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Convolution and polynomial multiplication - MATLAB conv - MathWorks Nordic
Polynomial Multiplication via Convolution
Vector Convolution
Central Part of Convolution
w = conv(u,v)
w = conv(u,v,shape)
w = conv(u,v) returns the convolution of vectors u and v. If u and v are vectors of polynomial coefficients, convolving them is equivalent to multiplying the two polynomials.
w = conv(u,v,shape) returns a subsection of the convolution, as specified by shape. For example, conv(u,v,'same') returns only the central part of the convolution, the same size as u, and conv(u,v,'valid') returns only the part of the convolution computed without the zero-padded edges.
Create vectors u and v containing the coefficients of the polynomials
{x}^{2}+1
2x+7
u = [1 0 1];
Use convolution to multiply the polynomials.
w contains the polynomial coefficients for
2{x}^{3}+7{x}^{2}+2x+7
Create two vectors and convolve them.
The length of w is length(u)+length(v)-1, which in this example is 9.
Create two vectors. Find the central part of the convolution of u and v that is the same size as u.
u = [-1 2 3 -2 0 1 2];
v = [2 4 -1 1];
w = conv(u,v,'same')
15 5 -9 7 6 7 -1
w has a length of 7. The full convolution would be of length length(u)+length(v)-1, which in this example would be 10.
u,v — Input vectors
Input vectors, specified as either row or column vectors. The vectors u and v can be different lengths or data types.
When u or v are of type single, then the output is of type single. Otherwise, conv converts inputs to type double and returns type double.
Subsection of the convolution, specified as 'full', 'same', or 'valid'.
Full convolution (default).
Central part of the convolution of the same size as u.
Only those parts of the convolution that are computed without the zero-padded edges. Using this option, length(w) is max(length(u)-length(v)+1,0), except when length(v) is zero. If length(v) = 0, then length(w) = length(u).
The convolution of two vectors, u and v, represents the area of overlap under the points as v slides across u. Algebraically, convolution is the same operation as multiplying polynomials whose coefficients are the elements of u and v.
Let m = length(u) and n = length(v) . Then w is the vector of length m+n-1 whose kth element is
w\left(k\right)=\sum _{j}u\left(j\right)v\left(k-j+1\right).
The sum is over all the values of j that lead to legal subscripts for u(j) and v(k-j+1), specifically j = max(1,k+1-n):1:min(k,m). When m = n, this gives
w(1) = u(1)*v(1)
w(2) = u(1)*v(2)+u(2)*v(1)
w(3) = u(1)*v(3)+u(2)*v(2)+u(3)*v(1)
w(n) = u(1)*v(n)+u(2)*v(n-1)+ ... +u(n)*v(1)
w(2*n-1) = u(n)*v(n)
The inputs u and v must be column vectors.
If shape is 'full' (default), then only one of u or v can be a tall array.
If shape is 'same' or 'valid', then v cannot be a tall array.
For information about C/C++ code generation limitations, see Variable-Sizing Restrictions for Code Generation of Toolbox Functions (MATLAB Coder).
conv2 | convn | deconv | filter | convmtx (Signal Processing Toolbox) | xcorr |
Talk:Tensor field - formulasearchengine
Talk:Tensor field
Template:Maths rating Template:Physics
The section on "twisting by a line bundle" and "half-density" is quite opaque as it currently is written. linas 00:21, 5 May 2005 (UTC)
I was redirected to this page from half-form (by clicking the link in the article on geometric quantization) but on this page there is no explanation to what a half-form is, if it is the same as a tensor field or something else. I have not found information on this anyewhere else. Erik J 10:41, 5 October 2005 (UTC)
I'm planning on fixing this, but at the rate I'm going, it'll take another 6 months. Mean time line bundle and Hopf bundle may help. linas 14:02, 5 October 2005 (UTC)
OK, a half-form would be something like a square root of an n-form on an n-dimensional manifold. The question is, what is the geometric meaning of such a thing? It is not obvious that geometrically taking square roots is harmless (it isn't for complex numbers). But you are right that it is time to support this all by references (Guillemin-Sternberg, Geometric Asymptotics, p. 251). Charles Matthews 21:28, 5 October 2005 (UTC)
Sounds like a mangled volume form, a jacobian-like thingy. Anyway, the difficulty of understanding geometric quantization will be far beyond resolving this issue. linas 23:17, 5 October 2005 (UTC)
1 discussion at Wikipedia talk:WikiProject Mathematics/related articles
3 Poor lead
5 Tensor field graphic
6 Amazing Introduction
7 The C∞(M) module explanation a bit conservative?
discussion at Wikipedia talk:WikiProject Mathematics/related articles
This article is part of a series of closely related articles for which I would like to clarify the interrelations. Please contribute your ideas at Wikipedia talk:WikiProject Mathematics/related articles. --MarSch 14:12, 12 Jun 2005 (UTC)
WikiProject class rating
Poor lead
The lead doesn't actually say what it is. It talks around it, but without any attempt at definition. Leads are supposed to define.- (User) Wolfkeeper (Talk) 23:39, 27 April 2009 (UTC)
I made an image demonstrating the concept if anyone wants to use it. LokiClock (talk) 23:04, 4 September 2009 (UTC)
Tensor field graphic
By analogy with the tensor field arrows, shouldn't the vector field illustration have 2 arrows for each cell. One arrow pointing out of the cell and one in the cell and their product is the resulting arrow? I think that is the sense of the final illustration. Each pair of arrows defines a product. YouRang? (talk) 02:23, 23 February 2010 (UTC)
My understanding (and this is more from intuition) is that in a plot of tensors any number of arrows greater than one represent a 2nd-order tensor of that dimension. Note that there are three arrows with three degrees of freedom; the tensors in question can be represented by 3x3 matrices, with each vector representing a separate three indices. To represent a 3rd-order tensor requires groups of such arrows (multiple 2nd-order tensors). To differentiate you could use color coding or make them the vertices of a polytope (with multiple polytopes being assigned to each point). Further methods are available, but every visualization method must make compromises, and in general higher-order tensors are notoriously difficult to visualize. I can understand why this would be the case much easier if the geometric complexity of the units of representation grows as the order grows. But if a 3rd-order tensor can be parametrized by three vectors, what then is the point of using the higher-order form? ᛭ LokiClock (talk) 23:46, 14 September 2010 (UTC)
I agree with User:YouRang?, and as you can see I have removed the graphic. For one thing, 3rd-order tensors are not parametrized by three vectors (you are confusing the tensor product with the direct sum). Anyway, the graphic was supposed to illustrate a 0th order, 1st order, and 2nd order tensor in two dimensions. But 2nd order tensors in two dimensions are represented in a basis by 2×2 matrices, not 3×3 matrices, and so it would be more natural to display a pair of vectors at each point. In any event, if there is to be such a graphic, then it would be better to also explain the meaning of the representation of a 2nd order tensor as a pair of vectors, lest readers leave with the false impression that an nth order tensor field is nothing more than an association of n vectors to each point of space. Sławomir Biały (talk) 13:04, 21 September 2010 (UTC)
I am trying to teach myself the basics of fluid mechanics in order to implement wind flow in a video game, and I have been intimidated all night by the constant references to tensors. The sentence in the introduction about how a tensor is a generalization of a scalar field and a vector field was exactly what I needed. I'm not scared anymore. That is a much simpler and more concise explanation than is found in the actual article on tensors. —Preceding unsigned comment added by 65.50.39.118 (talk) 05:31, 7 September 2010 (UTC)
The C∞(M) module explanation a bit conservative?
Is there any particular reason that it only assumes that
{\displaystyle {\mathcal {T}}(M)}
is a module over a ring, instead of a vector space over a field? It seems to me that C∞(M) would definitely qualify as a field (just using point-wise addition and multiplication), and that
{\displaystyle {\mathcal {T}}(M)}
could just as easily be a vector space. It could be that I'm overlooking some subtleties, but if so, I'd love to hear what they are. (Perhaps it would be good in general to add one or two references.) --Jaspervdg (talk) 10:29, 28 October 2011 (UTC)
The function f(x)=x is a C∞ function on the real numbers. Its inverse, if it had one, would be g(x)=1/x. But this is not even continuous on the real numbers, i.e. C∞(R) is not a field. RobHar (talk) 14:13, 28 October 2011 (UTC)
Retrieved from "https://en.formulasearchengine.com/index.php?title=Talk:Tensor_field&oldid=311525" |
Forward contract | Brilliant Math & Science Wiki
A forward contract (forward) is a non-standardized contract between two parties, to trade an asset at a specified price, and at a specified future date. The seller will deliver the underlying and the buyer will take delivery of the underlying. The price that is agreed on, is known as the forward price or the delivery price, and is determined when the contract is entered into. Since the price of the forward is dependent on the price of the asset, this is a derivative instrument.
A forward contract is very similar to a Futures contract. A forward does not trade on a centralized exchange, hence is regarded as an over-the-counter (OTC) instrument. While this allows the contract to be customized, it also results in a higher degree of default risk. Hence, forward contracts are not usually available to retail investors.
A forward contract is easily traded.
Features of a Forward
Pricing Forward Contracts
Each contract can be set up with it's own terms, and thus may be tailored to the individual.
A forward contract should state:
The amount and quality of the underlying asset. For example, 1000 bushels of Grade 4 corn.
The time of maturity
The date at which the underlying is to be delivered. This could be a specific date, or a time range.
The place of delivery for physical underlying The place(s) at which the underlying is to be delivered. As the contract is customizable, this is often determined by the buyer.
Mark-to-market or daily margin calls
This helps to reduce the risk that either party might default on the trade, as collateral (cash, the underlying, securities, etc) is provided as insurance. The margin calls can also be based upon trigger events, like credit worthiness or stock price.
The delivery price
K
The price that is to be paid. This could be quoted as a per unit amount ($3.00 per bushel) or a total amount ($3000).
A forward contract is similar to a Futures contract, in that they agree to trade an underlying at a specified future date. However, forward contracts are
1. not cleared through an exchange,
2. not standardized,
3. might not require posting a margin,
4. have significant counter-party risk.
Nothing. Your contract is with the exchange. The SEC will guarantee your contract The bankruptcy court will decide a suitable settlement Your futures contract is invalidated You sell it back for $0 because they are bankrupt.
You buy a May Gold Forward contract from a Lehman Brothers trader. What happens when Lehman collapsed?
The value of a forward position at maturity is the difference between the delivery price
K
and the underlying price
S_T
at the time of maturity.
For a long position, the payoff is
S_T - K
, and it will benefit from a higher underlying price.
For a short position, the payoff is
K - S_T
, and it will benefit from a lower underlying price.
For further details, see Pricing Forward and Futures.
The value of underlying at today's prices (receiving delivery today) can be priced as
S_0 = F e ^ { - rt }.
However, the counter-party risk would also need to be priced in, as these contracts are generally not liquid.
Cite as: Forward contract. Brilliant.org. Retrieved from https://brilliant.org/wiki/forward-contract/ |
Braid monodromy of univariate fewnomials
Alexander Esterov and Lionel Lang
{\mathsc{𝒞}}_{d}\subset {ℂ}^{d+1}
be the space of nonsingular, univariate polynomials of degree
d
. The Viète map
\mathsc{𝒱}:{\mathsc{𝒞}}_{d}\to {Sym}_{d}\left(ℂ\right)
sends a polynomial to its unordered set of roots. It is a classical fact that the induced map
{\mathsc{𝒱}}_{\ast }
at the level of fundamental groups realises an isomorphism between
{\pi }_{1}\left({\mathsc{𝒞}}_{d}\right)
and the Artin braid group
{B}_{d}
. For fewnomials, or equivalently for the intersection
\mathsc{𝒞}
{\mathsc{𝒞}}_{d}
with a collection of coordinate hyperplanes in
{ℂ}^{d+1}
, the image of the map
{\mathsc{𝒱}}_{\ast }:{\pi }_{1}\left(\mathsc{𝒞}\right)\to {B}_{d}
is not known in general.
We show that the map
{\mathsc{𝒱}}_{\ast }
is surjective provided that the support of the corresponding polynomials spans
ℤ
as an affine lattice. If the support spans a strict sublattice of index
b
, we show that the image of
{\mathsc{𝒱}}_{\ast }
is the expected wreath product of
ℤ∕bℤ
{B}_{d∕b}
. From these results, we derive an application to the computation of the braid monodromy for collections of univariate polynomials depending on a common set of parameters.
https://projecteuclid.org/gt
braid group, monodromy, fewnomial, tropical geometry
Primary: 20F36, 55R80, 14T05
Proposed: Mladen Bestvina
Seconded: Jim Bryan, Benson Farb |
De Broglie Hypothesis | Brilliant Math & Science Wiki
Akshay Yadav, Abhiram Rao, and Jimin Khim contributed
Today we know that every particle exhibits both matter and wave nature. This is called wave-particle duality. The concept that matter behaves like wave is called the de Broglie hypothesis, named after Louis de Broglie, who proposed it in 1924.
Explanation of Bohr's Quantization Rule
De Broglie gave the following equation which can be used to calculate de Broglie wavelength,
\lambda
, of any massed particle whose momentum is known:
\lambda = \frac{h}{p},
h
is the Plank's constant and
p
is the momentum of the particle whose wavelength we need to find.
With some modifications the following equation can also be written for velocity
(v)
or kinetic energy
(K)
of the particle (of mass
m
\lambda = \frac{h}{mv} = \frac{h}{\sqrt{2mK}}.
Notice that for heavy particles, the de Broglie wavelength is very small, in fact negligible. Hence, we can conclude that though heavy particles do exhibit wave nature, it can be neglected as it's insignificant in all practical terms of use.
Calculate the de Broglie wavelength of a golf ball whose mass is 40 grams and whose velocity is 6 m/s.
\lambda = \frac{h}{mv} = \frac{6.63 \times 10^{-34}}{40 \times 10^{-3} \times 6} \text{ m}=2.76 \times 10^{-33} \text{ m}.\ _\square
One of the main limitations of Bohr's atomic theory was that no justification was given for the principle of quantization of angular momentum. It does not explain the assumption that why an electron can rotate only in those orbits in which the angular momentum of the electron,
mvr,
is a whole number multiple of
\frac{h}{2\pi}
De Broglie successfully provided the explanation to Bohr's assumption by his hypothesis.
Cite as: De Broglie Hypothesis. Brilliant.org. Retrieved from https://brilliant.org/wiki/debroglie-hypothesis/ |
Write the expression shown on each of the Expression Mats below. Then simplify them by making zeros and combining like terms.
Notice that a grey tile and a white tile of the same value cancel each other out and make a zero. Circle around the positive x and the negative x tiles, with the tiles crossed out, labeled 0. A second circle around the positive unit and one of the negative unit tiles, with the tiles crossed out, labeled 0.
Now write an expression for the values that remain. A circle is added around the remaining 2 unit tiles.
−2
2x^2+3x-1
Use the eTool below to solve part (a). |
Import Frequency-Domain Data into the App - MATLAB & Simulink - MathWorks América Latina
Importing Frequency-Domain Input/Output Signals into the App
Importing Frequency-Response Data into the App
Importing Complex-Valued Frequency-Response Data
Importing Amplitude and Phase Frequency-Response Data
Frequency-domain data consists of Fourier transforms of time-domain data (a function of frequency).
Before you can import frequency-domain data into the System Identification app, you must import the data into the MATLAB® workspace, as described in Frequency-Domain Input/Output Signal Representation.
The input and output signals must have the same number of data samples.
In the System Identification app window, select Import data > Freq. domain data. This action opens the Import Data dialog box.
Frequency — Enter the MATLAB variable name of a vector or a MATLAB expression that represents the frequencies. The expression must evaluate to a column vector.
The frequency vector must have the same number of rows as the input and output signals.
Frequency unit — Enter Hz for Hertz or keep the rad/s default value.
Sample time — Enter the actual sample time in the experiment. For continuous-time data, enter 0. For more information about this setting, see Specifying the Data Sample Time.
(Optional) In the Data Information area, click More to expand the dialog box and enter the following optional settings:
foh (first-order hold) indicates that the output was piecewise-linear during data acquisition.
bl (bandwidth-limited behavior) specifies that the continuous-time input signal has zero power above the Nyquist frequency (equal to the inverse of the sample time).
See the d2c and c2d reference page for more information about transforming between discrete-time and continuous-time models.
Naming channels helps you to identify data in plots. For multivariable input and output signals, you can specify the names of individual Input and Output channels, separated by commas.
Before you can import frequency-response data into the System Identification app, you must import the data into the MATLAB workspace, as described in Frequency-Response Data Representation.
To import frequency-response data consisting of complex-valued frequency values at specified frequencies:
In the Data Format for Signals list, select Freq. Function (Complex).
Response — Enter the MATLAB variable name or a MATLAB expression that represents the complex frequency-response data G(eiw).
To import frequency-response data consisting of amplitude and phase values at specified frequencies:
In the Data Format for Signals list, select Freq. Function (Amp/Phase).
Amplitude — Enter the MATLAB variable name or a MATLAB expression that represents the amplitude
|G|
Phase (deg) — Enter the MATLAB variable name or a MATLAB expression that represents the phase
\phi =\mathrm{arg}G |
Control system design of multi-dimensional lumbar traction treatment bed | JVE Journals
Yanying Luo1 , Liang Liu2 , Yunjia Liu3 , Jingjing Ma4
Copyright © 2019 Yanying Luo, et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
A multi-dimensional lumbar traction treatment bed is designed with two degrees of freedom, which can realize controllable traction treatment of lumbar through flexion, extension and rotation motion. Two linear actuators are used to provide motion. Building a mathematical model of the device by least squares identification. PID controller and Kalman filter constitute two groups of control modes: (i) speed control; (ii) position control. Using MATLAB to perform simulation experiments. The results show that the designed controller can achieve high control accuracy. The motion speed of lumbar platform is stable and the position of traction treatment set by user is approached exactly, which ensuring the security and stability of this device.
Keywords: lumbar traction treatment, least squares identification, PID control, Kalman filter.
Sedentary and sports injuries in modern people are leading to a younger lumbar disease trend [1]. In addition to the elderly, teenagers under the age of 18 also have a risk of disease [2]. Non-surgical treatments are thought to achieve effective therapeutic effects in the early stages of the illness. Traction is a common method of conservative treatment for lumbar disease. Several studies show that the swing traction therapy facilitated the patient's improvement in pain and relieve the lumbar disease [3-5]. What's more, the effect of multi-directional traction is probably superior to that of longitudinal traction in improving the symptoms and clinical findings of patients with lumbar disc herniation [6].
Compared with surgical treatment, traction therapy won't destroy the intervertebral disc. The risk of this method is much lower. Therefore, most patients will prefer such conservative treatment. Many medical device companies have committed to the study of traction beds. Lojer from Finland has designed Manuthera 242 [7]. This device use the traction, flexion, lateral flexion and rotation of table to treat the key positions of patient. However, Manuthera 242 is purely manual. Hill Laboratories has produced AIRFLEX [8], which can provide both manual and motorized flexion. But this kind of production can only realize limited freedom. None of the above devices have achieved precise control of traction speed and position.
The multi-dimensional lumbar traction treatment bed presented in this paper has two degrees of freedom, which can realize controllable traction treatment of lumbar through flexion, extension and rotation motion. Six-axis motion attitude sensor is used to evaluate the speed and position of lumbar platform. The position controller allows the platform to reach the set position stably and safely. Speed controller keeps the platform speed near the set speed during steady motion.
The design goal of the lumbar traction treatment bed is to provide controlled motion in selected spine portions. In order to achieve the above functions, this device includes a buckling and extending component and a rotating component as shown in Fig. 1. Both components use linear actuator as the active member to form a linkage mechanism which provides controlled motion of lumbar platform. The linear actuator has a peak speed of 20 mm/s. The stroke length is 100 mm. A six-axis motion attitude sensor is placed below the platform which is used to feedback the current position and speed. The speed and position signal is sent to the control borad. The motor of linear actuator is driven at 24 V using a 20 kHz PWM signal through driver.
Fig. 1. 3-D model of traction treatment bed
Fig. 2. Parameterization of platform without mattress
3.1. Motion model
As shown in Fig. 2, the device implements flexion, extension and rotation motion trough two crank-and-slide mechanisms in series. When the patient is lying down, the
x
axis is the coronal axis and the
y
axis is the vertical axis. The subscript
b
indicates the buckling and extending component with swing around
x
. The subscript r indicates the rotating component which swing around
y
{l}_{1}
is the distance between the rotating joint of platform and the bottom mounting position of linear actuator.
{l}_{2}
is the distance between the rotating joint of platform and the top mounting position of linear actuator.
l
the current length of linear actuator.
\theta
is the current position of platform.
\omega
is the current speed of platform.
For current length
l
l={l}_{c}+s,
{l}_{c}
means the initial installation distance of linear actuator.
s
means the current stroke of linear actuator.
From the triangle cosine theorem, there are:
\theta =\frac{{{l}_{1}}^{2}+{{l}_{2}}^{2}-{l}^{2}}{2{l}_{1}{l}_{2}}-{\theta }_{0},
{l}_{1}
{l}_{2}
{\theta }_{0}
is the initial angle of platform in the horizontal position.
Eq. (2) derives the time. Obtain the platform angular velocity
\omega
\omega =\frac{d\theta }{dt}=\frac{l}{{l}_{1}{l}_{2}\mathrm{sin}\theta }\bullet \frac{dl}{dt}.
In order to make the traction motion smooth, the angular velocity
\omega
of platform should be kept as constant as possible to improve patient comfort.
3.2. System model of the linear actuator
The principle of a linear actuator is that after DC motor is geared down, the screw nut converts the motor rotational motion into the linear motion. It is known that the transfer function between DC motor speed and voltage.:
\frac{N\left(s\right)}{U\left(s\right)}=\frac{1/{C}_{e}}{{T}_{m}{T}_{a}{s}^{2}+{T}_{m}s+1},
{C}_{e}
is the motor potential coefficient.
{T}_{m}
{T}_{a}
is the electrical time constant. From Eq. (4) and the formula of linear actuator, it can be deduced that the transfer function between the speed and voltage of linear actuator is:
G\left(s\right)=\frac{V\left(s\right)}{U\left(s\right)}=\frac{V\left(s\right)}{N\left(s\right)}\bullet \frac{N\left(s\right)}{U\left(s\right)}=\frac{{P}_{s}}{{Q}_{s}}\bullet \frac{1/{C}_{e}}{{T}_{m}{T}_{a}{s}^{2}+{T}_{m}s+1},
{P}_{s}
is the screw lead and
{Q}_{s}
is the gear reduction ratio. Deduced from Eq. (5), the transfer function between the linear actuator stroke and the voltage is:
\frac{X\left(s\right)}{U\left(s\right)}=\frac{V\left(s\right)}{U\left(s\right)}\bullet \frac{1}{s}=\frac{{P}_{s}}{{Q}_{s}}\bullet \frac{1/{C}_{e}}{{T}_{m}{T}_{a}{s}^{2}+{T}_{m}s+1}\bullet \frac{1}{s}.
When building a system model, the traditional method is to calculate the specific parameters using empirical formulas. However, the transfer function obtained through this method will course a large error. Therefore, the direct identification method [9] is used to estimate the transfer function parameters. The transfer function of linear actuators is second-order no-lag according to Eq. (5). The model is:
G\left(s\right)=\frac{K}{{\left(T}_{1}s+1\right)\bullet {\left(T}_{2}s+1\right)}=\frac{K}{{T}_{1}{T}_{2}{s}^{2}+{\left(T}_{1}+{T}_{2}\right)s+1}.
{T}_{a}\ll {T}_{m}
, approximate
{T}_{a}+{T}_{m}\approx {T}_{m}
. The problem translates into solving
{T}_{1}
{T}_{2}
K
. Add 10 V, 14 V, 18 V, 22 V step voltage to the linear actuators. The Hall sensor is used to obtain the speed of the linear actuators. The speed in 0-5s is sampled at a sampling frequency of 100 Hz. 500 points were totally sampled. Using least squares method to identify the transfer function. Obtain that
{T}_{a}=
{T}_{m}=
K=
0.8913. Simulating the speed-time curve by using Matlab. Fig. 3. Since the simulation results are similar to the sample data, the transfer function is accurate.
Fig. 3. Step response curve of linear actuator based on direct identification
The control system designed in this paper implements two control modes: (i) speed control; (ii) position control. A six-axis motion attitude sensor is used to detect the current position and speed of platform. PID controller features short transients and high stability. Both control modes designed in this paper use PID control.
The control structure of speed control is shown in Fig. 4. Speed control is enabled when the platform accelerates to the set speed
{\omega }_{c}
. The error is the difference between set speed
{\omega }_{c}
and actual speed
\omega
. These errors are caused by external disturbances of the driver and measurement noise of the sensor. To obtain an accurate speed signal, the Kalman filter is used before feedback. The PID controller generates a PWM signal which drives the linear actuator and adjusts the controller gain. The formula of PID model is:
u\left(t\right)={K}_{p}\left[e\left(t\right)+\frac{1}{{T}_{i}}{\int }_{0}^{t}e\left(t\right)dt+{T}_{d}\frac{de\left(t\right)}{dt}\right].
The control structure of speed control is shown in Fig. 5. Position control is used during the acceleration phase and deceleration phase. The set position of platform is
{\theta }_{c}
, and the current position is
\theta
. At the start-up phase, position control is used to acceleration stability. Disable the position control and enable the speed control when the set speed
{\omega }_{c}
is reached. When it’s 1° away from the distance setting position
{\theta }_{c}
, disable the speed control and enable the position control. Slow down the platform until set position
{\theta }_{c}
Fig. 4. The PID control structure for speed control
Fig. 5. The PID control structure for position control
Taking the extension motion as an example, the motion of one period
T
is: Firstly, the linear actuator in the buckling and extending component is pushed out to swing the platform upward. When the setting position
{\theta }_{c}
is reached, the linear actuator is driven to contract. The platform swing downward until return to the initial position.
For experiment, the limit position of platform
{\theta }_{c}
is set to +18°, and the angular velocity
{\omega }_{c}
is set to ±5°/s. An experimental system was built in Matlab using the mathematical model described in Section 3. Adjust the PID parameters to obtain the experimental results are shown in Fig. 6.
Thereafter, introducing Gaussian white noise into the controller and sensor. Obtain position-time curve and speed-time curve of platform without using the Kalman filter as shown in Fig. 7. Finally, the Kalman filter is added. Adjust the parameters of Kalman filter to obtain the experimental results under this condition which shown in Fig. 8.
Fig. 6. Simulation results in ideal status
Fig. 7. Simulation results without Kalman filter
a) Position-time curve of platform
b) Speed-time curve of platform
Fig. 8. Simulation results with Kalman filter
Analyze the error of experimental results. As shown in Table 1, using the Kalman filter can significantly improve the accuracy of speed control. Speed error reduced from 0.063 % to 0.037 %. For position control, the actual limit position is more approach to set position after using Kalman filter. What's more, the position-time curve is smoother.
Thereafter, simulation experiment of flexion, clockwise rotation and counter-clockwise rotation is performed. The error of each group of motion is shown in Table 2. Analyze Table 1 and Table 2. The designed controller can achieve high control accuracy. The motion speed of platform is stable and the position of traction treatment set by user is approached exactly.
Table 1. The error of platform in extension motion
The limit 𝜃 of platform (°)
Error of 𝜃 (%)
𝜔 of platform (°/s)
Error of 𝜔 (%)
Without Kalman filter
Table 2. The error of platform in other three motion
This paper presents a multi-dimensional lumbar traction treatment bed which provides flexion, extension, clockwise rotation and counter-clockwise rotation motion. Each treatment motion is controllable. Two linear actuators are used to provide motion. PID controller and Kalman filter constitute two groups of control modes: (i) speed control; (ii) position control. The simulation experiments are based on the mathematical model of this device. Experimental results show that the designed controller has good performance and achieve precise control, which ensuring the safety and stability of this device. In further studies, the efficacy of the device in actual spinal diseases should be tested.
Klang E., et al. Prevalence and awareness of sacroiliac joint alterations on lumbar spine CT in low back pain patients younger than 40 years. Acta Radiologica, Vol. 58, Issue 4, 2016, p. 449-455. [Publisher]
Durham S. R., Sun P. P., Sutton L. N. Surgically treated lumbar disc disease in the pediatric population: an outcome study. Journal of Neurosurgery: Spine, Vol. 92, Issue 1, 2000, p. 1-6. [Search CrossRef]
Pin X., et al. Biomechanical effects of different traction modes on lumbar spine. Journal of Medical Biomechanics, Vol. 29, Issue 5, 2014, p. 399-404. [Search CrossRef]
Gagne A. R., Hasson S. M. Lumbar extension exercises in conjunction with mechanical traction for the management of a patient with a lumbar herniated disc. Physiotherapy Theory and Practice, Vol. 26, Issue 4, 2010, p. 256-266. [Publisher]
Kim H. S., Yun D. H., Huh K. Y. Effect of spinal decompression therapy compared with intermittent mechanical traction in lumbosacral disc herniation. Annals of Rehabilitation Medicine, Vol. 32, Issue 3, 2008, p. 319-323. [Search CrossRef]
Zhang Y., Yue S., Yue Y. A comparison between multi-directional mechanical traction and longitudinal traction for treatment of lumbar disc herniation: A randomized clinical trial with parallel-group design. Chinese Journal of Rehabilitation Medicine, Vol. 26, Issue 7, 2011, p. 638-643. [Search CrossRef]
Manuthera 242 Mobilisation Table. Lojer, http://www.lojer.com/product/manuthera-242-mobilisation-table [Search CrossRef]
Hill AIRFLEX II Flexion & Distraction Table. Hill Laboratories, https://hilllabs.com/chiropractic/Hill-Air-Flex-Table.php. [Search CrossRef]
Zhou R. X., Zhang Z., Qi Y. C. Direct identification of DC electromotor model parameter. Computer Simulation, Vol. 23, Issue 6, 2006, p. 31-33. [Search CrossRef] |
M-ary amplitude phase shift keying (APSK) modulation - Simulink - MathWorks Switzerland
M-APSK Modulator Baseband
M-ary amplitude phase shift keying (APSK) modulation
The M-APSK Modulator Baseband block modulates the input signal using M-ary amplitude phase shift keying (APSK) modulation. The output is a baseband representation of the modulated signal. M, the Modulation Order for M-APSK, equals the sum of the elements in Constellation points per circle. For a description of M-APSK modulation, see Algorithms.
M-APSK Modulator Baseband specifically applies to multiple ring PSK constellations. For a single ring PSK constellation, use M-PSK Modulator Baseband.
Input signal, specified as a scalar, vector, or matrix. The input signal must consist of binary values or integers in the range [0, (M – 1)], where M is the Modulation Order for M-APSK. This port is unnamed on the block.
To process the input signal as binary elements, set the Input type parameter value to Bit. For binary inputs, the number of rows must be an integer multiple of log2(M). Groups of log2(M) bits in a column are mapped onto a symbol, with the first bit representing the MSB and the last bit representing the LSB.
Out — APSK modulated signal
APSK modulated signal, returned as a complex scalar, vector, or matrix. The output signal dimensions depend on the specified Input type value. This port is unnamed on the block.
Dimensions of Output Signal
Integer The output signal has the same dimensions as the input signal.
Bit The number of rows in the output signal equals the number of rows in the input signal divided by log2(M), where M is the Modulation Order for M-APSK.
Radius per PSK ring, specified as a vector with the same length as Constellation points per circle. Each vector element indicates the radius of its corresponding PSK ring. The first element corresponds to the innermost circle, and so on until the last element, which corresponds to the outermost circle. These element values must be positive and arranged in increasing order.
Gray — Uses Gray mapping along the contour in both the amplitude and phase dimensions. For Gray symbol mapping, all the values for Constellation points per circle must be equal and all the values for Phase offset of each circle (rad) must be equal. For a description of the Gray mapping used, see [2].
The default symbol mapping depends on Constellation points per circle and Phase offset of each circle (rad). When all the elements of Constellation points per circle are equal and all the elements of Phase offset of each circle (rad) are equal, the default is Gray. For all other cases, the default is Contourwise-gray.
Input type, specified as one of these options.
Integer –– The input signal must consist of integers in the range [0, (M – 1)].
Bit –– The input signal must contain binary values, and the number of rows must be an integer multiple of log2(M), where M is the Modulation Order for M-APSK. Binary input signals are assumed to be left-MSB aligned and specified column-wise. Groups of log2(sum(M)) bits in a column are mapped onto a symbol, with the first bit representing the MSB and the last bit representing the LSB.
View Constellation — Plot reference constellation
To plot the reference constellation, click the View Constellation button.
Click Apply before clicking the View Constellation to view latest parameter values.
Plot Noisy 16-APSK Constellation
Apply 16-APSK modulation to a signal of random data. Pass the modulated signal through an AWGN channel. Plot the signal constellation.
The modulation order, M, for M-APSK is the total number of points in the signal constellation. M equals the sum of the elements in Constellation points per circle. The element values in Constellation points per circle must be multiples of four. M must be a power of two.
The block implements a pure APSK constellation.
A pure M-APSK constellation is composed of NC concentric rings or contours, each with uniformly spaced PSK points. The M-APSK constellation set is
\chi =\left\{\begin{array}{cc}{R}_{1}\mathrm{exp}\left(j\left(\frac{2\pi }{{M}_{1}}i+{\theta }_{1}\right)\right),& i=0,\dots ,{M}_{1}-1,\\ {R}_{2}\mathrm{exp}\left(j\left(\frac{2\pi }{{M}_{2}}i+{\theta }_{2}\right)\right),& i=0,\dots ,{M}_{2}-1,\\ ⋮& ⋮\\ {R}_{{N}_{\text{C}}}\mathrm{exp}\left(j\left(\frac{2\pi }{{M}_{{N}_{\text{C}}}}i+{\theta }_{\text{Nc}}\right)\right),& i=0,\dots ,{M}_{{N}_{\text{C}}}-1,\end{array}
The modulation order is equal to the sum of all Ml for l = 1, 2, ... , NC.
NC is the number of concentric rings. NC ≥ 2.
Ml is the number of constellation points in the lth ring.
Rl is the radius of the lth ring.
θl is the phase offset of the lth ring.
j=\sqrt{-1}
[1] Corazza, Giovanni E. Digital Satellite Communications. New York: Springer Science Business Media, LLC, 2007.
M-APSK Demodulator Baseband | DVBS-APSK Modulator Baseband | MIL-188 QAM Modulator Baseband | M-PSK Modulator Baseband
apskmod |
{\frac{1}{5}}^{\mathrm{th}}
P and G were partners in a firm sharing profits in the ratio of 7:4. On 1-1-2016 their firm was dissolved. After transferring assets (other than cash) and outsiders liabilities to realization account you are given the following information:
(a) Kumar, a creditor for Rs 3,90,000 accepted building at Rs 7,00,000 and paid the balance to the firm by cheque.
(b) Karan, a second creditor for Rs 2,83,000 accepted machinery of the books value of Rs 3,00,000 at Rs 2,80,000 in full settlement of his claim.
(c) Kishor, a third creditor for Rs 5,00,000 accepted investments of Rs 4,10,000 and a bank draft of Rs 89,000 in full settlement of his claim.
S, T and U were partners in a firm sharing profits and losses in the ratio of 4:3:3. On 31-3-2015 their Balance Sheet was as follows:
Balance Sheet S, T and U
From 1-4-2015 they decided to share future profits equally. For this purpose it was decided that
(i) Goodwill of the firm be valued at Rs 90,000.
(ii) Land be revalued at Rs 3,00,000 and building by depreciated by 10%.
(iii) Creditors Rs 7,500 were not likely to be claimed and hence be written-off.
On 1.4.2013 Kaka Ltd. had Rs 50,00,000, 9% debentures of Rs 100 each outstanding.
(i) On 1-4-2014 the company purchased in the open market 25,000 of its own debentures of Rs 99 each and cancelled the same immediately.
(ii) On 1-4-2015 the company redeemed at par debentures of Rs 10,00,000 by draw of lot.
(iii) On 1-1-2016 the remaining debentures were purchased for Rs 14,99,000 and were cancelled immediately.
Ignoring debenture redemption reserve and interest on debentures pass necessary journal entries for the above transactions in the books of the company. VIEW SOLUTION
(a) List any four items that are shown under the heading 'Non-current Assets' as per Schedule III of the Companies Act, 2013.
(b) State any two limitations of financial statements analysis of a company. VIEW SOLUTION
(a) What is meant by 'Liquidity' of business? |
The simplified form of the expressiondisplaystyle{sqrt[{{3}}]{{a}}}{sqrt[{{6}}]{{a}}} text{in radicalnotation
The simplified form of the expressiondisplaystyle{sqrt[{{3}}]{{a}}}{sqrt[{{6}}]{{a}}} text{in radicalnotation is} sqrt{a}.
The simplified form of the expression
\sqrt[3]{a}\sqrt[6]{a}\text{ }\text{in radicalnotation is}\text{ }\sqrt{a}.
Fatema Sutton
Steps to simplify radical expression:
Step 1: First convert the radical expression into exponential expressions.
Step 2: Simplify the exponential expression by adding exponents or subtracting exponents.
Step 3: Now, convert the exponential expression into radical expression.
Positive rational exponents:
\sqrt[m]{{a}^{n}}={a}^{n\text{/}m}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}{a}^{n\text{/}m}=\sqrt[m]{{a}^{n}}
Power of product:
{\left({x}^{m}\right)}^{n}={x}^{m\cdot n}
Consider the provided expression,
\sqrt[3]{a}\sqrt[6]{a}
Now, convert the provided radical expression into the exponential notationusing formula of positive rational exponents
\sqrt[m]{{a}^{n}}={a}^{n\text{/}m}
\sqrt[3]{a}\sqrt[6]{a}={a}^{1\text{/}3}\cdot {a}^{1\text{/}6}
Now, solve the exponents.
\frac{1}{3}+\frac{1}{6}=\frac{2+1}{6}
=\frac{3}{6}
=\frac{1}{2}
Now put equation (2) in equation (1),
{a}^{1\text{/}3}\cdot {a}^{1\text{/}6}={a}^{1\text{/}2}
Now convert the exponential expression into radical expression using formula
{a}^{n\text{/}m}=\sqrt[m]{{a}^{n}}.
{a}^{1\text{/}2}=\sqrt{a}
Therefore, the simplified form of the expression
\sqrt[3]{a}\sqrt[6]{a}\text{ }\text{in radical notation is}\text{ }\sqrt{a}.
Prove that if x is rational and
x\ne 0
\frac{1}{x}
cot -(
\sqrt{ }
17/8)
Use rational exponents to write a single radical expression.
\sqrt[7]{11}×\sqrt[6]{13}
a) Find the rational zeros and then the other zeros of the polynomial function
\left(x\right)={x}^{3}-4{x}^{2}+2x+4,\text{that is, solve}\text{ }f\left(x\right)=0.
b) Factor
f\left(x\right)
into linear factors.
\sqrt[4]{c{d}^{2}}×\sqrt[3]{{c}^{2}d}.
Use a right triangle to write the following expression as an algebraic expression. Assume that x is positive and in the domain of the given inverse trigonometric function.
\mathrm{tan}\left({\mathrm{cos}}^{-1}5x\right)=?
To rationalize:
Each numerator. Assume that variables represent positive real numbers.
Given: An expression :
\frac{\sqrt[3]{9}}{7}. |
Give the meaning of 'Debenture'. VIEW SOLUTION
Why heirs of a retiring/deceased partner are entitled to a share of goodwill of the firm? VIEW SOLUTION
\frac{1}{2},\frac{2}{5}
\frac{1}{10}
Give any one purpose for which the amount received as 'Securities Premium' may be utilised. VIEW SOLUTION
What is meant by 'Reconstitution of a Partnership Firm' ? VIEW SOLUTION
{\left(\frac{1}{6}\right)}^{{}^{th}}
(i) Kay Ltd. converted 3,000, 12% debentures of Rs 100 each issued at a premium of 10% into equity shares of Rs 100 each issued at a premium of 25%.
(ii) Jay Ltd. redeemed 1,500, 12% debentures of Rs 1,000 each issued at a discount of 10% by converting them into equity shares of Rs 50 each issued at par.
2\frac{1}{2}
On 1st April, 2012, Mayank Ltd. was formed with an authorised capital of Rs 25,00,000 divided into 50,000 equity shares of Rs 50 each. The company issued prospectus inviting applications for 45,000 shares. The issue price was payable as under :
Amar, Karan and Varun were partners in a firm manufacturing garments. They were sharing profits in the ratio of 5 : 3 : 2. On 1st April, 2012 their capitals were Rs 3,00,000, Rs 4,00,000 and Rs 5,00,000 respectively. After the flood in Uttaranchal, all partners decide to personally help the flood victims. For this Amar withdrew Rs 30,000 from the firm on 1st September, 2012, Karan instead of withdrawing cash from the firm took garments amounting to Rs 36,000 from the firm and distributed to the flood victims. On the other hand, Varun withdrew Rs 1,50,000 from his capital on 1st January, 2013 and started a school to provide elementary education in the flood affected area.
Kumar and Gaurav were partners in a firm sharing profits in the ratio of their capitals. On 31-3-2013 their Balance Sheet was as follows:
Balance Sheet of Kumar and Gaurav as on 31st Marck, 2013
Gaurav’s Current Account
(i) Kumar took over 50% of stock at 10% less than its book value. The remaining stock was sold for Rs 10,000.
(ii) Debtors were realized at a discount of 5%.
(iii) An unrecorded asset was sold for Rs 9,000 and machinery was sold for Rs 18,000.
(iv) Creditors were paid in full.
(v) There was an outstanding bill for repairs for amounting to Rs 14,000 which was settled at Rs 12,000.
\frac{{2}^{\mathrm{th}}}{5}
\frac{{2}^{\mathrm{th}}}{5}
\frac{{1}^{\mathrm{th}}}{5}
What is meant by 'Cash Equivalents' while preparing Cash Flow Statement? VIEW SOLUTION
From the following 'Statement of Profit & Loss' for the year ended 31st March, 2013, prepare a 'Comparative Statement of Profit & Loss' of Better Sales Ltd.
Expenses 7,00,000
Non-current-Assets 14,40,000
(b) The Quick Ratio of Z Ltd. is 1 : 1. State with reason which of the following transactions would (i) increase; (ii) decrease or (iii) not change the ratio :
(2) Debentures of Rs 50,000 were converted into Equity shares. |
1 Department of Bioresource Engineering, McGill University, Ste-Anne-de-Bellevue, QC, Canada.
2 WorldFish, Lusaka, Zambia.
Abstract: This study presents field data and results on local fish processing, its energy supply, and consumption in the Northern Province of Zambia. The study also evaluates the impact of processing conditions on fish quality and cost for the different processing systems available in the selected communities. The result shows that fuelwood is the primary source of energy for fish processing used either in a modified three-stone fire (MTSF) system or a recently developed kiln. The charcoal smoking alternative had the least fuel consumption, yet was not considered as the preferred option due to the high cost of the fuel, a smaller quantity of fish processed per batch and longer processing time. The result also revealed that irrespective of the system used, the type of fish being processed and the pre-smoking drying time had a significant impact on the total energy consumption. Overall, the smoking kiln was found to increase the quantity of fish processed by five folds, reduce fuel use by 48% and reduce the overall smoking time per kilogram fish processed by 39%. A relatively higher fish quality was obtained with the use of the smoking kiln in comparison with the MTSF and the charcoal smoking system. The use of the smoking kiln as a fish processing system is recommended because it improves the quality of the fish, can scale up fish processing due to its capacity and reduces energy use with its associated costs.
Keywords: Fish, Energy Dynamics, Fish Processing, Fuel Supply, Smoking Systems
\begin{array}{c}\text{HHV}=20.7999-0.3214\frac{\text{VM}}{\text{FC}}+0.0051{\left(\frac{\text{VM}}{\text{FC}}\right)}^{2}-11.2277\frac{\text{ASH}}{\text{VM}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+4.4953{\left(\frac{\text{ASH}}{\text{VM}}\right)}^{2}-0.7223{\left(\frac{\text{ASH}}{\text{VM}}\right)}^{3}+0.0383{\left(\frac{\text{ASH}}{\text{VM}}\right)}^{4}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+0.0076\frac{\text{FC}}{\text{ASH}}\end{array}
{\text{HHV}}_{\text{fuel}}=\frac{{\displaystyle \underset{i=1}{\overset{i=4}{\sum }}\left({m}_{i}\times {\text{HHV}}_{i}\right)}}{4}
{\text{HHV}}_{\text{fuel}}
represent the heating value of the bulk fuel,
{m}_{i}
{\text{HHV}}_{i}
represent the mass fraction and the heating value, respectively, of a wood species.
Cite this paper: Kwofie, E. , Ellis, E. , Genschick, S. , Ngadi, M. and Thilsted, S. (2019) Fish Processing and Its Energy Dynamics in Zambia. Journal of Sustainable Bioenergy Systems, 9, 44-63. doi: 10.4236/jsbs.2019.92004.
[1] FAO (2006) Zambia Fishery Country Profile. Food and Agriculture Organization of the United Nations, Roma.
[2] Desai, M.A., Mehta, S. and Smith, K.R. (2004) Indoor Smoke from Solid Fuels: Assessing the Environmental Burden of Disease at National and Local Levels. WHO Environmental Burden of Disease Series No. 4, World Health Organization, Geneva.
[3] Johnson, N.G. (2005) Risk Analysis and Safety Evaluation of Household Stoves in Developing Nations.
[4] Wickramasinghe, A. (2003) Gender and Health Issues in the Biomass Energy Cycle: Impediments to Sustainable Development. Energy for Sustainable Development, 7, 51-61.
[5] Ramanathan, V. and Carmichael, G. (2008) Global and Regional Climate Changes Due to Black Carbon. Nature Geoscience, 1, 221-227.
[6] Lim, J.S., Abdul Manan, Z., Wan Alwi, S.R. and Hashim, H. (2012) A Review on Utilization of Biomass from Rice Industry as a Source of Renewable Energy. Renewable and Sustainable Energy Reviews, 16, 3084-3094.
[7] CSO (2010) Census of Population and Housing. National Descriptive Tables, Vol. 11, Central Statistics Office, Lusaka.
[8] Bhattacharya, S.C., Albina, D.O. and Abdul Salam, P. (2002) Emission Factors of Wood and Charcoal-Fired Cookstoves. Biomass and Bioenergy, 23, 453-469.
[9] Bond, T. and Sun, H. (2005) Can Reducing Black Carbon Emissions Counteract Global Warming? Environmental Science & Technology, 39, 5921-5926.
[10] Adamu, I.G., Kabri, H.U., Hussaini, I.D. and Mada, A.D. (2013) Design and Construction of Smoking Kiln. Journal of Engineering and Technology Research, 5, 15-20.
[11] ASTM (2010) Standard E1131-08. In: Test Method for Compositional Analysis by Thermogravimetry, American Society for Testing and Materials, West Conshohocken, 17.
[12] Nhuchhen, D.R. and Abdul Salam, P. (2012) Estimation of Higher Heating Value of Biomass from the Proximate Analysis: A New Approach. Fuel, 99, 55-63.
[13] WHO (2014) Household Air Pollution and Health. World Health Organization, Geneva.
[14] Bruce, N., Perez-Padilla, R. and Albalak, R. (2000) Indoor Air Pollution in Developing Countries: A Major Environmental and Public Health Challenge. Bulletin of the World Health Organization, 78, 1078-1092. |
On the Development of a Low Cost Pyrheliometer | J. Sol. Energy Eng. | ASME Digital Collection
Michael Gnos,
Brenton Greska,
Brenton Greska
V. P. of Research Sustainable Energy Technologies
, 1404 Hamlin Avenue Unit G, St. Cloud, FL 34771
e-mail: bgreska@sustaintech.com
Eminent Scholar Professor and Director
Energy and Sustainability Center, Department of Mechanical Engineering,
, Tallahassee, FL 32306
e-mail: akrothapalli@esc.fsu.edu
Michael Gnos Project Engineer
Anjaneyulu Krothapalli Eminent Scholar Professor and Director
Gnos, M., Greska, B., and Krothapalli, A. (August 17, 2011). "On the Development of a Low Cost Pyrheliometer." ASME. J. Sol. Energy Eng. August 2011; 133(3): 034501. https://doi.org/10.1115/1.4004266
A low cost pyrheliometer, based on a thermoelectric sensor, was developed at the Energy and Sustainability Center at the Florida State University. In addition, an inexpensive double-axis tracking device, capable of autonomous operation, enables the pyrheliometer to operate as a stand-alone system. Widely available off-the-shelf components were used and compromises in accuracy and time responsiveness were made in order to keep the cost low. The obtained data was compared with an Eppley Normal Incidence Pyrheliometer (NIP) using model ST-1 solar tracker. Steady state values of irradiance were measured with an accuracy better than
±2%
. Transient measurements are time delayed by a thermal lag of about 2 min, which leads to a high error for instantaneous measured values. However, the integrated irradiance over the course of any given day yields irradiation values with accuracy better than
±2%
, even on days when the sun and clouds quickly alternate. Based on a manufacturing cost analysis, the prototype pyrheliometer system is anticipated to cost an order of magnitude less than commercially available products if mass-produced.
atmospheric measuring apparatus, electric sensing devices, solar radiation, star trackers, temperature sensors, thermoelectric devices
Disks, Errors, Irradiation (Radiation exposure), Lenses (Optics), Radiation (Physics), Sensors, Solar energy, Solar radiation, Steady state, Sustainability, Temperature, Transients (Dynamics), Cooling, Design, Manufacturing, Sunspots, Engineering prototypes, Temperature sensors, Thermoelectric devices, Heating, Instrumentation
Solar Resource and Meteorological Assessment Project (Solrmap). Solar and Meteorological Stations Options: Configurations and Specifications.
Technical Report, National Renewable Energy Laboratories, December.
,” M.Sc. thesis, Florida State University, Tallahassee, FL.
Fourier Series Representation of the Position of the Sun
The Absorption of Radiation in Solar Stills
. Tech. Rep. 7th ed., Part I, World Meteorological Organization, Chap. VII.
On the Design and Analysis of a Pyrheliometer Comprising a Convex Lens |
If z is a complex function and z^{3}=8i, what are
Pam Stokes 2022-01-18 Answered
If z is a complex function and
{z}^{3}=8i
, what are all the values of z?
{z}^{3}=8i
{z}^{3}-8i=0
{z}^{3}+{\left(2i\right)}^{3}=0
\left(z+2i\right)\left({z}^{2}+{\left(2i\right)}^{2}-2iz\right)=0
\left(z+2i\right)\left({z}^{2}-2iz-4\right)=0
z=-2i
{z}^{2}-2iz-4=0
z=\frac{2i±\sqrt{-4+16}}{2}
z=i±\sqrt{3}
So roots are
-2i,i+\sqrt{3},i-\sqrt{3}
aquariump9
{z}^{3}=8i=8{e}^{i\left(\frac{\pi }{2}+2n\pi \right)}
n\in Z
z=2{e}^{i\left(\frac{\pi }{6}+2n\frac{\pi }{3}\right)}
z=2{e}^{i\frac{\pi }{6}},2{e}^{i\left(5\frac{\pi }{6}\right)},2{e}^{i\left(3\frac{\pi }{2}\right)}
z=\sqrt{3}+i,-\sqrt{3}+i,-2i
\omega ={e}^{\frac{2i\pi }{3}}=-\frac{1}{2}+i\frac{\sqrt{3}}{2}
is a root of unit. So the roots of
{Z}^{3}=1\text{ }\text{are}:{\omega }^{0}=1;\omega \text{ }\text{and}\text{ }{\omega }^{2}=\stackrel{―}{\omega }
{z}^{3}=8i⇔{z}^{3}=\left(-2i{\right)}^{3}⇔\left(\frac{z}{-2i}{\right)}^{3}=1
\frac{z}{-2i}={\omega }^{k}k\in 0;1;2
. Thus The root of the equation
{z}^{3}=8i
z=-2i{\stackrel{˙}{\omega }}^{k}k\in 0;1;2\text{ }i.e.\text{ }z=-2i\text{ }\text{or}\text{ }z=\sqrt{3}+i\text{ }\text{or}\text{ }z=-\sqrt{3}+i
\left(6+5i\right)÷\left(7+3i\right)=
\frac{918327461923874632928}{327861038736208137463408473640}-\frac{82736287346219348}{83618273619736876376287616919128736973}
How to prove that
{\mathrm{cos}}^{2}\left(z\right)+{\mathrm{sin}}^{2}\left(z\right)=1
, where z is a complex variable (if it is true)?
\left(5+4i\right)×\left(7+5i\right)
f\left(z\right)=\frac{1}{{z}^{2}\left(1-z\right)}
I am to find two Laurent series expansions. There are two singularities, z=0 and z=1. So for the first expansion, I used the region 0<|z|<1 and I got
\sum _{n=0}^{\mathrm{\infty }}{z}^{n}+\frac{1}{z}+\frac{1}{{z}^{2}}
. The second expansion is for the region
1<|z|<\mathrm{\infty }
. I don't know how to approach this, the explanation in my book is confusing. Any help?
How to express
{e}^{2-i}
a+ib
{z}_{1}
and are complex numbers then
\stackrel{―}{\frac{{z}_{1}}{{z}_{2}}}=\frac{\stackrel{―}{{z}_{1}}}{\stackrel{―}{{z}_{2}}} |
Conditional Probability Distribution | Brilliant Math & Science Wiki
Sharky Kesa, Adam Strandberg, Japjot Singh, and
Conditional probability is the probability of one thing being true given that another thing is true, and is the key concept in Bayes' theorem. This is distinct from joint probability, which is the probability that both things are true without knowing that one of them must be true.
For example, one joint probability is "the probability that your left and right socks are both black," whereas a conditional probability is "the probability that your left sock is black if you know that your right sock is black," since adding information alters probability. This can be high or low depending on how frequently your socks are paired correctly. An Euler diagram, in which area is proportional to probability, can demonstrate this difference.
X
be the probability that your left sock is black, and let
Y
be the probability that your right sock is black. On the left side of the diagram, the yellow area represents the probability that both of your socks are black. This is the joint probability
P(X,Y)
Y
is definitely true (e.g., given that your right sock is definitely black), then the space of everything not
Y
is dropped and everything in
Y
is rescaled to the size of the original space. The rescaled yellow area is now the conditional probability of
X
Y
P(X \mid Y)
. In other words, this is the probability that your left sock is black if you know that your right sock is black. Note that the conditional probability of
X
Y
is not in general equal to the conditional probability of
Y
X
. That would be the fraction of
X
that is yellow, which in this picture is slightly smaller than the fraction of
Y
that is yellow.
Philosophically, all probabilities are conditional probabilities. In the Euler diagram,
X
Y
are conditional on the box that they are in, in the same way that
X | Y
is conditional on the box
Y
that it is in. Treating probabilities in this way makes chaining together different types of reasoning using Bayes' theorem easier, allowing for the combination of uncertainties about outcomes ("given that the coin is fair, how likely am I to get a head") with uncertainties about hypotheses ("given that Frank gave me this coin, how likely is it to be fair?"). Historically, conditional probability has often been misinterpreted, giving rise to the famous Monty Hall problem and Bayesian mistakes in science.
For discrete random variables, the conditional probability mass function of
Y
given the occurrence of the value
x
X
can be written according to its definition as
P(Y = y \mid X = x) = \dfrac{P(X=x \cap Y=y)}{P(X=x)}.
P(X=x)
rescales the joint probability to the conditional probability, as in the diagram in the introduction. Since
P(X=x)
is in the denominator, this is defined only for non-zero (hence strictly positive)
P(X=x)
P(X = x) \leq 1
P(Y = y \mid X = x) \geq P(X=x \cap Y=y)
, and that they are only equal in the case where
P(X = x) = 1
. In any other case, it is more likely that
X = x
Y = y
if it is already known that
X = x
than if that is not known.
The relation with the probability distribution of
X
Y
P(Y=y \mid X=x) P(X=x) = P(X=x\ \cap Y=y) = P(X=x \mid Y=y)P(Y=y).
As a concrete example, the picture below shows a probability tree, breaking down the conditional distribution over the binary random variables
A
B
. The four nodes on the right-hand side are the four possible events in the space. The leftmost node has value one. The intermediate nodes each have a value equal to the sum of their children. The edge values are the nodes to their right divided by the nodes to their left. This reflects the idea that all probabilities are conditional.
P(A)
P(B)
are conditional on the assumptions of the whole probability space, which may be something like "
A
B
are the outcomes of flipping fair coins."
0.3
0.7
0.4
Similarly, for continuous random variables, the conditional probability density function of
Y
x
X
f_Y(y \mid X=x) = \dfrac{f_{X, Y}(x, y)}{f_X(x)},
f_{X, Y} (x, y)
gives the joint density of
X
Y
f_X(x)
gives the marginal density for
X
. Also in this case it is necessary that
f_{X} (x) > 0
. The relation with the probability distribution of
X
Y
f_Y(y \mid X=x)f_X(x) = f_{X,Y}(x, y) = f_X(x \mid Y=y)f_Y(y).
Conditional distributions and marginal distributions are related using Bayes' theorem, which is a simple consequence of the definition of conditional distributions in terms of joint distributions.
Bayes' theorem for discrete distributions states that
\begin{aligned} P(Y=y \mid X=x) P(X=x) &= P(X=x\ \cap Y=y) \\ &= P(X=x \mid Y=y)P(Y=y)\\\\ \Rightarrow P(Y=y \mid X=x) &= \dfrac{P(X=x \mid Y=y)}{P(X=x)} P(Y=y). \end{aligned}
This can be interpreted as a rule for turning the marginal distribution
P(Y=y)
into the conditional distribution
P(Y=y \mid X=x)
by multiplying by the ratio
\frac{P(X=x \mid Y=y)}{P(X=x)}
. These functions are called the prior distribution, posterior distribution, and likelihood ratio, respectively.
For continuous distributions, a similar formula holds relating conditional densities to marginal densities:
f_{Y} (y \mid X = x) = \dfrac{f_{X}(x \mid Y = y)}{f_{X} (x)} f_{Y} (y).
0.4.
0.7
0.2
Two variables are independent if knowing the value of one gives no information about the other. More precisely, random variables
X
Y
are independent if and only if the conditional distribution of
Y
X
is, for all possible realizations of
X,
equal to the unconditional distribution of
Y
. For discrete random variables this means
P(Y = y | X = x) = P(Y = y)
for all relevant
x
y
. For continuous random variables
X
Y
, having a joint density function, it means
f_Y(y | X=x) = f_Y(y)
x
y
\dfrac{121}{243}
\dfrac{122}{243}
\dfrac{124}{243}
\dfrac{125}{243}
A biased coin is tossed repeatedly. Assume that the outcomes of different tosses are independent and the probability of heads is
\frac{2}{3}
for each toss. What is the probability of obtaining an even number of heads in 5 tosses?
Seen as a function of
y
x
P(Y = y | X = x)
is a probability, so the sum over all
y
(or integral if it is a conditional probability density) is 1. Seen as a function of
x
y
, it is a likelihood function, so that the sum over all
x
need not be 1.
(\Omega, \mathcal{F}, P)
be a probability space,
\mathcal{G} \subseteq \mathcal{F}
\sigma
-field in
\mathcal{F}
X : \Omega \to \mathbb{R}
a real-valued random variable
\big(
measurable with respect to the Borel
\sigma
\mathcal{R}^1
\mathbb{R}\big).
It can be shown that there exists a function
\mu : \mathcal{R}^1 \times \Omega \to \mathbb{R}
\mu(\cdot, \omega)
\mathcal{R}^1
\omega \in \Omega
(i.e., it is regular) and
\mu(H, \cdot) = P(X \in H | \mathcal{G})
(almost surely) for every
H \in \mathcal{R}^1
\omega \in \Omega
\mu(\cdot, \omega) : \mathcal{R}^1 \to \mathbb{R}
is called a conditional probability distribution of
X
\mathcal{G}
E[X | \mathcal{G}] = \int_{-\infty}^\infty x \, \mu(d x, \cdot)
A \in \mathcal{A} \supseteq \mathcal B
, define the indicator function
\mathbf{1}_A (\omega) = \begin{cases} 1 \; &\text{if } \omega \in A \\ 0 \; &\text{if } \omega \notin A, \end{cases}
which is a random variable. Note that the expectation of this random variable is equal to the probability of
A
itself:
\operatorname{E}(\mathbf{1}_A) = \operatorname{P}(A).
Then the conditional probability given
\mathcal B
\operatorname{P}(\cdot|\mathcal{B}):\mathcal{A} \times \Omega \to (0,1)
\operatorname{P}(A|\mathcal{B})
is the conditional expectation of the indicator function for
A
\operatorname{P}(A|\mathcal{B}) = \operatorname{E}(\mathbf{1}_A|\mathcal{B}).
\operatorname{P}(A|\mathcal{B})
\mathcal B
-measurable function satisfying
\int_B \operatorname{P}(A|\mathcal{B}) (\omega) \, \operatorname{d} \operatorname{P}(\omega) = \operatorname{P} (A \cap B) \quad \text{for all} \quad A \in \mathcal{A}, B \in \mathcal{B}.
A conditional probability is regular if
\operatorname{P}(\cdot|\mathcal{B})(\omega)
is also a probability measure for all
\omega ∈ \Omega
. An expectation of a random variable with respect to a regular conditional probability is equal to its conditional expectation.
For a trivial sigma algebra
\mathcal B= \{\emptyset,\Omega\}
the conditional probability is a constant function,
\operatorname{P}\!\left( A| \{\emptyset,\Omega\} \right) \equiv\operatorname{P}(A)
A\in \mathcal{B}
, as outlined above,
\operatorname{P}(A|\mathcal{B})=1_A.
Cite as: Conditional Probability Distribution. Brilliant.org. Retrieved from https://brilliant.org/wiki/conditional-probability-distribution/ |
Aubrey Hendricks 2022-01-21 Answered
SaphypycleDapc5
The vector cross product of two 3-dimesnional vectors in the vector space
{\mathbb{R}}^{3}
may be computed as a matrix determinant
\left(3,1,-4\right)×\left(1,1,18\right)=\left[\begin{array}{ccc}\stackrel{\to }{i}& \stackrel{\to }{j}& \stackrel{\to }{k}\\ 3& 1& -4\\ 1& 1& 18\end{array}\right]
=\stackrel{\to }{i}\left(18+4\right)-\stackrel{\to }{j}\left(54-1\right)+\stackrel{\to }{k}\left(3-1\right)
=22\stackrel{\to }{i}-\stackrel{\to }{j}53+2\stackrel{\to }{k}
=\left(22,-53,2\right)
r\left(t\right)=<{t}^{2},\frac{2}{3}{t}^{3},t>
<4,-\frac{16}{3},-2>
\left(1,3,0\right),\left(-2,0,2\right),\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left(-1,3,-1\right)
These are the planes and the result is gonna be a line in
{\mathbb{R}}^{3}
x+2y+z-1=0
2x+3y-2z+2=0
What are the standard three-dimensional unit vectors?
How do you find the inner product and state whether the vectors are perpendicular given
<8,4>\cdot <2,4>
Find the best approximation to z by vectors of the form
{c}_{1}{v}_{1}+{c}_{2}{v}_{2}
z=\left[\begin{array}{c}3\\ -7\\ 2\\ 3\end{array}\right],\text{ }{v}_{1}=\left[\begin{array}{c}2\\ -1\\ -3\\ 1\end{array}\right]\text{ }{v}_{2}=\left[\begin{array}{c}1\\ 1\\ 0\\ -1\end{array}\right] |
Thales' Theorem | Brilliant Math & Science Wiki
Arkajyoti Banerjee, Sravanth C., Akshay Yadav, and
If a triangle is inscribed inside a circle, where one side of the triangle is the diameter of the circle, then the angle opposite to that side is a right angle. The converse of this is also true.
_\square
There are many ways to prove this theorem. One of the most classic proofs is as follows:
AO=BO=CO
as all of them are the radii of the circle. Hence,
\angle OAB=\angle OBA
\angle OBC=\angle OCB
because angles opposite to equal sides are equal.
\begin{aligned} \angle OAB&=\angle OBA=\alpha \\ \angle OBC&=\angle OCB=\beta. \end{aligned}
\begin{aligned} \angle ABC + \angle BCA + \angle CAB &=180^\circ \\ \angle OAB+(\angle OBA+\angle OBC)+\angle OCB&=180^\circ \\ \alpha+(\alpha+\beta)+\beta&=180^\circ \\ \alpha+\beta&=90^\circ \\ \angle ABC&= 90^\circ.\ _\square \end{aligned}
There's another way of proving this, which is based on another theorem called the alternate segment theorem which states that
The angle subtended by a chord (or two radii) at the center of a circle is two times the angle subtended by it on the remaining part of the circle.
_\square
Let us now try to prove Thales' theorem with the help of the above theorem.
According to the angle segment theorem, we have the following diagram:
\angle AOB = 2 \angle ADB.
Let us say that the radii
AO
BO
form the diameter, then the figure would look like this:
\angle AOB = 180^{\circ} \implies \angle ADB = \frac{180^{\circ}}{2} = 90^{\circ}.\ _\square
60 ^ \circ
360 ^ \circ
180 ^ \circ
90 ^ \circ
AB
is the diameter of a semi-circle.
C
is a point on the circumference.
What is the measure of angle
ACB?
The longest chord in a circle is its diameter.
Let a circle with center
O
have a chord
AB
OA
OB
P
AB
OP\perp AB
We know that the perpendicular from the center of a circle to a chord bisects the chord. Then
PA=PB= \dfrac12 AB
Let the radius be
r
, the length of chord
AB
c
, and the perpendicular distance between the center and the chord
AB
x
Now, by the Pythagorean theorem, in
\triangle OPA
\begin{aligned} OA^2 &=OP^2 +AP^2\\ r^2 &= x^2 + \left( \dfrac c2\right)^2 \\ \Rightarrow c &= 2\sqrt{r^2-x^2}. \end{aligned}
So, we need to find the maximum possible value of
c
, right? We know that a function is maximized when its derivative is equal to zero. Then, since radius
r
c
is maximized when
\begin{aligned} \dfrac{dc}{dx} &=& 0 \\ \dfrac d{dx} \Big (2\sqrt{r^2-x^2} \Big) &=& 0 \\ 2 \dfrac d{dx} \Big (\sqrt{r^2-x^2} \Big ) &=& 0 \\ 2 \left( \dfrac x{\sqrt{r^2-x^2}} \right) &=& 0 \\ \Rightarrow x &=& 0. \end{aligned}
c
x=0
, implying that the maximum value of
c
c = 2\sqrt{r^2 - 0^2} = 2\sqrt{r^2} = 2r = 2 \times \text{(radius)} = \text{(diameter)}.\ _\square
The length of the largest chord possible in a circle is always equal to twice its radius.
Cite as: Thales' Theorem. Brilliant.org. Retrieved from https://brilliant.org/wiki/thales-theorem/ |
Hodrick-Prescott Filter - MATLAB & Simulink
The Hodrick-Prescott (HP) filter is a specialized filter for trend and business cycle estimation (no seasonal component). Suppose a time series yt can be additively decomposed into a trend and business cycle component. Denote the trend component gt and the cycle component ct. Then
{y}_{t}={g}_{t}+{c}_{t}.
The HP filter finds a trend estimate,
{\stackrel{^}{g}}_{t}
, by solving a penalized optimization problem. The smoothness of the trend estimate depends on the choice of penalty parameter. The cycle component, which is often of interest to business cycle analysts, is estimated as
{\stackrel{^}{c}}_{t}={y}_{t}-{\stackrel{^}{g}}_{t}
hpfilter returns the estimated trend and cycle components of a time series. |
Can Someone Write My fluid mechanics using matlab Flawlessly?
Get In Touch With Our PhD Writers At matlabassignmentexperts.com For Excellence
Fluid mechanics, Bernoulli’s approach, Hardy Cross & Monte Carlo methods
Applications of fluid mechanics.
Fluid mechanics using Matlab
In this Matlab sample solution, the expert has demonstrated his capabilities to solve problems on topic fluid mechanics. The solution has been provided using Matlab. The given problem showcases an application of concepts like Bernoulli approach, the minimum pressure in vena contracta, Cavitation, Hardy cross method, and Monte Carlo method, etc.
Before we go ahead and define what fluid mechanics is, let’s understand what a fluid is. A Fluid as you probably know means a liquid. But in physics, it has a different meaning. It is defined as any substance that changes its shape when some external forces are applied to it. They include liquids and gases. Fluids are said to have shear modulus, which is a physical term that implies zero resistance to any force applied to it. Fluids and solids are all states of matter. Therefore fluids are substances that can flow and do not resist any change to its shape.
Fluid mechanics is a branch of physics that focuses on fluids, i.e., liquid, air, and plasma.
This branch of physics has lots of applications in many disciplines, which is not restricted only to real-life scenarios. The engineers are also interested in fluid mechanics because they can use the forces used in fluids for other meaningful purposes. Meteorologists also study fluid mechanics as it can help them in forecasting the weather patterns of a place. Let’s not forget that physicists study fluid mechanics to discover the movement of particles through magnetic fields, which help them in search of an acceptable means of harnessing energy for nuclear fusions.
Fluid mechanics has lots of applications that make it an important topic in most engineering fields. Here we will show you some of the practical applications of fluid mechanics.
Branches of fluid mechanics
Fluid mechanics is further subdivided into two branches which are:-
Fluid Statics -Fluid statics involves studying fluids which are at rest. It carries out a study of fluids that are in an equilibrium state. Fluid statistics has specific applications in real life. It can be used to explain the real-life phenomena such as the changes in atmospheric pressure with changes in altitude or why a block of wood floats in water while a stone cannot. It also finds its way in astrophysics, geophysics, and meteorology.
Fluid dynamics- Fluid dynamics is in contrast to fluid statistics. It studies the fluids in motion. The various empirical laws have been developed to tackle the different fluid movement problems. Sometimes, determining the properties of a fluid such as viscosity and density could be the solutions. Fluid dynamics can be subdivided further into aerodynamics and hydrodynamics—the latter deals with the movement of liquids while the former with the movement of gases. Fluid dynamics has wide-ranging applications, which include traffic engineering, predicting weather patterns, and aircraft movements.
History of fluid mechanics in a nutshell
The origin of fluid mechanics can not be traced specifically to any point in time. But it is said to be used in the ancient civilizations that flourished in Egypt and Babylon. The Egyptians used hydraulic pressure to water their fields, which were situated close to the river beds. As civilizations developed, scientists like Archimedes contributed to it by coming up with the Archimedes principle. Other great scientists, such as Blaise Pascal and Isaac Newton, contributed to the formulation of current laws that we study today.
Fluid mechanic’s common terms, and governing laws.
If you are studying fluid mechanics, there are some of the terms and governing laws that you will encounter from time to time. They form the basis of this topic. Some of these terms include: –
Viscosity. This is a term that implies the resistance of a fluid to deformation. It could also be defined as the inability of fluid to refuse to flow. A fluid with high viscosity is said to be very dense which could achieve the shape of a solid. High viscosity always happens in low temperatures. A fluid with low viscosity is known as an ideal fluid.
Surface tension. It is a phenomenon that is commonly seen on the surface of liquids. Mostly, these surfaces shrink together to occupy a minimum surface area. In water, this surface allows for other insets to float on water. Surface tension can be explained in terms of cohesive forces. In a liquid, all the molecules are attracted to each other by cohesive forces. But the issue is different at the surface as all the molecules do not have other molecules to hold them at the top. As a result, they shrink inwards.
According to this explanation, we can say that surface tension also affects the shape of droplets.
Archimedes principle. Archimedes principle is named after an ancient Greek named Archimedes who discovered it. It states that a body that is submerged in water is rested upon by an equal force even if it’s fully or partially submerged. It’s also known as the law of buoyancy, and one of its common applications is in ships. A ship floats on the sea once its weight is equal to the one it displaces in water.
Pascal’s law. This is a law about the transmission of pressure in a fluid. It states that in a confined and incompressible fluid, a pressure change on one side will lead to an equal transmission of the fluid on another part of the confined area. This law is very important and has its applications in real life.
Bernoulli’s principle. This is a law that was proposed by Daniel Bernoulli in 1738. It states that as the flow of the liquid increases, the pressure in the liquid decreases.
The real-life applications of the topic are numerous. Some of its uses are listed as follows: –
Hydroelectric power. Here, fluid mechanics is applied to determine the power capabilities of a certain area.
Housing. Fluid mechanics is responsible for the distribution of water in cities and towns. It can be used to calculate the amount of water that is needed in these cities. We can even know their monthly water consumption.
Biology. In biology, we can apply fluid mechanics to know the rate at which blood flows in the veins.
Pumps. Pumps are devices that are made for fluid movements. No matter what pump you are using, they apply the fluid mechanic's laws, such as Bernoulli’s principle and Pascal’s law.
Matlab is one of the finest mathematical computing software that exits in the world. It was built to help engineers and scientists in their research assignments. Matlab was developed during the internet age in the 1980s, but its use has grown exponentially since that time. Today, most institutions have incorporated it into their curriculum, while most science and engineering practitioners prefer to use Matlab. It’s therefore, a widely accepted programing language.
For fluid mechanics, Matlab has the CFD toolbox, which is an app for all computational fluid dynamics problems. With it, you can model fluid dynamics, simulate it, and use it for predictions. In other cases, you could use the symbolic math toolbox for calculations involving fluid mechanics.
If you want help with fluid mechanics using Matlab, then you can contact us. We are highly trained Matlab professionals with years of experience doing Matlab related assignments. We have been serving our clients for close to a decade and are still providing them with consistent results. We put the priorities of our clients first and ensure that we meet the deadlines and help them secure high grades.
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Task 1.1: Use Bernoulli and textbook approach to estimate minimum pressure in vena contra(i.e. location in a fluid flow where the diameter of the flow is the smallest). If needed, review section 6.9 in the textbook by White to learn more about the vena contra. Assume that at the narrowest location effectively defective /D ~ 0.23 due to vena contra.
Explain/sketch geometry.
Recall Bernoulli equation – see sketch below
With p1 given in the assignment, estimated minimum pressure
assuming 20C water, what are cavitation numbers based on both upstream and vena contra pressures. – leave the further discussion on if this might cavitate to task 1.3
Task 1.2: Evaluate the CFD results provided to you. Determine the margin to the onset of cavitation. If you are choosing your own case of interest, you will need to perform computations with a CFD software that you have available.
For this task, plot contours of pressure and identify and label minima. Plot velocity vectors, streamlines, and vorticity contours. Calculate the mass flow rate in and out as a check.
Take great care to not blindly trust the results. CFD is an indispensable tool for modern engineers, but the case you are working on highlights the need for critical thinking and proper problem setup, and the sample data provided will be imperfect on purpose.
Start with either your own code or with sample code “SampleCapstone_part1_ForStudentsToComplete.m”
streamlines and
vorticity contours
Calculate the mass flow rate in and out as a check.
For example, here is a streamline plot. Your plots should be formatted to be easier to read with proper labels, etc. Note for example that in this plot axis labels are too small to even read.
Task 1.3: Consider the limitations of both theory and numerical model used. Also, reviewing the literature, explore the effect of fluid temperature and nuclei contents to point where incipient cavitation may be expected to occur. Discuss how such cavitation could be detected, qualitatively observed, quantified, and what impact it may have in pipeline performance overall.
Explain what assumptions may be faulty and what physical phenomena are or are not accounted for that may lead to differences in the results of the inviscid theory, CFD vs. experiment.
Discuss/list how both Bernoulli and the time-averaged axisymmetric CFD were idealizations and may have been limited
Based on cavitation numbers from tasks 1.1 and 1.2, do you think cavitation might occur?
How does this conclusion match with data from Testud et al. (2007)?
Task 1.4: What measurement techniques discussed in the course could you use to observe the flow in this case, and succinctly discuss what would limit their suitability for the case in question.
List or make a table of suitable techniques and why they are useful, and up to what point. At least 5 should be easy to recall and discuss. See slides¬es from days 3&4 to recall techniques discussed.
Technique Why Limitation
Set Q at all outlets, some of them must equal our single inlet.
Calc pump power assuming lowest pressure node at 300kPa absolute, (note all up in solution are relative and since we treat fluid as incompressible absolute pressure within HC solution can be adjusted afterward if needed – pressure differences are still what the solution yield), the reservoir at 100kPa (1 atm absolute).
Task 2.1: Write your own code using the Hardy-Cross method to calculate pressure distribution in the sample pipe network. (Contact SSA for sample Hardy-Cross – at which time SSA will want to schedule a virtual meeting with you to discuss through the example. Also, another similar sample will be discussed by Gabriel.) Use the Haaland explicit approximation to compute you the friction factor – just like we did in class!(Make code flexible and see steps below to enable evaluation of temperature etc. effects.) Compute the power required to pump water through the \textbf{ideal} network at given flow rates.
Setup Hardy-Cross solution for pipe network – start either from sample code SSA will provide and discuss with you, or from example, Gabriel will discuss
Use equations covered in class to estimate pumping power
Optional: assume something for pump efficiency and discuss succinctly
Task 2.2: Use your code and Monte-Carlo method for propagating uncertainty (section
\text{???}
\ref{section:MC}
) and consider i) uncertainty in pipe roughness (see the table in handout), ii) uncertainty in pipe length, ii) uncertainty in temperature and iv) uncertainty in pipe diameter. To estimate uncertainty in roughness, use data from White table 6.5. (Consider, for example, if water temperature goes from 5C to 30C, what effect will this have?)
Modify code to loop over HC calculation and run MC for at least 100,000 cases – more if histogram looks ‘rough’. Plot histogram of needed pumping power – just like we did during Day 5 for White example 6.16
Task 2.3: Within each pipe, calculate viscous length scale (SSA will provide example upon request), wall shear stress and estimate if a superhydrophobic coating with a damage threshold of 50 Pa of shear and manufacturable with RMS roughness from 5 to 200 microns might result in drag reduction in each pipe segment (Requires having roughness below ~5 viscous length scales. For simplicity: assume that there would be a way to avoid entrainment of gas from the surface into the flow and a way to supply gas to the surface at pipeline pressure. – not necessarily achievable presently).
For each pipe, knowing diameter and average flow rate, calculate viscous length scale
How smooth is the pipe compared to the viscous length scale (epsilon over length scale, if<5 smooth, if >70 fully rough and in-between transitional – see White’s book)?
Calculate shear in each pipe (average at the wall)
Discuss if SHS may survive given damage threshold
Bonus task 2.1: add the cost of components and energy to try to improve your pipe network. Explain the cost basis you selected and your results.
Assume a cost for pipes with smoother and more durable (e.g. stainless) being a higher price. Compare energy cost vs. initial cost for different options you chose.
Bonus task 2.2: Compare your code’s predictions to the prediction from the free code, EPANET. EPANET can be downloaded from here.
Setup problems and run EPA net. The solution may look like something close to that below (Values WILL differ – sample below ran with different values than you will use.).
Compare node pressures and pipe flow rates to your MatLab solution from task 2.1
Bonus task 2.3: What would be the achievable reduction in pumping power requirements if the pipeline were to be one standard size larger have smoother surfaces owing to initial material selection or maintenance, or if fittings causing minor losses would have been chosen such that they have the lowest available minor loss coefficient presently available for a commercial product?
Run simulation with larger pipes, etc. |
Modulate using phase modulation - Simulink - MathWorks Switzerland
Modulate using phase modulation
The PM Modulator Passband block modulates using phase modulation. The output is a passband representation of the modulated signal. The output signal's frequency varies with the input signal's amplitude. Both the input and output signals are real scalar signals.
If the input is u(t) as a function of time t, then the output is
\mathrm{cos}\left(2\pi {f}_{c}t+{K}_{c}u\left(t\right)+\theta \right)
fc represents the Carrier frequency parameter
θ represents the Initial phase parameter
Kc represents the Phase deviation parameter
An appropriate Carrier frequency value is generally much higher than the highest frequency of the input signal. By the Nyquist sampling theorem, the reciprocal of the model's sample time (defined by the model's signal source) must exceed twice the Carrier frequency parameter.
The phase deviation of the carrier frequency in radians. This is sometimes referred to as the "variation" in the phase. |
Physical World, Popular Questions: CBSE Class 11-science PHYSICS, Physics Part I - Meritnation
Jacob Ambati & 4 others asked a question
how to make any working model in physics?
thrown with the
help of a sling
velocity v at an
angle theta from
a) Working of sling is based on...........
law of vector addition. (1)
b) With the help of a vector diagram,
state this law. (1)
c) Derive the expression for the
maximum height reached by
Arpitpatni asked a question
i want sl arora physics solution
A dielectric cylinder of a radius is infinite long. Its volume charge density varies directly as the distance from the cylinder. If P is zero at the axis and Ps on the surface, the electric intensity due to it is :
A. Psa^3/ 3E0a at a point distant 'r' outside it
B. Psr^2/ 3E0a at a point distant 'r' inside it
C. Ps^a/3E0 at a point on its surface
D. Ps/3E0a^2 at a point on its surface
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plsss!!!!!! suggest me a working model on any topic of physics for class xi bt a gud one.
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What is meaning of 13K in rank
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Write the dimensions of a and b in the relation P=b-x2/at where P is power, x is distance and t is time.
There is a cubicle cavity inside a conducting sphere of radius R A positive point charge Q is placed at the center of the cube and another charge q is placed at a distance l (>R)from the center of Sphere. The sphere is earthed.
28. Charge induced on the inner surface of cavity is
A. -Q, uniformly distributed B. -Q, non-uniformly distributed C. -(Q+q), non-uniformly distributed D. None
29. Net charge on the outer surface of conducting surface is
A. + Q B. B. Q- qR/l C. -qR/l D.None
30. Potential at a point inside the cavity is:
A. A. Zero B. Positive C. negative D. can not be determine
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WHAT IS MEANING OF INERTIA? plz explain me in hindi.
Q 2 electrostatic
Q2. Two isolated spherical shell having charges Q1 and Q2 and radii r1 and r2 are kept at very large separation. If they are joined by a conducting wire, then choose incorrect statement:
(A) Electrostatic potential energy of system must decrease.
(B) Electrostatic potential energy of system may decrease.
(C) If Q1r2 = Q2 r1, then no charge flow through the conducting wire.
(D) If Q1r2= Q2 r1, then electrostatic potential energy of system already minimum.
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differentiation of sin x by ab initio method
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what is unification and reduction in physics?
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examples for inertia of rest,motion and direction?
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Pavenesh Chaturvedi & 2 others asked a question
1.The moon is observed from 2 diametrically opposite points A & B on earth. The angle teta subtended at the moon by the 2 directions of observation is 1°54min. Given that the diameter of earth is 1.276*10power 7 metre.Calculate the distance of moon from earth.
2.The Suns angular diameter is 1920sec. The distance of Sun from earth is 1.496*10power11metre. What is the diameter of Sun
3. The escape velocity from surface of earth is given by V=sq root of 2GM/R , where M is mass, R is radius, G is universal gravitational constant. Check the correctness of this equation by using the method of dimension
The current gain in common base configuration of n-p-n transistor is 0.8. The current amplification factor in common emitter configuration is
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Take the Inverse Laplace Transform of the function.F(s)=\frac{1}{s(s^2+2s+2)}
Take the Inverse Laplace Transform of the function.
F\left(s\right)=\frac{1}{s\left({s}^{2}+2s+2\right)}
Khribechy
F\left(s\right)=\frac{1}{s\left({s}^{2}+2s+2\right)}=\frac{1}{s\left({\left(s+1\right)}^{2}+{\left(1\right)}^{2}\right)}=\frac{1}{s}\frac{1}{\left({\left(s+1\right)}^{2}+{\left(1\right)}^{2}\right)}
⇒F\left(s\right)=L\left\{1\right\}L\left\{{e}^{-t}\mathrm{sin}\left(t\right)\right\}
⇒{L}^{-1}\left\{F\left(s\right)\right\}=f\left(t\right)=1\cdot \left({e}^{-t}\mathrm{sin}t\right)
⇒f\left(t\right)={\int }_{0}^{t}{e}^{-u}\mathrm{sin}\left(u\right)1du={\left(\frac{{e}^{-k}}{2}\left[-1\mathrm{sin}\left(u\right)-1\mathrm{cos}\left(u\right)\right]\right)}_{0}^{t}
={\left(-\frac{{e}^{-4}}{2}\left[\mathrm{sin}\left(u\right)+\mathrm{cos}\left(u\right)\right]\right)}_{0}^{t}
=-\frac{1}{2}\left[{e}^{-t}\left(\mathrm{sin}\left(t\right)+\mathrm{cos}\left(t\right)\right)-1\right]
⇒f\left(t\right)=\frac{1}{2}\left[1-{e}^{-t}\left(\mathrm{sin}\left(t\right)+\mathrm{cos}\left(t\right)\right)\right]
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MetricSearch - Maple Help
Home : Support : Online Help : Mathematics : DifferentialGeometry : Library : MetricSearch
Library[MetricSearch] - a Maplet for searching the DifferentialGeometry libraries of metrics
MetricSearch()
MetricSearch(PropList)
PropList - a list of spacetime metric properties to search for.
The DifferentialGeometry Library package contains an extensive database of spacetime metrics and matter fields which give solutions to the Einstein equations of general relativity. The command MetricSearch allows the Maple user to search the database for metrics with specified properties. The first calling sequence launches an easy to use Maplet. The second calling sequence provides the same search capabilities in a command line format.
The command MetricSearch() initializes a maplet which searches the DifferentialGeometry Library for metrics with user-specified properties. The current search criteria are summarized in the following table.
Primary Description. Search for solutions to the Einstein equations with a prescribed class of energy-momentum tensors.
Secondary Description. Search for metrics with a prescribed class of mathematical or physical properties.
Keyword. Keyword descriptions include metric names or authors.
Petrov Type. Algebraic classification of the Weyl tensor.
Plebanski-Petrov Type. Algebraic classification of the Ricci tensor. This is the Petrov type of the Plebanski tensor.
Segre Type. Algebraic classification of the Ricci tensor. The Segre type specifies the normal form of a linear transformation which is self-adjoint with respect to a 4-dimensional Lorentz signature metric.
Isometry Properties
Isometry Dimension. The number of Killing vectors. KillingVectors
Orbit Dimension. The number of pointwise independent Killing vectors; or the dimension of the orbit of the group of isometries.
Orbit Type. The signature of the induced metric on the orbits. SubspaceType
Isotropy Type. For any Lorentz signature 4-dimensional spacetime, the infinitesimal isotropy representation defines a subalgebra of
\mathrm{so}\left(3,1\right)
. These subalgebras have been classified and are labeled
\mathrm{F1}, \mathrm{F2} , ..., \mathrm{F15}
IsotropySubalgebra, IsotropyType
Once properties are selected by checking appropriate boxes, pressing the Search button will return all metrics in the DifferentialGeometry library which possess all of the indicated properties. The format of the result is a string representing the source reference and a sequence of lists indicating the equation numbers in the reference where the metrics appear. All the information in the DifferentialGeometry library for each metric can be obtained with the Retrieve command.
Information regarding the metric found by a search of the DifferentialGeometry library can be retrieved in one of two ways. One method is to enter the reference name (string), equation number, and the name of an initialized manifold (created in the calling worksheet using DGsetup) into the text boxes in the Retrieve section of the MetricSearch Maplet. Then pressing the Retrieve button will return a list of the spacetime fields defining the solution (e.g., metric, electromagnetic field, etc.) to the calling worksheet. All the information in the DifferentialGeometry library for each metric can be obtained with the Retrieve command.
The Clear button resets all check boxes and clears all text boxes. The Close button will close the Maplet.
The second calling sequence accepts a list of desired properties, each specified by an equation
\mathrm{metricproperty} = \mathrm{propertyvalue}
. The possibilities are:
"PrimaryDescription"
"Dust", "Einstein", "EinsteinMaxwell", "PerfectFluid", "PureRadiation", "Vacuum"
EinsteinTensor, EnergyMomentumTensor
"SecondaryDescription"
"Homogeneous", "PlaneWave", "PPWave", "PureRadiation", "RobertsonWalker", "SimplyTransitive", "Static"
"KeywordDescription"
a list of strings: author names, metric names, etc.
"PetrovType"
"I", "II", "III", "D", "N", "O"
"PlebanskiPetrovType"
"SegreType"
For example, "[1,(111)]"
"IsometryDimension"
"OrbitDimension"
"OrbitType"
"Null", "Riemannian", "PseudoRiemannian"
SubspaceType, QuadraticFormSignature
"IsotropyType
"F1", "F2", "F3", "F4", "F5", "F6", "F7", "F8", "F9", "F10", "F11", "F12", "F13", "F14", "F15"
Currently the DifferentialGeometry library contains selected metrics from:
2. Hawking, Stephen; and Ellis, G. F. R. (1973). The Large Scale Structure of Space-Time. Cambridge: Cambridge University Press.
\mathrm{with}\left(\mathrm{DifferentialGeometry}\right):
\mathrm{with}\left(\mathrm{Tensor}\right):
\mathrm{with}\left(\mathrm{LieAlgebras}\right):
\mathrm{with}\left(\mathrm{Library}\right):
We find examples of metrics which are homogeneous Einstein metrics of Petrov type III. First initialize a manifold with coordinates, e.g.,
\left(\mathrm{x1}, \mathrm{x2}, \mathrm{x3}, \mathrm{x4}\right).
\mathrm{DGsetup}\left([\mathrm{x1},\mathrm{x2},\mathrm{x3},\mathrm{x4}],M\right)
\textcolor[rgb]{0,0,1}{\mathrm{frame name: M}}
Start the MetricSearch Maplet.
F≔\mathrm{MetricSearch}\left(\right):
Check the Einstein box in the PhysicalProperties-Primary Description section, check type III in the Algebraic Properties-Petrov Type section, check 4 in the Orbit Dimension section.
The result is "Stephani": [12, 35, 1].
Enter "Stephani" into the Reference textbox, enter [12, 35, 1] into the equation number textbox, and enter M into the manifold textbox. Press the Retrieve button. The metric is assigned to
F.
(If matter fields were present, they would also be assigned to
F
F
We calculate some properties of a given metric and identify the metrics with the same properties in the library database.
\mathrm{DGsetup}\left([t,x,y,\mathrm{\phi }],M\right)
\textcolor[rgb]{0,0,1}{\mathrm{frame name: M}}
g≔\mathrm{evalDG}\left(\frac{1}{{x}^{2}}\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}\right)+{x}^{2}\mathrm{dphi}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dphi}-\left(\mathrm{dt}-2y\mathrm{dphi}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\left(\mathrm{dt}-2y\mathrm{dphi}\right)\right)
\textcolor[rgb]{0,0,1}{g}\textcolor[rgb]{0,0,1}{:=}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{dt}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dt}}\textcolor[rgb]{0,0,1}{+}\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}{\mathrm{dt}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dphi}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{\mathrm{dx}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dx}}}{{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{\mathrm{dy}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dy}}}{{\textcolor[rgb]{0,0,1}{x}}^{\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}{y}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dphi}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dt}}\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dphi}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dphi}}
We use the command RainichConditions to see if the space-time is a solution of the Einstein-Maxwell equations (electrovac spacetime).
\mathrm{RainichConditions}\left(g\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
We use the command KillingVectors to determine the dimension of the group of isometries.
\mathrm{KV}≔\mathrm{KillingVectors}\left(g\right)
\textcolor[rgb]{0,0,1}{\mathrm{KV}}\textcolor[rgb]{0,0,1}{:=}\left[\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{4}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{D_x}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{4}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{D_y}}\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{4}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{φ}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{D_phi}}\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{φ}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{D_t}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{4}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{D_y}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{4}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{D_phi}}\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{D_t}}\right]
\mathrm{nops}\left(\mathrm{KV}\right)
\textcolor[rgb]{0,0,1}{4}
Next we find the Petrov type of the metric.
\mathrm{PetrovType}\left(g\right)
\textcolor[rgb]{0,0,1}{"I"}
Search for this metric in the data-base using the command line version of MetricSearch. We find that this is the metric in Stephani, Kramer et al. equation [12, 21, 1].
\mathrm{MetricSearch}\left(["PrimaryDescription"="EinsteinMaxwell","PetrovType"="I","IsometryDimension"=4]\right)
\left[\left[\textcolor[rgb]{0,0,1}{"Stephani"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\left[\textcolor[rgb]{0,0,1}{12}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{21}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\right]\right]\right]
\mathrm{Retrieve}\left("Stephani",1,[12,21,1],\mathrm{manifoldname}=M,\mathrm{output}=["Metric"]\right)
\left[\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{dt}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dt}}\textcolor[rgb]{0,0,1}{+}\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}{\mathrm{dt}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dphi}}\textcolor[rgb]{0,0,1}{+}\frac{{\textcolor[rgb]{0,0,1}{\mathrm{_a}}}^{\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{dx}}}{{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}}\textcolor[rgb]{0,0,1}{+}\frac{{\textcolor[rgb]{0,0,1}{\mathrm{_a}}}^{\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{dy}}}{{\textcolor[rgb]{0,0,1}{x}}^{\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}{y}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dphi}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dt}}\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dphi}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dphi}}\right]
Find the Goedel metric in the data-base and then retrieve it:
S≔\mathrm{MetricSearch}\left(["Keywords"=["Goedel"]]\right)
\textcolor[rgb]{0,0,1}{S}\textcolor[rgb]{0,0,1}{:=}\left[\left[\textcolor[rgb]{0,0,1}{"Stephani"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\left[\textcolor[rgb]{0,0,1}{12}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{26}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\right]\right]\right]
g≔\mathrm{Retrieve}\left(\mathrm{op}\left(S[1]\right),\mathrm{manifoldname}=M,\mathrm{output}=["Metric"]\right)[1]
\textcolor[rgb]{0,0,1}{g}\textcolor[rgb]{0,0,1}{:=}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{\mathrm{_a}}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dt}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dt}}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{\mathrm{_a}}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dt}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dphi}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{\mathrm{_a}}}^{\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{dx}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{\mathrm{_a}}}^{\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{dy}}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{\mathrm{_a}}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dphi}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dt}}\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{_a}}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\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}{\mathrm{dphi}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{dphi}}
DifferentialGeometry, Tensor, IsotropySubalgebra, IsotropyType, KillingVectors, PetrovType, PlebanskiTensor, SegreType, SubspaceType |
ARX model estimation using instrumental variable method with arbitrary instruments - MATLAB ivx - MathWorks Nordic
ARX model estimation using instrumental variable method with arbitrary instruments
sys = ivx(data,[na nb nk],x)
sys = ivx(data,[na nb nk],x,max_size)
sys = ivx(data,[na nb nk],x) estimates an ARX polynomial model, sys, using the instrumental variable method with arbitrary instruments. The model is estimated for the time series data data. [na nb nk] specifies the ARX structure orders of the A and B polynomials and the input to output delay, expressed in the number of samples.
An ARX model is represented as:
A\left(q\right)y\left(t\right)=B\left(q\right)u\left(t-nk\right)+v\left(t\right)
sys = ivx(data,[na nb nk],x,max_size) specifies the maximum size of matrices formed during estimation.
When using frequency-domain data, the number of outputs must be 1.
ARX model orders.
For more details on the ARX model structure, see arx.
Instrument variable matrix.
x is a matrix containing the arbitrary instruments for use in the instrumental variable method.
x must be of the same size as the output data, data.y. For multi-experiment data, specify x as a cell array with one entry for each experiment.
The instruments used are analogous to the regression vector, with y replaced by x.
ARX model that fits the estimation data, returned as a discrete-time idpoly object. This model is created using the specified model orders, delays, and estimation options. ivx does not return any estimated covariance information for sys.
Use iv4 first for IV estimation to identify ARX polynomial models where the instruments x are chosen automatically. Use ivx for nonstandard situations. For example, when there is feedback present in the data, or, when other instruments need to be tried. You can also use iv to automatically generate instruments from certain custom defined filters.
[1] Ljung, L. System Identification: Theory for the User, page 222, Upper Saddle River, NJ, Prentice-Hall PTR, 1999.
arx | arxstruc | idpoly | iv4 | ivar | polyest |
Evaluate the definite integral. \int_{0}^{1}\frac{7}{1+x^{2}}dx
ArcactCatmedeq8 2021-11-19 Answered
{\int }_{0}^{1}\frac{7}{1+{x}^{2}}dx
Refer to the question, we have to solve the definite integral a=0 and b=1
\frac{7}{1}+{x}^{2}dx
{\int }_{0}^{1}\frac{1}{1+{x}^{2}}
Use the formula of integration of arctanx (x) to solve the provided integral
\int \frac{1}{1+{x}^{2}}dx={\mathrm{tan}}^{-1}\left(x\right)
{\int }_{0}^{1}\frac{7}{1+{x}^{2}}dx=7{\left[{\mathrm{tan}}^{-1}\left(x\right)\right]}_{0}^{1}
=7×\frac{\pi }{4}
=\frac{7\pi }{4}
Step 1: If f(x) is a continuous function from a to b, and if F(x) is its integral, then:
{\int }_{a}^{b}f\left(x\right)dx=F\left(x\right){\mid }_{a}^{b}=F\left(b\right)-F\left(a\right)
Step 2: In this case,
f\left(x\right)=\frac{7}{1+{x}^{2}}
. Find its integral.
7{\mathrm{tan}}^{-1}\left(x\right){\mid }_{0}^{1}
F\left(x\right){\mid }_{a}^{b}=F\left(b\right)-F\left(a\right)
, expand the above into F(1)−F(0):
7{\mathrm{tan}}^{-1}\left(1\right)-7{\mathrm{tan}}^{-1}\left(0\right)
\frac{7\pi }{4}
For the integral state the number of the integration formula and the values of the constans a and b so that the formula fits the integral
\int \frac{1}{{x}^{2}\left(4x-3\right)}dx
{\int }_{0}^{\mathrm{\infty }}\frac{dx}{{e}^{x}+{e}^{-x}}
{\int }_{0}^{3}\frac{1}{{\left(x-2\right)}^{3}}dx
\int \left(6{x}^{5}-18{x}^{2}+7\right)dx
Evaluate the integral or state that it is divergent.
{\int }_{\frac{1}{2}}^{2}\frac{dx}{x\mathrm{ln}x}
\int \frac{3-x}{\sqrt{x}}dx
Use properties of the indefinite integral to express the following integral in terms of the simplest integrals:
\int \left(-4{x}^{2}+7x\right)dx |
Complex-number inequality |z_{1}z_{2}\cdots z_{m}-1|\leq e^{z_{1}-1|+\cdots+|z_{m}-1|}-1 Can anybody tell me how to prove
Complex-number inequality
|{z}_{1}{z}_{2}\cdots {z}_{m}-1|\le {e}^{{z}_{1}-1|+\cdots +|{z}_{m}-1\mid }-1
Can anybody tell me how to prove the following inequality?
Step 1 The inequality in question bounds how far you can get from 1 by multiplying several complex numbers that may individually not be far from 1. So it makes sense to try to derive a bound for the product of just two complex numbers, and then proceed by induction. Lemma: Suppose
|{z}_{1}-1|={\alpha }_{1}
|{z}_{2}-1|={\alpha }_{2}
|{z}_{1}{z}_{2}-1|\le \left(1+{\alpha }_{1}\right)\left(1+{\alpha }_{2}\right)-1
{\alpha }_{1}{\alpha }_{2}=|{z}_{1}{z}_{2}-{z}_{1}-{z}_{2}+1|
. By triangle inequality on the three points
{z}_{1}{z}_{2},\text{ }{z}_{1}+{z}_{2}-1
, and 1, we have
|{z}_{1}{z}_{2}-1|\le |{z}_{1}{z}_{2}-{z}_{1}-{z}_{2}+1|+|{z}_{1}+{z}_{2}-2|
\le {\alpha }_{1}{\alpha }_{2}+{\alpha }_{1}+{\alpha }_{2}
=\left(1+{\alpha }_{1}\right)\left(1+{\alpha }_{2}\right)-1
Now, for several numbers,
|{z}_{1}{z}_{2}\cdots {z}_{m}-1|\le \left(1+{\alpha }_{1}\right)\left(1+{\alpha }_{2,\cdots ,m}\right)-1
\le \left(1+{\alpha }_{1}\right)\left(1+{\alpha }_{2}\right)\left(1+{\alpha }_{3,\cdots ,m}\right)-1
⋮
\le \left(1+{\alpha }_{1}\right)\left(1+{\alpha }_{2}\right)\cdots \left(1+{\alpha }_{m}\right)-1
{a}_{2,\cdots ,m}
, for example, is my hopefully transparent abuse of notation to denote
|{z}_{2}\cdots {z}_{m}-1|
1+x\le {e}^{x}
for real x, the desired inequality follows.
EDIT: This answer is wrong. It all boils down to the inequality
|xy-1|\le |x-1|+|y-1|
, which I expect to be true. Given this inequality, prove by induction that
|{z}_{1}\cdots {z}_{m}-1|\le |{z}_{1}-1|+\cdots +|{z}_{m}-1|
Now use
{e}^{x}\ge 1+x
12\left({\mathrm{cos}60}^{\circ }+i{\mathrm{sin}60}^{\circ }\right)
Graph the complex numbers in the complex plane.
-1+3i
-4-5i
How do you find the power
{\left(-2+2i\right)}^{3}
and express the result in rectangular form?
How do you find the conjugate of
-4+5i
\frac{4-4i}{5+3i}
How to solve that quadratic complex equation?
\frac{{z}^{2}-1}{{z}^{2}+z+1}
So, there cant
How do you express the complex number in trigonometric form:
1-\left(\sqrt{3}\right)i |
Cryptogram Warmup Practice Problems Online | Brilliant
\Large \begin{array} {c c c } & 3 & \color{#EC7300}{X} \\ + & & 7 \\ \hline & 4 & 1 \\ \end{array}
\color{#EC7300}{X}
would make the above summation true?
\Large \begin{array} {c c c } & 2 & \color{#3D99F6}{A} \\ \times & & 4 \\ \hline & 9 & 6 \\ \end{array}
\color{#3D99F6}{A}
would make this multiplication true?
\Large \begin{array} {c c c } & 1 & \color{#3D99F6}{E} \\ \times & & \color{#3D99F6}{E} \\ \hline & 9 & \color{#3D99F6}{E} \\ \end{array}
\color{#3D99F6}{E}
\Large \begin{array} {c c c } & & \color{#20A900}{C} \\ + & \color{#20A900}{C} & \color{#20A900}{C} \\ \hline & \color{#69047E}{D} & 4 \\ \end{array}
In this cryptogram,
\color{#20A900}{C}
\color{#69047E}{D}
represent two different digits. What is the value of
\color{#20A900}{C}
\Large \begin{array} { c c c } & 1 & \color{#EC7300}{B} \\ + & \color{#EC7300}{B} & 6 \\ \hline & 7 & 1 \\ \end{array}
\color{#EC7300}{B} |
Analytical Chemistry: Uses Of Ammonium Hydroxide And Sodium Hydroxide, Popular Questions: ICSE Class 10 CHEMISTRY, Concise Chemistry 10 - Meritnation
What is the white precipitate that is insoluble in NH4OH but soluble in NAOH
Give the molecular structural formula of (i)ethyl propanoate and(ii) propyl ethanoate
(i) Insoluble, (ii) Soluble,
in (a) Caustic soda solution (b) Ammonium hydroxide solution.
Name the acid on mixing with Lead Nitrate produces a white precipitate which remains insoluble on heating?
Two gases which react in the 1)presence of copper catalyst at the high temperature to give an alcohol 2)a gas liberated when the product of reaction of bromoethane and AQUEOUS KOH react with sodium metal
What happens.at cathode and anode during electrolytic reduction of sodium.chloride
Lekha asked a question
The pale green solid turns reddish brown on heating.Its aqueous gives a white precipitate with barium chloride solution. The precipate is insoluble in mineral acid. Give the answer and state the reason why your answer was that..
Plzz help to find
Why are the alkalis added drop by drop to the salt solution?
What is the physical state of gas carbon?
The substance which react with hot conc.NaOH soln.& undergoes a neutralization reaction
Al(OH)?
how to solve numbers 4 and 5?
In which periodic properties does fluorine and chlorine show exception?
Plz EXPLAIN no.1
Commercial sodium hydroxide weighing 30 g has some sodium chloride in it.
The mixture on dissolving in water and subsequent treatment with excess silver nitrate solution formed a precipitate weighing 14.3 g. What is the percentage of sodium chloride in commercial sample of sodium hydroxide? The equation for the reaction is
NaCl + AgNO3 ----> AgCl + NaNO3
[Relative molecular mass of NaCl = 58 ; AgCl = 143]
A yellow yellow solution of salt yields a reddish brown precipitate with caustic soda solution. Thr precipitate does not dissolve in excess of alkali. The reddish brown precipitate on strong heating leaves behind the red powder, insoluble in water but soluble in dilute HCl. 1. Identify the metal of salt. 2. Write the equation of reaction involved, assuming the salt to be a sulphate salt.
What is the most appropriate method for the preparation of silver chloride
Armav asked a question
What is colour of pbo2
Please solve all of them fast.....
Plzz find it
Plzzz find it
The metal which does not react with water or dilute sulphuric acid but reacts with concentrated sulphuric acid
Add some suitable words which are missing in each of the three sentences
Name the following .i)an amphoteric oxide reduced by a basic gas.ii) the oxide which do not produce an acid when reacted with water.
Plz answer me the questions
State ur observation when sodium hydroxide is added in excess to magnesium chloride solution.
Answer question number 3 a by filling the tabular column
Give the balanced equation for action of hydrochloric acid on lead dioxide
Name the solution which on reacting with caustic soda gives a light green ppt and on reacting with barium chloride soln. gives a white ppt.
Fuzzikins Divya asked a question
BRAINLIEST answer this question
Aachal Milind Awasare asked a question
Name the salt solution that does not form a precipitate with NH4OH and give its reason
Question) What is the mass of pure ethanoic acid required to neutralise 280 mL of 0.5 molar pure lime water completely?
But, c6h12 belongs to alkene. So there must be a DOUBLE bond between 2 carbon atoms?
what happens when ammonium salt is heated with caustic soda sollution
Nilaruna Roy asked a question
Liquefied HCl and hydrochloric acid are the same?
Liquefied means water should be present. Then why do we say that liquefied HCL gas has no effect on Litmus?
Name an aqueous salt solution used for testing sulphate radical?
If x is a metal with valency 2, Y is a non metal with valency 5, and Y is a triatomic gas, write an equation for the direct combination of X and Y to form a compound.
Plz Answer Q. no 2
Ayush Chandra asked a question
Please tell me this question..
(b) Sodium hydroxide is added to the solutions containing the ions X. List Y give of the precipitate. Match the ions with their coloured precipitated :-
List X List Y
(1) Pb2+ (a) Reddish brown
(2) Fe2+ (b) Blue
(3) Zn2+ (c) Dirty green
(4) Fe3+ (d) White soluble in excess
(5) Cu2+ (e) White soluble in excess
Plz say yes or no only
Please tell me the solution of Q 2004 and Q 2005.
Q2004. Sodium hydroxide solution is added first in a small amount, then in excess to the aqueous salt solutions of
a] copper [II] sulphate b] zinc nitrate c] lead nitrate d] iron [III] sulphate. State in each case-
i] the colour of the precipitate when NaOH is added in a small quantity;
ii) the nature of precipitate [i.e. soluble or insoluble ] when NaOH is added in excess.
Write balanced equations for - a] Aluminium b] Zinc - is warmed with NaOH [ caustic soda ] soln.
Q2005. The questions below refers to the following salt solutions listed A to F: A :- Copper nitrate.
B: Iron [II] sulphate, C : Iron [III] chloride, D : Lead nitrate, E : Magnesium sulphate, F : Zinc chloride.
i] Which soln. becomes a deep / inky blue colour when excess of ammonium hydroxide is added to it.
ii] Which solution gives a white precipitate with excess ammonium hydroxide solution.
please solve this question of fill in the blanks with the choices given in brackets as fast as possible as it is very urgent
What is the difference between methoxy benzene, methoxy hexene, methoxy hexane structurally?
Calculate the percentage of nitrogen and oxygen in ammonium nitrate.
[Relative molecular mass of ammonium nitrate is 80, H=1, N=14 , O=16]
Differentiate between using nh4oh
1. Pb(NO3)2 and Zn(NO3)2
2. CaCl2 and Cucl2
Write the reaction and observations
1. Inert gases have high IP
2. Anion has a bigger atomic radius than its parent atom
At first tell me which one is odd one out and then why the others are similar?
55. The reaction that differs from the rest of the reactions given is
(A) Formation of calcium oxide from limestone
(B) Formation of aluminium from aluminium oxide
(C) Formation of sodium carbonate from sodium hydrogen carbonate
(D) Formation of mercury from mercuric oxide
Q. What is the meaning of this line?
A balanced equation must be molecular.
Identify the compound
It gives white ppt with less NaOH and white ppt with barium chloride solution insoluble in concentrated HCl.
Write the balance equations for this.
Q. Write balanced equations for the following conversions
\left(a\right) ZnS{O}_{4} \stackrel{A}{\to } Zn\left(OH{\right)}_{2} \stackrel{B}{\to } N{a}_{2}Zn{O}_{2} .\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(b\right) CuS{O}_{4} \stackrel{A}{\to } Cu\left(OH{\right)}_{2} \stackrel{B}{\to } \left[Cu\left(N{H}_{3}{\right)}_{4}\right]S{O}_{4} .
Does non metal react with dilute nitric acid?
The salt which does not react with ammonium hydroxide solution
ZnCl?
CuCl?
NH?Cl
FeCl?
ferrous salts on oxidation form ___ salt
Solve number 53 plz
53. The mass of sodium chloride formed when 5.3 g of sodium carbonate is dissolved in 250 ml of
\frac{1}{2}
(A) 5.85 g (B) 7.32 g
(C) 11.7 g (D) 58.5 g
Shreya Mandal asked a question
Aqua regia is a mixture of dilute hydrochloric acid and concentrated nitric acid
Voormika asked a question
Why metal oxides and hydroxides neutralize acids but do not react with bases?
What is the difference between NaHCO3 and NaCO3 with reason
Last 10 years papers for icse boards
Balwant Mehta asked a question
Identify the compounds
Please explain 3rd and 4th reactions
Nitrate which does not give nitrate test
Q17 Which among the following substances posses only vanderwall's forces of attraction?
How will you distinguiah between NH4OH solution from NaOH solution
Why alkalis are very good reagents? |
Basic Syllogism Logic Practice Problems Online | Brilliant
Based on the diagram above, which of these statements is true?
All Aliens are Martians. All Martians are Aliens.
If the statement "all bats are nocturnal" is represented on the diagram above, which region of the diagram will be empty? (Conventional definitions for the words used in these problems do not apply).
The intersection between Meeps and Blue is crossed out; nothing is in that section. This means
\text{\_\_\_\_\_\_\_\_\_\_}.
No Meeps are Blue All Meeps are Blue All Blue things are Meeps
The statement "some A are B" means which section of the diagram above can be crossed out because it will empty?
1 2 3 None of them
All mammals are animals. All animals are cats. All mammals are cats. |
Find an answer to any inferential statistics problems
Get help with inferential statistics problems
Recent questions in Inferential Statistics
The incomplete dot plot shows the result of a survey in which each student was asked how many dimes were in their pockets or wallets. The results for “4 dimes” are not shown. Each dot represents one student. It is known that 12.5% of the students had one dime.
Find the number of students surveyed. Then complete the dot plot.
What percent of the students had either 0 or 6 dimes?
Briefly describe the distribution of the data
\rho
, the pearson correlation, is a measure for the linear dependence of two random variables say X, Y. But can't you say just transform X and Y such that we have,
{\rho }_{X,Y}\left(f\left(X\right),g\left(Y\right)\right)
where f, g are non-linear functions such that it measures other kinds of dependce (take for example
f\left(s\right)=g\left(s\right)={s}^{2}
for quadratic dependence).
X=
Lot &
Y=
Give a broken line linear model with a breakpoint at
250
Y={B}_{0}+{B}_{1}{X}_{1}+{B}_{2}{X}_{2}+{B}_{3}{X}_{3}+e
{X}_{2}=0
1
depending on whether the lot size is
\ge 250
<250
{X}_{3}={X}_{1}\cdot {X}_{2}.
Which hypothesis statement is equivalent to the statement: The two regression lines have the same intercept term?
{H}_{0}:{B}_{0}=0\phantom{\rule{0ex}{0ex}}{H}_{0}:{B}_{1}=0\phantom{\rule{0ex}{0ex}}{H}_{0}:{B}_{2}=0\phantom{\rule{0ex}{0ex}}{H}_{0}:{B}_{3}=0
{B}_{0}
is obviously the intercept for simple linear regression models, however, the broken line phrasing is causing me to be confused on this. My initial instinct was to assume
{B}_{0}
hypothesis was appropriate, but now I'm wondering if
{B}_{2}=0
makes more sense.
If I have built two linear regression models over sets
A
B
, and now want a linear regression over set
A\cup B
.Is there a way to reuse what I already have?
Is the total sum of squares for multiple regression the same as the total sum of squares for anova?
Is anova a test for bivariate correlation or multiple regression?
What do you mean by a distribution is homoscedastic (ie,
\sigma \left(Y|X=x\right)=\sigma
) in the context of simple linear regression? Why do we need this assumption in simple linear regression? What will happen to the regession if a distribution is not homoscedastic?
Carley Haley 2022-05-08 Answered
In Linear regression, we have
\theta ={\left({X}^{T}X\right)}^{-1}{X}^{T}y
In Ridge regression, we have
\theta ={\left(\lambda I+{X}^{T}X\right)}^{-1}{X}^{T}y
I learnt somewhere that while
{X}^{T}X
is not guaranteed to be invertible,
\lambda I+{X}^{T}X
is guaranteed to be invertible.
Porter Mccullough 2022-05-03 Answered
\begin{array}{cccccc}x& 1& 2& 3& 4& 5\\ y& 3& 6& 8& 9& 0\\ y& 4& 6& 1& 2& 4\end{array}
and know that it is a simple linear regression model, what is the value of
n
? I think it is either
5
10
but am not sure which one.
juniorychichoa70 2022-05-02 Answered
I am getting
{f}_{X,Y}\left(x,y\right)={f}_{X}\left(x\right){f}_{Y}\left(y\right)
even if the correlation coefficient
\rho \ne 0
znacimavjo 2022-05-01 Answered
Suppose you have two sequences of complex numbers
{a}_{i}
{b}_{i}
indexed over the integer numbers such that they are convergent in
{l}^{2}
norm and a has norm greater than b in the sense
\mathrm{\infty }>\sum _{i}|{a}_{i}{|}^{2}\ge \sum _{i}|{b}_{i}{|}^{2}.
Suppose moreover they are uncorrelated over any time delay, meaning
\sum _{i}{a}_{i}\overline{{b}_{i-n}}=0\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\in \mathbb{Z}.
Is it true that the polinomial
a\left(z\right)=\sum _{i}{a}_{i}{z}^{-i}
is greater in absolute value than
b\left(z\right)=\sum _{i}{b}_{i}{z}^{-i}
for any unit norm complex number z?
Two random variables, X and Y, have the joint density function:
f\left(x,y\right)=\left\{\begin{array}{ll}2& 0<x\le y<1\\ 0& ioc\end{array}
Calculate the correlation coefficient between X and Y.
We have a situation where we have pairs of points where x=heights of fathers and y=heights of the sons of these fathers. The mean of
X
Y
SD\left(Y\right)=SD\left(X\right)
0<R<1
. Now I am supposed to show that the expected height of a son whose father is shorter than avg. is also less than average, but by a smaller degree. First off I need help understanding the "regression effect" i.e. "Regression Towards the Mean." If the standard deviations are the same why is this happening? |
List of Common Misconceptions | Brilliant Math & Science Wiki
Lu Chee Ket, Pranshu Gaba, Rohit Gupta, and
Raph Pane
Zee Ell
Xiaoying Qin
This is part of an ongoing wiki collaboration. Contributions are greatly appreciated. You can
add a common misconception to this page;
create a wiki page explaining the misconception, copy and paste the basic structure;
read these wiki pages and provide feedback for improvement; in particular, if you are still not convinced, ask!
This page highlights common misconceptions that people have when understanding concepts. Try your best to answer these questions. If you find that you are stuck, click on the link and read on.
0.999 \ldots = 1?
\frac{1}{0}?
\frac{0}{0}?
ab = ac
, is it true that
b = c?
\frac{ \left(\frac{ a}{b}\right) } { c} = \frac{ a } { \left(\frac{b}{c}\right) }?
\frac{a}{b+c} = \frac{a}{b} + \frac{a}{c}?
in progress (Sravanth Chebrolu)
\frac{a}{c} + \frac{b}{d} = \frac{a+b}{c+d}?
in progress (Sharky)
\frac{a}{bc} = \frac{a}{b} \times \frac{a}{c}
? in progress
If ax + by = ap + bq, must we have x = p and y = q?
0^0?
How are exponent towers evaluated? Is
a^{b^c}
\left(a^b \right)^c?
a^n = b^n
a = b?
in progress (Sachin Vishwakarma)
a^2 + a^3 = a^5?
(ab)^2 = a^2b^2?
Do square roots multiply? Is
\sqrt{a} \times \sqrt{b} = \sqrt{ab}?
\sqrt{a^2+b^2} = a+b?
\sqrt{x^2} = \pm x?
When can we multiply inequalities without changing their directions? If
a > b
c > d
a c > b d?
Does cross multiply always work for inequalities? Does
\frac{ x+2}{x+4} > 1
x + 2 > x + 4?
a > b
a^2 > b^2?
ab > b ?
a > b,
a^n > b^n?
1. Can the graph of a function cross the horizontal asymptote?
\log (a+b) = \log a + \log b?
Answer: No. Logarithm is about doing multiplication in a method of addition but not a distributive law. The statement is not valid at all unless we are looking for a form of
-\infty
a = b = 0.
\log(a×b ) = \log a \times \log b?
\log (a \times b) = \log a + \log b
on another hand is true.
e^{\ln (a + b)} = a + b \neq e^{\ln a \times \ln b} = a^{\ln b}
b^{\ln a}.
Is i<0?
Is 1+i>-1+i?
Is 1+i rational?
A
B
are square matrices of the same order, is
(A+B)^2 = A^2 + 2AB + B^2?
Is -3 a multiple of 3?
Is 0 a prime number? in progress
Since 2 is even, it is not prime!
a \equiv b \pmod m
a \equiv b \pmod n,
a \equiv b \pmod {mn}?
a \equiv b \pmod m,
a = b + m?
na \equiv nb \pmod {m},
a \equiv b \pmod{m}?
a \equiv b \pmod{m},
na \equiv nb \pmod {nm}?
na \equiv nb \pmod{nm},
a\equiv b \pmod m?
\log 0
doesn't exist, does that mean
\int_0^{\frac{\pi}{2}} \log(\sin x)\, dx
doesn't converge?
Answer: No. In fact,
-\frac{\pi}{2} \log 2
is its value. The infinite form at an edge of an integral does not reveal its resultant effect as its area tends to zero and it is dominated by the whole trend of sum of all others within the range.
Do local extrema occur if and only if
f'(x) = 0?
Do global extrema occur if and only if
f'(x) = 0?
Answer: Finding the roots of
f'(x) = 0
will only give us the local extrema. They may or may not be the global extrema. The actual use of derivatives is to find the tangent of the curve. If the tangent has a slope equal to 0, it indicates that it is a maxima/minima. But there are many more points on the curve that have derivative equal to 0. Hence, we can conclude that
f'(x)=0
doesn't give the global maxima/minima always.
A differentiable function
f
is strictly increasing if and only if
f'(x) > 0
x^2 = 1
, then we can differentiate both sides to conclude that
2x = 0
. Reference explanation.
\frac{dy}{dx} = 2x
, can we say that
dy = 2x \, dx?
(
Can we multiply and divide by
dx?)
Answer: Effectively yes but traditionally no. For integration, it is
\int \frac{dy}{dx} dx = \int 2x\, dx
which makes the consequence. It is said that
dy
dx
when separated from each other or from
\int
cannot be meaningful as each of them would become value of indeterminate in extreme limit of an unknown destination. Only
\delta x
\delta y
can be written alone as expected proximity to some exact figures. However, to stay in an equation, the tradition to refuse could be improved.
Are all continuous functions differentiable?
Answer: No, one example is
y=|x|
, it is continuous, but is not differentiable at
x=0
Are all differentiable functions continuous?
True or False: Since
\int x^2 \, dx = \frac{1}{3} x^3 + C
\int \sin^2 x \, dx = \frac{1}{3} \sin^3 x + C
Answer: False.
d x
ought to be
d \sin x
instead, to be true, which is
d u
as introduced usually.
If the limit of a sequence is 0, does the series converge? (in progress)
F(x)
f(x)
\int_a^b f(x) \, dx = F(b) - F(a)?
\frac{\infty}{\infty}?
(Aareyan)
Why can we have
x = x + 1?
(Calvin)
Are there infinitely many numbers in the interval [0,1]? (Pranshu)
Infinity is the number at the end of the real number line. / Infinity is a real number at the end of the real number line. (Kaito)
All figures with finite area have finite perimeter. (Zandra)
1-1+1-1+1-\cdots = ?
(Grandi's Series) (Nihar)
\infty + 1 > \infty
(+ calvin's followup of 'closest to infinity') (Pranshu)
\infty - \infty?
\lim{x\rightarrow \infty} \frac{f(x)}{g(x)} = 1 , \lim_{x\rightarrow \infty} f(x) - g(x) = \infty
(Vighnesh)
\frac{1}{0}?
(in progress)
\infty + \infty > \infty?
(Evan)
\infty \times 0 = ?
0 ^{\infty}
an indeterminate form?
1 \text{ m}^2 = 100 \text{ cm}^2?
2
triangles have the same angles (AAA), does it mean they are congruent?
Answer: No. Both triangles can be of different scales. AAA, however, means that both triangles are similar.
2
triangles pass the similarity test ASS, does it mean they are congruent? Or does it mean they are similar?
Answer: No to both. Consider the following diagram:
\angle ABC
is acute, both triangles
ABC
ABD
pass the similarity test ASS. Hence, the triangles do not necessary or congruent.
What happens when we reflect along a line that is not vertical or horizontal?
Do angles in a spherical triangle sum to 180 degrees?
Can a right triangle with hypotenuse 10 have a height of 6?
Is the area of a trapezium equal to base times height?
Is the area of a rhombus equal to half the product of the diagonals?
For all triangles, do we have
a + b \geq c?
a^2 + b^2 \geq c^2?
- Relate to cosine rule
In a triangle, is it true that the angle opposite the longer leg must be larger than the angle opposite the smaller leg?
Surface area of a slanted cylinder in progress
Is every cyclic polygon a regular polygon?
Is every equiangular polygon a regular polygon?
Is every equilateral polygon a regular polygon?
After I flip a coin, it is either heads or tails. Thus the probability of heads is either 0 or 1.
The Gambler's Fallacy: Believing that randomness keeps some variables due.
When you toss a coin six times and get consecutive heads, tails aren't due. The next toss is as likely to provide heads as tails.
Suppose that we have events A and B in chronological order. The outcome of event B does not affect the outcome of event A. Does knowledge of the outcome of event B affect the probability of event A's outcome?
Everything that moves must eventually come to a stop. In order for continuous motion, there must always be a force acting on it.
If an object experiences no force, then its acceleration is 0.
Answer: The statement is true, According to Newton's law of motion Net force equals the product of mass and acceleration. Thus, if force is zero then acceleration is zero. The catch here can be, even if the forces are applied, still the acceleration can be zero.
If an object is moving, then a net force must be acting on it. In progress
Since every action has an equal and opposite reaction, nothing will be moved. In progress
If a body experiences a force in a direction, then it must be moving in that direction.
A car's gas mileage is independent of the number of passengers. In progress
Does force of friction between two objects depend on the area of contact?
Acceleration is the rate of change of speed.
An object performing uniform circular motion does not accelerate.
If two particles are thrown horizontally with different speeds, the one with a higher speed reaches the ground first.
When a ball is thrown upwards from the surface, at the highest point of its motion, it is at rest.
Average speed = (initial speed + final speed)/2
Law of conservation of kinetic energy
A rolling ball comes to a stop even when no external force acts on it. This is a violation of Newton's first law of motion.
A larger planet will have a larger gravitational pull.
Outer space is a complete vacuum.
Planetary motion is completely circular.
Does temperature depend upon the frame of reference?
A black body is black in color.
We can travel faster than the speed of light.
Relative speeds no longer approximate a vector sum when it approaches the speed of light.
Neutrinos travel faster than the speed of light.
Are photons mass-less?
If an object has positive charge, has it gained protons? in progress
Optically denser medium means it is denser by mass too.
A \Rightarrow B
\neg A \Rightarrow \neg B?
Let statement
A
be the statement "I am a boy" and statement
B
be the statement "I am a human."
A
B,
A
does not imply not
B
(i.e. "I am not a boy" does not imply "I am not a human"). Of course, I may not be human, but I could be, and thus
\text{not } A
\text{not } B.
Do the following two different codes in C++ do the same thing?
cout<<++i;
cout<<i++;
\color{#D61F06}{\textbf{No!}}
++
is a unary operator which increments the operand
i.
++i
i++
are valid statements but the difference is that
++i
increments
i
before the function is performed such as cout where as
i++
i
after performing the function (like cout). More clearly, the first code statement is equivalent to
whereas the second one is equivalent to
Bottomline: Do not increment and use the same variable in the same statement. This is considered a bad coding standard (except in competitive programming).
Cite as: List of Common Misconceptions. Brilliant.org. Retrieved from https://brilliant.org/wiki/common-misconceptions/ |
APX - Wikipedia
Complexity class of approximable problems
For other uses, see APX (disambiguation).
In computational complexity theory, the class APX (an abbreviation of "approximable") is the set of NP optimization problems that allow polynomial-time approximation algorithms with approximation ratio bounded by a constant (or constant-factor approximation algorithms for short). In simple terms, problems in this class have efficient algorithms that can find an answer within some fixed multiplicative factor of the optimal answer.
An approximation algorithm is called an
{\displaystyle f(n)}
-approximation algorithm for input size
{\displaystyle n}
if it can be proven that the solution that the algorithm finds is at most a multiplicative factor of
{\displaystyle f(n)}
times worse than the optimal solution. Here,
{\displaystyle f(n)}
is called the approximation ratio. Problems in APX are those with algorithms for which the approximation ratio
{\displaystyle f(n)}
{\displaystyle c}
. The approximation ratio is conventionally stated greater than 1. In the case of minimization problems,
{\displaystyle f(n)}
is the found solution's score divided by the optimum solution's score, while for maximization problems the reverse is the case. For maximization problems, where an inferior solution has a smaller score,
{\displaystyle f(n)}
is sometimes stated as less than 1; in such cases, the reciprocal of
{\displaystyle f(n)}
is the ratio of the score of the found solution to the score of the optimum solution.
A problem is said to have a polynomial-time approximation scheme (PTAS) if for every multiplicative factor of the optimum worse than 1 there is a polynomial-time algorithm to solve the problem to within that factor. Unless P = NP there exist problems that are in APX but without a PTAS, so the class of problems with a PTAS is strictly contained in APX. One such problem is the bin packing problem.
1 APX-hardness and APX-completeness
2.1 PTAS
2.2 APX-intermediate
2.3 f(n)-APX
APX-hardness and APX-completeness[edit]
A problem is said to be APX-hard if there is a PTAS reduction from every problem in APX to that problem, and to be APX-complete if the problem is APX-hard and also in APX. As a consequence of P ≠ NP ⇒ PTAS ≠ APX, if P ≠ NP is assumed, no APX-hard problem has a PTAS. In practice, reducing one problem to another to demonstrate APX-completeness is often done using other reduction schemes, such as L-reductions, which imply PTAS reductions.
One of the simplest APX-complete problems is MAX-3SAT-3, a variation of the boolean satisfiability problem. In this problem, we have a boolean formula in conjunctive normal form where each variable appears at most 3 times, and we wish to know the maximum number of clauses that can be simultaneously satisfied by a single assignment of true/false values to the variables.
Other APX-complete problems include:
Max independent set in bounded-degree graphs (here, the approximation ratio depends on the maximum degree of the graph, but is constant if the max degree is fixed).
Min vertex cover. The complement of any maximal independent set must be a vertex cover.
Min dominating set in bounded-degree graphs.
The travelling salesman problem when the distances in the graph satisfy the conditions of a metric. TSP is NPO-complete in the general case.
The token reconfiguration problem, via L-reduction from set cover.
PTAS[edit]
Main article: Polynomial-time approximation scheme
PTAS (polynomial time approximation scheme) consists of problems that can be approximated to within any constant factor besides 1 in time that is polynomial to the input size, but the polynomial depends on such factor. This class is a subset of APX.
APX-intermediate[edit]
Unless P = NP, there exist problems in APX that are neither in PTAS nor APX-complete. Such problems can be thought of as having a hardness between PTAS problems and APX-complete problems, and may be called APX-intermediate. The bin packing problem is thought to be APX-intermediate. Despite not having a known PTAS, the bin packing problem has several "asymptotic PTAS" algorithms, which behave like a PTAS when the optimum solution is large, so intuitively it may be easier than problems that are APX-hard.
One other example of a potentially APX-intermediate problem is min edge coloring.
f(n)-APX[edit]
One can also define a family of complexity classes
{\displaystyle f(n)}
-APX, where
{\displaystyle f(n)}
-APX contains problems with a polynomial time approximation algorithm with a
{\displaystyle O(f(n))}
approximation ratio. One can analogously define
{\displaystyle f(n)}
-APX-complete classes; some such classes contain well-known optimization problems. Log-APX-completeness and poly-APX-completeness are defined in terms of AP-reductions rather than PTAS-reductions; this is because PTAS-reductions are not strong enough to preserve membership in Log-APX and Poly-APX, even though they suffice for APX.
Log-APX-complete, consisting of the hardest problems that can be approximated efficiently to within a factor logarithmic in the input size, includes min dominating set when degree is unbounded.
Poly-APX-complete, consisting of the hardest problems that can be approximated efficiently to within a factor polynomial in the input size, includes max independent set in the general case.
There also exist problems that are exp-APX-complete, where the approximation ratio is exponential in the input size. This may occur when the approximation is dependent on the value of numbers within the problem instance; these numbers may be expressed in space logarithmic in their value, hence the exponential factor.
Max/min CSP/Ones classification theorems - a set of theorems that enable mechanical classification of problems about boolean relations into approximability complexity classes
MaxSNP - a closely related subclass
Complexity Zoo: APX
C. Papadimitriou and M. Yannakakis. Optimization, approximation and complexity classes. Journal of Computer and System Sciences, 43:425–440, 1991.
Pierluigi Crescenzi, Viggo Kann, Magnús Halldórsson, Marek Karpinski and Gerhard Woeginger. Maximum Satisfiability Archived 2007-04-13 at the Wayback Machine. A compendium of NP optimization problems Archived 2007-04-05 at the Wayback Machine.
Retrieved from "https://en.wikipedia.org/w/index.php?title=APX&oldid=1040375348" |
How to solve \ln(x+1)^{2}=2?
Irvin Dukes 2021-12-10 Answered
{\mathrm{ln}\left(x+1\right)}^{2}=2
Natalie Yamamoto
{\mathrm{ln}\left(x+1\right)}^{2}=2
{\left(x+1\right)}^{2}={e}^{2}
x+1=e
x=e-1
{\mathrm{ln}\left(x+1\right)}^{2}=2
\mathrm{ln}\left(x+1\right)=1
x+1={e}^{1}=e
x=e-1
2\mathrm{log}\left(a\right)\mathrm{log}\left(b\right)=\mathrm{log}\left(ab{\right)}^{2}-\mathrm{log}\left(a{\right)}^{2}-\mathrm{log}\left(b\right)2
for a>1and b>1?
I try to solve the following equation:
{\left(N+1\right)}^{{\mathrm{log}}_{N}125}=216
I know the answer is 5 here but how could I rewrite the equations so I can solve it?
I tried to take the log of both sides but that didn't help me because I got stuck. Could anyone please explain me how to do this?
{\mathrm{log}}_{3}\left(x\sqrt{y}\right)
{e}^{-2x+1}=13
Proof of a closed form of
{\int }_{0}^{1}{\left(-\mathrm{ln}x\right)}^{n}dx
Find intersection of linear and logarithmic lines
I have equations for two lines, one of which is linear and the other is logarithmic, ie:
y={m}_{1}x+{c}_{1}
y={m}_{2}\cdot \mathrm{ln}\left(x\right)+{c}_{2}
..and I need to find out where (if at all) these lines intersect. I realise that I need to solve:
{m}_{1}\cdot x+{c}_{1}={m}_{2}\cdot \mathrm{ln}\left(x\right)+{c}_{2}
..for x, but apart from shuffling the constants around I'm not sure how to do this
Is there a general solution to this problem?
How does this simplify to
\mathrm{log}\sqrt{x}
\mathrm{log}x-\mathrm{log}\sqrt{3}x-\mathrm{log}\sqrt{6}x
\mathrm{log}\sqrt{x}
I've tried to get Bagatrix Algebra Solved! to solve it, but it even got the wrong answer... (I checked it by replacing x with 5 and typing it out on a calculator..)
No matter what I do, I end up with an answer that is correct and a bit simplified, but not as simplified as
\mathrm{log}\sqrt{x} |
Root Functions - Maple Help
Home : Support : Online Help : Science and Engineering : Units : Environments : Natural : Root Functions
the square root function in the Natural Units environment
the nth root function in the Natural Units environment
the non-principal root function in the Natural Units environment
\sqrt{x}
{\mathrm{root}}_{n}\left(x\right)
The sqrt(x) function takes the square root the unit-free portion of x and multiplies it by the unit raised to the power 1/2.
The root(x, n) and root[n](x) functions take the nth root of the unit-free portion of x and multiply it by the unit raised to the power 1/n.
The surd(x, n) function takes the nth root of the unit-free portion of x, whose (complex) argument is closest to the unit-free portion of x, and multiplies it by the unit raised to the power 1/n.
For other properties, see the global functions sqrt, root, and surd.
\mathrm{with}\left(\mathrm{Units}[\mathrm{Natural}]\right):
\mathrm{sqrt}\left(3.532{m}^{2}\right)
\textcolor[rgb]{0,0,1}{1.879361594}\textcolor[rgb]{0,0,1}{}⟦\textcolor[rgb]{0,0,1}{m}⟧
\mathrm{root}\left(3.532{m}^{3},3\right)
\textcolor[rgb]{0,0,1}{1.522907638}\textcolor[rgb]{0,0,1}{}⟦\textcolor[rgb]{0,0,1}{m}⟧
\mathrm{root}[3]\left(-3.532{m}^{3}\right)
\left(\textcolor[rgb]{0,0,1}{0.7614538193}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1.318876702}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{I}\right)\textcolor[rgb]{0,0,1}{}⟦\textcolor[rgb]{0,0,1}{m}⟧
\mathrm{surd}\left(-3.532{m}^{3},3\right)
\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1.522907638}\textcolor[rgb]{0,0,1}{}⟦\textcolor[rgb]{0,0,1}{m}⟧
\mathrm{surd}\left(16.532{m}^{4},4\right)
\textcolor[rgb]{0,0,1}{2.016421638}\textcolor[rgb]{0,0,1}{}⟦\textcolor[rgb]{0,0,1}{m}⟧ |
Spacetime symmetries - formulasearchengine
Revision as of 11:23, 1 October 2014 by en>Michael H 34 (Grammar)
Spacetime symmetries are features of spacetime that can be described as exhibiting some form of symmetry. The role of symmetry in physics is important in simplifying solutions to many problems, spacetime symmetries are used in the study of exact solutions of Einstein's field equations of general relativity.
1 Physical motivation
3 Killing symmetry
4 Homothetic symmetry
5 Affine symmetry
6 Conformal symmetry
7 Curvature symmetry
8 Matter symmetry
9 Local and global symmetries
10.1 Spacetime classifications
Physical problems are often investigated and solved by noticing features which have some form of symmetry. For example, in the Schwarzschild solution, the role of spherical symmetry is important in deriving the Schwarzschild solution and deducing the physical consequences of this symmetry (such as the non-existence of gravitational radiation in a spherically pulsating star). In cosmological problems, symmetry finds a role to play in the cosmological principle which restricts the type of universes that are consistent with large-scale observations (e.g. the Friedmann-Lemaître-Robertson-Walker (FLRW) metric). Symmetries usually require some form of preserving property, the most important of which in general relativity include the following:
preserving geodesics of the spacetime
preserving the metric tensor
preserving the curvature tensor
These and other symmetries will be discussed in more detail later. This preservation feature can be used to motivate a useful definition of symmetries.
A rigorous definition of symmetries in general relativity has been given by Hall (2004). In this approach, the idea is to use (smooth) vector fields whose local flow diffeomorphisms preserve some property of the spacetime. This preserving property of the diffeomorphisms is made precise as follows. A smooth vector field X on a spacetime M is said to preserve a smooth tensor T on M (or T is invariant under X) if, for each smooth local flow diffeomorphism ϕt associated with X, the tensors T and ϕt*(T) are equal on the domain of ϕt. This statement is equivalent to the more usable condition that the Lie derivative of the tensor under the vector field vanishes:
{\displaystyle {\mathcal {L}}_{X}T=0}
on M. This has the consequence that, given any two points p and q on M, the coordinates of T in a coordinate system around p are equal to the coordinates of T in a coordinate system around q. A symmetry on the spacetime is a smooth vector field whose local flow diffeomorphisms preserve some (usually geometrical) feature of the spacetime. The (geometrical) feature may refer to specific tensors (such as the metric, or the energy-momentum tensor) or to other aspects of the spacetime such as its geodesic structure. The vector fields are sometimes referred to as collineations, symmetry vector fields or just symmetries. The set of all symmetry vector fields on M forms a Lie algebra under the Lie bracket operation as can be seen from the identity:
{\displaystyle {\mathcal {L}}_{[X,Y]}T={\mathcal {L}}_{X}({\mathcal {L}}_{Y}T)-{\mathcal {L}}_{Y}({\mathcal {L}}_{X}T)}
the term on the right usually being written, with an abuse of notation, as
{\displaystyle [{\mathcal {L}}_{X},{\mathcal {L}}_{Y}]T}
Killing symmetry
{{#invoke:main|main}} A Killing vector field is one of the most important types of symmetries and is defined to be a smooth vector field that preserves the metric tensor:
{\displaystyle {\mathcal {L}}_{X}g_{ab}=0}
This is usually written in the expanded form as:
{\displaystyle X_{a;b}+X_{b;a}\,=0}
Killing vector fields find extensive applications (including in classical mechanics) and are related to conservation laws.
Homothetic symmetry
{{#invoke:main|main}} A homothetic vector field is one which satisfies:
{\displaystyle {\mathcal {L}}_{X}g_{ab}=2cg_{ab}}
where c is a real constant. Homothetic vector fields find application in the study of singularities in general relativity.
Affine symmetry
{{#invoke:main|main}} An affine vector field is one that satisfies:
{\displaystyle ({\mathcal {L}}_{X}g_{ab})_{;c}=0}
An affine vector field preserves geodesics and preserves the affine parameter.
The above three vector field types are special cases of projective vector fields which preserve geodesics without necessarily preserving the affine parameter.
{{#invoke:main|main}} A conformal vector field is one which satisfies:
{\displaystyle {\mathcal {L}}_{X}g_{ab}=\phi g_{ab}}
where ϕ is a smooth real-valued function on
{\displaystyle M}
Curvature symmetry
{{#invoke:main|main}} A curvature collineation is a vector field which preserves the Riemann tensor:
{\displaystyle {\mathcal {L}}_{X}R^{a}{}_{bcd}=0}
where Rabcd are the components of the Riemann tensor. The set of all smooth curvature collineations forms a Lie algebra under the Lie bracket operation (if the smoothness condition is dropped, the set of all curvature collineations need not form a Lie algebra). The Lie algebra is denoted by CC(M) and may be infinite-dimensional. Every affine vector field is a curvature collineation.
Matter symmetry
{{#invoke:main|main}} A less well-known form of symmetry concerns vector fields that preserve the energy-momentum tensor. These are variously referred to as matter collineations or matter symmetries and are defined by:
{\displaystyle {\mathcal {L}}_{X}T_{ab}=0}
where Tab are the energy-momentum tensor components. The intimate relation between geometry and physics may be highlighted here, as the vector field X is regarded as preserving certain physical quantities along the flow lines of X, this being true for any two observers. In connection with this, it may be shown that every Killing vector field is a matter collineation (by the Einstein field equations, with or without cosmological constant). Thus, given a solution of the EFE, a vector field that preserves the metric necessarily preserves the corresponding energy-momentum tensor. When the energy-momentum tensor represents a perfect fluid, every Killing vector field preserves the energy density, pressure and the fluid flow vector field. When the energy-momentum tensor represents an electromagnetic field, a Killing vector field does not necessarily preserve the electric and magnetic fields.
As mentioned at the start of this article, the main application of these symmetries occur in general relativity, where solutions of Einstein's equations may be classified by imposing some certain symmetries on the spacetime.
Spacetime classifications
Classifying solutions of the EFE constitutes a large part of general relativity research. Various approaches to classifying spacetimes, including using the Segre classification of the energy-momentum tensor or the Petrov classification of the Weyl tensor have been studied extensively by many researchers, most notably Stephani et al. (2003). They also classify spacetimes using symmetry vector fields (especially Killing and homothetic symmetries). For example, Killing vector fields may be used to classify spacetimes, as there is a limit to the number of global, smooth Killing vector fields that a spacetime may possess (the maximum being 10 for 4-dimensional spacetimes). Generally speaking, the higher the dimension of the algebra of symmetry vector fields on a spacetime, the more symmetry the spacetime admits. For example, the Schwarzschild solution has a Killing algebra of dimension 4 (3 spatial rotational vector fields and a time translation), whereas the Friedmann-Lemaître-Robertson-Walker (FLRW) metric (excluding the Einstein static subcase) has a Killing algebra of dimension 6 (3 translations and 3 rotations). The Einstein static metric has a Killing algebra of dimension 7 (the previous 6 plus a time translation).
The assumption of a spacetime admitting a certain symmetry vector field can place restrictions on the spacetime.
|CitationClass=book }} See Section 10.1 for a definition of symmetries.
|CitationClass=book }} See Chapter 3 for properties of the Lie derivative and Section 3.10 for a definition of invariance.
Retrieved from "https://en.formulasearchengine.com/index.php?title=Spacetime_symmetries&oldid=240970" |
Mr. Anderson’s doctor has advised him to go on a diet. He must reduce his caloric intake by
15\%
. He currently eats
2800
Calories per day. Calculate his new daily caloric intake rate in two different ways, using two different multipliers.
What is the ''discounted'' number of Calories (the number that is subtracted from the original)? What is the ''sale price'' (the new daily caloric intake rate)?
2380 |
How do you solve \sin x\cos x =\frac{1}{2} for
How do you solve \sin x\cos x =\frac{1}{2} for x in the interval [0,2
\mathrm{sin}x\mathrm{cos}x=\frac{1}{2}
\left[0,2\pi \right)
dieseisB
\mathrm{sin}2x=2\mathrm{sin}x\mathrm{cos}x
\mathrm{sin}x\mathrm{cos}x=\left(\frac{1}{2}\right)\mathrm{sin}2x=\frac{1}{2}⇒\mathrm{sin}2x=1
\mathrm{sin}2x=1⇒\mathrm{arcsin}2x=\frac{\pi }{4}⇒x=\frac{\pi }{4}
\mathrm{sin}\frac{\pi }{4}\mathrm{cos}\frac{\pi }{4}=\left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{2}}{2}\right)=\frac{2}{4}=\frac{1}{2}
\mathrm{sin}t=,\mathrm{cos}t=,\text{ }\text{and}\text{ }\mathrm{tan}t=
{e}^{\mathrm{sin}x}+{e}^{\mathrm{cos}x}=e+1
\sum _{n=1}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{n+1}\mathrm{sin}\left(n\right)}{n}=\frac{1}{2}
Limit of
\mathrm{sin}\left(\sqrt{x+2}\right)-\mathrm{sin}\left(\sqrt{x+4}\right)\text{ }\text{ as }\text{ }x\to \mathrm{\infty }
\int \frac{\mathrm{cos}\theta d\theta }{\sqrt{2-9{\mathrm{sin}}^{2}\theta }}
\alpha +\beta =\frac{\pi }{2}
\beta +\varphi =\alpha
\mathrm{tan}\alpha
equals.
How do I find out the value of p for
\mathrm{sin}p+\mathrm{cos}p=0 |
Find an answer to any physics question
Get help with physics questions
{I}_{\nu }=\frac{8\pi {\nu }^{2}}{{c}^{3}}\frac{h\nu }{{e}^{h\nu /{k}_{b}T}-1}
\frac{d{I}_{\nu }}{d\nu }=0:\text{ }0=\frac{\mathrm{\partial }}{\mathrm{\partial }\nu }\left(\frac{{\nu }^{3}}{{e}^{h\nu /{k}_{b}T}-1}\right)=\frac{3{\nu }^{2}\left({e}^{h\nu /{k}_{b}T}-1\right)-{\nu }^{3}h/{k}_{b}T\cdot {e}^{h\nu /{k}_{b}T}}{\left({e}^{h\nu /{k}_{b}T}-1{\right)}^{2}}
0
\nu >0
3\left({e}^{h\nu /{k}_{b}T}-1\right)-h\nu /{k}_{b}T\cdot {e}^{h\nu /{k}_{b}T}=0
\gamma =h\nu /{k}_{b}T
3\left({e}^{\gamma }-1\right)-\gamma {e}^{\gamma }=0\to \gamma =2.824
\lambda =c/\nu :\text{ }\lambda =\frac{hc}{\gamma {k}_{b}}\frac{1}{T}
\lambda T=b
hc/\gamma {k}_{b}
b
\frac{hc}{\gamma {k}_{b}}=0.005099
b=0.002897
\frac{dI}{d\lambda }=\frac{dI}{d\nu }\frac{d\nu }{d\lambda }=\frac{c}{{\nu }^{2}}\frac{dI}{d\nu }
c/{\nu }^{2}
d{I}_{\lambda }/d\lambda
{\lambda }_{\text{max}}T=\text{a constant}
{\lambda }_{\text{max}}
{\lambda }_{\text{max}}
\phantom{\rule{mediummathspace}{0ex}}\mathrm{W}/{\mathrm{m}}^{2}\mathrm{s}\mathrm{r}
R=1.097×{10}^{7}\text{ }{m}^{-1}
X\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{V}
X\prime \phantom{\rule{thinmathspace}{0ex}}\mathrm{V}
\left(X+X\prime \right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{V}
A
B
A
B
C
C
3
A
B
C
How many kilograms of waters are needed to obtain the 198.8 mol of deuterium, assuming that deiterium is 0.01500% of natural hydrogen?
E
\lambda =\frac{h}{\sqrt{2mE}}
\lambda
= wavelength
m |
Natural Resources and Intangible Assets - Course Hero
Principles of Accounting/Long-Term Assets/Natural Resources and Intangible Assets
Learn all about natural resources and intangible assets in just a few minutes! Fabio Ambrosio, CPA, instructor of accounting at the Central Washington University, walks through accounting for depletion of natural resources and four common types of intangible assets: patents, copyrights, trademarks, and goodwill.
Natural resources are defined as assets that are formed naturally over time. Some examples of natural resources include timber, minerals, and oil. Like other long-term assets, natural resources are reported on the balance sheet at book value, and all costs associated with getting the asset ready for its intended purpose or use are included in the cost of the asset.
Depletion is the process of transferring the cost of a natural resource to an expense as it is consumed. As natural resources are consumed, they are considered to be depleted. Depletion, like depreciation, allows for the transfer of the cost of the asset to an expense account over time. In order to determine the rate of depletion, the cost of the resource is divided by the estimated total units of the resource available. Then, to determine the current period depletion expense, the depletion rate is multiplied by the quantity extracted. This approach is similar to the units-of-production method. To record the depletion expense for the period, debit the Depletion Expense account and credit Accumulated Depletion, a contra asset account.
To illustrate, Cassie Enterprises purchased mining rights at $600,000.
Cassie Enterprises Mining Rights
Cost of mineral deposit $600,000
Estimated total units of resource 1,500,000 tons
Tons mined during the year 80,000 tons
The annual depletion expense of $32,000 is computed using a formula.
\begin{aligned}\text{Depletion Rate}&=\frac{\text{Cost of Resource}}{\text{Estimated Total Units of Resource}}\\\\&=\frac{\$600\text{,}000}{1\text{,}500\text{,}000\;\text{Tons}}\\\\&=\$0.40\;\text{per Ton}\end{aligned}
\begin{aligned}\text{Current Period Depletion Expense}&=\$0.40\;\text{per Ton}\times80\rm{,}000\;\text{Tons}\\&=\$32\rm{,}000\end{aligned}
Cassie Enterprises then makes an entry to record depletion expense as a debit of $32,000 for the current period.
Depletion Expense Journal Entry
Dec 31 Depletion Expense $32,000
Accumulated Depletion $32,000
To record depletion of mineral deposit
An intangible asset is an asset that is nonphysical and provides business entities with long-term rights, privileges, or competitive advantages. Characterized by the lack of physical substance, intangible assets are recorded at cost. The most common intangible assets are patents, copyrights, trademarks, and goodwill. Intangible assets can be categorized by their useful life: limited or unlimited. Intangible assets with limited life should be amortized over its useful life. Intangible assets with an indefinite life, such as goodwill, should not be amortized. Very similar to depreciation, amortization is the process of spreading or allocating a cost or payment over a period of time. Amortizing an intangible asset spreads the asset costs over the life of the asset. Unlike depreciation, the straight-line method is, with few exceptions, the only method of amortization used for intangibles. Amortization can be computed by dividing the intangible assets' costs by the estimated useful life.
Defined as an exclusive right granted to use a process or to produce or sell an item, a patent has a maximum useful life of 20 years. To properly account for patents, the patent account should be debited at acquisition for the cost to acquire the rights. Because of the fact that a patent has a limited life, it should be amortized over its estimated useful life not to exceed 20 years. The amortization entry is a debit to Amortization Expense-Patents and a credit to Accumulated Amortization-Patents.
Hollywood Movies purchases a patent for $35,000. To record the patent, Hollywood Movies will make a journal entry.
Patent Purchase Journal Entry
June 1 Patents $35,000
To record patent purchase
Hollywood Movies estimates the useful life of the patent to be 20 years. Each year, it will record amortization expense for its patents of $1,750, calculated as
(\$35\rm{,}000/20\;\text{Years})=\$1\rm{,}750\;\text{Annual Amortization Amount}
Patent Amortization Journal Entry
June 30 Amortization Expense Patents $1,750
Accumulated Amortization Patents $1,750
To record patent amortization
A copyright provides exclusive publishing rights for performing arts, literary works, visual arts, digital content, photographs, and motion pictures. A copyright has a useful life that extends 70 years beyond the death of said author of the musical, literary, or artistic work. The United States Copyright Office provides copyrights. Copyrights are amortized over the life of the intangible asset. To determine the amortization for a copyright, divide the cost of the copyright by the estimated useful life.
Hollywood Movies purchases a copyright for $140,000. It makes a journal entry to record the purchase.
Copyright Purchase Journal Entry
June 1 Copyrights $140,000
To record copyright purchase
Hollywood Movies estimates the copyright to have a useful life of 25 years. Each year it will record amortization expense for its copyright of $7,000, calculated as
(\$140\rm{,}000/25\;\text{Years})=\$5\rm{,}600\;\text{Annual Amortization Amount}
Copyright Amortization Journal Entry
June 30 Amortization Expense Copyrights $5,600
Accumulated Amortization Copyrights $5,600
To record copyright amortization
A trademark is a unique symbol, name, phrase, or jingle used to identify a business entity, product, or service. Some examples of trademarks include the "golden arches" representing McDonald's, "swoosh" for Nike, and "the siren" for Starbucks. Trademark ownership is established by registering with the United States Patent Office. Trademark development and enhancement costs are charged to an expense account when incurred. However, if a trademark is purchased, the cost of the asset is debited to trademarks and should be amortized over the life of the intangible asset. Trademarks can be renewed indefinitely. The cost associated with the renewal is not amortized.
Tom's Burgers purchases a trademark for $150,000 for cash. It makes a journal entry to record the purchase.
Trademark Purchase Journal Entry
June 1 Trademarks $150,000
To record trademark purchase
Tom's Burgers estimates the trademark to have a useful life of 30 years. Each year, it will record amortization expense for its trademark of $5,000, calculated as
(\$150\rm{,}000/30\;\text{Years})=\$5\rm{,}000\;\text{Annual Amortization Amount}
June 30 Amortization Expense Trademarks $5,000
Accumulated Amortization Trademarks $5,000
To record trademark amortization
Goodwill is the premium paid over the market value of net assets (market value of identifiable assets minus market value of identifiable liabilities) in a completed acquisition. The implication behind goodwill is that the acquired business entity has a superior-quality product or service, reputation, or other attribute that demands paying a premium to acquire it. Unlike other intangible assets, goodwill is not recorded unless the entire business is acquired. To determine the amount of goodwill in the event that the entire entity is acquired, one would take the purchase price minus the market value of the net assets. The net assets are the total assets less goodwill minus the total liabilities. Goodwill is then recorded as an asset. Goodwill is not subject to amortization. However, it must be tested each year to determine if it has been impaired.
<Disposal of Fixed Assets>Financial Reporting of Long-Term Assets |
Convert quaternion to Euler-Rodrigues vector - Simulink - MathWorks Australia
Quaternions to Rodrigues
Convert quaternion to Euler-Rodrigues vector
The Quaternions to Rodrigues block converts the 4-by-1 quaternion to the three-element Euler-Rodrigues vector. Aerospace Blockset™ uses quaternions that are defined using the scalar-first convention. This block normalizes all quaternion inputs. The rotation used in this function is a passive transformation between two coordinate systems. For more information on Euler-Rodrigues vectors, see Algorithms.
Quaternion — Quaternion
Quaternion from which to determine Euler-Rodrigues vector. Quaternion scalar is the first element.
Euler-Rodrigues vector determined from the quaternion.
\stackrel{⇀}{b}
\stackrel{\to }{b}=\left[\begin{array}{ccc}{b}_{x}& {b}_{y}& {b}_{z}\end{array}\right]
\begin{array}{l}{b}_{x}=\mathrm{tan}\left(\frac{1}{2}\theta \right){s}_{x},\\ {b}_{y}=\mathrm{tan}\left(\frac{1}{2}\theta \right){s}_{y},\\ {b}_{z}=\mathrm{tan}\left(\frac{1}{2}\theta \right){s}_{z}\end{array}
\stackrel{⇀}{s}
Direction Cosine Matrix to Rodrigues | Rodrigues to Direction Cosine Matrix | Rodrigues to Quaternions | Rodrigues to Rotation Angles | Rotation Angles to Rodrigues |
Using factorization homology, we realize the rational homology of the unordered configuration spaces of an arbitrary manifold
M
, possibly with boundary, as the homology of a Lie algebra constructed from the compactly supported cohomology of
M
. By locating the homology of each configuration space within the Chevalley–Eilenberg complex of this Lie algebra, we extend theorems of Bödigheimer, Cohen and Taylor and of Félix and Thomas, and give a new, combinatorial proof of the homological stability results of Church and Randal-Williams. Our method lends itself to explicit calculations, examples of which we include.
configuration spaces, factorization homology, Lie algebras, homological stability
http://scholar.harvard.edu/knudsen/ |
Compare accuracies of two classification models using new data - MATLAB compareHoldout - MathWorks Australia
{\stackrel{^}{\pi }}_{ijk}
{\stackrel{^}{\pi }}_{ijk}
\sum _{i,j,k}{n}_{ijk}={n}_{test}.
\sum _{i,j,k}{\pi }_{ijk}=\sum _{i,j,k}{\stackrel{^}{\pi }}_{ijk}=1.
\delta =\sum _{i=1}^{K}\sum _{j=1}^{K}\sum _{k=1}^{K}\left({c}_{ki}-{c}_{kj}\right){\pi }_{ijk}.
\begin{array}{c}{H}_{0}:\delta =0\\ {H}_{1}:\delta \ne 0\end{array}.
{t}_{{\chi }^{2}}^{\ast }=\sum _{i\ne j}\sum _{k}\frac{{\left({n}_{ijk}+1-\left({n}_{test}+{K}^{3}\right){\stackrel{^}{\pi }}_{ijk}^{\left(1\right)}\right)}^{2}}{{n}_{ijk}+1}.
1-{F}_{{\chi }^{2}}\left({t}_{{\chi }^{2}}^{\ast };1\right)<\alpha
{\stackrel{^}{\pi }}_{ijk}^{\left(1\right)}
{t}_{{\chi }^{2}}^{\ast }
{F}_{{\chi }^{2}}\left(x;1\right)
{t}_{LRT}^{\ast }=2\mathrm{log}\left[\frac{P\left(\underset{i,j,k}{\cap }{N}_{ijk}={n}_{ijk};{n}_{test},{\stackrel{^}{\pi }}_{ijk}={\stackrel{^}{\pi }}_{ijk}^{\left(2\right)}\right)}{P\left(\underset{i,j,k}{\cap }{N}_{ijk}={n}_{ijk};{n}_{test},{\stackrel{^}{\pi }}_{ijk}={\stackrel{^}{\pi }}_{ijk}^{\left(3\right)}\right)}\right].
1-{F}_{{\chi }^{2}}\left({t}_{LRT}^{\ast };1\right)<\alpha ,
{\stackrel{^}{\pi }}_{ijk}^{\left(2\right)}=\frac{{n}_{ijk}}{{n}_{test}}
{\stackrel{^}{\pi }}_{ijk}^{\left(3\right)}=\frac{{n}_{ijk}}{{n}_{test}+\lambda \left({c}_{ki}-{c}_{kj}\right)}
\sum _{i,j,k}\frac{{n}_{ijk}\left({c}_{ki}-{c}_{kj}\right)}{{n}_{test}+\lambda \left({c}_{ki}-{c}_{kj}\right)}=0.
{F}_{{\chi }^{2}}\left(x;1\right)
{\stackrel{^}{\pi }}_{2•}={n}_{2•}/n
{\stackrel{^}{\pi }}_{•2}={n}_{•2}/n
\begin{array}{c}{H}_{0}:{\pi }_{•2}={\pi }_{2•}\\ {H}_{1}:{\pi }_{•2}\ne {\pi }_{2•}\end{array}.
{H}_{0}:{\pi }_{12}={\pi }_{21}.
{t}_{a1}^{\ast }=\frac{{n}_{12}-{n}_{21}}{\sqrt{{n}_{12}+{n}_{21}}}.
1-\Phi \left(|{t}_{1}^{\ast }|\right)<\alpha ,
{t}_{a2}^{\ast }=\frac{{\left({n}_{12}-{n}_{21}\right)}^{2}}{{n}_{12}+{n}_{21}}.
1-{F}_{{\chi }^{2}}\left({t}_{2}^{\ast };m\right)<\alpha
{F}_{{\chi }^{2}}\left(x;m\right)
{n}_{d}={n}_{12}+{n}_{21}
{t}_{1}^{\ast }={n}_{12}.
{F}_{\text{Bin}}\left({t}_{1}^{\ast };{n}_{d},0.5\right)<\alpha
{F}_{\text{Bin}}\left(x;n,p\right)
{t}_{2}^{\ast }=\mathrm{min}\left({n}_{12},{n}_{21}\right).
{F}_{\text{Bin}}\left({t}_{2}^{\ast };{n}_{d},0.5\right)<\alpha /2
{t}_{1}^{\ast }={n}_{12}.
{F}_{\text{Bin}}\left({t}_{1}^{\ast }-1;{n}_{12}+{n}_{21},0.5\right)+0.5{f}_{\text{Bin}}\left({t}_{1}^{\ast };{n}_{12}+{n}_{21},0.5\right)<\alpha
{F}_{\text{Bin}}\left(x;n,p\right)
{f}_{\text{Bin}}\left(x;n,p\right)
{t}_{2}^{\ast }=\mathrm{min}\left({n}_{12},{n}_{21}\right).
{F}_{\text{Bin}}\left({t}_{2}^{\ast }-1;{n}_{12}+{n}_{21}-1,0.5\right)+0.5{f}_{\text{Bin}}\left({t}_{2}^{\ast };{n}_{12}+{n}_{21},0.5\right)<\alpha /2
{\stackrel{^}{\pi }}_{ijk}=\frac{{n}_{ijk}}{{n}_{test}}.
\sum _{i,j,k}{n}_{ijk}={n}_{test}.
{e}_{1}=\sum _{j=1}^{K}\sum _{k=1}^{K}\sum _{i\ne k}^{}{\stackrel{^}{\pi }}_{ijk}.
{e}_{1}=\sum _{j=1}^{K}\sum _{k=1}^{K}\sum _{i\ne k}^{}{\stackrel{^}{\pi }}_{ijk}{c}_{ki}, |
Dynamics of tripod drive with elastic self-sustaining transmission | JVE Journals
I. A. Nesmiyanov1 , V. V. Zhoga2 , N. S. Vorobieva3 , V. V. Dyashkin-Titov4
1, 3, 4Volgograd State Agrarian University, Volgograd, Russia
2Volgograd State Technical University, Volgograd, Russia
The article describes a mathematical model of dynamics of electromechanical drive with elastic transmission gear. A motor is connected to the tripod actuating links by self-sustaining transmission. The analytical conditions of drive dynamic seizure absence have been obtained.
Keywords: tripod, electric drive, worm reducer, self-sustaining, elasticity, dynamic seizure.
The tripod shown in Fig. 1 is designed to carry out handling operations. The tripod mechanism includes three controlled actuating links 1, 2, 3 ends of which are fixed with the help of special two-movable hinges on a swiveling base 5. Opposite ends of the links are connected with a special hinge joint 6 allowing geometric axis of these links to cross in one point that prevents bending moments from external loads from appearing in them. The handled item is attached to a rigid suspension which is fastened to a joint 6 [1] via cylindrical hinge.
Tripod design is reinforced that ensures high accuracy of program motions [2]. However, deformations in drive mechanisms can have a significant impact on parameters of dynamic accuracy and dynamic loads values.
Fig. 1. Tripod on a swiveling base
Fig. 2. Design model of a tripod on a swiveling base
1.1. Tripod dynamics
To simplify the differential equations describing the tripod dynamics the real mechanism is replaced with dynamically equivalent one which contains two lumped masses:
m
in a grip attachment point (point
M
{m}_{A}
in point
A
of a swiveling base (Fig. 2) [3].
In this case a differential equations system describing tripod motion established through Lagrange`s equations of the second kind [4] is written as follows:
m\stackrel{¨}{x}={F}_{1}\frac{x}{{l}_{1}}+{F}_{2}\frac{x-OB}{{l}_{2}}+{F}_{3}\frac{x+OB}{{l}_{3}},
m\stackrel{¨}{y}={F}_{1}\frac{y+OA\cdot \mathrm{s}\mathrm{i}\mathrm{n}\phi }{{l}_{1}}+{F}_{2}\frac{y}{{l}_{2}}+{F}_{3}\frac{y}{{l}_{3}},
m\stackrel{¨}{z}={F}_{1}\frac{z-OA\cdot \mathrm{c}\mathrm{o}\mathrm{s}\phi }{{l}_{1}}+{F}_{2}\frac{z}{{l}_{2}}+{F}_{3}\frac{z}{{l}_{3}}-mg,
\stackrel{¨}{\varphi }\cdot {m}_{A}O{A}^{2}={F}_{1}\frac{OA\left(y\cdot \mathrm{c}\mathrm{o}\mathrm{s}\varphi +z\cdot \mathrm{s}\mathrm{i}\mathrm{n}\varphi \right)}{{l}_{1}}+{F}_{4}\frac{OA\left(DK\cdot \mathrm{s}\mathrm{i}\mathrm{n}\varphi -OK\cdot \mathrm{c}\mathrm{o}\mathrm{s}\varphi \right)}{{l}_{4}}+{m}_{A}gOA\mathrm{s}\mathrm{i}\mathrm{n}\varphi .
where Cartesian coordinates of point
M
(Fig. 2)
x\left(t\right)
y\left(t\right)
z\left(t\right)
in absolute coordinates system
Oxyz
(Fig. 2) and base rotation angle φ were taken as generalized coordinates of the tripod;
{F}_{k}
k=
1,…,4 – drive control force in tripod links.
Length of actuating links
{l}_{k}\left(t\right)
k=
1,…,4 are expressed in terms of generalized coordinates of the tripod as follows [5, 6]:
{l}_{1}\left(t\right)=\sqrt{{x}^{2}+\left(y+OA\cdot \mathrm{s}\mathrm{i}\mathrm{n}\varphi {\right)}^{2}+\left(z-OA\cdot \mathrm{c}\mathrm{o}\mathrm{s}\varphi {\right)}^{2}},
{l}_{2}\left(t\right)=\sqrt{\left(x-OB{\right)}^{2}+{y}^{2}+{z}^{2}},
{l}_{3}\left(t\right)=\sqrt{\left(x+OB{\right)}^{2}+{y}^{2}+{z}^{2}},
{l}_{4}\left(t\right)=\sqrt{\left(OK-\mathrm{О}\mathrm{А}\cdot \mathrm{s}\mathrm{i}\mathrm{n}\varphi {\right)}^{2}+\left(OA\cdot \mathrm{c}\mathrm{o}\mathrm{s}\varphi +DK{\right)}^{2}}.
Geometrical parameters
OA
OB
OK
DK
of the tripod base and points of its attachment on a swiveling base are shown on Fig. 2.
1.2. Drive dynamics
Each actuating link of the tripod (Fig. 3) consists of separately excited DC electric motor, irreversible worm reducer and screw gear.
Fig. 3. Structural diagram of a driving mechanism of the tripod actuating link; 1 – rod; 2 – program controller; 3 – electric motor; 4 – worm reducer; 5 – screw gear; 6 – feedback transmitter; 7 – feedback controller
Mechanism of motion transmission from a motor to moved mass is considered to be a three-mass system consisting of an electric motor rotor with a worm, worm wheel
c/w
a screw and a nut with a rod. In this case a model has three degrees of freedom. As generalized coordinates
k
of a drive the following is selected: rotation angle
{\alpha }_{1k}
of an output motor shaft (worm), rotation angle
{\alpha }_{2k}
of a worm wheel and rotation angle
{\alpha }_{3k}
of a screw at the inlet of a slave cylinder nut. Due to transmission yielding they differ by the value of angular deformation.
Dynamic characteristic of a separately excited DC drive motor is written as [7]:
\tau {\stackrel{˙}{T}}_{k}+{T}_{k}=r{u}_{k}\left(t\right)-s{\omega }_{1k},
{T}_{k}
– motor shaft torque;
\tau
– electromagnetic time constant;
r
s
– motor parameters-dependent coefficients;
{u}_{k}\left(t\right)
– control voltage;
{\omega }_{1k}
– angular velocity to a motor shaft.
Differential equations of rotational motion of a worm with electric motor rotor, a worm wheel with an electro cylinder screw of and forward motion of a nut with an actuating link rod are written as:
{I}_{1}{\stackrel{˙}{\omega }}_{1k}={T}_{k}-{c}_{1\phi }\left(\frac{{\alpha }_{1k}}{i}-{\alpha }_{2k}\right)-{T}_{c1}\cdot \mathrm{s}\mathrm{i}\mathrm{g}\mathrm{n}{\omega }_{1k},
{I}_{2}{\stackrel{˙}{\omega }}_{2k}={c}_{1\phi }\left(\frac{{\alpha }_{1k}}{i}-{\alpha }_{2k}\right)i\eta -{c}_{2\phi }\left({\alpha }_{2k}-{\alpha }_{3k}\right)-{T}_{c2}\mathrm{s}\mathrm{i}\mathrm{g}\mathrm{n}{\omega }_{2k},
{m}_{3}{\stackrel{¨}{l}}_{k}={F}_{k}^{ax}-{F}_{k},
{I}_{1}
{I}_{2}
– inertia moments of a worm with a rotor and a worm wheel with a screw;
{\stackrel{˙}{\omega }}_{1k}
{\stackrel{˙}{\omega }}_{2k}
– angular acceleration of an electric motor rotor and a worm wheel;
{T}_{1k}
{T}_{2k}
– torques applied to a worm and to a reducer worm wheel;
{T}_{c1}
{T}_{c2}
– constant torques of resistance forces applied to a worm and a worm wheel shaft;
\eta
– reducer efficiency factor;
i
– gear ratio of a worm reducer;
{c}_{1\phi }
– unit stiffness of elastic linkage between motor shaft and worm wheel;
{c}_{2\phi }
– screw rigidity in torsion;
{m}_{3}
– mass of a nut with an actuating link rod;
{F}_{k}^{ax}
– axial moving force;
{F}_{k}
– rod force determined by load when tripod is in operation Eq. (2).
In case of a lead worm
{c}_{1\varphi }\left({\alpha }_{1k}/i-{\alpha }_{2k}\right)>0
and efficiency factor equals to
\eta =\mathrm{t}\mathrm{g}{\beta }_{1}/\mathrm{t}\mathrm{g}\left({\beta }_{1}+{\rho }_{1}\right)
{\beta }_{1}
– ascending angle of worm screw line;
{\rho }_{1}
– modified friction angle. If
{c}_{1\varphi }\left({\alpha }_{1k}/i-{\alpha }_{2k}\right)<0
, motion is impossible due to reducer irreversibility. In this case dynamic seizure in mechanical transmission will occur [8].
Torque on a guide rotating screw is connected to axial force acting on a nut with a rod by the expression:
{c}_{2\phi }\left({\alpha }_{2k}-{\alpha }_{3k}\right)=0.9{F}_{k}^{ax}\frac{{d}_{3}}{2}\mathrm{t}\mathrm{g}\left({\beta }_{3}+{\rho }_{3}\right),
{\beta }_{3}
{\rho }_{3}
– ascending angle of a screw line and modified friction angle respectively;
{d}_{3}
– average diameter of screw thread.
Angular screw speed is connected to linear speed of actuating link motion by the relation:
{\stackrel{˙}{\alpha }}_{3k}=\frac{2\pi }{pn}{\stackrel{˙}{l}}_{k}.
From the system consisting of 32 algebraic and differential Eq. (1-8) based on the laws of control voltage variation
{u}_{k}\left(t\right)
in energizing winding of drive electric motors and initial conditions 32 unknown functions can be found: generalized coordinates of the tripod
x
y
z
\phi
, length of actuating links
{l}_{k}\left(t\right)
, force in these links
{F}_{k}
{F}_{k}^{ax}
, as well as electric motor shaft torque moments
{T}_{k}
and angular coordinates
{\alpha }_{1k}
{\alpha }_{2k}
{\alpha }_{3k}
k=
Torque moment on an electric motor shaft can be calculated via simultaneous solution of Eq. (4-8):
{T}_{k}={I}_{1}{\stackrel{˙}{\omega }}_{1k}+\frac{0.45{d}_{3}\mathrm{t}\mathrm{g}\left({\beta }_{3}+{\varphi }_{3}\right){m}_{3}{I}_{2}}{{c}_{2\varphi }i\eta }{\stackrel{⃛}{l}}_{k}\left(t\right)+\left(\frac{2\pi }{pni\eta }{I}_{2}+\frac{0.45{d}_{3}\mathrm{t}\mathrm{g}\left({\beta }_{3}+{\mathrm{\varphi }}_{3}\right){m}_{3}}{i\eta }\right){\stackrel{¨}{l}}_{k}\left(t\right)
\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }+\frac{0.45{d}_{3}\mathrm{t}\mathrm{g}\left({\beta }_{3}+{\varphi }_{3}\right)}{i\eta }{F}_{k}+\frac{0.45{d}_{3}\mathrm{t}\mathrm{g}\left({\beta }_{3}+{\varphi }_{3}\right){I}_{2}}{{c}_{2\varphi }i\eta }{\stackrel{¨}{F}}_{k}+{T}_{c1}\mathrm{s}\mathrm{i}\mathrm{g}\mathrm{n}{\omega }_{1k}+\frac{{T}_{c2}}{i\eta }\mathrm{s}\mathrm{i}\mathrm{g}\mathrm{n}{\omega }_{2k}.
Given program motion equations
{l}_{k}^{pr}\left(t\right)
and introducing dynamic error
\psi ={l}_{k}\left(t\right)-{l}_{k}^{pr}\left(t\right)
, allows for going over to a disturbed motion equation.
No-dynamic seizure conditions are obtained from Eq. (4):
{F}_{k}\ge -\frac{{T}_{c2}}{0.45{d}_{3}\mathrm{t}\mathrm{g}\left({\beta }_{3}+{\varphi }_{3}\right)}-{\stackrel{¨}{l}}_{k}\left({m}_{3}+{I}_{2}\frac{2\pi }{0.45pn{d}_{3}\mathrm{t}\mathrm{g}\left({\beta }_{3}+{\varphi }_{3}\right)}\right),\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{\omega }_{k}>0,
{F}_{k}\le \frac{{T}_{c2}}{0.45{d}_{3}\mathrm{t}\mathrm{g}\left({\beta }_{3}+{\varphi }_{3}\right)}-{\stackrel{¨}{l}}_{k}\left({m}_{3}+{I}_{2}\frac{2\pi }{0.45pn{d}_{3}\mathrm{t}\mathrm{g}\left({\beta }_{3}+{\varphi }_{3}\right)}\right),\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{\omega }_{k}\le 0.
Therefore, expression Eq. (9) determines a permissible value of electric motor shaft torque. When there is a violation of conditions Eqs. (10, 11) dynamic seizure will occur in transmission that can entail unstable operation of a drive and even its malfunction.
In order to determine model parameters experimental and numerical research were conducted to study dependence of tripod grip motion in plane
yOz
when one (the first) slave cylinder is in operation while the control voltage was varied according to rectangular waveform
{u}_{1}\left(t\right)=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}
{t}_{1}>t\ge 0
{u}_{1}\left(t\right)=0
t\ge {t}_{1}
For out of service actuating links worm rotation angle
{\alpha }_{1k}
was taken as zero and warm shaft torque was taken as
{T}_{k}={c}_{1\phi }{\alpha }_{2k}
. Grip load amounts to 420 N. In the process of calculations, the no-dynamic seizure conditions were monitored. A tripod control system is equipped with current sensor ACS712 designed for measuring electric drive current consumption proportional to torque developed by an electric motor [9].
In Fig. 4 presents experimental and theoretical dependences of the moment equivalent
{T}_{1}\left(t\right)
, on an electric motor shaft for two modes of tripod grip motion, namely the first cylinder extension (curve 1, 2), the first cylinder retraction (curve 3, 4), as well as experimental dependence of the first cylinder stroke length change while extending (curve 5) and retracting (curve 6).
Fig. 4. Dependence of torque equivalent developed by electric motor of tripod actuating link drive: 1, 3 – experimental curves, 2, 4 – theoretical and experimental dependence of change of the first link stroke length at extending (curve 5) and retracting (curve 6)
The developed mathematical model of tripod dynamics allows for solving the issues regarding determination of electric motors torques required to perform program motions and dynamic loads in kinematic pairs of the tripod, as well as for determining dynamic errors caused by influence transmission elasticity and mechanism dynamic exert on electric motor operation.
The work is completed under financial support from the Russian Foundation for Basic Research (RFBR) (Contract No. 15-01-04577-а, No. 16-48-340395 р-а and No. 16-38-00485).
Zhoga V., Gavrilov A., Gerasun V., Nesmianov I., Pavlovsky V., Skakunov V., Bogatyrev V., Golubev D., Dyashkin-Titov V., Vorobieva N. Walking mobile robot with manipulator-tripod. Proceedings of Romansy 20th CISM-IFToMM Symposium on Theory and Practice of Robots and Manipulators, Vol. 22, 2014, p. 463-471. [Search CrossRef]
Glazunov V. A., Koliskor A. W., Krainev A. F. Spatial Mechanisms of Parallel Structure. Nauka, Moscow, 1991, p. 95, (in Russian). [Search CrossRef]
Nesmiyanov I., Zhoga V., Skakunov V., Terekhov S., Vorob’eva N., Dyashkin-Titov V., Fares Ali Hussein Al-hadsha Synthesis of control algorithm and computer simulation of robotic manipulator-tripod. Communications in Computer and Information Science, Vol. 535, 2015, p. 392-404. [Search CrossRef]
Zhoga V., Gerasun V., Nesmiyanov I., Vorob'eva N., Dyashkin-Titov V. Dynamic creation of the optimum program motion of a manipulator-tripod. Journal of Machinery Manufacture and Reliability, Vol. 44, Issue 2, 2015, p. 181-186. [Search CrossRef]
Kolovskii M. Z., Sloushch A. V. Foundations of Industrial Robot Dynamics. Nauka, Moscow, 1998, (in Russian). [Search CrossRef]
Korendesev A. I., Salamandra B. L., Tyves L. I. Theoretical Basis of Robotics. Institute of Machines Science named after A.A. Blagonravov of the Russian Academy of Sciences, Nauka, Moscow, Vol. 1, 2006, p. 383, (in Russian). [Search CrossRef]
Korendesev A. I., Salamandra B. L., Tyves L. I., et al. Manipulation Systems for Robots. Mashinostroenie, Moscow, 1989, p. 427, (in Russian). [Search CrossRef]
Vejts V. L., Vasil’kov D. V., Gidaspov I. A., Shneerson E. S. Dynamics of Drives of Process Units with Non-Reversible Devices. Monograph in 5 Parts, St. Petersburg, 2002, (in Russian). [Search CrossRef]
Zhoga V. V., Skakunov V. N., Terehov S. E. Adaptive control system of electric driver walking robot. Journal Izvestiya VSTU, Vol. 157, Issue 2, 2015, p. 180-183, (in Russian). [Search CrossRef] |
Metallurgy, Popular Questions: ICSE Class 10 CHEMISTRY, Concise Chemistry 10 - Meritnation
Q.61. The turmeric solution will turn red by an aqueous solution of -
(A) Potassium acetate
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2. Can form2+and 3+ions
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Bronze-------hard, brittle
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Sn — lowers melting point of alloy
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Iron and zinc oxide
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Create 2-DOF PID controller in parallel form, convert to parallel-form 2-DOF PID controller - MATLAB pid2 - MathWorks India
Discrete-Time 2-DOF PI Controller
Convert 2-DOF PID Controller from Standard to Parallel Form
Convert Dynamic System to 2-DOF Parallel-Form PID Controller
Discretize a Continuous-Time 2-DOF PID Controller
Create 2-DOF PID controller in parallel form, convert to parallel-form 2-DOF PID controller
C2 = pid2(Kp,Ki,Kd,Tf,b,c)
C2 = pid2(Kp,Ki,Kd,Tf,b,c,Ts)
C2 = pid2(sys)
Two-degree-of-freedom (2-DOF) PID controllers include setpoint weighting on the proportional and derivative terms. A 2-DOF PID controller can achieve fast disturbance rejection without significant increase of overshoot in setpoint tracking. 2-DOF PID controllers are also useful to mitigate the influence of changes in the reference signal on the control signal. The following illustration shows a typical control architecture using a 2-DOF PID controller.
C2 = pid2(Kp,Ki,Kd,Tf,b,c) creates a continuous-time 2-DOF PID controller with proportional, integral, and derivative gains Kp, Ki, and Kd and first-order derivative filter time constant Tf. The controller also has setpoint weighting b on the proportional term, and setpoint weighting c on the derivative term. The relationship between the 2-DOF controller output (u) and its two inputs (r and y) is given by:
u={K}_{p}\left(br-y\right)+\frac{{K}_{i}}{s}\left(r-y\right)+\frac{{K}_{d}s}{{T}_{f}s+1}\left(cr-y\right).
This representation is in parallel form. If all coefficients are real-valued, then the resulting C2 is a pid2 controller object. If one or more of these coefficients is tunable (realp or genmat), then C2 is a tunable generalized state-space (genss) model object.
C2 = pid2(Kp,Ki,Kd,Tf,b,c,Ts) creates a discrete-time 2-DOF PID controller with sample time Ts. The relationship between the controller output and inputs is given by:
u={K}_{p}\left(br-y\right)+{K}_{i}IF\left(z\right)\left(r-y\right)+\frac{{K}_{d}}{{T}_{f}+DF\left(z\right)}\left(cr-y\right).
IF\left(z\right)=DF\left(z\right)=\frac{{T}_{s}}{z-1}.
To choose different discrete integrator formulas, use the IFormula and DFormula properties. (See Properties for more information). If DFormula = 'ForwardEuler' (the default value) and Tf ≠ 0, then Ts and Tf must satisfy Tf > Ts/2. This requirement ensures a stable derivative filter pole.
C2 = pid2(sys) converts the dynamic system sys to a parallel form pid2 controller object.
To create a discrete-time pid2 controller, provide a positive real value (Ts > 0). pid2 does not support discrete-time controllers with unspecified sample time (Ts = -1).
Ts must be a scalar value. In an array of pid2 controllers, each controller must have the same Ts.
SISO dynamic system to convert to parallel pid2 form.
sys must be a two-input, one-output system. sys must represent a valid 2-DOF PID controller that can be written in parallel form with Tf ≥ 0.
Use Name,Value syntax to set the numerical integration formulas IFormula and DFormula of a discrete-time pid2 controller, or to set other object properties such as InputName and OutputName. For information about available properties of pid2 controller objects, see Properties.
2-DOF PID controller, returned as a pid2 controller object, an array of pid2 controller objects, a genss object, or a genss array.
If all the coefficients have scalar numeric values, then C2 is a pid2 controller object.
If one or more coefficients is a numeric array, C2 is an array of pid2 controller objects. The controller type (such as PI, PID, or PDF) depends upon the values of the gains. For example, when Kd = 0, but Kp and Ki are nonzero, C2 is a PI controller.
Setpoint weights on the proportional and derivative terms, respectively. b and c values are real, finite, and positive. When you use the pid2 command to create a 2-DOF PID controller, the b, and c input arguments, respectively, set the initial values of these properties.
Proportional, integral, and derivative gains, respectively. Kp, Ki, and Kd values are real and finite. When you use the pid2 command to create a 2-DOF PID controller, the Kp, Ki, and Kd input arguments, respectively, set the initial values of these properties.
The Tf property stores the derivative filter time constant of the pid2 controller object. Tf is real, finite, and greater than or equal to zero. When you create a 2-DOF PID controller using the pid2 command, the Tf input argument sets the initial value of this property.
Discrete integrator formula IF(z) for the integrator of the discrete-time pid2 controller C2. The relationship between the inputs and output of C2 is given by:
u={K}_{p}\left(br-y\right)+{K}_{i}IF\left(z\right)\left(r-y\right)+\frac{{K}_{d}}{{T}_{f}+DF\left(z\right)}\left(cr-y\right).
\frac{{T}_{s}}{z-1}.
\frac{{T}_{s}z}{z-1}.
\frac{{T}_{s}}{2}\frac{z+1}{z-1}.
Discrete integrator formula DF(z) for the derivative filter of the discrete-time pid2 controller C2. The relationship between the inputs and output of C2 is given by:
u={K}_{p}\left(br-y\right)+{K}_{i}IF\left(z\right)\left(r-y\right)+\frac{{K}_{d}}{{T}_{f}+DF\left(z\right)}\left(cr-y\right).
\frac{{T}_{s}}{z-1}.
\frac{{T}_{s}z}{z-1}.
\frac{{T}_{s}}{2}\frac{z+1}{z-1}.
The Trapezoidal value for DFormula is not available for a pid2 controller with no derivative filter (Tf = 0).
Time delay on the system input. InputDelay is always 0 for a pid2 controller object.
Time delay on the system Output. OutputDelay is always 0 for a pid2 controller object.
Create a continuous-time 2-DOF controller with proportional and derivative gains and a filter on the derivative term. To do so, set the integral gain to zero. Set the other gains and the filter time constant to the desired values.
u = Kp (b*r-y) + Kd -------- (c*r-y)
with Kp = 1, Kd = 3, Tf = 0.1, b = 0.5, c = 0.5
Continuous-time 2-DOF PDF controller in parallel form.
Create a discrete-time 2-DOF PI controller using the trapezoidal discretization formula. Specify the formula using Name,Value syntax.
C2 = pid2(Kp,Ki,Kd,Tf,b,c,Ts,'IFormula','Trapezoidal')
u = Kp (b*r-y) + Ki -------- (r-y)
with Kp = 5, Ki = 2.4, b = 0.5, Ts = 0.1
Discrete-time 2-DOF PI controller in parallel form.
Setting Kd = 0 specifies a PI controller with no derivative term. As the display shows, the values of Tf and c are not used in this controller. The display also shows that the trapezoidal formula is used for the integrator.
Create a 2-DOF PID controller, and set the dynamic system properties InputName and OutputName. Naming the inputs and the output is useful, for example, when you interconnect the PID controller with other dynamic system models using the connect command.
C2 = pid2(1,2,3,0,1,1,'InputName',{'r','y'},'OutputName','u')
with Kp = 1, Ki = 2, Kd = 3, b = 1, c = 1
Create a 2-by-3 grid of 2-DOF PI controllers with proportional gain ranging from 1–2 across the array rows and integral gain ranging from 5–9 across columns.
When you pass these arrays to the pid2 command, the command returns the array of controllers.
pi_array = pid2(Kp,Ki,0,0,0.5,0,'Ts',0.1,'IFormula','BackwardEuler');
If you provide scalar values for some coefficients, pid2 automatically expands them and assigns the same value to all entries in the array. For instance, in this example, Kd = Tf = 0, so that all entries in the array are PI controllers. Also, all entries in the array have b = 0.5.
u = Kp (b*r-y) + Ki ------ (r-y)
with Kp = 2, Ki = 9, b = 0.5, Ts = 0.1
C2 = pid2(1,5,0.1,0,0.5,0.5); % PID controller
C2f = pid2(1,5,0.1,0.5,0.5,0.5); % PID controller with filter
pid_array = stack(2,C2,C2f); % stack along 2nd array dimension
Convert a standard-form pidstd2 controller to parallel form.
Standard PID form expresses the controller actions in terms of an overall proportional gain Kp, integrator and derivative time constants Ti and Td, and filter divisor N. You can convert any 2-DOF standard-form controller to parallel form using the pid2 command. For example, consider the following standard-form controller.
C2_std = pidstd2(Kp,Ti,Td,N,b,c)
with Kp = 2, Ti = 3, Td = 4, N = 50, b = 0.1, c = 0.5
Convert this controller to parallel form using pid2.
C2_par = pid2(C2_std)
with Kp = 2, Ki = 0.667, Kd = 8, Tf = 0.08, b = 0.1, c = 0.5
A response plot confirms that the two forms are equivalent.
H\left(s\right)=\frac{3\left(s+1\right)\left(s+2\right)}{s}.
Convert a discrete-time dynamic system that represents a 2-DOF PID controller with derivative filter to parallel pid2 form.
The following state-space matrices represent a discrete-time 2-DOF PID controller with a sample time of 0.1 s.
A = [1,0;0,0.99];
B = [0.1,-0.1; -0.005,0.01];
C = [3,0.2];
D = [2.6,-5.2];
C2fe = pid2(sys)
Ts 1
u = Kp (b*r-y) + Ki ------ (r-y) + Kd ----------- (c*r-y)
z-1 Tf+Ts/(z-1)
with Kp = 5, Ki = 3, Kd = 2, Tf = 10, b = 0.5, c = 0.5, Ts = 0.1
Discrete-time 2-DOF PIDF controller in parallel form.
C2trap = pid2(sys,'IFormula','Trapezoidal','DFormula','Trapezoidal')
Ts*(z+1) 1
u = Kp (b*r-y) + Ki -------- (r-y) + Kd ------------------- (c*r-y)
2*(z-1) Tf+Ts/2*(z+1)/(z-1)
with Kp = 4.85, Ki = 3, Kd = 2, Tf = 9.95, b = 0.485, c = 0.5, Ts = 0.1
Discretize a continuous-time 2-DOF PID controller and specify the integral and derivative filter formulas.
C2con = pid2(10,5,3,0.5,1,1); % continuous-time 2-DOF PIDF controller
with Kp = 10, Ki = 5, Kd = 3.31, Tf = 0.552, b = 1, c = 1, Ts = 0.1
However, these commands do not compute new PID gains for the discretized controller. To see this, examine C2dis2 and compare the coefficients to C2con and C2dis1.
Ts*z 1
u = Kp (b*r-y) + Ki ------ (r-y) + Kd ------------- (c*r-y)
z-1 Tf+Ts*z/(z-1)
with Kp = 10, Ki = 5, Kd = 3, Tf = 0.5, b = 1, c = 1, Ts = 0.1
Create arrays of pid2 controller objects by:
Specifying array values for one or more of the coefficients Kp, Ki, Kd, Tf, b, and c.
In an array of pid2 controllers, each controller must have the same sample time Ts and discrete integrator formulas IFormula and DFormula.
To create or convert to a standard-form controller, use pidstd2. Standard form expresses the controller actions in terms of an overall proportional gain Kp, integral and derivative times Ti and Td, and filter divisor N. For example, the relationship between the inputs and output of a continuous-time standard-form 2-DOF PID controller is given by:
u={K}_{p}\left[\left(br-y\right)+\frac{1}{{T}_{i}s}\left(r-y\right)+\frac{{T}_{d}s}{\frac{{T}_{d}}{N}s+1}\left(cr-y\right)\right].
There are two ways to discretize a continuous-time pid2 controller:
If you require different discrete integrator formulas, you can discretize the controller by directly setting Ts, IFormula, and DFormula to the desired values. (See Discretize a Continuous-Time 2-DOF PID Controller.) However, this method does not compute new gain and filter-constant values for the discretized controller. Therefore, this method might yield a poorer match between the continuous- and discrete-time pid2 controllers than using c2d.
pidstd2 | pid | piddata2 | getComponents | make1DOF | pidtune | pidTuner | tunablePID2 | genss | realp |
Problems on Numbers | Quantitative Aptitude
Problems on Numbers Questions
FACTS AND FORMULAE FOR PROBLEMS ON NUMBERS
I. Numeral : In Hindu Arabic system, we use ten symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 called digits to represent any number. A group of digits, denoting a number is called a numeral.
\begin{array}{cccccc}Ten-Crores& Crores& Ten-Lacs& Lacs& Ten-Thousands& Thousands\\ 6& 8& 9& 7& 4& 5\end{array}\begin{array}{ccc}Hundreds& Tens& Units\\ 1& 3& 2\end{array}
We read it as : 'Sixty-eight crores, ninety-seven lacs, forty-five thousand, one hundred and thirty-two'.
II. PlaceValue or Local Value of a Digit in a Numeral: In the above numeral : Place value of 2 is (2 x 1) = 2 Place value of 3 is (3 x 10) = 30 Place value of 1 is (1 x 100) = 100 and so on. Place value of 6 is
6×{10}^{8}
III. Face Value : The face value of a digit in a numeral is the value of the digit itself at whatever place it may be. In the above numeral, the face value of 2 is 2; the face value of 3 is 3 and so on.
1. Natural Numbers : Counting numbers 1, 2, 3, 4, 5,..... are called natural numbers.
2. Whole Numbers : All counting numbers together with zero form the set of whole numbers. Thus,
3. Integers : All natural numbers, 0 and negatives of counting numbers i.e., {…, -3,-2,-1, 0, 1, 2, 3,…..} together form the set of integers.
(ii) Negative Integers : {- 1, - 2, - 3,…..} is the set of all negative integers.
(iii) Non-Positive and Non-Negative Integers : 0 is neither positive nor negative. So, {0, 1, 2, 3,….} represents the set of non-negative integers, while{0, -1,-2,-3,…..} represents the set of non-positive integers.
6. Prime Numbers : - A number greater than 1 is called a prime number, if it has exactly two factors, namely 1 and the number itself. - Prime numbers upto 100 are : 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43,47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. - Prime numbers Greater than 100 : Let p be a given number greater than 100. To find out whether it is prime or not, we use the following method : Find a whole number nearly greater than the square root of p. Let k > sqrt{p}. Test whether p is divisible by any prime number less than k. If yes, then p is not prime. Otherwise, p is prime. e.g., We have to find whether 191 is a prime number or not. Now, 14 > sqrt{191}.Prime numbers less than 14 are 2, 3, 5, 7, 11, 13. 191 is not divisible by any of them. So, 191 is a prime number.
7. Composite Numbers : Numbers greater than 1 which are not prime, are known as composite numbers, e.g., 4, 6, 8, 9, 10, 12.
(i) 1 is neither prime nor composite.
8. Co-primes :Two numbers a and b are said to be co-primes, if their H.C.F. is 1. e.g., (2, 3), (4, 5), (7, 9), (8, 11), etc. are co-primes
V. TESTS OF DIVISIBILITY
1. Divisibility By 2 : A number is divisible by 2, if its unit's digit is any of 0, 2, 4, 6, 8.
2. Divisibility By 3 :A number is divisible by 3, if the sum of its digits is divisible by 3.
Ex.592482 is divisible by 3, since sum of its digits = (5 + 9 + 2 + 4 + 8 + 2) = 30, which is divisible by 3. But, 864329 is not divisible by 3, since sum of its digits =(8 + 6 + 4 + 3 + 2 + 9) = 32, which is not divisible by 3.
3. Divisibility By 4 : A number is divisible by 4, if the number formed by the last two digits is divisible by 4.
Ex. 892648 is divisible by 4, since the number formed by the last two digits is 48, which is divisible by 4. But, 749282 is not divisible by 4, since the number formed by the last tv/o digits is 82, which is not divisible by 4.
4. Divisibility By 5 : A number is divisible by 5, if its unit's digit is either 0 or 5.
Thus, 20820 and 50345 are divisible by 5, while 30934 and 40946 are not.
Ex. The number 35256 is clearly divisible by 2. Sum of its digits = (3 + 5 + 2 + 5 + 6) = 21, which is divisible by 3. Thus, 35256 is divisible by 2 as well as 3. Hence, 35256 is divisible by 6.
Ex. 953360 is divisible by 8, since the number formed by last three digits is 360, which is divisible by 8. But, 529418 is not divisible by 8, since the number formed by last three digits is 418, which is not divisible by 8.
Ex. 60732 is divisible by 9, since sum of digits * (6 + 0 + 7 + 3 + 2) = 18, which is divisible by 9. But, 68956 is not divisible by 9, since sum of digits = (6 + 8 + 9 + 5 + 6) = 34, which is not divisible by 9.
8. Divisibility By 10 : A number is divisible by 10, if it ends with 0. Ex. 96410, 10480 are divisible by 10, while 96375 is not.
9. Divisibility By 11 : A number is divisible by 11, if the difference of the sum of its digits at odd places and the sum of its digits at even places, is either 0 or a number divisible by 11.
Ex. The number 4832718 is divisible by 11, since : (sum of digits at odd places) - (sum of digits at even places) (8 + 7 + 3 + 4) - (1 + 2 + 8) = 11, which is divisible by 11.
10. Divisibility By 12 : A number is divisible by 12, if it is divisible by both 4 and 3.
(i) The number formed by last two digits is 32, which is divisible by 4
(ii) Sum of digits = (3 + 4 + 6 + 3 + 2) = 18, which is divisible by 3. Thus, 34632 is divisible by 4 as well as 3.
Hence, 34632 is divisible by 12.
13. Divisibility By 16 : A number is divisible by 16, if the number formed by the last4 digits is divisible by 16.
14. Divisibility By 24 : A given number is divisible by 24, if it is divisible by both3 and 8.
15. Divisibility By 40 : A given number is divisible by 40, if it is divisible by both 5 and 8.
16. Divisibility By 80 : A given number is divisible by 80, if it is divisible by both 5 and 16.
Ex. 36 is divisible by both 4 and 6, but it is not divisible by (4x6) = 24, since 4 and 6 are not co-primes.
VI . MULTIPLICATION BY SHORT CUT METHODS
1. Multiplication By Distributive Law :
(i) a x (b + c) = a x b + a x c (ii) a x (b-c) = a x b-a x c.
(i) 567958 x 99999 = 567958 x (100000 - 1) = 567958 x 100000 - 567958 x 1 = (56795800000 - 567958) = 56795232042.
(ii) 978 x 184 + 978 x 816 = 978 x (184 + 816) = 978 x 1000 = 978000.
2. Multiplication of a Number By
{5}^{n}
: Put n zeros to the right of the multiplicand and divide the number so formed by
{2}^{n}
Ex. 975436 x 625 =
975436×{5}^{4}
\frac{9754360000}{{2}^{4}}
VII. DIVISION ALGORITHM OR EUCLIDEAN ALGORITHM : If we divide a given number by another number, then:Dividend = (Divisor x Quotient) + Remainder
(i).
\left({x}^{n}-{a}^{n}\right)
is divisible by (x - a) for all values of n
(ii).
\left({x}^{n}-{a}^{n}\right)
is divisible by (x + a) for all even values of n.
(iii).
\left({x}^{n}+{a}^{n}\right)
is divisible by (x + a) for all odd values of n.
IX. PROGRESSION A succession of numbers formed and arranged in a definite order according to certain definite rule, is called a progression.
1. Arithmetic Progression (A.P) : If each term of a progression differs from its preceding term by a constant, then such a progression is called an arithmetical progression. This constant difference is called the common difference of the A.P.
An A.P. with first term 'a' and common difference 'd' is given by a, (a + d), (a + 2d),(a + 3d),.....
1. nth term = a + (n - 1)d
2. Sum of n terms =
\frac{n}{2}\left[2a+\left(n-1\right)d\right]
\frac{n}{2}\left[a+l\right]
, Where l is the last term.
\left(1+2+3+...+n\right)=\frac{1}{2}n\left(n+1\right)
\left({1}^{2}+{2}^{2}+{3}^{2}+...+{n}^{2}\right)=\frac{1}{6}n\left(n+1\right)\left(2n+1\right)
\left({1}^{3}+{2}^{3}+{3}^{3}+...+{n}^{3}\right)=\frac{1}{4}{n}^{2}\left(n+1\right){ }^{2}
2. Geometrical Progression (G.P) : A progression of numbers in which every term bears a constant ratio with its preceding term, is called a geometrical progression. The constant ratio is called the common ratio of the G.P.
A G.P. with first term 'a' and common ratio 'r' is :
a,ar,a{r}^{2},a{r}^{3},...
In this G.P.
I. nth term =
a{r}^{n-1}
II. Sum of n terms=
\left\{\begin{array}{ll}\frac{a\left(1-{r}^{n}\right)}{\left(1-r\right)}& where r<1\\ \frac{a\left({r}^{n}-1\right)}{\left(r-1\right)}& where r>1\end{array}\right\
Then, xy = 9375 and x/y = 15.
=> x = 15y = 15 x 25 = 375.
Sum of the numbers = 375 + 25 = 400.
â— Each digit has a fixed position called its place.
â— Each digit has a value depending on its place called the place value of the digit.
â— The face value of a digit for any place in the given number is the value of the digit itself
â— Place value of a digit = (face value of the digit) × (value of the place).
Hence, the place value of 6 in 64 = 6 x 10 = 60.
(3x + 2) (2x - 5) = ax² + kx + n .
What is the value of a - n + k ?
This is a quadratic equation. To find n we multiply the last terms in each bracket together (2 times -5) = -10
To find ‘a’ we multiply the first terms in each bracket together (3x times 2x) = 6x2, So a = 6.
Thus a - n = 16
To find k we multiply the first term in the first bracket by the second term in the second bracket and add the result to the product of the second term in the first bracket and the first term in the second. (-15x + 4x) = -11x. So k = -11
x and y are integers
The smallest value will be obtained by giving x the smallest possible value while y gets the largest possible value. The smallest value for x = 7. This means that y will have to be less than or equal to 3 (so that x + y <11), but we want the largest value of y which is 3.
The difference, x - y will be 4.
What approximate value will come in place of the question mark(?) in the below question?
(47% of 1442 - 36% of 1412) + 63 = ?
Three numbers are in the ratio 4 : 5 : 6 and their average is 25. The largest number is :
Let the numbers be 4x, 5x and 6x, Then, (4x + 5x + 6x ) / 3 = 25
Largest number 6x = 30.
In a two-digit number, if it is known that its unit's digit exceeds its ten's digit by 2 and that the product of the given number and the sum of its digits is equal to 144, then the number is :
Number = 10x + (x + 2) = 11x + 2
(11x + 2) (2x + 2) = 144
=> 22x^2+26x-140=0
Hence, Required Number = 11x + 2 = 24 |
Frost beamformer - MATLAB - MathWorks Deutschland
Apply Frost Beamforming to ULA
Find Frost Beamforming Weights
Frost beamformer
The phased.FrostBeamformer object implements a Frost beamformer. A Frost beamformer consists of a time-domain MVDR beamformer combined with a bank of FIR filters. The beamformer steers the beam towards a given direction while the FIR filters preserve the input signal power.
To compute the beamformed signal:
Create the phased.FrostBeamformer object and set its properties.
beamformer = phased.FrostBeamformer
beamformer = phased.FrostBeamformer(Name,Value)
beamformer = phased.FrostBeamformer creates a Frost beamformer System object™, beamformer, with default property values.
beamformer = phased.FrostBeamformer(Name,Value) creates a Frost beamformer object, beamformer, with each specified property Name set to the specified Value. You can specify additional name-value pair arguments in any order as (Name1,Value1,...,NameN,ValueN). Enclose each property name in single quotes.
Example: beamformer = phased.FrostBeamformer('SensorArray',phased.ULA('NumElements',20),'SampleRate',300e3) sets the sensor array to a uniform linear array (ULA) with default ULA property values except for the number of elements. The beamformer has a sample rate of 300 kHz.
Sensor array, specified as a Phased Array System Toolbox array System object. The array cannot contain subarrays.
FilterLength — FIR filter length
Length of FIR filter for each sensor element, specified as a positive integer.
DirectionSource — Source of beamforming direction
Source of beamforming direction, specified as 'Property' or 'Input port'. Specify whether the beamforming direction comes from the Direction property of this object or from the input argument, ANG. Values of this property are:
'Property' Specify the beamforming direction using the Direction property.
'Input port' Specify the beamforming direction using the input argument, ANG.
Direction — Beamforming directions
[0;0] (default) | real-valued 2-by-1 vector | real-valued 2-by-L matrix
Beamforming directions, specified as a real-valued 2-by-1 vector or a real-valued 2-by-L matrix. For a matrix, each column specifies a different beamforming direction. Each column has the form [AzimuthAngle;ElevationAngle]. Azimuth angles must lie between –180° and 180° and elevation angles must lie between –90° and 90°. All angles are defined with respect to the local coordinate system of the array. Units are in degrees.
To enable this property, set the DirectionSource property to 'Property'.
Y = beamformer(X,ANG)
Y = beamformer(X,XT,ANG)
Y = beamformer(X) performs Frost beamforming on the input, X, and returns the beamformed output, Y. This syntax uses the input data, X, as training samples to calculate the beamforming weights.
Y = beamformer(X,ANG) uses ANG as the beamforming direction. To use this syntax, set the DirectionSource property to 'Input port'.
Y = beamformer(X,XT,ANG) combines all input arguments. To use this syntax, set the TrainingInputPort property to true and set the DirectionSource property to 'Input port'.
[Y,W] = beamformer(___) returns the beamforming weights, W. To use this syntax, set the WeightsOutputPort property to true.
Input signal, specified as a complex-valued M-by-N matrix. M is the signal length and N is the number of array elements specified in the SensorArray property. M must be larger than the length of the filter specified by the FilterLength property.
Training data, specified as a complex-valued M-by-N matrix. M and N are equal to the values for X.
ANG — Beamforming directions
[0;0] (default) | real-valued 2-by-1 column vector | real-valued 2-by-L matrix
Beamforming directions, specified as a real-valued 2-by-1 column vector The vector has the form [AzimuthAngle;ElevationAngle]. Units are in degrees. The azimuth angle must lie between –180° and 180°, and the elevation angle must lie between –90° and 90°.
To enable this argument, set the DirectionSource property to 'Input port'.
complex-valued 1-by-M vector
Beamformed output, returned as a complex-valued 1-by-Mvector, where M is the number of rows of the input X.
complex-valued L-by-1 vector.
Beamforming weights, returned as a complex-valued L-by-1 vector where L is the number of degrees of freedom of the beamformer. The number of degrees of freedom is given by the product of the number of elements specified by the SensorArray property and the FIR filter length specified by FilterLength property.
To enable this output, set the WeightsOutputPort property to true.
Apply Frost beamforming to an 11-element acoustic ULA array. The incident angle of the incoming signal is -50 degrees in azimuth and 30 degrees in elevation. The speed of sound in air is assumed to be 340 m/sec. The signal has added Gaussian white noise.
Simulate the signal.
array = phased.ULA('NumElements',11,'ElementSpacing',0.04);
array.Element.FrequencyRange = [20 20000];
'PropagationSpeed',c,'SampleRate',fs,...
'ModulatedInput',false,'NumSubbands',8192);
incidentAngle = [-50;30];
x = collector(x.',incidentAngle);
noise = 0.2*randn(size(x));
Beamform the signal.
beamformer = phased.FrostBeamformer('SensorArray',array,...
'Direction',incidentAngle,'FilterLength',5);
Plot the beamformed output.
plot(t,rx(:,6),'r:',t,y)
legend('Original','Beamformed')
Find the beamformer weights of a Frost beamforming applied to signals received at a 7-element acoustic ULA array. The incident angle of the incoming signal is
-2{0}^{\circ }
in azimuth and
3{0}^{\circ }
in elevation. The signal has added Gaussian white noise. The speed of sound in air is assumed to be 340 m/s. Use a filter length of 15.
First, create the signal.
element = phased.OmnidirectionalMicrophoneElement('FrequencyRange',[50,10000]);
array = phased.ULA('Element',element,'NumElements',numelements,'ElementSpacing',0.04);
Create a beamformer with a filter length of 15. Then, beamform the arriving signal and obtain the beamformer weights.
filterlength = 15;
beamformer = phased.FrostBeamformer('SensorArray',array, ...
'PropagationSpeed',c,'SampleRate',fs,'WeightsOutputPort',true, ...
'Direction',incidentAngle,'FilterLength',filterlength);
[y,wt] = beamformer(rx);
size(wt)
There are 7*15 = 105 weights computed as expected.
Compare the beamformed output with the signal arriving at the middle element of the array.
plot(1000*t,rx(:,4),'r:',1000*t,y)
xlabel('time (msec)')
legend('Middle Element','Beamformed')
phased.FrostBeamformer uses a beamforming algorithm proposed by Frost. It can be considered the time-domain counterpart of the minimum variance distortionless response (MVDR) beamformer. The algorithm does the following:
Steers the array to the beamforming direction.
Applies an FIR filter to the output of each sensor to achieve the distortionless response constraint. The filter is specific to each sensor.
For more information about Frost beamforming, see [1].
[1] Frost, O. “An Algorithm For Linearly Constrained Adaptive Array Processing”, Proceedings of the IEEE. Vol. 60, Number 8, August, 1972, pp. 926–935.
phased.PhaseShiftBeamformer | phased.SubbandPhaseShiftBeamformer | phased.TimeDelayBeamformer | phased.TimeDelayLCMVBeamformer |
Solve the quadratic equation below twice: once using the Quadratic Formula and once by completing the square. You should get the same result using both methods. What happens?
x^2 + 6x + 11 = 0
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
a = 1, b = 6, c = 11
No solution, you cannot take a square root of a negative. You should get no solution for completing the square, as well. |
Home : Support : Online Help : Mathematics : Discrete Mathematics : Ordinals : Eval
substitute values for parameters in an ordinal
Eval(o, x=v)
Eval(o, l)
integer or polynomial with integer coefficients
list or set of equations of type x=v
The Eval(o, x=v) calling sequence substitutes the value
v
for the parameter
x
in the ordinal
o
, returning either an ordinal data structure, a nonnegative integer, or a polynomial with positive integer coefficients.
It is possible for
v
to be a negative integer or a polynomial with some negative integer coefficients, provided that the result is a valid ordinal, which means it does not have any negative integer coefficients.
The resulting ordinal is simplified, namely, any coefficients that become zero are removed, and if only a single term with exponent
0
is left after that, a nonnegative integer or a polynomial with positive integer coefficients is returned.
The Eval(o, l) calling sequence performs all the substitutions in l simultaneously.
This command can also be applied to a polynomial with positive integer coefficients representing a nonnegative integer ordinal.
\mathrm{with}\left(\mathrm{Ordinals}\right)
[\textcolor[rgb]{0,0,1}{\mathrm{`+`}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{`.`}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{`<`}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{<=}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Add}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Base}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Dec}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Decompose}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Div}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Eval}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Factor}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Gcd}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Lcm}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{LessThan}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Log}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Max}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Min}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Mult}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Ordinal}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Power}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Split}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Sub}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{`^`}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{degree}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{lcoeff}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{log}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{lterm}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{\omega }}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{quo}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{rem}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{tcoeff}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{tdegree}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{tterm}}]
\mathrm{o1}≔\mathrm{Ordinal}\left([[\mathrm{\omega },x],[2,3],[1,y+1],[0,4]]\right)
\textcolor[rgb]{0,0,1}{\mathrm{o1}}\textcolor[rgb]{0,0,1}{≔}{\textcolor[rgb]{0,0,1}{\mathbf{\omega }}}^{\textcolor[rgb]{0,0,1}{\mathbf{\omega }}}\textcolor[rgb]{0,0,1}{\cdot }\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{\mathbf{\omega }}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{\cdot }\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathbf{\omega }}\textcolor[rgb]{0,0,1}{\cdot }\left(\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{4}
\mathrm{Eval}\left(\mathrm{o1},x=0\right)
{\textcolor[rgb]{0,0,1}{\mathbf{\omega }}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{\cdot }\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathbf{\omega }}\textcolor[rgb]{0,0,1}{\cdot }\left(\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{4}
Several substitutions can be done at once. It is also possible to substitute a polynomial for a parameter and not just an integer.
\mathrm{Eval}\left(\mathrm{o1},[x={x}^{2}+1,y=4]\right)
{\textcolor[rgb]{0,0,1}{\mathbf{\omega }}}^{\textcolor[rgb]{0,0,1}{\mathbf{\omega }}}\textcolor[rgb]{0,0,1}{\cdot }\left({\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\right)\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{\mathbf{\omega }}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{\cdot }\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathbf{\omega }}\textcolor[rgb]{0,0,1}{\cdot }\textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{4}
The result need not be an ordinal data structure.
\mathrm{o2}≔\mathrm{Ordinal}\left([[2,{x}^{2}+x],[1,x],[0,4]]\right)
\textcolor[rgb]{0,0,1}{\mathrm{o2}}\textcolor[rgb]{0,0,1}{≔}{\textcolor[rgb]{0,0,1}{\mathbf{\omega }}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{\cdot }\left({\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathbf{\omega }}\textcolor[rgb]{0,0,1}{\cdot }\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{4}
\mathrm{Eval}\left(\mathrm{o2},x=0\right)
\textcolor[rgb]{0,0,1}{4}
\mathrm{Eval}\left(\mathrm{\omega }·x,x=0\right)
\textcolor[rgb]{0,0,1}{0}
The attempt to substitute a negative integer or a polynomial with negative coefficients may result in an error if the result has negative coefficients.
\mathrm{o3}≔\mathrm{Ordinal}\left([[1,2x+2],[0,3]]\right)
\textcolor[rgb]{0,0,1}{\mathrm{o3}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathbf{\omega }}\textcolor[rgb]{0,0,1}{\cdot }\left(\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}{2}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{3}
\mathrm{Eval}\left(\mathrm{o3},x=-1\right)
\textcolor[rgb]{0,0,1}{3}
\mathrm{Eval}\left(\mathrm{o3},x=-2\right)
Error, (in Ordinals:-Eval) invalid substitution; result is not a valid ordinal
\mathrm{Eval}\left(\mathrm{o3},x=x-1\right)
\textcolor[rgb]{0,0,1}{\mathbf{\omega }}\textcolor[rgb]{0,0,1}{\cdot }\left(\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{3}
\mathrm{Eval}\left(\mathrm{o3},x=x-2\right)
The Eval command can also be applied to a polynomial with positive integer coefficients representing a constant ordinal.
\mathrm{Eval}\left({x}^{2}+1,x=3\right)=\mathrm{eval}\left({x}^{2}+1,x=3\right)
\textcolor[rgb]{0,0,1}{10}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{10}
The Ordinals[Eval] command was introduced in Maple 2015. |
Consider the following rational functions: r(x)=(2x−1)/((x^2)−x−2) , s(x)=((x^3)+27)/((x^2)+4) , t(x)=((x^3)−9x)/(x+2)
Consider the rational functions: r(x)=(2x−1)/((x^2)−x−2) , s(x)=((x^3)+27)/((x^2)+4) , t(x)=((x^3)−9x)/(x+2)
Consider the following rational functions:
r\left(x\right)=\frac{2x-1}{\left({x}^{2}\right)-x-2}
s\left(x\right)=\frac{\left({x}^{3}\right)+27}{\left({x}^{2}\right)+4}
t\left(x\right)=\frac{\left({x}^{3}\right)-9x}{x+2}
u\left(x\right)=\frac{\left({x}^{2}\right)+x-6}{\left({x}^{2}\right)-25}
w\left(x\right)=\frac{\left({x}^{3}\right)+\left(6{x}^{2}\right)+9x}{x+3}
Which of these rational functions has a horizontal asymptote?
r\left(x\right)=\frac{2x-1}{\left({x}^{2}\right)-x-2}=\frac{2x-1}{\left(x-2\right)\left(x+1\right)}
s\left(x\right)=\frac{\left({x}^{3}\right)+27}{\left({x}^{2}\right)+4}=\frac{\left(x+3\right)\left(x+9+3x\right)}{\left({x}^{2}\right)+4}
t\left(x\right)=\frac{\left({x}^{3}\right)-9x}{x+2}=\frac{x\left(x-3\right)\left(x+3\right)}{x+2}
u\left(x\right)=\frac{\left({x}^{2}\right)+x-6}{\left({x}^{2}\right)-25}=\frac{\left(x-2\right)\left(x+3\right)}{\left(x-5\right)\left(x+5\right)}
w\left(x\right)=\frac{\left({x}^{3}\right)+\left(6{x}^{2}\right)+9x}{x+3}=x\left(x+3\right)
r,t and u have horizontal asymptotes
A space traveler weights 540 N on earth. what will the traveler weigh on another planet whose radius is three times that of earth and whose mass is twice that of earth ?
The heaviest invertebrate is the giant squid, which is estimated to have a weight of about 2 tons spread out over its length of 70 feet. What is its weight in newtons.
I am at work right now and i was told to post this questionagain. Please help. Question # 4:The distribution coefficient, k= (conc. In ligroin/conc. In water),between ligroin and water for solute A is 7.5. What weight of Awould be removed from a solution of 10 grams of A in 100 mL ofwater by single extraction with 100 mL of ligroin? What weight of Awould be removed by four successive extractions with 25 mL portionsof ligroin? How much ligroin would be required to remove 98.5 % ofA in a single extraction?
The lens and mirror in the figure below are separated by 1.00 m and have focal lengths of +79.6 cm and -50.2 cm, respectively
How many seconds are in a 24 hour day? |
Find your quantum mechanics solutions
Get help with nuclear quantum mechanics
Recent questions in Quantum Mechanics
Wien's displacement law in frequency domain
When I tried to derive the Wien's displacement law I used Planck's law for blackbody radiation:
{I}_{\nu }=\frac{8\pi {\nu }^{2}}{{c}^{3}}\frac{h\nu }{{e}^{h\nu /{k}_{b}T}-1}
Asking for maximum:
\frac{d{I}_{\nu }}{d\nu }=0:\text{ }0=\frac{\mathrm{\partial }}{\mathrm{\partial }\nu }\left(\frac{{\nu }^{3}}{{e}^{h\nu /{k}_{b}T}-1}\right)=\frac{3{\nu }^{2}\left({e}^{h\nu /{k}_{b}T}-1\right)-{\nu }^{3}h/{k}_{b}T\cdot {e}^{h\nu /{k}_{b}T}}{\left({e}^{h\nu /{k}_{b}T}-1{\right)}^{2}}
It follows that numerator has to be
0
and looking for
\nu >0
3\left({e}^{h\nu /{k}_{b}T}-1\right)-h\nu /{k}_{b}T\cdot {e}^{h\nu /{k}_{b}T}=0
\gamma =h\nu /{k}_{b}T
3\left({e}^{\gamma }-1\right)-\gamma {e}^{\gamma }=0\to \gamma =2.824
Now I look at the wavelength domain:
\lambda =c/\nu :\text{ }\lambda =\frac{hc}{\gamma {k}_{b}}\frac{1}{T}
but from Wien's law
\lambda T=b
I expect that
hc/\gamma {k}_{b}
b
which is not:
\frac{hc}{\gamma {k}_{b}}=0.005099
b=0.002897
Why the derivation from frequency domain does not correspond the maximum in wavelength domain?
I tried to justify it with chain rule:
\frac{dI}{d\lambda }=\frac{dI}{d\nu }\frac{d\nu }{d\lambda }=\frac{c}{{\nu }^{2}}\frac{dI}{d\nu }
where I see that
c/{\nu }^{2}
does not influence where
d{I}_{\lambda }/d\lambda
Uses of Wien's law of displacement
Wiens's displacement law says
{\lambda }_{\text{max}}T=\text{a constant}
So if I have the
{\lambda }_{\text{max}}
, I can find the temperature of a star. But if I have the temperature, is there any point in calculating
{\lambda }_{\text{max}}
? What information does that give us of the star, besides temperature?
Deeper underlying explanation for color-shift in Wien's Law?
Suppose we have a blackbody object, maybe a star or a metal (although I understand neither of these are actually blackbody objects, to some extent my understanding is they can approximate one). According to Wien's Law, as temperature increases, peak wavelength will decrease, so the color we observe will "blueshift." Despite having reached Wien's Law in my research, my understanding is that this is in fact not an answer to why the blueshift occurs, but just a statement that the blueshift does occur.
So, is there some deeper reason why blackbodies peak wavelength emitted decreases with temperature?
If an electron which already possesses some kinetic energy of
X\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{V}
is further accelerated through a potential difference of
X\prime \phantom{\rule{thinmathspace}{0ex}}\mathrm{V}
, then is it correct to state that the electron now has a total kinetic energy of
\left(X+X\prime \right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{V}
? Using this information, am I allowed to substitute this value in the de Broglie equation to find the wavelength of the electron?
E
in below equation? Does it represent total energy or kinetic energy(De-broglie equation for uncharged particle like neutron.)
De-broglie equation for uncharged particle:
\lambda =\frac{h}{\sqrt{2mE}}
\lambda
= wavelength
= planks constant
m
= mass of uncharged particles
Wien's Displacement Law and reradiation of LWIR by
\mathrm{C}{\mathrm{O}}_{2}
Consider Wien's Displacement Law (if I understand correctly): λ = b / T Where, λ = Peak Wavelength b = 0.028977 mK (Wien's constant) T = Temperature
According to this law
\mathrm{C}{\mathrm{O}}_{2}
molecules can only absorb and reradiate the Long Wave I.R. frequency (radiated from Earth) in their 15 micrometre (μm) spectrum at a temperature of -80 degrees Celcius. The link with temperature seems crucial to me since the only part of our athmosphere cold enough for this to happen is the Mesosphere at about 50-80 km from Earth. Troposphere and Stratosphere are not cold enough (coldest temperature is about -55 degrees Celsius). In these parts of our atmosphere it is just not cold enough for
\mathrm{C}{\mathrm{O}}_{2}
to reradiate I.R. wavelengths back to Earth. Then why is
\mathrm{C}{\mathrm{O}}_{2}
considered a greenhouse gas in our Tropo- or Stratosphere? Am I missing out on something here?
In the Mesosphere, however, temperature can drop to over -100 degrees Celsius. In this part of the atmosphere
\mathrm{C}{\mathrm{O}}_{2}
can reradiate LWIR back to Earth. But the problem is that the air is so thin, there are hardly any molecules of
\mathrm{C}{\mathrm{O}}_{2}
What am I missing here in my reasoning?
Is the relation c=νλ valid only for Electromagnetic waves?
What is the validity of the relation
c=\nu \lambda
? More specifically, is this equation valid only for Electromagnetic waves?
I read this statement in a book, which says:
"de Broglie waves are not electromagnetic in nature, because they do not arise out of accelerated charged particle."
This seems correct, but arises a doubt in my mind.
Suppose I find out the wavelength of a matter wave (or de Broglie wave) using de Broglie's wave equation:
\lambda =\frac{h}{p}
Now, can I use
c=\nu \lambda
to find out the frequency of the wave?
In a quantum mechanics script I'm reading, the Schrödinger equation is "derived" (more precisely, motivated) by the De Broglie hypothesis. It starts at
\lambda =\frac{2\pi h}{p}
\omega =\frac{E}{h}
then takes the partial derivatives of the wave
\mathrm{\Psi }\left(x,t\right)=\mathrm{\Psi }\left(0,0\right){e}^{\frac{2\pi ix}{\lambda }-it\omega }
\begin{array}{}\text{(1)}& \frac{\mathrm{\partial }}{\mathrm{\partial }x}\mathrm{\Psi }\left(x,t\right)=\frac{ip}{h}\mathrm{\Psi }\left(x,t\right)\end{array}
\frac{\mathrm{\partial }}{\mathrm{\partial }t}\mathrm{\Psi }\left(x,t\right)=-i\frac{E}{h}\mathrm{\Psi }\left(x,t\right)
With the non-relativistic free particle
E=\frac{1}{2m}{p}^{2}
\begin{array}{}\text{(2)}& ih\frac{\mathrm{\partial }}{\mathrm{\partial }t}\mathrm{\Psi }\left(x,t\right)=E\mathrm{\Psi }\left(x,t\right)=\frac{{p}^{2}}{2m}\mathrm{\Psi }\left(x,t\right)\end{array}
From there, they miraculously get the time-dependent Schrödinger equation. I cannot understand this step. If I insert formula (1) for
{p}^{2}
in (2), I get something with
\left(\frac{\mathrm{\partial }}{\mathrm{\partial }x}\mathrm{\Psi }{\right)}^{2}
, which is not the second partial derivative of
\mathrm{\Psi }
x
What is the most intense wavelength and frequency in the spectrum of a black body?
Planck's Law is commonly stated in two different ways:
{u}_{\lambda }\left(\lambda ,T\right)=\frac{2h{c}^{2}}{{\lambda }^{5}}\frac{1}{{e}^{\frac{hc}{\lambda kT}}-1}
{u}_{\nu }\left(\nu ,T\right)=\frac{2h{\nu }^{3}}{{c}^{2}}\frac{1}{{e}^{\frac{h\nu }{kT}}-1}
We can find the maximum of those functions by differentiating those equations with respect to
\lambda
\nu
, respectively. We get two ways to write Wien's Displacement Law:
{\lambda }_{\text{peak}}T=2.898\cdot {10}^{-3}m\cdot K
\frac{{\nu }_{\text{peak}}}{T}=5.879\cdot {10}^{10}Hz\cdot {K}^{-1}
{\lambda }_{\text{peak}}\ne \frac{c}{{\nu }_{\text{peak}}}
. So what frequency or wavelength is actually detected by an optical instrument most intensely when analyzing a black body? If they are
{\lambda }_{\text{peak}}
{\nu }_{\text{peak}}
, how is
{\lambda }_{\text{peak}}\ne \frac{c}{{\nu }_{\text{peak}}}
According to de Broglie's wave-particle duality, the relation between electron's wavelength and momentum is
\lambda =h/mv
The proof of this is given in my textbook as follows:
1.De Broglie first used Einstein's famous equation relating matter and energy,
E=m{c}^{2},
E=
energy,
m=
c=
speed of light.
2.Using Planck's theory which states every quantum of a wave has a discrete amount of energy given by Planck's equation,
E=h\nu ,
E=
h=
Plank's constant (
6.62607×{10}^{-34}\phantom{\rule{mediummathspace}{0ex}}\mathrm{J}\phantom{\rule{mediummathspace}{0ex}}\mathrm{s}
\nu =
frequency.
3.Since de Broglie believes particles and wave have the same traits, the two energies would be the same:
m{c}^{2}=h\nu .
m{c}^{2}=h\nu .
4.Because real particles do not travel at the speed of light, de Broglie substituted
v
, velocity, for
c
, the speed of light:
m{v}^{2}=h\nu .
I want a direct proof without substituting
v
c
. Is it possible to prove directly
\lambda =h/mv
without substituting
v
c
\lambda =h/mc
Jaylene Duarte 2022-05-14 Answered
c
needs to be added in numerator and denominator while solving a de Broglie equation word problem?
Calculate the de Broglie wavleength of a 0.05 eV ("theremal") neutron.Making a nonrelativistic calculation,
\lambda =\frac{h}{p}=\frac{h}{\sqrt{2{m}_{0}K}}=\frac{hc}{\sqrt{2\left({m}_{0}{c}^{2}\right)K}}=\frac{12.4×{10}^{3}\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{V}\cdot \text{Å}}{\sqrt{2\left(940×{10}^{6}\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{V}\right)\left(0.05\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{V}\right)}}=1.28\phantom{\rule{thinmathspace}{0ex}}\text{Å}
I am confused as to why it is necessary to add
c
in the numerator and denominator (red equation) while solving a word problem like this?
What is the difference between Wien's Displacement Law for peak frequency vs peak wavelength?
While doing research for a high school report I came across the fact that WDL actually has two forms, one for peak frequency and one for peak wavelength, and that these two forms are not the same and can not be used interchangeably.
My question is why peak frequency isn't the same as peak wavelength? That is, since wavelength is directly determined by frequency (since frequency = speed of light divided by wavelength), there is a one-to-one correspondence between a given wavelength and its frequency. Therefore why doesn't a peak in frequency correspond to a peak in wavelength, and visa versa (meaning that the two forms of WDL could be used interchangeably)?
I know that this question was already posted elsewhere, but I did not understand the answers. Since I am a high school student, complicated terminology can fly right over my head, so I would greatly appreciate it if someone could take the time to explain it clearly and simply (i.e. no monster equations).
Blackbody radiation, de Broglie equation, and lightwaves to be shifted left
I'm having a hard time figuring this out.
1.Say we heated a lead ball to 1,000 Kelvin. Not all of the particles are at the exact same temperature--some parts are a little hotter, some are a little cooler. But for now, let’s assume that it follows a normal distribution that is centered at 1,000K.
2.The heat (or movement of the molecules) causes it to emit light.
3.Because the light photons have to be discrete (you can't have a 1/2 photon), this causes the observed light wavelengths to be shifted left.
4.This means we observe more red light than we might otherwise expect.
5.This is a long wind up to my specific question--does a particle vibrating at a specific frequency emit light at the same frequency? (i.e. a particle vibrating at 4.3 MhZ emits light at 4.3 Mhz). Because it seems like the whole thing hinges on that.
I mention this because I asked a physics teacher this, and he said, “No, the particles emit light following the de Broglie equation.” This would mean that the light emitted ignores the frequency and instead is based solely on its momentum. But, if this were true, then I would assume it would emit light in a standard distribution of frequencies as opposed to the left-skewed distribution that is actually observed.
De-Broglie equation for other particles except photon
If E=hf is applicable for electron and other particles, the De Broglie wavelength should be λ=hv/pc. Because, mc^2=hf which implies mc^2=hc/λ which implies m=h/λc and thus λ=hv/pc. But I have found in my text book that λ=h/p is applicable not only for photon but also for all particle. But how can λ=h/p=h/mv be applicable for all particle?
Wavelength and relativity
From de Broglie equation λ=h/p. But p=mv and velocity is a relativistic quantity so also wavelength is relative ? In other words does wavelength depends on the reference frame ?
Special Relativity, Louis de Broglie Equation Dilemma
While learning atomic structure I stumbled upon a very unusual doubt.
As we know that the energy of a wave is given by the equation:
E=\frac{hc}{\lambda }
and Louis de broglie wave equation is given by the equation
{\lambda }_{B}=\frac{h}{p}
. My doubt is that, that is
{\lambda }_{B}=\lambda
. Do the
{\lambda }_{B},\lambda
represent the same thing
?
My teacher equated
E=\frac{hc}{\lambda }
E=mc²
\frac{hc}{\lambda }=mc²
and rearranged to form
\lambda =\frac{h}{mc}
and then replaced
\lambda
{\lambda }_{Β}
c
v
for general formula and derived the Louis de broglie equation. This created my doubt in first place and I created another doubt that whether the equation for energy of wave is valid for relativistic equation of
E=mc²
E=mc²
is for particles while the former is for waves.
Is my understanding correct
?
Please help and thanks in advance
!
Which formula for the de Broglie wavelength of an electron is correct?
So, I have my exams in physics in a week, and upon reviewing I was confused by the explanation of de Broglie wavelength of electrons in my book. Firstly, they stated that the equation was:
\lambda =\frac{h}{p}
h
p
is the momentum of the particle. Later, however, when talking about electron diffraction and finding the angles of the minima, the author gave the formula equivalent to that for light:
\lambda =\frac{hc}{E}
. Now, what I don't understand is if it is simply a mistake made by the author, or whether a different formula have to be used for electron diffraction, as the two formulae are very clearly not equivalent. In the latter case, I don't understand why the formula would be different. I greatly appreciate the help, as the exams are really close, and I would like to make sure I get this right!
Edit: I was told that pictures of text are taking away from the readability of the posts, and thus they were removed. Essentially, the difference between the two cases are that in the first case, the proton did not have any significantly large kinetic energy, while in the second example, the kinetic energy was
400\text{ }MeV
Wien's fifth power Law and Stephan Boltzmann's fourth power laws of emissive power
Wien's fifth power law says that emissive power is proportional to the temperature raised to the fifth power. On the other hand, the Stefan–Boltzmann law says emissive power is proportional to the temperature raised to the fourth power. How can both of these be true?
Wien's Displacement Law for real bodies
It is known that for perfect blackbodies,
\lambda T=c
\lambda =\text{peak wavelength}
T=\text{Absolute temperature}
c=\text{Wien's constant}
But this is for perfect blackbodies only, which have no theoretical existence. Does a similar formula exist for real bodies, which expresses
\lambda T
in terms of its emissitivity
ϵ
? I googled it, but found no relevant results.
Area under Wien's displacement graph
Why does the area under Wien's displacement graph give Stefan-Boltzmann law for a black body?
I couldn't find any proof of this. (I could just find this expression). I am not aware of the function of Wien's displacement graph as well (I just know that it is between Intensity and wavelength emitted by a black body).
Is there a mathematical way to prove this? |
Solve the systems of equations by substitute 4x - 5y
Solve the systems of equations by substitute 4x - 5y = -17, 3x + 2y = -7
Solve the systems of equations by substitute
\left\{\begin{array}{l}4x-5y=-17\\ 3x+2y=-7\end{array}
Answered on photo
Find the weighted average of a data set where 10 has a weight of 5, 30 has a weight of 3, and 50 has a weight of 2.
Fill in the bla
so the resulting statement is true.
3{x}^{2}+2{y}^{2}=35
4{x}^{2}+3{y}^{2}=48
by the addition method, we can eliminate
{x}^{2}
by the multiplying the first equation by -4 and the second equation by __________ and then adding the equations
Katrina has a 120 dollar gift card. She paid 60 dollar for a new suitcase and wants to spend the rest on 10 dollars pairs of flip flops. What is the maximum number of pairs of flip flops she could purchase?
Including 5% sales tax, an inn charges $252 per night.
Find the inn’s nightly cost before the tax is added.
\frac{-2+3x}{1+5x}
{x}^{2}=-1
Solving systems of non-linear equation
\left\{\begin{array}{l}x+{y}^{2}+3=0\\ 2{x}^{2}+{y}^{2}-4=0\end{array}
Solve the equations and inequalities. Write the answers to the inequalities in interval notation if possible. (A)
|7x+4|+11=2
|7x+4|+11<2
|7x+4|+11>2 |
Relativity questions, solved and explained
Relativity questions and answers
Recent questions in Relativity
We have two particles
A
B
flying with different speed. Now the rest frame for
A
B
is not the same. But is the inertial frame the same? Or would be the inertial frame the same if we have a third fix point
C
and we would observe them both from
C
For me it seems logical to have
3
frame of references. From the point of view of
A
, from the point of view of
B
and from an independent point of view
C
. Is this correct?
What is a homogeneous and isotropic frame of reference?
I have heard that inertial frames of reference in the context of special relativity are both isotropic and homogeneous. I know what isotropic and homogeneous mean in a general context, but what do they mean when relating to a frame of reference?
If a body
A
is moving with a constant velocity v and an observer on that body
A
observes another body
B
to be at rest then the kinetic energy of
B
is zero. So is energy dependent on the frame of reference if so then how is the conservation of energy stated?
It is said that we can't study quantum gravity because gravity is a weak force. But gravity and acceleration are the same. Why can't we study quantum gravity in a strongly accelerated frame of reference?
The relativistic energy-momentum equation is:
{E}^{2}=\left(pc{\right)}^{2}+\left(m{c}^{2}{\right)}^{2}.
pc=Ev/c
, so we get:
E=m{c}^{2}/\left(1-{v}^{2}/{c}^{2}{\right)}^{1/2}.
Now, accelerating a proton to near the speed of light:
\begin{array}{}0.990000000000000& c=>& 0.0000000011& J=& 0.01& TeV\\ 0.999000000000000& c=>& 0.0000000034& J=& 0.02& TeV\\ 0.999900000000000& c=>& 0.0000000106& J=& 0.07& TeV\\ 0.999990000000000& c=>& 0.0000000336& J=& 0.21& TeV\\ 0.999999000000000& c=>& 0.0000001063& J=& 0.66& TeV\\ 0.999999900000000& c=>& 0.0000003361& J=& 2.10& TeV\\ 0.999999990000000& c=>& 0.0000010630& J=& 6.64& TeV\\ 0.999999999000000& c=>& 0.0000033614& J=& 20.98& TeV\\ 0.999999999900000& c=>& 0.0000106298& J=& 66.35& TeV\\ 0.999999999990000& c=>& 0.0000336143& J=& 209.83& TeV\\ 0.999999999999000& c=>& 0.0001062989& J=& 663.54& TeV\\ 0.999999999999900& c=>& 0.0003360908& J=& 2,097.94& TeV\\ 0.999999999999990& c=>& 0.0010634026& J=& 6,637.97& TeV\\ 0.999999999999999& c=>& 0.0033627744& J=& 20,991.10& TeV\end{array}
If the LHC is accelerating protons to
7TeV
it means they're traveling with a speed of
0.99999999c
Is everything above correct?
A rotating frame of reference (since you rotate your BEC to generate these), where the energy is given by:
\stackrel{~}{E}\left[\mathrm{\Psi }\right]=E\left[\mathrm{\Psi }\right]-L\left[\mathrm{\Psi }\right]\cdot \mathrm{\Omega },
\mathrm{\Omega }
is the rotational velocity which you apply to the BEC.
Now the argument that the term
L\left[\mathrm{\Psi }\right]\cdot \mathrm{\Omega }
should be substracted comes from the fact that in a rotating frame your system loses that fraction of rotational energy. Now I was wondering if anyone knew where the form of
L\left[\mathrm{\Psi }\right]\cdot \mathrm{\Omega }
came from?
I know that in a rotating frame of reference you have that
\stackrel{\to }{v}={\stackrel{\to }{v}}_{r}+\stackrel{\to }{\mathrm{\Omega }}×\stackrel{\to }{r}
. If you fill this in into the kinetic energy and use some basic definitions, you get that the extra effect of rotation is given by:
\frac{1}{2}I{\mathrm{\Omega }}^{2}=\frac{1}{2}J\mathrm{\Omega }.
This yields half of the value that is used in the book, is there something that I'm missing or not seeing right ?
instead of assuming that the velocity
c
is a maximal velocity, proving that while assuming
E=m{c}^{2}
What is the coordinate in this system then and how do they all connect to each other? I've read the mentioning of an observer and the observer's state of motion and I don't understand how that relates to a frame of reference.
If the light velocity is a vector quantity, why vector addition cannot be applied to it? Or the light velocity is not a vector quantity?
An inertial frame of reference is a frame of reference which is not accelerating. All laws of physics are the same measured from an inertial frame of reference. A rest frame is a frame of reference where a particle is at rest.
Does this mean that a rest frame could possibly be non-inertial (that is, accelerating), but the particle with respect to his rest frame would have a velocity of
0
? What kind of velocity? And what exactly would it mean to be at rest with respect to a possibly accelerating frame of reference?
What are the differences and relations between rest frame and inertial reference frame?
Take the following gedankenexperiment in which two astronauts meet each other again and again in a perfectly symmetrical setting - a hyperspherical (3-manifold) universe in which the 3 dimensions are curved into the 4. dimension so that they can travel without acceleration in straight opposite directions and yet meet each other time after time.
On the one hand this situation is perfectly symmetrical - even in terms of homotopy and winding number. On the other hand the Lorentz invariance should break down according to GRT, so that one frame is preferred - but which one?
So the question is: Who will be older? And why?
And even if there is one prefered inertial frame - the frame of the other astronaut should be identical with respect to all relevant parameters so that both get older at the same rate. Which again seems to be a violation of SRT in which the other twin seems to be getting older faster/slower...
How should one find out what the preferred frame is when everything is symmetrical - even in terms of GRT
In physics problems, the earth is usually considered to be an inertial frame. The earth has a gravitational field and the second postulate of the general theory of relativity says:
"In the vicinity of any point, a gravitational field is equivalent to an accelerated frame of reference in gravity-free space (the principle of equivalence)."
Does this mean that accelerating frames of reference can be inertial?
Observer A and B are at the same "depth" in a gravity well. Observer B then descends into the well. A will observe B's time as going slower than their own. B will observe A's time as going faster than their own.
What happens if B were to ascend the well back to A's depth, would B's local time speed back up to the same rate as A's, but B would be younger (relative to A)?
What about the paradox caused by relative motion (ignoring gravity)? If A is moving relative to B, A and B will both observe the other's time as going slower. If A and B were together initially, then B moves away and returns, do their clocks agree?
Formula for the Bekenstein bound
S\le \frac{2\pi kRE}{\hslash c}
E
is the total mass-energy. That seems to imply that the presence of a black hole in the region is dependent on an observer's frame of reference. Yet, my understanding is that the Bekenstein bound is the maximum entropy that any area can withstand before collapsing into a black hole.
Does this mean that the existence of black holes is observer dependent? Or that even if an observer does not report a black hole in their frame, one is guaranteed to form there in the future?
For a car that is accelerating linearly, the non-inertial frame of reference is the driver in the car where from his reference frame, the car is stationary. It is so called stationary because the non-inertial frame of reference has the same acceleration as the car. Is like the car's acceleration "transform" the driver frame of reference into a non-inertial. That's why in the non-inertial frame of reference, there is no force acting on the car.
But when the car is driving in circles at a constant speed, in the non-inertial frame of reference there is a force acting on the car, which is the centripetal force. Why isn't this frame of reference like the above, not having the acceleration found in their each respective inertial reference frame? Why can't we have a non-inertial reference frame(due to rotation) whereby there is no centripetal force, subsequently eradicating the need for a centrifugal force?
Why is it hopeless to view differential geometry as the limit of a discrete geometry?
Classical mechanics can be understood as the limit of relativistic mechanics
R{M}_{c}
c\to \mathrm{\infty }
Classical mechanics can be understood as the limit of quantum mechanics
Q{M}_{h}
h\to 0
As a limit of which discrete geometry
{\mathrm{\Gamma }}_{\lambda }
can classical mechanics be understood for
\lambda \to 0
If there were a light cone centered at some point P, and you were to look at that light cone from different reference frames, would it change its shape? I know that points inside and outside the light cone would remain inside/outside of the light cone in every frame, but does the light cone itself shift? If it does, how would it shift?
Imaging that the speed of that spacecraft is almost the speed of light "
{c}^{-1}\text{ }\text{m/s}
". But no acceleration. Meaning it's an inertial frame of reference right? Will you be able to throw a ball in the direction of movement? What about the direction inverse to the movement? Does that means that there is a direction of movement in an inertial frame of reference?
Is this possible to create a inertial frame of reference in the earth? How it is possible?
lasquiyas5loaa 2022-05-08 Answered
The reason the galaxy does not spin around me (relative to me) at a trillion times the speed of light when I do a pirouette is the inertial frame of reference to which I am confined along with the rest of humanity.
Where does this frame of reference end?
To the best of my limited understanding, gravity doesn't just stop at some point. Anything that has mass has gravity; gravitational waves (or whatever) extend from the mass spherically. They weaken as they get further away from the source, they get pushed around by other, more powerful, gravity waves, they become progressively weaker as the distance increases, but do they ever vanish completely, i.e. get reduced to absolute zero? Or do they become weaker and weaker forever, without zeroing out? And if so, where does a frame of reference end? |
Gold candlesticks - OSRS Wiki
The RuneScape Wiki also has an article on: rsw:Gold candlesticks
This article is about the construction item. For the useless item for the Priest in Peril quest, see Golden candle.
The smoke goes up to the gods.
Gold candlesticks? (edit) ? (edit)File:Gold candlesticks built.pngFile:Gold candlesticks (lit) built.png File:Gold candlesticks icon.pngTorch31 May 2006 (Update)575746 xp? (edit)ChapelLamp spaceNo? (edit)The smoke goes up to the gods.Construction? (edit)1320613207? (edit)132061320780728072ConstructionVersions: 2
SMW Subobject for unlitIcon: File:Gold candlesticks icon.pngUses infobox: ConstructionObject ID: 13206Image: File:Gold candlesticks built.pngConstruction level: 57Item ID: 8072Uses skill: ConstructionIs variant of: Gold candlesticks
SMW Subobject for litIcon: File:Gold candlesticks icon.pngUses infobox: ConstructionObject ID: 13207Image: File:Gold candlesticks (lit) built.pngConstruction level: 57Item ID: 8072Uses skill: ConstructionIs variant of: Gold candlesticks
The gold candlesticks can be built in the Lamp space of the Chapel in a player-owned house. It requires 57 Construction to build and when built, it gives 46 experience. The player must have a hammer and a saw in their inventory to build it.
Gold candlesticks do not improve Prayer experience gained when using bones with it, so they are purely cosmetic.
2 Burn duration
Gold bar 6 510
Candle 6 1,764
Gold candlesticks 1 N/A
Burn duration[edit | edit source]
The duration in which burners stay lit are random and depend on the player's Firemaking level. The duration is
{\displaystyle 200+firemakingLevel+A}
ticks, in which
{\displaystyle A}
is a random number between 0 and
{\displaystyle firemakingLevel-1}
At level 50, a burner would last between 250 and 299 ticks (150-179 seconds).
Burners will not stay lit when all players have left the house. However, if there are other players in the house when the owner leaves through the entrance portal; they will stay lit as long as the owner returns to the house within 5 seconds.
^ Jagex. Mod Ash's Twitter account. 16 October 2018. (Archived from the original on 28 May 2020.) Mod Ash: "200 cycles + your FM level + random number between 0 and your FM level, inclusive of 0 but not inclusive of your FM level. Really. So at level 70 that'd be 200 + 70 + (random number 0-69 inclusive), i.e. between 270 and 339."
Oak burners
Gnome child icon
Bob icon
Retrieved from ‘https://oldschool.runescape.wiki/w/Gold_candlesticks?oldid=14253120’ |
S-Function Concepts - MATLAB & Simulink - MathWorks Switzerland
Setting Sample Times and Offsets
Valid C MEX S-Function Sample Times
Valid Level-2 MATLAB S-Function Sample Times
Guidelines for Choosing a Sample Time
Direct feedthrough means that the output (or the variable sample time for variable sample time blocks) is controlled directly by the value of an input port signal. Typically, an S-function input port has direct feedthrough if
The output function (mdlOutputs) is a function of the input u. That is, there is direct feedthrough if the input u is accessed by mdlOutputs. Outputs can also include graphical outputs.
The “time of next hit” function (mdlGetTimeOfNextVarHit) of a variable sample time S-function accesses the input u.
An example of a system that requires its inputs (that is, has direct feedthrough) is the operation
y=k×u,
where u is the input, k is the gain, and y is the output.
An example of a system that does not require its inputs (that is, does not have direct feedthrough) is the simple integration algorithm
y=x,
\stackrel{˙}{x}=u,
where x is the state,
\stackrel{˙}{x}
is the state derivative with respect to time, u is the input, and y is the output. Simulink® integrates the variable
\stackrel{˙}{x}.
It is very important to set the direct feedthrough flag correctly because it affects the execution order of the blocks in your model and is used to detect algebraic loops (see Algebraic Loop Concepts in Using Simulink). If the simulation results for a model containing your S-function do not converge, or the simulation fails, you may have the direct feedthrough flag set incorrectly. Try turning on the direct feedthrough flag and setting the Algebraic loop solver diagnostic to warning (see the Algebraic loop option on the Model Configuration Parameters: Diagnostics reference page in Simulink Graphical User Interface). Subsequently running the simulation displays any algebraic loops in the model and shows if the engine has placed your S-function within an algebraic loop.
You can write your S-function to support arbitrary input dimensions. In this case, the Simulink engine determines the actual input dimensions when the simulation is started by evaluating the dimensions of the input vectors driving the S-function. Your S-function can also use the input dimensions to determine the number of continuous states, the number of discrete states, and the number of outputs.
A dynamically sized input can have a different size for each instance of the S-function in a particular model or during different simulations, however the input size of each instance of the S-function is static over the course of a particular simulation.
A C MEX S-function and Level-2 MATLAB® S-function can have multiple input and output ports and each port can have different dimensions. The number of dimensions and the size of each dimension can be determined dynamically.
For example, the following illustration shows two instances of the same S-Function block in a model.
The upper S-Function block is driven by a block with a three-element output vector. The lower S-Function block is driven by a block with a scalar output. By specifying that the S-Function block has dynamically sized inputs, the same S-function can accommodate both situations. The Simulink engine automatically calls the block with the appropriately sized input vector. Similarly, if other block characteristics, such as the number of outputs or the number of discrete or continuous states, are specified as dynamically sized, the engine defines these vectors to be the same length as the input vector.
See Input and Output Ports for more information on configuring S-function input and output ports.
Both Level-2 MATLAB and C MEX S-functions provide the following sample time options, which allow for a high degree of flexibility in specifying when an S-function executes:
Continuous sample time — For S-functions that have continuous states and/or nonsampled zero crossings (see Simulation Phases in Dynamic Systems for an explanation of zero crossings). For this type of S-function, the output changes in minor time steps.
Continuous, but fixed in minor time step sample time — For S-functions that need to execute at every major simulation step, but do not change value during minor time steps.
Discrete sample time — If the behavior of your S-function is a function of discrete time intervals, you can define a sample time to control when the Simulink engine calls the S-function. You can also define an offset that delays each sample time hit. The value of the offset cannot exceed the corresponding sample time.
A sample time hit occurs at time values determined by the formula
where the integer n is the current simulation step. The first value of n is always zero.
If you define a discrete sample time, the engine calls the S-function mdlOutputs and mdlUpdate routines at each sample time hit (as defined in the previous equation).
Variable sample time — A discrete sample time where the intervals between sample hits can vary. At the start of each simulation step, S-functions with variable sample times are queried for the time of the next hit.
Inherited sample time — Sometimes an S-function has no inherent sample time characteristics (that is, it is either continuous or discrete, depending on the sample time of some other block in the system). In this case, you can specify that the sample time is inherited. A simple example of this is a Gain block that inherits its sample time from the block driving it.
An S-function can inherit its sample time from
The driving block
The destination block
The fastest sample time in the system
To specify an S-function sample time is inherited, use -1 in Level-2 MATLAB S-functions and INHERITED_SAMPLE_TIME in C MEX S-functions as the sample time. For more information on the propagation of sample times, see How Propagation Affects Inherited Sample Times in the Simulink User's Guide.
S-functions can be either single or multirate; a multirate S-function has multiple sample times.
Sample times are specified in pairs in this format: [sample_time, offset_time].
The valid sample time pairs for a C MEX S-function are
[discrete_sample_time_period, offset]
CONTINUOUS_SAMPLE_TIME = 0.0
FIXED_IN_MINOR_STEP_OFFSET = 1.0
VARIABLE_SAMPLE_TIME = -2.0
and variable names in italics indicate that a real value is required.
Alternatively, you can specify that the sample time is inherited from the driving block. In this case, the C MEX S-function has only one sample time pair, either
INHERITED_SAMPLE_TIME = -1.0
The valid sample time pairs for a Level-2 MATLAB S-function are
[0 offset] % Continuous sample time
[discrete_sample_time_period, offset] % Discrete sample time
[-1, 0] % Inherited sample time
[-2, 0] % Variable sample time
where variable names in italics indicate that a real value is required. When using a continuous sample time, an offset of 1 indicates the output is fixed in minor integration time steps. An offset of 0 indicates the output changes at every minor integration time step.
Use the following guidelines for help with specifying sample times:
A continuous S-function that changes during minor integration steps should register the [CONTINUOUS_SAMPLE_TIME, 0.0] sample time.
A continuous S-function that does not change during minor integration steps should register the [CONTINUOUS_SAMPLE_TIME, FIXED_IN_MINOR_STEP_OFFSET] sample time.
A discrete S-function that changes at a specified rate should register the discrete sample time pair, [discrete_sample_time_period, offset], where
0.0 ≤ offset < discrete_sample_period
A discrete S-function that changes at a variable rate should register the variable-step discrete sample time.
In a C MEX S-function, the mdlGetTimeOfNextVarHit routine is called to get the time of the next sample hit for the variable-step discrete task. In a Level-2 MATLAB S-function, the NextTimeHit property is set in the Outputs method to set the next sample hit.
If your S-function has no intrinsic sample time, you must indicate that your sample time is inherited. There are two cases:
An S-function that changes as its input changes, even during minor integration steps, should register the [INHERITED_SAMPLE_TIME, 0.0] sample time.
An S-function that changes as its input changes, but does not change during minor integration steps (that is, remains fixed during minor time steps), should register the [INHERITED_SAMPLE_TIME, FIXED_IN_MINOR_STEP_OFFSET] sample time.
The Scope block is a good example of this type of block. This block runs at the rate of its driving block, either continuous or discrete, but never runs in minor steps. If it did, the scope display would show the intermediate computations of the solver rather than the final result at each time point.
See Specify S-Function Sample Times for information on implementing different types of sample times in S-functions. |
3-D bar graph - MATLAB bar3 - MathWorks América Latina
Create 3-D Bar Graph from Vector Data
Create 3-D Bar Graph from Matrix Data
Specify Bar Width and Style
Customize Series Colors
3-D bar graph
bar3(y,z)
bar3(___,width)
bar3(___,style)
bar3(___,color)
bar3(ax,___)
b = bar3(___)
bar3(z) creates a 3-D bar graph for the elements of z. Each bar corresponds to an element in z.
To plot a single series of bars, specify z as a vector. For a vector of length m, the function plots the bars on a y-axis ranging from 1 to m.
To plot multiple series of bars, specify z as a matrix with one column for each series. For an m-by-n matrix, the function plots the bars on an x-axis ranging from 1 to n and a y-axis ranging from 1 to m.
bar3(y,z) creates a bar graph of the elements in z at the y-values specified in y. If z is a matrix, elements from the same row in z appear at the same location along the y-axis.
bar3(___,width) sets the width of the bars along the x- and y-axes and controls the separation of bars within a group. By default, width is 0.8 and the bars have a slight separation. If width is 1, the bars within a group touch one another.
bar3(___,style) specifies the style of the bars, where style can be 'detached', 'grouped', or 'stacked'. The default mode of display is 'detached'.
bar3(___,color) displays all bars using the color specified by color. For example, use 'r' to specify all red bars.
bar3(ax,___) plots into the axes specified by ax instead of into the current axes (gca). The option ax can precede any of the input argument combinations in the previous syntaxes.
b = bar3(___) returns one or more Surface objects. If z is a vector, then bar3 creates one Surface object. If z is a matrix, then bar3 returns a Surface object for each series. Use b to set properties of the bars after displaying the bar graph. For a list of properties, see Surface Properties.
Specify z as a vector of five values. Plot these values as a series of 3-D bars, with the height of each bar corresponding to a value in z and its y-axis location corresponding to the index of that value.
z = [50 40 30 20 10];
bar3(z);
Specify z as a matrix. Create a 3-D bar graph of z with each series corresponding to a column in z.
z = [1 4 7; 2 5 8; 3 6 9; 4 7 10];
Specify y as a vector of y-axis locations for the bars in z. Plot the bars at the specified y-axis locations.
y = [1950 1960 1970 1980 1990];
z = [16 8 4 2 1];
Specify z as a matrix with three series. Plot z with the data for each row grouped together by using the 'grouped' style. To eliminate space between bars of the same group, set width to 1.
z = [70 50 33 10; 75 55 35 15; 80 60 40 20];
bar3(z,1,'grouped')
Create a set of y-coordinates ranging from 0 to
\pi
. Plot sine functions of the y-values as a 3-D bar graph. Use the color specification 'r' to make the bars red.
y = 0:pi/16:pi;
z = [sin(y')/4 sin(y')/2 sin(y')];
bar3(y,z,1,'r')
Plot data matrix z with the bar style 'stacked' and bar width 0.5. Store the returned Surface objects as b. In this case, b has 4 elements, with one for each column in z.
Note that bar3 colors each series based on the default colormap.
z = [19 30 21 30; 40 16 32 12];
b = bar3(z,0.5,'stacked');
Customize the color of each series by setting the FaceColor of the corresponding object in b.
b(2).FaceColor = 'white';
Starting in R2019b, you can display a tiling of plots using the tiledlayout and nexttile functions. Call the tiledlayout function to create a 1-by-2 tiled chart layout. Call the nexttile function to create the axes objects ax1 and ax2. Create separate line plots in the axes by specifying the axes object as the first argument to bar3.
Display a stacked 3-D bar graph in the left axes. In the right axes, display a grouped bar graph of the same data.
z = [64 56 48; 49 42 35; 36 30 24];
% Left bar graph
bar3(ax1,z,0.5,'stacked')
% Right bar graph
bar3(ax2,z,1,'grouped')
z-coordinates, specified as a vector or matrix. The dimensions of z determine how the function displays your data. This table describes the most common situations.
How to Specify Y and Z
Specify y and z as vectors that are the same length. The values in y must be unique, but the values in z do not need to be unique.
Alternatively, specify just the z values.
Define vectors y and z, then create a 3-D bar chart.
z = [10 20 30 40];
Display multiple series of bars stacked along the x-axis.
Specify y as a vector and z as a matrix. The number of elements in y must match the number of rows in z.
Alternatively, specify just the z matrix.
Define vector y and matrix z, then create a 3-D bar chart with three series of bars.
z = [1 5 9;
y-coordinates, specified as a vector. The values of y do not need to be in order, but the size of y depends on the size of z and how you want to display your data. This table describes the most common situations.
Bar width, specified as a scalar representing a fraction of the total space available for each bar. The default of 0.8 means the bar width is 80% of the space from the previous bar to the next bar, with 10% of that space on each side. If the width is 1, then the bars within a group touch one another.
Example: bar3([1 2 3],0.5) creates bars that use 50% of the available space.
'detached' (default) | 'grouped' | 'stacked'
Group style, specified as 'detached', 'grouped', or 'stacked'. The group style determines how multiple series of bars display with respect to one another. If z is a vector, style does not affect graph appearance.
This table shows the result of each group style option.
Display each bar at its corresponding x and y value location.
Display each group as adjacent bars that are centered around their corresponding y value.
Axes object. If you do not specify an axes, then bar3 uses the current axes for the bar graph.
b — Displayed bars
vector of Surface objects
Displayed bars, returned as a vector of Surface objects. Use the elements in b to access and modify properties of a specific series of bars after it has been created. The number of Surface objects returned depends on the size of z. If z is a vector, then b is one Surface object. If z is a matrix, then b is a vector containing a Surface object for each column in z.
A series of bars consists of all the bars that correspond to a specific column in z. The bar3 function displays each series at a different location along the x-axis. By default, each series is a different color.
In the following image, the first series is highlighted.
A group consists of all the bars at a particular y location.
In the following image, the third group is highlighted.
The plotted bars can be modified by altering the properties of the returned Surface objects. For a list of Surface properties, see Surface Properties.
The returned surfaces have their XData, YData, and ZData properties configured to represent a set of vertices. To change the color or transparency of an individual bar or bar face, set the desired property for the appropriate vertices. For a series containing n bars, the resulting Surface object contains 6n vertices.
bar | barh | bar3h | histogram2 | tiledlayout | nexttile |
The RuneScape Wiki also has an article on: rsw:Combat levelThe RuneScape Classic Wiki also has an article on: classicrsw:Combat level
A player's, monster's, or non-player character's combat level indicates the difficulty in defeating them in combat.
Normally, aggressive non-player characters stop attacking you when you reach higher than double their combat level. There are exceptions, however - for example, any monsters in the Wilderness will be aggressive no matter what your combat level is.
By default, monsters with a higher combat level than a player cannot be attacked by using a left mouse click, requiring a right click to select the attack monster option. So, to be able to attack a Blue dragon with combat level 111 with a simple left click, a player would need a combat level of at least 111. However, this behaviour can be changed in the controls section of the Settings tab by changing the "NPC 'Attack' Options" setting.
All new players start out at combat level 3, and the maximum level for players in Old School RuneScape is level 126.
Calculating combat level[edit | edit source]
Simply[edit | edit source]
{\displaystyle {\begin{aligned}{\text{Base}}&={\frac {1}{4}}\times \left({\text{Defence}}+{\text{Hitpoints}}+\left\lfloor {\text{Prayer}}\times {\frac {1}{2}}\right\rfloor \right)\\{\text{Melee}}&={\frac {13}{40}}\times ({\text{Attack}}+{\text{Strength}})\\{\text{Range}}&={\frac {13}{40}}\times \left\lfloor {\text{Ranged}}\times {\frac {3}{2}}\right\rfloor \\{\text{Mage}}&={\frac {13}{40}}\times \left\lfloor {\text{Magic}}\times {\frac {3}{2}}\right\rfloor \\{\text{Final}}&=\lfloor {\text{Base}}+{\textrm {max}}({\text{Melee}},{\text{Range}},{\text{Mage}})\rfloor \end{aligned}}}
Combined formulae[edit | edit source]
{\displaystyle {\begin{aligned}{\text{Melee Combat Level}}&=\left\lfloor {\frac {1}{4}}\times \left({\text{Defence}}+{\text{Hitpoints}}+\left\lfloor {\text{Prayer}}\times {\frac {1}{2}}\right\rfloor \right)+{\frac {13}{40}}\times ({\text{Attack}}+{\text{Strength}})\right\rfloor \\{\text{Ranged Combat Level}}&=\left\lfloor {\frac {1}{4}}\times \left({\text{Defence}}+{\text{Hitpoints}}+\left\lfloor {\text{Prayer}}\times {\frac {1}{2}}\right\rfloor \right)+{\frac {13}{40}}\times \left\lfloor {\text{Ranged}}\times {\frac {3}{2}}\right\rfloor \right\rfloor \\{\text{Magic Combat Level}}&=\left\lfloor {\frac {1}{4}}\times \left({\text{Defence}}+{\text{Hitpoints}}+\left\lfloor {\text{Prayer}}\times {\frac {1}{2}}\right\rfloor \right)+{\frac {13}{40}}\times \left\lfloor {\text{Magic}}\times {\frac {3}{2}}\right\rfloor \right\rfloor \end{aligned}}}
Wiki combat calculator[edit | edit source]
This calculator can be used to experiment how combat skills affect combat level.
template = Calculator:Combat level/Template
form = combatCalcForm
result = combatCalcResult
param = playername|Player name||hs|attack,1,1;strength,3,1;ranged,5,1;magic,7,1;defence,2,1;hitpoints,4,1;prayer,6,1
param = attack|Attack|1|int|1-99
param = strength|Strength|1|int|1-99
param = ranged|Ranged|1|int|1-99
param = magic|Magic|1|int|1-99
param = defence|Defence|1|int|1-99
param = hitpoints|Hitpoints|10|int|10-99
param = prayer|Prayer|1|int|1-99
Combat levels that are different than your own display as a different colour. These colours range from ±10 of your combat level, from shades of green for levels that are slightly lower than the player's to red for levels slightly higher than the player's; 9 distinct colours exist for this purpose.
+10 (level+10) #ff0000
+9 (level+9) #ff3000
+3 (level+3) #ffb000
±0 (level±0) #ffff00
-1 (level-1) #c0ff00
-4 (level-4) #80ff00
-10 (level-10) #00ff00
In RuneScape Classic, the maximum combat level is 123, this is due to classic using an alternative combat level formula.
In RuneScape 3, the maximum combat level is 138 due to the introduction of the Summoning skill.
Retrieved from ‘https://oldschool.runescape.wiki/w/Combat_level?oldid=14286195’
The Jormungand |
Titu's Lemma | Brilliant Math & Science Wiki
Sandeep Bhardwaj, Jubayer Nirjhor, Hua Zhi Vee, and
Titu's lemma (also known as T2 Lemma, Engel's form, or Sedrakyan's inequality) states that for positive reals
a_1, a_2, \ldots, a_n
b_1, b_2, \ldots, b_n
\frac{ a_1^2 } { b_1 } + \frac{ a_2 ^2 } { b_2 } + \cdots + \frac{ a_n ^2 } { b_n } \geq \frac{ (a_1 + a_2 + \cdots+ a_n ) ^2 } { b_1 + b_2 + \cdots+ b_n }.
It is a direct consequence of Cauchy-Schwarz inequality. This form is especially helpful when the inequality involves fractions where the numerator is a perfect square.
a_i= \frac{x_i}{ \sqrt{y_i} }
b_i = \sqrt{y_i}
into the Cauchy-Schwarz inequality. Equality holds if and only if
a_i = k b_i
for a non-zero real constant
k
\begin{aligned} \left( \frac{ x_1^2 }{ y_1 } + \frac{ x_2^2 }{ y_2 } + \cdots + \frac{ x_n^2 }{ y_n } \right) (y_1 + y_2 + \cdots + y_n ) &\geq ( x_1 + x_2 + \cdots + x_n )^2 \\ \frac{ x_1^2 }{ y_1 } + \frac{ x_2^2 }{ y_2 } + \cdots + \frac{ x_n^2 }{ y_n } &\geq \frac{ ( x_1 + x_2 + \cdots + x_n )^2 }{ (y_1 + y_2 + \cdots + y_n ) }. \end{aligned}
a_1, a_2, \ldots, a_n
b_1, b_2, \ldots, b_n
\frac{ a_1^2 } { b_1 } + \frac{ a_2 ^2 } { b_2 } + \cdots + \frac{ a_n ^2 } { b_n } \geq \frac{ (a_1 + a_2 + \cdots+ a_n ) ^2 } { b_1 + b_2 + \cdots+ b_n }.
a_i= \frac{x_i}{ \sqrt{y_i} }
b_i = \sqrt{y_i}
\begin{aligned} \left( \frac{ x_1^2 }{ y_1 } + \frac{ x_2^2 }{ y_2 } + \cdots + \frac{ x_n^2 }{ y_n } \right) (y_1 + y_2 + \cdots + y_n ) &\geq ( x_1 + x_2 + \cdots + x_n )^2 \\ \frac{ x_1^2 }{ y_1 } + \frac{ x_2^2 }{ y_2 } + \cdots + \frac{ x_n^2 }{ y_n } &\geq \frac{ ( x_1 + x_2 + \cdots + x_n )^2 }{ (y_1 + y_2 + \cdots + y_n ) }. \ _\square \end{aligned}
Prove that for positive reals
a,b,c
summing up to
1
\dfrac{1}{a+b}+\dfrac{16}{c}+\dfrac{81}{a+b+c}\ge 98.
The hypothesis implies
a+b+c=1
. We have square terms in the numerator, so
\begin{aligned} \dfrac{1}{a+b}+\dfrac{16}{c}+\dfrac{81}{a+b+c} &=& \dfrac{1}{a+b}+\dfrac{4^2}{c}+\dfrac{9^2}{a+b+c} \\ \\ &\ge &\dfrac{(1+4+9)^2}{(a+b)+c+(a+b+c)} \\ \\ & = & \dfrac{(14)^2}{2(a+b+c)} = \dfrac{196}{2} = 98, \end{aligned}
which is exactly what we wanted.
_\square
Note that the equality condition
a+b = k, c = 16k, a+b+c = 81k
In fact, we have the stronger statement:
\dfrac{1}{a+b}+\dfrac{16}{c}+\dfrac{81}{a+b+c}\ge 106.
Can you prove this using Titu's lemma?
Over all triplets of positive reals
x,y,z
x+y+z=3
, find the minimum value of
f(x,y,z)=\dfrac{yz+4zx+9xy}{xyz}.
Dividing out simplifies
f(x,y,z)
\frac{1}{x}+\frac{4}{y}+\frac{9}{z}
. Now we have square terms on the numerator. We get
f(x,y,z)=\dfrac{1^2}{x}+\dfrac{2^2}{y}+\dfrac{3^2}{z} \ge \dfrac{(1+2+3)^2}{x+y+z} = \dfrac{36}{3}=12.
Equality holds when
\dfrac 1 x = \dfrac 2 y = \dfrac 3 z~\Longleftrightarrow ~ y=2x, ~z=\dfrac{3y}{2}=3x,
x
using the constraint
x+y+z=3
, we have equality at
x=\frac{1}{2}
y=1
z=\frac{3}{2}
. We check that equality indeed occurs.
Therefore, the minimum value is 12.
_\square
x,y,z\in\mathbb{R^{+}}
\frac{2}{x+y}+\frac{2}{y+z}+\frac{2}{z+x}≥\frac{9}{x+y+z}.
\frac{2}{x+y}+\frac{2}{y+z}+\frac{2}{z+x}=\frac{\big(\sqrt{2}\big)^{2}}{x+y}+\frac{\big(\sqrt{2}\big)^{2}}{y+z}+\frac{\big(\sqrt{2}\big)^{2}}{z+x}.
\frac{\big(\sqrt{2}\big)^{2}}{x+y}+\frac{\big(\sqrt{2}\big)^{2}}{y+z}+\frac{\big(\sqrt{2}\big)^{2}}{z+x}≥\frac{\big(3\sqrt{2}\big)^{2}}{2(x+y+z)}=\frac{9}{x+y+z}. \ _\square
a,b\in\mathbb{R^{+}}
8\big(a^{4}+b^{4}\big)\geq(a+b)^{4}.
a^{4}+b^{4}=\dfrac{a^{4}}{1}+\dfrac{b^{4}}{1}\geq\dfrac{\left(a^{2}+b^{2}\right)^{2}}{2}\geq\dfrac{\Big(\frac{(a+b)^{2}}{2}\Big)^{2}}{2}=\dfrac{(a+b)^{4}}{8}. \ _\square
a,b,c\in\mathbb{R^{+}}
\frac{a^{2}+b^{2}}{a+b}+\frac{b^{2}+c^{2}}{b+c}+\frac{c^{2}+a^{2}}{c+a}≥a+b+c.
Write the left side of the given inequality as
\dfrac{a^{2}}{a+b}+\dfrac{b^{2}}{a+b}+\dfrac{b^{2}}{b+c}+\dfrac{a^{2}}{c+a}+\dfrac{c^{2}}{a+c}+\dfrac{c^{2}}{c+b},
≥\frac{(2a+2b+2c)^{2}}{4(a+b+c)},
_\square
Now you can try to prove the Nesbitt's inequality.
a, b, c \in \mathbb{R}^+
\frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{b + a} \ge \frac{3}{2}.
The numerator currently isn't a perfect square. Let's make it a perfect square:
\frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{b + a} = \frac{ a^2 } { ab+ac} + \frac{ b^2 } { ab + bc } + \frac{ c^2 } { bc + ac } .
We can then apply Titu's lemma to obtain
\frac{ a^2 } { ab+ac} + \frac{ b^2 } { ab + bc } + \frac{ c^2 } { bc + ac } \geq \frac{ (a+b+c) ^2 } { 2 ( ab+bc+ca) } .
\begin{array} { c r c l } & \frac{ (a+b+c) ^2 } { 2 ( ab+bc+ca) } & \geq & \frac{3}{2} \\ \Leftrightarrow & (a+b+c)^2 & \geq & 3 (ab+bc+ca) \\ \Leftrightarrow & a^2 + b^2 + c^2 & \geq & ab + bc + ca \\ \Leftrightarrow & (a-b)^2+(b-c)^2+(c-a)^2 & \geq & 0. \end{array}
Note that if we tried to apply Titu's lemma directly, we end up with
\frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{b + a} \geq \frac{ \big( \sqrt{a} + \sqrt{b} + \sqrt{c} \big)^2 } { 2 ( a + b + c ) }.
Then, the right-hand side actually has a maximum value of
\frac{3}{2}
, so we cannot proceed any further.
_\square
Another approach is to apply Titu's lemma to show that:
\frac{1}{ b+c} + \frac{1}{ c+a} + \frac{1}{a+b} \geq \frac{ (1 + 1 + 1)^2 } { 2(a+b+c) }.
\frac{a+b+c}{ b+c} + \frac{a+b+c}{ c+a} + \frac{a+b+c}{a+b} \geq \frac{9}{2}.
\frac{a}{ b+c} + \frac{b}{ c+a} + \frac{c}{a+b} \geq \frac{9}{2} - 3 = \frac{3}{2} .
Here is the generalized form:
m
\frac{x_1^m}{a_1^{m-1}}+\frac{x_2^m}{a_2^{m-1}}+\cdots+\frac{x_n^m}{a_n^{m-1}}\ge\frac{(x_1+x_2+\cdots+x_n)^m}{(a_1+a_2+\cdots+a_n)^{m-1}}.
To learn how we reached this generalized form, follow the process below.
\displaystyle \bigg(\sum_{i=1}^n\sqrt[\frac{m}{m-1}]{a_i}^{\frac{m}{m-1}}\bigg)^{\frac{1}{\ \frac{m}{m-1}\ }}\left(\sum_{i=1}^n\dfrac{x_i^m}{\left(\frac{a_i}{\sqrt[m]{a_i}}\right)^m}\right)^{\frac{1}{m}} \ge \sum_{i=1}^n x_i,
which is true by Hölder's inequality.
Then by simplifying, we get
\begin{aligned}\displaystyle \bigg(\sum_{i=1}^n a_i\bigg)^{\frac{m-1}{m}}\bigg(\sum_{i=1}^n\dfrac{x_i^m}{a_i^{m-1}}\bigg)^{\frac{1}{m}}&\ge \sum_{i=1}^n x_i \\\bigg(\sum_{i=1}^n a_i\bigg)^{m-1}\bigg(\sum_{i=1}^n\dfrac{x_i^m}{a_i^{m-1}}\bigg)&\ge \bigg(\sum_{i=1}^n x_i\bigg)^m \\ \bigg(\sum_{i=1}^n\dfrac{x_i^m}{a_i^{m-1}}\bigg)&\ge \dfrac{\bigg(\sum_{i=1}^n x_i\bigg)^m}{\bigg(\sum_{i=1}^n a_i\bigg)^{m-1}}.\end{aligned}
{\frac{x_1^m}{a_1^{m-1}}+\frac{x_2^m}{a_2^{m-1}}+\cdots+\frac{x_n^m}{a_n^{m-1}}\ge\frac{(x_1+x_2+\cdots+x_n)^m}{(a_1+a_2+\cdots+a_n)^{m-1}}}
m \ge 2.
When we put
m=2
\frac{x_1^2}{a_1}+\frac{x_2^2}{a_2}+\cdots+\frac{x_n^2}{a_n}\ge\frac{(x_1+x_2+\cdots+x_n)^2}{(a_1+a_2+\cdots+a_n)}.
We start with some examples followed by several problems to try.
[IMO/1995] Given
a,b,c\in\mathbb{R^{+}} \text{ and } abc=1
\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(a+c)}+\frac{1}{c^{3}(a+b)}≥\frac{3}{2}.
We write the left side of the given inequality as
\dfrac{\frac{1}{a^{2}}}{ab+ac}+\dfrac{\frac{1}{b^{2}}}{ba+bc}+\dfrac{\frac{1}{c^{2}}}{ca+cb}
\ge \dfrac{\big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\big)^{2}}{2(ab+bc+ca)}
abc=1
\big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\big)^{2}=(ab+bc+ca)^{2}
Hence, the left-hand side of the given inequality is
≥\dfrac{(ab+bc+ca)}{2}≥3\dfrac{\sqrt[3]{(abc)^{2}}}{2}=\dfrac{3}{2}
_\square
a,b,c\in\mathbb{R^{+}}
\frac{a}{2a+b}+\frac{b}{2b+c}+\frac{c}{2c+a}≤1.
Let's modify the inequality that we have to prove as follows:
\begin{aligned} \left(\frac{a}{2a+b}-\frac{1}{2}\right)+\left(\frac{b}{2b+c}-\frac{1}{2}\right)+\left(\frac{c}{2c+a}-\frac{1}{2}\right) &\leq 1-\frac{3}{2} \\ \frac{2a-(2a+b)}{2(2a+b)}+\frac{2b-(2b+c)}{2(2b+c)}+\frac{2c-(2c+a)}{2(2c+a)} &\leq -\frac{1}{2} \\ -\frac{1}{2}\left( \frac{b}{2a+b}+\frac{c}{2b+c}+\frac{a}{2c+a}\right) &\leq -\frac{1}{2} \\ \frac{b}{2a+b}+\frac{c}{2b+c}+\frac{a}{2c+a} &\geq 1. \end{aligned}
\frac{b^2}{2ab+b^2}+\frac{c^2}{2bc+c^2}+\frac{a^2}{2ca+a^2} \geq \frac{(a+b+c)^2}{a^2+b^2+c^2+2(ab+bc+ca)}=1. \ _\square
For the positive real numbers
x, y,
z,
x+y+z=1,
\displaystyle \sum_{cyc} \frac{x^3}{(x+y)^2} \ge \dfrac14.
What we should actually prove is that
\frac{x^3}{(x+y)^2}+\frac{y^3}{(y+z)^2}+\frac{z^3}{(z+x)^2}\ge \frac14
. Thus by applying the above inequality on the LHS, we get
\begin{aligned}\dfrac{x^3}{(x+y)^2}+\dfrac{y^3}{(y+z)^2}+\dfrac{z^3}{(z+x)^2}&\ge \dfrac{(x+y+z)^3}{(x+y+y+z+z+x)^2} \\&=\dfrac{(x+y+z)^3}{4(x+y+z)^2} \\&=\dfrac{1}{4}.\end{aligned}
_\square
Try the following examples.
x,y,z
x+y+z=3
, then find the minimum value of
E=\frac { 4yz+9xz+16xy }{ xyz }.
x,y
z
x+y+z=6
. Find the minimum value of the expression
\dfrac{16yz+36xz+64xy}{xyz}.
\frac { { x }^{ 2 }+{ y }^{ 2 } }{ z } +\frac { { z }^{ 2 }+{ x }^{ 2 } }{ y } +\frac { { y }^{ 2 }+{ z }^{ 2 } }{ x },
x + y + z = 2015
x, y ,z > 0
a,b,c,d,e,f
are positive real numbers which satisfy
a+b+c+d+e+f=7,
\frac { 1 }{ a } +\frac { 4 }{ b } +\frac { 9 }{ c } +\frac { 16 }{ d } +\frac { 25 }{ e } +\frac { 36 }{ f } .
There exists a smallest possible positive integer
N
\dfrac{(x_1+2x_2+\cdots +2014x_{2014})^2}{x_1^2+x_2^2+\cdots +x_{2014}^2}\le N
for all real sequences
\large \{x_i\}_{i=1}^{2014}
Find the sum of digits of
N
a,b
c
are three positive real numbers such that
a+b+c=n
n
(ab+bc+ca)\left(\dfrac{a}{2b+3c}+\dfrac{b}{2c+3a}+\dfrac{c}{2a+3b}\right)\ge 20
for all possible ordered triplets
(a,b,c)?
and
b
a^{2} + b^{2} = 1,
and always satisfy
\frac{1}{1 + a^{2}}+ \frac{1}{1 + b^{2}} + \frac{1}{1 + ab} \geq \frac{x}{1 + \frac{(a+b)^{2}}{z}}.
\left\lfloor \frac{x}{z} \right\rfloor
when the value of the LHS of the above inequality is the least.
x
z
\lfloor \cdot \rfloor
denotes the greatest integer function.
Cite as: Titu's Lemma. Brilliant.org. Retrieved from https://brilliant.org/wiki/titus-lemma/ |
Abstract Algebra | Brilliant Math & Science Wiki
Alexander Katz, Prince Loomba, Eli Ross, and
Abstract algebra is a broad field of mathematics, concerned with algebraic structures such as groups, rings, vector spaces, and algebras. On the 12-hour clock,
9+4=1
, rather than 13 as in usual arithmetic
Roughly speaking, abstract algebra is the study of what happens when certain properties of number systems are abstracted out; for instance, altering the definitions of the basic arithmetic operations result in a structure known as a ring, so long as the operations are consistent.
For example, the 12-hour clock is an example of such an object, where the arithmetic operations are redefined to use modular arithmetic (with modulus 12). An even further level of abstraction--where only one operation is considered--allows the clock to be understood as a group. In either case, the abstraction is useful because many properties can be understood without needing to consider the specific structure at hand, which is especially important when considering the relationship(s) between structures; the concept of a group isomorphism is an example.
Levels of Abstraction in Abstract Algebra
Other Applications of Abstract Algebra
It is possible to abstract away practically all of the properties found in the "usual" number systems, the tradeoff being that the resulting object--known as a magma (which consists of a set and a binary operation, that need not satisfy any properties other than closure)--is simply too general to be interesting. On the other extreme, it is possible to abstract out practically no properties, which allows for many results to be found, but the resulting object (the usual number systems) is too specific to solve more general problems.
Most of abstract algebra is dedicated to objects that have a reasonable balance between generality and structure, most notably groups and rings (discussed in more detail below) in which most of the basic properties of arithmetic are maintained, but their specifics are left free. Still, some higher levels of abstraction are occasionally useful; quasigroups, for instance, are related to Latin squares, and monoids are often used in computer science and are simple examples of categories.
Main article: Group theory The possible moves on a Rubik's cube form a (very large) group. Group theory is useful as an abstract notion of symmetry, which makes it applicable to a wide range of areas: the relationship between the roots of a polynomial (as in Galois theory) and the solution methods to the Rubik's cube are both prominent examples.
Informally, a group is a set equipped with a binary operation
\circ
, so that operating on any two elements of the group also produces an element of the group. For example, the integers form a group under addition, and the nonzero real numbers form a group under multiplication. The
\circ
operation needs to satisfy a number of properties analogous to the ones it satisfies for these "normal" number systems: it should be associative (which essentially means that the order of operations doesn't matter), and there should be an identity element (0 in the first example above, and 1 in the second). More formally, a group is a set equipped with an operation
\cdot
such that the following axioms hold; note that
\cdot
does not necessarily refer to multiplication; rather, it should be viewed as a function on two variables (indeed,
\cdot
can even refer to addition):
x, y, z \in G
(x \cdot y) \cdot z = x \cdot (y \cdot z)
e \in G
e \cdot x = x \cdot e = x
x \in G
e
G
x \in G
y \in G
x \cdot y = e = y \cdot x
y
x
It is also worth noting the closure axiom for emphasis, as it is important to verify closure when working with subgroups (groups contained entirely within another):
x, y \in G
x*y
G
Additional examples of groups include
\mathbb{Z}_n
\{0, 1, \ldots, n-1\}
with the operation addition modulo
n
S_n
, the set of permutations of
\{1, 2, \ldots, n\}
with the operation of composition.
S_3
is worth special note as an example of a group that is not commutative, meaning that
a \cdot b = b \cdot a
does not generally hold. Formally speaking,
S_3
is nonabelian (an abelian group is one in which the operation is commutative). When the operation is not clear from context, groups are written in the form
(\text{set}, \text{op})
; e.g. the nonzero reals equipped with multiplication can be written as
(\mathbb{R}^*, \cdot)
Much of group theory (and abstract algebra in general) is centered around the concept of a group homomorphism, which essentially means a mapping from one group to another that preserves the structure of the group. In other words, the mapping of the product of two elements should be the same as the product of the two mappings; intuitively speaking, the product of two elements should not change under the mapping. Formally, a homomorphism is a function
\phi: G \rightarrow H
\phi(g_1) \cdot_H \phi(g_2) = \phi(g_1 \cdot_G g_2),
\cdot_H
is the operation on
H
\cdot_G
G
\phi(g) = g \pmod n
is an example of a group homomorphism from
\mathbb{Z}
\mathbb{Z}_n
. The concept of potentially differing operations is necessary; for example,
\phi(g)=e^g
(\mathbb{R},+)
(\mathbb{R}^{*},\cdot)
Rings are one of the lowest level of abstraction, essentially obtained by overwriting the addition and multiplication functions simultaneously (compared to groups, which uses only one operation). Thus a ring is--in some sense--a combination of multiple groups, as a ring can be viewed as a group over either one of its operations. This means that the analysis of groups is also applicable to rings, but rings have additional properties to work with (the tradeoff being that rings are less general and require more conditions).
The definition of a ring is similar to that of a group, with the extra condition that the distributive law holds as well:
R
together with two operations
+
\cdot
satisfying the following properties (ring axioms):
R
is an abelian group under addition. That is,
R
is closed under addition, there is an additive identity (called
0
), every element
a\in R
has an additive inverse
-a\in R
, and addition is associative and commutative.
R
is closed under multiplication, and multiplication is associative:
\forall a,b\in R\quad a.b\in R\\ \forall a,b,c\in R\quad a\cdot ( b\cdot c ) =( a\cdot b ) \cdot c.
(3) Multiplication distributes over addition:
\forall a,b,c\in R\\ a\cdot \left( b+c \right) =a\cdot b+a\cdot c\quad \text{ and }\quad \left( b+c \right) \cdot a=b\cdot a+c\cdot a.
A ring is usually denoted by
\left( R,+,. \right)
and often it is written only as
R
when the operations are understood.
For example, the integers
\mathbb{Z}
form a ring, as do the integers modulo
n
(
\mathbb{Z}_n).
Less obviously, the square matrices of a given size also form a ring; this ring is noncommutative. Commutative ring theory, or commutative algebra, is much better understood than noncommutative rings are.
As in groups, a ring homomorphism can be defined as a mapping preserving the structure of both operations.
Rings are used extensively in algebraic number theory, where "integers" are reimagined as slightly different objects (for example, Gaussian integers), and the effect on concepts such as prime factorization is analyzed. Of particular interest is the fundamental theorem of arithmetic, which involves the concept of unique factorization; in other rings, this may not hold, such as
6 = 2 \cdot 3 = \big(1+\sqrt{-5}\big)\big(1-\sqrt{-5}\big).
Theory developed in this field solves problems ranging from sum of squares theorems to Fermat's last theorem.
Abstract algebra also has heavy application in physics and computer science through the analysis of vector spaces. For example, the Fourier transform and differential geometry both have vector spaces as their underlying structures; in fact, the Poincare conjecture is (roughly speaking) a statement about whether the fundamental group of a manifold determines if the manifold is a sphere.
Related to vector spaces are modules, which are essentially identical to vector spaces but defined over a ring rather than over a field (and are thus more general). Modules are heavily related to representation theory, which views the elements of a group as linear transformations of a vector space; this is desirable to make an abstract object (a group) somewhat more concrete, in the sense that the group is better understood by translating it into a well-understood object in linear algebra (as matrices can be viewed as linear transformations, and vice versa).
The relationships between various algebraic structures are formalized using category theory.
Cite as: Abstract Algebra. Brilliant.org. Retrieved from https://brilliant.org/wiki/abstract-algebra/ |
Nonlinear Constraint Solver Algorithm - MATLAB & Simulink - MathWorks í•œêµ
\underset{x}{\mathrm{min}}f\left(x\right)
\begin{array}{c}{c}_{i}\left(x\right)â¤0,\text{â}\text{â}i=1â¦m\\ ce{q}_{i}\left(x\right)=0,\text{â}\text{â}\text{â}i=m+1â¦mt\\ Aâ
xâ¤b\\ Aeqâ
x=beq\\ lbâ¤xâ¤ub,\end{array}
\mathrm{Î}\left(x,\mathrm{λ},s,\mathrm{Ï}\right)=f\left(x\right)â\underset{i=1}{\overset{m}{â}}{\mathrm{λ}}_{i}{s}_{i}\mathrm{log}\left({s}_{i}â{c}_{i}\left(x\right)\right)+\underset{i=m+1}{\overset{mt}{â}}{\mathrm{λ}}_{i}ce{q}_{i}\left(x\right)+\frac{\mathrm{Ï}}{2}\underset{i=m+1}{\overset{mt}{â}}ce{q}_{i}{\left(x\right)}^{2},
The pattern search minimizes a sequence of subproblems, each of which is an approximation of the original problem. Each subproblem has a fixed value of λ, s, and Ï. When the subproblem is minimized to a required accuracy and satisfies feasibility conditions, the Lagrangian estimates are updated. Otherwise, the penalty parameter is increased by a penalty factor (PenaltyFactor). This results in a new subproblem formulation and minimization problem. These steps are repeated until the stopping criteria are met.
[1] Kolda, Tamara G., Robert Michael Lewis, and Virginia Torczon. “A generating set direct search augmented Lagrangian algorithm for optimization with a combination of general and linear constraints.†Technical Report SAND2006-5315, Sandia National Laboratories, August 2006. |
Fixed-Point Concepts and Terminology - MATLAB & Simulink - MathWorks India
In digital hardware, numbers are stored in binary words. A binary word is a fixed-length sequence of bits (1's and 0's). The way hardware components or software functions interpret this sequence of 1's and 0's is defined by the data type.
Binary numbers are represented as either floating-point or fixed-point data types. In this section, we discuss many terms and concepts relating to fixed-point numbers, data types, and mathematics.
A fixed-point data type is characterized by the word length in bits, the position of the binary point, and the signedness of a number which can be signed or unsigned. Signed numbers and data types can represent both positive and negative values, whereas unsigned numbers and data types can only represent values that are greater than or equal to zero.
The position of the binary point is the means by which fixed-point values are scaled and interpreted.
wl is the number of bits in a binary word, also known as word length.
bwl–1 is the location of the most significant, or highest, bit (MSB). In signed binary numbers, this bit is the sign bit which indicates whether the number is positive or negative.
b0 is the location of the least significant, or lowest, bit (LSB). This bit in the binary word can represent the smallest value. The weight of the LSB is given by:
weigh{t}_{LSB}={2}^{-fractionlength}
where, fractionlength is the number of bits to the right of the binary point.
Bits to the left of the binary point are integer bits and/or sign bits, and bits to the right of the binary point are fractional bits. Number of bits to the left of the binary point is known as the integer length. The binary point in this example is shown four places to the left of the LSB. Therefore, the number is said to have four fractional bits, or a fraction length of four.
Fixed-point data types can be either signed or unsigned.
Signed binary fixed-point numbers are typically represented in one of three ways:
Sign/magnitude –– Representation of signed fixed-point or floating-point numbers. In the sign/magnitude representation, one bit of a binary word is always the dedicated sign bit, while the remaining bits of the word encode the magnitude of the number. Negation using sign/magnitude representation consists of flipping the sign bit from 0 (positive) to 1 (negative), or from 1 to 0.
Two's complement –– Two's complement is the most common representation of signed fixed-point numbers. See Two's Complement for more information.
Unsigned fixed-point numbers can only represent numbers greater than or equal to zero.
In [Slope Bias] representation, fixed-point numbers can be encoded according to the scheme
real\text{-}worldvalue=\left(slope×integer\right)+bias
slope=slope\text{ }adjustment×{2}^{exponent}
The term slope adjustment is sometimes used as a synonym for fractional slope.
In the trivial case, slope = 1 and bias = 0. Scaling is always trivial for pure integers, such as int8, and also for the true floating-point types single and double.
The integer is sometimes called the stored integer. This is the raw binary number, in which the binary point assumed to be at the far right of the word. In System Toolboxes, the negative of the exponent is often referred to as the fraction length.
The slope and bias together represent the scaling of the fixed-point number. In a number with zero bias, only the slope affects the scaling. A fixed-point number that is only scaled by binary point position is equivalent to a number in the Fixed-Point Designer™ [Slope Bias] representation that has a bias equal to zero and a slope adjustment equal to one. This is referred to as binary point-only scaling or power-of-two scaling:
real\text{-}world\text{ }value={2}^{exponent}×integer
real\text{-}world\text{ }value={2}^{-fractionlength}×integer
In System Toolbox software, you can define a fixed-point data type and scaling for the output or the parameters of many blocks by specifying the word length and fraction length of the quantity. The word length and fraction length define the whole of the data type and scaling information for binary-point only signals.
All System Toolbox blocks that support fixed-point data types support signals with binary-point only scaling. Many fixed-point blocks that do not perform arithmetic operations but merely rearrange data, such as Delay and Matrix Transpose, also support signals with [Slope Bias] scaling.
You must pay attention to the precision and range of the fixed-point data types and scalings you choose for the blocks in your simulations, in order to know whether rounding methods will be invoked or if overflows will occur.
The range is the span of numbers that a fixed-point data type and scaling can represent. The range of representable numbers for a two's complement fixed-point number of word length wl, scaling S, and bias B is illustrated below:
For example, in two's complement, negative numbers must be represented as well as zero, so the maximum value is 2wl–1. Because there is only one representation for zero, there are an unequal number of positive and negative numbers. This means there is a representation for -2wl–1 but not for 2wl–1:
The full range is the broadest range for a data type. For floating-point types, the full range is –∞ to ∞. For integer types, the full range is the range from the smallest to largest integer value (finite) the type can represent. For example, from -128 to 127 for a signed 8-bit integer.
Overflow Handling. Because a fixed-point data type represents numbers within a finite range, overflows can occur if the result of an operation is larger or smaller than the numbers in that range.
System Toolbox software does not allow you to add guard bits to a data type on-the-fly in order to avoid overflows. Guard bits are extra bits in either a hardware register or software simulation that are added to the high end of a binary word to ensure that no information is lost in case of overflow. Any guard bits must be allocated upon model initialization. However, the software does allow you to either saturate or wrap overflows. Saturation represents positive overflows as the largest positive number in the range being used, and negative overflows as the largest negative number in the range being used. Wrapping uses modulo arithmetic to cast an overflow back into the representable range of the data type. See Modulo Arithmetic for more information.
The precision of a fixed-point number is the difference between successive values representable by its data type and scaling, which is equal to the value of its least significant bit. The value of the least significant bit, and therefore the precision of the number, is determined by the number of fractional bits. A fixed-point value can be represented to within half of the precision of its data type and scaling. The term resolution is sometimes used as a synonym for this definition.
For example, a fixed-point representation with four bits to the right of the binary point has a precision of 2-4 or 0.0625, which is the value of its least significant bit. Any number within the range of this data type and scaling can be represented to within (2-4)/2 or 0.03125, which is half the precision. This is an example of representing a number with finite precision.
Rounding Modes. When you represent numbers with finite precision, not every number in the available range can be represented exactly. If a number cannot be represented exactly by the specified data type and scaling, it is rounded to a representable number. Although precision is always lost in the rounding operation, the cost of the operation and the amount of bias that is introduced depends on the rounding mode itself. To provide you with greater flexibility in the trade-off between cost and bias, DSP System Toolbox™ software currently supports the following rounding modes:
Ceiling rounds the result of a calculation to the closest representable number in the direction of positive infinity.
Convergent rounds the result of a calculation to the closest representable number. In the case of a tie, Convergent rounds to the nearest even number. This is the least biased rounding mode provided by the toolbox.
Floor, which is equivalent to truncation, rounds the result of a calculation to the closest representable number in the direction of negative infinity. The truncation operation results in dropping of one or more least significant bits from a number.
Nearest rounds the result of a calculation to the closest representable number. In the case of a tie, Nearest rounds to the closest representable number in the direction of positive infinity.
Round rounds the result of a calculation to the closest representable number. In the case of a tie, Round rounds positive numbers to the closest representable number in the direction of positive infinity, and rounds negative numbers to the closest representable number in the direction of negative infinity.
Simplest rounds the result of a calculation using the rounding mode (Floor or Zero) that adds the least amount of extra rounding code to your generated code. For more information, see Rounding Mode: Simplest (Fixed-Point Designer).
Zero rounds the result of a calculation to the closest representable number in the direction of zero.
To learn more about each of these rounding modes, see Rounding (Fixed-Point Designer).
For a direct comparison of the rounding modes, see Choosing a Rounding Method (Fixed-Point Designer). |
Resource Grid Indexing - MATLAB & Simulink - MathWorks í•œêµ
Initialize Required Parameters
Create Empty Resource Arrays
Compare Resource Arrays
Linear Indices and Subscripts
Converting Between Linear Indices and Subscripts
Multi-Antenna Linear Indices
The LTE Toolbox™ provides facilities to generate sequences of symbols corresponding to the physical channels and signals. Indices for the mapping of these sequences to the resource grid are also generated. For convenience, the LTE Toolbox uses the MATLAB® linear indexing style to represent these indices.
Before applying OFDM modulation (IFFT), the physical channels and signals in LTE are mapped to different portions of the subframe resource grid. The subframe resource grid is represented in the LTE Toolbox as a multidimensional array of the following size.
12{N}_{RB}Ã2{N}_{symb}ÃP
In the preceding expression,
{N}_{RB}
is the number of resource blocks spanning the available bandwidth,
{N}_{symb}
is the number of OFDM (or SC-FDMA in the uplink) symbols per slot, and
P
is the number of antenna ports. Therefore, the resource grid represents a subframe (two slots) and whole bandwidth, since there are 12 subcarriers per resource block. For the single antenna case, a resource grid can be a two-dimensional matrix of the following size.
12{N}_{RB}Ã2{N}_{symb}
Create an empty downlink resource array by using two different methods. You can create valid and equivalent subframe resource arrays by using the lteDLResourceGrid function or the zeros function.
Create a parameter structure for a normal cyclic prefix, nine downlink resource blocks, and one transmit antenna.
enb = struct('CyclicPrefix','Normal','NDLRB',9,'CellRefP',1);
Define seven symbols per slot for use in the zeros function.
symbolsPerSlot = 7;
Create an empty subframe resource array using each method.
resourceGrid1 = lteDLResourceGrid(enb);
resourceGrid2 = zeros(enb.NDLRB*12,symbolsPerSlot*2,enb.CellRefP);
Compare the two resource arrays. Because both approaches generate the same result, you can use either to create an empty downlink resource array. Similarly, you can create an empty uplink, sidelink, or narrowband resource array by using the lteULResourceGrid, lteSLResourceGrid, or lteNBResourceGrid function, respectively, or the zeros function.
isequal(resourceGrid1,resourceGrid2)
Generate a reference signal and map it to an empty resource grid for the single antenna case. The LTE Toolbox™ has been designed to facilitate the mapping of physical channels and signals in the resource grid.
Configure cell-wide settings. Create a structure and specify the cell-wide settings as its fields.
antPort = 0;
Create an empty subframe resource grid by using the lteDLResourceGrid function, then populate the grid with reference symbolsby using the lteCellRSIndices and lteCellRS functions.
resourceGrid = lteDLResourceGrid(enb);
ind = lteCellRSIndices(enb,antPort);
rs = lteCellRS(enb,antPort);
resourceGrid(ind) = rs;
To generates a list of indices identifying where to map the reference signal use the function lteCellRSIndices. To generate the reference signal symbols, use the lteCellRS function.
Generate indices in linear and subscript form. All of the LTE Toolbox™ index generation functions can produce linear or subscript formats by setting the appropriate options. The default is linear indexing, which allows access to any element of a matrix with a single index value. Using subscripted indexing in a 2-D matrix, you can access each element with a set of two elements representing the row and column equivalents.
The linear indexing style allows you to conveniently map the reference sequence symbols to the appropriate location in the resource grid with just one line of code. Mapping reference symbols to the resource grid using subscripted indices would require more finesse.
Create a structure specifying the cell-wide settings as its fields. Assign zero as the antenna port number.
Create an empty subframe resource grid by using the lteDLResourceGrid function, then create reference signal symbols by using the lteCellRS function. View the empty resource grid.
mesh(abs(resourceGrid))
Generate linear indices.
ind_lin = lteCellRSIndices(enb,antPort);
Map the reference signal symbols to the resource grid.
resourceGrid(ind_lin) = rs;
Show the reference symbols active for the cell-wide settings assigned in enb by plotting the updated resource grid.
Alternatively, generate indices in subscript form by providing the 'sub' option string for lteCellRSIndices. In this case, the output indices represent the resource grid in the form [subcarrier, OFDM symbol, antenna port].
ind_sub = lteCellRSIndices(enb,antPort,'sub');
Conversion between linear indices and subscripts can be achieved using the MATLAB ind2sub and sub2ind functions. Alternatively, all index generation functions in the LTE Toolbox can produce both formats.
Generate indices in multi-antenna linear form. This form is a variant of the MATLAB® linear indexing style in which the indices corresponding for each antenna port are in a different column. However, all indices are still in linear form. Several toolbox functions return indices in multi-antenna linear form.
To illustrate this, call the function ltePDSCH for the four antenna case.
pdsch.TxScheme = 'TxDiversity';
pdsch.Modulation = 'QPSK';
pdsch.PRBSet = (0:5).';
data = ones(768,1);
symb = ltePDSCH(enb,pdsch,data);
size(symb)
symb(1:10,:)
ans = 10×4 complex
0.5000 - 0.5000i 0.0000 + 0.0000i -0.5000 + 0.5000i 0.0000 + 0.0000i
0.0000 + 0.0000i -0.5000 - 0.5000i 0.0000 + 0.0000i 0.5000 - 0.5000i
0.0000 + 0.0000i -0.5000 - 0.5000i 0.0000 + 0.0000i -0.5000 + 0.5000i
0.5000 - 0.5000i 0.0000 + 0.0000i 0.5000 + 0.5000i 0.0000 + 0.0000i
-0.5000 + 0.5000i 0.0000 + 0.0000i 0.5000 + 0.5000i 0.0000 + 0.0000i
0.0000 + 0.0000i 0.5000 + 0.5000i 0.0000 + 0.0000i -0.5000 + 0.5000i
0.0000 + 0.0000i 0.5000 + 0.5000i 0.0000 + 0.0000i 0.5000 - 0.5000i
-0.5000 + 0.5000i 0.0000 + 0.0000i -0.5000 + 0.5000i 0.0000 + 0.0000i
The output argument, symb, is a matrix with four columns, in which each column corresponds to each antenna port.
In a similar format, generate the indices for the PDSCH by calling ltePDSCHIndices.
pdschIndices = ltePDSCHIndices(enb,pdsch,pdsch.PRBSet);
size(pdschIndices)
pdschIndices(1:10,:)
Again, each column corresponds to each of the four antenna ports. The concatenation of all four columns produces a column vector of indices using the MATLAB linear indexing style.
Generate either zero-based or one-based indices. All mapping operations in the LTE technical specification (TS) documents refer to zero-based indexing. However, MATLAB® indices must be one-based. LTE Toolbox™ generates one-based indices by default, but you can generate zero-based indices by setting the appropriate options string.
Create a cell-wide setting structure and assign an antenna port number.
Since one-based indexing is the default, you can generate one-based indices by specifying the '1based' flag or leaving it out.
ind = lteCellRSIndices(enb,antPort,'1based');
Generate zero-based indices by specifying the '0based' flag.
The 3GPP documents describes a resource block to be a group of resource elements spanning 12 consecutive subcarriers in the frequency domain and one slot in the time domain. For processing efficiency, the LTE Toolbox™, operates on a subframe (two timeslot) basis and describes a resource block pair to represent 12 consecutive subcarriers spanning in the frequency domain and one subframe (two slots) in the time domain. For example, the command ltePDSCHIndices uses the parameter PRBSet to define the set of physical resource block (PRB) indices for a subframe of data.
Create the cell-wide settings structure and define the PDSCH configuration.
Create a set of PDSCH PRB indices for the initialized configuration.
pdsch = struct with fields:
pdsch.PRBSet can be either a column vector or a two-column matrix. If you provide a column vector, the resource allocation is the same in both slots of the subframe, which means the set of resource indices applies to both subframe time slots. On the other hand, if you provide a two-column matrix, the PRB indices refer to each slot individually.
lteCellRSIndices | lteCellRS | ltePDSCHIndices | ltePDSCH | lteDLResourceGrid | zeros |
Solve for G.S./P.S. for the following differential equations using the
kramtus51 2022-01-19 Answered
Solve for G.S./P.S. for the following differential equations using the solutions by a change in variables as suggested by the equations.
\left({t}^{2}{e}^{t}-4{x}^{2}\right)dt+8txdx=0
einfachmoipf
\left({t}^{2}{e}^{t}-4{x}^{2}\right)dt+8txdx=0
M={t}^{2}{e}^{t}-4{x}^{2}
N=8tx
{M}_{x}==8x
{N}_{t}=8x
\frac{{M}_{x}-{N}_{t}}{N}=\frac{-16x}{8tx}=\frac{-2}{t}
{e}^{-2\int \frac{1}{t}dt}={e}^{-2\mathrm{ln}t}=\frac{1}{{t}^{2}}
M={e}^{t}=\frac{4{x}^{2}}{{t}^{2}}
N=\frac{8x}{t}
{M}_{x}=-\frac{8x}{{t}^{2}}
{N}_{1}=\frac{-8x}{{t}^{2}}
\therefore \text{ }\text{ }{M}_{x}={N}_{t}
\int \left({e}^{t}-\frac{4{x}^{2}}{{t}^{2}}\right)dt+\int 0dx=C
⇒\text{ }{e}^{t}+\frac{4{x}^{2}}{t}=C
x\frac{dy}{dx}-2y={x}^{3}{\mathrm{sin}}^{2}\left(x\right)
dy/dx-12{x}^{3}y={x}^{3}
Are the following are linear equation?
1.\text{ }\frac{dP}{dt}+2tP=P+4t-2
2.\text{ }\frac{dy}{dx}={y}^{2}-3y
Solve the equation separable, linear, bernoulli, or homogenous
1.\frac{dy}{dx}=\frac{{x}^{4}+4x{y}^{2}}{2{x}^{3}+{x}^{2}y+{y}^{3}}
2.\left({e}^{-y}\mathrm{cos}\left(x\right)\right){y}^{\prime }={x}^{4}+6{x}^{2}{y}^{3}
3.{y}^{\prime }=\frac{y+y{x}^{3}}{x+{x}^{2}}\mathrm{cos}\left(\frac{{x}^{2}}{{y}^{2}}\right)
Write an equivalent first-order differential equation and initial condition for y.
y=-1+{\int }_{0}^{x}\left(2t-5y\left(t\right)\right)dt
What is the equivalent first-order differential equation?
{y}^{\prime }=?
What is the initial condition?
y\left({y}^{\prime }=?\right)=?
What method would you use to solve:
\left(1+{x}^{2}\right)\frac{\mathrm{d}y}{\mathrm{d}x}=1+{y}^{2}\phantom{\rule{thickmathspace}{0ex}};\phantom{\rule{2em}{0ex}}y\left(2\right)=3
I am asking this because I only know two methods of solving the DEs - separation of variables and integrating factor. Since the separation of variables does not work here, I tried integrating factor, however, I don't know what to do with the y2, because for the IF to work I need to get y on its own
\frac{\mathrm{d}y}{\mathrm{d}x}+P\left(x\right)y=Q\left(x\right)
What method do I use to solve this? |
EUDML | On Hermite interpolation in . EuDML | On Hermite interpolation in .
On Hermite interpolation in
{R}_{d}
Shekhtman, Boris. "On Hermite interpolation in .." ETNA. Electronic Transactions on Numerical Analysis [electronic only] 18 (2004): 65-72. <http://eudml.org/doc/128920>.
@article{Shekhtman2004,
author = {Shekhtman, Boris},
keywords = {Hermite interpolation; Lagrange interpolation},
title = {On Hermite interpolation in .},
AU - Shekhtman, Boris
TI - On Hermite interpolation in .
KW - Hermite interpolation; Lagrange interpolation
Hermite interpolation, Lagrange interpolation
Articles by Shekhtman |
Introduction to Linear Regression Practice Problems Online | Brilliant
Many data sets have an approximately linear relationship between variables. In these cases, we can predict one variable using a known value for another using a best-fit line, a line of the form
y = mx+b
that follows the trends in the data as closely as possible.
x
is called the predictor variable because it will be used to predict
y,
y
is often called the response variable.
This technique is known as linear regression, and although it is one of the simplest machine learning techniques, it is often surprisingly powerful.
Of course, linear regression isn’t limited to just one predictor variable. The key concept behind it is the idea that a change in one or more predictor variables will produce a linear change in the response variable.
We can think of this as a way to estimate the response variable by summing many weighted predictor variables, each of which has an impact on the final guess proportional to its importance.
For now, we'll limit ourselves to a single predictor variable in order to learn about linear regression more intuitively.
For ten years, Alfred has been planting trees in his exceedingly large backyard. Each year, he records how many seeds he planted in the spring, as well as how many new sprouts there are by fall.
With the data points graphed below, which line best represents the relationship between seeds planted and sprouts observed? Use your intuition! |
Home : Support : Online Help : Mathematics : Linear Algebra : LinearAlgebra Package : Standard : Normalize
Normalize(V, p, ip, conj, options)
(optional) equation of the form conjugate=true or false; specifies whether to use complex conjugates when computing the norm
The Normalize(V) function returns a Vector in which each component of V has been divided by the infinity-norm for V.
An alternate norm may be selected by using the Normalize(V, p) form of the calling sequence.
The inplace option (ip) determines where the result is returned. If given as inplace=true, the result overwrites the first argument. Otherwise, if given as inplace=false or if this option is not included in the calling sequence, the result is returned in a new Vector.
The inplace option must be used with caution since, if the operation fails, the original Vector argument may be corrupted.
The conjugate option (conj) determines whether complex conjugates are used when computing the norm of the input Vector, V. Note that the form conjugate=false has an effect only if a norm selection parameter is also given which specifies an even power norm (for example, Normalize(V, 2, conjugate=false).
This function is part of the LinearAlgebra package, and so it can be used in the form Normalize(..) only after executing the command with(LinearAlgebra). However, it can always be accessed through the long form of the command by using LinearAlgebra[Normalize](..).
\mathrm{with}\left(\mathrm{LinearAlgebra}\right):
\mathrm{Normalize}\left(〈3,0,4〉,\mathrm{Euclidean}\right)
[\begin{array}{c}\frac{\textcolor[rgb]{0,0,1}{3}}{\textcolor[rgb]{0,0,1}{5}}\\ \textcolor[rgb]{0,0,1}{0}\\ \frac{\textcolor[rgb]{0,0,1}{4}}{\textcolor[rgb]{0,0,1}{5}}\end{array}]
V≔〈1.55|1.56|1.53〉
\textcolor[rgb]{0,0,1}{V}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{ccc}\textcolor[rgb]{0,0,1}{1.55}& \textcolor[rgb]{0,0,1}{1.56}& \textcolor[rgb]{0,0,1}{1.53}\end{array}]
\mathrm{Normalize}\left(V,\mathrm{inplace}\right)
[\begin{array}{ccc}\textcolor[rgb]{0,0,1}{0.9935897436}& \textcolor[rgb]{0,0,1}{1.000000000}& \textcolor[rgb]{0,0,1}{0.9807692307}\end{array}]
V
[\begin{array}{ccc}\textcolor[rgb]{0,0,1}{0.9935897436}& \textcolor[rgb]{0,0,1}{1.000000000}& \textcolor[rgb]{0,0,1}{0.9807692307}\end{array}]
\mathrm{Normalize}\left(〈a,b,c〉,\mathrm{Euclidean},\mathrm{conjugate}=\mathrm{false}\right)
[\begin{array}{c}\frac{\textcolor[rgb]{0,0,1}{a}}{\sqrt{{\textcolor[rgb]{0,0,1}{a}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{b}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{c}}^{\textcolor[rgb]{0,0,1}{2}}}}\\ \frac{\textcolor[rgb]{0,0,1}{b}}{\sqrt{{\textcolor[rgb]{0,0,1}{a}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{b}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{c}}^{\textcolor[rgb]{0,0,1}{2}}}}\\ \frac{\textcolor[rgb]{0,0,1}{c}}{\sqrt{{\textcolor[rgb]{0,0,1}{a}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{b}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{c}}^{\textcolor[rgb]{0,0,1}{2}}}}\end{array}] |
We study optimal embeddings for the space of functions whose Laplacian Δu belongs to
{L}^{1}\left(\Omega \right)
\Omega \subset {ℝ}^{N}
is a bounded domain. This function space turns out to be strictly larger than the Sobolev space
{W}^{2,1}\left(\Omega \right)
in which the whole set of second-order derivatives is considered. In particular, in the limiting Sobolev case, when
N=2
, we establish a sharp embedding inequality into the Zygmund space
{L}_{\mathrm{𝑒𝑥𝑝}}\left(\Omega \right)
. On one hand, this result enables us to improve the Brezis–Merle (Brezis and Merle (1991) [13]) regularity estimate for the Dirichlet problem
\Delta u=f\left(x\right)\in {L}^{1}\left(\Omega \right)
u=0
on ∂Ω; on the other hand, it represents a borderline case of D.R. Adams' (1988) [1] generalization of Trudinger–Moser type inequalities to the case of higher-order derivatives. Extensions to dimension
N⩾3
are also given. Besides, we show how the best constants in the embedding inequalities change under different boundary conditions.
Classification : 46E35, 35B65
Mots clés : Sobolev embeddings, Pohožaev, Strichartz and Trudinger–Moser inequalities, Best constants, Elliptic equations, Regularity estimates in
{L}^{1}
, Brezis–Merle type results
author = {Cassani, Daniele and Ruf, Bernhard and Tarsi, Cristina},
title = {Best constants in a borderline case of second-order {Moser} type inequalities},
AU - Cassani, Daniele
AU - Ruf, Bernhard
AU - Tarsi, Cristina
TI - Best constants in a borderline case of second-order Moser type inequalities
Cassani, Daniele; Ruf, Bernhard; Tarsi, Cristina. Best constants in a borderline case of second-order Moser type inequalities. Annales de l'I.H.P. Analyse non linéaire, Tome 27 (2010) no. 1, pp. 73-93. doi : 10.1016/j.anihpc.2009.07.006. http://archive.numdam.org/articles/10.1016/j.anihpc.2009.07.006/
[1] D.R. Adams, A sharp inequality of J. Moser for higher order derivatives, Ann. of Math. 128 (1988), 385-398 | Zbl 0672.31008
[2] R.A. Adams, Reduced Sobolev inequalities, Canad. Math. Bull. 31 (1988) | Zbl 0662.46036
[3] R.A. Adams, J.J.F. Fournier, Sobolev Spaces, Pure Appl. Math. vol. 140, Elsevier (2003) | Zbl 0347.46040
[4] S. Agmon, A. Douglis, L. Nirenberg, Estimates near the boundary for solutions of elliptic partial differential equations satisfying general boundary conditions I, Comm. Pure Appl. Math. 12 (1959), 623-727 | Zbl 0093.10401
[5] A. Alvino, Sulla disuguaglianza di Sobolev in spazi di Lorentz, Boll. Unione Mat. Ital. 14 (1977), 148-156 | Zbl 0352.46020
[6] A. Alvino, V. Ferone, G. Trombetti, Estimates for the gradient of solutions of nonlinear elliptic equations with
{L}^{1}
data, Ann. Mat. Pura Appl. 178 (2000), 129-142 | Zbl 1220.35079
[7] A. Alvino, P.L. Lions, G. Trombetti, On optimization problems with prescribed rearrangements, Nonlinear Anal. 13 (1989), 185-220 | Zbl 0678.49003
[8] P. Bénilan, L. Boccardo, T. Gallouët, M. Pierre, J.L. Vazquez, An
{L}^{1}
-theory of existence and uniqueness of solutions of nonlinear elliptic equations, Ann. Sc. Norm. Super. Pisa Cl. Sci. 22 (1995), 241-273 | EuDML 84205 | Numdam | Zbl 0866.35037
[9] J. Bastero, M. Milman, F.J. Ruiz Blasco, A note on
L\left(\infty ,q\right)
spaces and Sobolev embeddings, Indiana Univ. Math. J. 52 (2003), 1215-1230 | Zbl 1098.46023
[10] C. Bennett, K. Rudnick, On Lorentz–Zygmund spaces, Dissertationes Math. 175 (1980), 1-72 | Zbl 0456.46028
[11] C. Bennett, R. Sharpley, Interpolation of Operators, Pure Appl. Math. vol. 129, Boston Academic Press, Inc. (1988) | Zbl 0647.46057
[12] L. Boccardo, Minimization problems with singular data, Milan J. Math. 74 (2006), 265-278 | Zbl 1214.35018
[13] H. Brezis, F. Merle, Uniform estimates and blow-up behavior for solutions of
-\Delta u=V\left(x\right){e}^{u}
in two dimensions, Comm. Partial Differential Equations 16 (1991), 1223-1253 | Zbl 0746.35006
[14] H. Brezis, W. Strauss, Semi-linear second order elliptic equations in
{L}^{1}
, J. Math. Soc. Japan 25 (1973), 565-590 | Zbl 0278.35041
[15] H. Brezis, S. Wainger, A note on limiting cases of Sobolev embeddings, Comm. Partial Differential Equations 5 (1980), 773-789 | Zbl 0437.35071
[16] V.I. Burenkov, Sobolev Spaces on Domains, Teubner-Texte Math. vol. 137, B.G. Teubner, Stuttgart (1998) | Zbl 0893.46024
[17] A. Cianchi, Higher-order Sobolev and Poincaré inequalities in Orlicz spaces, Forum Math. 18 (2006), 745-767 | Zbl 1120.46015
[18] A. Cianchi, Symmetrization and second order Sobolev inequalities, Ann. Mat. Pura Appl. 183 (2004), 45-77 | Zbl 1223.46033
[19] M. Cwikel, E. Pustylnik, Sobolev type embeddings in the limiting case, J. Fourier Anal. Appl. 4 (1998), 433-446 | EuDML 59575 | Zbl 0930.46027
[20] D.E. Edmunds, D. Fortunato, E. Jannelli, Critical exponents, critical dimensions and the biharmonic operator, Arch. Ration. Mech. Anal. 112 (1990), 269-289 | Zbl 0724.35044
[21] D.E. Edmunds, R. Kerman, L. Pick, Optimal Sobolev imbeddings involving rearrangement-invariant quasinorms, J. Funct. Anal. 170 (2000), 307-355 | Zbl 0955.46019
[22] D. Gilbarg, N.S. Trudinger, Elliptic Partial Differential Equations of Second Order, Classics Math., Springer-Verlag (1998) | Zbl 0691.35001
[23] K. Hansson, Imbedding theorems of Sobolev type in potential theory, Math. Scand. 45 (1979), 77-102 | EuDML 166665 | Zbl 0437.31009
[24] V.P. Ill'N, Conditions of validity of inequalities between
{L}^{p}
-norms of partial derivatives of functions of several variables, Proc. Steklov Inst. Math. 96 (1968), 259-305 | Zbl 0205.41801
[25] B. Kawohl, Rearrangements and Convexity of Level Sets in PDE, Lecture Notes in Math. vol. 1150, Springer-Verlag, Berlin (1985) | Zbl 0593.35002
[26] R. Kerman, L. Pick, Optimal Sobolev imbeddings, Forum Math. 18 (2006), 535-570 | Zbl 1120.46018
[27] S. Kesavan, Symmetrization & Applications, Ser. Anal. vol. 3, World Scientific (2006) | Zbl 1110.35002
[28] M. Krbec, H.-J. Shmeisser, Imbedding of Brezis–Wainger type. The case of missing derivatives, Proc. Roy. Soc. Edinburgh Sect. A 131 (2001), 667-700 | Zbl 0985.46018
[29] G.G. Lorentz, Some new functional spaces, Ann. of Math. 51 (1950), 37-55 | Zbl 0035.35602
[30] V.G. Maz'Ya, Some estimates for solutions of elliptic second-order equations, Soviet Math. Dokl. 2 (1961), 413-415 | Zbl 0115.08701
[31] M. Milman, E. Pustylnik, On sharp higher order Sobolev embeddings, Commun. Contemp. Math. 6 (2004), 495-511 | Zbl 1108.46029
[32] R. O'Neil, Convolution operators and
{L}^{p,q}
spaces, Duke Math. J. 30 (1963), 129-142 | Zbl 0178.47701
[33] D. Ornstein, A non-equality for differential operators in the
{L}_{1}
norm, Arch. Ration. Mech. Anal. 11 (1962), 40-49 | Zbl 0106.29602
[34] L. Orsina, Solvability of linear and semilinear eigenvalue problems with
{L}^{1}
data, Rend. Sem. Mat. Univ. Padova 90 (1993), 207-238 | EuDML 108304 | Numdam | Zbl 0822.35106
[35] J. Peetre, Espaces d'interpolation et théorème de Soboleff, Ann. Inst. Fourier 16 (1966), 279-317 | EuDML 73895 | Numdam | Zbl 0151.17903
[36] S.I. Pohožaev, The Sobolev embedding in the case
pl=n
, Proc. Tech. Sci. Conf. on Adv. Sci., Research 1964–1965, Moscow Energy Inst., Mathematics Section, Moscow (1965), 158-170
[37] S. Poornima, An embedding theorem for the Sobolev space
{W}^{1,1}
, Bull. Sci. Math. 107 no. 2 (1983), 253-259 | Zbl 0529.46025
[38] G. Stampacchia, Some limit cases of
{L}^{p}
-estimates for solutions of second order elliptic equations, Comm. Pure Appl. Math. 16 (1963), 505-510 | Zbl 0147.09202
[39] E. Stein, Singular Integrals and Differentiability Properties of Functions, Princeton Univ. Press (1970) | Zbl 0207.13501
[40] R.S. Strichartz, A note on Trudinger's extension of Sobolev's inequalities, Indiana Univ. Math. J. 21 (1972), 841-842 | Zbl 0241.46028
[41] C. Swanson, The best Sobolev constant, Appl. Anal. 47 (1992), 227-239 | Zbl 0739.46026
[42] G. Talenti, Elliptic equations and rearrangements, Ann. Sc. Norm. Super. Pisa Cl. Sci. 3 (1976), 697-718 | EuDML 83735 | Numdam | Zbl 0341.35031
[43] L. Tartar, Imbedding theorems of Sobolev spaces into Lorentz spaces, Boll. Unione Mat. Ital. (1998), 479-500 | EuDML 195538 | Zbl 0929.46028
[44] R.C.A.M. Van Der Vorst, Best constant for the embedding of the space
{H}^{2}\cap {H}_{0}^{1}\left(\Omega \right)
{L}^{2N/\left(N-4\right)}\left(\Omega \right)
, Differential Integral Equations 6 (1993), 259-276 | Zbl 0801.46033 |
Sigma and Pi Bonds | Brilliant Math & Science Wiki
Aditya Virani, Jordan Calmes, Christopher Williams, and
Rohit Ner
Sigma and pi bonds are chemical covalent bonds. Sigma and pi bonds are formed by the overlap of atomic orbitals. Sigma bonds are formed by end-to-end overlapping and Pi bonds are when the lobe of one atomic orbital overlaps another. Both acquired their names from the Greek letters and the bond when viewed down the bond axis. A sigma bond,
\sigma
, resembles a similar "s" atomic orbital, and a pi pond,
\pi
, has the same orbital symmetry of the p orbital (again, in both cases when viewed down the bond axis). Generally sigma bonds are stronger than pi bonds. Both are used extensively to predict the behavior of molecules in molecular orbital theory.
Sigma bonds
(\sigma)
are the strongest type of covalent bond, formed by head-on overlapping of atomic orbitals.
Note that sigma bond has been referred to as the strongest type of covalent bond because the extent of overlap is maximum in case of orbitals involved in the formation of the sigma bond.
Examples of sigma bonds with different types of overlap.[1]
Pi bonds
(\pi)
are a type of covalent bond formed by sideways or lateral overlapping of atomic orbitals.
Illustration of a pi bond forming from two p orbitals. Notice how the orientation of the atomic orbitals differs from the atomic orbitals involved in sigma bonds. The blue plane is the nodal plane. [2]
The figure below illustrates the sigma and pi bonds in an ethylene molecule (
C_2H_4
). Note that every single bond consists of one sigma bond, and that the double bond is made of one sigma bond and one pi bond. Likewise, a triple bond consists of one sigma bond and two pi bonds. Also observe how the sigma bonds are formed in between the atoms, and how the pi bonds are formed above, underneath, or beside the sigma bond. In general, the overlap of
n
atomic orbitals will create
n
molecular orbitals.
The following is the structure of acrylonitrile. Count the number of sigma and pi bonds in an acrylonitrile molecule.
Note that there are four single bonds, one double bond, and one triple bond in an acrylonitrile molecule. Hence there are
4+1+1=6
sigma bonds and
1+2=3
pi bonds in this molecule.
Sigma bond in C-H single bond Pi bond in C-C double bond Sigma bond in C-C double bond
2,2,5,5-tetramethyl-3-hexene
Which bond would you expect to break first in this molecule?
Many covalent compounds contain multiple bonds (double or triple bonds). One difference between single and multiple bonds is that single bonds only have a sigma bond, whereas multiple bonds have both sigma and pi bonds.
Like a double bond contains 1 sigma and 1 pi bond whereas a triple bond contains 1 sigma and 2 pi bonds.
Multiple bonds affect the electronic effects of a molecule and can alter physical properties like boiling point and melting point. Multiple bonds are also useful for deciphering spectra obtained via nuclear magnetic resonance (NMR).
CCoil, . σ bonds list. Retrieved August 24, 2016, from https://commons.wikimedia.org/wiki/File:Sigma-bonds-2D.svg
JoJan, ., & Vladsinger, . Two p-orbitals forming a π-bond.. Retrieved August 24, 2016, from https://en.wikipedia.org/wiki/Pi_bond#/media/File:Pi-Bond.svg
Cite as: Sigma and Pi Bonds. Brilliant.org. Retrieved from https://brilliant.org/wiki/sigma-and-pi-bonds/ |
Expansions of sin(nx) and cos(nx) | Brilliant Math & Science Wiki
Satyajit Mohanty and Jimin Khim contributed
If you have gone through double-angle formula or triple-angle formula, you must have learned how to express trigonometric functions of
2\theta
3\theta
\theta
only. In this wiki, we'll generalize the expansions of various trigonometric functions.
\sin(n\theta), \cos(n\theta),
\tan(n\theta)
Generalized Version of the above Expansion
\sin(\theta), \cos(\theta), \tan(\theta)
\theta
\theta
\sin^n(\theta), \cos^n(\theta),
\tan^n(\theta)
k\theta
\sin(n\theta), \cos(n\theta),
\tan(n\theta)
We have the following general formulas:
{\sin(n\theta) = \displaystyle \sum_{r=0 \atop 2r+1 \leq n} (-1)^r \binom{n}{2r+1} \cos^{n-2r-1}(\theta) \sin^{2r+1}(\theta)}
{\cos(n\theta) = \displaystyle \sum_{r=0 \atop 2r \leq n} (-1)^r \binom{n}{2r} \cos^{n-2r}(\theta) \sin^{2r}(\theta)}
{\tan(n\theta) = \frac{\displaystyle \sum_{r=0 \atop 2r+1 \leq n} (-1)^r \binom{n}{2r+1} \tan^{2r+1}(\theta)}{\displaystyle \sum_{r=0 \atop 2r \leq n} (-1)^r \binom{n}{2r} \tan^{2r}(\theta)}}
Display The Proof
Let's invoke De Moivre's theorem here. We have
\cos(n\theta) + i\sin(n\theta) = \big(\cos(\theta) + i\sin(\theta) \big)^n.
n
is a positive integer, the binomial theorem-N choose K holds for
\big(\cos(\theta) + i\sin(\theta) \big)^n
Hence, by expanding, we have
\big(\cos(\theta) + i\sin(\theta) \big)^n= \cos^n(\theta) + n\cos^{n-1}(\theta) \cdot i \sin(\theta) + \frac{n(n-1)}{1\cdot 2} \cos^{n-2}(\theta) \cdot i^2 \sin^2(\theta)+ \cdots.
i^2 = -1, i^3 = -i, i^4 = 1, i^5 = i, \ldots,
\begin{aligned} \big(\cos(\theta) + i\sin(\theta) \big)^n =&\left[\cos^n(\theta) - \dfrac{n(n-1)}{1\cdot 2} \cos^{n-2}(\theta) \cdot \sin^2(\theta) + \dfrac{n(n-1)(n-2)(n-3)}{1 \cdot 2 \cdot 3 \cdot 4} \cos^{n-4}(\theta) \cdot \sin^4(\theta) + \cdots\right] \\ &+ i \left[ n\cos^{n-1}(\theta) \cdot \sin(\theta) -\dfrac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3 } \cos^{n-3}(\theta) \cdot \sin^3(\theta) + \cdots \right]. \end{aligned}
By equating the real and imaginary parts, we obtain
\begin{aligned} \cos(n\theta) &= \cos^n(\theta) - \dfrac{n(n-1)}{1\cdot 2} \cos^{n-2}(\theta) \cdot \sin^2(\theta) + \dfrac{n(n-1)(n-2)(n-3)}{1 \cdot 2 \cdot 3 \cdot 4} \cos^{n-4}(\theta) \cdot \sin^4(\theta) + \cdots \\ \sin(n\theta) &= n\cos^{n-1}(\theta) \cdot \sin(\theta) -\dfrac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3 } \cos^{n-3}(\theta) \cdot \sin^3(\theta) + \cdots, \end{aligned}
where the terms of both the above series are alternately positive and negative. Also, each series continues till one of the factors in the numerator is zero and then ceases. Hence, proved.
To prove for
\tan(n\theta)
\tan(n\theta) = \dfrac{\sin(n\theta)}{\cos(n\theta)} = \dfrac{\displaystyle \sum_{r=0 \atop 2r+1 \leq n} (-1)^r \binom{n}{2r+1} \cos^{n-2r-1}(\theta) \sin^{2r+1}(\theta)}{\displaystyle \sum_{r=0 \atop 2r \leq n} (-1)^r \binom{n}{2r} \cos^{n-2r}(\theta) \sin^{2r}(\theta)}.
Divide the numerator and the denominator by
\cos^n(\theta)
{\tan(n\theta) = \frac{\displaystyle \sum_{r=0 \atop 2r+1 \leq n} (-1)^r \binom{n}{2r+1} \tan^{2r+1}(\theta)}{\displaystyle \sum_{r=0 \atop 2r \leq n} (-1)^r \binom{n}{2r} \tan^{2r}(\theta)}}.
Hence, our proof is complete.
_\square
For angles
\theta_1, \theta_2, \theta_3, \ldots
{\sin(\theta_1 + \theta_2 + \theta_3 + \cdots) = \cos(\theta_1) \cdot \cos(\theta_2) \cdot \cos(\theta_3) \dotsm [s_1 - s_3 + s_5 - s_7 + \cdots]}
{ \cos(\theta_1 + \theta_2 + \theta_3 + \cdots) = \cos(\theta_1) \cdot \cos(\theta_2) \cdot \cos(\theta_3) \dotsm [1 - s_2 + s_4 - s_6 + \cdots]}
{\tan(\theta_1 + \theta_2 + \theta_3 + \cdots) = \dfrac{s_1 - s_3 + s_5 - s_7 + \cdots}{1 - s_2 + s_4 - s_6 + \cdots}},
{s_n = \displaystyle \sum_\text{cyclic} \tan(\theta_1) \tan(\theta_2) \tan(\theta_3) \dotsm \tan(\theta_n)}.
Sorry. This section is temporarily empty. It will be completed as soon as possible.
\sin(\theta), \cos(\theta), \tan(\theta)
\theta
\theta
We'll show here, without using any form of Taylor's series, the expansion of
\sin(\theta), \cos(\theta), \tan(\theta)
\theta
\theta
Here are the generalized formulaes:
{\sin(\theta) = \displaystyle \sum_{r=0}^\infty (-1)^r \dfrac{\theta^{2r+1}}{(2r+1)!}}
{\cos(\theta) = \displaystyle \sum_{r=0}^\infty (-1)^r \dfrac{\theta^{2r}}{(2r)!}}
{\tan(\theta) = \theta + \dfrac{\theta^3}{3} + \dfrac{2\theta^5}{15} + \cdots}.
\sin^n(\theta), \cos^n(\theta),
\tan^n(\theta)
k\theta
n
{\cos^n(\theta) = \dfrac{1}{2^{n-1}} \left[ \displaystyle \sum_{r=0 \atop 2r < n} \binom{n}{2r} \cos \big( (n-2r)\theta \big) \right] + \dfrac{1}{2^n} \displaystyle\binom{n}{n/2}}
{\sin^n(\theta) = \dfrac{(-1)^{n/2}}{2^{n-1}} \left[ \displaystyle \sum_{r=0 \atop 2r < n} (-1)^r \binom{n}{2r} \cos \big( (n-2r)\theta \big)\right] + \dfrac{1}{2^n} \displaystyle\binom{n}{n/2} }.
n
{\cos^n(\theta) = \dfrac{1}{2^{n-1}} \left[ \displaystyle \sum_{r=0 \atop 2r < n} \binom{n}{2r} \cos \big( (n-2r)\theta \big) \right] }
{\sin^n(\theta) = \dfrac{(-1)^{n/2}}{2^{n-1}} \left[ \displaystyle \sum_{r=0 \atop 2r < n} (-1)^r \binom{n}{2r} \sin \big( (n-2r)\theta \big) \right] }.
Cite as: Expansions of sin(nx) and cos(nx). Brilliant.org. Retrieved from https://brilliant.org/wiki/expansions-of-certain-trigonometric-functions/ |
Units And Measurements, Popular Questions: CBSE Class 11-science PHYSICS, Physics Part I - Meritnation
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A dimensionless quantity --
a)Never has a unit
b)Always has a unit
c)May have a unit
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Energy = ½ MV2. How can you help him with the help of dimensional
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\frac{E}{B}
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Can anybody please tell me what does U refer to????? Please give my answer!!!!. I will surely give thumbs up!!!!.
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What is Personal Error, Instrumental Error and Random Error? Explain with giving an example?
PLEASE TELL ME............
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<W1 + W2
>W1 + W2
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derivation of three laws of motion
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i. 500.00kg.
ii. 0.0005kg.
iii. 6.00kg
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dimensional formula for newton
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v=root under E/P ,
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the displacement time graph for the two particles A and B are straight lines inclined at angles of 30 and 45 degrees with the time axis.what is the ratio of the velocities VA : VB ?
PLZ ANS AS SOON AS POSSIBLE ...........TOMMOROW IS MY EXAM
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Two resistors of resistances R1= 100+/-3 ohm and R2= 200+/-4 ohm are connected (a) in series, (b) in parallel. Find the equivalent resistance of the (a) series combination, (b) parallel combination. Use for (a) the realtion R = R1+ R2and for (b) 1/R ' = 1/R1+ 1/R2andΔR '/R '2=ΔR1/R12+ΔR2/R22.
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the period of oscillation of a simple pendulum is T=2pi√4g. measured value of l is 20.0 cm known to 1mm accuracy and time for 100 oscillations of the pendulum is found to be 90s using a wrist watch of 1s resolution. What is the determination of g.
a man of mass 60kg is standing on a spring balance inside a lift .if the lift falls freely downwards, then the reading of springbalance would be
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To evaluate: The simplified form of the expression 10.7-(-9.8).
William Collins 2021-12-07 Answered
The given expression is 10.7 - (-9.8).
Write the subtraction in terms of addition as,
The given expression becomes,
10.7-(-9.8) = 10.7 +9.8
Now, the two values are of same signs and for the addition of any two numbers of same signs, add their absolute values and then attach their common signs.
Absolute values of the numbers are,
|10.7| = 10.7
Now, add the absolute value.
10.7+9.8 = 20.5
As the numbers are positive, the answer is also positive
So, the difference of the given expression is 20.5.
Therefore, the simplified form of the expression 10.7 - (-9.8) is
\frac{5}{6}
Why is 1 not considered a prime number?
Or, why is the definition of prime numbers given for integers greater than 1?
Find the absolute maximum value and the absolute minimum value of the following function
f(x)=2|x| on [-1,6]
The sum of the numbers -11 and 7 by using absolute value.
f\left(x\right)=3-4{x}^{2},-4\le x\le 1
The absolute maximum value is
and this occurs at x=
The absolute minimum value is
Express the interval in terms of an inequality involving absolute value.
f\left(x\right)=2{x}^{3}-6{x}^{2}-18x+9
What is the absolute minimum of f(x) on [-2,4]?
What is the absolute maximum of f(x) on [-2,4]?
|x-4|\le 10 |
Will a Solution Form? - Course Hero
General Chemistry/Solutions and Colloids/Will a Solution Form?
Solutions will form only if solute and solvent particles have similar intermolecular forces.
Two substances are miscible if they are able to form a homogeneous (evenly distributed) mixture when added together in any proportion. The enthalpy of solution (
\Delta H_{\rm{soln}}
) indicates whether a solution forms when two substances are mixed. If the enthalpy of solution is less than zero,
\Delta H_{\rm{soln}}<0
, the attractive forces between solvent and solute molecules are greater than the sum of the attractive forces between like molecules, and the solution formation is exothermic. This can happen when hydrogen bonds (an attractive intermolecular force between a hydrogen atom on one molecule and a more electronegative atom on another molecule) exist between unlike molecules. When
\Delta H_{\rm{soln}}<0
, the solution will form. This is the case when two liquids are miscible or when a liquid solvent can dissolve a solid solute.
A solution for which
\Delta H_{\rm{soln}}=0
is an ideal solution. The attractive forces between solvent and solute molecules are approximately equal to the sum of the attractive forces between solvent-solvent molecules and solute-solute molecules. The solute molecules in an ideal solution are similar to the solvent molecules, as in the case of benzene (C6H6) and toluene (C6H5CH3).
Benzene and toluene have very similar structures. When mixed, they form an ideal solution with
\Delta H_{\rm{soln}}=0
Finally, when the heat needed to separate the solute-solute and solvent-solvent molecules is less than the heat released when solvent-solute molecules interact,
\Delta H_{\rm{soln}}>0
, and the solution formation is endothermic. A
\Delta H_{\rm{soln}}
that is both large and positive indicates that the solute-solvent attractive forces are much smaller than the attractive forces between the like molecules, and a solution will not form. This is the case for water and oil. These liquids are immiscible, which means they are not able to form a homogeneous (evenly distributed) mixture when added together. The nonpolar oil molecules are not attractive enough to the water molecules to break the hydrogen bonds between the water molecules. Just because
\Delta H_{\rm{soln}}>0
, however, does not necessarily mean that a solution will not form. The dissolution of sodium chloride in water has a
\Delta H_{\rm{soln}}
of approximately 3.88 kJ/mol, and yet table salt is readily soluble in water. This is because the entropy of the solution is greater than the entropy of the pure solute and solvent. Entropy is a measure of the disorder of a system, and the tendency of the universe is for entropy to increase, so any process that has a net increase of entropy will happen spontaneously, without external assistance. So, if
\Delta H_{\rm{soln}}
is positive but small, dissolution is likely to occur.
Water and octane are dissimilar and will not form a homogeneous mixture. The octane is less dense and forms a layer on top of the water. Water and ethyl alcohol form a homogeneous mixture because the
\rm{{-}OH}
group on the small carbon chain (two-carbon molecule) creates a small dipole that is similar to water's dipole. Water and octyl alcohol do not form a homogeneous mixture because the small dipole created by the
\rm{{-}OH}
group is not strong enough to overcome the nonpolar property of the long carbon chain (eight-carbon molecule).
Molecules with similar structures have similar intermolecular forces and are therefore soluble or miscible with each other, such as HOH (water) and CH3CH2OH (ethyl alcohol). The familiar H2O formula for water is written as HOH to emphasize the
\rm{{-}OH}
group in its structure. If the intermolecular forces are significantly different, they will be immiscible, such as HOH and CH3(CH2)6CH3 (octane). Both water and ethyl alcohol have
\rm{{-}OH}
groups that contribute to hydrogen bonding. They are both polar molecules. Octane, on the other hand, is nonpolar and has no
\rm{{-}OH}
group. Octane is immiscible in water.
The presence of an
\rm{{-}OH}
group in a compound's structure does not necessarily mean it is soluble in water. For example, octyl alcohol (CH3(CH2)6CH2OH) could be expected to be soluble in water because it has an
\rm{{-}OH}
group that could form hydrogen bonds with water. However, the carbon chain is so long that the alcohol's
\rm{{-}OH}
group cannot compensate for its nonpolar properties. Octyl alcohol is not soluble in water. The molecular structures of octyl alcohol and water are too dissimilar.
<How Solutions Form>Electrolytic Solutions |
z-test - MATLAB ztest - MathWorks Australia
z=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}},
\overline{x}
is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size. Under the null hypothesis, the test statistic has a standard normal distribution. |
Gray-Coded Binary Ordering - MATLAB & Simulink - MathWorks Switzerland
Compare Error Rate for Gray- and Binary-Coded Ordering
Gray coding is a technique that multilevel modulation schemes often use to minimize the bit error rate. It consists of ordering modulation symbols so that the binary representations of adjacent symbols differ by only one bit. This section shows a communications system with Gray-coded 8-ary phase shift keying (8-PSK) modulation to compare the error rate performance of Gray and natural binary coded bit ordering.
The example in this section modulates gray and natural binary coded data frames using the 8-PSK method. The data frames passes through an AWGN channel, and are demodulated using an 8-PSK demodulator. Error rate calculator System objects measure the symbol and bit error rates.
In this communications system, PSK Modulator System objects:
Accept binary-valued inputs that represent integers between 0 and M – 1. M is the modulation order and is equal to 8 for 8-PSK modulation.
Map binary representations to constellation points using Gray-coded and natural binary-coded ordering.
Produce unit-magnitude complex phasor outputs, with evenly spaced phases between 0 and 2π(M – 1)/M.
This table indicates the relationship between Gray-coded binary representations in the input and phasors in the output. The second column of the table is an intermediate representation that the System object uses in its computations.
000 0 exp(0)
011 2 exp(jπ/2) = exp(j2π/4)
110 4 exp(jπ) = exp(j4π/4)
This table sorts the first two columns from the previous table, according to the output values. This sorting makes it clearer that there is only a 1 bit difference between neighboring symbols. In the following figure, notice that the numbers in the second column of the table appear in counterclockwise order.
exp(jπ/2) = exp(j2π/4) 011
exp(jπ) = exp(j4π/4) 110
Compare Gray coding with natural binary coding by using appropriately configured PSK modulator and PSK demodulator System objects. This simulation iterates over a range of bit energy to noise power spectral density,
{\mathit{E}}_{\mathrm{b}}/{\mathit{N}}_{0}
, values and runs until either the specified maximum number of bit errors (maxNumErrs) or the maximum number of bits (maxNumBits) is reached for Gray coding for each
{\mathit{E}}_{\mathrm{b}}/{\mathit{N}}_{0}
Initialize the system variables and create System objects for modulation, demodulation, AWGN channel, and error rate operations. Since the comm.AWGNChannel System object™ and the randi function use the default random stream, set the random number generator seed to ensure repeatable results. The state of the random number generator is stored before setting the random stream seed, and restored at the end of the example.
M = 8; % Modulation order for 8-PSK
maxNumErrs=100;
maxNumBits=1e8;
prevState = rng;
Create PSK modulator and demodulator System objects to map the binary input data to 8-PSK Gray- and binary-coded constellations.
pskmod = comm.PSKModulator('ModulationOrder',M, ...
'SymbolMapping','Gray', ...
pskdemod = comm.PSKDemodulator('ModulationOrder',M, ...
'OutputDataType','uint8', ...
pskmodb = comm.PSKModulator('ModulationOrder',M, ...
'SymbolMapping','Binary', ...
pskdemodb = comm.PSKDemodulator('ModulationOrder',M, ...
Create an AWGN channel System object to add noise to the modulated signal. The noise method is configured to
{\mathit{E}}_{\mathrm{b}}/{\mathit{N}}_{0}
for the processing loop. The PSK modulator generates symbols with 1 W of power, so the signal power property of the AWGN channel object is set to 1 W also.
'BitsPerSymbol',log2(M),'SignalPower',1);
Create a symbol error rate and bit error rate calculator System objects to compare the demodulated integer and bit data with the original source data. This comparison yields symbol error and bit error statistics. The output of the error rate calculator System object is a three-element vector containing the calculated error rate, the number of errors observed, and the amount of data processed. The simulation uses the three-element vector to determine when to stop the simulation.
symerror = comm.ErrorRate;
biterror = comm.ErrorRate;
biterrorb = comm.ErrorRate;
Configure a frame processing loop where data is coded, modulated, and demodulated using 8-PSK modulation. The loop simulates the communications system for
{\mathit{E}}_{\mathrm{b}}/{\mathit{N}}_{0}
values in the range 0 dB to 12 dB in steps of 2 dB.
{\mathit{E}}_{\mathrm{b}}/{\mathit{N}}_{0}
value, the simulation stops when either the maximum number of errors (maxNumErrs) or the maximum number of bits (maxNumBits) processed by the bit error rate calculator System object is reached for the Gray coded bits.
EbNoVec = 0:2:12; % Eb/No values to simulate
SERVec = zeros(size(EbNoVec)); % SER history
BERVec = zeros(size(EbNoVec)); % BER history for Gray ordered
BERVecb = zeros(size(EbNoVec)); % BER history for binary ordered
for p = 1:length(EbNoVec)
reset(symerror);
reset(biterrorb);
awgnchan.EbNo = EbNoVec(p);
% Reset SER / BER for the current Eb/No value
SER = zeros(3,1);
BER = zeros(3,1);
while (BER(2)<maxNumErrs) && (BER(3)<maxNumBits)
txSym = randi([0 M-1],SamplesPerFrame,1,'uint8');
txBits = reshape(de2bi(txSym,log2(M),'left-msb')',[],1); % Convert symbols to bits
tx = pskmod(txBits);
txb = pskmodb(txBits);
rx = awgnchan(tx);
rxb = awgnchan(txb);
rxBits = pskdemod(rx);
rxBitsb = pskdemodb(rxb);
rxSym = bi2de(reshape(rxBits,log2(M),[])','left-msb');
SER = symerror(txSym,rxSym); % Symbol error rate for Gray-coded data
BER = biterror(txBits,rxBits); % Bit error rate for Gray-coded data
BERb = biterrorb(txBits,rxBitsb); % Bit error rate for natural binary-coded data
% Save history of SER and BER values
SERVec(p) = SER(1);
BERVec(p) = BER(1);
BERVecb(p) = BERb(1);
Restore the default stream.
rng(prevState)
Analyze the data from the example and compare theoretical performance with simulation performance. The theoretical symbol error probability of MPSK is
{\mathit{P}}_{\mathrm{E}}\left(\mathit{M}\right)=\mathrm{erfc}\left(\sqrt{\frac{{\mathit{E}}_{\mathit{S}}}{{\mathit{N}}_{0}}}\mathrm{sin}\left(\frac{\pi }{\mathit{M}}\right)\right)
where erfc is the complementary error function,
{\mathit{E}}_{\mathit{S}}/{\mathit{N}}_{0}
is the ratio of energy in a symbol to noise power spectral density, and M is the number of symbols.
To determine the bit error probability, convert the symbol error probability,
{\mathit{P}}_{\mathrm{E}}
, to its bit error equivalent. There is no general formula for the symbol to bit error conversion. Nevertheless, upper and lower limits are easy to establish. The actual bit error probability,
{\mathit{P}}_{\mathit{b}}
, can be shown to be bounded by
\frac{{\mathit{P}}_{\mathrm{E}}\left(\mathit{M}\right)}{{\mathrm{log}}_{2}\mathit{M}}\le {\mathit{P}}_{\mathit{b}}\le \frac{\mathit{M}/2}{\mathit{M}-1}{\mathit{P}}_{\mathrm{E}}\left(\mathit{M}\right)
Calculate theoretical error probabilities by using the use the berawgn function. Plot the simulated symbol error rate for Gray coding, bit error rate for Gray and natural binary coding, and the theoretical symbol error and bit error probabilities for Gray coding.
[theorBER,theorSER] = berawgn(EbNoVec,'psk',M,'nondiff');
semilogy(EbNoVec,SERVec,'o',EbNoVec,BERVecb,'x',EbNoVec,BERVec,'*', ...
EbNoVec,theorSER,'-',EbNoVec,theorBER,'-');
legend('Symbol error rate','Bit error rate (Binary)','Bit error rate (Gray)', ...
'Theoretical Symbol error rate','Theoretical Bit error rate', ...
xlabel('Eb/No (dB)'); ylabel('Error Probability');
title('Symbol and Bit Error Probability'); |
That parametric equations contain more information than just the shape of the cu
That parametric equations contain more information than just the shape of the curve. Write a short paragraph explaining this statement. Use the follow
That parametric equations contain more information than just the shape of the curve. Write a short paragraph explaining this statement. Use the following example and your answers to parts (a) and (b) below in your explanation. The position of a particle is given by the parametric equations
x=\text{ }\mathrm{sin}\text{ }\left(t\right)\text{ }and\text{ }y=\text{ }\mathrm{cos}\text{ }\left(t\right)
where t represents time. We know that the shape of the path of the particle is a circle. a) How long does it take the particle to go once around the circle? Find parametric equations if the particle moves twice as fast around the circle. b) Does the particle travel clockwise or counterclockwise around the circle? Find parametric equations if the particle moves in the opposite direction around the circle.
delilnaT
Step 1 (a) Note that, the position of the particle is given by the parametric equations
x=\text{ }\mathrm{sin}t,\text{ }and\text{ }y=\text{ }\mathrm{cos}\text{ }t
. The parametric equations contain more than just shape of the curve. They also represent the direction of curve as traveling. If a position of a particle is determined by the equation
x=\text{ }\mathrm{sin}\text{ }t,\text{ }y=\text{ }\mathrm{cos}\text{ }t,
this set of equations denotes which direction the particle is traveling based on different times t. For example, at
t=0,\text{ }\text{the particle is at the point}\text{ }\left(0,\text{ }1\right)\text{ }\text{but at time}\text{ }t=\text{ }\frac{\pi }{2}\text{ }\text{the particle has moved to the point}\text{ }\left(1,\text{ }0\right)
in a clockwise direction As the period of the parametric equations is
2\text{ }\pi
, to find for the particle to travel a full rotation around the circle. It will take the time
t=2\text{ }\pi
to traverse the circle in a clockwise direction. To travel the circle twice as fast simply double the coefficient inside each trigonometric function and the parametric equations are
x=\text{ }\mathrm{sin}\text{ }2t,\text{ }y=\text{ }\mathrm{cos}\text{ }2t.
Thus, the time that will be taken by the particle to go once around the circle is
t=2\text{ }\pi \text{ }\text{and the parametric equations, the particle moves twice as fast around the circle are}\text{ }x=\text{ }\mathrm{sin}\text{ }2t,\text{ }y=\text{ }\mathrm{cos}\text{ }2t.
Step 2 (b) Note that, the particle travels clockwise. For example, at
t=0,\text{ }\text{the particle is at the point}\text{ }\left(0,\text{ }1\right),\text{ }\text{but at the time}\text{ }t=\text{ }\frac{\pi }{2}\text{ }\text{the particle has moved to the point}\text{ }\left(1,\text{ }0\right)
in a clockwise direction. The parametric equations when the particle travels in the opposite direction, the parametric equations will be exchanged. That are,
x=\text{ }\mathrm{cos}\text{ }t,\text{ }y=\text{ }\mathrm{sin}\text{ }t.
Thus, the particle travels clockwise and if the particle travels in opposite direction around the circle, the parametric equations are
x=\text{ }\mathrm{cos}\text{ }t,\text{ }y=\text{ }\mathrm{sin}\text{ }t.
Find the area of the parallelogram with vertices A(-3, 0), B(-1, 5), C(7, 4), and D(5, -1).
Find an equation of the plane tangent to the following surface at the given point.
7xy+yz+4xz-48=0;\text{ }\left(2,2,2\right)
Parametric equations and a value for the parameter t are given
x={t}^{2}+3,y=6-{t}^{3}
.t=2
Find the coordinates of the point on the plane curve described by the parametric equations corresponding to the given value of t.
Two components of a minicomputer have the following joint pdf for their useful lifetimes X and Y:
f\left(x,y\right)=\left\{\begin{array}{ll}x{e}^{-x\left(1+y\right)}& x\ge 0\text{ }and\text{ }\ge 0\\ 0& otherwise\end{array}
a) What is the probability that the lifetime X of the first component exceeds 3?
b) What are the marginal pdf's of X and Y? Are the two lifetimes independent? Explain.
c) What is the probability that the lifetime of at least one component exceeds 3?
Working with parametric equations Consider the following parametric equations. a. Eliminate the parameter to obtain an equation in x and y. b. Describe the curve and indicate the positive orientation.
x=\text{ }-7\text{ }\mathrm{cos}\text{ }2t,\text{ }y=\text{ }-7\text{ }\mathrm{sin}\text{ }2t,\text{ }0\text{ }\le \text{ }t\text{ }\le \text{ }\pi
Find the points on the cone
{z}^{2}={x}^{2}+{y}^{2}
that are closest tothe point (2, 2,0).
The following data was collected about students in Mr. Rexingers |
Estimate frequency response with fixed frequency resolution using spectral analysis - MATLAB spa - MathWorks Nordic
\pi
y\left(t\right)=G\left(q\right)u\left(t\right)+v\left(t\right)
G\left(q\right)u\left(t\right)=\sum _{k=1}^{\infty }g\left(k\right)u\left(t-k\right)
G\left(q\right)=\sum _{k=1}^{\infty }g\left(k\right){q}^{-k}\text{ }{q}^{-1}u\left(t\right)=u\left(t-1\right)
{\stackrel{^}{\Phi }}_{v}\left(\omega \right)
{\stackrel{^}{G}}_{N}\left({e}^{i\omega }\right)=\frac{{\stackrel{^}{\Phi }}_{yu}\left(\omega \right)}{{\stackrel{^}{\Phi }}_{u}\left(\omega \right)}
{\stackrel{^}{\Phi }}_{v}\left(\omega \right)={\stackrel{^}{\Phi }}_{y}\left(\omega \right)-\frac{{|{\stackrel{^}{\Phi }}_{yu}\left(\omega \right)|}^{2}}{{\stackrel{^}{\Phi }}_{u}\left(\omega \right)}
y\left(t\right)=G\left(q\right)u\left(t\right)+v\left(t\right)
{\Phi }_{y}\left(\omega \right)={|G\left({e}^{i\omega }\right)|}^{2}{\Phi }_{u}\left(\omega \right)+{\Phi }_{v}\left(\omega \right)
{\Phi }_{yu}\left(\omega \right)=G\left({e}^{i\omega }\right){\Phi }_{u}\left(\omega \right)
{\Phi }_{v}\left(\omega \right)\equiv \sum _{\tau =-\infty }^{\infty }{R}_{v}\left(\tau \right){e}^{-iw\tau }
{\stackrel{^}{\Phi }}_{yu}\left(\omega \right)
{\stackrel{^}{\Phi }}_{u}\left(\omega \right)
v\left(t\right)=H\left(q\right)e\left(t\right)
\lambda
{\Phi }_{v}\left(\omega \right)=\lambda {|H\left({e}^{i\omega }\right)|}^{2}
\begin{array}{l}{\stackrel{^}{R}}_{y}\left(\tau \right)=\frac{1}{N}\sum _{t=1}^{N}y\left(t+\tau \right)y\left(t\right)\\ {\stackrel{^}{R}}_{u}\left(\tau \right)=\frac{1}{N}\sum _{t=1}^{N}u\left(t+\tau \right)u\left(t\right)\\ {\stackrel{^}{R}}_{yu}\left(\tau \right)=\frac{1}{N}\sum _{t=1}^{N}y\left(t+\tau \right)u\left(t\right)\end{array}
\begin{array}{l}{\stackrel{^}{\Phi }}_{y}\left(\omega \right)=\sum _{\tau =-M}^{M}{\stackrel{^}{R}}_{y}\left(\tau \right){W}_{M}\left(\tau \right){e}^{-i\omega \tau }\\ {\stackrel{^}{\Phi }}_{u}\left(\omega \right)=\sum _{\tau =-M}^{M}{\stackrel{^}{R}}_{u}\left(\tau \right){W}_{M}\left(\tau \right){e}^{-i\omega \tau }\\ {\stackrel{^}{\Phi }}_{yu}\left(\omega \right)=\sum _{\tau =-M}^{M}{\stackrel{^}{R}}_{yu}\left(\tau \right){W}_{M}\left(\tau \right){e}^{-i\omega \tau }\end{array}
{W}_{M}\left(\tau \right)
{\stackrel{^}{G}}_{N}\left({e}^{i\omega }\right)
{\stackrel{^}{\Phi }}_{v}\left(\omega \right)
{\stackrel{^}{G}}_{N}\left({e}^{i\omega }\right)=\frac{{\stackrel{^}{\Phi }}_{yu}\left(\omega \right)}{{\stackrel{^}{\Phi }}_{u}\left(\omega \right)}
{\Phi }_{v}\left(\omega \right)\equiv \sum _{\tau =-\infty }^{\infty }{R}_{v}\left(\tau \right){e}^{-iw\tau }
S=\sum _{m=-M}^{M}Ez\left(t+m\right)z{\left(t\right)}^{\prime }{W}_{M}\left({T}_{s}\right)\mathrm{exp}\left(-i\omega m\right) |
karlnapf | herr strathmann
Author: karlnapf
November 29, 2019 November 30, 2019 ~ karlnapf ~ Leave a comment
If six months ago I was told that I would climb El Capitan within the year I would not have believed it
First sentence of a blog post I read a few months ago. Had I been told I would write the same myself soon, I would not have believed it either.
The last year and a half has certainly been intense. Roughly: I stopped working, converted a van, drove to Spain, and climbed loads: adventure trad in Costa Blanca, sport in El Chorro, slabs and off-widths in La Pedriza, granite corner cracks in Galayos and Torozo, monstrous alpine classics in the Pyrenees, amazing limestone canalizos in the Picos de Europa.
And then finally Yosemite, confronting Lurking Fear VI 5.7 C2+ on the Captain. Stating this as a goal always felt somewhat ridiculous, almost like a joke. But in hindsight, all we did was to start climbing, and follow Andy Kirkpatrick’s advice: When things get hard don’t come down.
It is hard to summarise all these experiences in words — many firsts, many achievements, many failures and many I-will-never-climb-again-moments — so I will not. Below are some pictures.
Costa Blanca and El Chorro in southern Spain
La Pedriza <3, Galayos, and Torozo in central Spain
Pyrenees and Asturias
\varphi \left(x\right)
<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ϕ</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo></math>
\phi(x)
\varphi \left(x\right)
<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ϕ</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo></math>
\phi(x)
\log p(x)=\sum_{i=1}^n \alpha_i k(z_i, x)
k(x,y)=\exp\left(-\Vert x-y\Vert ^2 / \sigma \right)
{\mathrm{zi}}_{}
z_i
\sigma
{\mathrm{\alpha i}}^{}
<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>αi</mi></msub></math>
\alpha_i
{\mathrm{\alpha i}}^{}
<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>αi</mi></msub></math>
\alpha_i
Job opportunity with Shogun and the ATI
October 17, 2018 June 28, 2019 ~ karlnapf ~ Leave a comment
We, the community around the shogun.ml [1] open-source machine learning library, are looking for a developer for a paid 6 month pilot project (October 18 – March 19) on improving meta-learning capabilities (openml, coreml). The ideal candidate is a highly motivated MSc/PhD/postdoc with the desire to get involved in the open-source movement, who
is based on London
is able to start working in October
is flexible enough to spend full-time or at least 50% on the project
has a background in designing software in C++ (gcc, valgrind, C++11, etc)
(optional) has knowledge of openml [2] and coreml [3]
(optional) has experience with build management and dev-ops tools (git, cmake, travis, buildbot, linux, docker, etc)
(optional) has experience in computational sciences (ml, stats, etc)
(optional) has contributed to open-source before
The project is funded by with the Alan Turing institute, and at least part of the work will be located there. You will be supervised by the Shogun core development team, partly in person and partly remotely. This is a great opportunity to get involved in one of the oldest ML libraries out there, getting your hands dirty on a huge code-base, and tipping into the open-source community.
The project is currently in planning stage. After a successful pilot, there is the option for an extension. If you are interested, please get in touch via the developers, the mailing list, or even better, read how to get involved [4] and send us a pull-request for an entrance task [5] on github. See our website for contact details.
Shogun is a library aiming to offer unified and efficient machine learning methods. Its core is written in C++ and it interfaces to a large number of modern computing languages. The Shogun community is vibrant, diverse, and international. Shogun is a fiscally sponsored project of NumFOCUS, a nonprofit dedicated to supporting the open source scientific computing community.
[1] http://shogun.ml/
[2] https://www.openml.org/
[3] https://github.com/apple/coremltools
[4] https://github.com/shogun-toolbox/shogun/wiki/Getting-involved
[5] https://github.com/shogun-toolbox/shogun/issues
[6] https://numfocus.org/
DS3 summer school in Paris
June 30, 2018 June 30, 2018 ~ karlnapf ~ Leave a comment
I had the pleasure to run a (actually two) practical on “representing and comparing probabilities using kernels” at the DS3 summer school at the Polytechnique in Paris, following a lecture by Arthur Gretton. Thanks to Zoltan Szabo for organising the session.
We covered the implementation basics of two-sample testing, independence testing, and goodness-of-fit testing, with examples including testing the quality of GAN samples, detecting dependence across translated documents, and more. I even managed to sneak Shogun into the practical 😉 Good fun overall!
Slides 1, slides 2, and practical session notebook.
Bleau & Swanage
June 5, 2018 ~ karlnapf ~ Leave a comment
May 19, 2018 May 23, 2018 ~ karlnapf ~ Leave a comment
May 5, 2018 May 23, 2018 ~ karlnapf ~ Leave a comment
April 23, 2018 June 28, 2019 ~ karlnapf ~ Leave a comment
April 21, 2018 May 24, 2018 ~ karlnapf ~ Leave a comment |
Wile E. Coyote is so desperate to catch the Roadrunner that he builds himself a catapult to use on those darn vertical cliffs he is always running into. He figures he will need it to shoot him
500
meters straight up. How fast must the catapult propel him upwards to reach that height? (Use
–5
m/sec2 for the acceleration due to gravity. As you know, Cartoonland has a different gravitational constant, allowing cartoon characters to fall from great heights without being injured. Do not try this at home.)
\text{Set up a position function: }s(t)=\frac{a_{0}}{2}t^{2}+v_{0}t+s_{0}
a_{0} =
v_{0} =
s_{0} =
Notice that the question is asking, 'how fast?'.
Be mindful of units. |
Parametric design and analysis of a mortar base plate | JVE Journals
Fengfeng Wang1 , Guolai Yang2
1, 2School of Mechanical Engineering, Nanjing University of Science and Technology, Nanjing, P. R. China
Received 19 August 2019; accepted 5 September 2019; published 28 November 2019
Copyright © 2019 Fengfeng Wang, et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
In order to improve the efficiency of designing and analyzing a mortar base plate, parametric technology has been applied to the design and analysis of the base plate. By conducting characteristic analysis of the base plate, critical design parameters extracted have been driven as characteristics to perform secondary development on UG, so that the base plate could be automatically modeled in three dimensions. Meanwhile, Python has also been applied in the secondary development of Abaqus, so as to realize the automatic finite element modeling of the base plate. After conducting parametric analysis on the model established, sensitivities of design variables of the base plate have been analyzed using an approximate model in combination with the sample designed by the optimal Latin hypercube approach. Moreover, parametric optimization carrying out on results of sensitivity analysis could give feedback to parametric design for guiding parameter values of design variables to improve the design quality
Keywords: mortar base plate, parametric design, sensitivity analysis.
As the base plate is an integral part of the mortar, the rationality of the structure of the base plate has a significant influence on the shooting precision and stability of the mortar [1]. The base plate should be repeatedly tested and modified in the design stage before reaching the specific requirements, resulting in problems such as high research and development costs and long design cycles. The parametric technology is the best choice for improving the design and analysis efficiency. Specifically, parametric design is to restrain the sequence of structural dimensions in geometry with various parameters. In order to reach a new target geometry, means are provided for the variability, the reusability, and the parallel design of a product model [2]. Parametric analysis is a digital rapid analysis method based on the parametric technology, which can simplify analysis and improve the pre-processing efficiency of finite element analysis through automatically analyzing the model with relevant software [3].
The parametric technology has been utilized to design and analyze base plates in this paper. By conducting a characteristic analysis of the base plate, critical design parameters extracted have been driven as characteristics to perform secondary development on UG, so that the base plate could be automatically modeled in three dimensions. Meanwhile, Python has been applied in the secondary development of Abaqus, so as to realize the automatic finite element modeling of the base plate. By processing the model established, sensitivities of design variables of the base plate have been analyzed using an approximate model in combination with the sample designed by the optimal Latin hypercube approach. After optimizing results of sensitivity in a parametric way, optimized results could give feedback to parametric design for guiding parameter values of design variables to improve the design quality. The study method is shown in Fig. 1.
2. Parametric design of the base plate
2.1. Characteristic analysis of the base plate
A base plate of a mortar with the caliber of 120 mm is studied in this paper, which is a trapezoidal pyramid structure combining the advantages of the arched base plate and the trapezoidal base plate. Such structure has been mostly applied in active mortars worldwide [4]. It's mainly comprised of main metal plate, cone-shaped basin, stationary joint, wrapped rib, stud (inner), reinforcing rib and spade, as shown in Fig. 2.
Fig. 1. Study method
Fig. 2. Structure diagram of the base plate
As the base plate is composed of three identical components that are characterized by central symmetry at 120°. Each component can be decomposed by structural characteristics and appropriately simplified into five basic parts including main metal plate, stud, stationary joint, wrapped rib, and spade. The two-dimensional diagram of each component is shown in Fig. 3. Moreover, there is an intermediate region seamed by the formation of main metal plate and stationary joints in the cone-shaped basin.
Fig. 3. Two-dimensional diagram of main components of the base plate
a) Main metal plate
b) Wrapped rib
c) Stud
d) Stationary joint
2.2. Parametric design of the base plate
UG secondary development tools, such as UG/ Open MenuScript, UG/ Open UIStyler, UG/ Open API, and UG/ Open Grip, etc. are adopted to call the VC+ + program for implementing the parametric design of the base plate [5]. The basic flow of the parametric design of the base plate is presented in Fig. 4.
To be specific, MenuScript is a special module for UG to provide a customized menus; UIStyler is for generating different dialog boxes combining basic controls in the dialog box; API is an interface between UG and external applications; GRIP language is an engineer-oriented language which can implement various interacting operations with UG with the GRIP program.Parametric design is utilized to obtain a preliminary three-dimensional geometric model of a certain mortar base plate, as shown in Fig. 5.
Fig. 4. Parametric design flow of the base plate
Fig. 5. Geometric model of the base plate
2.3. Parametric analysis of the base plate
First and foremost, problem types to be simulated are modeled and generated into an input file. All operating commands are recorded by ABAQUS in the rpy file with Python scripts. And then, codes for the secondary development can be formed with simple modifications by using notepad++ to open the rpy file. A results file can be formed after the calculation. Next, accessing the results database with Python scripts can control, read and write the results file. Followed by that, data to be process are obtained by quoting object variables or using data objects in accordance with the actual problem and the storage path of data. Finally, results of data processed with proper are outputted in the form of a nephogram or a curve, contributing to subsequent analysis and view [6]. The call relationship between program modules of the system is presented in Fig. 6.
Numerical simulation is performed with the hard surface floor as the using environment at the elevation of 70° [7]. The bottom of the spade is set as a rigid restraint for shooting. The pressure curve of the mortar base with the caliber of 120 mm is applied to the spherical surface of the stationary joints. Dynamic Explicit is adopted for analyzing. The unit type is set as C3D10M, and models are made of titanium alloy. The finite element model established is shown in Fig. 7. The thicknesses of main metal plate
{d}_{1}
, cone-shaped basin
{d}_{2}
, stud
{d}_{3}
, wrapped rib
{d}_{4}
, and stationary joint
{d}_{5}
are input as parameters with value ranges shown in Table 1.
Table 1. The value range of parameters
{d}_{1}
{d}_{2}
{d}_{3}
{d}_{4}
{d}_{5}
Lower limit / mm
Upper limit / mm
After setting simulation resources and paths, submit job to generate an input file (.inp). And input files are submitted to the ABAQUS/Standard solver for numerical operation via the background scripting. On the premise of completing analysis, open the result data file to access steps object below odb. Meanwhile, search for and extract the maximum stress Smax and the maximum displacement Umax in each analysis step and frame.
Fig. 6. Call between program modules
Fig. 7. The finite element model of the base plate
3. Sensitivity analysis of parameters
3.1. Approximate model
Sensitivity analyses of designed parameters of the base plate are completed with the approximate model, so as to obtain the influence of the thickness parameters of each component in the parametric design on the maximum stress
{S}_{max}
and the maximum displacement
{U}_{max}
. RSM (response surface model) of is selected as the approximate model in this paper. The second-order RSM can be expressed as Eq. (1) [8]:
\stackrel{^}{y}\left(x\right)={a}_{0}+\sum _{i=1}^{n}{b}_{i}{x}_{i}+\sum _{i=1}^{n}{c}_{ii}{x}_{i}^{2}+\sum _{1\le i\le j\le n}^{n}{e}_{ij}{x}_{i}{x}_{j},
\stackrel{^}{y}\left(x\right)
is the approximate value of the response surface;
{x}_{i}
{x}_{j}
is the design variable,
n
is the number of design variables,
{a}_{0}
{b}_{i}
{c}_{ii}
{e}_{ij}
are undetermined coefficients of the constant term, the monomial term, the quadratic term, and the cross term, respectively. The multiple correlation coefficient
{R}^{2}
is adopted to predict the fitting precision of RSM.
3.2. Parametric analysis of sensitivity
The finite element modeling and post-processing of parameterization are implemented automatically to extract results of finite element calculation using Python on the parametric analysis system established. And data are processed with Python to obtain the value of the objective function and return it to Isight for optimized analysis. After calling the file, modified files are firstly ranked according to the sequence of operation time and gathered in a test flow. Sample points are filtered within the feasible domain of design variables with the optimal Latin hypercube sampling approach. Secondly, variables in the intermediate link of the 100 sets of samples are automatically calculated by a computer and provided to the downstream
R
is module for real-time parameter modification. Finally, the modified *.py is submitted to the Simcode module for operation, while the corresponding results obtained by the operation are read by the
R
is module. In this way, the response values obtained are the maximum stress
{S}_{max}
{U}_{max}
Regarding the maximum stress,
{d}_{5}
has the greatest influence, indicating that the thickness of the stationary joint is the most influential part over the maximum stress on the base plate. Regarding the maximum displacement,
{d}_{1}
has the greatest influence, indicating that the thickness of the main metal plate is the most influential part. By conducting sensitivity analysis on the parameters, providing feedback of optimized parameters to the parametric modelling of the base plate can help optimize the structural design of the base plate.
4. Parameter optimization of the base plate
4.1. Establishment of the parameter optimization model
Based on the above analysis,
{d}_{1}
{d}_{5}
in the base plates are optimized to ensure the establishment of a reasonable base plate model. Parametric optimization is aimed at searching for optimal parameters, so that calculated results can satisfy the following objective Eq. (2) [9]:
\left\{\begin{array}{l}\mathrm{m}\mathrm{i}\mathrm{n}:f\left(D\right)=f\left({d}_{1},{d}_{5}\right),\\ {S}_{2\mathrm{m}\mathrm{a}\mathrm{x}}\le {S}_{1\mathrm{m}\mathrm{a}\mathrm{x}},\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{U}_{2\mathrm{m}\mathrm{a}\mathrm{x}}\le {U}_{1\mathrm{m}\mathrm{a}\mathrm{x}},\\ {d}_{1}^{L}\le {d}_{1}\le {d}_{1}^{U},\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{d}_{5}^{L}\le {d}_{5}\le {d}_{5}^{U},\end{array}\right\
f\left(D\right)
represents the mass of the base plate;
{S}_{1max}
{S}_{2max}
present the maximum stresses of the base plate before and after optimizing, respectively;
{U}_{1max}
{U}_{2max}
show the maximum displacement of the base plate before and after optimizing, respectively;
{d}_{1}^{U}
{d}_{1}^{L}
are the upper and lower limits of the value
{d}_{1}
{d}_{5}^{U}
{d}_{5}^{L}
{d}_{5}
4.2. Optimizing results
The stress nephogram and the displacement nephogram of the base plate before and after optimization are shown in Fig. 8 and Fig. 9, respectively. The thickness of the main metal plate and the stationary joint of the base plate, the mass of the base plate, the maximum stress and the maximum displacement are compared in Table 2.
Table 2. Comparison of the base plate before and after optimizing
{d}_{1}
{d}_{5}
Max. stress/ MPa
Max. displacement/ mm
Fig. 8. Stress nephogram
a) Before optimizing
b) After optimizing
Based on the analysis, it can be seen from the analysis that the maximum stress and the maximum displacement of the base plate are decreased after optimizing in comparison to that before optimizing, satisfying the design requirements. What’s more, the total mass of the base plate after optimizing is lower than that before optimizing, achieving the purpose of optimized design. And, the optimal values of
{d}_{1}
{d}_{5}
feed back to the parametric design system studied previously, which can effectively enhance the design quality and efficiency of the base plate model at the initial design stage.
Fig. 9. Displacement nephogram
In this paper, parametric design and parametric analysis are applied to the design of the mortar base plate for the first time, and the following conclusions are given:
1) Critical design parameters extracted have been promoted as characteristics to perform secondary development on UG on the basis of conducting characteristic analysis of the base plate, so that the three-dimensional automatic modelling of the base plate can be implemented to enhance the design efficiency.
2) Meanwhile, applying Python into the secondary development of Abaqus can achieve the automatic finite element modeling and result extraction of the base plate.
3) Results obtained by parametric optimization is performed by sensitivity analysis, and which could give feedback to improve the design quality. It’s innovative in the parametric design of the base plate.
Tang Z. Mortar Design. Weapons Industry Press, 1994. [Search CrossRef]
Meng X., Xu Y. Research on parametric design. Journal of Computer-Aided Design and Computer Graphics, Vol. 14, Issue 11, 2002, p. 1086-1090. [Search CrossRef]
Lin Y., Ni Z., Liu X., et al. Research on digital fast analysis method based on parametric technology. Machinery Design and Manufacturing Engineering, Vol. 3, 2017, p. 8-53. [Search CrossRef]
Wang X., Ge J., Yang G., et al. Lightweight design of mortar base plate. Journal of Gun Launch and Control, Vol. 39, Issue 4, 2018, p. 56-61. [Search CrossRef]
Lu H., Liu X. Research and application of secondary development of UG software. Electronic Technology and Software Engineering, Vol. 149, Issue 3, 2019, p. 49. [Search CrossRef]
Ma C. H. Research on Framework Design Platform Based on ABAQUS Secondary Development. Xi’an University of Architecture and Technology, 2018. [Search CrossRef]
Zhou Z., Bu H., He Y. Analysis of the enthalpy-soil coupled launch dynamics of mortars. Journal of Ordnance Equipment, Vol. 37, Issue 2, 2016, p. 18-21. [Search CrossRef]
Wang L., Yang G., Xiao H., et al. Interval optimization for structural dynamic responses of an artillery system under uncertainty. Engineering Optimization, 2019, https://doi.org/10.1080/0305215X.2019.1590563. [Publisher]
Hong Q. Theoretical basis and engineering application of OptiStruct and HyperStudy. Mechanical Industry Press, 2013, p. 158-196. [Search CrossRef] |
Determine American call option prices or sensitivities using Roll-Geske-Whaley option pricing model - MATLAB optstocksensbyrgw - MathWorks Benelux
Compute American Call Option Prices and Sensitivities Using the Roll-Geske-Whaley Option Pricing Model
Determine American call option prices or sensitivities using Roll-Geske-Whaley option pricing model
PriceSens = optstocksensbyrgw(RateSpec,StockSpec,Settle,Maturity,OptSpec,Strike)
PriceSens = optstocksensbyrgw(___,Name,Value)
PriceSens = optstocksensbyrgw(RateSpec,StockSpec,Settle,Maturity,OptSpec,Strike) computes American call option prices or sensitivities using the Roll-Geske-Whaley option pricing model.
optstocksensbyrgw computes prices of American calls with a single cash dividend using the Roll-Geske-Whaley option pricing model. All sensitivities are evaluated by computing a discrete approximation of the partial derivative. This means that the option is revalued with a fractional change for each relevant parameter, and the change in the option value divided by the increment, is the approximated sensitivity value.
PriceSens = optstocksensbyrgw(___,Name,Value) adds an optional name-value pair argument for OutSpec.
This example shows how to compute American call option prices and sensitivities using the Roll-Geske-Whaley option pricing model. Consider an American stock option with an exercise price of $82 on January 1, 2008 that expires on May 1, 2008. Assume the underlying stock pays dividends of $4 on April 1, 2008. The stock is trading at $80 and has a volatility of 30% per annum. The risk-free rate is 6% per annum. Using this data, calculate the price and the value of delta and gamma of the American call using the Roll-Geske-Whaley option pricing model.
Maturity = 'May-01-2008';
DivDate = 'Apr-01-2008';
'EndDates', Maturity, 'Rates', Rate, 'Compounding', -1, 'Basis', 1);
% define the OutSpec
[Price, Delta, Gamma] = optstocksensbyrgw(RateSpec, StockSpec, Settle,...
Maturity, Strike,'OutSpec', OutSpec)
Example: [Delta,Gamma,Price] = optstocksensbyrgw(RateSpec,StockSpec,Settle,Maturity,OptSpec,Strike,'OutSpec',OutSpec)
\mathrm{max}\left(St-K,0\right)
\mathrm{max}\left(K-St,0\right)
impvbyrgw | intenvset | optstockbyrgw | stockspec |
Option value | Brilliant Math & Science Wiki
The option value is estimated through a predictive formula like Black Scholes or through a numerical method like Binomial Tree model. This price can often be interpreted as the expected value of the option on expiration, given the information in the market.
The value of an option can be divided into 2 parts:
- Intrinsic Value: The value of exercising an in the money option.
- Extrinsic (Time) Value The value of the price insurance obtained from the right to exercise (or not exercise) the option. This is always non-negative.
For American options, the intrinsic value is easily calculated, as the expected profit from (smartly) exercising the option.
Let's say we have a call option. When the stock price
S
X
, then we can buy the stock at
X
through exercising the call and sell the stock at
S
in the market. This gives us an intrinsic value of
S - X
. When the stock price
S
is lower then the strike price
X
, then we would choose to not exercise the call option, hence it has an intrinsic value of 0. Thus,
\text{Intrinsic value of Call} = \max( S - X , 0 ).
We can perform a similar analysis for the put option. When the stock price
S
is lower than the strike price
X
, then exercising the put will allow us to buy the stock at
S
and sell the stock at
X
through exercising. This gives us an intrinsic value of
X - S
S
is higher than the strike price
X
, then we would choose to not exercise the put option, hence it has an intrinsic value of 0. Thus,
\text{Intrinsic value of Put} = \max( X-S , 0 ).
For European options, because they can only be exercised on expiry, the intrinsic value of the option is calculated on the future price, which would include any dividends that are to be paid out.
For example, if the stock is worth $30, we would typically expect the call option on the $25 strike to be worth at least $5, reflecting the intrinsic value. However, if the stock will be paying a dividend of $3 before expiry, then the future value of the stock is only
\$ 30 - \$3 = \$ 27
(and account for interest), and thus the option could be trading for less than $5.
\text{Intrinsic value of Call} = \max( F - X , 0 ).
\text{Intrinsic value of Put} = \max( X-F , 0 ).
The time value of an option is the difference between the Option value and the Intrinsic Value.
\text{Time value} = \text{ Option value } - \text{ Intrinsic value}
Note that this is always non-negative, as it reflects the right (but not obligation) to perform at trade at the strike price.
For European options with no possibility of early exercise, the time value of the put and call on the same strike and expiry are the same.
For American options, the time value of the put and the call could be slightly different, to reflect the early exercise premium. For example, a deep-in-the-money call that will be early exercised may be trading at intrinsic value (and thus has no time value), while the put could be trading way above $0. Having said that, for options that are close to at-the-money or are expiring soon, the calls and puts will have the same time value.
Cite as: Option value. Brilliant.org. Retrieved from https://brilliant.org/wiki/option-value/ |
(-)-menthol,NADPH:oxygen oxidoreductase (8-hydroxylating) Wikipedia
(-)-menthol,NADPH:oxygen oxidoreductase (8-hydroxylating)
(−)-menthol monooxygenase
In enzymology, a (−)-menthol monooxygenase (EC 1.14.13.46) is an enzyme that catalyzes the chemical reaction
(−)-menthol + NADPH + H+ + O2
{\displaystyle \rightleftharpoons }
p-menthane-3,8-diol + NADP+ + H2O
The 4 substrates of this enzyme are (−)-menthol, NADPH, H+, and O2, whereas its 3 products are p-menthane-3,8-diol, NADP+, and H2O.
This enzyme belongs to the family of oxidoreductases, specifically those acting on paired donors, with O2 as oxidant and incorporation or reduction of oxygen. The oxygen incorporated need not be derived from O2 with NADH or NADPH as one donor, and incorporation of one atom o oxygen into the other donor. The systematic name of this enzyme class is (−)-menthol,NADPH:oxygen oxidoreductase (8-hydroxylating). This enzyme is also called l-menthol monooxygenase.
Use of (−)-menthol monooxygenase has been explored by several companies including Procter & Gamble for cleaning products.[1] [2] [3]
^ "Cleaning compositions comprising a specific oxygenase"|Alfons, Ivan Maurice et al.|https://patents.google.com/patent/EP1002040B2/en
^ "Methods for conducting assays for enzyme activity on protein microarrays"|Zhou, Fang X.; Schweitzer, Barry| https://patents.google.com/patent/US7635572B2/en?oq=US7635572B2
^ "Cleaning compositions contain a specific oxygenase"|Barnabas, Mary et al.|https://patents.google.com/patent/DE69727704T3/en
Madyastha KM, Srivatsan V (1988). "Studies on the metabolism of l-menthol in rats". Drug Metab. Dispos. 16 (5): 765–72. PMID 2906604. |
\frac{1}{{x}^{2}-14x+40}\le 0
\frac{1}{{x}^{2}-14x+40}\le 0
To determine the solution sets of the given inequality.
Factorising the term
{x}^{2}-14x+40
p\left(x\right)={x}^{2}-14x+40
p\left(x\right)={x}^{2}-14x+40
={x}^{2}-4x-10x+40
=x\left(x-4\right)-10\left(x-4\right)
=\left(x-10\right)\left(x-4\right)
x=10,4.p\left(x\right)=0
x\in \left(-\mathrm{\infty },4\right).p\left(x\right)>0
x\in \left(4,10\right).p\left(x\right)<0
x\in \left(10,\mathrm{\infty }\right).p\left(x\right)>0
\frac{1}{{x}^{2}-14x+40}\le 0
{x}^{2}-14x+40
should be less than 0.
{x}^{2}-14x+40<0
x\in \left(4,10\right)
Therefore, the solution set of the given inequality is (4,10)
-\sqrt{\frac{9}{25}}=
2x-4x+2?
3x+7h+y-\left(-8k\right)-11x+65y+\left(-40k\right)
Determine whether the series is convergent or divergent by expressing as a telescoping sum. If it is convergent, find its sum. summation
n=2
to infinity
\frac{2}{{n}^{2}-1}
Find two functions
f\left(x\right)
g\left(x\right)
h\left(x\right)=\left(f\circ g\right)\left(x\right)
h\left(x\right)={\left(2x+3\right)}^{2}
h\left(x\right)=|3-2x-{x}^{2}|
h\left(x\right)=h\left(x\right)=\frac{5}{\sqrt{3-x}}
h\left(x\right)=\sqrt{5}\left\{{x}^{3}+4{x}^{2}+1\right\}
How do you evaluate the expressions
\frac{2{y}^{2}}{-5y}
x=3
y=5
-3\left(a+5\right)+9 |
Specify variable or constant parameter value - MATLAB - MathWorks Australia
Specify variable or constant parameter value
The ConstantValue property indicates whether the value of a parameter can change during a simulation. Enter either true (value is constant) or false (value can change).
You can allow the value of the parameter to change during a simulation by specifying a rule that changes the Value property of the parameter object.
The property ConstantValue is for parameter objects; the property ConstantAmount is for species objects.
As an example, consider feedback inhibition of an enzyme such as aspartate kinase by threonine. Aspartate kinase has three isozymes that are independently inhibited by the products of downstream reactions (threonine, homoserine, and lysine). Although threonine is made through a series of reactions in the synthesis pathway, for illustration, the reactions are simplified as follows:
\begin{array}{l}Aspartic\text{ acid }\stackrel{\text{aspartate kinase}}{\to }\beta -Aspartylphosphate\\ \beta -Aspartylphosphate\text{ }\stackrel{}{\to }Threonine\end{array}
To model inhibition of aspartate kinase by threonine, you could use a rule like the algebraic rule below to vary the rate of the above reaction and simulate inhibition. In the rule, the rate constant for the above reaction is denoted by k_aspartate_kinase and the quantity of threonine is threonine.
k_aspartate_kinase -(1/threonine)
Applies to Object: parameter
Data values true or false. The default value is 'true'.
Add a parameter object.
parameterObj = addparameter (modelObj, 'kf');
Change the ConstantValue property of the parameter object from default (true) to false and verify.
MATLAB® returns 1 for true and 0 for false.
set (parameterObj, 'ConstantValue', false);
get(parameterObj, 'ConstantValue') |
EUDML | Continuous Trace Groupoid C*-Algebras, II. EuDML | Continuous Trace Groupoid C*-Algebras, II.
Continuous Trace Groupoid C*-Algebras, II.
Paul S. Muhly; Dana P. Williams
Muhly, Paul S., and Williams, Dana P.. "Continuous Trace Groupoid C*-Algebras, II.." Mathematica Scandinavica 70.1 (1992): 127-145. <http://eudml.org/doc/167191>.
@article{Muhly1992,
author = {Muhly, Paul S., Williams, Dana P.},
keywords = {principal groupoid; continuous trace -algebra; Haar system; 2- cocycle},
title = {Continuous Trace Groupoid C*-Algebras, II.},
AU - Muhly, Paul S.
AU - Williams, Dana P.
TI - Continuous Trace Groupoid C*-Algebras, II.
KW - principal groupoid; continuous trace -algebra; Haar system; 2- cocycle
principal groupoid, continuous trace
{C}^{*}
-algebra, Haar system, 2- cocycle
{C}^{*}
{W}^{*}
{C}^{*}
{C}^{*}
Articles by Paul S. Muhly
Articles by Dana P. Williams |
EUDML | The canonical endomorphism for infinite index inclusions. EuDML | The canonical endomorphism for infinite index inclusions.
The canonical endomorphism for infinite index inclusions.
Fidaleo, F.; Isola, T.
Fidaleo, F., and Isola, T.. "The canonical endomorphism for infinite index inclusions.." Zeitschrift für Analysis und ihre Anwendungen 18.1 (1999): 47-66. <http://eudml.org/doc/48368>.
@article{Fidaleo1999,
author = {Fidaleo, F., Isola, T.},
keywords = {quantum object; regular representation; endomorphism; von Neumann algebra; faithful conditional expectation; Woronowicz algebras; Jones' tower},
title = {The canonical endomorphism for infinite index inclusions.},
AU - Fidaleo, F.
AU - Isola, T.
TI - The canonical endomorphism for infinite index inclusions.
KW - quantum object; regular representation; endomorphism; von Neumann algebra; faithful conditional expectation; Woronowicz algebras; Jones' tower
quantum object, regular representation, endomorphism, von Neumann algebra, faithful conditional expectation, Woronowicz algebras, Jones' tower
{C}^{*}
{W}^{*}
Subfactors and their classification
Articles by Fidaleo
Articles by Isola |
EUDML | Solvability of quasilinear elliptic equations with strong dependence on the gradient. EuDML | Solvability of quasilinear elliptic equations with strong dependence on the gradient.
Solvability of quasilinear elliptic equations with strong dependence on the gradient.
Žubrinić, Darko. "Solvability of quasilinear elliptic equations with strong dependence on the gradient.." Abstract and Applied Analysis 5.3 (2000): 159-173. <http://eudml.org/doc/49754>.
@article{Žubrinić2000,
author = {Žubrinić, Darko},
keywords = {-Laplace operator; spherically symmetric; decreasing; weak and strong solutions; singular ordinary integro-differential equation; a posteriori regularity; -Laplace operator},
title = {Solvability of quasilinear elliptic equations with strong dependence on the gradient.},
AU - Žubrinić, Darko
TI - Solvability of quasilinear elliptic equations with strong dependence on the gradient.
KW - -Laplace operator; spherically symmetric; decreasing; weak and strong solutions; singular ordinary integro-differential equation; a posteriori regularity; -Laplace operator
p
-Laplace operator, spherically symmetric, decreasing, weak and strong solutions, singular ordinary integro-differential equation, a posteriori regularity,
p
Articles by Žubrinić |
Home : Support : Online Help : Mathematics : Algebra : Algebraic Numbers : Degree
formal degree of an algebraic extension
set of RootOfs of type algext
Given a set of RootOfs defining an algebraic extension, Degree determines the formal algebraic degree of that extension.
The formal algebraic degree of a non-nested RootOf is the degree of its first argument with respect to the variable _Z, i.e., the degree of its defining polynomial.
The formal algebraic degree of a set of RootOfs is the product of the degrees of all the defining polynomials.
This procedure assumes that all sub-RootOfs of a RootOf in S are elements of S as well.
If all RootOfs in S are independent, then the formal algebraic degree is equal to the actual degree of the field extension defined by the RootOfs in S over the ground field.
\mathrm{Algebraic}[\mathrm{Degree}]\left({\mathrm{RootOf}\left({x}^{2}-2,\mathrm{index}=1\right),\mathrm{RootOf}\left({y}^{2}-\mathrm{RootOf}\left({x}^{2}-2,\mathrm{index}=1\right),\mathrm{index}=1\right)}\right)
\textcolor[rgb]{0,0,1}{4}
\mathrm{Algebraic}[\mathrm{Degree}]\left(\varnothing \right)
\textcolor[rgb]{0,0,1}{1}
The RootOfs in the following example are dependent, and the formal algebraic degree is bigger than the actual degree, which is
4
\mathrm{Algebraic}[\mathrm{Degree}]\left({\mathrm{RootOf}\left({x}^{2}-2,\mathrm{index}=1\right),\mathrm{RootOf}\left({x}^{2}-3,\mathrm{index}=1\right),\mathrm{RootOf}\left({x}^{2}-6,\mathrm{index}=1\right)}\right)
\textcolor[rgb]{0,0,1}{8}
The set of RootOfs in the following example is not closed under sub-RootOfs, and Degree does not return the correct formal degree, which is
4
\mathrm{Algebraic}[\mathrm{Degree}]\left({\mathrm{RootOf}\left({y}^{2}-\mathrm{RootOf}\left({x}^{2}-2,\mathrm{index}=1\right),\mathrm{index}=1\right)}\right)
\textcolor[rgb]{0,0,1}{2} |
ARIMAX Model Specifications - MATLAB & Simulink - MathWorks 日本
Create ARIMAX Model Using Name-Value Pairs
Specify ARMAX Model Using Dot Notation
Specify ARIMAX or SARIMAX Model Using Econometric Modeler App
This example shows how to specify an ARIMAX model using arima.
Specify the ARIMAX(1,1,0) model that includes three predictors:
\left(1-0.1L\right)\left(1-L{\right)}^{1}{y}_{t}={x}_{t}^{â²}{\left[\begin{array}{ccc}3& -2& 5\end{array}\right]}^{â²}+{\mathrm{ε}}_{t}.
Mdl = arima('AR',0.1,'D',1,'Beta',[3 -2 5])
Beta: [3 -2 5]
The output shows that the ARIMAX model Mdl has the following qualities:
Property P in the output is the sum of the autoregressive lags and the degree of integration, i.e., P = p + D = 2.
Beta contains three coefficients corresponding to the effect that the predictors have on the response. Mdl does not store predictor or response data. You specify the required data when you operate on Mdl.
The rest of the properties are 0, NaN, or empty cells.
Be aware that if you specify nonzero D or Seasonality, then Econometrics Toolbox™ differences the response series
{y}_{t}
before the predictors enter the model. Therefore, the predictors enter a stationary model with respect to the response series
{y}_{t}
. You should preprocess the predictors
{x}_{t}
by testing for stationarity and differencing if any are unit root nonstationary. If any nonstationary predictor enters the model, then the false negative rate for significance tests of
\mathrm{β}
can increase.
This example shows how to specify a stationary ARMAX model using arima.
Specify the ARMAX(2,1) model
{y}_{t}=6+0.2{y}_{t-1}-0.3{y}_{t-2}+3{x}_{t}+{\mathrm{ε}}_{t}+0.1{\mathrm{ε}}_{t-1}
by including one stationary exogenous covariate in arima.
Mdl = arima('AR',[0.2 -0.3],'MA',0.1,'Constant',6,'Beta',3)
The output shows the model Mdl has NaN values or an empty cell ({}) for the Variance, SAR, and SMA properties. You can modify the model using dot notation at the command line. For example, you can introduce another exogenous, stationary covariate, and specify the variance of the innovations as 0.1
{y}_{t}=6+0.2{y}_{t-1}-0.3{y}_{t-2}+{x}_{t}^{â²}\left[\begin{array}{c}3\\ -2\end{array}\right]+{\mathrm{ε}}_{t}+0.1{\mathrm{ε}}_{t-1};\phantom{\rule{0.2777777777777778em}{0ex}}\phantom{\rule{0.2777777777777778em}{0ex}}{\mathrm{ε}}_{t}â¼N\left(0,0.1\right).
Modify the model by specifying an innovations variance of 0.1 and regression coefficient values.
Mdl.Beta = [3 -2];
Beta: [3 -2]
In the Econometric Modeler app, you can specify the seasonal and nonseasonal lag structure, presence of a constant, innovation distribution, and predictor variables of an ARIMA(p,D,q) or a SARIMA(p,D,q)×(ps,Ds,qs)s model by following these steps. All specified coefficients are unknown but estimable parameters.
On the Econometric Modeler tab, in the Models section, click the arrow to display the models gallery. For strictly nonseasonal models, click ARIMAX; for seasonal models, click SARIMAX. ARIMAX and SARIMAX models must contain at least one predictor variable.
The Type Model Parameters dialog box appears. This figure shows the SARIMAX Model Parameters dialog box. All variables in the Time Series pane, except the chosen response variable, appear in the Predictors section.
Specify the lag structure. Use the Lag Order tab to specify a model that includes:
All consecutive lags from 1 through their respective orders, in the seasonal polynomials
For the flexibility to specify the inclusion of particular lags, use the Lag Vector tab. For more details, see Specifying Univariate Lag Operator Polynomials Interactively. Regardless of the tab you use, you can verify the model form by inspecting the equation in the Model Equation section.
In the Predictors section, choose at least one predictor variable by selecting the Include? check box for the time series.
For example, suppose you are working with the Data_USEconModel.mat data set and its variables are listed in the Time Series pane.
To specify an ARIMAX(3,1,2) model for the unemployment rate containing a constant, all consecutive AR and MA lags from 1 through their respective orders, Gaussian-distributed innovations, and the predictor variables COE, CPIAUCSL, FEDFUNDS, and GDP:
In the Time Series pane, select the UNRATE time series.
In the ARIMAX Model Parameters dialog box in the Nonseasonal section of the Lag Order tab, set Degree of Integration to 1.
In the Predictors section, select the Include? check box for the COE, CPIAUCSL, FEDFUNDS, and GDP time series.
To specify an ARIMAX(3,1,2) model for the unemployment rate containing all AR and MA lags from 1 through their respective orders, Gaussian-distributed innovations, no constant, and the predictor variables COE and CPIAUCSL:
In the ARIMAX Model Parameters dialog box, in the Nonseasonal section of the Lag Order tab, set Degree of Integration to 1.
In the Predictors section, select the Include? check box for the COE and CPIAUCSL time series.
To specify an ARMA(8,1,4) model for the unemployment rate containing nonconsecutive lags
\left(1â{\mathrm{Ï}}_{1}Lâ{\mathrm{Ï}}_{4}{L}^{4}â{\mathrm{Ï}}_{8}{L}^{8}\right)\left(1âL\right){y}_{t}=\left(1+{\mathrm{θ}}_{1}L+{\mathrm{θ}}_{4}{L}^{4}\right){\mathrm{ε}}_{t}+{\mathrm{β}}_{1}CO{E}_{t}+{\mathrm{β}}_{2}CPIAUCS{L}_{t},
where εt is a series of IID Gaussian innovations:
In the ARIMAX Model Parameters dialog box, click the Lag Vector tab.
Set Autoregressive Lags to 1 4 8.
Set Moving Average Lags to 1 4.
To specify an ARIMA(3,1,2) model for the unemployment rate containing all consecutive AR and MA lags through their respective orders, a constant term, the predictor variables COE and CPIAUCSL, and t-distributed innovations:
Click the Innovation Distribution button, then select t.
In the Predictors section, select the Include? check box for COE and CPIAUCSL time series.
The degrees of freedom parameter of the t distribution is an unknown but estimable parameter. |
Option price by Merton76 model using finite differences - MATLAB optByMertonFD - MathWorks Benelux
Compute Price for an American Option Using the Merton Model
Option price by Merton76 model using finite differences
[Price,PriceGrid,AssetPrices,Times] = optByMertonFD(Rate,AssetPrice,Settle,ExerciseDates,OptSpec,Strike,Sigma,MeanJ,JumpVol,JumpFreq)
[Price,PriceGrid,AssetPrices,Times] = optByMertonFD(___,Name,Value)
[Price,PriceGrid,AssetPrices,Times] = optByMertonFD(Rate,AssetPrice,Settle,ExerciseDates,OptSpec,Strike,Sigma,MeanJ,JumpVol,JumpFreq) computes a vanilla European or American option price by the Merton76 model, using the Crank-Nicolson Adams-Bashforth (CNAB) IMEX method.
[Price,PriceGrid,AssetPrices,Times] = optByMertonFD(___,Name,Value) specifies options using one or more name-value pair arguments in addition to the input arguments in the previous syntax.
Define the option variables and Merton model parameters.
DividendYield = 0.1;
MeanJ = -0.10;
JumpFreq = 1.00;
Compute the American call option price using the finite differences method.
Price = optSensByMertonFD(Rate, AssetPrice, Settle, ExerciseDates, OptSpec, Strike,...
Sigma, MeanJ, JumpVol, JumpFreq, 'DividendYield', DividendYield, 'AmericanOpt', 1)
character vector with value of 'call' or 'put' | string array with value of "call" or "put"
Definition of the option, specified as a scalar using a character vector or string array with a value of 'call' or 'put'.
Volatility of the underling asset, specified as a scalar numeric.
Mean of the random percentage jump size (J), specified as a scalar decimal value where log(1+J) is normally distributed with the mean (log(1+MeanJ)-0.5*JumpVol^2) and the standard deviation JumpVol.
Example: [Price,PriceGrid] = optByMertonFD(Rate,AssetPrice,Settle,ExerciseDates,OptSpec,Strike,MeanJ,JumpVol,JumpFreq,'Basis',7)
Size of the asset grid for the finite difference grid, specified as the comma-separated pair consisting of 'AssetGridSize' and a scalar numeric.
Grid containing prices calculated by the finite difference method, returned as a two-dimensional grid with size AssetGridSize ⨉ TimeGridSize. The number of columns is not necessarily equal to the TimeGridSize because exercise and ex-dividend dates are added to the time grid. PriceGrid(:, :, end) contains the price for t = 0.
\mathrm{max}\left(St-K,0\right)
\mathrm{max}\left(K-St,0\right)
The Merton jump diffusion model [2] extends the Black-Scholes model by using the Poisson process to include jump diffusion parameters in the modeling of sudden asset price movements (both up and down).
\begin{array}{l}d{S}_{t}=\left(r-q-{\lambda }_{p}{\mu }_{j}\right){S}_{t}dt+\sigma {S}_{t}d{W}_{t}+J{S}_{t}d{P}_{t}\\ \text{prob(}d{P}_{t}=1\right)={\lambda }_{p}dt\end{array}
\mathrm{ln}\left(1+{\mu }_{J}\right)-\frac{{\delta }^{2}}{2}
\frac{1}{\left(1+J\right)\delta \sqrt{2\pi }}\mathrm{exp}\left\{{\frac{-\left[\mathrm{ln}\left(1+J\right)-\left(\mathrm{ln}\left(1+{\mu }_{J}\right)-\frac{{\delta }^{2}}{2}\right]}{2{\delta }^{2}}}^{2}\right\}
[1] Cont, R., and E. Voltchkova. “A Finite Difference Scheme for Option Pricing in Jump Diffusion and Exponential Lévy Models.” SIAM Journal on Numerical Analysis. Vol. 43, Number 4, 2005, pp. 1596-1626.
[2] Merton, R. "Option Pricing When Underlying Stock Returns Are Discontinuous." The Journal of Financial Economics. Vol 3. 1976, pp. 125-144.
optByLocalVolFD | optSensByLocalVolFD | optByHestonFD | optSensByHestonFD | optSensByMertonFD | optSensByBatesFD | optByBatesFD |
R
z=4\left( {x}^{2}+{y}^{2}\right)
z=9\left({x}^{2}+{y}^{2}\right)
z=2
z=3
x=\left(u/v\right)\mathrm{cos}\left(w\right)
y=\left(u/v\right)\mathrm{sin}\left(w\right)
z={u}^{2}
Figure 8.5.3(a) is a preliminary sketch of the region
R
, a sketch sufficient to suggest that in cylindrical coordinates, the iteration-order
\mathrm{dr} \mathrm{dz} d\mathrm{θ}
is one of the simpler approaches, even though it requires expressing the outer (red) surface as
r\left(z\right)=\sqrt{z}/2
, and the inner (green) surface as
r\left(z\right)=\sqrt{z}/3
Figure 8.5.3(b) depicts the actual region
R
whose volume is to be calculated.
Figure 8.5.3(a) Preliminary sketch
Figure 8.5.3(b) The region
R
The following iterated integral computes the volume of
R
in cylindrical coordinates.
{∫}_{0}^{2 \mathrm{π}}{∫}_{2}^{3}{∫}_{\sqrt{z}/3}^{\sqrt{z}/2}r \mathit{ⅆ}r \mathit{ⅆ}z \mathit{ⅆ}\mathrm{θ}
The Jacobian of the transformation
x=\left(u/v\right)\mathrm{cos}\left(w\right)
y=\left(u/v\right)\mathrm{sin}\left(w\right)
z={u}^{2}
\frac{∂\left(x,y,z\right)}{∂\left(u,v,w\right)}=\left|\begin{array}{ccc}{∂}_{u}\left[\begin{array}{c}x\\ y\\ z\end{array}\right]& {∂}_{v}\left[\begin{array}{c}x\\ y\\ z\end{array}\right]& {∂}_{w}\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\end{array}\right|
\left|\begin{array}{ccc}\frac{\mathrm{cos}\left(w\right)}{v}& -\frac{u \mathrm{cos}\left(w\right)}{{v}^{2}}& -\frac{u \mathrm{sin}\left(w\right)}{v}\\ \frac{\mathrm{sin}\left(w\right)}{v}& -\frac{u \mathrm{sin}\left(w\right)}{{v}^{2}}& \frac{u \mathrm{cos}\left(w\right)}{v}\\ 2 u& 0& 0\end{array}\right|
-2{\left(\frac{u}{v}\right)}^{3}
{x}^{2}+{y}^{2}={\left(u/v\right)}^{2}
, the outer surface
z=4\left({x}^{2}+{y}^{2}\right)
{u}^{2}=4{\left(u/v\right)}^{2}
v=2
. Similarly, the inner surface
z=9\left({x}^{2}+{y}^{2}\right)
{u}^{2}=9{\left(u/v\right)}^{2}
v=3
z={u}^{2}=2
describes the lower plane,
u=\sqrt{2}
on that surface, while
z={u}^{2}=3
u=\sqrt{3}
on the upper plane. Hence, the volume of
R
can also be calculated via the iterated integral
{∫}_{0}^{2 \mathrm{π}}{∫}_{2}^{3}{∫}_{{\sqrt{2}}_{}}^{{\sqrt{3}}_{}}2{\left(u/v\right)}^{3} \mathit{ⅆ}u \mathit{ⅆ}v \mathit{ⅆ}w
The integration in cylindrical coordinates can be implemented in the task template in Table 8.5.3(a). Note the iteration-order
\mathrm{dr} \mathrm{dz} d\mathrm{θ}
Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻Cylindrical
{∭}_{R}\mathrm{Ψ}\left(r,\mathrm{θ},z\right) \mathrm{dv}
R
\mathrm{dv}
r \mathrm{dz} \mathrm{dr} d\mathrm{θ}
r \mathrm{dz} d\mathrm{θ} \mathrm{dr}
r \mathrm{dr} d\mathrm{θ} \mathrm{dz}
r \mathrm{dr} \mathrm{dz} d\mathrm{θ}
r d\mathrm{θ} \mathrm{dr} \mathrm{dz}
r d\mathrm{θ} \mathrm{dz} \mathrm{dr}
\mathrm{Ψ}=
F=
G=
b=
f=
g=
a=
Table 8.5.3(a) Integration in cylindrical coordinates via a visualization task template
The change of variables and the integration in the resulting coordinate system is implemented in Table 8.5.3(b).
Assign the transformation equations the name
T
T
\left[x=\frac{u \mathrm{cos}\left(w\right)}{v},y=\frac{u \mathrm{sin}\left(w\right)}{v},z={u}^{2}\right]
\stackrel{\text{assign to a name}}{\to }
\textcolor[rgb]{0,0,1}{T}
Obtain the Jacobian
\frac{∂\left(x,y,z\right)}{∂\left(u,v,w\right)}
Context Panel: Jacobian Matrix
Standard Operations≻Determinant
\genfrac{}{}{0}{}{\left[x,y,z\right]}{\phantom{x=a}}|\genfrac{}{}{0}{}{\phantom{\mathrm{f\left(x\right)}}}{T}
\left[\frac{\textcolor[rgb]{0,0,1}{u}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{cos}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{w}\right)}{\textcolor[rgb]{0,0,1}{v}}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{u}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{sin}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{w}\right)}{\textcolor[rgb]{0,0,1}{v}}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{u}}^{\textcolor[rgb]{0,0,1}{2}}\right]
\stackrel{\text{Jacobian}}{\to }
\left[\begin{array}{ccc}\frac{\textcolor[rgb]{0,0,1}{\mathrm{cos}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{w}\right)}{\textcolor[rgb]{0,0,1}{v}}& \textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{u}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{cos}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{w}\right)}{{\textcolor[rgb]{0,0,1}{v}}^{\textcolor[rgb]{0,0,1}{2}}}& \textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{u}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{sin}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{w}\right)}{\textcolor[rgb]{0,0,1}{v}}\\ \frac{\textcolor[rgb]{0,0,1}{\mathrm{sin}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{w}\right)}{\textcolor[rgb]{0,0,1}{v}}& \textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{u}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{sin}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{w}\right)}{{\textcolor[rgb]{0,0,1}{v}}^{\textcolor[rgb]{0,0,1}{2}}}& \frac{\textcolor[rgb]{0,0,1}{u}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{cos}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{w}\right)}{\textcolor[rgb]{0,0,1}{v}}\\ \textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{u}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}\end{array}\right]
\stackrel{\text{determinant}}{\to }
\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{\mathrm{cos}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{w}\right)}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{\mathrm{sin}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{w}\right)}^{\textcolor[rgb]{0,0,1}{2}}\right)}{{\textcolor[rgb]{0,0,1}{v}}^{\textcolor[rgb]{0,0,1}{3}}}
\stackrel{\text{simplify}}{=}
\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}^{\textcolor[rgb]{0,0,1}{3}}}{{\textcolor[rgb]{0,0,1}{v}}^{\textcolor[rgb]{0,0,1}{3}}}
Apply the change of coordinates to the bounding surfaces
v
\genfrac{}{}{0}{}{z=9\left({x}^{2}+{y}^{2}\right)}{\phantom{x=a}}|\genfrac{}{}{0}{}{\phantom{\mathrm{f\left(x\right)}}}{T}
{\textcolor[rgb]{0,0,1}{u}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{=}\frac{\textcolor[rgb]{0,0,1}{9}\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}{\mathrm{cos}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{w}\right)}^{\textcolor[rgb]{0,0,1}{2}}}{{\textcolor[rgb]{0,0,1}{v}}^{\textcolor[rgb]{0,0,1}{2}}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{9}\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}{\mathrm{sin}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{w}\right)}^{\textcolor[rgb]{0,0,1}{2}}}{{\textcolor[rgb]{0,0,1}{v}}^{\textcolor[rgb]{0,0,1}{2}}}
\stackrel{\text{simplify}}{=}
{\textcolor[rgb]{0,0,1}{u}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{=}\frac{\textcolor[rgb]{0,0,1}{9}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}^{\textcolor[rgb]{0,0,1}{2}}}{{\textcolor[rgb]{0,0,1}{v}}^{\textcolor[rgb]{0,0,1}{2}}}
\stackrel{\text{solve for v}}{\to }
\left[\left[\textcolor[rgb]{0,0,1}{v}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{3}\right]\textcolor[rgb]{0,0,1}{,}\left[\textcolor[rgb]{0,0,1}{v}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{3}\right]\right]
\genfrac{}{}{0}{}{z=4\left({x}^{2}+{y}^{2}\right)}{\phantom{x=a}}|\genfrac{}{}{0}{}{\phantom{\mathrm{f\left(x\right)}}}{T}
{\textcolor[rgb]{0,0,1}{u}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{=}\frac{\textcolor[rgb]{0,0,1}{4}\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}{\mathrm{cos}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{w}\right)}^{\textcolor[rgb]{0,0,1}{2}}}{{\textcolor[rgb]{0,0,1}{v}}^{\textcolor[rgb]{0,0,1}{2}}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{4}\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}{\mathrm{sin}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{w}\right)}^{\textcolor[rgb]{0,0,1}{2}}}{{\textcolor[rgb]{0,0,1}{v}}^{\textcolor[rgb]{0,0,1}{2}}}
\stackrel{\text{simplify}}{=}
{\textcolor[rgb]{0,0,1}{u}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{=}\frac{\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}^{\textcolor[rgb]{0,0,1}{2}}}{{\textcolor[rgb]{0,0,1}{v}}^{\textcolor[rgb]{0,0,1}{2}}}
\stackrel{\text{solve for v}}{\to }
\left[\left[\textcolor[rgb]{0,0,1}{v}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{2}\right]\textcolor[rgb]{0,0,1}{,}\left[\textcolor[rgb]{0,0,1}{v}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}\right]\right]
\genfrac{}{}{0}{}{z=2}{\phantom{x=a}}|\genfrac{}{}{0}{}{\phantom{\mathrm{f\left(x\right)}}}{T}
{\textcolor[rgb]{0,0,1}{u}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{2}
\stackrel{\text{solve}}{\to }
\left\{\textcolor[rgb]{0,0,1}{u}\textcolor[rgb]{0,0,1}{=}\sqrt{\textcolor[rgb]{0,0,1}{2}}\right\}\textcolor[rgb]{0,0,1}{,}\left\{\textcolor[rgb]{0,0,1}{u}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{-}\sqrt{\textcolor[rgb]{0,0,1}{2}}\right\}
\genfrac{}{}{0}{}{z=3}{\phantom{x=a}}|\genfrac{}{}{0}{}{\phantom{\mathrm{f\left(x\right)}}}{T}
{\textcolor[rgb]{0,0,1}{u}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{3}
\stackrel{\text{solve}}{\to }
\left\{\textcolor[rgb]{0,0,1}{u}\textcolor[rgb]{0,0,1}{=}\sqrt{\textcolor[rgb]{0,0,1}{3}}\right\}\textcolor[rgb]{0,0,1}{,}\left\{\textcolor[rgb]{0,0,1}{u}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{-}\sqrt{\textcolor[rgb]{0,0,1}{3}}\right\}
Write and evaluate an appropriate iterated triple-integral
{∫}_{0}^{2 \mathrm{π}}{∫}_{2}^{3}{∫}_{{\sqrt{2}}_{}}^{{\sqrt{3}}_{}}2{\left(u/v\right)}^{3} \mathit{ⅆ}u \mathit{ⅆ}v \mathit{ⅆ}w
\frac{\textcolor[rgb]{0,0,1}{25}}{\textcolor[rgb]{0,0,1}{72}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{π}}
Note the use of the absolute value of the Jacobian.
Solution in cylindrical coordinates
Use the Int and int commands.
\mathrm{Int}\left(r,\left[r=\sqrt{z}/3..\sqrt{z}/2,z=2..3,\mathrm{θ}=0..2 \mathrm{π}\right]\right)=\phantom{\rule[-0.0ex]{0.0em}{0.0ex}} \mathrm{int}\left(r,\left[r=\sqrt{z}/3..\sqrt{z}/2,z=2..3,\mathrm{\theta }=0..2 \mathrm{\pi }\right]\right)\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}
{\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{∫}}_{\textcolor[rgb]{0,0,1}{0}}^{\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{π}}}{\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{∫}}_{\textcolor[rgb]{0,0,1}{2}}^{\textcolor[rgb]{0,0,1}{3}}{\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{∫}}_{\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{z}}}^{\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{z}}}\textcolor[rgb]{0,0,1}{r}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{ⅆ}\textcolor[rgb]{0,0,1}{r}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{ⅆ}\textcolor[rgb]{0,0,1}{z}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{ⅆ}\textcolor[rgb]{0,0,1}{\mathrm{θ}}\textcolor[rgb]{0,0,1}{=}\frac{\textcolor[rgb]{0,0,1}{25}}{\textcolor[rgb]{0,0,1}{72}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{π}}
Next, implement a solution in the coordinates defined by the transformation equations
x=\left(u/v\right)\mathrm{cos}\left(w\right)
y=\left(u/v\right)\mathrm{sin}\left(w\right)
z={u}^{2}
\mathrm{with}\left(\mathrm{Student}:-\mathrm{MultivariateCalculus}\right):
T
represent the transformation equations.
T≔\left[x=\frac{u \mathrm{cos}\left(w\right)}{v},y=\frac{u \mathrm{sin}\left(w\right)}{v},z={u}^{2}\right]:
\frac{∂\left(x,y,z\right)}{∂\left(u,v,w\right)}
Apply the simplify command to the result of a call to the Jacobian command.
\mathrm{simplify}\left(\mathrm{Jacobian}\left(\mathrm{eval}\left(\left[x,y,z\right],T\right),\left[u,v,w\right],\mathrm{output}=\mathrm{determinant}\right)\right)
\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}^{\textcolor[rgb]{0,0,1}{3}}}{{\textcolor[rgb]{0,0,1}{v}}^{\textcolor[rgb]{0,0,1}{3}}}
Use the eval command to apply the equations in
T
to the bounding surfaces.
Use the simplify and solve commands as needed.
{s}_{1}≔\mathrm{simplify}\left(\mathrm{eval}\left(z=4\left({x}^{2}+{y}^{2}\right),T\right)\right)
{\textcolor[rgb]{0,0,1}{u}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{=}\frac{\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}^{\textcolor[rgb]{0,0,1}{2}}}{{\textcolor[rgb]{0,0,1}{v}}^{\textcolor[rgb]{0,0,1}{2}}}
\mathrm{solve}\left({s}_{1},v\right)
\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}
{s}_{3}≔\mathrm{eval}\left(z=2,T\right)
{\textcolor[rgb]{0,0,1}{u}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{2}
\mathrm{solve}\left({s}_{3},u\right)
\sqrt{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}\sqrt{\textcolor[rgb]{0,0,1}{2}}
{s}_{2}≔\mathrm{simplify}\left(\mathrm{eval}\left(z=9\left({x}^{2}+{y}^{2}\right),T\right)\right)
{\textcolor[rgb]{0,0,1}{u}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{=}\frac{\textcolor[rgb]{0,0,1}{9}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}^{\textcolor[rgb]{0,0,1}{2}}}{{\textcolor[rgb]{0,0,1}{v}}^{\textcolor[rgb]{0,0,1}{2}}}
\mathrm{solve}\left({s}_{2},v\right)
\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{3}
{s}_{4}≔\mathrm{eval}\left(z=3,T\right)
{\textcolor[rgb]{0,0,1}{u}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{3}
\mathrm{solve}\left({s}_{4},u\right)
\sqrt{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}\sqrt{\textcolor[rgb]{0,0,1}{3}}
Write and evaluate the appropriate iterated triple-integral
\mathrm{Int}\left(2{\left(u/v\right)}^{3},\left[u=\sqrt{2}..\sqrt{3},v=2..3,w=0..2 \mathrm{π}\right]\right)=\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathrm{int}\left(2{\left(u/v\right)}^{3},\left[u=\sqrt{2}..\sqrt{3},v=2..3,w=0..2 \mathrm{\pi }\right]\right)\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}
{\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{∫}}_{\textcolor[rgb]{0,0,1}{0}}^{\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{π}}}{\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{∫}}_{\textcolor[rgb]{0,0,1}{2}}^{\textcolor[rgb]{0,0,1}{3}}{\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{∫}}_{\sqrt{\textcolor[rgb]{0,0,1}{2}}}^{\sqrt{\textcolor[rgb]{0,0,1}{3}}}\frac{\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}^{\textcolor[rgb]{0,0,1}{3}}}{{\textcolor[rgb]{0,0,1}{v}}^{\textcolor[rgb]{0,0,1}{3}}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{ⅆ}\textcolor[rgb]{0,0,1}{u}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{ⅆ}\textcolor[rgb]{0,0,1}{v}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{ⅆ}\textcolor[rgb]{0,0,1}{w}\textcolor[rgb]{0,0,1}{=}\frac{\textcolor[rgb]{0,0,1}{25}}{\textcolor[rgb]{0,0,1}{72}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{π}}
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Integer Function - Floor, Ceiling, Sawtooth | Brilliant Math & Science Wiki
Integer Function - Floor, Ceiling, Sawtooth
Christopher Boo, Daniel Xiang, and Jimin Khim contributed
An integer function maps a real number to an integer value. In this wiki, we're going to discuss three integer functions that are widely applied in number theory—the floor function, ceiling function, sawtooth function.
x
the floor function returns the largest integer less than or equal to
x
\lfloor x \rfloor ;
the ceiling function returns the smallest integer larger than or equal to
x
\lceil x \rceil ;
the sawtooth function returns the fractional part of
x
\{x\}
Note: The variable
n
in this section is assumed to be an integer.
The floor function has the following properties:
\lfloor x \rfloor + \{x\} = x
\lfloor x+y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor
\{x\} + \{y\} \geq \{x+y\}
\lfloor x + n \rfloor = \lfloor x \rfloor + n
\{x+n\}=\{x\}
\lfloor xy\rfloor \geq \lfloor x \rfloor \lfloor y\rfloor \quad \mathrm{for} \quad x, y \ge 0
\displaystyle \lfloor \sqrt[n]{x} \rfloor ^n \leq \lfloor x \rfloor
\displaystyle \bigg \lfloor \frac{nx}{y} \bigg \rfloor = n\bigg \lfloor \frac{x}{y} \bigg \rfloor
Solve the following equation for a non-zero solution:
x+2\{x\}=3\lfloor x \rfloor .
\begin{aligned} x+2\{x\}&=3\lfloor x \rfloor \\ \lfloor x \rfloor + \{x\} + 2\{x\} &= 3\lfloor x \rfloor \\ 3\{x\} &= 2\lfloor x \rfloor \\ \{x\} &= \frac{2}{3}\lfloor x \rfloor. \end{aligned}
0 \leq \{x\} < 1
0 \leq \lfloor x \rfloor < 1\frac{1}{2}
\lfloor x \rfloor = 1
\{x\} = \frac{2}{3}
. Then we finally have
x = \lfloor x \rfloor + \{x\} = 1 + \frac{2}{3} = 1\frac{2}{3}.\ _\square
Cite as: Integer Function - Floor, Ceiling, Sawtooth. Brilliant.org. Retrieved from https://brilliant.org/wiki/integer-function-floor-ceiling-sawtooth/ |
Loss of linear incremental learning model on batch of data - MATLAB loss - MathWorks Switzerland
Measure Model Performance During Incremental Learning
Compute Custom Loss on Incoming Chunks of Data
Loss of linear incremental learning model on batch of data
loss returns the regression or classification loss of a configured incremental learning model for linear regression (incrementalRegressionLinear object) or linear binary classification (incrementalClassificationLinear object).
To measure model performance on a data stream and store the results in the output model, call updateMetrics or updateMetricsAndFit.
L = loss(Mdl,X,Y) returns the loss for the incremental learning model Mdl using the batch of predictor data X and corresponding responses Y.
L = loss(Mdl,X,Y,Name,Value) uses additional options specified by one or more name-value pair arguments. For example, you can specify that the columns of the predictor data matrix correspond to observations, or specify the classification loss function .
The performance of an incremental model on streaming data is measured in three ways:
Cumulative metrics measure the performance since the start of incremental learning.
Window metrics measure the performance on a specified window of observations. The metrics are updated every time the model processes the specified window.
The loss function measures the performance on a specified batch of data only.
Create an incremental linear SVM model for binary classification. Configure the model for loss by specifying the class names, prior class distribution (uniform), and arbitrary coefficient and bias values. Specify a metrics window size of 1000 observations.
Mdl = incrementalClassificationLinear('Beta',Beta,'Bias',Bias, ...
Mdl is an incrementalClassificationLinear model. All its properties are read-only. Instead of specifying arbitrary values, you can take either of these actions to configure the model:
Train an SVM model using fitcsvm or fitclinear on a subset of the data (if available), and then convert the model to an incremental learner by using incrementalLearner.
Simulate a data stream, and perform the following actions on each incoming chunk of 50 observations:
Call updateMetrics to measure the cumulative performance and the performance within a window of observations. Overwrite the previous incremental model with a new one to track performance metrics.
Call loss to measure the model performance on the incoming chunk.
Store all performance metrics to see how they evolve during incremental learning.
Mdl is an incrementalClassificationLinear model object trained on all the data in the stream. During incremental learning and after the model is warmed up, updateMetrics checks the performance of the model on the incoming observations, then and the fit function fits the model to those observations. loss is agnostic of the metrics warm-up period, so it measures the classification error for all iterations.
To see how the performance metrics evolve during training, plot them.
The yellow line represents the classification error on each incoming chunk of data. After the metrics warm-up period, Mdl tracks the cumulative and window metrics. The cumulative and batch losses converge as the fit function fits the incremental model to the incoming data.
Fit an incremental learning model for regression to streaming data, and compute the mean absolute deviation (MAD) on the incoming data batches.
Load the robot arm data set. Obtain the sample size n and the number of predictor variables p.
Track the mean absolute deviation (MAD) to measure the performance of the model. Create an anonymous function that measures the absolute error of each new observation. Create a structure array containing the name MeanAbsoluteError and its corresponding function.
Configure the model to predict responses by specifying that all regression coefficients and the bias are 0.
Mdl = incrementalRegressionLinear('MetricsWarmupPeriod',1000,'MetricsWindowSize',500, ...
Mdl is an incrementalRegressionLinear model object configured for incremental learning.
Call updateMetrics to compute cumulative and window metrics on the incoming chunk of data. Overwrite the previous incremental model with a new one fitted to overwrite the previous metrics.
Call loss to compute the MAD on the incoming chunk of data. Whereas the cumulative and window metrics require that custom losses return the loss for each observation, loss requires the loss on the entire chunk. Compute the mean of the absolute deviation.
Call fit to fit the incremental model to the incoming chunk of data.
Store the cumulative, window, and chunk metrics to see how they evolve during incremental learning.
Mdl is an incrementalRegressionLinear model object trained on all the data in the stream. During incremental learning and after the model is warmed up, updateMetrics checks the performance of the model on the incoming observations, and the fit function fits the model to those observations.
Plot the performance metrics to see how they evolved during incremental learning.
The plot suggests the following:
updateMetrics computes the performance metrics after the metrics warm-up period only.
updateMetrics computes the cumulative metrics during each iteration.
updateMetrics computes the window metrics after processing 500 observations
Because Mdl was configured to predict observations from the beginning of incremental learning, loss can compute the MAD on each incoming chunk of data.
You must configure Mdl to compute its loss on a batch of observations.
If Mdl is a converted, traditionally trained model, you can compute its loss without any modifications.
If Mdl is an incrementalClassificationLinear model, its model coefficients Mdl.Beta and bias Mdl.Bias must be nonempty arrays, the class names Mdl.ClassNames must contain two classes, and the prior class distribution Mdl.Prior must contain known values.
Batch of predictor data with which to compute the loss, specified as a floating-point matrix of n observations and Mdl.NumPredictors predictor variables. The value of the ObservationsIn name-value argument determines the orientation of the variables and observations. The default ObservationsIn value is "rows", which indicates that observations in the predictor data are oriented along the rows of X.
loss supports only floating-point input predictor data. If your input data includes categorical data, you must prepare an encoded version of the categorical data. Use dummyvar to convert each categorical variable to a numeric matrix of dummy variables. Then, concatenate all dummy variable matrices and any other numeric predictors. For more details, see Dummy Variables.
Y — Batch of responses (labels)
Batch of responses (labels) with which to compute the loss, specified as a categorical, character, or string array, logical or floating-point vector, or cell array of character vectors for classification problems; or a floating-point vector for regression problems.
loss supports binary classification only.
If Y contains a label that is not a member of Mdl.ClassNames, loss issues an error.
Example: 'ObservationsIn','columns','Weights',W specifies that the columns of the predictor matrix correspond to observations, and the vector W contains observation weights to apply.
string vector | function handle | cell vector | structure array | ...
Classification problems: The following table lists the available loss functions when Mdl is an incrementalClassificationLinear model. Specify one using its corresponding character vector or string scalar.
"classiferror" (default) Misclassification rate in decimal
"exponential" Exponential loss
"hinge" Hinge loss
"logit" Logistic loss
"quadratic" Quadratic loss
For more details, see Classification Loss.
Logistic regression learners return posterior probabilities as classification scores, but SVM learners do not (see predict).
The output argument lossval is an n-by-1 floating-point vector, where lossval(j) is the classification loss of observation j.
You specify the function name (lossfcn).
Regression problems: The following table lists the available loss functions when Mdl is an incrementalRegressionLinear model. Specify one using its corresponding character vector or string scalar.
"mse" (default) Weighted mean squared error 'svm' and 'leastsquares'
For more details, see Regression Loss.
Y is a length n numeric vector of observed responses.
Example: 'LossFun',"mse"
Example: 'LossFun',@lossfcn
Weights — Batch of observation weights
Batch of observation weights, specified as the comma-separated pair consisting of 'Weights' and a floating-point vector of positive values. loss weighs the observations in the input data with the corresponding values in Weights. The size of Weights must equal n, which is the number of observations in the input data.
For more details, see Observation Weights.
L — Classification or regression loss
Classification or regression loss, returned as a numeric scalar. The interpretation of L depends on Weights and LossFun.
The weight for observation j is wj.
Binomial deviance "binodeviance"
L=\sum _{j=1}^{n}{w}_{j}\mathrm{log}\left\{1+\mathrm{exp}\left[-2{m}_{j}\right]\right\}.
Exponential loss "exponential"
L=\sum _{j=1}^{n}{w}_{j}\mathrm{exp}\left(-{m}_{j}\right).
Misclassification rate in decimal "classiferror"
L=\sum _{j=1}^{n}{w}_{j}I\left\{{\stackrel{^}{y}}_{j}\ne {y}_{j}\right\},
{\stackrel{^}{y}}_{j}
is the class label corresponding to the class with the maximal score, and I{·} is the indicator function.
Hinge loss "hinge"
L=\sum _{j=1}^{n}{w}_{j}\mathrm{max}\left\{0,1-{m}_{j}\right\}.
Logit loss "logit"
L=\sum _{j=1}^{n}{w}_{j}\mathrm{log}\left(1+\mathrm{exp}\left(-{m}_{j}\right)\right).
Quadratic loss "quadratic"
L=\sum _{j=1}^{n}{w}_{j}{\left(1-{m}_{j}\right)}^{2}.
This figure compares the loss functions over the score m for one observation. Some functions are normalized to pass through the point (0,1).
Regression loss functions measure the predictive inaccuracy of regression models. When you compare the same type of loss among many models, a lower loss indicates a better predictive model.
yj is the observed response of observation j.
f(Xj) is the predicted value of observation j of the predictor data X.
Epsilon insensitive loss "epsiloninsensitive"
L=\mathrm{max}\left[0,|y-f\left(x\right)|-\epsilon \right].
Mean squared error "mse"
L={\left[y-f\left(x\right)\right]}^{2}.
For classification problems, if the prior class probability distribution is known (in other words, the prior distribution is not empirical), loss normalizes observation weights to sum to the prior class probabilities in the respective classes. This action implies that observation weights are the respective prior class probabilities by default.
For regression problems or if the prior class probability distribution is empirical, the software normalizes the specified observation weights to sum to 1 each time you call loss.
Use saveLearnerForCoder, loadLearnerForCoder, and codegen (MATLAB Coder) to generate code for the loss function. Save a trained model by using saveLearnerForCoder. Define an entry-point function that loads the saved model by using loadLearnerForCoder and calls the loss function. Then use codegen to generate code for the entry-point function.
To generate single-precision C/C++ code for loss, specify the name-value argument "DataType","single" when you call the loadLearnerForCoder function.
This table contains notes about the arguments of loss. Arguments not included in this table are fully supported.
The specified function cannot be an anonymous function.
If you configure Mdl to shuffle data (Mdl.Shuffle is true, or Mdl.Solver is 'sgd' or 'asgd'), the loss function randomly shuffles each incoming batch of observations before it fits the model to the batch. The order of the shuffled observations might not match the order generated by MATLAB®. Therefore, if you fit Mdl before computing the loss, the loss computed in MATLAB and those computed by the generated code might not be equal. |
Extended Kalman filter for object tracking - MATLAB - MathWorks France
Kalman filter state, specified as a real-valued M-element vector, where M is the size of the filter state. The value of M is determined based on the motion model you use. For example, if you use a 2-D constant velocity model specified by constvel, in which the state is [x;vx;y;vy], M is four.
Setting this property to a positive integer enables the OOSM retrodiction capability of the filter object. This option requires a Sensor Fusion and Tracking Toolbox license. With OOSM enabled, the filter object saves the past state and state covariance history. You can use the OOSM and the retrodict and retroCorrect (or retroCorrectJPDA for multiple OOSMs) object functions to reduce the uncertainty of the estimated state.
{x}_{k+1}=f\left({x}_{k},{u}_{k},{w}_{k},t\right)
{x}_{k+1}=f\left({x}_{k},{u}_{k},t\right)+{w}_{k}
{z}_{k}=h\left({x}_{k},{v}_{k},t\right)
{z}_{k}=h\left({x}_{k},t\right)+{v}_{k}
trackingKF | trackingUKF | trackingCKF | trackingGSF | trackingPF | trackingIMM | trackingABF | trackingMSCEKF | trackerTOMHT | trackerGNN |
Receiver operating characteristic (ROC) curve or other performance curve for classifier output - MATLAB perfcurve - MathWorks Italia
score\left(:,2\right)-\mathrm{max}\left(score\left(:,1\right),score\left(:,3\right)\right)
score\left(:,2\right)-score\left(:,3\right)
S=\frac{\text{Cost}\left(P|N\right)-\text{Cost}\left(N|N\right)}{\text{Cost}\left(N|P\right)-\text{Cost}\left(P|P\right)}*\frac{N}{P} |
Sketch the following conic sections:a) y^2-4y+8-4x^2=8 b)y^2-4y-4x^2=8 c)3(y-5)^2-7(x+1)^2=1
{y}^{2}-4y+8-4{x}^{2}=8
{y}^{2}-4y-4{x}^{2}=8
3{\left(y-5\right)}^{2}-7{\left(x+1\right)}^{2}=1
The answer is given below in the photo
{y}^{2}-4y-4{x}^{2}+8=0
Hint: Solve the quadratic equation by completing the square.
-4{x}^{2}+8
{y}^{2}-4y=4{x}^{2}-8
Hint: Take one half of the coefficient of y and square it, then add it to both sides.
{y}^{2}-4y+4=4{x}^{2}-4
Hint: Factor the left hand side.
\left(y-2{\right)}^{2}=4{x}^{2}-4
Hint: Eliminate the exponent on the left hand side.
y-2=\sqrt{4{x}^{2}-4}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}y-2=-\sqrt{4{x}^{2}-4}
Hint: Look at the first equation: Solve for y.
y=\sqrt{4{x}^{2}-4}+2\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}y-2=-\sqrt{4{x}^{2}-4}
Hint: Look at the second equation: Solve for y.
y=\sqrt{4{x}^{2}-4}+2\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}y=2-\sqrt{4{x}^{2}-4}
{y}^{2}-4y-4{x}^{2}=8
4{x}^{2}
{y}^{2}-4y=4{x}^{2}+8
{y}^{2}-4y+4=4{x}^{2}+12
\left(y-2{\right)}^{2}=4{x}^{2}+12
y-2=\sqrt{4{x}^{2}+12}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}y-2=-\sqrt{4{x}^{2}+12}
y=\sqrt{4{x}^{2}+12}+2\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}y-2=-\sqrt{4{x}^{2}+12}
y=\sqrt{4{x}^{2}+12}+2\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}y=2-\sqrt{4{x}^{2}+12}
3{\left(y-5\right)}^{2}-7{\left(x+1\right)}^{2}=1
Hint: Isolate terms with y to the left hand side.
7\left(x+1{\right)}^{2}
3{\left(y-5\right)}^{2}=7{\left(x+1\right)}^{2}+1
Hint: Divide both sides by a constant to simplify the equation.
{\left(y-5\right)}^{2}=\frac{7}{3}{\left(x+1\right)}^{2}+\frac{1}{3}
y-5=\sqrt{\frac{7}{3}{\left(x+1\right)}^{2}+\frac{1}{3}}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}y-5=-\sqrt{\frac{7}{3}{\left(x+1\right)}^{2}+\frac{1}{3}}
Hint: | Look at the first equation: Solve for y.
At one point in a pipeline the water’s speed is 3.00 m/s and the gauge pressure is
5.00×{10}^{4}Pa
. Find the gauge pressure at a second point in the line, 11.0 m lower than the first, if the pipe diameter at the second point is twice that at the first.
e=2,\text{ }x=4
Show that for eccentricity equal to one in Equation 13.10 for conic sections, the path is a parabola. Do this by substituting Cartesian coordinates, x and y, for the polar coordinates, r and θ , and showing that it has the general form for a parabola,
x=a{y}^{2}+by+c
9{x}^{2}+9{y}^{2}+18x-18y+14=0
5{x}^{2}–4{y}^{2}=20x+24y+36
{r}^{2}=\mathrm{sec}\text{ }2\text{ }\theta |
If A has
m
n
columns, and B has
p
q
columns, then the output is a Matrix with
m
p
n
q
columns. The output can be thought of as a partitioned matrix with
m
n
blocks of size
p
q
. The (
i,j
)th block in the output is then
{A}_{i,j}·B
\mathrm{with}\left(\mathrm{LinearAlgebra}\right):
A≔\mathrm{Matrix}\left([[1,2,3],[4,5,6]]\right)
\textcolor[rgb]{0,0,1}{A}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{ccc}\textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{3}\\ \textcolor[rgb]{0,0,1}{4}& \textcolor[rgb]{0,0,1}{5}& \textcolor[rgb]{0,0,1}{6}\end{array}]
B≔\mathrm{Matrix}\left([[0,x],[-x,1]]\right)
\textcolor[rgb]{0,0,1}{B}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{x}\\ \textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{1}\end{array}]
\mathrm{KroneckerProduct}\left(A,B\right)
[\begin{array}{cccccc}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\\ \textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{3}\\ \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{6}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\\ \textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{4}& \textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{5}& \textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{6}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{6}\end{array}]
\mathrm{KroneckerProduct}\left(B,A\right)
[\begin{array}{cccccc}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\\ \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{6}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\\ \textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{3}\\ \textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{6}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{4}& \textcolor[rgb]{0,0,1}{5}& \textcolor[rgb]{0,0,1}{6}\end{array}]
P≔\mathrm{BandMatrix}\left([2,4,6],1,2\right)
\textcolor[rgb]{0,0,1}{P}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{4}& \textcolor[rgb]{0,0,1}{6}\\ \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{4}\end{array}]
Q≔\mathrm{BandMatrix}\left([x,2x,3x],1,2\right)
\textcolor[rgb]{0,0,1}{Q}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\\ \textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\end{array}]
\mathrm{Multiply}\left(P,Q\right)-\mathrm{Multiply}\left(Q,P\right)
[\begin{array}{cc}\textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}\\ \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}\end{array}]
B≔\mathrm{KroneckerProduct}\left(\mathrm{Matrix}\left([[1,0],[0,1]]\right),P\right)+\mathrm{KroneckerProduct}\left(\mathrm{Matrix}\left([[0,1],[1,0]]\right),Q\right)
\textcolor[rgb]{0,0,1}{B}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{cccc}\textcolor[rgb]{0,0,1}{4}& \textcolor[rgb]{0,0,1}{6}& \textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\\ \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{4}& \textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{2}\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}{x}& \textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{4}& \textcolor[rgb]{0,0,1}{6}\\ \textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{4}\end{array}]
\mathrm{Determinant}\left(B\right)
{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{4}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{8}\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}{16}
\mathrm{Determinant}\left(\mathrm{Multiply}\left(P,P\right)-\mathrm{Multiply}\left(Q,Q\right)\right)
{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{4}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{8}\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}{16} |
Kurs:Computation of tight closure (Ann Arbor 2012) – Wikiversity
Kurs:Computation of tight closure (Ann Arbor 2012)
This is a short course on the computation of tight closure held by Holger Brenner at the conference Computational workshop on Frobenius Singularities and invariants in Ann Arbor 2012 (organized by Karl Schwede, Kevin Tucker and Wenliang Zhang).
Lecture 1 - Basics
Lecture 2 - Applications of strong semistability
Lecture 3 - Deformations and generic results
Forcing algebras, syzygy bundles, and tight closure (Holger Brenner, survey article)
Some remarks on the affineness of
{\displaystyle {}{\mathbb {A} }^{1}}
-bundles (Holger Brenner, survey article)
A guide to closure operations in commutative algebra (Neil Epstein, survey article)
The theory of tight closure from the viewpoint of vector bundles (Holger Brenner, Habilitationsschrift)
Tight closure and plus closure in dimension two (Holger Brenner)
On the arithmetic of tight closure (Holger Brenner and Moty Katzman)
Tight closure does not commute with localization (Holger Brenner and Paul Monsky)
Continuous solutions to algebraic forcing equations (Holger Brenner)
Continuous closure, axes closure, and natural closure (Neil Epstein and Mel Hochster)
Continuous closure of sheaves (János Kollár)
John Baez explains the use of torsors in physics
Abgerufen von „https://de.wikiversity.org/w/index.php?title=Kurs:Computation_of_tight_closure_(Ann_Arbor_2012)&oldid=315214“
Theorie der Vektorbündel auf Schemata/Kurse |
LeftCoset - Maple Help
Home : Support : Online Help : Mathematics : Group Theory : LeftCoset
construct the left coset of a subgroup
construct the right coset of a subgroup
LeftCoset( g, H )
RightCoset( H, g )
an element of some group containing H
a subgroup of a permutation group
H
G
g
G
. The left coset
\mathrm{gH}
is defined to be the subset
\left\{\mathrm{gh}:h∈H\right\}
G
. Similarly, the right coset
\mathrm{Hg}
is the subset
\left\{\mathrm{hg}:h∈H\right\}
G
The LeftCoset( g, H ) command returns a data structure representing the left coset
\mathrm{gH}
of a subgroup H of a permutation group G. The RightCoset( H, g ) command returns a data structure representing the right coset
\mathrm{Hg}
of a subgroup H of a permutation group G.
The data structures representing (left or right) cosets respond to the following methods.
Representative( c )
returns the representative of the coset c
numelems( c )
returns the number of members of the coset c
member( x, c ) or x in c
returns true if x belongs to the coset c
Elements( c )
returns the elements of the coset c, as a set
Subgroup( c )
returns the subgroup of the coset c
\mathrm{with}\left(\mathrm{GroupTheory}\right):
G≔\mathrm{Alt}\left(4\right)
\textcolor[rgb]{0,0,1}{G}\textcolor[rgb]{0,0,1}{≔}{\textcolor[rgb]{0,0,1}{\mathbf{A}}}_{\textcolor[rgb]{0,0,1}{4}}
H≔\mathrm{Subgroup}\left({[[1,2],[3,4]]},G\right)
\textcolor[rgb]{0,0,1}{H}\textcolor[rgb]{0,0,1}{≔}〈\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\right)\left(\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\right)〉
C≔\mathrm{RightCoset}\left(H,[[1,2,3]]\right)
\textcolor[rgb]{0,0,1}{C}\textcolor[rgb]{0,0,1}{≔}〈\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\right)\left(\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\right)〉\textcolor[rgb]{0,0,1}{·}\left(\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)\right)
\mathrm{GroupOrder}\left(H\right)
\textcolor[rgb]{0,0,1}{2}
\mathrm{numelems}\left(C\right)
\textcolor[rgb]{0,0,1}{2}
\mathrm{Elements}\left(C\right)
{\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}{4}\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)}
The symmetric group on 3 letters.
M≔〈〈1|2|3|4|5|6〉,〈2|1|6|5|4|3〉,〈3|5|1|6|2|4〉,〈4|6|5|1|3|2〉,〈5|3|4|2|6|1〉,〈6|4|2|3|1|5〉〉
\textcolor[rgb]{0,0,1}{M}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{cccccc}\textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{3}& \textcolor[rgb]{0,0,1}{4}& \textcolor[rgb]{0,0,1}{5}& \textcolor[rgb]{0,0,1}{6}\\ \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{6}& \textcolor[rgb]{0,0,1}{5}& \textcolor[rgb]{0,0,1}{4}& \textcolor[rgb]{0,0,1}{3}\\ \textcolor[rgb]{0,0,1}{3}& \textcolor[rgb]{0,0,1}{5}& \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{6}& \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{4}\\ \textcolor[rgb]{0,0,1}{4}& \textcolor[rgb]{0,0,1}{6}& \textcolor[rgb]{0,0,1}{5}& \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{3}& \textcolor[rgb]{0,0,1}{2}\\ \textcolor[rgb]{0,0,1}{5}& \textcolor[rgb]{0,0,1}{3}& \textcolor[rgb]{0,0,1}{4}& \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{6}& \textcolor[rgb]{0,0,1}{1}\\ \textcolor[rgb]{0,0,1}{6}& \textcolor[rgb]{0,0,1}{4}& \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{3}& \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{5}\end{array}]
G≔\mathrm{CayleyTableGroup}\left(M\right)
\textcolor[rgb]{0,0,1}{G}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{< a Cayley table group with 6 elements >}}
2 is an involution.
H≔\mathrm{Subgroup}\left([2],G\right)
\textcolor[rgb]{0,0,1}{H}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{< a Cayley table group with 1 generator >}}
\mathrm{RC}≔\mathrm{RightCoset}\left(H,3\right)
\textcolor[rgb]{0,0,1}{\mathrm{RC}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{< a Cayley table group with 1 generator >}}\textcolor[rgb]{0,0,1}{·}\textcolor[rgb]{0,0,1}{3}
\mathrm{numelems}\left(H\right)=\mathrm{numelems}\left(\mathrm{RC}\right)
\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{2}
\mathrm{Elements}\left(\mathrm{RC}\right)
{\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{6}}
\mathrm{LC}≔\mathrm{LeftCoset}\left(3,H\right)
\textcolor[rgb]{0,0,1}{\mathrm{LC}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{·}\textcolor[rgb]{0,0,1}{\mathrm{< a Cayley table group with 2 elements >}}
\mathrm{numelems}\left(H\right)=\mathrm{numelems}\left(\mathrm{LC}\right)
\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{2}
\mathrm{Elements}\left(\mathrm{LC}\right)
{\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}}
The GroupTheory[LeftCoset] and GroupTheory[RightCoset] commands were introduced in Maple 17. |
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