, number of channels, and number of frames of the video, respectively.
video = zeros(H,W,C,numFrames,'uint8');
The modelGradients function takes as input a dlnetwork object dlnet and a mini-batch of input data dlX with corresponding labels Y, and returns the gradients of the loss with respect to the learnable parameters in dlnet, the network state, and the loss. To compute the gradients automatically, use the dlgradient function.
function [gradients,state,loss] = modelGradients(dlnet,dlX,Y)
The modelPredictions function takes as input a dlnetwork object dlnet, a minibatchqueue object of input data mbq, and the network classes, and computes the model predictions by iterating over all data in the mini-batch queue. The function uses the onehotdecode function to find the predicted class with the highest score. The function returns the predicted labels.
function [predictions] = modelPredictions(dlnet,mbq,classes)
% Extract a mini-batch from the minibatchqueue and pass it to the
% network for predictions
[dlXTest] = next(mbq);
dlYPred = predict(dlnet,dlXTest);
% To obtain categorical labels, one-hot decode the predictions
YPred = onehotdecode(dlYPred,classes,1)';
Labeled Sequence Data Preprocessing Function
The preprocessLabeledSequences function preprocesses the sequence data using the following steps:
Use the padsequences function to pad the sequences in the time dimension and concatenate them in the batch dimension.
Extract the label data from the incoming cell array and concatenate into a categorical array.
One-hot encode the categorical labels into numeric arrays.
Transpose the array of one-hot encoded labels to match the shape of the network output.
function [X, Y] = preprocessLabeledSequences(XCell,YCell)
% Pad the sequences with zeros in the second dimension (time) and concatenate along the third
% dimension (batch)
X = padsequences(XCell,2);
% Transpose the encoded labels to match the network output
Unlabeled Sequence Data Preprocessing Function
The preprocessUnlabeledSequences function preprocesses the sequence data using the padsequences function. This function pads the sequences with zeros in the time dimension and concatenates the result in the batch dimension.
function [X] = preprocessUnlabeledSequences(XCell)
Unlabeled Video Data Preprocessing Function
The preprocessUnlabeledVideos function preprocesses unlabeled video data using the padsequences function. This function pads the videos with zero in the time dimension and concatenates the result in the batch dimension.
function [X] = preprocessUnlabeledVideos(XCell)
% Pad the sequences with zeros in the fourth dimension (time) and
% concatenate along the fifth dimension (batch)
lstmLayer | sequenceInputLayer | dlfeval | dlgradient | dlarray |
Since classical times, it has been known that some materials, such as amber, attract lightweight particles after rubbing. The Greek word for amber, ἤλεκτρον (ḗlektron), was thus the source of the word 'electricity'. Electrostatic phenomena arise from the forces that electric charges exert on each other. Such forces are described by Coulomb's law.
Coulomb's lawEdit
{\displaystyle r}
{\displaystyle q}
{\displaystyle Q}
{\displaystyle F={\frac {1}{4\pi \varepsilon _{0}}}{\frac {qQ}{r^{2}}}=k_{0}{\frac {qQ}{r^{2}}}\,,}
{\displaystyle \varepsilon _{0}\approx 8.854\ 187\ 817\times 10^{-12}\;\;\mathrm {C^{2}\ N^{-1}\ m^{-2}} .}
{\displaystyle k_{0}={\frac {1}{4\pi \varepsilon _{0}}}\approx 8.987\ 551\ 792\times 10^{9}\;\;\mathrm {N\ m^{2}\ C} ^{-2}.}
{\displaystyle e=1.602\ 176\ 634\times 10^{-19}\;\;\mathrm {C} .}
{\displaystyle {\vec {E}}}
, in units of newtons per coulomb or volts per meter, is a vector field that can be defined everywhere, except at the location of point charges (where it diverges to infinity).[6] It is defined as the electrostatic force
{\displaystyle {\vec {F}}\,}
{\displaystyle q\,}
{\displaystyle {\vec {E}}={{\vec {F}} \over q}}
{\displaystyle N}
{\displaystyle Q_{i}}
{\displaystyle {\vec {r}}_{i}}
{\displaystyle {\vec {r}}}
(called the field point) is:[6]
{\displaystyle {\vec {E}}({\vec {r}})={\frac {1}{4\pi \varepsilon _{0}}}\sum _{i=1}^{N}{\frac {{\widehat {\mathcal {R}}}_{i}Q_{i}}{\left\|{\mathcal {\vec {R}}}_{i}\right\|^{2}}},}
{\displaystyle {\vec {\mathcal {R}}}_{i}={\vec {r}}-{\vec {r}}_{i},}
{\displaystyle {\vec {r}}_{i}}
{\displaystyle {\vec {r}}}
{\displaystyle {\widehat {\mathcal {R}}}_{i}={\vec {\mathcal {R}}}_{i}/\left\|{\vec {\mathcal {R}}}_{i}\right\|}
{\displaystyle E=k_{e}Q/{\mathcal {R}}^{2},}
{\displaystyle \rho ({\vec {r}})}
{\displaystyle {\vec {E}}({\vec {r}})={\frac {1}{4\pi \varepsilon _{0}}}\iiint {\frac {{\vec {r}}-{\vec {r}}\,'}{\left\|{\vec {r}}-{\vec {r}}\,'\right\|^{3}}}\rho ({\vec {r}}\,')\,\mathrm {d} ^{3}r\,'}
Gauss' lawEdit
{\displaystyle \oint _{S}{\vec {E}}\cdot \mathrm {d} {\vec {A}}={\frac {1}{\varepsilon _{0}}}\,Q_{\text{enclosed}}=\int _{V}{\rho \over \varepsilon _{0}}\cdot \mathrm {d} ^{3}r,}
{\displaystyle \mathrm {d} ^{3}r=\mathrm {d} x\ \mathrm {d} y\ \mathrm {d} z}
{\displaystyle \rho \,\mathrm {d} ^{3}r}
{\displaystyle \sigma \,\mathrm {d} A}
{\displaystyle \lambda \,\mathrm {d} \ell }
{\displaystyle {\vec {\nabla }}\cdot {\vec {E}}={\rho \over \varepsilon _{0}}.}
{\displaystyle {\vec {\nabla }}\cdot }
Poisson and Laplace equationsEdit
{\displaystyle {\nabla }^{2}\phi =-{\rho \over \varepsilon _{0}}.}
{\displaystyle {\nabla }^{2}\phi =0,}
Electrostatic approximationEdit
{\displaystyle {\vec {\nabla }}\times {\vec {E}}=0.}
{\displaystyle {\partial {\vec {B}} \over \partial t}=0.}
Electrostatic potentialEdit
{\displaystyle \phi }
{\displaystyle E}
{\displaystyle {\vec {E}}=-{\vec {\nabla }}\phi .}
{\displaystyle a}
{\displaystyle b}
{\displaystyle -\int _{a}^{b}{{\vec {E}}\cdot \mathrm {d} {\vec {\ell }}}=\phi ({\vec {b}})-\phi ({\vec {a}}).}
Electrostatic energyEdit
{\displaystyle U_{\mathrm {E} }^{\text{single}}}
{\displaystyle q_{n}{\vec {E}}\cdot \mathrm {d} {\vec {\ell }}}
{\displaystyle N}
{\displaystyle Q_{n}}
{\displaystyle {\vec {r}}_{i}}
{\displaystyle U_{\mathrm {E} }^{\text{single}}=q\phi ({\vec {r}})={\frac {q}{4\pi \varepsilon _{0}}}\sum _{i=1}^{N}{\frac {Q_{i}}{\left\|{\mathcal {{\vec {R}}_{i}}}\right\|}}}
{\displaystyle {\vec {\mathcal {R_{i}}}}={\vec {r}}-{\vec {r}}_{i}}
{\displaystyle Q_{i}}
{\displaystyle q}
{\displaystyle {\vec {r}}}
{\displaystyle \phi ({\vec {r}})}
{\displaystyle {\vec {r}}}
{\displaystyle k_{e}Q_{1}Q_{2}/r}
{\displaystyle U_{\mathrm {E} }^{\text{total}}={\frac {1}{4\pi \varepsilon _{0}}}\sum _{j=1}^{N}Q_{j}\sum _{i=1}^{j-1}{\frac {Q_{i}}{r_{ij}}}={\frac {1}{2}}\sum _{i=1}^{N}Q_{i}\phi _{i},}
{\displaystyle \phi _{i}={\frac {1}{4\pi \varepsilon _{0}}}\sum _{\stackrel {j=1}{j\neq i}}^{N}{\frac {Q_{j}}{r_{ij}}}.}
{\displaystyle \phi _{i}}
{\displaystyle {\vec {r}}_{i}}
{\displaystyle Q_{i}}
{\textstyle \sum (\cdots )\rightarrow \int (\cdots )\rho \,\mathrm {d} ^{3}r}
{\displaystyle U_{\mathrm {E} }^{\text{total}}={\frac {1}{2}}\int \rho ({\vec {r}})\phi ({\vec {r}})\,\mathrm {d} ^{3}r={\frac {\varepsilon _{0}}{2}}\int \left|{\mathbf {E} }\right|^{2}\,\mathrm {d} ^{3}r,}
{\textstyle {\frac {1}{2}}\rho \phi }
{\textstyle {\frac {1}{2}}\varepsilon _{0}E^{2}}
Electrostatic pressureEdit
{\displaystyle P={\frac {\varepsilon _{0}}{2}}E^{2},}
Triboelectric seriesEdit
Electrostatic generatorsEdit
Charge neutralizationEdit
Electrostatic inductionEdit
Static electricityEdit
Static electricity and chemical industryEdit
Applicable standardsEdit
Electrostatic induction in commercial applicationsEdit
Retrieved from "https://en.wikipedia.org/w/index.php?title=Electrostatics&oldid=1080133454" |
Multiple Choice For Exercises 1−6, choose the correct letter. For Exercises 1 an
Multiple Choice For Exercises 1−6, choose the correct letter. For Exercises 1 and 2 , use the diagram at the right. The figure at the right is a Ferri
davonliefI
Since the octagon is regular, then it is divided into 8 congruent isosceles triangles. Hence, we divide 360° by 8 to find
m\angle 1m
m\angle 1=360.8=45
which is choice B.
If a plane contains two distinct points P1 and P2 then show that it contains every point on the line through P1 and P2.
a,b,c\in \mathbb{R}
a+b+c=0
\left(a,{a}^{3}\right),\left(b,{b}^{3}\right),\left(c,{c}^{3}\right)
\left(z>0\right)\text{ }\text{of the sphere}\text{ }{x}^{2}+{y}^{2}+{z}^{2}=1
\frac{{\partial }^{2}}{\partial {x}^{2}}f+\frac{{\partial }^{2}}{\partial {y}^{2}}f=0
The distance between point
{P}_{1}\left({\rho }_{1},\text{ }{\theta }_{1},\text{ }{\varphi }_{1}\right)
{P}_{2}\left({\rho }_{2},\text{ }{\theta }_{2},\text{ }{\varphi }_{2}\right)
given in spherical coordinates extract the formula.
4{x}^{2}+{y}^{2}=4
that are the farthest from the point (1;0) |
Predict state and state estimation error covariance at next time step using extended or unscented Kalman filter, or particle filter - MATLAB predict - MathWorks Benelux
\stackrel{^}{x}\left[k|k\right]
\stackrel{^}{x}\left[k+1|k\right]
\stackrel{^}{x}\left[k+1|k\right]
\stackrel{^}{x}\left[k+1|k\right]
\underset{}{\overset{ˆ}{x}}\left[k|k-1\right]
\underset{}{\overset{ˆ}{x}}\left[k|k-1\right]
\underset{}{\overset{ˆ}{x}}\left[k|k\right]
\underset{}{\overset{ˆ}{x}}\left[k+1|k\right]
\underset{}{\overset{ˆ}{x}}\left[k|k\right]
\underset{}{\overset{ˆ}{x}}\left[k|k-1\right]
\underset{}{\overset{ˆ}{x}}\left[k-1|k-1\right]
x\left[k\right]=\sqrt{x\left[k-1\right]+u\left[k-1\right]}+w\left[k-1\right]
y\left[k\right]=x\left[k\right]+2*u\left[k\right]+v\left[k{\right]}^{2}
\stackrel{^}{x}\left[k|k-1\right]
\stackrel{^}{x}\left[k|k\right]
\stackrel{^}{x}\left[k+1|k\right]
\stackrel{^}{x}\left[k-1|k-1\right]
\stackrel{^}{x}\left[k|k-1\right]
\stackrel{^}{x}\left[k|k\right]
\stackrel{^}{x}\left[k|k\right] |
Suppose that A and B are diagonalizable matrices. Prove or disprove that A is si
Consider A and B are matrices that are diagonalizable.
The objective is to prove or disprove that A is similar to B only if A and B are unitarily equivalent.
Consider the two diagonalizable matrices are,
A=\left[\begin{array}{cc}1& -1\\ 0& 0\end{array}\right]
B=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]
Consider that there is an invertible matrix,
P=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]
The matrix A is similar to B as shown below,
\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]={\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]}^{-1}\left[\begin{array}{cc}1& -1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]
=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]
=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]
The A and B are similar matrices.
But A and B are not unitary because B is symmetric, but A is not.
Also, recall the result, If B and A are unitarily equivalent.
\sum _{i,j=1}^{n}|{b}_{ij}{|}^{2}=\sum _{i,j=1}^{n}|{a}_{ij}{|}^{2}
Since, 2 is not equal then,
\sum _{i,j=1}^{n}|{b}_{ij}{|}^{2}\ne \sum _{i,j=1}^{n}|{a}_{ij}{|}^{2}
A and B are not unitarily equivalent.
Therefore, it is not necessary that A is similar to B if and only if A and B are unitarily equivalent.
2×2
\text{Basis }=\left\{\left[\begin{array}{cc}& \\ & \end{array}\right],\left[\begin{array}{cc}& \\ & \end{array}\right]\right\}
Find values for the variables so that the matrices are equal.
\left[\begin{array}{c}x\\ 7\end{array}\right]=\left[\begin{array}{c}11\\ y\end{array}\right]
Find the determinant of the following matrices.
\left[\begin{array}{ccc}4& 1& 0\\ -1& 7& -2\\ 2& 3& 5\end{array}\right]
write B as a linear combination of the other matrices, if possible.
B=\left[\left[2,-2,3\right],\left[0,0,-2\right],\left[0,0,2\right]\right]
{A}_{1}=\left[\left[1,0,0\right],\left[0,1,0\right],\left[0,0,1\right]\right]
{A}_{2}=\left[\left[0,1,1\right],\left[0,0,1\right],\left[0,0,0\right]\right]
{A}_{3}=\left[\left[-1,0,-1\right],\left[0,1,0\right],\left[0,0,-1\right]\right]
{A}_{4}=\left[\left[1,-1,1\right],\left[0,-1,-1\right],\left[0,0,1\right]\right]
compute the indicated matrices . FE
A=\left[\begin{array}{cc}3& 0\\ -1& 5\end{array}\right],B=\left[\begin{array}{ccc}4& -2& 1\\ 0& 2& 3\end{array}\right],C=\left[\begin{array}{cc}1& 2\\ 3& 4\\ 5& 6\end{array}\right],D=\left[\begin{array}{cc}0& -3\\ -2& 1\end{array}\right],E=\left[\begin{array}{cc}4& 2\end{array}\right],F=\left[\begin{array}{c}-1\\ 2\end{array}\right]
Find the characteristic polynomial of the matrices
\left[\begin{array}{ccc}1& 2& 1\\ 0& 1& 2\\ -1& 3& 2\end{array}\right] |
DGsolve - Maple Help
Home : Support : Online Help : Mathematics : DifferentialGeometry : DGsolve
DifferentialGeometry[DGsolve] - solve a list of tensor equations for an unknown list of tensors
DGsolve(Eq, T, options)
a vector, differential form or tensor constructed from the objects in the 2nd argument; or list of such. The vanishing of these tensors defines the equations to be solved.
a vector, differential form, or tensor, depending upon a number of arbitrary parameters or functions; or a list of such
auxiliaryequations
(optional) a keyword argument to specify a set of auxiliary equations, to be solved in conjunction with the equations specified by the first argument
(optional) list of parameters and functions, explicitly specifying the unknowns to be solved for.
(optional) a Maple procedure which will be used to solve the equations
(optional) additional arguments to be passed to the procedure used the solve the equations
T
be a vector, a differential form, or a tensor which depends upon a number of parameters
\left\{{f}_{1}, {f}_{2} ..., {f}_{n}\right\}
. These parameters may be constants or functions. Now let
\mathrm{ℰ}
be a differential-geometric construction depending upon
T
which can be implemented in Maple by a sequence of commands in the DifferentialGeometry package. For example,
T
could be a metric tensor and
\mathrm{ℰ}\mathit{ }
the Einstein tensor constructed from
g
. The command DGsolve will solve the equations obtained by setting to zero all the components of
\mathrm{ℰ}\mathit{ }
for the unknowns
\left\{{f}_{1}, {f}_{2} ..., {f}_{n}\right\}
. The output is a set containing those
T
\mathrm{ℰ}
=0 (obtainable by Maple).
Additional constraints (for example, initial conditions or inequalities) can be imposed upon the unknowns
\left\{{f}_{1}, {f}_{2} ..., {f}_{n}\right\}
with the keyword argument auxiliaryequations.
The command DGsolve uses the general purpose solver PDEtools:-Solve to solve the system
\mathrm{ℰ}\mathit{ }
=0 for the unknowns
\left\{{f}_{1}, {f}_{2} ..., {f}_{n}\right\}
. The keyword argument method can be used to specify a particular Maple solver (for example, solve, pdsolve, dsolve) or a customized solver created by the user.
If the equations defined by
\mathrm{ℰ}\mathit{ }
=0 are homogenous linear algebraic equations, then the command DGNullSpace can also be used.
\mathrm{with}\left(\mathrm{DifferentialGeometry}\right):
\mathrm{with}\left(\mathrm{Tensor}\right):
M
be a 4-dimensional space. We define a metric tensor depending upon an arbitrary function. We find the metrics which have vanishing Einstein tensor, and vanishing Bach tensor.
\mathrm{DGsetup}\left([x,y,u,v],M\right)
\textcolor[rgb]{0,0,1}{\mathrm{frame name: M}}
g≔\mathrm{evalDG}\left(\mathrm{dx}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dx}+\mathrm{dy}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dy}+\mathrm{du}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&s\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dv}+f\left(x,u\right)\mathrm{du}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{du}\right)
\textcolor[rgb]{0,0,1}{g}\textcolor[rgb]{0,0,1}{:=}\textcolor[rgb]{0,0,1}{\mathrm{dx}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dx}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{dy}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dy}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{f}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{u}\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{du}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{du}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{du}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dv}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dv}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{du}}
Here are the metrics of the form (4.2) with vanishing Einstein tensor.
\mathrm{DGsolve}\left(\mathrm{EinsteinTensor}\left(g\right),g\right)
\left\{\textcolor[rgb]{0,0,1}{\mathrm{dx}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dx}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{dy}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dy}}\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{\mathrm{_F1}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{u}\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{_F2}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{u}\right)\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{du}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{du}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{du}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dv}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dv}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{du}}\right\}
Here are the metrics of the form (4.2) with vanishing Bach tensor.
\mathrm{DGsolve}\left(\mathrm{BachTensor}\left(g\right),g\right)
\left\{\textcolor[rgb]{0,0,1}{\mathrm{dx}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dx}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{dy}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dy}}\textcolor[rgb]{0,0,1}{+}\left(\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{6}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{_F1}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{u}\right)\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{_F2}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{u}\right)\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{_F3}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{u}\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{_F4}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{u}\right)\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{du}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{du}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{du}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dv}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dv}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{du}}\right\}
In this example we define a 2-form
\mathrm{α}
which depends upon parameters
\left\{r, s\right\}
. We find those values of the parameters for which
\mathrm{α} ∧\mathrm{α} = 0.
\mathrm{DGsetup}\left([x,y,u,v],M\right)
\textcolor[rgb]{0,0,1}{\mathrm{frame name: M}}
\mathrm{\alpha }≔\mathrm{evalDG}\left(\mathrm{dx}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&w\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dy}+r\mathrm{dx}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&w\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{du}+s\mathrm{dy}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&w\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dv}\right):
\mathrm{DGsolve}\left(\mathrm{\alpha }\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&wedge\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{\alpha },\mathrm{\alpha },{r,s}\right)
\left\{\textcolor[rgb]{0,0,1}{\mathrm{dx}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{⋀}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dy}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{r}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dx}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{⋀}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{du}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{dx}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{⋀}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dy}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{s}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dy}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{⋀}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dv}}\right\}
We define a connection
\mathrm{Γ}
and calculate the parallel transport of a vector
X\left(t\right)
C\left(t\right)
\mathrm{DGsetup}\left([x,y],M\right)
\textcolor[rgb]{0,0,1}{\mathrm{frame name: M}}
\mathrm{Gamma}≔\mathrm{Connection}\left(-\left(\mathrm{D_x}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dx}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dy}+\left(\mathrm{D_y}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dy}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dx}\right)
\textcolor[rgb]{0,0,1}{\mathrm{Γ}}\textcolor[rgb]{0,0,1}{:=}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{D_x}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dx}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dy}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{D_y}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dy}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dx}}
C≔[\mathrm{cos}\left(t\right),\mathrm{sin}\left(t\right)]
\textcolor[rgb]{0,0,1}{C}\textcolor[rgb]{0,0,1}{:=}\left[\textcolor[rgb]{0,0,1}{\mathrm{cos}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{sin}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)\right]
X≔\mathrm{evalDG}\left(A\left(t\right)\mathrm{D_x}+B\left(t\right)\mathrm{D_y}\right)
\textcolor[rgb]{0,0,1}{X}\textcolor[rgb]{0,0,1}{:=}\textcolor[rgb]{0,0,1}{A}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{D_x}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{B}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{D_y}}
\mathrm{DGsolve}\left(\mathrm{ParallelTransportEquations}\left(C,X,\mathrm{Gamma},t\right),X\right)
\left\{\textcolor[rgb]{0,0,1}{\mathrm{_C2}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{\mathrm{sin}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{D_x}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{_C1}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{cos}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{D_y}}\right\}
We can use the keyword argument auxiliaryequations to specify an initial position for the vector
X.
\mathrm{DGsolve}\left(\mathrm{ParallelTransportEquations}\left(C,X,\mathrm{Gamma},t\right),X,\mathrm{auxiliaryequations}={A\left(0\right)=1,B\left(0\right)=0}\right)
\left\{{\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{\mathrm{sin}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{D_x}}\right\}
The source-free Maxwell equations may be expressed in terms of a 2-form
F
\mathrm{dF} =0
d*F =0
d
is the exterior derivative and
*
is the Hodge star operator. In this example we define a 2-form
F
depending on 2 functions of 4 variables and solve the Maxwell equations for
F.
\mathrm{DGsetup}\left([x,y,z,t],M\right)
\textcolor[rgb]{0,0,1}{\mathrm{frame name: M}}
g≔\mathrm{evalDG}\left(\mathrm{dx}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dx}+\mathrm{dy}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dy}+\mathrm{dz}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dz}-\mathrm{dt}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dt}\right)
\textcolor[rgb]{0,0,1}{g}\textcolor[rgb]{0,0,1}{:=}\textcolor[rgb]{0,0,1}{\mathrm{dx}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dx}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{dy}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dy}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{dz}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dz}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{dt}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dt}}
F≔\mathrm{evalDG}\left(A\left(x,y,z,t\right)\mathrm{dx}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&w\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dy}+B\left(x,y,z,t\right)\mathrm{dx}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&w\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dt}\right)
\textcolor[rgb]{0,0,1}{F}\textcolor[rgb]{0,0,1}{:=}\textcolor[rgb]{0,0,1}{A}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{t}\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dx}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{⋀}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dy}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{B}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{t}\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dx}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{⋀}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dt}}
\mathrm{DGsolve}\left([\mathrm{ExteriorDerivative}\left(F\right),\mathrm{ExteriorDerivative}\left(\mathrm{HodgeStar}\left(g,F,\mathrm{detmetric}=-1\right)\right)],F\right)
\left\{\left(\textcolor[rgb]{0,0,1}{\mathrm{_F1}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{y}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{_F2}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{y}\right)\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dx}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{⋀}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dy}}\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{\mathrm{_F1}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{y}\right)\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{_F2}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{y}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{_C1}}\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dx}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{⋀}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dt}}\right\} |
In compressible Neohookean elasticity one minimizes functionals which are composed by the sum of the
{L}^{2}
norm of the deformation gradient and a nonlinear function of the determinant of the gradient. Non-interpenetrability of matter is then represented by additional invertibility conditions. An existence theory which includes a precise notion of invertibility and allows for cavitation was formulated by Müller and Spector in 1995. It applies, however, only if some
{L}^{p}
-norm of the gradient with
p>2
is controlled (in three dimensions). We first characterize their class of functions in terms of properties of the associated rectifiable current. Then we address the physically relevant
p=2
case, and show how their notion of invertibility can be extended to
p=2
. The class of functions so obtained is, however, not closed. We prove this by giving an explicit construction.
Classification : 74B20, 35D05, 46E35, 49J45
author = {Conti, Sergio and de Lellis, Camillo},
title = {Remarks on the theory of elasticity},
AU - Conti, Sergio
TI - Remarks on the theory of elasticity
Conti, Sergio; de Lellis, Camillo. Remarks on the theory of elasticity. Annali della Scuola Normale Superiore di Pisa - Classe di Scienze, Série 5, Tome 2 (2003) no. 3, pp. 521-549. http://www.numdam.org/item/ASNSP_2003_5_2_3_521_0/
[1] E. Acerbi - G. Dal Maso, New lower semicontinuity results for polyconvex integrals, Calc. Var. Partial Differential Equations 2 (1994), 329-371. | MR 1385074 | Zbl 0810.49014
[2] L. Ambrosio - N. Fusco - D. Pallara, “Functions of bounded variation and free discontinuity problems", Oxford Mathematical Monographs, Clarendon Press, Oxford, 2000. | MR 1857292 | Zbl 0957.49001
[3] J. M. Ball, Convexity conditions and existence theorems in nonlinear elasticity, Arch. Rational Mech. Anal. 63 (1977), 337-403. | MR 475169 | Zbl 0368.73040
[4] J. M. Ball, Discontinuous equilibrium solutions and cavitation in nonlinear elasticity, Philos. Trans. Roy. Soc. London 306 A (1982), 557-611. | MR 703623 | Zbl 0513.73020
[5] P. Bauman - D. Phillips - N. C. Owen, Maximal smoothness of solutions to certain Euler-Lagrange equations from nonlinear elasticity, Proc. Roy. Soc. Edinburgh 119 A (1991), 241-263. | MR 1135972 | Zbl 0744.49008
[6] H. Brezis - L. Nirenberg, Degree theory and BMO: Part 1, compact manifolds without boudaries, Selecta Math. (N.S.) 1 (1995), 197-263. | MR 1354598 | Zbl 0852.58010
[7] P. G. Ciarlet - J. Nečas, Injectivity and self-contact in nonlinear elasticity, Arch. Rational Mech. Anal. 97 (1987), 171-188. | MR 862546 | Zbl 0628.73043
[8] B. Dacorogna - J. Moser, On a partial differential equation involving the jacobian determinat, Ann. IHP Anal. Non Lin. 7 (1990), 1-26. | Numdam | MR 1046081 | Zbl 0707.35041
[9] C. De Lellis, Some fine properties of currents and applications to distributional Jacobians, Proc. Roy. Soc. Edinburgh 132 A (2002), 815-842. | MR 1926918 | Zbl 1025.49029
[10] H. Federer, “Geometric measure theory", Classics in Mathematics, Springer Verlag, Berlin, 1969. | MR 257325 | Zbl 0874.49001
[11] I. Fonseca - W. Gangbo, “Degree theory in analysis and applications", Oxford Lecture Series in Mathematics and its Applications, 2, Clarendon Press, Oxford, 1995. | MR 1373430 | Zbl 0852.47030
[12] M. Giaquinta - G. Modica - J. Souček, “Cartesian currents in the calculus of variations”, Vol. 1, 2, Springer Verlag, Berlin, 1998. | MR 1645086 | Zbl 0914.49001
[13] J. Malý, Weak lower semicontinuity of polyconvex integrals, Proc. Roy. Soc. Edinburgh 123 A (1993), 681-691. | MR 1237608 | Zbl 0813.49017
[14] J. Malý, Lower semicontinuity of quasiconvex integrals, Manuscripta Math. 85 (1994), 419-428. | MR 1305752 | Zbl 0862.49017
[15] S. Müller - S. Spector, An existence theory for nonlinear elasticity that allows for cavitation, Arch. Rat. Mech. Anal. 131 (1995), 1-66. | MR 1346364 | Zbl 0836.73025
[16] J. Sivaloganathan - S. Spector, On the optimal location of singularities arising in variational problems of nonlinear elasticity, J. of Elast. 58 (2000), 191-224. | MR 1816651 | Zbl 0977.74005
[17] V. Šverák, Regularity properties of deformations with finite energy, Arch. Rat. Mech. Anal. 100 (1988), 105-127. | MR 913960 | Zbl 0659.73038 |
What is thorey of relativity in mode median and mean - Maths - Statistics - 12548567 | Meritnation.com
What is thorey of relativity in mode median and mean
For moderately skewed or assymetrical distributions, a relationship exists between the mean, median and mode.
Mode = Mean -3\left[Mean - Median\right]\phantom{\rule{0ex}{0ex}}Mode = 3Median - 2Mean\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Median = Mode + \frac{2}{3}\left[Mean - Mode\right]
``sakshi** Jaiswal`` answered this
Mode=3median-2mean. |
Define absolute refractive index of a medium Find its value for glass in which speed of light is 2 10 - Science - Light - Reflection and Refraction - 12676517 | Meritnation.com
Please find below the solution of your asked queries:
Definition- The absolute refractive index of any medium is the refractive index of the medium with respect to vacuum. It is always greater than 1.
Calculation of Absolute refractive index of glass-
For solving this numerical I am assuming that speed of light in glass is 2 * 108 m/sec as the speed given by you seems to be incorrect.
\mathrm{n} \left(\mathrm{absolute} \mathrm{refractive} \mathrm{index}\right)=\frac{\mathrm{c} \left(\mathrm{speed} \mathrm{of} \mathrm{light} \mathrm{in} \mathrm{vacuum}\right)}{\mathrm{speed} \mathrm{of} \mathrm{light} \mathrm{in} \mathrm{glass} }\phantom{\rule{0ex}{0ex}}\mathrm{n}=\frac{3×{10}^{8}}{2×{10}^{8}}\phantom{\rule{0ex}{0ex}}\mathrm{n}=\frac{3}{2}\phantom{\rule{0ex}{0ex}}\mathbf{n}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{5} |
Shunting Yard Algorithm | Brilliant Math & Science Wiki
Beakal Tiliksew, Josh Thelwall, Jimin Khim, and
The way we write mathematical expressions is infix notation. Operators have precedence and brackets override this precedence. Many programs require the parsing calculation on the fly. One very good way to do this is to convert from infix notation to some intermediate format. In this case, we will be using a very common and simple format called reverse polish notation.
The shunting yard algorithm is a simple technique for parsing infix expressions containing binary operators of varying precedence. In general, the algorithm assigns to each operator its correct operands, taking into account the order of precedence. It can, therefore, be used to evaluate the expression immediately, to convert it into postfix, or to construct the corresponding syntax tree.
The shunting yard algorithm is not your basic algorithm like mergesort, string search, etc. It is quite advanced as stacks, queues, and arrays are all contained in the same algorithm. Although the algorithm itself is very simple, a solid flexible implementation might be thousands of lines of code.
The Shunting Algorithm
Before we have our program calculate expressions, we need to convert them into an intermediate notation where the operators are in the order they must be performed. Unlike humans who look at infix expressions in their head, computers must be told explicitly what the order of the operations and parameters should be. The most common intermediate format is the reverse polish.
The procedure used is as follows:
Expressions are parsed left to right.
Each time a number or operand is read, we push it to the stack.
Each time an operator comes up, we pop the required operands from the stack, perform the operations, and push the result back to the stack.
We are finished when there are no tokens (numbers, operators, or any other mathematical symbol) to read. The final number on the stack is the result.
Consider the following infix notations:
4+18/(9-3).
Now we know that the answer to this from the rule or order of operations is
7
We have not seen it yet, given the above infix notation, that the shunting yard algorithm will output the reverse polish notation as
4, 18, 9, 3, -, /, +.
Note that the commas are not part of the reverse polish, but used to separate each token.
Using the procedure for reverse polish,
a
d
, we push the numbers in reverse polish expression into the stack. In step
e
, where we have reached the operator sign
(-)
, we pop the two numbers involved, perform the operation
9-3=6
, and push it back to the stack. Next is the operator sign
(/)
, where we pop the
6
18
and perform the operation
18/6=3
and push it to the stack. We continue with the procedure until we are left with a single number that is
7
To build the algorithm, we will need
1 stack for operations
1 queue of the output
1 array (or other list) of tokens.
A pseudocode of the algorithm is as follows:
1. While there are tokens to be read:
2. Read a token
3. If it's a number add it to queue
4. If it's an operator
5. While there's an operator on the top of the stack with greater precedence:
6. Pop operators from the stack onto the output queue
7. Push the current operator onto the stack
8. If it's a left bracket push it onto the stack
9. If it's a right bracket
10. While there's not a left bracket at the top of the stack:
11. Pop operators from the stack onto the output queue.
12. Pop the left bracket from the stack and discard it
13. While there are operators on the stack, pop them to the queue
The algorithm is fairly simple. Consider the infix notation again. To solve it, let us set up an array of list, a stack, and a queue.
The list of tokens in the left is filled from bottom to top and is the same as the infix expression stated earlier. We are now going to fill the stack and the queue bottom up, according to the shunting yard algorithm.
We read the tokens bottom up, so the number
4
goes to the output queue. The next element which is an operator sign
(+)
, according to the algorithm in lines
4, 5
6,
will be pushed onto the stack because an empty stack does not have any element of precedence.
Since the addition operator has less precedence than division, we ignore lines
5
6
. When we get a left bracket, we push it onto the stack, and when we get a right bracket, according to lines
10
11,
we pop operators from the stack into output queue until the left bracket is reached. Then we discard the left bracket.
Finally, when we are left with operators in the stack, we pop them to the queue.
That is it: the output queue contains the reverse polish notation, which as we have already seen can be computed to give the final answer.
Cite as: Shunting Yard Algorithm. Brilliant.org. Retrieved from https://brilliant.org/wiki/shunting-yard-algorithm/ |
Stack effect - 3D BIM Objects - 3D BIM Components
Stack effect (7624 views - Architecure & BIM & MEP)
Stack effect or chimney effect is the movement of air into and out of buildings, chimneys, flue-gas stacks, or other containers, resulting from air buoyancy. Buoyancy occurs due to a difference in indoor-to-outdoor air density resulting from temperature and moisture differences. The result is either a positive or negative buoyancy force. The greater the thermal difference and the height of the structure, the greater the buoyancy force, and thus the stack effect. The stack effect helps drive natural ventilation, air infiltration, and fires (e.g. the Kaprun tunnel fire and King's Cross underground station fire).
PARTcloud - chimney, air, heating
1 Stack effect in buildings
2 Stack effect in flue gas stacks and chimneys
3 Cause for the stack effect
4 Induced flow
Since buildings are not totally sealed (at the very minimum, there is always a ground level entrance), the stack effect will cause air infiltration. During the heating season, the warmer indoor air rises up through the building and escapes at the top either through open windows, ventilation openings, or unintentional holes in ceilings, like ceiling fans and recessed lights. The rising warm air reduces the pressure in the base of the building, drawing cold air in through either open doors, windows, or other openings and leakage. During the cooling season, the stack effect is reversed, but is typically weaker due to lower temperature differences.
In a modern high-rise building with a well-sealed envelope, the stack effect can create significant pressure differences that must be given design consideration and may need to be addressed with mechanical ventilation. Stairwells, shafts, elevators, and the like, tend to contribute to the stack effect, while interior partitions, floors, and fire separations can mitigate it. Especially in case of fire, the stack effect needs to be controlled to prevent the spread of smoke and fire, and to maintain tenable conditions for occupants and firefighters.[1]
The Grenfell Tower fire, as a result of which 71 people died[2], was in part exacerbated by the stack effect. A cavity between the outer aluminium cladding and the inner insulation formed a chimney and drew the fire upwards.[3][4]
Stack effect in flue gas stacks and chimneys
Large temperature differences between the outside air and the flue gases can create a strong stack effect in chimneys for buildings using a fireplace for heating. Fireplace chimneys can sometimes draw in more cold outside air than can be heated by the fireplace, resulting in a net heat loss.
Cause for the stack effect
{\displaystyle \Delta P=Cah{\bigg (}{\frac {1}{T_{o}}}-{\frac {1}{T_{i}}}{\bigg )}}
The draft (draught in British English) flow rate induced by the stack effect can be calculated with the equation presented below.[5][6][7] The equation applies only to buildings where air is both inside and outside the buildings. For buildings with one or two floors, h is the height of the building and A is the flow area of the openings. For multi-floor, high-rise buildings, A is the flow area of the openings and h is the distance from the openings at the neutral pressure level (NPL) of the building to either the topmost openings or the lowest openings. Reference[5] explains how the NPL affects the stack effect in high-rise buildings.
{\displaystyle Q=CA{\sqrt {2gh{\frac {T_{i}-T_{o}}{T_{i}}}}}}
Air conditioningArchitectural engineeringBlast furnaceChimneyConstruction engineeringCooling towerFurnaceHeat exchangerInduction furnaceInduction heatingStructural engineeringVentilation (architecture)
This article uses material from the Wikipedia article "Stack effect", which is released under the Creative Commons Attribution-Share-Alike License 3.0. There is a list of all authors in Wikipedia |
Stat | PoE Wiki
\color [rgb]{0.6392156862745098,0.5529411764705883,0.42745098039215684}{\begin{aligned}a&{\text{ the total amount of added stats }}&a&\in \mathbb {N} _{0}\\b&{\text{ the total amount of increased stats }}&b&\in \mathbb {N} _{0}\\c&{\text{ the total amount of reduced stats }}&c&\in \mathbb {N} _{0}\\d&{\text{ the total amount of more stats }}&d&\in \mathbb {N} _{0}\\e&{\text{ the total amount of less stats }}&e&\in \mathbb {N} _{0}\\\end{aligned}}
\color [rgb]{0.6392156862745098,0.5529411764705883,0.42745098039215684}{\begin{aligned}{\text{stat}}_{total}=&\sum _{i=1}^{a}{\text{added stat}}_{i}\times \\&\left(1+\sum _{j=1}^{b}{\text{increased stat}}_{j}-\sum _{k=1}^{c}{\text{reduced stat}}_{k}\right)\times \\&\prod _{l=1}^{d}(1+{\text{more stat}}_{l})\times \prod _{m=1}^{e}(1-{\text{less stat}}_{m})\end{aligned}}
Projectile Speed: 750Infuses your melee weapon with molten energies to attack with physical and fire damage. This attack causes balls of molten magma to launch forth from the enemies you hit, divided amongst all enemies hit by the strike. These will explode, causing AoE attack damage to enemies where they land.Per 1% Quality:1Superior2Anomalous3Divergent4Phantasmal1% increased Fire Damage1% increased Projectile Speed+0.1 to Melee Strike Range2% increased Damage over Time60% of Physical Damage Converted to Fire Damage
\color [rgb]{0.6392156862745098,0.5529411764705883,0.42745098039215684}1+{x \over 100}
1% reduced Area of EffectBase duration is 3 seconds
Beams deal 40% less DamagePlace into an item socket of the right colour to gain this skill. Right click to remove from a socket. )
ru:Показатель de:Statistik
Retrieved from ‘https://www.poewiki.net/w/index.php?title=Stat&oldid=259747’ |
Julie Kneller contributed
Stoichiometry is the numerical relationship between the reactants and products of a chemical reaction. In fact, the word ‘stoichiometry’ is derived from the Ancient Greek words stoicheion "element" and metron "measure". Stoichiometric techniques are used to calculate the required quantities of chemical reactants – or the substances in the reaction – needed to generate the desired amount of product using balanced chemical equations. A stoichiometric chemical reaction occurs when all reagents are consumed as a result of the reaction.
Solids, liquids, and gases react together to form chemical substances whose properties and behaviors shape practical applications. Understanding how the atoms and molecules making up these substances work together at the molecular level to create new compounds is fundamental to chemistry. At the basis of this work is stoichiometry.
Laws governing chemical reactions
Avogadro's law - from grams to moles
Molarity, molality, and normality
Stoichiometry is based on some of the most fundamental principles in chemistry. The most central being the law of conservation of mass that stipulates that mass is neither created nor destroyed.
Collect a cup of water from the faucet and find its mass. Now freeze the water and find the mass of the ice. The mass of liquid water is the same as the mass of the ice showing that mass is conserved.
The chemical properties of compounds are established by the assortment and arrangement of its atoms. When compounds and their elements rearrange in a reaction they do so in integer proportions. Based on these observations, other laws, the so-called law of definite proportions, law of multiple proportions, and law of reciprocal proportions were articulated. These laws all convey the fundamental idea that in a chemical reaction, the reactant molecules combine in definite ratios, e.g., one molecule of glucose always reacts with six molecules of oxygen. Matter cannot be created or destroyed; one element cannot be changed into another; and the number of atoms of each element must remain the same from the beginning to the end of the reaction (assuming no nuclear reactions such as fission, fusion, or radioactive decay are taking place).
You are familiar with the molecular description of water as
\ce{H_{2}O}
; two atoms of hydrogen for every atom of oxygen. No matter what amount of water you have, whether it is the cup you are drinking from or a river you are rafting on, that ratio of hydrogen to oxygen remains the same; just as the ratio of eggs to butter to flour to sugar in a cake must always be the same to maintain taste and consistency.
Again, hydrogen and oxygen combine to form water
\ce{H_{2}O}
\ce{H_{2}O_{2}}
. In water, there are two atoms of hydrogen with one atom of oxygen and in hydrogen peroxide there are two atoms of hydrogen with two atoms of oxygen. The ratio of oxygen atoms to hydrogen atoms in these two compounds follows small, whole number ratios.
Consider hydrogen, oxygen, and carbon. These elements combine separately to form methane
\ce{CH_{4}}
and carbon dioxide
\ce{CO_{2}}
. As we will see shortly when we discuss balancing chemical reactions, the ratios at which the elements combine remain simple multiples of each other.
\ce{carbon\, dioxide + hydrogen\, ->[{heat\, required}] methane + water}
\ce{methane + oxygen -> carbon\, dioxide + water}
Stoichiometry is used in chemistry to conveniently represent quantitative relationships between reactants and products in a chemical reaction. Chemical reactions are usually written in equation form with reactants/reagents on the left, products on the right, and an arrow in the middle indicating the direction of the reaction. Such as,
\ce{X + Y\, (reactants) -> XY\, (product)}
where XY is a new chemical entity distinct from the reactants. To ensure that the number of atoms of each type are conserved these equations must be balanced so that the number of atoms on the left is equivalent to the number of atoms on the right; the equation is balanced.
This reaction, the Haber process, makes ammonia
\ce{NH_{3(g)}}
by reacting nitrogen gas
\ce{N_{2(g)}}
with hydrogen gas
\ce{H_{2(g)}}
(where the subscript
_{(g)}
denotes the gaseous state):
N_{2(g)} + H_{2(g)} → NH_{3(g)}
This equation describes what you need to make ammonia but not how much – you’ve got to ensure that the number of atoms on the left is equal to the number on the right. In this reaction,
N_{2}
H_{2}
are both gases, there are always two atoms of nitrogen and two atoms of hydrogen, as such N—N and H—H; this can’t be changed as it would change the compounds. To balance the equation you can add coefficients, whole numbers in front of the reactants to tell you have many atoms or molecules you have.
This is a simple reaction that can be balanced by inspection. It is generally helpful to balance oxygen and hydrogen last as they are usually plentiful. So, consider nitrogen first. There are 2 nitrogen atoms on the left side, the reactant side, and 1 nitrogen atom on the right side, the product side. To balance the nitrogen atoms, place a coefficient of 2 in front of the ammonia on the right:
N_{2(g)} + H_{2(g)} → 2NH_{3(g)}
Now the number of nitrogen atoms on each side of the equation is equivalent. To balance hydrogen, you see there are 2 hydrogen atoms on the left and 6 hydrogen atoms on the right. So place a coefficient of 3 in front of the hydrogen atoms on the left and you have:
N_{2(g)} + 3H_{2(g)} → 2NH_{3(g)}
Final check: 2 nitrogen atoms on the left and 2 nitrogen atoms on the right 6 hydrogen atoms on the left and 6 hydrogen atoms on the right. This balanced equation can now be read as 1 nitrogen molecule reacting with 3 hydrogen molecules yields 2 ammonia molecules.
When elements are combined with correct stoichiometric ratios, all reagents are converted into product. When a reactant is consumed in a reaction, it is called a stoichiometric reactant. This is opposed to a catalytic reactant that is not consumed in the overall reaction but reacts in one step and is regenerated in another step.
The coefficients in the balanced equation put strict demands on the relative abundance of reactants. If the number of molecules of one far exceeds the number of molecules of the other (after taking their coefficients into account), we can have so-called limiting reagents that limit the amount of product that can be formed. If we mix reagents out of stoichiometric proportion, one of them will run out first and bring the reaction to a halt. An excess reagent is the reactant that is left over once the reaction has stopped because the limiting reactant has been consumed. If a reagent participates in, but isn’t changed by, a reaction it’s called a catalyst.
\underline{Here\, are\, a\, few\, steps\, to\, keep\, in\, mind\, when\, balancing\, chemical\, equations}
• Chemical equations can get complicated when there are many possible reactants and products. Choose an element to begin with – preferably an element in only one reactant and one product. Adjust the coefficients so the number of atoms of this element is the same on both sides of the equation.
• If polyatomic ions (ions comprised of many atoms) are on both sides of the equation, balance them as a unit.
• Continue balancing the remaining components in order of the most complicated substance remaining to the least complicated; using fractional coefficients, if necessary. Fractional coefficients can be made whole at the end by multiplying both sides of the equation by the denominator to obtain whole number coefficients.
• Do a final check on the numbers of atoms of each kind on both sides of the equation to make sure you have balanced the equation correctly.
try a non-trivial equation: photosynthesis
Photosynthesis is the process by which plants, and a few organisms, use energy from the sun to convert carbon dioxide and water into glucose (a sugar) and oxygen:
CO_{2(g)} + H_{2}O_{(l)} → sunlight → C_{6}H_{12}O_{6} + O_{2(g)}
\ce{CO_{2(g)}}
is carbon dioxide, gas;
\ce{H_{2}O_{(l)}}
is water, liquid;
\ce{C_{6}H_{12}O_{6}}
is glucose; and
\ce{O_{2(g)}}
is oxygen, gas.
\underline{balance\, the\, equation}
It is evident that the equation above is not balanced. To do so, first, let’s choose carbon C to start. There is 1 C on the left and 6 on the right so we add the coefficient 6 to the left and we now have:
6 CO_{2(g)} + H_{2}O_{(l)} → sunlight → C_{6}H_{12}O_{6} + O_{2(g)}
Second, consider hydrogen H. There are 2 on the left and 12 on the right. If we put a coefficient of 6 in front of
\ce{H_{2}O_{(l)}}
we can balance the H as such:
6 CO_{2(g)} + 6 H_{2}O_{(l)} → sunlight → C_{6}H_{12}O_{6} + O_{2(g)}
Finally, only oxygen O remains to be done. We now have 18 on the left and 8 on the right. A coefficient of 6 in front of the
\ce{O_{2(g)}}
should do it. And the final overall balanced chemical equation for photosynthesis is:
6 CO_{2(g)} + 6 H_{2}O_{(l)} → sunlight → C_{6}H_{12}O_{6} + 6 O_{2(g)}
Each element is made up of different numbers of protons, neutrons, and electrons, and thus has a different atomic mass. Molecules, as collections of single atoms, have a molar mass (by definition, oxygen-16 has a molar mass of 16 g/mol). Molar mass is measured in terms of a unit mole which is
6.02\,×\,10^{23}
individual molecules. This number is known as Avogadro's constant.
To know stoichiometric relationships by mass, the number of molecules required for each reactant can be expressed in moles and multiplied by the molar mass of each reactant to give the mass of each reactant per mole of reaction. Stoichiometry is used to converting from units from grams to moles. This conversion is required because atoms and molecules are too small to count in a meaningful way so we use molar mass find the number of moles of atoms or molecules. This number can then be used in a chemical equation.
Converting grams to moles. How many moles in 100.0 grams of
\ce{H_{2}O}
First, lets determine how many grams are in 1 mole of
\ce{H_{2}O}
. Look at the Periodic Table of Elements and you will see that the atomic mass of hydrogen = 1.01 g/mol and the atomic mass of oxygen is 16.0 g/mol.
Now, find the molecular mass of
\ce{H_{2}O}
= 2 (1.01 g/mol) + (16.0 g/mol)
= 2.02 g/mol + 16.0 g/mol
The number of moles in 100.0 g of
\ce{H_{2}O}
is then found by:
\ce{H_{2}O}
× (1 mol/18.02 g) = 5.5 moles
\ce{H_{2}}O
Out of curiosity, the number of
{H_{2}O}
molecules would be:
5.5 moles ×
6.02\,×\,10^{23}
3.31\,×\,10^{24}
\ce{H_{2}O}
molecules – and that’s a number that’s hard to work with!
When dealing with liquids, other units are commonly employed. Molarity (M) is the number of moles of solute dissolved in one liter of a solution and is expressed in moles/L. or M. Molality (m) is the number of moles per kilogram of solvent and is expressed in moles/kg, or m.
While not used extensively, normality (N), or equivalent concentration, of a solution is defined as the molar concentration
c_{i}
divided by an equivalence factor
f_{eq}
. To calculate normality, you would need a defined equivalence factor, which in turn depends on the definition of equivalents. This implies that the same solution can possess different normalities for different reactions. Due to normality’s ambiguity as a measure of concentration, molarity and molality are used for the majority of applications.
Cite as: Stoichiometry. Brilliant.org. Retrieved from https://brilliant.org/wiki/stoichiometry-in-progress/ |
Write the equation of the circle described. a. Center at the origin, containing the point
Write the equation of the circle described. a. Center at the origin, containing the point (-6, -8) b. Center (7, 5), containing the point (3, -2).
a. Center at the origin, containing the point (-6, -8)
b. Center (7, 5), containing the point (3, -2).
Use the standard equation of a circle with center (h,k) and radius r:
{\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={r}^{2}
a.Using
\left(h,k\right)=\left(0,0\right)
\left(x,y\right)=\left(-6,-8\right)
{r}^{2}
{\left(-6-0\right)}^{2}+{\left(-8-0\right)}^{2}={r}^{2}
36+64={r}^{2}
100={r}^{2}
So, the equation of the circle is:
{\left(x-0\right)}^{2}+{\left(y-0\right)}^{2}=100
{x}^{2}+{y}^{2}=100
b.Using
\left(h,k\right)=\left(7,5\right)
\left(x,y\right)=\left(3,-2\right)
{r}^{2}
{\left(3-7\right)}^{2}+{\left(-2-5\right)}^{2}={r}^{2}
16+49={r}^{2}
65={r}^{2}
{\left(x-7\right)}^{2}+{\left(y-5\right)}^{2}=65
Find the linear equations that can be used to convert an (x, y) equation to a (x, v) equation using the given angle of rotation θ.
\theta ={\mathrm{tan}}^{-1}\left(5/12\right)
f\left(x\right)=\sqrt{2x-{x}^{2}}
, graph the following functions in the viewing rectangle [-5,5] by [-4,4] . How is each graph related to the graph in part (a)?
By using the transformations of function
y={x}^{2}
sketch the function of
y=x\left(6+x\right)\right)
Sketch a graph of the function. Use transformations of functions when ever possible.
f\left(x\right)=1\text{ }-\text{ }\sqrt{x\text{ }+\text{ }2}
Graph the function by hand, not by plotting points, but by starting with the graph of one of the standard functions and then applying the appropriate transformations.
y=1-2x+{x}^{2}
What transformations of the parent graph of
f\left(x\right)=\sqrt{c}
produce the graphs of the following functions?
m\left(x\right)=\sqrt{7x-3.5}-10
j\left(x\right)=-2\sqrt{12x}+4 |
17 Pseudo-Random Numbers
17.1 Setting the seed (replacements for G05CBF and G05CCF)
17.2 Data type of the random numbers
17.3 Replacements for G05CAF and G05DAF
17.4 Replacement for G05DBF
17.5 Replacements for G05DDF, G05DRF, and G05FFF
The routines for creating pseudo-random numbers in this library all have a period of 2
{}^{26}
and 6–7 digits accuracy. They are based upon code by Ahrens, Dieter, & Grube. They use a multiplicative congruential generator which is certainly not the state of the art and may not be suitable for critical or sophisticated use.
PDA_RAND (NETLIB/TOMS599)
Returns uniform pseudo-random numbers in the range 0 to 1.
PDA_RNEXP (NETLIB/TOMS599)
Draws pseudo-random numbers from an exponential distribution.
PDA_RNGAM (NETLIB/TOMS599)
Draws pseudo-random numbers from a Gamma-function distribution.
PDA_RNNOR (NETLIB/TOMS599)
Draws pseudo-random numbers from a Normal distribution of specified mean and standard deviation.
PDA_RNPOI (NETLIB/TOMS599)
Draws pseudo-random numbers from a Poisson distribution of specified mean.
PDA_RNSED (NETLIB/TOMS599)
Sets the seed. This must be called before any of the other random-number routines.
Before any random numbers can be selected, a seed must be set using PDA_RNSED. The integer seed should satisfy the relationship
seed=4\ast k+1
k
is a non-negative integer. A fixed seed gives rise to a reproducible sequence of pseudo-random numbers.
For a non-repeatable sequence, there is no equivalent to NAG routine G05CCF because the system clock used to create the seed is not accessible portably in Fortran, and PDA is independent of other libraries. However, the following code has the desired effect.
INTEGER SEED, STATUS, TICKS, PID
INCLUDE ’PRM_PAR’
CALL PSX_TIME( TICKS, STATUS )
CALL PSX_GETPID( PID, STATUS )
SEED = TICKS + PID
SEED = MOD( SEED, VAL__MAXI / 4 ) * 4 + 1
SEED = MOD( SEED, 2**28 )
PSX_TIME returns the time in units of clock ticks since some arbitrary time. See SUN/121 for more details and linking instructions. The above code also permits storage of the chosen seed.
There is a major difference between the PDA random-number routines and those provided in the standard NAG library: in general the former are single-precision functions, whereas the latter are double precision. However, PDA_RNPOI and the corresponding G05DRF are both integer functions.
Like G05CAF, PDA_RAND has a dummy argument demanded by the Fortran standard. It is convenient to set it to zero. Here is an example where two random numbers are drawn from a uniform distribution between 0 and 1. In this example a fixed seed is used, but you could use the computer’s clock to create a random seed (see Section 17.1).
EXTERNAL PDA_RAND
REAL PDA_RAND, VALUES( 2 )
* Use a fixed seed of 1.
* Obtain two random numbers from a uniform distribution between 0
* and 1.
VALUE( 1 ) = PDA_RAND( 0.0 )
The EXTERNAL statement is recommended, although in many cases it will be unnecessary. To obtain in the range [
a
b
] as provided by G05DAF, merely apply the following relationship.
\text{random value}=\left(b-a\right)\ast \text{ PDA_RAND}\left(0.0\right)+a
PDA_RNEXP is only a partial replacement for G05DBF in that it computes pseudo-random numbers from
{e}^{-x}
, whereas G05DBF uses the function
\frac{1}{a}{e}^{-x/a}
. Thus its argument is also a dummy mandated by the Fortran standard.
The following code shows the remaining three routines in action.
EXTERNAL PDA_RNGAM, PDA_RNNOR, PDA_RNPOI
REAL PDA_RNGAM, PDA_RNNOR, PDA_RNPOI, VALUES( 3 )
* Use a fixed seed of 1001.
* Obtain a random number from a Normal distribution of mean 4.2 and
* standard deviation 0.15
VALUE( 1 ) = PDA_RNNOR( 4.2, 0.15 )
* Obtain a random number from a Poisson distribution of mean 3.4.
VALUE( 2 ) = PDA_RNPOI( 3.4 )
* Obtain a random number from a Gamma-function distribution of mean
VALUE( 3 ) = PDA_RNGAM( 1.2 )
Apart from the change of data type, calls to G05DDF can be replaced with PDA_RNNOR using the same arguments. PDA_RNPOI is in effect a renamed G05DRF.
PDA_RNGAM only has one argument—the mean—of the Gamma function, whereas G05FFF has a second scaling parameter similar in role to the
argument of G05DBF. G05FFF also generates a vector of pseudo-random numbers. |
Intersection of Lines | Brilliant Math & Science Wiki
div n, Sahil Silare, Tarun Singh, and
Rohith M.Athreya
Lines that are non-coincident and non-parallel intersect at a unique point. Lines are said to intersect each other if they cut each other at a point. By Euclid's lemma two lines can have at most
1
point of intersection. In the figure below lines
L1
L2
intersect each other at point
P.
Three or more lines when met at a single point are said to be concurrent and the point of intersection is point of concurrency.
In the figure above, point
P=(p, q)
To find the intersection of two lines, you first need the equation for each line. At the intersection,
x
y
have the same value for each equation. This means that the equations are equal to each other. We can therefore solve for
x
x
in one of the equations (it does not matter which) and solve for
y
Find the intersection of the lines
y = 3x - 3
y = 2.3x + 4
\begin{aligned} 3x - 3 &= 2.3x + 4\\ 3x - 2.3x &= 4 + 3\\ 0.7x &= 7\\ \Rightarrow x &= 10\\ \Rightarrow y &= 3(10) - 3\\ &= 27. \end{aligned}
Thus, the intersection point is
(10, 27)
_\square
Ange between the lines is given by
\tan(\theta )=\frac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } ,
{m}_{1}
is the slope of the first line,
{m}_{2}
is the slope of the second line, and
\theta
is the angle between them.
For two lines intersecting at right angle,
{ m }_{ 1 }{ m }_{ 2 } =-1.
Second-degree equation representing a pair of straight lines:
Homogeneous equations (theorem):
A second-degree homogeneous equation in
x
y
always represents a pair of straight lines (real or imaginary) passing through the origin.
h^2 \geq ab
ax^2+2hxy+by^2=0
represents a pair of straight lines passing through the origin. This equation can be considered a quadratic in
y
and can be solved to obtain two equations (of degree
1
y=mx
y=nx
However, general equation in degree
2
ax^2+2hxy+by^2+2gx+2fy+c=0
will represent a pair of straight lines if and only if
abc+ 2fgh- af^2 -bg^2 -ch^2 =0\quad \text{ and }\quad h^2 - ab > 0.
\theta
between these lines satisfies
\tan \theta=\frac{\sqrt{h^2-ab}}{|a-b|}.
x^2 -2y^2 +axy+3y-1=0
a
for which this equation represents a pair of straight lines.
Comparing the above equation with the general one and substituting in the
2
conditions, we find that
\begin{aligned} 1(-2)(-1) +3(0)(a) - 1\frac{9}{4}- 2(0) - (-1)\frac{a^2}{4}&=0\\ 2-\frac{9}{4} + \frac{a^2}{4}&=0\\ a^2 &= 1. \end{aligned}
Checking if
h^2>ab
\frac{a^2}{2} > (-2)
\frac{1}{4} >(-2).
a=1
a = -1. \ _\square
ax^2+2hxy+by^2+2gx+2fy+c=0
lx+my+n=0.
\begin{aligned} lx+my&=-n\\ \frac{lx+my}{-n}&=1. \end{aligned}
A
B
be the points of intersection of the curve and the line. In order to make the pair of lines homogeneous with the help of
\frac{lx+my}{-n}=1
, we write the pair of lines as
\begin{aligned} ax^2+2hxy+by^2+\left(2gx+2fy\right)\left(1\right)+c\left(1\right)^2&=0\\ ax^2+2hxy+by^2+\frac{\left(2gx+2fy\right)\left(lx+my\right)}{\left(-n\right)}+c\left(\frac{\left(lx+my\right)}{\left(-n\right)}\right)^2&=0. \end{aligned}
So, this is locus through points
A
B
. Also it represents the homogeneous equation of second degree in
x
y
through origin.
Cite as: Intersection of Lines. Brilliant.org. Retrieved from https://brilliant.org/wiki/linear-equations-intersection-of-lines/ |
Find the linear regression line for a scatterplot formed by
Find the linear regression line for a scatterplot formed by the points (10, 261), (21, 252), (42, 209), (33, 163), and (52, 98).
Find the linear regression line for a scatterplot formed by the points (10, 261), (21, 252), (42, 209), (33, 163), and (52, 98). Round to the nearest tenth.
SchepperJ
Unusual points Each of the four scatterplots that follow shows a cluster of points and one “stray” point. For each, answer these questions:
1) In what way is the point unusual? Does it have high leverage, a large residual, or both?
2) Do you think that point is an influential point?
3) If that point were removed, would the correlation be- come stronger or weaker? Explain.
4) If that point were removed, would the slope of the re- gression line increase or decrease? Explain
Sketch a scatterplot where the association is nonlinear, but the correlation is close to r = -1.
Make a scatterplot of the data. Use 87 for 1987.
\begin{array}{|cc|}\hline \text{ Year }& \text{ Sales }\\ & \text{ (millions of dollars) }\\ 1987& 300\\ 1988& 345\\ 1989& 397\\ 1990& 457\\ 1991& 510\\ 1992& 587\\ 1993& 664\\ 1994& 700\\ 1995& 770\\ 1996& 792\\ 1997& 830\\ 1998& 872\\ 1999& 915\\ \hline\end{array}
Scatterplots Which of the scatterplots below show
a) little or no association?
b) a negative association?
c) a linear association?
d) a moderately strong association?
e) a very strong association?
Match each verbal statement with the value o fQCR U best matches. Drawing sample scatter plots might help you decide QCR values: -1 0 0.33 0.81 1. a. Mrs. A Every student who was below average on test 1 was also below average on test 2. Every student who was above average on test 1 was also above average on test 2." b. Ms. B: "Most of the students who were below average on test 1 were below average on test 2. Similarly, most of the students scoring above average on test 1 were also above average on test 2. There were a few exceptions, but the trend was clear." c. Mr. C: "Wow, there was really no correlation between test 1 and test 2! half the students who were below average on the first test were also below average on the second test, but half were above average! The same was true for the above average students!" d. Mr. D: "This is so weird. Every student who was below average on test 1 was above average on test 2, and vice versa." e. Mr. E: "Of those scoring above average on test 1, about 60% were above average on test 2, but 40$ were below average."
Predicting Land Value Both figures concern the assessed value of land (with homes on the land), and both use the same data set. (a). Which do you think has a stronger relationship with value of the land-the number of acres of land or the number of rooms n the homes? Why? b. Il you were trying to predict the value of a parcel of land in e arca (on which there is a home), would you be able to te a better prediction by knowing the acreage or the num- ber of rooms in the house? Explain. (Source: Minitab File, Student 12. "Assess.")
Answer true or false to the following statements and explain your answers.
a. In multiple linear regression, we can determine whether we are extrapolating in predicting the value of the response variable for a given set of predictor variable values by determining whether each predictor variable value falls in the range of observed values of that predictor.
b. Irregularly shaped regions of the values of predictor variables are easy to detect with two-dimensional scatterplots of pairs of predictor variables, and thus it is easy to determine whether we are extrapolating when predicting the response variable. |
Simplify each expression. 1. -n+9n+3-8-8n 2. 3(-4x+5y)-3x(2+4) 5. 5-4y+x+9y 7. -2x+3y-5x-(-8y)
Simplify each expression.1. -n+9n+3-8-8n2. 3(-4x+5y)-3x(2+4)5. 5-4y+x+9y7. -2x+3y-5x-(-8y)
-n+9n+3-8-8n
3\left(-4x+5y\right)-3x\left(2+4\right)
5-4y+x+9y
-2x+3y-5x-\left(-8y\right)
-n+9n+3-8-8n
=\left(-n+9n-8n\right)+\left(3-8\right)
=n\left(-1+9-8\right)+\left(3-8\right)
=n\left(8-8\right)+\left(-5\right)
=nx0+\left(-5\right)=0+\left(-5\right)=-5
3\left(-4x+5y\right)-3x\left(2+4\right)
=3\left(-4x+5y\right)+\left(-3x\right)\left(2+4\right)
=3\left(-4x+5y\right)+\left(-3x\right)\left(6\right)
=\left(-12x\right)+15y+\left(-18x\right)
=15y+\left(-12x+-18x\right)
=15y+\left(-30x\right)=15y-30x=15\left(y-2x\right)
5-4y+x+9y
=\left(9y-4y\right)+x+5
=y\left(9-4\right)+x+5
=5y+x+5
=\left(5y+5\right)+x=5\left(y+1\right)+x
-2x+3y-5x-\left(-8y\right)
=-2x+3y-5x+8y
=\left(3y+8y\right)+\left(-2x+-5x\right)
=y\left(3+8\right)+x\left(-2+-5\right)
=11y+\left(-7x\right)=11y-7x
-5
15\left(y-2x\right)
5\left(y+1\right)+x
11y-7x
Video of first expression
Video of second expression
Video of third expression
Video of fourth expression
y=5x+2
Use the given graph off to find a number
\delta
|x-1|<\delta \text{ then }|f\left(x\right)-1|<0.2
The total costs for music record label are given by C(x)=2400+10x+x and the total revenues are given by R(x)=120x. Find the break-even points.
Contain linear equations with constants in denominators. Solve each equation.
\frac{x+1}{4}=\frac{1}{6}+\frac{2-x}{3}
What would a person look for while solving a quadratic equation on a graph?
\frac{x}{5}-\frac{1}{2}=\frac{x}{6} |
Determine whether the study is an observational study or an experiment. Explain. In a stud
Determine whether the study is an observational study or an experiment. Explain. In a study designed to research the effect of music on driving habits
Determine whether the study is an observational study or an experiment. Explain. In a study designed to research the effect of music on driving habits, 1000 motorists ages 17- 25 years old were asked whether the music they listened to influenced their driving.
Lacey-May Snyder
It is an observational study since researchers observed motorcyclist's opinion.
\begin{array}{|ccc|}\hline \text{ }& \text{White Welcome Screen}& \text{Red Welcome Screen}\\ \text{Number of Web users}& 190& 183\\ \text{Number who break off survey}& 49& 37\\ \text{Break-off rate}& 0.258& 0.202\\ \hline\end{array}
c) Conduct the test, part b, at
\alpha =0.10.
What do you conclude?
In a study designed to investigate the effects of a strong magnetic field on the early development of mice, ten cages, each containing three 30-day-old albino female mice, were subjected for a period of 12 days to a magnetic field having an average strength of 80 Oe/cm. Thirty other mice, housed in ten similar cages, were not put in the magnetic field and served as controls. Listed in the table are the weight gains, in grams, for each of the twenty sets of mice.
\overline{)\begin{array}{cccc}\text{ Cage }& \text{ Weight Gain (g) }& \text{ Cage }& \text{ Weight Gain (g) }\\ 1& 22.8& 11& 23.5\\ 2& 10.2& 12& 31.0\\ 3& 20.8& 13& 19.5\\ 4& 27.0& 14& 26.2\\ 5& 19.2& 15& 26.5\\ 6& 9.0& 16& 25.2\\ 7& 14.2& 17& 24.5\\ 8& 19.8& 18& 23.8\\ 9& 14.5& 19& 27.8\\ 10& 14.8& 20& 22.0\end{array}}
Test whether the variances of the two sets of weight gains are significantly different. Let α=0.05. For the mice in the magnetic field,
{s}_{X}=5.67
. for the other mice,
{s}_{Y}=3.18
Is the gift you purchased for that special someone really appreciated? This was the question investigated in the Journal of Experimental Social Psychology (Vol. 45, 2009). Toe researchers examined the link between engagement ring price (dollars) and level of appreciation of the recipient
\left(\text{measured on a 7-point scale where}\text{ }1=\text{ }\text{"not at all" and}\text{ }7=\text{ }\text{to a great extent"}\right).
Participants for the study were those who used a popular Web site for engaged couples. The Web site's directory was searched for those with "average" American names (e.g., "John Smith," "Sara Jones"). These individuals were then invited to participate in an online survey in exchange for a $10 gift certificate. Of the respondents, those who paid really high or really low prices for the ring were excluded, leaving a sample size of 33 respondents.a) Identify the experimental units for this study.b) What are the variables of interest? Are they quantitative or qualitative in nature?c) Describe the population of interest.d) Do you believe the sample of 33 respondents is representative of the population? Explain. e. In a second, designed study, the researchers investigated whether the link between gift price and level of appreciation was stronger for birthday gift givers than for birthday gift receivers. Toe participants were randomly assigned to play the role of gift-giver or gift-receiver. Assume that the sample consists of 50 individuals. Use a random number generator to randomly assign 25 individuals to play the gift-receiver role and 25 to play the gift-giver role.
In a study designed to determine whether babies have an innate sense of morality, babies were shown two puppet shows in a random order: one of them had a puppet being nice, and the other had a different puppet being mean. The babies were then given the opportunity to reach for either the nice puppet or the mean puppet, and the researchers recorded which puppet the babies reached for. Suppose that out of 23 babies in the study, 15 of them reached for the nice puppet. Using the distribution you picked and observed 15 out of 23 babies reaching for the nice puppet, what conclusion should be drawn, and why?
Use either the critical-value approach or the P-value approach to perform the required hypothesis test. Approximately 450,000 vasectomies are performed each year in the United States. In this surgical procedure for contraception, the tube carrying sperm from the testicles is cut and tied. Several studies have been conducted to analyze the relationship between vasectomies and prostate cancer. The results of one such study by E. Giovannucci et al. appeared in the paper “A Retrospective Cohort Study of Vasectomy and Prostate Cancer in U.S. Men” (Journal of the American Medical Association, Vol. 269(7), pp. 878-882). Of 21,300 men who had not had a vasectomy, 69 were found to have prostate cancer, of 22,000 men who had had a vasectomy, 113 were found to have prostate cancer. a. At the 1% significance level, do the data provide sufficient evidence to conclude that men who have had a vasectomy are at greater risk of having prostate cancer? b. Is this study a designed experiment or an observational study? Explain your answer. c. In view of your answers to parts (a) and (b), could you reasonably conclude that having a vasectomy causes an increased risk of prostate cancer? Explain your answer.
A bank wants to know which of two incentive plans will most increase the use of its credit cards. It offers each incentive to a group of current credit card customers, determined at random, and compares the amount charged during the following six months. What type of study design is being used to produce data? |
The rectangular coordinates of a point are (4, −4). Plot the point and find two
The rectangular coordinates of a point are (4, −4). Plot the point and find two sets of polar coordinates for the point for 0 < 2.
The rectangular coordinates of a point are (4, −4). Plot the point and find two sets of polar coordinates for the point for
0<2
\left(x,y\right)=\left(4,-4\right)
For the polar coordinates
r=\sqrt{{x}^{2}+{y}^{2}}\text{ }and\text{ }\theta =ta{n}^{-}1\left(\frac{y}{x}\right)
r=\sqrt{16+16}
r=\sqrt{32}
r=4\sqrt{2}
\theta =ta{n}^{-}1\left(\frac{-4}{4}
\theta =ta{n}^{-}1\left(-1\right)
\theta =\frac{\pi }{4}
The two sets of polar coordinates are
\left(-4\sqrt{2},\frac{\pi }{4}\right).\left(4\sqrt{2},\frac{\pi }{4}\right)
{z}^{2}={x}^{2}+{y}^{2}
f\left(x,y\right)=-\frac{xy}{{x}^{2}+{y}^{2}}
Find limit of
f\left(x,y\right)\text{ }\text{as}\text{ }\left(x,y\right)\text{ }\to \left(0,0\right)\text{ }\text{i)Along y axis and ii)along the line}\text{ }y=x.\text{ }\text{Evaluate Limes}\text{ }\underset{x,y\to \left(0,0\right)}{lim}y\mathrm{log}\left({x}^{2}+{y}^{2}\right)
,by converting to polar coordinates.
At what value of t does the curve
x=2t-3{t}^{2},y={t}^{2}-3t
have a vertical tangent?
The rectangular coordinates of a point are given (2, -2). Find polar coordinates of the point. Express theta in radians.
Find, correct to four decimal places, the length of the curve of intersection of the cylinder
4{x}^{2}+{y}^{2}=4
x+y+z=5.
Express as a trigonometric function of one angle.
\mathrm{cos}2\mathrm{sin}\left(-9\right)-\mathrm{cos}9\mathrm{sin}2
\mathrm{sin}\left(\mathrm{arcsin}\frac{\sqrt{3}}{2}+\mathrm{arccos}0\right) |
The true stomach in ruminants where most of digestion takes place is
In ruminants (e.g., cow, goat and camel) the stomach is 4 chambered as follows
(a) Rumen (cellulose is digested)
(b) Reticulum (cellulose is digested)
(c) Omasum (absorb water)
(d) Abomasum (true stomach)
Enzymes are absent in
Virus is without necessary metabolic enzymes, hence, free virus are inert particles incapable of any vital activities and uses host machinery. Virus is regarded as obligate parasite and have characteristic of both living and non-livings.
Arachidonic acid is
Arachidonic acid is polyunsaturated (i.e., have more than one double bond) and essential fatty acids. Essential fatty acids cannot be synthesised by the body of animal and has to be taken from outside to fulfil the body requirement. Linoleic acid and linolenic acid are also essential fatty acids.
The vasopressin hormone, secreted by neurohypophysis of pituitary gland promotes the reabsorption of water from the distal convoluted tubules of nephrons, reducing excretion of water in urine (diuresis). Hence, it is called Antidiuretic Hormone (ADH). Its release in blood is controlled by an 'osmoregulatory centre' located in hypothalamus. Hyposecretion of ADH causes diabetes incipidus.
At high altitude, RBC's of human blood will
At altitude the partial pressure of the oxygen decreases in atmosphere so there is less oxygen available to carry out respiration. In order to compensate the cellular oxygen demand, the body increases the amount of RBC present to trap as many oxygen molecules as possible.
Striped muscles are
Striped (striated or skeletal or voluntary) muscles are syncytial. Nuclei are spindle shaped, peripheral in position and lie near the sarcolemma. They are multi nucleate because each fibre is formed by the fusion of a number of embryonic stem cells (myoblasts), hence, regarded multicellular syncytial body.
The basic unit of chitin is
The chitin (polyglycosamine) is an acetate of mucopolysaccharide called glycosamine which is formed by the combination of polysaccharide with small peptide molecules. The basic unit (monomer) of chitin is N-acetylglucosamine, monomers are joined by 1-4
\beta
linkages.
What type of enzyme are present in lysosome
Lysosomes or suicidal bags are bounded by single unit membrane. It contains hydrolytic enzymes which help to digest the nucleic acids, proteins and polysaccharide etc. (i.e., extracellular material). It also helps in autolysis.
Ventricles are related to
Ventricles are related to both heart and brain. Mammalian heart is four chambered. The upper two chambers known as right and left auricle and the lower as right and left ventricles. In brain there are four ventricles. Ventricle Ist and II (lateral ventricles) are cavities of two cerebral hemispheres. IIIrd ventricle (diocoel) is cavity of diencephalon and IV ventricle (metacoel) is cavity of medulla oblongata.
Which of the following amino acids are present in ornithine cycle
Valine and cystine
Ornithine cycle takes place in liver. The amino acid arginine and citrulline are formed during this cycle. Therefore it is referred as ornithine, arginine cycle and also Kreb-Henseleit cycle. The products of this cycle are urea and ornithine. The substances excreted through this cycle are CO2 and NH2 |
Distinct Objects into Distinct Bins | Brilliant Math & Science Wiki
Andy Hayes, Pranshu Gaba, Canwen Jiao, and
Distinct objects into distinct bins is a type of problem in combinatorics in which the goal is to count the number of possible distributions of objects into bins.
A distribution of objects into bins is an arrangement of those objects such that each object is placed into one of the bins. In this type of problem, the objects and bins are distinct. This means that it matters which objects go into which bin when counting distributions.
Derrick is eating lunch with his friends, Edward and Francine. Derrick has an apple, a banana, and a cherry in his lunch that he is thinking about sharing. Derrick can give some, all, or none of his fruit away (e.g., he can keep fruit for himself). In how many ways can Derrick distribute his fruit?
In this problem, the "distinct objects" are the fruit, and the "distinct bins" they are being distributed to are the people.
This problem can be solved by listing out all the possibilities, but a more efficient way to solve this problem is to use the rule of product. There are 3 possible destinations for the apple, 3 possible destinations for the banana, and 3 possible destinations for the cherry. Because these objects are distributed at the same time, the rule of product is used, and thus there are
3\times 3\times 3=\boxed{27}
possible distributions of the fruit.
_\square
Base Case for Distinct Objects into Distinct Bins
Distributing Part of a Set of Objects
In the previous example, there was no regard for fairness when determining the possible distributions of fruit. One of the possible distributions had Derrick keeping all the fruit for himself. Another possible distribution had Derrick giving all the fruit to Francine. In the base case of the "distinct objects into distinct bins" problem, each object is placed independently, and this allows for the distributions to be counted efficiently using the rule of product.
n
distinct objects that are to be distributed among
r
r^n
For each object, there are
r
bins it can be placed into. This placement occurs for each of the
n
objects. By the rule of product, this can be done in
\underbrace{r \times r \times r \times \cdots \times r} _{n \text{ times}} = r^n
_\square
We have 8 carrots of different sizes and 2 cute bunnies. In how many ways can we feed the bunnies? (The bunnies are very hungry, so they will eat all 8 carrots.)
In this question, we can model the carrots as objects and the bunnies as bins. Since they are all distinct, the above formula holds true. Therefore, the total number of ways in which the bunnies can be fed is
2^8 = 256
_\square
Note: If you are familiar with binary numbers, you can think of it this way. Suppose we have an 8-digit binary number--the number of digits corresponds to the number of carrots, and the base of the number corresponds to the number of bunnies. If a digit is 0, it means that a carrot is given to the first bunny; if a digit is 1, it means that a carrot is given to the second bunny. For example, the number 01100001 means that the
2^\text{nd}, 3^\text{rd},
8^\text{th}
carrots are given to the second bunny. There are a total of
2^8
8-digit binary numbers, so the answer is 256.
5
balls, each of a different color, be distributed among
3
distinct urns?
In this question, the balls are the objects and the urns are the bins. Since each ball has a different color, they are distinct, and the urns are distinct as well. Therefore, we can use the above formula and see this can be done in
3^5 = 243
_\square
There are 20 distinct people signed up for a raffle. There are 3 prizes that can be won: a 100 cm HDTV, a signed football jersey, and an envelope filled with gift cards. It is possible for a person to win more than one prize.
How many possible prize distributions are there?
Discount Al's Thrift Shop is having a one-day going-out-of-business sale in which everything must go!
Al only has four distinct things left in his shop.
All the years that Al has been in business, he's only ever had four distinct customers, and he does not expect that to change today
At the end of the day, the bank representative will come to foreclose the shop, and he will take whatever is left.
How many ways can Al's things be sold to his customers (or taken by the bank)?
The previous examples and problems covered situations in which all of the objects were distributed. However, what if only some of the objects are distributed? This opens up some interesting possibilities.
n
distinct objects, of which
k
are to be distributed among
r
\binom{n}{k}r^k
\binom{n}{k}
is notation for the binomial coefficient.
\binom{n}{k}
combinations of
k
n
distinct objects. These
k
objects are then distributed among
r
distinct bins. For each combination, there are
r^k
distributions of the
k
objects among the
r
bins. Thus, there are a total of
\binom{n}{k}r^k
k
objects, chosen from
n
distinct objects, into
r
_\square
Grandpa Joe has
8
different presents that he is considering giving to his
5
grandchildren. However, he decides that he'd like to keep
2
presents for himself. How many ways can he distribute the remaining presents?
In this problem, there are
8
6
5
distinct bins. Using the above theorem, this can be done in precisely
\binom{8}{6}5^6=28\times15625=\boxed{437500}
_\square
This result is actually more than the number of distributions of
8
5
distinct bins,
5^8=390625
At a farm auction, various pieces of farm equipment and livestock are bid on and sold.
Today, there are 7 distinct items up for auction, and there are 3 farmers bidding on them. One of the farmers declares that he will be bid on at least 2 of the items (not specifying which ones), and he won't be outbid.
If this farmer is telling the truth, then how many ways can the items be distributed among farmers?
Assume that all items are sold.
More interesting and challenging problems can be made by imposing additional conditions. Generic formula for this type,
5
3
distinct urns such that no urn remains empty?
The restriction that no urn is left empty seems harmless, but it makes the problem much more complicated than previous problems and examples in this wiki.
If we use the generic formula we directly get
3^5-3*(3-1)^5+3+0
243-3*32+3 = \boxed{150}
Other way is to solve this problem is by using the principle of inclusion and exclusion.
First, consider what the problem would be without the restriction. Let
U
be the set of all distributions of
5
3
|U|=3^5=243
A
5
distinct objects into the
1^\text{st}
2^\text{nd}
bins. Likewise, let
B
be the set of all distributions into the
1^\text{st}
3^\text{rd}
bins, and let
C
2^\text{nd}
3^\text{rd}
bins.
A\cup B\cup C
will be the set of all distributions in which at least one bin is left empty.
Our goal is to find the number of distributions in which no bin is left empty. This will be
|U|-|A\cup B\cup C|
\text{The goal area is shaded in blue}
By the principle of inclusion and exclusion,
|A\cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|.
|A|
is the number of distributions of
5
2
distinct bins. This is
|A|=2^5=32
. Without loss of generality,
|A|=|B|=|C|
|A\cap B|
5
1
distinct bin. This is
|A\cap B|=1
|A\cap B|=|A\cap C|=|B\cap C|
It is not possible for all three bins to be empty, so
|A\cap B\cap C|=0
Using the principle of inclusion and exclusion formula,
|A\cup B \cup C| =32+32+32-1-1-1+0=93.
Therefore, the number of distributions of
5
distinct balls into
3
distinct urns in which no urn is left empty is
|U|-|A\cup B\cup C|=243-93=\boxed{150}.\ _\square
Sri and Godfrey are marooned on a desert island.
Together, they have a toothbrush, a calculator, a volleyball, an mp3 player, a kite, and a shovel.
They decide to distribute these objects randomly among themselves. However, they agree that each person should get at least one object.
How many ways can the objects be distributed among Sri and Godfrey?
It is the end of the school year, and a teacher is giving out awards to her 3 students. She has 6 distinct awards (for grades, attendance, generosity, etc.) to give out, and she will give each award to the student who is most deserving.
However, she knows that her students can be rather immature, and one of them might throw a fit if he or she doesn't get an award. She secretly decides to make sure that each student gets at least one award (even if he or she doesn't deserve it).
How many ways can the awards be distributed among the students if all of the awards are given?
Cite as: Distinct Objects into Distinct Bins. Brilliant.org. Retrieved from https://brilliant.org/wiki/distinct-objects-into-distinct-bins/ |
Units in plot Ranges
The gcdex Command
Display of Examples in 2-D Math in the Help System
Function Assignments in 2-D Input
The plot and plot3d commands accept ranges with units, such as
x=2⟦m⟧..5⟦m⟧
\mathrm{pressure}=10⟦\mathrm{kPa}⟧..1000⟦\mathrm{kPa}⟧
. Before Maple 2018, you could use a range that had a unit on only the left- or the right-hand side, such as
x=2..5⟦m⟧
\mathrm{pressure}=10⟦\mathrm{kPa}⟧..1000
. This functionality was considered more confusing than helpful and therefore removed; you now need to supply a unit on both sides of a range.
For the calling sequence with 6 arguments, the gcdex command now displays an error message when the gcd of the first two polynomials does not divide the third polynomial.
In the Examples section of help pages, the display of the 2-D math inputs has changed in Maple 2018. To better match the calling sequences that are being demonstrated, examples are shown using the Maple language format, rather than the typeset format. For example,
\mathrm{int}\left(\mathrm{sin}\left(x\right),x=0..\phantom{\rule[-0.0ex]{0.5ex}{0.0ex}}\mathrm{\pi }\right)
{\int }_{0}^{\mathrm{\pi }}\mathrm{sin}\left(x\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}ⅆx
on the int/details help page.
When the functionassign setting (see Typesetting:-Settings) is true (its default value) and
f
is a procedure, then
f\left(\mathrm{args}\right)≔\mathrm{expression}
is now parsed as a remember-table assignment unless
f
has option function_assign, in which case it is parsed as a new function assignment. This change does not affect any procedures that are initially created via the function-assignment syntax, as option function_assign is now automatically added to these procedures.
In previous Maple releases, the kernel would automatically expand the product of a complex number with rational coefficients by
±\mathrm{\infty }
±I\mathrm{\infty }
s\mathrm{\infty }+t\mathrm{\infty }I
s,t∈\left\{-1,0,1\right\}
. As of Maple 2018, this returns a product
\left(p+qI\right)\mathrm{\infty }
p,q
are coprime integers. For example, both of the following examples used to return
\mathrm{\infty }+\mathrm{\infty }I
(2+2*I)*infinity;
\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{I}\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{\infty }}
a := 4/3-2*I;
\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{≔}\frac{\textcolor[rgb]{0,0,1}{4}}{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{I}
b := I*infinity;
\textcolor[rgb]{0,0,1}{b}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{\infty }}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{I}
\left(\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{I}\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{\infty }}
The previous behavior can be restored by applying the expand command:
\textcolor[rgb]{0,0,1}{\mathrm{\infty }}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{\infty }}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{I}
\textcolor[rgb]{0,0,1}{\mathrm{\infty }}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{\infty }}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{I} |
Engineering Acoustics/Attenuation of Sound Waves - Wikibooks, open books for an open world
Engineering Acoustics/Attenuation of Sound Waves
2 Types of Attenuation
2.1 Viscosity and Heat conduction
2.2 Boundary Layer Losses
3 Modeling of losses
When sound travels through a medium, its intensity diminishes with distance. This weakening in the energy of the wave results from two basic causes, scattering and absorption. The combined effect of scattering and absorption is called attenuation. For small distances or short times the effects of attenuation in sound waves can usually be ignored. Yet, for practical reasons it should be considered. So far in our discussions, sound has only been dissipated by the spreading of the wave, such as when we consider spherical and cylindrical waves. However this dissipation of sound in these cases is due to geometric effects associated with energy being spread over an increasing area and not actually to any loss of total energy.
Types of AttenuationEdit
As mentioned above, attenuation is caused by both absorption and scattering. Absorption is generally caused by the media. This can be due to energy loss by both viscosity and heat conduction. Attenuation due to absorption is important when the volume of the material is large. Scattering, the second cause of attenuation, is important when the volume is small or in cases of thin ducts and porous materials.
Viscosity and Heat conductionEdit
Whenever there is a relative motion between particles in a media, such as in wave propagation, energy conversion occurs. This is due to stress from viscous forces between particles of the medium. The energy lost is converted to heat. Because of this, the intensity of a sound wave decreases more rapidly than the inverse square of distance. Viscosity in gases is dependent upon temperature for the most part. Thus as you increase the temperature you increase the viscous forces.
Boundary Layer LossesEdit
A special type of absorption occurs when a sound wave travels over a boundary, such as a fluid flowing over a solid surface. In such a situation, the fluid in immediate contact with the surface must be at rest. Subsequent layers of fluid will have a velocity that increases as the distance from the solid surface increases such as in the figure below.
The velocity gradient causes an internal stress associated with viscosity, that leads to a loss of momentum. This loss of momentum leads to a decrease in the amplitude of a wave close to the surface. The region over which the velocity of the fluid decreases from its nominal velocity to that of zero is called the acoustic boundary layer. The thickness of the acoustic boundary layer due to viscosity can be expressed as
{\displaystyle \delta _{visc}={\sqrt {\left({\frac {2*\mu }{\omega *\rho _{o}}}\right)}}}
{\displaystyle \mu \,}
is the shear viscosity number. Ideal fluids would not have a boundary layer thickness since
{\displaystyle \mu =0\,}
Attenuation can also occur by a process called relaxation. One of the basic assumptions prior to this discussion on attenuation was that when a pressure or density of a fluid or media depended only on the instantaneous values of density and temperature and not on the rate of change in these variables. However, whenever a change occurs, equilibrium is upset and the media adjusts until a new local equilibrium is achieved. This does not occur instantaneously, and pressure and density will vary in the media. The time it takes to achieve this new equilibrium is called the relaxation time,
{\displaystyle \theta \,}
. As a consequence the speed of sound will increase from an initial value to that of a maximum as frequency increases. Again the losses associated with relaxation are due to mechanical energy being transformed into heat.
Modeling of lossesEdit
The following is done for a plane wave. Losses can be introduced by the addition of a complex expression for the wave number
{\displaystyle k=\ \beta -j\alpha }
which when substituted into the time-solution yields
{\displaystyle \ p=Ae^{-\alpha x}e^{jwt-j\beta x}}
with a new term of
{\displaystyle \ e^{-\alpha x}}
which resulted from the use of a complex wave number. Note the negative sign preceding
{\displaystyle \alpha }
to denote an exponential decay in amplitude with increase values of
{\displaystyle x}
{\displaystyle \ \alpha }
is known as the absorption coefficient with units of nepers per unit distance and
{\displaystyle \ \beta }
is related to the phase speed. The absorption coefficient is frequency dependent and is generally proportional to the square of sound frequency. However, its relationship does vary when considering the different absorption mechanisms as shown below.
The velocity of the particles can be expressed as
{\displaystyle \ u={\frac {k}{w*\rho _{o}}}p={\frac {1}{\rho _{o}c}}\left(1-j{\frac {\alpha }{k}}\right)p}
The impedance for this traveling wave would be given by
{\displaystyle \ z=\rho _{o}c{\frac {1}{1-j{\frac {\alpha }{k}}}}}
From this we can see that the rate of decrease in intensity of an attenuated wave is
{\displaystyle \ a=8.7\alpha }
Wood, A. A Textbook of Sound. London: Bell, 1957.
Blackstock, David. Fundamentals of Physical Acoustics. New York: Wiley, 2000.
Attenuation considerations in Ultrasound
Retrieved from "https://en.wikibooks.org/w/index.php?title=Engineering_Acoustics/Attenuation_of_Sound_Waves&oldid=3442538" |
Ensemble averaging (machine learning) - Wikipedia
Ensemble averaging (machine learning)
(Redirected from Ensemble Averaging)
For other uses, see Ensemble averaging (disambiguation).
In machine learning, particularly in the creation of artificial neural networks, ensemble averaging is the process of creating multiple models and combining them to produce a desired output, as opposed to creating just one model. Frequently an ensemble of models performs better than any individual model, because the various errors of the models "average out."
Ensemble averaging is one of the simplest types of committee machines. Along with boosting, it is one of the two major types of static committee machines.[1] In contrast to standard network design in which many networks are generated but only one is kept, ensemble averaging keeps the less satisfactory networks around, but with less weight.[2] The theory of ensemble averaging relies on two properties of artificial neural networks:[3]
Ensemble averaging creates a group of networks, each with low bias and high variance, then combines them to a new network with (hopefully) low bias and low variance. It is thus a resolution of the bias-variance dilemma.[4] The idea of combining experts has been traced back to Pierre-Simon Laplace.[5]
The theory mentioned above gives an obvious strategy: create a set of experts with low bias and high variance, and then average them. Generally, what this means is to create a set of experts with varying parameters; frequently, these are the initial synaptic weights, although other factors (such as the learning rate, momentum etc.) may be varied as well. Some authors recommend against varying weight decay and early stopping.[3] The steps are therefore:
Generate N experts, each with their own initial values. (Initial values are usually chosen randomly from a distribution.)
Train each expert separately.
Combine the experts and average their values.
Alternatively, domain knowledge may be used to generate several classes of experts. An expert from each class is trained, and then combined.
A more complex version of ensemble average views the final result not as a mere average of all the experts, but rather as a weighted sum. If each expert is
{\displaystyle y_{i}}
, then the overall result
{\displaystyle {\tilde {y}}}
{\displaystyle {\tilde {y}}(\mathbf {x} ;\mathbf {\alpha } )=\sum _{j=1}^{p}\alpha _{j}y_{j}(\mathbf {x} )}
{\displaystyle \mathbf {\alpha } }
is a set of weights. The optimization problem of finding alpha is readily solved through neural networks, hence a "meta-network" where each "neuron" is in fact an entire neural network can be trained, and the synaptic weights of the final network is the weight applied to each expert. This is known as a linear combination of experts.[2]
It can be seen that most forms of neural networks are some subset of a linear combination: the standard neural net (where only one expert is used) is simply a linear combination with all
{\displaystyle \alpha _{j}=0}
and one
{\displaystyle \alpha _{k}=1}
. A raw average is where all
{\displaystyle \alpha _{j}}
are equal to some constant value, namely one over the total number of experts.[2]
A more recent ensemble averaging method is negative correlation learning,[6] proposed by Y. Liu and X. Yao. Now this method has been widely used in evolutionary computing.
The resulting committee is almost always less complex than a single network that would achieve the same level of performance[7]
The resulting committee can be trained more easily on smaller input sets[1]
The resulting committee often has improved performance over any single network[2]
The risk of overfitting is lessened, as there are fewer parameters (weights) which need to be set[1]
^ a b c Haykin, Simon. Neural networks : a comprehensive foundation. 2nd ed. Upper Saddle River N.J.: Prentice Hall, 1999.
^ a b c d Hashem, S. "Optimal linear combinations of neural networks." Neural Networks 10, no. 4 (1997): 599–614.
^ a b Naftaly, U., N. Intrator, and D. Horn. "Optimal ensemble averaging of neural networks." Network: Computation in Neural Systems 8, no. 3 (1997): 283–296.
^ Geman, S., E. Bienenstock, and R. Doursat. "Neural networks and the bias/variance dilemma." Neural computation 4, no. 1 (1992): 1–58.
^ Clemen, R. T. "Combining forecasts: A review and annotated bibliography." International Journal of Forecasting 5, no. 4 (1989): 559–583.
^ Y. Liu and X. Yao, Ensemble Learning via Negative Correlation Neural Networks, Volume 12, Issue 10, December 1999, pp. 1399-1404. doi:10.1016/S0893-6080(99)00073-8
^ Pearlmutter, B. A., and R. Rosenfeld. "Chaitin–Kolmogorov complexity and generalization in neural networks." In Proceedings of the 1990 conference on Advances in neural information processing systems 3, 931. Morgan Kaufmann Publishers Inc., 1990.
Perrone, M. P. (1993), Improving regression estimation: Averaging methods for variance reduction with extensions to general convex measure optimization
Wolpert, D. H. (1992), "Stacked generalization", Neural Networks, 5 (2): 241–259, CiteSeerX 10.1.1.133.8090, doi:10.1016/S0893-6080(05)80023-1
Hashem, S. (1997), "Optimal linear combinations of neural networks", Neural Networks, 10 (4): 599–614, doi:10.1016/S0893-6080(96)00098-6, PMID 12662858
Hashem, S. and B. Schmeiser (1993), "Approximating a function and its derivatives using MSE-optimal linear combinations of trained feedforward neural networks", Proceedings of the Joint Conference on Neural Networks, 87: 617–620
Retrieved from "https://en.wikipedia.org/w/index.php?title=Ensemble_averaging_(machine_learning)&oldid=1057851906" |
The Littlewood-Offord problem for Markov chains
2021 The Littlewood-Offord problem for Markov chains
Shravas Rao
Shravas Rao1
1Northwestern University, United States of America
The celebrated Littlewood-Offord problem asks for an upper bound on the probability that the random variable
{\mathit{\epsilon }}_{1}{v}_{1}+\cdots +{\mathit{\epsilon }}_{n}{v}_{n}
lies in the Euclidean unit ball, where
{\mathit{\epsilon }}_{1},\dots ,{\mathit{\epsilon }}_{n}\in \left\{-1,1\right\}
are independent Rademacher random variables and
{v}_{1},\dots ,{v}_{n}\in {\mathbb{R}}^{d}
are fixed vectors of at least unit length. We extend some known results to the case that the
{\mathit{\epsilon }}_{i}
are obtained from a Markov chain, including the general bounds first shown by Erdős in the scalar case and Kleitman in the vector case, and also under the restriction that the
{v}_{i}
are distinct integers due to Sárközy and Szemeredi. In all extensions, the upper bound includes an extra factor depending on the spectral gap and additional dependency on the dimension. We also construct a pseudorandom generator for the Littlewood-Offord problem using similar techniques.
This material is based upon work supported by the National Science Foundation Graduate Research Fellowship Program under Grant No. DGE-1342536.
Shravas Rao. "The Littlewood-Offord problem for Markov chains." Electron. Commun. Probab. 26 1 - 11, 2021. https://doi.org/10.1214/21-ECP410
Received: 18 December 2020; Accepted: 16 June 2021; Published: 2021
Keywords: Littlewood-Offord , Markov chain , pseudorandom generator
Shravas Rao "The Littlewood-Offord problem for Markov chains," Electronic Communications in Probability, Electron. Commun. Probab. 26(none), 1-11, (2021) |
NLPSolve (Matrix Form) - Maple Help
Home : Support : Online Help : Mathematics : Optimization : Optimization Package : NLPSolve (Matrix Form)
Optimization[NLPSolve](Matrix Form)
solve a nonlinear program in Matrix Form
NLPSolve(n, p, nc, nlc, lc, bd, opts)
NLPSolve(n, p, lc, bd, opts)
\mathrm{posint}
\mathrm{nonnegint}
(optional) equation(s) of the form option = value where option is one of assume, constraintjacobian, feasibilitytolerance, infinitebound, initialpoint, iterationlimit, maximize, method, objectivegradient, optimalitytolerance, or output; specify options for the NLPSolve command
The NLPSolve command solves a nonlinear program (NLP), which involves computing the minimum (or maximum) of an objective function, possibly subject to constraints. Generally, a local minimum is returned unless the problem is convex. However, global search is available in limited situations, as described in the following Notes section. An NLP has the following form:
f\left(x\right)
v\left(x\right)\le 0
w\left(x\right)=0
A·x\le b
\mathrm{Aeq}·x=\mathrm{beq}
\mathrm{b1}\le x\le \mathrm{bu}
x
f\left(x\right)
x
v\left(x\right)
w\left(x\right)
x
b
\mathrm{beq}
\mathrm{bl}
\mathrm{bu}
A
\mathrm{Aeq}
are matrices. The relations involving matrices and vectors are element-wise.
Most of the algorithms used by the NLPSolve command assume that the objective function and the constraints are twice continuously differentiable. NLPSolve will sometimes succeed even if these conditions are not met.
This help page describes how to specify the problem in Matrix form. For details about the exact format of the objective function and the constraints, see the Optimization/MatrixForm help page. The algebraic and operator forms for specifying an NLP are described in the Optimization[NLPSolve] help page. The Matrix form is more complex, but leads to more efficient computation.
It is recommended that you use the Optimization[LPSolve] command for linear programs (problems with linear objective functions and linear constraints). Use the Optimization[QPSolve] command for quadratic programs (problems with quadratic objective functions and linear constraints). The Optimization[LSSolve] command is available for objective functions that can be put into least-squares form.
Consider the first calling sequence. The first parameter n is the number of problem variables. The second parameter p is a procedure that takes one input Vector parameter of size n, representing
x
, and returns the value of
f\left(x\right)
\mathrm{proc}\left(x,y\right)\mathrm{...}\mathrm{end proc}
x
v\left(x\right)
w\left(x\right)
y
The fifth parameter lc is an optional list of linear constraints. The most general form is
[A,b,\mathrm{Aeq},\mathrm{beq}]
, where A and Aeq are Matrices, and b and beq are Vectors. This parameter can take other forms if either inequality or equality constraints do not exist. For a full description of how to specify general linear constraints, see the Optimization/MatrixForm help page.
The sixth parameter
\mathrm{bd}
[\mathrm{bl},\mathrm{bu}]
n
If there are no nonlinear constraints, the second calling sequence, in which parameters nc and nlc are omitted, can be used.
initialpoint = Vector -- Use the provided initial point, which is an n-dimensional Vector of numeric values. The initial point is ignored when the quadratic interpolation method is used. For more information, see the Optimization/Methods help page.
iterationlimit = posint -- Set the maximum number of iterations performed by the algorithm. This option is only available when the method option is set to pcg or sqp.
objectivegradient = procedure -- Use the provided procedure to compute the gradient of the objective function. The form required for the procedure is described in the Nonlinear Objective section of the Optimization/MatrixForm help page.
The computation is performed in floating-point. Therefore, all data provided must have type realcons and all returned solutions are floating-point, even if the problem is specified with exact values. For best performance, Vectors and Matrices should be constructed with the datatype = float option and all procedures should work with evalhf. Because the solver fails when a complex value is encountered, it is sometimes necessary to add additional constraints to ensure that the objective function and constraints always evaluate to real values. For more information about numeric computation in the Optimization package and suggestions on how to obtain the best performance using the Matrix form of input, see the Optimization/Computation help page.
For certain methods, it is highly recommended that you provide derivatives of the objective function and constraints through the objectivegradient and constraintjacobian options, because NLPSolve performs more efficiently when this information is available. For information on the methods that use derivatives, see the Optimization/Methods help page.
The following example demonstrates how to specify a nonlinear program in Matrix form and solve it using the NLPSolve command.
Consider the objective function
{w}^{3}{\left(v-w\right)}^{2}+{\left(w-x-1\right)}^{2}+{\left(x-y-2\right)}^{2}+{\left(y-z-3\right)}^{2}
and constraints
w+x+y+z\le 5
3z+2v-3=0
Express the objective function as a procedure with the single parameter
V
representing the Vector with
v
x
w
y
z
as components.
As recommended previously, provide the gradient of the objective function using the objectivegradient option. Other Maple commands such as VectorCalculus[Gradient] can be helpful in constructing such procedures.
objgrd := proc (V, W)
Express the linear constraints in Matrix form.
A≔\mathrm{Matrix}\left([[0,1,1,1,1]],\mathrm{datatype}=\mathrm{float}\right):
b≔\mathrm{Vector}\left([5],\mathrm{datatype}=\mathrm{float}\right):
\mathrm{Aeq}≔\mathrm{Matrix}\left([[2,0,0,0,3]],\mathrm{datatype}=\mathrm{float}\right):
\mathrm{beq}≔\mathrm{Vector}\left([3],\mathrm{datatype}=\mathrm{float}\right):
\mathrm{lc}≔[A,b,\mathrm{Aeq},\mathrm{beq}]:
Solve the problem with NLPSolve, specifying that all variables must be non-negative. The second calling sequence is used because there are no nonlinear constraints.
\mathrm{Optimization}[\mathrm{NLPSolve}]\left(5,p,\mathrm{lc},\mathrm{objectivegradient}=\mathrm{objgrd},\mathrm{assume}=\mathrm{nonnegative}\right)
[\textcolor[rgb]{0,0,1}{6.43845963876504790}\textcolor[rgb]{0,0,1}{,}[\begin{array}{c}\textcolor[rgb]{0,0,1}{1.50000000000000}\\ \textcolor[rgb]{0,0,1}{1.93714208200858}\\ \textcolor[rgb]{0,0,1}{1.68857375382855}\\ \textcolor[rgb]{0,0,1}{1.37428416416287}\\ \textcolor[rgb]{0,0,1}{0.}\end{array}]] |
Probability Practice Problems Online | Brilliant
Probability and statistics are the foundation of most quantitative financial models. They allow us to interpret large amounts of data and model future outcomes in a way that humans could not do on their own.
Financial firms often need to determine how various assets relate to each other, creating potentially profitable opportunities to trade both simultaneously. Which of the following pairs of stocks would you expect to have the greatest correlation?
Bank of America (BA) and McDonald's (MCD) General Motors (GM) and Ford (F) Walmart (WMT) and Coca-Cola (KO)
Models of stocks (or other assets) often include a consideration of how erratically the price moves. Using your intuition, which of these stocks had the highest average daily percent change in 2015?
Coca-Cola (KO), a beverage company McDonald’s (MCD), a fast-food company Tesla (TSLA), an electric vehicle company
A biotechnology stock is currently trading at $12, and the company is releasing the results of a drug test tomorrow. An analyst tells you that if the test was successful, the stock should rise to $20; otherwise, it will fall to $8.
If the current stock price is equal to its future expected value, what is the probability that the drug test will be announced as successful as implied by the analyst?
\frac{1}{6}
\frac{1}{3}
\frac{2}{5}
\frac{1}{2}
Over time, you have found that the price of a certain asset seems to follow a Markov model in which it will increase or decrease minute-to-minute according to the model above. For example, if its price increases one minute, it is 80% likely to increase again in the next minute. In the long run, in what portion of all minutes does the stock price increase?
Hint: If it increases with probability
p,
it decreases with probability
1-p.
p
would provide a steady state in this chain?
\frac{2}{3}
\frac{3}{4}
\frac{4}{5}
\frac{5}{6}
Certain types of traders attempt to repeatedly buy and sell the same asset for a profit over a short time period, such as high-frequency “market makers”. For example, if you can repeatedly sell a stock for $8.50 and buy it for $8.49, you will make $0.01 each time.
If this transaction succeeds with probability 99%, about how many times can this transaction be executed before the probability of at least one failure exceeds 50%? |
How to level DexArm manually? - Rotrics Manual
This method only applied to users who can NOT level DexArm with Rotrics Studio or Touchscreen. If you've done the level procedure, please skip this section.
A. Open Rotrics Studio and click Terminal, send commandM891 X0 Y0to reset the slope value. Then reboot DexArm.
B. Place a piece of A4 paper between the print plate and the nozzle.
C. Adjust Z-axis for 4 points:
Open Rotrics Studio and click "Terminal", send commands "G0 X0 Y350 Z0" to move the arm to Point A, switch to "Basic" panel, and adjust module height with "Z-" button, keep adjusting until there is slight resistance on the A4 paper from the nozzle.
Send commandM114to read and record down the current Z-axis value. We will refer to this value as "ZA".
Open Rotrics Studio and click "Terminal", send commands "G0 X0 Y250 Z0" to move the arm to Point B, switch to "Basic" panel, and adjust module height with "Z-" button, keep adjusting until there is slight resistance on the A4 paper from the nozzle.
Send commandM114 to read and record down this Z-axis value. We will refer to this value as "ZB".
b) Point C
Open Rotrics Studio and click "Terminal", send commands "G0 X50 Y300 Z0" to move the arm to Point C, switch to "Basic" panel, and adjust module height with "Z-" button, keep adjusting until there is slight resistance on the A4 paper from the nozzle.
Send commandM114 to read and record down this Z-axis value. We will refer to this value as "ZC".
b) Point D
Open Rotrics Studio and click "Terminal", send commands "G0 X-50 Y300 Z0" to move the arm to Point C, switch to "Basic" panel, and adjust module height with "Z-" button, keep adjusting until there is slight resistance on the A4 paper from the nozzle.
Send commandM114 to read and record down this Z-axis value. We will refer to this value as "ZD".
D. Configure and set the slope value
Please calculate according to the following formula, keep the negative sign of Z-axis value.
Slope value of
Y axis=(ZA-ZB )/100
X axis=(ZC-ZD )/100
Then open "Terminal", and set slope values with command M891 XSlope YSlope, for example: M891 X0.02 Y0.02.
Reboot the Arm and the manual leveling is completed. |
On the Uniqueness of Solutions for the Identification of Linear Structural Systems | J. Appl. Mech. | ASME Digital Collection
Guillermo Franco,
The Earth Institute,
, 634A S. W. Mudd Building, New York, NY 10027
e-mail: franco@civil.Columbia.edu.
Raimondo Betti,
, 640 S. W. Mudd Building, New York, NY 10027
e-mail: betti@civil.Columbia.edu
e-mail: RWL14@Columbia.edu
Guillermo Franco Postdoctoral Research Fellow
Raimondo Betti Professor
Richard W. Longman Professor
J. Appl. Mech. Jan 2006, 73(1): 153-162 (10 pages)
Franco, G., Betti, R., and Longman, R. W. (July 30, 2005). "On the Uniqueness of Solutions for the Identification of Linear Structural Systems." ASME. J. Appl. Mech. January 2006; 73(1): 153–162. https://doi.org/10.1115/1.2062829
This work tackles the problem of global identifiability of an undamped, shear-type,
N
degrees of freedom linear structural system under forced excitation without any prior knowledge of its mass or stiffness distributions. Three actuator/sensor schemes are presented, which guarantee the existence of only one solution for the mass and stiffness identification problem while requiring a minimum amount of instrumentation (only 1 actuator and 1 or 2 sensors). Through a counterexample for a 3DOF system it is also shown that fewer measurements than those suggested result invariably in non-unique solutions.
elastic constants, actuators, sensors
Actuators, Degrees of freedom, Polynomials, Sensors, Stiffness, Shear (Mechanics), Excitation
Review and Unification of Linear Identifiability Concepts
Proceedings of the U. S. National Workshop on Structural Control Research
Some Uniqueness Results Related to Building Structural Identification
Uniqueness of Damping and Stiffness Distributions in the Identification of Soil and Structural Systems
Parameter Identification of Large Structural Systems in Time Domain
Genetic Algorithm in Structural Damage Detection
Identification of Structural Systems Using in Evolutionary Strategy
,” ASCE
Chelsea Publishing Co.
Solving Polynomial Systems for the Kinematic Analysis and Synthesis of Mechanism and Robot Manipulators
Electromechanical Coupling in Ionic Polymer-Metal Composites
Piezothermoelastic Characteristics and Control of Active Piezoelectric Structures |
Jordan–Hölder Theorem | Brilliant Math & Science Wiki
The Jordan–Hölder theorem is a theorem about composition series of finite groups. A composition series is a chain of subgroups
1 = H_0 \triangleleft H_1 \triangleleft H_2 \triangleleft \cdots \triangleleft H_{k-1} \triangleleft H_k = G,
H_i
is a maximal proper normal subgroup of
H_{i+1}.
By the third isomorphism theorem, this is equivalent to the statement that the quotient
H_{i+1}/H_i
is a simple group. This quotient is called a composition factor.
It is not hard to show that every finite group
G
has a composition series. The Jordan–Hölder theorem states that any two composition series of the same group have the same length and the same composition factors (up to permutation).
The cyclic group
{\mathbb Z}_6
has two composition series:
1 \triangleleft {\mathbb Z}_3 \triangleleft {\mathbb Z}_6 \, \, \text{and} \, \, 1 \triangleleft {\mathbb Z}_2 \triangleleft {\mathbb Z}_6.
{\mathbb Z}_3
is isomorphic to the subgroup
\{0,2,4\}
{\mathbb Z}_6,
{\mathbb Z}_2
\{0,3\}.
Note that both composition series have the same length, and both have the same composition factors, but in a different order:
{\mathbb Z}_6/{\mathbb Z}_3 \cong {\mathbb Z}_2
{\mathbb Z}_6/{\mathbb Z}_2 \cong {\mathbb Z}_3.
Here is a discussion of two assertions made in the introduction: that every finite group has a composition series, and that
H_i
being a maximal normal subgroup of
H_{i+1}
is equivalent to saying that
H_{i+1}/H_i
It is clear that every finite group has a composition series. To show that a maximal normal subgroup exists, the trivial subgroup is normal in every group, and if it is not maximal, there must be a larger normal subgroup containing it. Continue looking for larger normal subgroups until finding one that is maximal. The process is finite because the group is finite. The maximal normal subgroup is
H_{k-1}.
Run the same process on subgroups of
H_{k-1}.
The third isomorphism theorem states that normal subgroups of
G/N
are in one-to-one correspondence with normal subgroups of
G
N,
via the natural correspondence coming from the standard homomorphism
\pi \colon G \to G/N.
H_i
is maximal in
H_{i+1}
if and only if there is no normal subgroup strictly containing
H_i
and strictly contained in
H_{i+1},
which corresponds to there being no normal subgroup in
H_{i+1}/H_i
which is strictly bigger than
\{1\}
and strictly smaller than the whole quotient. This is the same as saying that
H_{i+1}/H_i
Warning: Normality is not a transitive relation. That is, if
K
H
H
G,
it is not necessarily true that
K
G.
Each of the subgroups in a composition series is normal in the succeeding one, but not necessarily normal in
G
itself. For example,
S_4
has a composition series
1 \triangleleft {\mathbb Z}_2 \triangleleft V_4 \triangleleft A_4 \triangleleft S_4
(see the simple group wiki for a derivation), but
{\mathbb Z}_2,
the subgroup generated by a double transposition, is not normal in
S_4.
G
be a finite group. Consider two composition series
\begin{aligned} 1 = H_0 \triangleleft H_1 \triangleleft H_2 \triangleleft \cdots \triangleleft H_{k-1} \triangleleft H_k &= G \\ 1 = K_0 \triangleleft K_1 \triangleleft K_2 \triangleleft \cdots \triangleleft K_{\ell-1} \triangleleft K_\ell &= G. \end{aligned}
k=\ell
and the list of composition factors is unique up to permutation; that is, the lists
\{H_{i+1}/H_i\}
\{K_{j+1}/K_j\}
are the same, after rearranging one of the lists suitably.
Unique factorization: The Jordan–Hölder theorem can be viewed as a generalization of the fundamental theorem of arithmetic that every integer can be factored as a product of prime numbers, essentially uniquely (up to permutation of the factors).
In fact, it is not hard to show that the fundamental theorem of arithmetic follows from Jordan–Hölder: consider the cyclic group
{\mathbb Z}_n.
Every subgroup is normal, and a maximal subgroup in a cyclic group is precisely one which has prime index. (This is because the quotient, being abelian, is simple if and only if it contains no nontrivial proper subgroups, which is only true if the quotient has prime order.) So a composition series for
{\mathbb Z}_n
must have the form
1 \triangleleft {\mathbb Z}_{p_1} \triangleleft {\mathbb Z}_{p_1p_2} \triangleleft \cdots \triangleleft {\mathbb Z}_{p_1p_2\cdots p_k} = {\mathbb Z}_n,
p_i
are primes. Since
{\mathbb Z}_n
has a composition series,
n
can be factored as a product of primes, and by Jordan–Hölder, a different factorization of
n
leads to a different composition series
1 \triangleleft {\mathbb Z}_{q_1} \triangleleft {\mathbb Z}_{q_1q_2} \triangleleft \cdots \triangleleft {\mathbb Z}_{q_1q_2\cdots q_\ell} = {\mathbb Z}_n,
which must have the same composition factors up to permutation
(
k=\ell).
But the orders of the composition factors are just the primes
p_i\ (\text{and } q_j),
so these lists are the same up to permutation, which is precisely the statement of the fundamental theorem of arithmetic.
Normal subgroups of the symmetric group: For
n\ge 5,
A_n
is simple, so there is a composition series
1 \triangleleft A_n \triangleleft S_n.
Since the composition factor sizes are
2
\frac{n!}{2},
this implies that the only other possible nontrivial normal subgroup of
S_n
2,
which can easily be ruled out by a simple analysis of the nontrivial element of the subgroup. So, for
n \ge 5,
the only nontrivial proper normal subgroup of
S_n
A_n.
This is related to solvability of polynomial equations by radicals.
Algebraic geometry: A slight generalization of Jordan–Hölder (to modules) allows for an elegant and robust definition of the multiplicity of the intersection of two algebraic curves. This is used in Bezout's theorem, which predicts the number of intersection points of two projective curves. The details of the definition are beyond the scope of this wiki, but the idea is that the intersection multiplicity is defined as the length of a certain module, which is defined to be the length of its composition series. The composition series are not unique, but they all have the same number of terms, thanks to Jordan–Hölder.
This proof is fairly technical. It will help to compare with the proof of the fundamental theorem of arithmetic, and to understand the second isomorphism theorem.
As with the fundamental theorem of arithmetic, the proof proceeds by induction, on
|G|.
|G|=1
is trivial. Now suppose the theorem has been proven for all groups strictly smaller than
G.
Take two composition series
(H_1,H_2,\ldots,H_k)
(K_1,K_2,\ldots,K_\ell)
G.
The theorem is true for
H = H_{k-1}
K = K_{\ell-1}.
H=K,
then we are done, as the composition series must be rearrangements of each other. If
H \ne K,
L = H \cap K.
L
has a composition series consisting of groups
L_j,
by the inductive hypothesis. Then there are two composition series for
H,
the one involving the
H_i
and the following one:
1 \triangleleft L_1 \triangleleft L_2 \triangleleft \cdots \triangleleft L_{t-1} \triangleleft L \triangleleft H.
(L=H \cap K
is a maximal subgroup of
H
H/(H\cap K) \cong G/K,
by the second isomorphism theorem.
)
By induction, this composition series must be a rearrangement of the other one:
(H_1/H_0,H_2/H_1,\ldots,H/H_{k-2}) \sim (L_1/L_0,L_2/L_1,\ldots,L/L_{t-1},H/L),
\sim
means "is the same up to permutation." Note that the lengths being the same implies that
t+1=k.
Similarly we get two composition series for
K,
using the same
L_i
series for the second one. That is,
(K_1/K_0,K_2/K_1,\ldots,K/K_{\ell-2}) \sim (L_1/L_0,L_2/L_1,\ldots,L/L_{t-1},K/L).
t+1=\ell.
k=\ell.
Now append
G/H
to the first pair of lists, and append
G/K
to the second pair of lists. This gives
\begin{aligned} (H_1/H_0,H_2/H_1,\ldots,H/H_{k-2},G/H) &\sim (L_1/L_0,L_2/L_1,\ldots,L/L_{t-1},H/L,G/H)\\ (L_1/L_0,L_2/L_1,\ldots,L/L_{t-1},K/L,G/K) &\sim (K_1/K_0,K_2/K_1,\ldots,K/K_{\ell-2},G/K). \end{aligned}
We want that the outer two lists are the same up to permutation. The inner two lists are the same except for the last two entries. But
(H/L,G/H)
(K/L,G/K)
are the same as
(G/K,G/H)
(G/H,G/K),
again by the second isomorphism theorem. So the inner two lists are the same up to permutation (a transposition of the last two factors), and the result follows.
_\square
Cite as: Jordan–Hölder Theorem. Brilliant.org. Retrieved from https://brilliant.org/wiki/jordan-holder/ |
Withdrawal Amount Calculations - DeFiner.org
How to calculate the withdrawal amount?
Interest (I): the interest earned by the user from the protocol and has not been withdrawn yet
Total Withdrawal Amount (TWA): the amount deducted from the deposit balance
Interest Reserve (IR): the amount of interest reserved for the protocol and only be deducted when the user withdraws the interest
Withdrawal Amount (WA): the final amount transferred to the user's wallet after the deduction of interest reserve from the total withdrawal amount
Interest Reserve Factor (IRF): the ratio between interest reserve and the total withdrawal amount.
When users withdraw funds from the savings contract, the interest will always be deducted first from the users' deposit balance. And there is a portion of the interest that will be reserved from users' deposit balance for the protocol upon the withdrawal of any interest. If the user didn't withdraw any interest, the reserved interest would be kept in the user's account and continue to generate interest. The interest reserve is only a portion of interest earned through the platform.
Total Withdrawal Amount =Withdrawal Amount +Interest Reserve
Total Withdrawal Amount <= Interest
Interest Reserve=Total Withdrawal Amount*Interest Reserve Factor
Total Withdrawal Amount > Interest,
Interest Reserve=Interest*Interest Reserve Factor
0<=IFR<=1 |
Tell whether the function represents exponential growth or exponential decay. Then graph the function. y=3e^-x
y=3{e}^{-x}
y=3{e}^{-x}
y=a{e}^{-kt}
y=3{e}^{-x}
Tell whether the function represents exponential growth or decay. z(x)=47(0.55)^x
Decide if the equation represents exponential growth or decay. Explain your answer.
y=\left(3{\right)}^{-2x}
State whether the equation represents exponential growth, exponential decay,f(x)=14⋅12x
Tell whether the function represents exponential growth or exponential decay.
f\left(x\right)=\frac{3}{5}{\left(\frac{5}{4}\right)}^{x}
Identify the function as exponential growth or exponential decay.Then identify the growth or decay factor.
y=1.3{\left(\frac{1}{4}\right)}^{x}
Determine whether the function represents exponential growth or exponential decay. Identify the percent rate of change. Then graph the function.
f\left(x\right)={\left(0.2\right)}^{x}
y=0.6\left(1/10{\right)}^{x} |
Find the maximum rate of change off at the given point and the direction in which it occur
Find the maximum rate of change off at the given point and the direction in which it occurs. f(x, y) = 4y \sqrt{x}, (4,1)
f\left(x,y\right)=4y\sqrt{x},\left(4,1\right)
a=\left(4,7,-4\right),b=\left(3,-1,1\right)
\left(a\cdot b\right)\cdot c
\left(a\cdot b\right)\cdot c
\left(a\cdot b\right)c
\left(a\cdot b\right)c
|a|\left(b\cdot c\right)
|a|\left(b\cdot c\right)
a\cdot \left(b+c\right)
a\cdot \left(b+c\right)
a\cdot b+c
a\cdot b+c
|a|\cdot \left(b+c\right)
|a|\cdot \left(b+c\right)
{M}_{2×4}
{M}_{3×4}\right)
{M}_{2×4}
Imagine a rope tied around the Earth at the equator. Show that you need to add only feet of length to the rope in order to lift it one foot above the ground around the entire equator. (You do NOT need to know the radius of the Earth to show this.)
\overline{A}=\left(2,1,-4\right),\overline{B}=\left(-3,0,1\right),\text{ and }\overline{C}=\left(-1,-1,2\right)
\overline{A}\cdot \overline{B}
\overline{A}\cdot \overline{B}\cdot \overline{C}
\overline{A}\left(\overline{B}\cdot \overline{C}\right)
\overline{A}\left(\overline{B}+\overline{C}\right)
3\overline{A}
{v}_{1}
{v}_{1}\text{ and }{v}_{2}
{v}_{1}\text{ and }{v}_{2}
v is a set of ordered pairs (a, b) of real numbers. Sum and scalar multiplication are defined by:
\left(a,b\right)+\left(c,d\right)=\left(a+c,b+d\right)k\left(a,b\right)=\left(kb,ka\right)
(attention in this part) show that V is not linear space. |
RTableSparseCompact - Maple Help
Home : Support : Online Help : Connectivity : Calling External Routines : ExternalCalling : C Application Programming Interface : RTableSparseCompact
remove zero entries of a NAG sparse rtable in external code
retrieve a NAG sparse rtable's index vector in external code
sort a NAG sparse rtable's index vectors in external code
set the number of stored elements of a NAG sparse rtable in external code
find the size of the data-block of a NAG sparse rtable in external code
resize the data-block of a NAG sparse rtable in external code
RTableSparseCompact(kv, rt)
RTableSparseIndexRow(kv, rt, dim)
RTableSparseIndexSort(kv, rt, by_dim)
RTableSparseSetNumElems(kv, rt, num)
RTableSparseSize(kv, rt)
RTableSparseResize(kv, rt, size)
type ALGEB rtable object
specify dimension of the rtable
by_dim
number of stored elements in the rtable
number of storable elements in the existing rtable data block
RTableSparseCompact removes zero entries from a NAG sparse rtable. Such rtables can be manipulated in external code, but on return to Maple, they must not contain zero entries or duplicates.
RTableSparseIndexRow retrieves the ith index vector from a NAG sparse rtable. NAG sparse rtables have one index vector for every dimension, plus a data vector. The ith entries of the index vectors combine to form the index that specifies the ith data entry.
RTableSparseIndexSort sorts the index vectors in a NAG sparse rtable. The by_dim parameter indicates which vector is sorted first. A value of 1 indicates that the row vector in a 2-D rtable is sorted first. A value of 2 indicates the column vector of the same rtable is sorted first. Internally Maple maintains the index vectors in two sections, the first part is assumed to be sorted, and the second part is unsorted. A sort is automatically triggered when the unsorted section becomes too large. A value of by_dim = -1 indicates that the order of the first part may have changed, so the entire data vector must be resorted. Otherwise, sorting by row assumes the first block is sorted so only sorts the second block, and then merges the two blocks. Changing the order without resorting results in unpredictable rtable access from Maple.
NAG sparse rtables usually have space for inserting new elements without reallocating the index and data vectors. If external code makes use of this space, then RTableSparseSetNumElems must be called to update the internal structure with the new number of elements stored in the rtable. The size of the data block can be retrieved by calling RTableSparseSize. To increase or reduce the size of the data block, use RTableSparseResize.
ALGEB M_DECL MyFillRight( MKernelVector kv, ALGEB *args )
M_INT argc, i, j, colbound, size, numelems;
ALGEB rt;
NAG_INT *row, *col;
if( rts.num_dimensions != 2 ) {
MapleRaiseError(kv,"2D rtable expected");
if( rts.storage != RTABLE_SPARSE ) {
MapleRaiseError(kv,"sparse rtable expected");
if( rts.data_type != RTABLE_FLOAT64 ) {
MapleRaiseError(kv,"float[8] rtable expected");
size = RTableSparseSize(kv,rt);
if( numelems == 0 )
RTableSparseIndexSort(kv,rt,1);
if( 2*numelems > size ) {
RTableSparseResize(kv,rt,2*numelems);
data = (FLOAT64*)RTableDataBlock(kv,rt);
row = RTableSparseIndexRow(kv,rt,1);
col = RTableSparseIndexRow(kv,rt,2);
colbound = RTableUpperBound(kv,rt,1);
for( i=0, j=numelems; i<numelems; ++i ) {
if( col[i] == colbound )
if( i < numelems-1 && row[i+1] == row[i] && col[i+1] == col[i]+1 )
row[j] = row[i];
col[j] = col[i]+1;
data[j++] = data[i];
RTableSparseSetNumElems(kv,rt,j);
\mathrm{with}\left(\mathrm{ExternalCalling}\right):
\mathrm{dll}≔\mathrm{ExternalLibraryName}\left("HelpExamples"\right):
\mathrm{dup_right}≔\mathrm{DefineExternal}\left("MyFillRight",\mathrm{dll}\right):
M≔\mathrm{Matrix}\left(5,5,\mathrm{storage}=\mathrm{sparse}[10],\mathrm{datatype}=\mathrm{float}[8]\right):
M[1,1]≔1:
M[2,5]≔2:
M[3,3]≔3:
M[3,4]≔4:
M[5,2]≔5:
M
[\begin{array}{ccccc}\textcolor[rgb]{0,0,1}{1.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}\\ \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{2.}\\ \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{3.}& \textcolor[rgb]{0,0,1}{4.}& \textcolor[rgb]{0,0,1}{0.}\\ \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}\\ \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{5.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}\end{array}]
\mathrm{dup_right}\left(M\right)
[\begin{array}{ccccc}\textcolor[rgb]{0,0,1}{1.}& \textcolor[rgb]{0,0,1}{1.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}\\ \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{2.}\\ \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{3.}& \textcolor[rgb]{0,0,1}{4.}& \textcolor[rgb]{0,0,1}{4.}\\ \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}\\ \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{5.}& \textcolor[rgb]{0,0,1}{5.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}\end{array}]
\mathrm{dup_right}\left(M\right)
[\begin{array}{ccccc}\textcolor[rgb]{0,0,1}{1.}& \textcolor[rgb]{0,0,1}{1.}& \textcolor[rgb]{0,0,1}{1.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}\\ \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{2.}\\ \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{3.}& \textcolor[rgb]{0,0,1}{4.}& \textcolor[rgb]{0,0,1}{4.}\\ \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}\\ \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{5.}& \textcolor[rgb]{0,0,1}{5.}& \textcolor[rgb]{0,0,1}{5.}& \textcolor[rgb]{0,0,1}{0.}\end{array}]
\mathrm{dup_right}\left(M\right)
[\begin{array}{ccccc}\textcolor[rgb]{0,0,1}{1.}& \textcolor[rgb]{0,0,1}{1.}& \textcolor[rgb]{0,0,1}{1.}& \textcolor[rgb]{0,0,1}{1.}& \textcolor[rgb]{0,0,1}{0.}\\ \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{2.}\\ \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{3.}& \textcolor[rgb]{0,0,1}{4.}& \textcolor[rgb]{0,0,1}{4.}\\ \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{0.}\\ \textcolor[rgb]{0,0,1}{0.}& \textcolor[rgb]{0,0,1}{5.}& \textcolor[rgb]{0,0,1}{5.}& \textcolor[rgb]{0,0,1}{5.}& \textcolor[rgb]{0,0,1}{5.}\end{array}] |
Physics Study Guide/Gravity - Wikibooks, open books for an open world
Physics Study Guide/Gravity
Newtonian Gravity (simplified gravitation) is an apparent force (a.k.a. pseudoforce) that simulates the attraction of one mass to another mass. Unlike the three fundamental (real) forces of electromagnetism and the strong and weak nuclear forces, gravity is purely attractive. As a force it is measured in newtons. The distance between two objects is measured between their centers of mass.
{\displaystyle F=G\!\cdot \!{\frac {m_{1}m_{2}}{r^{2}}}}
Gravitational force is equal to the product of the universal gravitational constant and the masses of the two objects, divided by the square of the distance between their centers of mass.
{\displaystyle g=G\cdot {\frac {m_{1}}{r^{2}}}}
The value of the gravitational field which is equivalent to the acceleration due to gravity caused by an object at a point in space is equal to the first equation about gravitational force, with the effect of the second mass taken out.
{\displaystyle U=-G\cdot {\frac {Mm}{r}}}
Gravitational potential energy of a body to infinity is equal to the universal gravitational constant times the mass of a body from which the gravitational field is being created times the mass of the body whose potential energy is being measured over the distance between the two centers of mass. Therefore, the difference in potential energy between two points is the difference of the potential energy from the position of the center of mass to infinity at both points. Near the earth's surface, this approximates:
{\displaystyle \Delta U_{g}=mgh}
Potential energy due to gravity near the earth's surface is equal to the product of mass, acceleration due to gravity, and height (elevation) of the object.
If the potential energy from the body's center of mass to infinity is known, however, it is possible to calculate the escape velocity, or the velocity necessary to escape the gravitational field of an object. This can be derived based on utilizing the law of conservation of energy and the equation to calculate kinetic energy as follows:
{\displaystyle {\begin{aligned}&{\boldsymbol {ke}}_{initial}=\Delta U\\\\&{\boldsymbol {ke}}_{initial}=U_{infinity}-U_{initial}\\\\&{\tfrac {1}{2}}mv^{2}=G\cdot {\frac {Mm}{r}}\\\\&v_{esc}={\sqrt {\frac {2GM}{r}}}\end{aligned}}}
F: force (N)
G: universal constant of gravitation, (6.67x10-11 N•m2/kg2)
m1: mass of the first body
m2: mass of the second body
r: the distance between the point at which the force or field is being taken, and the center of mass of the first body
g: acceleration due to gravity (on the earth’s surface, this is 9.8 m/s2)
U: potential energy from the location of the center of mass to infinity (J)
ΔUg: Change in potential energy (J)
m and M: mass (kg)
vesc: escape velocity (m/s)
Universal constant of gravitation (G): This is a constant that is the same everywhere in the known universe and can be used to calculate gravitational attraction and acceleration due to gravity.
6.67x10-11 N·m2/kg2
Mass one (m1): One of two masses that are experiencing a mutual gravitational attraction. We can use this for the mass of the Earth (1023 kg).
Mass two (m2): One of two masses that are experiencing a mutual gravitational attraction. This symbol can represent the mass of an object on or close to earth.
Units: kilograms (kg)
Acceleration due to gravity (g): This is nearly constant near the earth's surface because the mass and radius of the earth are essentially constant. At extreme altitudes the value can vary slightly, but it varies more significantly with latitude. This is also equal to the value of the gravitational field caused by a body at a particular point in space
(9.8 m/s2)
Escape velocity (vesc): The velocity necessary to completely escape the gravitational effects of a body.
A black hole is a geometrically defined region of space time exhibiting such large centripetal gravitational effects that nothing such as particles and electromagnetic radiation such as light may escape from inside of it. That is the escape velocity upon the event horizon is equivalent to the speed of light. General relativity is a metric theory of gravitation generalizing space time and Newton's law of universal gravitational attraction as a geometric property of space time.
Retrieved from "https://en.wikibooks.org/w/index.php?title=Physics_Study_Guide/Gravity&oldid=3288921" |
Find the NORMALIZED Gaussian factorization of 666. Also, find the
Find the NORMALIZED Gaussian factorization of 666. Also, find the number of ways 666 can be written as a sum of 2 integer squares.
66=2×3×3×37
=\left(1+i\right)\left(1-i\right)\left[\left(1+\sqrt{2i}\right)\left(1-\sqrt{2i}\right){\right]}^{2}\left(1+6i\right)\left(1-6i\right)
N\left(a+bi\right)={a}^{2}+{b}^{2}=666
This looks like a circle equation where the radius is
\sqrt{666}\approx 25.81
Thus, a,b < 25.81 as a and b are to be integers.
On close observation, the values of a,b are
(22,15),(-22,15),(22,-15),(-22,-15)
Therefore, the number of ways in which 666 can be written in sum of 2 integer squares is 8.
g\left(x\right)=3-\frac{{x}^{2}}{4}
Which of the following polynomials in
{Z}_{3}\left[x\right]
is irreducible?
p\left(x\right)={x}^{3}+x+1
p\left(x\right)={x}^{4}+1
(c) Factorize the polynomials that are not irreducible.
Zeros:
−2,
2,
8;
degree: 3
Type a polynomial with integer coefficients and a leading coefficient of 1 in the box below.
How do you find the product of
-2x\left({x}^{2}-3\right)
\left(\frac{x}{5}\right)\left(\frac{x}{3}\right)
How do you write a polynomial in standard form, then classify it by degree and number of terms
-1+2{x}^{2}
\left(2{x}^{2}+{y}^{2}\right)\left(x-2y\right) |
Home : Support : Online Help : Statistics and Data Analysis : Statistics Package : Tests : OneSampleTTest
OneSampleTTest(X, mu0, test_options)
OneSampleTTest[SampleSize](width, sigma, samplesize_options)
(optional) equation(s) of the form option=value where option is one of alternative, confidence, ignore, output, summarize or weights; specify options for the OneSampleTTest function
realcons; a worst-case estimate on the value of the standard deviation
(optional) equation(s) of the form option=value where option is one of confidence or iterations; specify options for the OneSampleTTest[SampleSize] utility function
The OneSampleTTest function computes the one sample t-test on a dataset X. This calculation is used to determine the significance of the difference between the sample mean and an assumed population mean when the standard deviation of the population is unknown.
The OneSampleTTest[SampleSize] utility computes the number of samples required in a data set in order to get a confidence interval with the specified width using this test.
The second parameter of the utility, sigma, specifies a worst-case estimate on the standard deviation of the sample size.
Vector of weights (one-dimensional rtable). If weights are given, the OneSampleTTest function will scale each data point to have given weight. Note that the weights provided must have type realcons and the results are floating-point, even if the problem is specified with exact values. Both the data array and the weights array must have the same number of elements.
confidence=float -- This option is used to specify the confidence level of the interval and must be a floating-point value between 0 and 1. By default this is set to 0.95.
iterations=posint -- This option specifies the maximum number of iterations to process when attempting to calculate the number of samples required. By default this is set to 100.
\mathrm{with}\left(\mathrm{Statistics}\right):
X≔\mathrm{Array}\left([9,10,8,4,8,3,0,10,15,9]\right):
\mathrm{Mean}\left(X\right)
\textcolor[rgb]{0,0,1}{7.60000000000000}
Calculate the one sample t-test on an array of values.
\mathrm{OneSampleTTest}\left(X,5,\mathrm{confidence}=0.95,\mathrm{summarize}=\mathrm{embed}\right):
\textcolor[rgb]{0,0,1}{10.}
\textcolor[rgb]{0,0,1}{7.60000}
\textcolor[rgb]{0,0,1}{4.24788}
\textcolor[rgb]{0,0,1}{\mathrm{StudentT}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{9}\right)
\textcolor[rgb]{0,0,1}{1.93554}
\textcolor[rgb]{0,0,1}{0.0849151}
\textcolor[rgb]{0,0,1}{4.56125}\textcolor[rgb]{0,0,1}{..}\textcolor[rgb]{0,0,1}{10.6387}
\mathrm{OneSampleTTest}\left(X,5,\mathrm{confidence}=0.95,\mathrm{alternative}='\mathrm{lowertail}',\mathrm{summarize}=\mathrm{true}\right)
Sample drawn from population with mean greater than 5
Sample drawn from population with mean less than 5
\textcolor[rgb]{0,0,1}{\mathrm{hypothesis}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{true}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{confidenceinterval}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{\infty }}\textcolor[rgb]{0,0,1}{..}\textcolor[rgb]{0,0,1}{10.0624132658958}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{distribution}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{StudentT}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{9}\right)\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{pvalue}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0.957542459363412}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{statistic}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{1.93553750114025}
As an alternative to using the summarize option, setting infoleve[Statistics] := 1 also returns the printed summary.
\mathrm{infolevel}[\mathrm{Statistics}]≔1:
\mathrm{OneSampleTTest}\left(X,5,\mathrm{confidence}=0.95,\mathrm{alternative}='\mathrm{uppertail}'\right)
Confidence Interval: 5.1375867341042 .. infinity
\textcolor[rgb]{0,0,1}{\mathrm{hypothesis}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{false}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{confidenceinterval}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{5.13758673410420}\textcolor[rgb]{0,0,1}{..}\textcolor[rgb]{0,0,1}{\mathrm{\infty }}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{distribution}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{StudentT}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{9}\right)\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{pvalue}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0.0424575406365881}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{statistic}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{1.93553750114025}
Calculate the number of samples required to produce a confidence interval of width 3, given a worst case standard deviation of 5.
\mathrm{OneSampleTTest}[\mathrm{SampleSize}]\left(3,5\right)
\textcolor[rgb]{0,0,1}{46}
The Statistics[OneSampleTTest] command was updated in Maple 2016. |
General Chemistry/Gases - Wikibooks, open books for an open world
General Chemistry/Gases
← Liquids ·Phase Changes →
← Liquids · General Chemistry · Phase Changes →
1 Characteristics of Gases
2.2 Ideal Gas Constant
3 Kinetic Molecular Theory
3.1 Kinetic Energy and Temperature
3.2 Pressure and Collisions
Characteristics of GasesEdit
Gases have a number of special characteristics that differentiate them from other states of matter. Here is a list of characteristics of gases:
Gases have low density, unless compressed. Being made of tiny particles in a large, open space, gases are very compressible.
Standard Temperature and PressureEdit
Wikipedia has related information at Standard conditions for temperature and pressure
Standard Temperature and Pressure, or STP, is 0 °C and 1 atmosphere of pressure. Expressed in other units, STP is 273 K and 760 torr. The Kelvin and torr are useful units of temperature and pressure respectively that we will discuss later in the following sections.
Avogadro's LawEdit
Amedeo Avogadro, the Italian chemist. Avogadro's Law is named after him and his discoveries about the behavior of gases
Wikipedia has related information at Avogadro's Law
Avogadro's Law states that equal volumes of gases at the same temperature and pressure contain the same number of molecules. So both one mole of Xenon at STP (131.3 grams) and one mole of helium at STP (4.00 grams) take up 22.4 liters. Even 1 mole of air, which is a mixture of several gases, takes up 22.4 liters of volume. 22.4 L is the standard molar volume of a gas.
[Avogadro's Law]
{\displaystyle {\frac {V}{n}}=k\,}
The most important consequence of Avogadro's law is that the ideal gas constant has the same value for all gases. This means that the constant
{\displaystyle {\frac {p_{1}\cdot V_{1}}{T_{1}\cdot n_{1}}}={\frac {p_{2}\cdot V_{2}}{T_{2}\cdot n_{2}}}=const}
has the same value for all gases, independent of the size or mass of the gas molecules.
Gases exert pressure on their containers and all other objects. Pressure is measured as force per unit area. A barometer is a device that measures pressure. There are a number of different units to measure pressure:
torr, equal to millimeters of mercury (mm Hg): if a glass cylinder with no gas in it is placed in a dish of liquid mercury, the mercury will rise in the cylinder to a certain number of millimeters.
atmosphere (atm), the pressure of air at sea level.
pascal (Pa), equal to one newton (N) per square meter. A newton is the force necessary to accelerate one kilogram by one meter per second squared.
You should know that 1 atm = 760 torr = 101.3 kPa.
Ideal GasesEdit
Wikipedia has related information at Ideal gas
Gases are complicated things composed a large numbers of tiny particles zipping around at high speeds. There are a number of complex forces governing the interactions between molecules in the gas, which in turn affect the qualities of the gas as a whole. To get around these various complexities and to simplify our study, we will talk about ideal gases.
An ideal gas is a simplified model of a gas that follows several strict rules and satisfies several limiting assumptions. Ideal gases can be perfectly modeled and predicted with a handful of equations.
Ideal gases follow, among others, these important rules:
Rules of Ideal Gases
The molecules that make up a gas are point masses, meaning they have no volume.
Gas particles are spread out with very great distance between each molecule. Thus, intermolecular forces are essentially zero, meaning they neither attract nor repel each other.
If collisions do occur between gas particles, these collisions are elastic, meaning there is no loss of kinetic (motion) energy.
Gas molecules are in continuous random motion.
Temperature is directly proportionate to kinetic energy.
Note: Ideal gases never truly exist (because the nature of gases is so complicated), but gases are often close enough to an ideal gas that the equations still hold fairly accurate.
Ideal Gas LawEdit
Ideal gases can be completely described using the ideal gas law:
[Ideal Gas Law]
{\displaystyle \ pV=nRT}
{\displaystyle \ p}
{\displaystyle \ V}
{\displaystyle \ n}
is the number of moles of gas,
{\displaystyle \ R}
{\displaystyle \ T}
is the absolute temperature, in Kelvin.
Ideal Gas ConstantEdit
The ideal gas constant, R, is a constant from the ideal gas equation, above, that helps to relate the various quantities together. The gas constant represents the same value, but the exact numerical representation of it may be different depending on the units used for each term. The table at right shows some values of R for different units. Here is the value of R using Joules for energy, Kelvin for temperature, and moles for quantity:
[Ideal Gas Constant]
{\displaystyle R=8.314472\ \ JK^{-1}mol^{-1}}
Real GasesEdit
All real gases (or non-ideal gases) deviate from the ideal gas laws that we discussed above. These deviations can occur for several reasons:
Real molecules have mass and volume. They are too big and no longer behave like ideal point masses
Low volumes and high pressures cause molecules to be close enough for intermolecular forces. Polar molecules exaggerate the problem.
Low temperature means low kinetic energy. At lower temperatures, intermolecular forces become significant and cannot be ignored like they are in ideal gasses
Other complicated factors may prevent ideal behavior.
When these issues are present, gas molecules attract each other, and may even condense into a liquid. Gases act most like ideal gases when the molecules have low mass (small volume), are not polar, and are at high temperature and low pressure. Noble gases like Xenon or Argon act the most like ideal gases because they are mostly electrical neutral and non-interactive.
Kinetic Molecular TheoryEdit
This theory describes why gases exhibit their properties. It only applies accurately to ideal gases. Because there is no such thing as an ideal gas, the Kinetic Molecular Theory can only approximate gas behavior. It is still very useful to chemists.
Wikipedia has related information at Kinetic theory
The Kinetic Molecular Theory explains the pressure, temperature, kinetic energy, and speed of gases and their molecules. See Wikipedia for the exact equations of the Kinetic Molecular Theory, as well as detailed explanations. What is most important is understanding the general concepts, not the specific equations.
Kinetic Energy and TemperatureEdit
Kinetic energy is the mechanical, or movement, energy. It is given by the equation:
{\displaystyle KE={\frac {1}{2}}mv_{rms}^{2}}
{\displaystyle m}
{\displaystyle v_{rms}}
is the average velocity
Explained with words, kinetic energy is dependent on the product of a particle's mass and its velocity squared. The more kinetic energy, the faster a particle moves. Conversely, the faster a particle moves, the more kinetic energy it has.
The Kinetic Molecular Theory states that kinetic energy and temperature are directly proportionate. Thus, a double in temperature will result in a double in kinetic energy and an increase in velocity by a factor of 1.4 (the square root of 2, see the KE equation). This means that the higher the temperature of a gas, the faster the individual particles in that gas are moving.
A hotter gas has more kinetic energy than a colder gas. If two gases are at the same temperature, they will have the same kinetic energy. The lighter-massed gas will have a higher average speed for its particles at the same energy level. It is important to know that gas temperature must be measured in kelvin. Zero degrees Celsius is 273 kelvin. One Celsius degree is equal to one kelvin, but the kelvin scale has water's freezing point at 273 and boiling point at 373. It is necessary to use kelvin because temperatures must always be positive when using the Kinetic Molecular Theory.
A gas's temperature is increased from 20 °C to 40° C. What factor does its kinetic energy increase? Velocity?
Keep in mind that gases are all about averages. For a temperature increase, there will be an average increase in the kinetic energy of the particles in that gas. Even in a very hot gas there will be some particles moving very slowly. However, the average will be high.
Pressure and CollisionsEdit
Pressure exists because the gas molecules are in continuous random motion, and they will constantly strike the walls of their container. Pressure will increase as the speed of the molecules increases, due to greater forces of collision. Pressure will also increase as the mass of the molecules increase. A small, slow molecule has less momentum than a large, fast molecule, which explains their difference in pressure.
There are two jars of ideal gas, for example. In Jar A there is nitrogen gas (N2). In Jar B there is methane gas (CH4). Both jars are at the same temperature. Which will have greater pressure?
Now, Jars A and B both have propane gas (C3H8). Jar A is at 300 K and Jar B is at 500 K. Which will have greater pressure?
Retrieved from "https://en.wikibooks.org/w/index.php?title=General_Chemistry/Gases&oldid=3596010" |
I have the following 2 samples data 1 : 59.09 59.17 59.27 59.13 59.10 59.14 59.54 59.
I have the following 2 samples
data 1 : 59.09 59.17 59.27 59.13 59.10 59.14 59.54 59.90
data 2: 59.06 59.40 59.00 59.12 59.01 59.25 59.23 59.564
And I need to check whether there is differentiation regarding the mean and the variance between the 2 data samples at significance level a=0.05
I think that the first thing I need to do is to check whether the samples come from a normal distribution in order to infer whether i should proceed using parametric or non marametric tests...
However using lillietest in matlab returned that both samples do not follow the normal distribution...
Any ideas on how should I proceed with checking the differentiation tests ? Should I perform ttest ? Or should I proceed by using something like Wilcoxon ? (p.s please confirm that that both data samples do not follow normal distribution...)
For a t-test, you need the samples to follow a normal distribution, so you're right to check this assumption first. I did a Shapiro-Wilk to test the normality, and it is rejected for the first sample, but it is not for the second sample. Thus you can't use a t-test.
The alternative is to use the Wilcoxon test, which is non-parametric.
Here is the code I have used with R :
data1 <- c(59.09, 59.17, 59.27, 59.13, 59.1, 59.14, 59.54, 59.9)
data2 <- c(59.06, 59.4, 59, 59.12, 59.01, 59.25, 59.23, 59.564)
shapiro.test(data1)
# P-value = 0.007987. Normality is rejected.
# P-value = 0.3873. Normality is not rejected.
wilcox.test(data1, data2)
# P-value = 0.5054. No significative difference.
A college professor randomly selects 25 freshmen to study the mathematical background of the incoming freshman class. The average SAT score of these 25 students is 565 and the standard deviation is estimated to be 40. Using this information, can this professor fully believe that the average test score of all incoming students is larger than 550 at a 5% level of significance?
Let K be the number of heads in 200 flips of a coin. The null hypothesis H is that the coin is fair. Devise significance tests with the folowng properties.
Note: Your answers below must be integers
a) The signficance level is
\alpha
=0.08 and the rejection set R has the form
\left\{|K-E\left[K\right]|>c\right\}.
Use the Central Limit theorem to find the accsptane ce set A.
A=\left\{\dots ,\dots ,\dots \right\}
b) Now tho significance level is
\alpha
= 0.016 and the rejection set R has the form
\left\{K>{c}^{1}\right\}
Again, use tho Central Limit Theorem to find tho accaptance set A.
A=\left\{\dots ,\dots ,\dots \right\}
Find the margin of error for the given values of c,s, and n. c=0.95, s=2.2, n=64
For a test of
{H}_{o}:p=0.5
, the z test statistic equals
1.52
. Find the p-value for
{H}_{a}:p>0.5
(a)Find the F-statistic from ANOVA table.
(b)Explain about three p-values for the three tests in parts(a),(b),and(c).
If a report states that certain data were used to reject a given hypothesis, would it be a good idea to know what type of test (one-tailed or two-tailed) was used? Explain. |
\textcolor[rgb]{0.470588235294118,0,0.0549019607843137}{\mathbf{ω}}
\mathrm{π}:E→M
be a fiber bundle, with base dimension
m
{\mathrm{π}}^{\mathrm{∞}}:{J}^{\mathrm{∞}}\left(E\right) → M
E
({x}^{i}, {u}^{\mathrm{α}}, {u}_{{i}_{}}^{\mathrm{α}}, {u}_{{i}_{}j}^{\mathrm{α}}
{u}_{\mathrm{ij} \cdot \cdot \cdot k}^{\mathrm{α}}, ....)
{J}^{\infty }\left(E\right)
{\mathrm{dx}}^{i}
M
{\mathrm{Θ}}^{\mathrm{α}} = {\mathrm{du}}^{\mathrm{α}}-{u}_{\mathrm{ℓ}}^{\mathrm{α}}{\mathrm{dx}}^{\mathrm{ℓ}},
{\mathrm{Θ}}_{i}^{\mathrm{α}} = {\mathrm{du}}_{i}^{\mathrm{α}}-{u}_{i\mathrm{ℓ}}^{\mathrm{α}}{\mathrm{dx}}^{\mathrm{ℓ}\mathit{ }}, .... ,
{\mathrm{Θ}}_{\mathrm{ij}\cdot \cdot \cdot k}^{\mathrm{α}} = {\mathrm{du}}_{\mathrm{ij}\cdot \cdot \cdot k}^{\mathrm{α}}-{u}_{\mathrm{ij}\cdot \cdot \cdot \mathrm{kℓ}}^{\mathrm{α}} {\mathrm{dx}}^{\mathrm{ℓ}} , ....
{\mathrm{dΘ}}^{\mathrm{α}} = {\mathrm{dx}}^{\mathrm{ℓ}} ∧ {\mathrm{Θ}}_{\mathrm{ℓ}\mathit{ }}^{\mathrm{\alpha }},
{\mathrm{Θ}}_{i}^{\mathrm{α}} = {\mathrm{dx}}^{\mathrm{ℓ}} ∧ {\mathrm{\Theta }}_{\mathrm{iℓ}\mathit{ }}^{\mathrm{α}}, .... ,
{\mathrm{dΘ}}_{\mathrm{ij}\cdot \cdot \cdot k}^{\mathrm{α}} = {\mathrm{dx}}^{\mathrm{ℓ}} ∧ {\mathrm{\Theta }}_{\mathrm{ij}\cdot \cdot \cdot \mathrm{kℓ}\mathit{ }}^{\mathrm{α}}.
p-
\mathrm{ω} ∈ {\mathrm{Ω}}^{p}\left({J}^{\mathrm{∞}}\right)
\left(r,s\right)
r 1
M
s
\mathrm{ω} = {A}_{{i}_{1}{i}_{2}\cdot \cdot \cdot {i}_{r} {a}_{1} \cdot \cdot \cdot {a}_{s}}^{ }{\mathrm{dx}}^{{i}_{1}}∧{\mathrm{dx}}^{{i}_{2} }∧ \cdot \cdot \cdot ∧{\mathrm{dx}}^{{i}_{r}} ∧ {C}^{{a}_{1}}∧{C}^{{a}_{2}} \cdot \cdot \cdot ∧{C}^{{a}_{s}},
{C}^{{a}_{k}}
p
{\mathrm{Ω}}^{p}\left({J}^{\mathrm{∞}} \right) = \underset{r+s =p}{\overset{}{⨁}} {\mathrm{Ω}}^{\left(r,s\right)}\left({J}^{\mathrm{∞}}\left(E\right)\right)
d:{\mathrm{\Omega }}^{\left(r,s\right)}\left({J}^{\infty }\left(E\right)\right)→ {\mathrm{\Omega }}^{\left(r+1,s\right)}\left({J}^{\infty }\left(E\right)\right) ⊕{\mathrm{\Omega }}^{\left(r,s+1\right)}\left({J}^{\infty }\left(E\right)\right)
d = {d}_{H } + {d}_{V},
{d}_{H }:{\mathrm{\Omega }}^{\left(r,s\right)}\left({J}^{\infty }\left(E\right)\right)→ {\mathrm{\Omega }}^{\left(r+1,s\right)}\left({J}^{\infty }\left(E\right)\right)
{d}_{V }:{\mathrm{\Omega }}^{\left(r,s\right)}\left({J}^{\infty }\left(E\right)\right)→ {\mathrm{\Omega }}^{\left(r,s+1\right)}\left({J}^{\infty }\left(E\right)\right).
{d}_{H }
{d}_{V}
{d}_{H}∘{d}_{H} =0, {d}_{H}∘{d}_{V} + {d}_{V}∘{d}_{H} =0,
{d}_{V}∘{d}_{V} =0
{d}_{H}\left({x}^{i}\right) = {\mathrm{dx}}^{i }, {d}_{H}\left({u}_{\mathrm{ij} \cdot \cdot \cdot k}^{\mathrm{\alpha }}\right) = {u}_{\mathrm{ij} \cdot \cdot \cdot k\mathrm{ℓ}}^{\mathrm{α}} {\mathrm{dx}}^{\mathrm{ℓ}}, {d}_{H}\left({\mathrm{dx}}^{i}\right) = 0, {d}_{H}\left({\mathrm{Θ}}_{\mathrm{ij}\cdot \cdot \cdot k}^{\mathrm{\alpha }}\right) = {\mathrm{dx}}^{\mathrm{ℓ}} ∧ {\mathrm{Θ}}_{\mathrm{ij}\cdot \cdot \cdot k\mathrm{ℓ}}^{\mathrm{\alpha }}
{d}_{V}\left({x}^{i}\right) =0, {d}_{V}\left({u}_{\mathrm{ij} \cdot \cdot \cdot k}^{\mathrm{\alpha }}\right) = {\mathrm{Θ}}_{\mathrm{ij} \cdot \cdot \cdot k}^{\mathrm{α}} , {d}_{V}\left({\mathrm{dx}}^{i}\right) = 0, {d}_{V}\left({\mathrm{Θ}}_{\mathrm{ij}\cdot \cdot \cdot k}^{\mathrm{\alpha }}\right) = 0.
\textcolor[rgb]{0.470588235294118,0,0.0549019607843137}{\mathrm{ω}}
{d}_{H}\left(\mathrm{ω}\right)
\mathrm{ω}
M
\mathrm{with}\left(\mathrm{DifferentialGeometry}\right):
\mathrm{with}\left(\mathrm{JetCalculus}\right):
{J}^{2}\left(E\right)
E
\left(x, y, u, v\right) → \left(x, y\right)
\mathrm{DGsetup}\left([x,y],[u,v],E,2\right):
F≔f\left(x,y,u[],u[1],u[2]\right):
\mathrm{PDEtools}[\mathrm{declare}]\left(F,\mathrm{quiet}\right):
\mathrm{HorizontalExteriorDerivative}\left(F\right)
\left({\textcolor[rgb]{0,0,1}{f}}_{{\textcolor[rgb]{0,0,1}{u}}_{[]}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{f}}_{{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{f}}_{{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{2}}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{f}}_{\textcolor[rgb]{0,0,1}{x}}\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{Dx}}\textcolor[rgb]{0,0,1}{+}\left({\textcolor[rgb]{0,0,1}{f}}_{{\textcolor[rgb]{0,0,1}{u}}_{[]}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{f}}_{{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{f}}_{{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{2}}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{f}}_{\textcolor[rgb]{0,0,1}{y}}\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{Dy}}
\mathrm{ω1}≔A\left(x,y,u[],u[1],u[2]\right)\mathrm{Dx}+B\left(x,y,u[],u[1],u[2]\right)\mathrm{Dy}
\textcolor[rgb]{0,0,1}{\mathrm{ω1}}\textcolor[rgb]{0,0,1}{:=}\textcolor[rgb]{0,0,1}{A}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{[}\textcolor[rgb]{0,0,1}{]}}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{2}}\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{Dx}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{B}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{[}\textcolor[rgb]{0,0,1}{]}}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{2}}\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{Dy}}
\mathrm{HorizontalExteriorDerivative}\left(\mathrm{ω1}\right)
\textcolor[rgb]{0,0,1}{-}\left({\textcolor[rgb]{0,0,1}{A}}_{{\textcolor[rgb]{0,0,1}{u}}_{[]}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{A}}_{{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{A}}_{{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{2}}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{B}}_{{\textcolor[rgb]{0,0,1}{u}}_{[]}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{B}}_{{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{B}}_{{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{2}}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{A}}_{\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{B}}_{\textcolor[rgb]{0,0,1}{x}}\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{Dx}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{⋀}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{Dy}}
\mathrm{ω2}≔\mathrm{Cu}[2]\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&wedge\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{Cv}[2]
\textcolor[rgb]{0,0,1}{\mathrm{ω2}}\textcolor[rgb]{0,0,1}{:=}{\textcolor[rgb]{0,0,1}{\mathrm{Cu}}}_{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{⋀}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{Cv}}}_{\textcolor[rgb]{0,0,1}{2}}
\mathrm{HorizontalExteriorDerivative}\left(\mathrm{ω2}\right)
\textcolor[rgb]{0,0,1}{\mathrm{Dx}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{⋀}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{Cu}}}_{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{⋀}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{Cv}}}_{\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{Dx}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{⋀}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{Cv}}}_{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{⋀}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{Cu}}}_{\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{Dy}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{⋀}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{Cu}}}_{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{⋀}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{Cv}}}_{\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{Dy}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{⋀}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{Cv}}}_{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{⋀}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{Cu}}}_{\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}} |
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