text
stringlengths 256
16.4k
|
|---|
\int_{0}^\infty {\cos(xt)\over 1+t^2}dt I'm supposed to solve this using Laplace Transformations.
Jazmine Sweeney 2022-04-30 Answered
{\int }_{0}^{\mathrm{\infty }}\left\{\mathrm{cos}\left(xt\right)over1+{t}^{2}\right\}dt
Let f(x) denote the integral, and assume temporarily that x>0. This makes no difference since f(x) is even by definition. Then its Laplace transform Lf(s) defines a continuous function on
s\in \left(0,\mathrm{\infty }\right)
(in fact, it defines an analytic function on R(s)>0). Thus we may assume further that
s\ne 1
and then rely on the continuity argument. Then
\mathcal{L}f\left(s\right)={\int }_{0}^{\mathrm{\infty }}f\left(x\right){e}^{-sx}dx
={\int }_{0}^{\mathrm{\infty }}{\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{cos}\left(xt\right)}{1+{t}^{2}}{e}^{-sx}dtdx
\stackrel{\cdot }{=}{\int }_{0}^{\mathrm{\infty }}{\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{cos}\left(xt\right)}{1+{t}^{2}}{e}^{-sx}dxdt
={\int }_{0}^{\mathrm{\infty }}\frac{1}{1+{t}^{2}}\frac{s}{{s}^{2}+{t}^{2}}dt
=\frac{s}{1-{s}^{2}}{\int }_{0}^{\mathrm{\infty }}\left(\frac{1}{{s}^{2}+{t}^{2}}-\frac{1}{1+{t}^{2}}\right)dt
=\frac{s}{1-{s}^{2}}\frac{\pi }{2}\left(\frac{1}{s}-1\right)=\frac{\pi }{2}\frac{1}{1+s}
Here, the change of order of integration (∗) is justified by the dominated convergence theorem. Though we proved this for
s\ne 1
, it remains valid by continuity argument as mentioned above. Then by the uniqueness of the Laplace transform, we find that
f\left(x\right)=\frac{\pi }{2}{e}^{-x}
Find that solution of
{y}^{\prime }=2\left(2x-y\right)
which passes through the point (0, 1).
Use the method of Laplace transformation to solve initial value problem.
\frac{dx}{dt}=x-2y,x\left(0\right)=-1,y\left(0\right)=2
\frac{dy}{dt}=5x-y
t{e}^{-4t}\mathrm{sin}3t
Determine a)
L\le ft\left\{2{t}^{4}{e}^{3t}right\right\}
L\le ft\left\{4{e}^{3t}\mathrm{cos}5tright\right\}
Find the general solution of the given differential equation.y” + 2y' + y = 2e−t
x\mathrm{sin}y-y\mathrm{cos}x=c
Find the general solutions of the differential equations
6{y}^{4}+11y4y=0
|
t^2 - k^2 t + k 4 if t and k are (+)ve integers above and tk, find the - Maths - Linear Inequalities - 8744781 | Meritnation.com
t^2 - k^2 t + k 4
if t and k are (+)ve integers above and tk, find the value of t
(SAT ques; expert help required)
\mathrm{Hi}\mathrm{Shivangi},\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Assuming}\mathrm{your}\mathrm{question}\mathrm{as},\phantom{\rule{0ex}{0ex}}{\mathrm{t}}^{2}-{\mathrm{k}}^{2}\mathrm{t}+\mathrm{k}=4\phantom{\rule{0ex}{0ex}}{\mathrm{t}}^{2}-{\mathrm{k}}^{2}\mathrm{t}+\mathrm{k}-4=0\phantom{\rule{0ex}{0ex}}\mathrm{Now}\mathrm{using}\mathrm{quadratic}\mathrm{formula},\phantom{\rule{0ex}{0ex}}\mathrm{t}=\frac{{\mathrm{k}}^{2}±\sqrt{{\mathrm{k}}^{4}-4\left(\mathrm{k}-4\right)}}{2}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{k}}^{2}±\sqrt{{\mathrm{k}}^{4}-4k+16}}{2}
|
Pictures of Julia and Mandelbrot Sets/Favourite formulas - Wikibooks, open books for an open world
Pictures of Julia and Mandelbrot Sets/Favourite formulas
{\displaystyle z^{2}/2+z^{4}/4+c}
{\displaystyle 1-z^{2}+z^{5}/(2+4z)+c}
{\displaystyle z^{2}/2+z^{4}/4+c}
{\displaystyle 1-z^{2}+z^{5}/(2+4z)+c}
Retrieved from "https://en.wikibooks.org/w/index.php?title=Pictures_of_Julia_and_Mandelbrot_Sets/Favourite_formulas&oldid=1989035"
|
Home : Support : Online Help : Connectivity : Maple T.A. : MapleTA Package : Builtin : lt
less-than test
The lt command returns true if a is less than b. Otherwise false is returned.
This is slightly different than in Maple T.A. proper, which understands 1.0 and 0.0 as true and false. Because Maple does not equate numbers as boolean conditionals, it is more appropriate to have this function return true and false inside Maple.
\mathrm{MapleTA}:-\mathrm{Builtin}:-\mathrm{lt}\left(2,1\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
\mathrm{MapleTA}:-\mathrm{Builtin}:-\mathrm{lt}\left(1,1\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
\mathrm{MapleTA}:-\mathrm{Builtin}:-\mathrm{lt}\left(1,2\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
The MapleTA[Builtin][lt] command was introduced in Maple 18.
MapleTA,Builtin,eq
MapleTA,Builtin,ge
MapleTA,Builtin,le
MapleTA,Builtin,lt
MapleTA,Builtin,ne
|
Solve system of linear equations — symmetric LQ method - MATLAB symmlq - MathWorks América Latina
Using symmlq with Preconditioner
Solve system of linear equations — symmetric LQ method
x = symmlq(A,b)
x = symmlq(A,b,tol)
x = symmlq(A,b,tol,maxit)
x = symmlq(A,b,tol,maxit,M)
x = symmlq(A,b,tol,maxit,M1,M2)
x = symmlq(A,b,tol,maxit,M1,M2,x0)
[x,flag] = symmlq(___)
[x,flag,relres] = symmlq(___)
[x,flag,relres,iter] = symmlq(___)
[x,flag,relres,iter,resvec] = symmlq(___)
[x,flag,relres,iter,resvec,resveccg] = symmlq(___)
x = symmlq(A,b) attempts to solve the system of linear equations A*x = b for x using the Symmetric LQ Method. When the attempt is successful, symmlq displays a message to confirm convergence. If symmlq fails to converge after the maximum number of iterations or halts for any reason, it displays a diagnostic message that includes the relative residual norm(b-A*x)/norm(b) and the iteration number at which the method stopped.
x = symmlq(A,b,tol) specifies a tolerance for the method. The default tolerance is 1e-6.
x = symmlq(A,b,tol,maxit) specifies the maximum number of iterations to use. symmlq displays a diagnostic message if it fails to converge within maxit iterations.
x = symmlq(A,b,tol,maxit,M) specifies a preconditioner matrix M and computes x by effectively solving the system
{H}^{-1}A\text{\hspace{0.17em}}{H}^{-T}y={H}^{-1}b
y={H}^{T}x
H={M}^{1/2}={\left({M}_{1}{M}_{2}\right)}^{1/2}
x = symmlq(A,b,tol,maxit,M1,M2) specifies factors of the preconditioner matrix M such that M = M1*M2.
x = symmlq(A,b,tol,maxit,M1,M2,x0) specifies an initial guess for the solution vector x. The default is a vector of zeros.
[x,flag] = symmlq(___) returns a flag that specifies whether the algorithm successfully converged. When flag = 0, convergence was successful. You can use this output syntax with any of the previous input argument combinations. When you specify the flag output, symmlq does not display any diagnostic messages.
[x,flag,relres] = symmlq(___) also returns the residual error in the computed solution. If flag is 0, then relres <= tol.
[x,flag,relres,iter] = symmlq(___) also returns the iteration number iter at which x was computed.
[x,flag,relres,iter,resvec] = symmlq(___) also returns a vector of the residual norm at each iteration, including the first residual norm(b-A*x0).
[x,flag,relres,iter,resvec,resveccg] = symmlq(___) also returns a vector of the conjugate gradients residual norms at each iteration.
Solve a square linear system using symmlq with default settings, and then adjust the tolerance and number of iterations used in the solution process.
Create a sparse tridiagonal matrix A as the coefficient matrix. Use the dense row sums of
\mathit{A}
as the vector for the right-hand side of
\mathrm{Ax}=\mathit{b}
\mathit{x}
\mathrm{Ax}=\mathit{b}
using symmlq. The output display includes the value of the relative residual error
\frac{‖\mathit{b}-\mathrm{Ax}‖}{‖\mathit{b}‖}
x = symmlq(A,b);
symmlq stopped at iteration 20 without converging to the desired tolerance 1e-06
By default symmlq uses 20 iterations and a tolerance of 1e-6, and the algorithm is unable to converge in those 20 iterations for this matrix. Since the residual is on the order of 1e-2, it is a good indicator that more iterations are needed. You also can use a larger tolerance to make it easier for the algorithm to converge.
x = symmlq(A,b,1e-4,250);
symmlq converged at iteration 199 to a solution with relative residual 1.5e-14.
Examine the effect of using a preconditioner matrix with symmlq to solve a linear system.
\mathrm{Ax}=\mathit{b}
Use symmlq to find a solution at the requested tolerance and number of iterations. Specify six outputs to return information about the solution process:
‖\mathit{b}-\mathrm{Ax}‖
‖{\mathit{A}}^{\mathit{T}}\mathit{A}\text{\hspace{0.17em}}\mathit{x}-{\mathit{A}}^{\mathit{T}}\mathit{b}‖
[x,fl0,rr0,it0,rv0,rvcg0] = symmlq(A,b,tol,maxit);
fl0 is 1 because symmlq does not converge to the requested tolerance 1e-12 within the requested 100 iterations.
\mathit{M}=\mathit{L}\text{\hspace{0.17em}}{\mathit{L}}^{\mathit{T}}
. Specify the 'ict' option to use incomplete Cholesky factorization with threshold dropping, and specify a diagonal shift value of 1e-6 to avoid nonpositive pivots. Solve the preconditioned system by specifying L and L' as inputs to symmlq.
[x1,fl1,rr1,it1,rv1,rvcg1] = symmlq(A,b,tol,maxit,L,L');
The use of an ichol preconditioner greatly improves the numerical properties of the problem, and symmlq is able to converge quickly. The output rv1(1) is norm(b), and the output rv1(end) is norm(b-A*x1).
You can follow the progress of symmlq by plotting the relative residuals at each iteration. Plot the conjugate gradient residual history of each solution with a line for the specified tolerance.
semilogy(0:length(rvcg0)-1,rvcg0/norm(b),'-o')
legend('No preconditioner','ICHOL preconditioner','Tolerance','Location','SouthEast')
Examine the effect of supplying symmlq with an initial guess of the solution.
\mathrm{Ax}=\mathit{b}
\mathit{x}
Use symmlq to solve
\mathrm{Ax}=\mathit{b}
x1 = symmlq(A,b,[],maxit);
symmlq converged at iteration 34 to a solution with relative residual 9.5e-07.
x2 = symmlq(A,b,[],maxit,[],[],x0);
symmlq converged at iteration 6 to a solution with relative residual 8.7e-07.
In this case supplying an initial guess enables symmlq to converge more quickly.
You also can use the initial guess to get intermediate results by calling symmlq in a for-loop. Each call to the solver performs a few iterations and stores the calculated solution. Then you use that solution as the initial vector for the next batch of iterations.
[x,flag,relres] = symmlq(A,b,tol,maxit,[],[],x0);
Solve a linear system by providing symmlq with a function handle that computes A*x in place of the coefficient matrix A.
\mathrm{Ax}
\mathrm{Ax}=\left[\begin{array}{cccccccccc}10& 1& 0& \cdots & & \cdots & & \cdots & 0& 0\\ 1& 9& 1& 0& & & & & & 0\\ 0& 1& 8& 1& 0& & & & & ⋮\\ ⋮& 0& 1& 7& 1& 0& & & & \\ & & 0& 1& 6& 1& 0& & & ⋮\\ ⋮& & & 0& 1& 5& 1& 0& & \\ & & & & 0& 1& 4& 1& 0& ⋮\\ ⋮& & & & & 0& 1& 3& \ddots & 0\\ 0& & & & & & 0& \ddots & \ddots & 1\\ 0& 0& \cdots & & \cdots & & \cdots & 0& 1& 10\end{array}\right]\left[\begin{array}{c}{\mathit{x}}_{1}\\ {\mathit{x}}_{2}\\ {\mathit{x}}_{3}\\ {\mathit{x}}_{4}\\ {\mathit{x}}_{5}\\ ⋮\\ \\ ⋮\\ \\ {\mathit{x}}_{21}\end{array}\right]=\left[\begin{array}{c}10{\mathit{x}}_{1}+{\mathit{x}}_{2}\\ {\mathit{x}}_{1}+9{\mathit{x}}_{2}+{\mathit{x}}_{3}\\ {\mathit{x}}_{2}+8{\mathit{x}}_{3}+{\mathit{x}}_{4}\\ ⋮\\ {\mathit{x}}_{19}+9{\mathit{x}}_{20}+{\mathit{x}}_{21}\\ {\mathit{x}}_{20}+10{\mathit{x}}_{21}\end{array}\right]
\mathrm{Ax}=\left[\begin{array}{c}0+10{\mathit{x}}_{1}+{\mathit{x}}_{2}\\ {\mathit{x}}_{1}+9{\mathit{x}}_{2}+{\mathit{x}}_{3}\\ {\mathit{x}}_{2}+8{\mathit{x}}_{3}+{\mathit{x}}_{4}\\ ⋮\\ {\mathit{x}}_{19}+9{\mathit{x}}_{20}+{\mathit{x}}_{21}\\ {\mathit{x}}_{20}+10{\mathit{x}}_{21}+0\end{array}\right]
\left[\begin{array}{c}0\\ {\mathit{x}}_{1}\\ ⋮\\ {\mathit{x}}_{20}\end{array}\right]+\left[\begin{array}{c}10{\mathit{x}}_{1}\\ 9{\mathit{x}}_{2}\\ ⋮\\ 10{\mathit{x}}_{21}\end{array}\right]+\left[\begin{array}{c}{\mathit{x}}_{2}\\ ⋮\\ {\mathit{x}}_{21}\\ 0\end{array}\right]
\mathrm{Ax}=\mathit{b}
by providing symmlq with the function handle that calculates A*x. Use a tolerance of 1e-12 and 50 iterations.
x1 = symmlq(@afun,b,tol,maxit)
Method tolerance, specified as a positive scalar. Use this input to trade-off accuracy and runtime in the calculation. symmlq must meet the tolerance within the number of allowed iterations to be successful. A smaller value of tol means the answer must be more precise for the calculation to be successful.
Maximum number of iterations, specified as a positive scalar integer. Increase the value of maxit to allow more iterations for symmlq to meet the tolerance tol. Generally, a smaller value of tol means more iterations are required to successfully complete the calculation.
Preconditioner matrices, specified as separate arguments of matrices or function handles. You can specify a preconditioner matrix M or its matrix factors M = M1*M2 to improve the numerical aspects of the linear system and make it easier for symmlq to converge quickly. You can use the incomplete matrix factorization functions ilu and ichol to generate preconditioner matrices. You also can use equilibrate prior to factorization to improve the condition number of the coefficient matrix. For more information on preconditioners, see Iterative Methods for Linear Systems.
symmlq treats unspecified preconditioners as identity matrices.
Initial guess, specified as a column vector with length equal to size(A,2). If you can provide symmlq with a more reasonable initial guess x0 than the default vector of zeros, then it can save computation time and help the algorithm converge faster.
Whenever the calculation is not successful (flag ~= 0), the solution x returned by symmlq is the one with minimal residual norm computed over all the iterations.
Success — symmlq converged to the desired tolerance tol within maxit iterations.
Failure — symmlq iterated maxit iterations but did not converge.
Failure — symmlq stagnated after two consecutive iterations were the same.
Failure — One of the scalar quantities calculated by the symmlq algorithm became too small or too large to continue computing.
Relative residual error, returned as a scalar. The relative residual error is an indication of how accurate the returned answer x is. symmlq tracks the relative residual and conjugate gradients residual at each iteration in the solution process, and the algorithm converges when either residual meets the specified tolerance tol. The relres output contains the value of the residual that converged, either the relative residual or the conjugate gradients residual:
The relative residual error is equal to norm(b-A*x)/norm(b) and is generally the residual that meets the tolerance tol when symmlq converges. The resvec output tracks the history of this residual over all iterations.
The conjugate gradients residual error is equal to norm(A'*A*x - A'*b). This residual causes symmlq to converge less frequently than the relative residual. The resveccg output tracks the history of this residual over all iterations.
The MINRES and SYMMLQ methods are variants of the Lanczos method that underpins the conjugate gradients method PCG. Like PCG, the coefficient matrix still needs to be symmetric, but MINRES and SYMMLQ allow it to be indefinite (not all eigenvalues need to be positive). This is achieved by avoiding the implicit LU factorization normally present in the Lanczos method, which is prone to breakdowns when zero pivots are encountered.
MINRES minimizes the residual in the 2-norm, while SYMMLQ solves a projected system using an LQ factorization and keeps the residual orthogonal to all previous ones [1].
[2] Paige, C. C. and M. A. Saunders, "Solution of Sparse Indefinite Systems of Linear Equations." SIAM J. Numer. Anal., Vol.12, 1975, pp. 617-629.
bicg | bicgstab | cgs | gmres | lsqr | minres | pcg | qmr | mldivide
|
Utilitarian rule - Wikipedia
Utilitarian rule
(Redirected from Utilitarian social choice rule)
Not to be confused with Rule utilitarianism.
In social choice and operations research, the utilitarian rule (also called the max-sum rule) is a rule saying that, among all possible alternatives, society should pick the alternative which maximizes the sum of the utilities of all individuals in society.[1]: sub.2.5 It is a formal mathematical representation of the utilitarian philosophy.
2 Tangible utility functions
3 Abstract utility functions
4 Relative utilitarianism
5 The utilitarian rule and Pareto-efficiency
6 The utilitarian rule in specific contexts
{\displaystyle X}
be a set of possible `states of the world' or `alternatives'. Society wishes to choose a single state from
{\displaystyle X}
{\displaystyle X}
{\displaystyle X}
may represent all possible allocations of the resource.
{\displaystyle I}
{\displaystyle i\in I}
{\displaystyle u_{i}:X\longrightarrow \mathbb {R} }
be a utility function, describing the amount of happiness an individual i derives from each possible state.
{\displaystyle (u_{i})_{i\in I}}
{\displaystyle X}
which are `best' for society (the question of what 'best' means is the basic problem of social choice theory).
The utilitarian rule selects an element
{\displaystyle x\in X}
which maximizes the utilitarian sum
{\displaystyle U(x):=\sum _{i\in I}u_{i}(x).}
Tangible utility functionsEdit
The utilitarian rule is easy to interpret and implement when the functions ui represent some tangible, measurable form of utility. For example:[1]: 44
Consider a problem of allocating wood among builders. The utility functions may represent their productive power –
{\displaystyle u_{i}(y_{i})}
is the number of buildings that agent
{\displaystyle i}
can build using
{\displaystyle y_{i}}
units of wood. The utilitarian rule then allocates the wood in a way that maximizes the number of buildings.
Consider a problem of allocating a rare medication among patient. The utility functions may represent their chance of recovery –
{\displaystyle u_{i}(y_{i})}
is the probability of agent
{\displaystyle i}
to recover by getting
{\displaystyle y_{i}}
doses of the medication. The utilitarian rule then allocates the medication in a way that maximizes the expected number of survivors.
Abstract utility functionsEdit
When the functions ui represent some abstract form of "happiness", the utilitarian rule becomes harder to interpret. For the above formula to make sense, it must be assumed that the utility functions
{\displaystyle (u_{i})_{i\in I}}
are both cardinal and interpersonally comparable at a cardinal level.
The notion that individuals have cardinal utility functions is not that problematic. Cardinal utility has been implicitly assumed in decision theory ever since Daniel Bernoulli's analysis of the Saint Petersburg Paradox. Rigorous mathematical theories of cardinal utility (with application to risky decision making) were developed by Frank P. Ramsey, Bruno de Finetti, von Neumann and Morgenstern, and Leonard Savage. However, in these theories, a person's utility function is only well-defined up to an `affine rescaling'. Thus, if the utility function
{\displaystyle u_{i}:X\longrightarrow \mathbb {R} }
is valid description of her preferences, and if
{\displaystyle r_{i},s_{i}\in \mathbb {R} }
are two constants with
{\displaystyle s_{i}>0}
, then the `rescaled' utility function
{\displaystyle v_{i}(x):=s_{i}\,u_{i}(x)+r_{i}}
is an equally valid description of her preferences. If we define a new package of utility functions
{\displaystyle (v_{i})_{i\in I}}
using possibly different
{\displaystyle r_{i}\in \mathbb {R} }
{\displaystyle s_{i}>0}
{\displaystyle i\in I}
, and we then consider the utilitarian sum
{\displaystyle V(x):=\sum _{i\in I}v_{i}(x),}
then in general, the maximizer of
{\displaystyle V}
will not be the same as the maximizer of
{\displaystyle U}
. Thus, in a sense, classic utilitarian social choice is not well-defined within the standard model of cardinal utility used in decision theory, unless a mechanism is specified to `calibrate' the utility functions of the different individuals.
Relative utilitarianismEdit
Relative utilitarianism proposes a natural calibration mechanism. For every
{\displaystyle i\in I}
, suppose that the values
{\displaystyle m_{i}\ :=\ \min _{x\in X}\,u_{i}(x)\quad {\text{and}}\quad M_{i}\ :=\ \max _{x\in X}\,u_{i}(x)}
are well-defined. (For example, this will always be true if
{\displaystyle X}
is finite, or if
{\displaystyle X}
{\displaystyle u_{i}}
is a continuous function.) Then define
{\displaystyle w_{i}(x)\ :=\ {\frac {u_{i}(x)-m_{i}}{M_{i}-m_{i}}}}
{\displaystyle x\in X}
{\displaystyle w_{i}:X\longrightarrow \mathbb {R} }
is a `rescaled' utility function which has a minimum value of 0 and a maximum value of 1. The Relative Utilitarian social choice rule selects the element in
{\displaystyle X}
{\displaystyle W(x):=\sum _{i\in I}w_{i}(x).}
As an abstract social choice function, relative utilitarianism has been analyzed by Cao (1982),[2] Dhillon (1998),[3] Karni (1998),[4] Dhillon and Mertens (1999),[5] Segal (2000),[6] Sobel (2001)[7] and Pivato (2008).[8] (Cao (1982) refers to it as the `modified Thomson solution'.)
The utilitarian rule and Pareto-efficiencyEdit
Every Pareto efficient choice is necessarily utilitarian. This is because every Pareto improvement necessarily increases the sum of utilities. But the utilitarian choice is not the only Pareto-efficient one. In fact, every weighted utilitarian choice (maximizing a weighted sum of utilities) is Pareto-optimal, whenever the weights of all individuals are positive.
The utilitarian rule in specific contextsEdit
In the context of voting, the utilitarian rule leads to several voting methods:
Range voting (also called score voting or utilitarian voting) implements the relative-utilitarian rule by letting voters explicitly express their utilities to each alternative on a common normalized scale.
Implicit utilitarian voting tries to approximate the utilitarian rule while letting the voters express only ordinal rankings over candidates.
A related voting rule is Nanson's method.
In the context of resource allocation, the utilitarian rule leads to:
A particular rule for division of a single homogeneous resource;
Several rules and algorithms for utilitarian cake-cutting – dividing a heterogeneous resource;
A particular rule for fair item allocation.[9]
Egalitarian rule – a different rule, that emphasizes the welfare of the worst-off individual rather than the sum of utilities.
Proportional-fair rule – a rule that tries to balance the efficiency of the utilitarian rule and the fairness of the egalitarian rule.
Utility maximization problem – a problem solved by an individual consumer (rather than by society).
^ a b Herve Moulin (2004). Fair Division and Collective Welfare. Cambridge, Massachusetts: MIT Press. ISBN 9780262134231.
^ Cao, Xiren (1982-12-01). "Preference functions and bargaining solutions". 1982 21st IEEE Conference on Decision and Control. IEEE: 164–171. doi:10.1109/cdc.1982.268420. S2CID 30395654.
^ Dhillon, Amrita (1998), "Extended Pareto rules and relative utilitarianism", Social Choice and Welfare, 15 (4): 521–542, doi:10.1007/s003550050121, S2CID 54899024
^ Karni, Edi (1998), "Impartiality: definition and representation", Econometrica, 66 (6): 1405–1415, doi:10.2307/2999622, JSTOR 2999622
^ Dhillon, Amrita; Mertens, Jean-Francois (1999), "Relative utilitarianism", Econometrica, 67 (3): 471–498, doi:10.1111/1468-0262.00033
^ Segal, Uzi (2000), "Let's agree that all dictatorships are equally bad", Journal of Political Economy, 108 (3): 569–589, doi:10.1086/262129, S2CID 154610036
^ Sobel, Joel (2001), "Manipulation of preferences and relative utilitarianism", Games and Economic Behavior, 37: 196–215, CiteSeerX 10.1.1.395.509, doi:10.1006/game.2000.0839
^ Pivato, Marcus (2008), "Twofold optimality of the relative utilitarian bargaining solution", Social Choice and Welfare, 32 (1): 79–92, CiteSeerX 10.1.1.537.5572, doi:10.1007/s00355-008-0313-0, S2CID 15475740
^ Aziz, Haris; Huang, Xin; Mattei, Nicholas; Segal-Halevi, Erel (2021-06-01). "Computing Welfare-Maximizing Fair Allocations of Indivisible Goods". arXiv:2012.03979 [cs.GT].
Retrieved from "https://en.wikipedia.org/w/index.php?title=Utilitarian_rule&oldid=1073417290"
|
Existence of Multiple Solutions for a Singular Elliptic Problem with Critical Sobolev Exponent
Zonghu Xiu, "Existence of Multiple Solutions for a Singular Elliptic Problem with Critical Sobolev Exponent", Abstract and Applied Analysis, vol. 2012, Article ID 806397, 15 pages, 2012. https://doi.org/10.1155/2012/806397
Zonghu Xiu 1
We consider the existence of multiple solutions of the singular elliptic problem , as , where , , , , , , . By the variational method and the theory of genus, we prove that the above-mentioned problem has infinitely many solutions when some conditions are satisfied.
In this paper, we consider the existence of multiple solutions for the singular elliptic problem where . and are nonnegative functions in .
In recent years, the existence of multiple solutions on elliptic equations has been considered by many authors. In [1], Assunção et al. considered the following quasilinear degenerate elliptic equation: where . When , where and ; the authors proved that problem (1.2) has at least two positive solutions. Rodrigues in [2] studied the following critical problem on bounded domain : By the variational method on Nehari manifolds [3, 4], the author proved the existence of at least two positive solutions and the nonexistence of solutions when some certain conditions are satisfied. When and , Miotto and Miyagaki in [5] considered the semilinear Dirichlet problem in infinite strip domains The authors also proved that problem (1.4) has at least two positive solutions by the methods of Nehari manifold. For other references, we refer to [6–11] and the reference therein. In fact, motivated by [1, 2, 5], we consider the problem (1.1). Since our problem is singular and is studied in the whole space , the loss of compactness of the Sobolev embedding renders a variational technique that is more delicate. By the variational method and the theory of genus, we prove that problem (1.1) has infinitely many solutions when some suitable conditions are satisfied.
In order to state our result, we introduce some weighted Sobolev spaces. For and in , we define the spaces and as being the set of Lebesgue measurable functions , which satisfy Particularly, when , we have We denote the completion of by with the norm of where and . It is easy to find that is a reflexive and separable Banach space with the norm .
The following Hardy-Sobolev inequality is due to Caffarelli et al. [12], which is called Caffarelli-Kohn-Nirenberg inequality. There exist constants such that where is called the Sobolev critical exponent.
In the present paper, we make the following assumptions: for , where ; for , where . for , where .
Then, we give some basic definitions.
Definition 1.1. is said to be a weak solution of (1.1) if for any there holds
Let be the energy functional corresponding to problem (1.1), which is defined as for all . Then the functional and for all , there holds It is well known that the weak solutions of problem (1.1) are the critical points of the functional , see [13]. Thus, to prove the existence of weak solutions of (1.1), it is sufficient to show that admits a sequence of critical points in .
Our main result in this paper is the following.
Theorem 1.2. Let , , , . Assume – are fulfilled. Then problem (1.1) has infinitely many solutions in .
Our proof is based on variational method. One important aspect of applying this method is to show that the functional satisfies condition which is introduced in the following definition.
Definition 2.1. Let and be a Banach space. The functional satisfies the condition if for any such that contains a convergent subsequence in .
The following embedding theorem is an extension of the classical Rellich-Kondrachov compactness theorem, see [14].
Lemma 2.2. Suppose is an open bounded domain with boundary and . . Then the embedding is continuous if and , and is compact if and .
Now we prove an embedding theorem, which is important in our paper.
Lemma 2.3. Assume - and . Then the embedding is compact.
Proof. We split our proof into two cases.
(i) Consider .
By the Hölder inequality and (1.9) we have that where . Then the embedding is continuous. Next, we will prove that the embedding is compact.
Let be a ball center at origin with the radius . For the convenience, we denote by , that is, . Assume is a bounded sequence in . Then is bounded in . We choose in Lemma 2.2, then there exist and a subsequence, still denoted by , such that as . We want to prove that where . In fact, we obtain from (2.2) that The fact shows that Then (2.4) and (2.5) imply that which gives (2.3).
In the following, we will prove that strongly in .
Since is a reflexive Banach space and is bounded in . Then we may assume, up to a subsequence, that In view of (2.3), we get that for any there exists large enough such that On the other hand, due to the compact embedding in Lemma 2.2, we have that Therefore, there is such that for . Thus, the inequalities (2.8) and (2.10) show that This shows that is convergent in .
(ii) Consider .
It follows from (1.8) and the Hölder inequality that where . Thus, the fact of and (2.12) imply that the embedding is continuous. Similar to the proof of (i) we can also prove that the embedding is compact for .
Similarly, we have the following result of compact embedding.
Lemma 2.4. Assume and , then the embedding is compact.
The following concentration compactness principle is a weighted version of the Concentration Compactness Principle II due to Lions [15–18], see also [19, 20].
Lemma 2.5. Let . Suppose that is a sequence such that where are measures supported on and is the space of bounded measures in . Then there are the following results.(1) There exists some at most countable set , a family of distinct points in , and a family of positive numbers such that where is the Dirac measure at .(2) The following equality holds for some family satisfying (3)There hold where
Lemma 2.6. Let . Then satisfies the condition with , where is as in (1.8).
Proof. We will split the proof into three steps.
Step 1. is bounded in .
Let be a sequence of in , that is, Then, we have Since , (2.20) shows that is bounded in .
Step 2. There exists in such that in .
The inequality (1.8) shows that is bounded in . Then the above argument and the compactness embedding in Lemma 2.2 mean that the following convergence hold: It follows from Lemma 2.5 that there exist nonnegative measures and such that Thus, in order to prove it is sufficient to prove that .
For the proof of , we define the functional such that where belongs to the support of . It follows from (2.1) that Since is bounded, we can get from (1.8)-(1.9), Lemmas 2.3 and 2.5 that On the other hand, where . Then ; furthermore, (2.16) implies that or . We will prove that the later does not hold. Suppose otherwise, there exists some such that . Then (2.19) and Lemma 2.4 show that which contradicts the hypothesis of . Then .
Similarly, we define the functional as Then, the similar proof as above shows that . Thus, we can deduce from (2.22) that which implies that in .
Step 3. converges strongly in .
The following inequalities [21] play an important role in our proof: Our aim is to prove that is a Cauchy sequence of . In fact, let in (1.12), it follows from (2.19) that where Using the inequalities (2.31), we can get by direct computation that with some constant , independent of and .
Then the Hölder inequality together with (1.8) and (2.30) yield that Similarly, we have from the Hölder inequality, Lemmas 2.3 and 2.4 that Therefore, the above estimates imply that , that is, is a Cauchy sequence of . Then converges strongly in and we complete the proof.
Similarly, we have the following lemma.
Lemma 2.7. Let . Then satisfies the condition with , where are as in (1.8), and (1.9) respectively.
Proof. Step 1. is bounded in .
Let be a sequence of in . Then we have from Lemma 2.3 that Since , (2.37) shows that is bounded in .
Similar to the proof of Lemma 2.5, we can get that or by applying the functional . Now we prove that there is no such that . Suppose otherwise, then Let Then has the unique minimum point at Then it follows from (2.38) that which contradicts the hypothesis of .
By Lemma 2.4, this result can be similarly obtained by the method in Lemma 2.6, so we omit the proof.
3. Existence of Infinitely Solutions
In this section, we will use the minimax procedure to prove the existence of infinity many solutions of problem (1.1). Let denotes the class of such that is closed in and symmetric with respect to the origin. For , we recall the genus which is defined by If there is no mapping as above for any , then , and . The following proposition gives some main properties of the genus, see [13, 22].
Proposition 3.1. Let . Then(1)if there exists an odd map , then ,(2)if , then ,(3).(4)if is a sphere centered at the origin in , then ,(5)if is compact, then and there exists such that and , where .
Lemma 3.2. Assume (A1)–(A3). Then for any , there exists such that
Proof. For given , let be a -dimensional subspace of . If , then for we have The fact that all the norms on finite dimensional space are equivalent implies that for all for some constant . Then there exist large and small such that Denote Then is a sphere centered at the origin with radius of and Therefore, Proposition 3.1 shows that .
If , we have Since is also a norm and all norms on the finite dimensional space are equivalent, we have Then there exist large and small such that Denote Then is a sphere centered at the origin with radius of and
Therefore, Proposition 3.1 shows that .
Let . It is easy to check that . We define It is not difficult to find that and for any since is coercive and bounded below. Furthermore, we define the set
Then, is compact and we have the following important lemma, see [22].
Lemma 3.3. All the are critical values of . Moreover, if , then .
Proof of Theorem 1.2. In view of Lemmas 2.6 and 2.7, satisfies the condition in . Furthermore, as the standard argument of [13, 22, 23], Lemma 3.3 gives that has infinity many critical points with negative values. Thus, problem (1.1) has infinitely many solutions in , and we complete the proof.
The author would like to express his sincere gratitude to the anonymous reviewers for the valuable comments and suggestions.
R. B. Assunção, P. C. Carrião, and O. H. Miyagaki, “Critical singular problems via concentration-compactness lemma,” Journal of Mathematical Analysis and Applications, vol. 326, no. 1, pp. 137–154, 2007. View at: Publisher Site | Google Scholar | Zentralblatt MATH
R. D. S. Rodrigues, “On elliptic problems involving critical Hardy-Sobolev exponents and sign-changing function,” Nonlinear Analysis: Theory, Methods & Applications, vol. 73, no. 4, pp. 857–880, 2010. View at: Publisher Site | Google Scholar | Zentralblatt MATH
Z. Nehari, “On a class of nonlinear second-order differential equations,” Transactions of the American Mathematical Society, vol. 95, pp. 101–123, 1960. View at: Publisher Site | Google Scholar | Zentralblatt MATH
K. J. Brown, “The Nehari manifold for a semilinear elliptic equation involving a sublinear term,” Calculus of Variations and Partial Differential Equations, vol. 22, no. 4, pp. 483–494, 2005. View at: Publisher Site | Google Scholar | Zentralblatt MATH
M. L. Miotto and O. H. Miyagaki, “Multiple positive solutions for semilinear Dirichlet problems with sign-changing weight function in infinite strip domains,” Nonlinear Analysis: Theory, Methods & Applications, vol. 71, no. 7-8, pp. 3434–3447, 2009. View at: Publisher Site | Google Scholar | Zentralblatt MATH
R. B. Assunção, P. C. Carrião, and O. H. Miyagaki, “Subcritical perturbations of a singular quasilinear elliptic equation involving the critical Hardy-Sobolev exponent,” Nonlinear Analysis: Theory, Methods & Applications, vol. 66, no. 6, pp. 1351–1364, 2007. View at: Publisher Site | Google Scholar
J. V. Gonçalves and C. O. Alves, “Existence of positive solutions for
m
-Laplacian equations in
{ℝ}^{N}
involving critical Sobolev exponents,” Nonlinear Analysis: Theory, Methods & Applications, vol. 32, no. 1, pp. 53–70, 1998. View at: Publisher Site | Google Scholar
T.-F. Wu, “On semilinear elliptic equations involving concave-convex nonlinearities and sign-changing weight function,” Journal of Mathematical Analysis and Applications, vol. 318, no. 1, pp. 253–270, 2006. View at: Publisher Site | Google Scholar | Zentralblatt MATH
T.-F. Wu, “On semilinear elliptic equations involving critical Sobolev exponents and sign-changing weight function,” Communications on Pure and Applied Analysis, vol. 7, no. 2, pp. 383–405, 2008. View at: Publisher Site | Google Scholar | Zentralblatt MATH
L. Iturriaga, “Existence and multiplicity results for some quasilinear elliptic equation with weights,” Journal of Mathematical Analysis and Applications, vol. 339, no. 2, pp. 1084–1102, 2008. View at: Publisher Site | Google Scholar | Zentralblatt MATH
H. Q. Toan and Q.-A. Ngô, “Multiplicity of weak solutions for a class of nonuniformly elliptic equations of
p
-Laplacian type,” Nonlinear Analysis: Theory, Methods & Applications, vol. 70, no. 4, pp. 1536–1546, 2009. View at: Publisher Site | Google Scholar
L. Caffarelli, R. Kohn, and L. Nirenberg, “First order interpolation inequalities with weights,” Compositio Mathematica, vol. 53, no. 3, pp. 259–275, 1984. View at: Google Scholar | Zentralblatt MATH
B. Xuan, “The solvability of quasilinear Brezis-Nirenberg-type problems with singular weights,” Nonlinear Analysis: Theory, Methods & Applications, vol. 62, no. 4, pp. 703–725, 2005. View at: Publisher Site | Google Scholar | Zentralblatt MATH
P.-L. Lions, “The concentration-compactness principle in the calculus of variations. The locally compact case. I,” Annales de l'Institut Henri Poincaré Analyse Non Linéaire, vol. 1, no. 2, pp. 109–145, 1984. View at: Google Scholar | Zentralblatt MATH
P.-L. Lions, “The concentration-compactness principle in the calculus of variations. The locally compact case. II,” Annales de l'Institut Henri Poincaré Analyse Non Linéaire, vol. 1, no. 4, pp. 223–283, 1984. View at: Google Scholar
P.-L. Lions, “The concentration-compactness principle in the calculus of variations. The limit case. I,” Revista Matemática Iberoamericana, vol. 1, no. 1, pp. 145–201, 1985. View at: Publisher Site | Google Scholar | Zentralblatt MATH
P.-L. Lions, “The concentration-compactness principle in the calculus of variations. The limit case. II,” Revista Matemática Iberoamericana, vol. 1, no. 2, pp. 45–121, 1985. View at: Google Scholar
G. Bianchi, J. Chabrowski, and A. Szulkin, “On symmetric solutions of an elliptic equation with a nonlinearity involving critical Sobolev exponent,” Nonlinear Analysis: Theory, Methods & Applications, vol. 25, no. 1, pp. 41–59, 1995. View at: Publisher Site | Google Scholar | Zentralblatt MATH
A. K. Ben-Naoum, C. Troestler, and M. Willem, “Extrema problems with critical Sobolev exponents on unbounded domains,” Nonlinear Analysis: Theory, Methods & Applications, vol. 26, no. 4, pp. 823–833, 1996. View at: Publisher Site | Google Scholar | Zentralblatt MATH
J. I. Díaz, Nonlinear Partial Differential Equations and Free Boundaries, vol. 106, Pitman, Boston, Mass, USA, 1985, Elliptic Equations.
M. Struwe, Variational Methods, vol. 34, Springer, New York, NY, USA, 3rd edition, 2000.
I. Kuzin and S. Pohozaev, Entire Solutions of Semilinear Elliptic Equations, Birkhäuser, 1997.
Copyright © 2012 Zonghu Xiu. 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.
|
You didn't specify limits, so I took 1 and 0
A 95% CI is a numerical interval within which we are 95% confident that the true mean \mu
\overline{x}
The true mean \mu
For a sample size n=29
, the number of degrees of freedom is n=30
If we repeat an experiment 100 times (with 100 different samples) and construct a 95% CI each time, then approximately 5 of those 100 CIs would not
functions pegeonhole principle
How many multiples of 6 are in the set
\left\{171,172,173,..,286\right\}
How many strings are there of lowercase letters of length four or less, not counting the empty string?
5) Find the least integer n such that f (x) is O(n^x) for each of these functions.
c. f(x) = (x4 + x2 + 1)/(x3 + 1)
|
Calculus/Centre of mass - Wikibooks, open books for an open world
Calculus/Centre of mass
← Work Calculus Kinematics →
{\displaystyle {\vec {r}}_{G}={\frac {\displaystyle \sum \limits _{k=1}^{n}m_{k}{\vec {r_{k}}}}{\displaystyle \sum \limits _{k=1}^{n}m_{k}}}}
Retrieved from "https://en.wikibooks.org/w/index.php?title=Calculus/Centre_of_mass&oldid=3178439"
|
Title: Topical Diversification Over Time In The Royal Society Corpus
Authors: Peter Fankhauser, Jörg Knappen, Elke Teich
Category: Paper:Short Paper
Keywords: topic models, historical corpora, history of science, information theory
Fankhauser, P., Knappen, J., Teich, E. (2016). Topical Diversification Over Time In The Royal Society Corpus. In Digital Humanities 2016: Conference Abstracts. Jagiellonian University & Pedagogical University, Kraków, pp. 496-500.
Science gradually developed into an established sociocultural domain starting from the mid-17 th century onwards. In this process it became increasingly specialized and diversified. Here, we investigate a particular aspect of specialization on the basis of probabilistic topic models. As a corpus we use the Royal Society Corpus (Khamis et al. 2015), which covers the period from 1665 to 1869 and contains 9015 documents 1.
We follow the overall approach of applying topic models to diachronic corpora (Blei and Lafferty 2006, Hall et al. 2008, Griffiths and Steyvers 2004, McFarland et al. 2013, Newman and Block 2006, Yang et al. 2011) to map documents to topics. Probabilistic topic models (Steyvers and Griffiths 2007) have become a popular means to summarize and analyze the content of text corpora. The principle idea is to model the generation of documents with a randomized two-stage process: For every word
{w}_{i}
in a document d select a topic
{z}_{k}
from the document-topic distribution
P\left({z}_{k}|d\right)
and then select the word from the topic-word distribution
P\left({w}_{i}|{z}_{k}\right).
Consequently, the document-word distribution is factored as follows:
P\left({w}_{i}|d\right)={\sum }_{k}P\left({w}_{i}|{z}_{k}\right)P\left({z}_{k}|d\right).
This factorization effectively reduces the dimensionality of the model for documents, improving their interpretability: Whereas
P\left({w}_{i}|d\right)
requires one dimension for each distinct word (10s of thousands) per document,
P\left({z}_{k}|d\right)
only requires one dimension for each topic (typically in the range of 20-100). Topics are thus not given explicitly for each document, but constitute latent variables: A variety of approaches exist to estimate the document-topic and topic-word distributions from the observable document-word distributions. We use Gibbs-Sampling as implemented in Mallet (McCallum 2002).
For the preliminary analysis in this paper, we process documents as is, without segmenting them further into pages, only excluding stop words but not performing lemmatization or normalization in order to stay reasonably close to the original source. We experimented with the number of topics ranging between 20 and 30, reporting here results on 24 topics. Cursory analysis of multiple runs with different seeds (Steyvers and Griffiths 2007) shows that the resulting topics are rather stable.
Table 1 displays the top words for the topics with manually assigned labels and their overall percentage of occurrence. We can roughly distinguish four groups of topics; three non-thematic groups and one thematic. The first group comprises topics arising from documents in Latin and French, some of which are also translated into English. The second group Formulae and Tables relates to highly formalized modes of information presentation. The third group of topics is also clearly non-thematic but relates to general scientific processes: Observation and Experiment both contain rather general verbs and adjectives in addition to nouns. Events contains words describing remarkable events. Headmatter includes formulaic expressions typically occurring at the beginning and end of documents that are letters. All topics in this group are relatively frequent. Finally, the topics in the fourth group ( Geography through Chemistry), consisting mainly of nouns, indeed have a fairly clear thematic interpretation.
Label Words %
Latin quae quam sed ab sit vero hoc ac sunt esse qui etiam autem pro erit inter quo aut sive 6.4
French la le les des en du par dans qui il une qu pour ou ce sur ne au je 1.3
Formulae cos equation sin equal series point equations number line terms form values curve 4.7
Tables weight water oo oz parts gr grain io grains fat increase weights grs passed urine specific 1.7
Observation great made make parts found body time small part water nature long good put find 10.4
Experiment present general subject case results similar nature author state result cases fact 7.3
Events great time account stone ground house fire letter place miles found side stones 5.9
Headmatter years year author society age number time royal life great letter account part letters 5.4
Geography water sea tide high found river coast north land tides miles height surface great level 3.1
Meteorology day ditto rain wind cloudy weather fair clear april year days night march july june 3.2
Botany leaves plant plants tree tab bark folio foliis trees seeds seed flowers species fruit leaf 2.9
Reproduction cells animal blood fluid eggs membrane found egg part animals ova size young 3.0
Cells fibres structure form surface portion cells anterior part section side posterior 2.7
Paleontology part bone bones teeth surface upper side lower anterior length posterior tooth large 2.5
Physiology blood heart muscles part animal nerves vessels left parts stomach bladder body 5.5
Galaxy distance position stars star obs small hill double equatorial vf diff st magnitudes nebula 1.6
Terrestrial Magn. observations needle ship magnetic direct force made variation observed north diurnal 2.6
Solar System sun time observations moon made observed difference observation clock latitude 5.5
Thermodyn. air water heat temperature experiments tube experiment glass made time mercury 4.2
Mechanical Eng. made length weight end diameter iron instrument experiments brass part point line 4.5
Electromagn. force electricity current wire action body power direction fluid motion surface effect 3.8
Optics light rays glass eye red colours spectrum colour surface lines angle white blue object 3.7
Metallurgy water acid salt grains quantity iron found solution colour substance experiments gold 4.8
Chemistry acid water solution gas oxygen hydrogen carbonic cent action obtained salt potash 3.4
Table 1: Top words and percentages for topics
To investigate topical trends in the corpus we follow the approach in (Hall et al. 2008), by averaging the document-topic distributions for each year
y
P\left({z}_{k}|y\right)=1/n{\sum }_{{d}_{j}\in y}P\left({z}_{k}|{d}_{j}\right)
n
the number of documents in a year. Figure 1 shows a selection of five topics with the most pronounced change over time. Interestingly, some of the major changes occur for non-thematic topics: The topic Observation declines sharply from over 30% to less than 1 %. The topics Experiment and Formulae on the other hand increase starting around 1750. This indicates a substantial paradigm shift over time. Indeed, as Gleick (2010) vividly describes, the early stages of the Royal Society were largely devoted to observing and reporting about natural phenomena. The non-thematic topic Latin reaches its peak in the early 18 th century, and the thematic topics Cells and Chemistry show a clear increase with the beginning of the 19 th century.
Figure 1: Major topical trends for selected topics
To gain a better understanding about the correlation of topics, we cluster them hierarchically on the basis of the Jensen-Shannon divergence between the topic-document distributions:
P\left(d|z\right)=P\left(z|d\right)/{\sum }_{j}P\left(z|{d}_{j}\right)
Topics that typically co-occur in documents have similar topic-document distributions, and thus will be placed close in the tree.
Figure 2: Hierarchical clustering of topics by their topic-document distribution
The resulting tree in Figure 2 indeed identifies meaningful subgroups. Cutting the tree into six groups - Nature, Latin, Medicine, Astronomy, Engineering, Matter - allows us to investigate the overall topic distribution over time ( Figure 3 with Latin left out):
Figure 3: Distribution of topic groups over time
The topic group Nature comprising reports of all kinds of natural phenomena (Gleick 2010) clearly decreases over time, which is partially to be attributed to the strong decrease of the topic Observation in this group. The topic groups Medicine and Astronomy increase over time, whereas the topic groups Engineering and Matter also generally increase but with some intermediate peaks. Similar to the overall trends at the level of individual topics ( Figure 1), the biggest overall change occurs in the 2 nd half of the 18 th century.
Looking at the individual trends together, Figure 3 clearly indicates topical diversification: Until around 1770, the dominance of the topic group Nature leads to a highly skewed distribution of topic groups, whereas after 1770 topic groups are distributed much more evenly. The amount of skew can be characterized by the Shannon-Entropy:
H\left({P}_{y}\right)=-{\sum }_{k}P\left({z}_{k}|y\right){log}_{2}P\left({z}_{k}|y\right)
of the year-topic distributions
P\left({z}_{k}|y\right)
(Hall et al. 2010), with highly skewed distributions having low entropy. Indeed as can be seen in Figure 4 (left), the entropy ( ent) increases fairly consistently during the 18 th century and levels out during the 19 th century, reflecting a general increase of topical diversity over time.
It is interesting to compare this with the mean entropy of the individual document-topic distributions ( ment):
{H}_{mean}\left({P}_{y}\right)≔1/n\sum _{{d}_{j}\in y}H\left({P}_{{d}_{i}}\right)
n
the number of documents
{d}_{j}
y
. This measure decreases over time, i.e., while the overall topical diversity increases, the individual documents become more specific in terms of their topic distributions.
The difference between the entropy of year-topic distributions and mean entropy of individual document-topic distributions,
JS\left({P}_{y}\right)=H\left({P}_{y}\right)-{H}_{mean}\left({P}_{y}\right)
is the Jensen-Shannon divergence, which is usually applied to two distributions, generalized to the
n
topic distributions of all documents published in year y. The two opposing trends of these quantities lead to a constantly increasing Jensen-Shannon divergence, with a particularly sharp increase between 1750 and 1800. Figure 4 (right) depicts similar trends based on the 24 individual topic distributions. At this level, the year-topic entropy ( ent) shows less of a clear trend, but mean entropy ( ment) also clearly decreases, and consequently the Jensen-Shannon divergence clearly increases. Thus, at both levels of abstraction we can observe a clear diversification of the topics assigned to the individual documents. This strongly indicates a growing separation of individual scientific disciplines over time.
Figure 4: Entropy (ent), mean Entropy (ment), and Jensen-Shannon Divergence for topic groups (left) and individual topics (right)
As an alternative perspective on topical entropy Table 2 gives examples of authors with more than 20 papers. The first three authors have the lowest entropy. The dominating topics for Cayley and Owen clearly characterize their main theme of work. Conversely, Rev. John Swinton’s top topic Headmatter (62%) does not really reflect the overall theme of his publications (Orientalism), but rather their style as letters to members of the Royal Society – the dominant form of publication in this period. The second three authors have the highest entropy, their three top topics together amount for less than 50% of their overall topic distribution. However, they do characterize the main line of work of the authors in question fairly well.
author Papers Ent Ment Jsd Years Top Topics
Arthur Cayley 30 1.26 1.12 0.14 1850-1866 Formulae
Richard Owen 26 1.83 1.58 0.25 1843-1869 Paleontology
John Swinton 35 2.50 2.05 0.45 1753-1774 Headmatter
John Davy 58 4.05 3.33 0.72 1800-1856 Experiment, Chemistry, Physiology
William Watson 39 4.03 3.09 0.95 1739-1778 Events, Observation, Botany
Edmond Halley 65 3.93 2.75 1.18 1683-1731 Solar System, Observation, Latin
Table 2: Authors with minimum entropy (top) and maximum entropy (bottom)
In this paper we have analyzed the progression of topics in a corpus of the Royal Society of London. Our main result is the observation that the overall mixture of topics becomes more diverse over time, while the topics of individual documents become more specialized. These two opposing trends lead to a topical fragmentation of scientific discourse, which can be quantified by means of the generalized Jensen-Shannon divergence between the topic distributions of individual documents per time period. We are currently working on consolidating our analysis, experimenting with documents segmented into pages, focusing the analysis on different text types, and more carefully evaluating the resulting topic models (McFarland et al. 2013).
Of course, topic models only provide one, rather broad perspective on diversification of domain specific language. We plan to apply our approach also to other levels of linguistic analysis, such as terminology or grammar.
Blei, D. and Lafferty, J.D. (2006). Dynamic topic models. ICML.
Gleick, J. (2010). At the Beginning: More Things in Heaven and Earth: Bryson, B. (Ed.), Seeing Further. The Story of Science and The Royal Society. Harper Press, pp. 17-36.
Griffiths, T. L. and Steyvers M. (2004). Finding scientific topics. PNAS, 101 Suppl 1:5228–35.
Hall, D., Jurafsky, D., and Manning, C.D. (2008). Studying the history of ideas using topic models. Proceedings of the Conference on Empirical Methods in Natural Language Processing (EMNLP '08). Association for Computational Linguistics, Stroudsburg, PA, USA, pp. 363-71.
Khamis, A., et al. (2015). A resource for the diachronic study of scientific English: Introducing the Royal Society Corpus. Corpus Linguistics 2015. Lancaster.
McFarland, D. A., et al. (2013). Differentiating language usage through topic models. Poetics, 41(6): 607-25. http://dx.doi.org/10.1016/j.poetic.2013.06.004.
Newman, D. J. and Block, S. (2006). Probabilistic topic decomposition of an eighteenth-century American newspaper. J. Am. Soc. Inf. Sci. Technol., 57(6): 753-67. DOI=http://dx.doi.org/10.1002/asi.v57:6
Steyvers, M. and Griffiths, T. (2007). Probabilistic topic models. Landauer, T., et al.(Eds.), Handbook of Latent Semantic Analysis. Hillsdale, NJ: Erlbaum.
Yang, T., Torget, A. J., and Mihalcea, R. (2011). Topic modeling on historical newspapers. Proceedings of the 5th ACL-HLT Workshop on Language Technology for Cultural Heritage, Social Sciences, and Humanities (LaTeCH '11). Association for Computational Linguistics, Stroudsburg, PA, USA, pp. 96-104.
Of these 205 years, 159 years actually contain documents (mean = 56.7, median=36, sd=61.6, min=12, max=444)
|
2.1 Representation and encoding in memory
2.2 Basic and interchange formats
2.3 Extended and extendable precision formats
2.4 Interchange formats
3.1 Roundings to nearest
3.2 Directed roundings
4 Required operations
4.1 Comparison predicates
4.2 Total-ordering predicate
6.1 Alternate exception handling
6.2 Recommended operations
7 Character representation
7.1 Hexadecimal literals
Standard development[edit]
The international standard ISO/IEC/IEEE 60559:2011 (with content identical to IEEE 754-2008) has been approved for adoption through ISO/IEC JTC1/SC 25 under the ISO/IEEE PSDO Agreement[2][3] and published.[4]
The international standard ISO/IEC 60559:2020 (with content identical to IEEE 754-2019) has been approved for adoption through ISO/IEC JTC1/SC 25 and published.[7]
Representation and encoding in memory[edit]
Basic and interchange formats[edit]
Extended and extendable precision formats[edit]
Interchange formats[edit]
Rounding rules[edit]
Roundings to nearest[edit]
Directed roundings[edit]
Required operations[edit]
Comparison predicates[edit]
Total-ordering predicate[edit]
Alternate exception handling[edit]
Recommended operations[edit]
{\displaystyle e^{x}}
{\displaystyle 2^{x}}
{\displaystyle 10^{x}}
{\displaystyle e^{x}-1}
{\displaystyle 2^{x}-1}
{\displaystyle 10^{x}-1}
{\displaystyle \ln x}
{\displaystyle \log _{2}x}
{\displaystyle \log _{10}x}
{\displaystyle \ln(1+x)}
{\displaystyle \log _{2}(1+x)}
{\displaystyle \log _{10}(1+x)}
{\displaystyle {\sqrt {x^{2}+y^{2}}}}
{\displaystyle {\sqrt {x}}}
{\displaystyle (1+x)^{n}}
{\displaystyle x^{\frac {1}{n}}}
{\displaystyle x^{n}}
{\displaystyle x^{y}}
{\displaystyle \sin x}
{\displaystyle \cos x}
{\displaystyle \tan x}
{\displaystyle \arcsin x}
{\displaystyle \arccos x}
{\displaystyle \arctan x}
{\displaystyle \operatorname {atan2} (y,x)}
{\displaystyle \operatorname {sinPi} x=\sin \pi x}
{\displaystyle \operatorname {cosPi} x=\cos \pi x}
{\displaystyle \operatorname {tanPi} x=\tan \pi x}
{\displaystyle \operatorname {asinPi} x={\frac {\arcsin x}{\pi }}}
{\displaystyle \operatorname {acosPi} x={\frac {\arccos x}{\pi }}}
{\displaystyle \operatorname {atanPi} x={\frac {\arctan x}{\pi }}}
{\displaystyle \operatorname {atan2Pi} (y,x)={\frac {\operatorname {atan2} (y,x)}{\pi }}}
{\displaystyle \sinh x}
{\displaystyle \cosh x}
{\displaystyle \tanh x}
{\displaystyle \operatorname {arsinh} x}
{\displaystyle \operatorname {arcosh} x}
{\displaystyle \operatorname {artanh} x}
Expression evaluation[edit]
Character representation[edit]
See also: Floating-point arithmetic § Binary-to-decimal conversion with minimal number of digits
{\displaystyle 1+\lceil p\log _{10}(2)\rceil ,}
Hexadecimal literals[edit]
Retrieved from "https://en.wikipedia.org/w/index.php?title=IEEE_754&oldid=1089181894"
|
Quantitative Aptitude Exam for CPT | Online Exam
Home Quantitative Aptitude Basic concepts of Permutations and combinations Basic Concept of Permutations and Combinations-Test 1
Chapter :- Basic Concept of Permutations and Combinations -Test 1
The number of triangles that can be formed by choosing the vertices from a set of 12 points ,seven of which lie on the same straight line is
The number of triangles that can be formed from a set of 12 points \phantom{\rule{0ex}{0ex}}=12{C}_{3}\phantom{\rule{0ex}{0ex}}\mathrm{sin}ce 7 points are on the same line,therefore no triangle can be formed from these points\phantom{\rule{0ex}{0ex}}i.e.number of triangles=12{C}_{3}–7{C}_{3}=220–35=185
The number of triangles that can be formed from a set of 12 points \phantom{\rule{0ex}{0ex}}=12{C}_{3}\phantom{\rule{0ex}{0ex}}\mathrm{sin}ce 7 points are on the same line,therefore no triangle can be formed from these points\phantom{\rule{0ex}{0ex}}i.e.number of triangles=12{C}_{3}–7{C}_{3}=220–35=185
A code word is to consist of two English alphabets followed by 2 distinct numbers between 1 and 9.How many such code words are there?
The number of ways of filling first two places with english alphabets=26×25=650
The number of ways of filling last two places with distinct numbers=9×8=72
The number of code words that can be formed are=650×72=46800
A boy has 3 library tickets and 8 books of his interest in the library of these 8,he does not want to borrow mathematics part-II unless mathematics part-I is also borrowed?In how many ways can he choose the three books to be borrowed?
CASE 1:- When mathematics part-II is borrowed(i.e it means part-I has also been borrowed )
number of ways=6{C}_{1}=6 ways
CASE 2:-when mathematics part-II is not borrowed(i.e.3 3 books are to be selected out of 7)
number of ways=7{C}_{3}=35 ways
hence,total number of ways =35+6=41 ways
number of ways=6{C}_{1}=6 ways
number of ways=7{C}_{3}=35 ways
An exam paper consist of 12 questions divided into two parts A and B.Part A contains 7 questions and Part B contains 5 questions.A candidate is required to attempt 8 questions selecting at least 3 from each part.In how many ways can the candidate select the questions?
The candidate can select 8Q by selecting at least 3 from each part in the following ways :
\left(a\right)3Q from part A and 5Q from part B=7{C}_{3}×5{C}_{5}=35 ways\phantom{\rule{0ex}{0ex}}\left(b\right)4Q from part A and part B each=7{C}_{3}×5{C}_{4}=175 ways\phantom{\rule{0ex}{0ex}}\left(c\right)5Q from part A and 3Q from part B=7{C}_{5}×5{C}_{3}=210 ways \phantom{\rule{0ex}{0ex}}
hence,the total number of ways in which the candidate can select the question will be=35+175+210=420 ways
\left(a\right)3Q from part A and 5Q from part B=7{C}_{3}×5{C}_{5}=35 ways\phantom{\rule{0ex}{0ex}}\left(b\right)4Q from part A and part B each=7{C}_{3}×5{C}_{4}=175 ways\phantom{\rule{0ex}{0ex}}\left(c\right)5Q from part A and 3Q from part B=7{C}_{5}×5{C}_{3}=210 ways \phantom{\rule{0ex}{0ex}}
A supreme court bench consist of 5 judges. in how many ways , the bench can give a majority division?
majority decision can be taken by a bench of 5 judges in \phantom{\rule{0ex}{0ex}}=5{C}_{3}+5{C}_{4}+5{C}_{5}=10+5+1=16 ways
majority decision can be taken by a bench of 5 judges in \phantom{\rule{0ex}{0ex}}=5{C}_{3}+5{C}_{4}+5{C}_{5}=10+5+1=16 ways
Given:p(7,k)=60 ; p(7,k-3).then:
p\left(7,k\right)=60p\left(7,k–3\right)\phantom{\rule{0ex}{0ex}}7{p}_{k}=60x7{p}_{k–3}\phantom{\rule{0ex}{0ex}}\frac{7!}{\left(7–k\right)!}=60x\frac{7!}{\left[7–\left(k–3\right)\right]!}\phantom{\rule{0ex}{0ex}}\frac{1}{\left(7–k\right)!}=60x\frac{1}{\left[7–\left(k–3\right)!}\phantom{\rule{0ex}{0ex}}\frac{\left(10–k\right)!}{\left(7–k\right)!}=60\phantom{\rule{0ex}{0ex}}\frac{\left(10–k\right)\left(9–k\right)\left(8–k\right)\left(7–k\right)!}{\left(7–k\right)!}=60\phantom{\rule{0ex}{0ex}}{k}^{3}–27{k}^{2}+24k–660=0\phantom{\rule{0ex}{0ex}}\left(k–5\right)\left({k}^{2}–22k+132\right)=0\phantom{\rule{0ex}{0ex}}k=5,\mathrm{sin}ce {k}^{2}–22k+132=0 gives imaginary roots
p\left(7,k\right)=60p\left(7,k–3\right)\phantom{\rule{0ex}{0ex}}7{p}_{k}=60x7{p}_{k–3}\phantom{\rule{0ex}{0ex}}\frac{7!}{\left(7–k\right)!}=60x\frac{7!}{\left[7–\left(k–3\right)\right]!}\phantom{\rule{0ex}{0ex}}\frac{1}{\left(7–k\right)!}=60x\frac{1}{\left[7–\left(k–3\right)!}\phantom{\rule{0ex}{0ex}}\frac{\left(10–k\right)!}{\left(7–k\right)!}=60\phantom{\rule{0ex}{0ex}}\frac{\left(10–k\right)\left(9–k\right)\left(8–k\right)\left(7–k\right)!}{\left(7–k\right)!}=60\phantom{\rule{0ex}{0ex}}{k}^{3}–27{k}^{2}+24k–660=0\phantom{\rule{0ex}{0ex}}\left(k–5\right)\left({k}^{2}–22k+132\right)=0\phantom{\rule{0ex}{0ex}}k=5,\mathrm{sin}ce {k}^{2}–22k+132=0 gives imaginary roots
The number of ways in which n books can be arranged on a shelf so that two particular books are not together is
(n-2)x(n-1)!
(n-2)x(n+1)!
In how many ways can the letters of the word FAILURE be arranged so that the consonants may occupy only odd positions?
The word FAILURE have 7 letters out of which 3 letters F,L and R are consonants and 4 letters A,I,U and E are vowels.
The four positions to be filled up with consonants are indicated below:
F L R not filled
Since,there are only 3 consonants ,the total number of permutations are
=4{P}_{3}=24
Again,for the arrangement described above ,the four vowels can occupy the 4 remaining positions not occupied by consonants =4!=24 ways
hence the total number of arrangements are =24×24=576
=4{P}_{3}=24
5 bulbs of which 3 are defective are to be tried in two lights points in a dark room.In how many trials the room shall be lighted?
Total number of trials =5{C}_{2}=10 ways \phantom{\rule{0ex}{0ex}}no.of trials for no light in the room =3{C}_{2}=3\phantom{\rule{0ex}{0ex}}The room shall be lighted in=10–3=7 ways
Total number of trials =5{C}_{2}=10 ways \phantom{\rule{0ex}{0ex}}no.of trials for no light in the room =3{C}_{2}=3\phantom{\rule{0ex}{0ex}}The room shall be lighted in=10–3=7 ways
In how many ways can a party of 4 men and 4 women be seated at a circular table ,so that no 2 women are adjacent?
The no.of ways in which 4 men can be seated at the circular table so that there is a vacant seat between every pair of men is=(4-1)!=3!=6 ways
hence,no.of ways in which 4 vacant seats can be occupied by 4 women =4!=24 ways
Required no of ways =6×24=144 ways
The value of \sum _{r=1}^{5}=5{C}_{r} is
\sum _{r=1}^{5}5{C}_{r}=5{C}_{1}+5{C}_{2}+5{C}_{3}+5{C}_{4}+5{C}_{5}\phantom{\rule{0ex}{0ex}} =5+10+10+5+1=31
\sum _{r=1}^{5}5{C}_{r}=5{C}_{1}+5{C}_{2}+5{C}_{3}+5{C}_{4}+5{C}_{5}\phantom{\rule{0ex}{0ex}} =5+10+10+5+1=31
If 6{P}_{r}=24x6{C}_{r} ,then find r
6{P}_{r}=24x6{C}_{r}\phantom{\rule{0ex}{0ex}}\frac{6!}{\left(6–r\right)!}=24x\frac{6!}{r!x\left(6–r\right)!}\phantom{\rule{0ex}{0ex}}4!=\frac{24}{r!}\phantom{\rule{0ex}{0ex}}r!=\frac{24}{4!}\phantom{\rule{0ex}{0ex}}r!=4!\phantom{\rule{0ex}{0ex}}r=4
6{P}_{r}=24x6{C}_{r}\phantom{\rule{0ex}{0ex}}\frac{6!}{\left(6–r\right)!}=24x\frac{6!}{r!x\left(6–r\right)!}\phantom{\rule{0ex}{0ex}}4!=\frac{24}{r!}\phantom{\rule{0ex}{0ex}}r!=\frac{24}{4!}\phantom{\rule{0ex}{0ex}}r!=4!\phantom{\rule{0ex}{0ex}}r=4
Find the number of combinations of the letters of the word COLLEGE taken four together
How many words can be formed with the letters of the word ORIENTAL so that A and E always occupy odd places
There are 4 odd places and 2 letters,hence this can be done in =4{P}_{2}=12 ways
The remaining 6 letters i.e.O,R,I,N.T,L can be arranged in 6! ways
hence,the total no.of arrangements=6!x12=8640
There are 4 odd places and 2 letters,hence this can be done in =4{P}_{2}=12 ways
If 1000{C}_{98}=999{C}_{97}+X{C}_{901},find X
1000{C}_{98}=999{C}_{97}+X{C}_{901}\phantom{\rule{0ex}{0ex}}\mathrm{sin}ce ,n+1{C}_{r}=n{C}_{r}+n{C}_{r–1}\phantom{\rule{0ex}{0ex}}999{C}_{98}=999{C}_{97}+X{C}_{901}\phantom{\rule{0ex}{0ex}}999{C}_{98}+999{C}_{97}=999{C}_{97}+X{C}_{901}\phantom{\rule{0ex}{0ex}}999{C}_{98}+999{C}_{97}=999{C}_{97}+X{C}_{98}\phantom{\rule{0ex}{0ex}}X=999
1000{C}_{98}=999{C}_{97}+X{C}_{901}\phantom{\rule{0ex}{0ex}}\mathrm{sin}ce ,n+1{C}_{r}=n{C}_{r}+n{C}_{r–1}\phantom{\rule{0ex}{0ex}}999{C}_{98}=999{C}_{97}+X{C}_{901}\phantom{\rule{0ex}{0ex}}999{C}_{98}+999{C}_{97}=999{C}_{97}+X{C}_{901}\phantom{\rule{0ex}{0ex}}999{C}_{98}+999{C}_{97}=999{C}_{97}+X{C}_{98}\phantom{\rule{0ex}{0ex}}X=999
How many numbers greater than a million can be formed with the digits 4,5,5,0,4,5,3?
total no.of numbers that can be formed from the digits 4,5,5,0,4,5,3=7!/2!x3!=420 ways
out of these 420 numbers some will begin with 0 and are less than one million ,so they are to be rejected
number of numbers beginning with
0=6!/2!x3!=60 ways
hence required number of numbers=420-60=360 ways
A building contractor needs 3 helpers and 10 men apply.In how many ways can these selections take place?
since there is no regard for order,the contractor can select any of the 3 helpers out of 10 men.
10{C}_{3 } ways =120 ways
10{C}_{3 } ways =120 ways
There are 3 blue balls,4 red balls and 5 green balls.in how many ways can they be arranged in a row ?
No.of way\mathrm{sin} which these balls can be arranged \phantom{\rule{0ex}{0ex}}=\frac{12!}{3!x4!x5!}\phantom{\rule{0ex}{0ex}}=27720 ways
No.of way\mathrm{sin} which these balls can be arranged \phantom{\rule{0ex}{0ex}}=\frac{12!}{3!x4!x5!}\phantom{\rule{0ex}{0ex}}=27720 ways
If C\left(n,r\right):C\left(n,r+1\right)=1:2 and C\left(n,r+1\right):C\left(n,r+2\right)=2:3,determine the value of n and r
6 seats if articled clerks are vacant in a Chartered Accountant Firm.how many different batches of candidates can be chosen out of 10 candidates?
The no.of ways in which 6 articles clerks can be selected out of 10 candidates
10{C}_{6}=210 ways
10{C}_{6}=210 ways
6 persons A,B,C,D,E and F are to be seated at a circular table.in how many ways can this be done ,if A must always have either B or C on his right and B must always have either C or D on his right ?
Using the given restrictions ,we must have AB or AC and BC or BD.
therefore,we have the following alternatives:
(i)ABC,D,E,F which gives (4-1)! or 31 ways
(ii)ABD,C,e,f which gives (4-1)! or 31 ways
(iii)AC,BD,E,F which gives (4-1)! or 31 ways
hence,the total number of ways are =3!+3!+3!=6+6+6=18 ways
If n{p}_{r}=n{p}_{r+1} and n{c}_{r}=n{c}_{r–1} then find the value of n
n{p}_{r}=n{p}_{r+1}⇒n=r+1...........\left(1\right)\phantom{\rule{0ex}{0ex}}n{c}_{r}=n{c}_{r–1}⇒n=2r–1..........\left(2\right)\phantom{\rule{0ex}{0ex}}from 1&2 we get\phantom{\rule{0ex}{0ex}}r+1=2r–1\phantom{\rule{0ex}{0ex}}r=2\phantom{\rule{0ex}{0ex}}hence,n=2+1=3
n{p}_{r}=n{p}_{r+1}⇒n=r+1...........\left(1\right)\phantom{\rule{0ex}{0ex}}n{c}_{r}=n{c}_{r–1}⇒n=2r–1..........\left(2\right)\phantom{\rule{0ex}{0ex}}from 1&2 we get\phantom{\rule{0ex}{0ex}}r+1=2r–1\phantom{\rule{0ex}{0ex}}r=2\phantom{\rule{0ex}{0ex}}hence,n=2+1=3
In how many ways a committee of 6 members cab be formed from a group of 7 boys and 4 girls having at least 2 girls in the committee.
Number of ways of painting a face of a cube by 6 colours is_____
number of ways of painting a face of a cube by 6 colours is 6,since any of the 6 colours can be used to paint the face of the cube.
if_____18{C}_{r}=18{C}_{r+2} find the value of r{c}_{5}
Previous articleonline exam
Next articleBasic Concept of Permutations and Combinations-Test 2
|
{\displaystyle f\!\,}
is a continuous real-valued function with domain
{\displaystyle (-\infty ,\infty )\!\,}
{\displaystyle f\!\,}
is absolutely continuous on every finite interval
{\displaystyle [a,b]\!\,}
Prove: I{\displaystyle f\!\,}
{\displaystyle f^{\prime }\!\,}
are both integrable on
{\displaystyle (-\infty ,\infty )\!\,}
{\displaystyle \int _{-\infty }^{\infty }f^{\prime }=0\!\,}
{\displaystyle f\!\,}
is absolutely continuous for all
{\displaystyle [a,b]\subset R^{1}\!\,}
{\displaystyle \int _{a}^{b}f^{\prime }(x)dx=f(b)-f(a)\!\,}
{\displaystyle \int _{-\infty }^{\infty }f^{\prime }(x)dx=\lim _{a,b\rightarrow \infty }\int _{a}^{b}f^{\prime }(x)=\lim _{a,b\rightarrow \infty }[f(b)-f(a)]\!\,}
{\displaystyle f^{\prime }\!\,}
is integrable i.e.
{\displaystyle f^{\prime }\in L^{1}(-\infty ,\infty )\!\,}
{\displaystyle \lim _{b\rightarrow \infty }f(b)\!\,}
{\displaystyle \lim _{a\rightarrow \infty }f(a)\!\,}
{\displaystyle \lim _{b\rightarrow \infty }|f(b)|=\delta >0\!\,}
{\displaystyle M\!\,}
{\displaystyle x>M\!\,}
{\displaystyle |f(x)|>{\frac {\delta }{2}}\!\,}
{\displaystyle f\!\,}
is continuous. (At some point,
{\displaystyle |f|\!\,}
will either monotonically increase or decrease to
{\displaystyle \delta \!\,}
.) This implies
{\displaystyle {\begin{aligned}\int _{-\infty }^{\infty }|f(x)|dx&\geq \int _{M}^{\infty }|f(x)|dx\\&\geq \int _{M}^{\infty }{\frac {\delta }{2}}dx\\&=\infty \end{aligned}}}
which contradicts the hypothesis that
{\displaystyle f\!\,}
{\displaystyle f\in L^{1}(-\infty ,\infty )\!\,}
{\displaystyle \lim _{b\rightarrow \infty }f(b)=0\!\,}
Using the same reasoning as above,
{\displaystyle \lim _{a\rightarrow -\infty }f(a)=0\!\,}
{\displaystyle \int _{-\infty }^{\infty }f^{\prime }(x)dx=\lim _{b\rightarrow \infty }f(b)-\lim _{a\rightarrow -\infty }f(a)=0\!\,}
Alternate SolutionEdit
{\displaystyle \int _{-\infty }^{\infty }f'=c\neq 0}
(without loss of generality,
{\displaystyle c>0}
). Then for small positive
{\displaystyle \epsilon }
, there exists some real
{\displaystyle M}
{\displaystyle m>M}
{\displaystyle c-\epsilon <\int _{-m}^{m}f'<c+\epsilon }
. By the fundamental theorem of calculus, this gives
{\displaystyle f(-m)+c-\epsilon <f(m)<f(-m)+c+\epsilon }
{\displaystyle m>M}
{\displaystyle f}
is integrable, this means that for any small positive
{\displaystyle \delta }
{\displaystyle N}
{\displaystyle n>N}
{\displaystyle \int _{-\infty }^{-n}+\int _{n}^{\infty }f<\delta }
. But by the above estimate,
{\displaystyle \int _{-\infty }^{-n}f+\int _{n}^{\infty }f>\int _{-\infty }^{-n}f+\int _{-\infty }^{-n}f+(c-\epsilon )=\int _{-\infty }^{-n}2f+\int _{-\infty }^{-n}(c-\epsilon )=\infty }
This contradicts the integrability o{\displaystyle f}
. Therefore, we must have
{\displaystyle c=0}
{\displaystyle \{f_{n}\}\!\,}
is a sequence of real valued measurable functions defined on the interval
{\displaystyle [0,1]\!\,}
{\displaystyle f_{n}(x)\rightarrow f(x)\!\,}
{\displaystyle x\in [0,1]\!\,}
{\displaystyle p>1\!\,}
{\displaystyle M>0\!\,}
{\displaystyle \|f_{n}\|_{p}\leq M\!\,}
{\displaystyle n\!\,}
{\displaystyle \|f\|_{p}\leq M\!\,}
(b)Prove that
{\displaystyle \|f-f_{n}\|_{1}\rightarrow 0\!\,}
{\displaystyle n\rightarrow \infty \!\,}
By definition of norm,
{\displaystyle \|f\|_{p}=\left(\int |f(x)|^{p}\,\mathrm {d} x\right)^{\frac {1}{p}}\!\,}
{\displaystyle \|f_{n}\|_{p}\leq M\!\,}
{\displaystyle \|f_{n}\|_{p}^{p}\leq M^{p}\!\,}
By Fatou's Lemma,
{\displaystyle {\begin{aligned}\|f\|_{p}^{p}&=\int _{0}^{1}|f(x)|^{p}dx\\&=\int _{0}^{1}{\underset {n}{\lim \inf }}|f_{n}(x)|^{p}dx\\&\leq {\underset {n}{\lim \inf }}\int _{0}^{1}|f_{n}(x)|^{p}dx\\&\leq M^{p}\end{aligned}}}
which implies, by taking the
{\displaystyle p\!\,}
{\displaystyle \|f\|_{p}\leq M\!\,}
By Holder's Inequality, for all
{\displaystyle A\subset [0,1]\!\,}
that are measurable,
{\displaystyle {\begin{aligned}\int _{A}|(f(x)-f_{n}(x))\cdot 1|dx&\leq \left(\int _{A}|f(x)-f_{n}(x)|^{p}dx\right)^{\frac {1}{p}}\cdot \left(\int _{A}1^{q}dx\right)^{\frac {1}{q}}\\&\leq \left(\int _{A}|2f(x)|^{p}dx\right)^{\frac {1}{p}}\cdot \left(\int _{A}1^{q}dx\right)^{\frac {1}{q}}\\&\leq 2\left(\int _{A}|f(x)|^{p}dx\right)^{\frac {1}{p}}\cdot \left(\int _{A}1^{q}dx\right)^{\frac {1}{q}}\\&\leq 2M\cdot (m(A))^{\frac {1}{q}}\end{aligned}}}
{\displaystyle {\frac {1}{p}}+{\frac {1}{q}}=1\!\,}
{\displaystyle |f(x)-f_{n}(x)|\leq 2M(m(A))^{\frac {1}{q}}\!\,}
The Vitali Convergence Theorem then implies
{\displaystyle \lim _{n\rightarrow \infty }\int _{0}^{1}|f(x)-f_{n}(x)|dx=\int _{0}^{1}\lim _{n\rightarrow \infty }|f(x)-f_{n}(x)|dx=0\!\,}
{\displaystyle f(x),xf(x)\in L^{2}(R)\!\,}
{\displaystyle f(x)\in L^{1}(R)\!\,}
{\displaystyle \|f\|_{1}\leq {\sqrt {2}}(\|f\|_{2}+\|xf\|_{2})\!\,}
{\displaystyle {\begin{aligned}\int _{R}|f(x)|dx&=\int _{|x|>1}|xf(x|\cdot {\frac {1}{|x|}}dx+\int _{|x|<1}|f(x)|\cdot 1dx\\&\leq \left(\int _{|x|>1}|xf(x)|^{2}dx\right)^{\frac {1}{2}}\cdot \left(\int _{|x|>1}{\frac {1}{|x|^{2}}}dx\right)^{\frac {1}{2}}+\left(\int _{|x|<1}|f(x)|^{2}dx\right)^{\frac {1}{2}}\cdot \left(\int _{|x|<1}1^{2}dx\right)^{\frac {1}{2}}\end{aligned}}}
|
Determine the force in each member of the loaded truss
{y}_{c}
4{x}^{2}y{ }^{″}-4x{y}^{\prime }+3y=8{x}^{\frac{3}{4}}
f \left( x \right) = 7 {x}^{2 }+ 2
g \left( x \right) = 5 {x}^{3 }+ 7 {x}^{2 }+ 6 x
\frac{f \left( x \right) }{g \left( x \right) }
{X}_{1},{X}_{2},\dots ,{X}_{n}
p
p
\stackrel{^}{P}
p
Var\left[\stackrel{^}{P}\right]=\frac{1}{nE\left[\left(\left(\partial /\partial p\right) \mathrm{ln} {f}_{x}\left(X\right){\right)}^{2}\right]}
\stackrel{^}{P}
\stackrel{―}{X}
{f}_{x}\left(\overline{X}\right)=\left(\genfrac{}{}{0}{}{n}{\sum _{i=1}^{n}{X}_{i}}\right)*{p}^{\sum _{i=1}^{n}{X}_{i}}*\left(1-p{\right)}^{n-\sum _{i=1}^{n}{X}_{i}}
\mathrm{ln} {f}_{X}\left(X\right)=\mathrm{ln}\left(\left(\genfrac{}{}{0}{}{n}{\sum _{i=1}^{n}{X}_{i}}\right)\right)+\sum _{i=1}^{n}{X}_{i}\mathrm{ln}\left(p\right)+ \left(n-\sum _{i=1}^{n}{X}_{i}\right)\mathrm{ln}\left(1-p\right)
\frac{\partial \mathrm{ln} {f}_{X}\left(X\right)}{\partial p}=\frac{{\sum }_{i=1}^{n}{X}_{i}}{p}-\frac{\left(n-{\sum }_{i=1}^{n}{X}_{i}\right)}{\left(1-p\right)}=\frac{n\overline{X}}{p}-\frac{\left(n-n\overline{X}\right)}{\left(1-p\right)}
\left(\frac{\partial ln {f}_{X}\left(X\right)}{\partial p}{\right)}^{2}=\left(\frac{n\overline{X}}{p}-\frac{\left(n-n\overline{X}\right)}{\left(1-p\right)}{\right)}^{2}=\frac{{n}^{2}{p}^{2}-2{n}^{2}p\overline{X}+{n}^{2}{\overline{X}}^{2}}{{p}^{2}\left(1-p{\right)}^{2}}
E\left[{\stackrel{―}{X}}^{2}\right]={\mu }^{2}+\frac{{\sigma }^{2}}{n}
E\left[X\right]=\mu =p=E\left[\stackrel{―}{X}\right]
Var\left[X\right]={\sigma }^{2}=p\left(1-p\right)
E\left[\left(\frac{\partial \mathrm{ln} {f}_{X}\left(X\right)}{\partial p}{\right)}^{2}\right]=\frac{{n}^{2}{p}^{2}-2{n}^{2}pE\left[\overline{X}\right]+{n}^{2}E\left[{\overline{X}}^{2}\right]}{{p}^{2}\left(1-p{\right)}^{2}}=\frac{{n}^{2}{p}^{2}-2{n}^{2}{p}^{2}+{n}^{2}\left({p}^{2}+\frac{p\left(1-p\right)}{n}\right)}{{p}^{2}\left(1-p{\right)}^{2}}=\frac{np\left(1-p\right)}{{p}^{2}\left(1-p{\right)}^{2}}=\frac{n}{p\left(1-p\right)}
Var\left[\stackrel{^}{P}\right]=\frac{1}{nE\left[\left(\left(\partial /\partial p\right) \mathrm{ln} {f}_{x}\left(X\right){\right)}^{2}\right]}=\frac{1}{n\frac{n}{p\left(1-p\right)}}=\frac{p\left(1-p\right)}{{n}^{2}}
\frac{p\left(1-p\right)}{n}
\begin{array}{|cccc|}\hline \text{ Category }& \text{ Observed Frequency }& \text{ Expected Frequency }& \text{ (obs-exp)^2/exp }\\ \text{ A }& 15\\ \text{ B }& 19\\ \text{ C }& 23\\ \text{ D }& 18\\ \text{ E }& 5\\ \hline\end{array}
{\chi }^{2}=
|
Building With Logic Gates Practice Problems Online | Brilliant
Remember, XOR stands for "exclusive or," and for each gate if the input consists of a single
0
and a single
1,
then the output is
\bf1,
otherwise the output is
\bf0
You want this set of logic gates with inputs
A,
B,
C,
Z = 1
only if there's an odd number of inputs that equal
1.
What kind of gate should go at the question mark?
A palindrome reads the same forwards and backwards. (For example, 10101 and 1001 are palindromes.) Suppose you want a machine that takes a number with the four binary digits A, B, C, and D and outputs a 1 only if the number is a palindrome.
You can build this with two XOR gates, one NOT gate, and which other extra kind of gate?
Suppose you want a machine that takes a binary number with four digits
A,
B,
C,
D
1
if the number is even. How many logic gates does this require?
0
A binary number is a number written in base
2.
These numbers are similar to "normal" numbers, but they only use the digits
0
1
and are based on powers of
2
rather than powers of
10.
For example, a "normal" base-
10
arithmetic example would be something like
1234_{10} = 1 \cdot 10^3 + 2 \cdot 10^2 + 3 \cdot 10^1 + 4 \cdot 10^0.
Here's a different number, but in binary:
1110_2 = 1 \cdot 2^3 + 1 \cdot 2^2 + 1 \cdot 2^1 + 0 \cdot 2^0 = 14_{10}.
0
1
2
3
You are designing a calculator display with digits like the ones shown above.
Consider the logic gates leading to the portion of the display marked:
The inputs are binary four digit numbers represented by the digits
A ,
B ,
C ,
D .
Which gates go in the red boxes?
AND, OR XOR, AND XOR, OR
Logic gates are used by computers to perform all sorts of operations, but designing the circuits can be tricky. We'll tackle addition of numbers (which requires many gates) in a later quiz, but for now let's consider a simpler problem:
We would like a machine that takes three inputs A, B, and C and returns 1 if and only if all the inputs are 1.
This can be done with two logic gates that are identical. Which type of logic gate works?
|
Troubleshoot Frequency-Domain Identification of Transfer Function Models - MATLAB & Simulink - MathWorks Switzerland
Estimate the Model Without Preprocessing and Filtering
Specify Weighting Filter
Estimate Model Using Preprocessed and Filtered Data
This example shows how to perform and troubleshoot the identification of a SISO system using frequency-response data (FRD). The techniques explained here can also be applied to MIMO models and frequency-domain input-output data.
When you use the tfest command to estimate a SISO transfer function model from the frequency-response data, the estimation algorithm minimizes the following least-squares loss (cost) function:
\underset{G\left(\omega \right)}{minimize}\sum _{k=1}^{{N}_{f}}{|W\left({\omega }_{k}\right)\left(G\left({\omega }_{k}\right)-f\left({\omega }_{k}\right)\right)|}^{2}
Here W is a frequency-dependent weight that you specify, G is the transfer function that is to be estimated, f is the measured frequency-response data, and
w
is the frequency. Nf is the number frequencies at which the data is available.
G\left(w\right)-f\left(w\right)
is the frequency-response error.
In this example, you first estimate the model without preprocessing the data or using estimation options to specify a weighting filter. You then apply these troubleshooting techniques to improve the model estimation.
Load the measured continuous-time frequency response data.
load troubleshooting_example_data Gfrd;
Gfrd is an idfrd object that stores the data.
Estimate an initial transfer function model with 11 poles and 10 zeros by using the estimation data.
Gfit = tfest(Gfrd,11,10);
Plot the frequency-response magnitude of the estimated model and the measured frequency-response data.
bodemag(Gfrd,Gfit);
The estimated model contains spurious dynamics. The estimation algorithm misses the valley at 60 rad/s and the resonant peaks after that. Therefore, the model is not a good fit to the data.
The algorithm minimizes the squared error magnitude,
|W\left(w\right)\phantom{\rule{0.2777777777777778em}{0ex}}\left(G\left(w\right)-f\left(w\right)\right){|}^{2}
, in the loss function. Plot this quantity as a function of frequency. This error plot provides a view of which data points contribute the most to the loss function, and so are the likely limiting factors during estimation. The plot can help you identify why there are spurious or uncaptured dynamics.
Because you have not specified a frequency-dependent weight,
W\left(w\right)
w = Gfrd.Frequency;
r1 = squeeze(freqresp(Gfit,w));
r2 = squeeze(freqresp(Gfrd,w));
fitError = r1-r2;
semilogx(w,abs(fitError).^2)
title('Weighted Estimation Error');
xlabel('Frequency (rad/s)');
ylabel('Magnitude (abs)')
From the data, model, and error plots you can see that:
The largest fitting errors are below 10 rad/s.
The algorithm focusses on fitting the noisy high magnitude data points below 10 rad/s, which have a large contribution to the optimization loss function. As a result, the algorithm assigns spurious poles and zeros to this data region. To address this issue, you can preprocess the data to improve signal-to-noise ratio in this region. You can also use frequency-dependent weights to make the algorithm put less focus on this region.
Below approximately 40 rad/s, most variations in data are due to noise. There are no significant system modes (valleys or peaks) in the data. To address this issue, you can use a moving-average filter over the data to smooth the measured response.
The algorithm ignores the valley around 60 rad/s and the three lightly damped resonant peaks that follow it. These features contribute little to the loss function because the fitting error is small at these frequencies. To address this issue, you can specify frequency-dependent weights to make the algorithm pay more attention to these frequencies.
To improve the estimated model quality, preprocess the data. To do so, you truncate the low signal-to-noise portions of data below 1 rad/s and above 2e4 rad/s that are not interesting. Then you use a moving-average filter to smooth data in the low-frequency high-magnitude region below 40 rad/s. At these frequencies, the data has a low signal-to-noise ratio, but has dynamics that you are interested in capturing. Do not apply the filter at frequencies above 40 rad/s to avoid smoothing data where you see the valley and the three peaks that follow it.
Make a copy of the original idfrd data object.
GfrdProcessed = Gfrd;
Truncate the data below 1 rad/s and above 2e4 rad/s.
GfrdProcessed = fselect(GfrdProcessed,1,2e4);
Apply a three-point centered moving-average filter to smooth out the frequency-response data below 40 rad/s that contains spurious dynamics. The response data is stored in the ResponseData property of the object.
w = GfrdProcessed.Frequency;
f = squeeze(GfrdProcessed.ResponseData);
idx = w<40;
f(idx) = movmean(f(idx),3);
Here f(idx) is the frequency-response data at frequencies less than 40 rad/s.
Place the filtered data back into the original data object.
GfrdProcessed.ResponseData = f;
Plot the original and preprocessed data.
bodemag(Gfrd,GfrdProcessed);
ylim([-100 0]);
legend('Original data','Preprocessed data');
The plot shows that all desired dynamics are intact after preprocessing.
Use a low weight for the low frequency region under 10 rad/s where spurious dynamics exist. This low weight and the smoothing applied earlier to this data reduce the chance of spurious peaks in the estimated model response in this region.
Weight = ones(size(f));
Weight(idx) = Weight(idx)/10;
Use a high weight for data in the frequency range 40-6e3 rad/s where you want to capture the dynamics but the response data magnitude is low.
idx = w>40 & w<6e3;
Weight(idx) = Weight(idx)*30;
Specify the weights in the WeightingFilter option of the estimation option set.
tfestOpt = tfestOptions('WeightingFilter',Weight);
Note that Weight is a custom weighting filter. You can also specify WeightingFilter as 'inv' or 'invsqrt' for frequency-response data. These options specify the weight as
1/|f\left(w\right)|
1/\sqrt{|f\left(w\right)|}
, respectively. These options enable you to quickly test the effect of using a higher weight for low magnitude regions of data. 'invsqrt' is typically a good initial choice. If these weights do not yield good estimation results, you can provide custom weights as shown in this example.
Estimate a transfer function model with 11 poles and 10 zeros using the preprocessed data and specified weighting filter.
GfitNew = tfest(GfrdProcessed,11,10,tfestOpt);
Plot the original data, the initial model response, and the new model response.
bodemag(Gfrd,Gfit,GfitNew);
legend('Original Data','Original Model','New Model');
Plot the estimation error. Compute the estimation error by including the weighting filter Weight that you used for estimating GfitNew.
r1 = squeeze(freqresp(GfitNew,w));
r2 = squeeze(freqresp(GfrdProcessed,w));
fitErrorNew = Weight.*(r1-r2);
semilogx(w,abs(fitErrorNew).^2)
ylabel('Magnitude (abs)');
The new model successfully captures all system dynamics of interest.
You can use the weighted error plot for further troubleshooting if your initial choice of weights does not yield a satisfactory result.
tfest | tfestOptions
|
Home : Support : Online Help : Mathematics : Calculus of Variations : Jacobi
compute and solve Jacobi's equation for conjugate points
Jacobi(f, t, x(t), X(t), h, a)
integrand to be tested
dependent function or list of functions
expression for the extremal (found by solving the Euler-Lagrange equations)
name for the unknown function in Jacobi's equation
initial point (left end of the interval)
The Jacobi(f, t, x(t), X(t), h, a) command finds Jacobi's equation and tries to find solutions of Jacobi's equation, that is, conjugate points.
The routine returns an expression sequence consisting of Jacobi's equation and any solutions found by dsolve.
If dsolve encounters a problem, an error message is returned.
If dsolve fails to find a solution, only Jacobi's equation is returned.
If the solution of Jacobi's equation has a zero on the region of interest, the extremal is not optimal.
\mathrm{with}\left(\mathrm{VariationalCalculus}\right)
[\textcolor[rgb]{0,0,1}{\mathrm{ConjugateEquation}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Convex}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{EulerLagrange}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Jacobi}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Weierstrass}}]
f≔-\frac{{\mathrm{diff}\left(y\left(t\right),t\right)}^{2}}{2}+\frac{{y\left(t\right)}^{2}}{2}
\textcolor[rgb]{0,0,1}{f}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{-}\frac{{\left(\frac{\textcolor[rgb]{0,0,1}{ⅆ}}{\textcolor[rgb]{0,0,1}{ⅆ}\textcolor[rgb]{0,0,1}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)\right)}^{\textcolor[rgb]{0,0,1}{2}}}{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\frac{{\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)}^{\textcolor[rgb]{0,0,1}{2}}}{\textcolor[rgb]{0,0,1}{2}}
\mathrm{Jacobi}\left(f,t,y\left(t\right),\mathrm{sin}\left(t\right),h,0\right)
\frac{{\textcolor[rgb]{0,0,1}{ⅆ}}^{\textcolor[rgb]{0,0,1}{2}}}{\textcolor[rgb]{0,0,1}{ⅆ}{\textcolor[rgb]{0,0,1}{t}}^{\textcolor[rgb]{0,0,1}{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\textcolor[rgb]{0,0,1}{h}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{h}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{h}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{_C1}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{sin}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{t}\right)
|
Laplace transform of \(\displaystyle{f{{\left({t}\right)}}}={t}{e}^{{-{t}}}{\sin{{\left({2}{t}\right)}}}\)
Zane Decker 2022-04-14 Answered
f\left(t\right)=t{e}^{-t}\mathrm{sin}\left(2t\right)
maggionmoo
\mathcal{L}\left(\mathrm{sin}2t\right)=\frac{2}{{s}^{2}+{2}^{2}}=\frac{2}{{s}^{2}+4}
, using the table.
\mathcal{L}\left({e}^{-t}\mathrm{sin}2t\right)=\frac{2}{{\left(s+1\right)}^{2}+4}
, using frequency shifting.
\mathcal{L}\left(t{e}^{-t}\mathrm{sin}2t\right)=-\frac{d}{ds}\left(\frac{2}{{\left(s+1\right)}^{2}+4}\right)=\frac{4\left(s+1\right)}{{\left({\left(s+1\right)}^{2}+4\right)}^{2}}
, using frequency differentiation.
veselrompoikm
f\left(t\right)=t{e}^{-t}\mathrm{sin}\left(2t\right)=tg\left(t\right)
g\left(t\right)={e}^{-t}\mathrm{sin}\left(2t\right)={e}^{-t}h\left(t\right)
h\left(t\right)=\mathrm{sin}\left(2t\right)
F\left(s\right)=-{G}^{\prime }\left(s\right)
G\left(s\right)=H\left(s+1\right)
H\left(s\right)=\frac{2}{{s}^{2}+4}
Use the transforms in the table below to find the inverse Laplace transform of the following function.
F\left(s\right)=\frac{5}{{s}^{4}}
Consider the system of differential equations
\frac{dx}{dt}=-y\text{ }\text{ }\text{ }\frac{dy}{dt}=-x
.a)Convert this system to a second order differential equation in y by differentiating the second equation with respect to t and substituting for x from the first equation.
b)Solve the equation you obtained for y as a function of t; hence find x as a function of t.
Use the definition of Laplace Transforms to find
L\left\{f\left(t\right)\right\}
f\left(t\right)=\left\{\begin{array}{ll}-1& 0\le t<1\\ 1& t\ge 1\end{array}
Then rewrite f(t) as a sum of step functions,
{u}_{c}\left(t\right)
, and show that by taking Laplace transforms, this yields the same answer as your direct computation.
The Laplace transform of
L\left\{\frac{{d}^{3}x\left(t\right)}{d{t}^{3}}\right\}
sX\left(s\right)-x\left(0\right)
{s}^{3}X\left(s\right)-{s}^{2}x\left(0\right)-s\stackrel{˙}{x}\left(0\right)-\stackrel{¨}{x}\left(0\right)
{s}^{3}X\left(s\right)-{s}^{2}\stackrel{¨}{x}\left(0\right)-s\stackrel{˙}{x}\left(0\right)-x\left(0\right)
{s}^{2}X\left(s\right)-sx\left(0\right)-\stackrel{˙}{x}\left(0\right)
Obtaining Differential Equations from Functions
\frac{dy}{dx}={x}^{2}-1
is a first order ODE,
\frac{{d}^{2}y}{{dx}^{2}}+2{\left(\frac{dy}{dx}\right)}^{2}+y=0
is a second order ODE and so on. I am having trouble to obtain a differential equation from a given function. I could find the differential equation for
y={e}^{x}\left(A\mathrm{cos}x+B\mathrm{sin}x\right)
and the steps that I followed are as follows.
\frac{dy}{dx}={e}^{x}\left(A\mathrm{cos}x+B\mathrm{sin}x\right)+{e}^{x}\left(-A\mathrm{sin}x+B\mathrm{cos}x\right)
=y+{e}^{x}\left(-A\mathrm{sin}x+B\mathrm{cos}x\right)
\frac{{d}^{2}y}{{dx}^{2}}=\frac{dy}{dx}+{e}^{x}\left(-A\mathrm{sin}x+B\mathrm{cos}x\right)+{e}^{x}\left(-A\mathrm{cos}x-B\mathrm{sin}x\right)
=\frac{dy}{dx}+\left(\frac{dy}{dx}-y\right)-y
using the orginal function and (1). Finally,
\frac{{d}^{2}y}{{dx}^{2}}-2\frac{dy}{dx}+2y=0
Similarly, if the function is
y=\left(A\mathrm{cos}2t+B\mathrm{sin}2t\right)
, the differential equation that I get is
\frac{{d}^{2}y}{{dx}^{2}}+4y=0
following similar steps as above.
Solve the following manually using separable differential equations method.
mydx+nxdy=0
I need to calculate a taylor polynomial for a function
f:R\to R
where we know the following
f{}^{″}\left(x\right)+f\left(x\right)={e}^{-x}\mathrm{\forall }x
f\left(0\right)=0
{f}^{\prime }\left(0\right)=2
|
{s}^{-\frac{3}{2}}
\frac{2}{\sqrt{\pi }}\frac{\sqrt{\pi }}{2{s}^{\frac{3}{2}}}=2\sqrt{\frac{t}{\pi }}
\frac{2}{\sqrt{\pi }}
\mathrm{\Gamma }\left(\frac{3}{2}\right)
cloirdxti
{t}^{\frac{1}{2}}
\frac{\mathrm{\Gamma }\left(\frac{3}{2}\right)}{{s}^{\frac{3}{2}}}
, so the inverse transform of
{s}^{-\frac{3}{2}}
\frac{{t}^{\frac{1}{2}}}{\mathrm{\Gamma }\left(\frac{3}{2}\right)}
Elementary Applications of Differential Equations 1. A culture of bacteria consists of 1000 bacteria. After 2 hrs., the culture becomes 300 bacteria. a. How many bacteria remain after 4 hours? b. In how many hours will the culture become 50 bacteria?
Use laplace transform to solve the given system
a\right)\frac{dx}{dt}-2x-\frac{dx}{dt}-y=6{e}^{3t}
2\left(\frac{dx}{dt}\right)-3x+\frac{dx}{dt}-3y=6{e}^{3t}x\left(0\right)=3y\left(0\right)=0
Solve for the following problems using substitution suggested by the equation.
\left(3x-2y+1\right)dx+\left(3x-2y+3\right)dy=0
{y}^{‴}+3{y}^{″}-18{y}^{\prime }-40y=-120
y\left(0\right)=6
{y}^{\prime }\left(0\right)=45
y"\left(0\right)=-21
{x}^{2}{y}^{\prime }+x\left(x+7\right)y={e}^{x}
y=
|
Second Order Differential Equations ay''+by'+cy=0, without complex numbers
sexe4yo 2022-05-02 Answered
Second Order Differential Equations
ay{}^{″}+b{y}^{\prime }+cy=0
, without complex numbers
Let us show that there is a unique solution, given that
g\left(0\right)=A,
{g}^{\prime }\left(0\right)=B
A\mathrm{cos}x+B\mathrm{sin}x
is a solution with these initial conditions, so look at
h\left(x\right):g\left(x\right)-A\mathrm{cos}x+B\mathrm{sin}x
; this satisfies
h{}^{″}+h=0,\text{ }h\left(0\right)=0,\text{ }{h}^{\prime }\left(0\right)=0
h\left(x\right)=0
h{}^{″}+h=0
{h}^{\prime }
h{}^{″}{h}^{\prime }+{h}^{\prime }h=0
for all x. Note that
h{}^{″}{h}^{\prime }+{h}^{\prime }h={\left({\frac{12}{{h}^{\prime }}}^{2}+{\frac{12}{h}}^{2}\right)}^{\prime }
. Therefore, by the Mean Value Theorem,
{\frac{12}{h}}^{\prime }{\left(x\right)}^{2}+\frac{12}{h}{\left(x\right)}^{2}=\left(x-0\right)\left(\frac{12}{{h}^{\prime }{\left(\theta x\right)}^{2}+\frac{12}{h}{\left(\theta x\right)}^{2}}=0.
Now a real sum of squares can only be zero if each summand is zero. So we have, as required,
h\left(x\right)=0
[For those of an applied bent, we are just doing the classical thing of conserving the total energy.]
You can make the educated guess that the solution will be of the form
f\left(x\right)={e}^{kx}g\left(x\right)
a\left[{e}^{kx}g\left(x\right)\right){}^{″}+b{\left({e}^{kx}g\left(x\right)\right)}^{\prime }+c{e}^{kx}g\left(x\right)
=a\left[{e}^{kx}g{}^{″}\left(x\right)+2k{e}^{kx}{g}^{\prime }\left(x\right)+{k}^{2}{e}^{kx}g\left(x\right)\right]+b\left[{e}^{kx}{g}^{\prime }\left(x\right)+k{e}^{kx}g\left(x\right)\right]+c{e}^{kx}g\left(x\right)
=a{e}^{kx}g{}^{″}\left(x\right)+\left[2ak+b\right]{e}^{kx}{g}^{\prime }\left(x\right)+\left[a{k}^{2}+bk+c\right]{e}^{kx}g\left(x\right)
=0
{e}^{kx}
is never 0, this reduces to
ag{}^{″}+\left(2ak+b\right){g}^{\prime }+\left(a{k}^{2}+bk+c\right)g=0.
We are free to choose k as we wish, because if
{e}^{kx}g\left(x\right)
solves the ODE, then so does
{e}^{\stackrel{~}{k}x}{e}^{\left(k-\stackrel{~}{k}\right)x}g\left(x\right)
, and we simply get
\stackrel{~}{g\left(x\right)}={e}^{\left(k-\stackrel{~}{k}\right)}g\left(x\right)
as the solution for g instead. So we choose k as simple as possible: such that
2ak+b=0
k=-\frac{b}{2a}
. Then the equation reads (after multiplying by 4a)
4{a}^{2}g{}^{″}-\left({b}^{2}-4ac\right)g=0.
The solution to this, given that
{b}^{2}-4ac<0
, is known: it's of the form
g\left(x\right)=A\mathrm{sin}\left(\omega x\right)+B\mathrm{cos}\left(\omega x\right)
\omega \phantom{\rule{0.222em}{0ex}}=\sqrt{c-\frac{{b}^{2}}{4a}}
So in total we get
f\left(x\right)={e}^{kx}\left[A\mathrm{sin}\left(\omega x\right)+B\mathrm{cos}\left(\omega x\right)\right]
with k and
\omega
as specified.
\frac{dw}{dt}
\frac{dw}{dt}
w=x\mathrm{sin}y,\text{ }x={e}^{t},\text{ }y=\pi -t
t=0
\left\{\begin{array}{rl}& {x}^{2}{x}^{\prime }={\mathrm{sin}}^{2}\left({x}^{3}-3t\right)\\ & x\left(0\right)=1\end{array}.
\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{y}{20}=50\left(1+\mathrm{cos}x\right)
y{}^{″}+4y=\mathrm{cos}\left(2x\right)
\left(y+1\right)dx+\left(x+1\right)dy=0
\frac{1}{6}g{}^{″}=v{g}^{3}+k{g}^{3}-kg
{x}^{2}y{}^{″}+\left(3{x}^{2}+4x\right){y}^{\prime }+2\left({x}^{2}+3x+1\right)y=0
\mu
{\left(\mu {y}^{\prime }\right)}^{\prime }={\mu }^{\prime }{y}^{\prime }+\mu y{}^{″}=\mu y{}^{″}+\mu \frac{3{x}^{2}+4x}{{x}^{2}}{y}^{\prime }+\mu \frac{2\left({x}^{2}+3x+1\right)}{{x}^{2}}y⇒
{\mu }^{\prime }{y}^{\prime }=\mu \frac{3{x}^{2}+4x}{{x}^{2}}{y}^{\prime }+\mu \frac{2\left({x}^{2}+3x+1\right)}{{x}^{2}}y⇒\frac{{\mu }^{\prime }}{\mu }=\frac{3{x}^{2}+4x}{{x}^{2}}+\frac{2\left({x}^{2}+3x+1\right)}{{x}^{2}}\frac{y}{{y}^{\prime }}
|
Calculate the work done by a force F = (X-y)i + xyj in moving a particle counterclockwise along the circle x+ y =4 from the point (2,0) to the point (0,-2).
{\mathrm{End}}_{\mathrm{ℝ}\left[x\right]}\left(M\right)
M=\frac{\mathbb{R}\left[x\right]}{\left({x}^{2}+1\right)}
is a module over the ring
\mathbb{R}\left[x\right]
The set of units of
\frac{\mathbb{Z}}{10}\mathbb{Z}
\left\{\stackrel{―}{1},\stackrel{―}{3},\stackrel{―}{7},\stackrel{―}{9}\right\}
, how can I show that this group is cyclic?
My guess is that we need to show that the group can be generated by some element in the set, do I need to show that powers of some element can generate all elements in the other congruence classes?
{7}^{2}=49\equiv 9±\mathrm{mod}10
, i.e. using 7 we can generate an element in the congruence class of 9, but can not generate 29 for example from any power of 7, so is it sufficent to say that an element is a generator if it generates at least one element in all other congruence classes?
\alpha
\left\{|K-E\left[K\right]|>c\right\}.
A=\left\{\dots ,\dots ,\dots \right\}
\alpha
\left\{K>{c}^{1}\right\}
A=\left\{\dots ,\dots ,\dots \right\}
f \left( x \right) = 7 {x}^{2 }+ 2
g \left( x \right) = 5 {x}^{3 }+ 7 {x}^{2 }+ 6 x
, find the derivative of the quotient
\frac{f \left( x \right) }{g \left( x \right) }
|
electric conductance - Maple Help
Home : Support : Online Help : Science and Engineering : Units : Known Units : electric conductance
Units of Electric Conductance
Electric conductance has the dimension time cubed electric current squared per length squared mass. The SI derived unit of electric conductance is the siemens, which is defined as an inverse ohm.
Maple knows the units of electric conductance listed in the following table.
S, mho
mho, mhos
abS abmho
abmho, abmhos
statS statmho
statmho, statmhos
An asterisk ( * ) indicates the default context, an at sign (@) indicates an abbreviation, and under the prefixes column, SI indicates that the unit takes all SI prefixes, IEC indicates that the unit takes IEC prefixes, and SI+ and SI- indicate that the unit takes only positive and negative SI prefixes, respectively. Refer to a unit in the Units package by indexing the name or symbol with the context, for example, siemens[SI] or abS[EMU]; or, if the context is indicated as the default, by using only the unit name or symbol, for example, siemens or abS.
The units of electric conductance are defined as follows.
An absiemens is defined as
1.×{10}^{9}
siemens and is energy-equivalent to the unit second per centimeter (
\frac{s}{\mathrm{cm}}
A statsiemens is defined as
\frac{100000.}{{c}^{2}}
siemens where c is the magnitude of the speed of light, and is energy-equivalent to the unit centimeter per second (
\frac{\mathrm{cm}}{s}
\mathrm{convert}\left('\mathrm{siemens}','\mathrm{dimensions}','\mathrm{base}'=\mathrm{true}\right)
\frac{{\textcolor[rgb]{0,0,1}{\mathrm{time}}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{electric_current}}}^{\textcolor[rgb]{0,0,1}{2}}}{{\textcolor[rgb]{0,0,1}{\mathrm{length}}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{mass}}}
\mathrm{convert}\left(1.60217733×{10}^{-19},'\mathrm{units}','S','\mathrm{abS}'\right)
\textcolor[rgb]{0,0,1}{1.602177330}\textcolor[rgb]{0,0,1}{×}{\textcolor[rgb]{0,0,1}{10}}^{\textcolor[rgb]{0,0,1}{-28}}
\mathrm{convert}\left(1.60217733×{10}^{-19},'\mathrm{units}','S',\frac{'s'}{'\mathrm{cm}'},'\mathrm{energy}'\right)
\textcolor[rgb]{0,0,1}{1.602177330}\textcolor[rgb]{0,0,1}{×}{\textcolor[rgb]{0,0,1}{10}}^{\textcolor[rgb]{0,0,1}{-28}}
\mathrm{convert}\left(1.60217733×{10}^{-19},'\mathrm{units}','S','\mathrm{statS}'\right)
\textcolor[rgb]{0,0,1}{1.439965173}\textcolor[rgb]{0,0,1}{×}{\textcolor[rgb]{0,0,1}{10}}^{\textcolor[rgb]{0,0,1}{-7}}
\mathrm{convert}\left(1.60217733×{10}^{-19},'\mathrm{units}','S',\frac{'\mathrm{cm}'}{'s'},'\mathrm{energy}'\right)
\textcolor[rgb]{0,0,1}{1.439965173}\textcolor[rgb]{0,0,1}{×}{\textcolor[rgb]{0,0,1}{10}}^{\textcolor[rgb]{0,0,1}{-7}}
|
Home : Support : Online Help : Programming : Audio Processing : Resample
resample audio data to the specified sample rate
Resample(audArray, samplesPerSec, options)
Array or Matrix containing the audio data to resample
desired sampling rate
options modifying the resampling operation
The Resample command resamples audio data, producing a new audio object with the specified sample rate.
The audArray parameter specifies the audio to resample, and must be a dense, rectangular, one or two dimensional Array, Vector, or Matrix with datatype=float[8].
The samplesPerSec parameter specifies the desired sample rate. Typical values are 11025, 22050, and 44100 samples per second.
By default, resampling is done using B-spline interpolation. This produces excellent results reasonably fast.
If the method=nearest option is specified, resampling is done using the nearest-neighbor algorithm. This produces poorer results, but the algorithm is very fast. This method is often suitable for producing low-quality resamplings of audio data.
\mathrm{audiofile}≔\mathrm{cat}\left(\mathrm{kernelopts}\left(\mathrm{datadir}\right),"/audio/stereo.wav"\right):
\mathrm{with}\left(\mathrm{AudioTools}\right):
\mathrm{aud}≔\mathrm{Read}\left(\mathrm{audiofile}\right)
\textcolor[rgb]{0,0,1}{\mathrm{aud}}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{"Sample Rate"}& \textcolor[rgb]{0,0,1}{22050}\\ \textcolor[rgb]{0,0,1}{"File Format"}& \textcolor[rgb]{0,0,1}{\mathrm{PCM}}\\ \textcolor[rgb]{0,0,1}{"File Bit Depth"}& \textcolor[rgb]{0,0,1}{8}\\ \textcolor[rgb]{0,0,1}{"Channels"}& \textcolor[rgb]{0,0,1}{2}\\ \textcolor[rgb]{0,0,1}{"Samples/Channel"}& \textcolor[rgb]{0,0,1}{19962}\\ \textcolor[rgb]{0,0,1}{"Duration"}& \textcolor[rgb]{0,0,1}{0.90531}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{s}\end{array}]
\mathrm{attributes}\left(\mathrm{aud}\right)
\textcolor[rgb]{0,0,1}{22050}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{8}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}
\mathrm{small}≔\mathrm{Resample}\left(\mathrm{aud},11025\right)
\textcolor[rgb]{0,0,1}{\mathrm{small}}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{"Sample Rate"}& \textcolor[rgb]{0,0,1}{11025}\\ \textcolor[rgb]{0,0,1}{"File Format"}& \textcolor[rgb]{0,0,1}{\mathrm{PCM}}\\ \textcolor[rgb]{0,0,1}{"File Bit Depth"}& \textcolor[rgb]{0,0,1}{8}\\ \textcolor[rgb]{0,0,1}{"Channels"}& \textcolor[rgb]{0,0,1}{2}\\ \textcolor[rgb]{0,0,1}{"Samples/Channel"}& \textcolor[rgb]{0,0,1}{9981}\\ \textcolor[rgb]{0,0,1}{"Duration"}& \textcolor[rgb]{0,0,1}{0.90531}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{s}\end{array}]
\mathrm{attributes}\left(\mathrm{small}\right)
\textcolor[rgb]{0,0,1}{11025}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{8}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}
AudioTools[Scale]
|
I am having trouble finding the inverse Laplace transform of: \frac{1}{s^2-9s+20} I
Rowan Huynh 2022-04-27 Answered
I am having trouble finding the inverse Laplace transform of:
\frac{1}{{s}^{2}-9s+20}
I tried writing it in a different way:
{L}^{-1}\left\{\frac{1}{{\left(s-\frac{9}{2}\right)}^{2}-{\left(\frac{1}{2}\right)}^{2}}\right\}=2{L}^{-1}\left\{\frac{\frac{1}{2}}{{\left(s-\frac{9}{2}\right)}^{2}-{\left(\frac{1}{2}\right)}^{2}}\right\}
Yaretzi Odom
I'll give you a hint, because I think it might be useful for you to get practice. If you need more tell in comment: try rewriting this using partial fractions and then use the linearity of the inverse laplace.
amisayq6t
X\left(s\right)=\frac{1}{{s}^{2}-9s+20}=\frac{1}{\left(s-4\right)\left(s-5\right)}=\frac{-1}{s-4}+\frac{1}{s-5}
=>x\left(t\right)=-{e}^{4t}+{e}^{5t}
A population like that of the United States with an age structure of roughly equal numbers in each of the age groups can be predicted to A) grow rapidly over a 30-year-period and then stabilize B) grow little for a generation and then grow rapidly C) fall slowly and steadily over many decades D) show slow and steady growth for some time into the future
\left(x+2y-1\right)dx+\left(2x+4y-3\right)dy=0
\left[0,\mathrm{\infty }\right]
F\left(s\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt
f\left(t\right)={e}^{-4t}\mathrm{sin}9t
How to solve for the equation y′′′+4y′′+5y′+2y=10cost using laplace transform method given that y(0)=0,y′(0)=0,and y′′(0)=3
y\prime \prime +3y\prime +2y={e}^{-t},y\left(0\right)=0y\prime \left(0\right)=0
determine a function f(t) that has the given Laplace transform F(s).
F\left(s\right)=\frac{3}{\left({s}^{2}\right)}
\left({D}^{2}-2D+1\right)y=x{e}^{x}\mathrm{sin}x
|
Octave - New World Encyclopedia
Previous (Ockham's razor)
Next (Octave Mirbeau)
The octave is the span or intervallic space between one tone and another, the latter having twice or half as many vibrations per second. The octave has been and remains a principal tonal foundation in both western and non-western music. This foundation is due to the strength and consistency of the interval between its first and last tones. Many composers have clearly favored the use of the octave since the first and last tones of its interval fit so well together that the listener appears to hear the two tones as one.
The frequent use of the octave is a part of most music, though there are those ethnomusicologists who say that it is far from universal in formative and early music (e.g., Nettl, 1956; Sachs & Kunst, 1962). Yet the strong resemblances between the two analogous tones that are an octave apart explain why the octave was noticed in many musical cultures and at a time when music was rapidly developing. In the later Middle Ages, the Gregorian chant or plainsong of the Christian Church was followed by polyphonic music in the form called organum, which added a second voice to a single melody. The second voice was usually a fourth or fifth apart; however, when these two voices were duplicated an octave apart creating four voices, the richness of the organum created a full and complete harmonic sound, giving greater richness to the religious offerings. Another example is in the indigenous music of India, in which the octave span is composed of quarter tones. Octave relationships appear very frequently in the tones of the instruments, i.e. the tabla and baya (drums) in juxtaposition to the singer(s). Octaves were used to bring stability and grandeur to Indian classical and folk compositions. In addition, many musical cultures used the overtone in their instrumentation and vocalisms to create octaves for a more complex sound. An overtone is a higher partial note heard over the main tone which is created by a voice or musical instrument. Overtones were favored in various folk songs to add fullness and dexterity to the melodic line.
Technically, an octave (sometimes abbreviated 8ve or 8va) is the interval between one musical note and another with half or double the frequency or vibrations per second. For example, if one note is pitched at 400 Hz, the note an octave above it is at 800 Hz, and the note an octave below is at 200 Hz. The ratio of frequencies of two notes an octave apart is therefore two to one. Further octaves of a note occur at
{\displaystyle 2^{n}}
times the frequency of that note (where n is an integer), such as 2, 4, 8, 16, etc. and the reciprocal of that series. Moreover, 50 Hz and 400 Hz are one and two octaves away from 100 Hz because they are
{\displaystyle 1/2}
{\displaystyle 1/2^{1}}
{\displaystyle 2^{2}}
) times the frequency, respectively, however 300 Hz is not a whole number octave above 100 Hz, despite being a harmonic of 100 Hz.
The octave is the second simplest interval in music (after the unison). Since the human ear tends to hear both notes as being essentially "the same," notes played an octave apart are given the same note name in the Western system of music notation—for example, the name of a note an octave above A is also A. This is called octave equivalency, and is closely related to the concept of harmonics. This is similar to enharmonic equivalency, and less so transpositional equivalency and, less still, inversional equivalency, the latter two of which are generally used only in musical set theory or atonal theory. Thus all C#s, or all 1s (if C=0), in any octave are part of the same pitch class. As well as being used to describe the relationship between two notes, the word is also used when speaking of a range of notes that fall between a pair an octave apart. In the diatonic scale, this is eight notes if one counts both ends, hence the name "octave," from Italian for eight. In the chromatic scale, this is 13 notes counting both ends, although traditionally, one speaks of 12 notes of the chromatic scale, not counting both ends. In most Western music, the octave is divided into 12 semitones (see musical tuning). These semitones are usually equally spaced out in a method known as equal temperament. Other scales may have a different number of notes covering the range of an octave, but the "octave" is still used. In a scale which uses microtones usually occurring in non-western music, there would be a number of notes, for example 17, 22, or any other number of notes, between the two octave tones depending on the instrumental capabilities and the compositional sliding or "bending of the tones." Yet the octave would be intact and used as a musical guidepost.
Diatonic intervals edit
Perfect : unison (0) | fourth (5) | fifth (7) | octave (12)
Major : second (2) | third (4) | sixth (9)| seventh (11)
Minor : second (1) | third (3)| sixth (8)| seventh (10)
Augmented/Diminished : tritone (6)
semitones are given in brackets
The notation 8va is sometimes seen in sheet music, meaning "play this an octave higher than written." 8va stands for ottava, the Italian word for octave. Sometimes 8va will also be used to indicate when a passage is to be played an octave lower, although the similar notation 8vb (ottava bassa) is more common. Similarly, 15ma means "play two octaves higher than written." Coll'ottava means to play the passage in octaves. Any of these directions can be cancelled with the word loco, but often a dashed line or bracket indicates the extent of the music affected.
From Gregorian chants through the magnificent symphonies of the classical masters to the iridescent sounds of jazz and New Age music, the octave has been and will continually rule as one of the most important intervals in theory and composition.
Burns, Edward M. "Intervals, Scales, and Tuning," The Psychology of Music second edition. Deutsch, Diana, ed. San Diego: Academic Press, 1999. ISBN 0122135644.
Sachs, C. and Kunst, J. The wellsprings of music. ed. Kunst, J. The Hague: Marinus Nijhoff, 1962.
Anatomy of an Octave by Kyle Gann.
Octave history
History of "Octave"
Retrieved from https://www.newworldencyclopedia.org/p/index.php?title=Octave&oldid=1016461
|
Common-mode signal - Wikipedia
Common-mode signal is the voltage common to both input terminals of an electrical device. In telecommunication, the common-mode signal on a transmission line is also known as longitudinal voltage.
In most electrical circuits the signal is transferred by a differential voltage between two conductors. If the voltages on these conductors are U1 and U2, the common-mode signal is the half-sum of the voltages:
{\displaystyle U_{\text{cm}}={\frac {U_{1}+U_{2}}{2}}}
When referenced to the local common or ground, a common-mode signal appears on both lines of a two-wire cable, in phase and with equal amplitudes. Technically, a common-mode voltage is one-half the vector sum of the voltages from each conductor of a balanced circuit to local ground or common. Such signals can arise from one or more of the following sources:
Radiated signals coupled equally to both lines,
An offset from signal common created in the driver circuit, or
A ground differential between the transmitting and receiving locations.
Noise induced into a cable, or transmitted from a cable, usually occurs in the common mode, as the same signal tends to be picked up by both conductors in a two-wire cable. Likewise, RF noise transmitted from a cable tends to emanate from both conductors. Elimination of common-mode signals on cables entering or leaving electronic equipment is important to ensure electromagnetic compatibility. Unless the intention is to transmit or receive radio signals, an electronic designer generally designs electronic circuits to minimise or eliminate common-mode effects.
Methods of eliminating common-mode signals[edit]
Differential amplifiers or receivers that respond only to voltage differences, e.g. those between the wires that constitute a pair. This method is particularly suited for instrumentation where signals are transmitted through DC bias.
An inductor where a pair of signalling wires follow the same path through the inductor, e.g. in a bifilar winding configuration such as used in Ethernet magnetics.[1] Useful for AC and DC signals, but will filter only higher frequency common-mode signals.
A transformer, which is useful for AC signals only, and will filter any form of common-mode noise, but may be used in combination with a bifilar wound coil to eliminate capacitive coupling of higher frequency common-mode signals across the transformer. Used in twisted pair Ethernet.[2]
Common-mode filtering may also be used to prevent egress of noise for electromagnetic compatibility purposes.
High frequency common-mode signals, for example, RF noise from a computing circuit, may be blocked using a ferrite bead clamped to the outside of a cable. These are often observable on laptop computer power supplies near the jack socket, and good quality mouse or printer USB cables and HDMI cables.[3]
Switch mode power supplies include common and differential mode filtering inductors to block the switching signal noise returning into mains wiring.[4]
Common-mode rejection ratio is a measure of how well a circuit eliminates common-mode interference.
^ http://www.hottconsultants.com/pdf_files/APEC-2002.pdf[bare URL PDF]
Retrieved from "https://en.wikipedia.org/w/index.php?title=Common-mode_signal&oldid=1077027679"
|
ioBots - iob.fi DAO | docs
iob.fi DAO managed hedged pools are powered by ioBots.
What is ioBots?
In its simplest form, the ioBots can be expressed as:
ioBots=\frac{DecisionAlgos+SmartContractOrderEngine}{RiskControlAlgos}
A buying or selling signal is not a trading system. A strategy is not a trading system either. A well designed working trading system needs to have the following components:
Clearly set objectives: All trading starts with a clear objective that will not change during the entire trading process;
Mechanism that seeks the consistency: The key to consistency is to keep what we have earned;
Market Types determination: ioBots identifies which of the 5 Market Types -- Very Bullish, Bullish, Choppy, Bearish, or Very Bearish -- is underway every day for each market we are trading through a proprietary algorithm;
Different strategies for different Market Types: There is no one single "best strategy." Each very bullish, bullish, choppy, bearish, or very bearish market type calls for a different trading strategy for that market on that trading day;
Pre-determined risk level: Knowing how much to lose before every order entry.
What is the most important thing in investing and trading success?
Consistency is the #1 secret to be a winning trader, but most traders fail to achieve it. Long-term views always win in trading. Most failed traders inadvertently treating trading as an adventure, rather than a profit-making business, therefore fail to achieve consistency.
A successful trader would start with a clear trading objective first. All methods of wealth growth start with a clear objective and a detailed action plan. Simply wanting to "make lots of money" is neither. Same as running a successful business, trading strategies and setups are only meaningful after a clearly defined objective is formulated.
Why am I not achieving trading consistency?
The key to consistency is to keep what you have earned, and the key to that is to clearly define your Expected Max Drawdown BEFORE entering any trade.
How to choose the best strategy?
For any trading strategy to work well, you first must identify what the Market Type is at the moment.
What are the iob Market Types?
iob.fi identifies which of the 5 Market Types -- Very Bullish, Bullish, Choppy, Bearish, or Very Bearish -- is underway every day for each market we are trading through a proprietary algorithm.
Why is market-type awareness important?
Trading methods and strategies are not one-size-fits-all. A certain strategy may work well in the bullish market but loses badly in a bearish market. It is essential to have the right strategies for the corresponding market type.
Again, what is the best trading strategy?
There is no one single "best strategy." Each very bullish, bullish, choppy, bearish, or very bearish market type calls for a different trading strategy for that market on that trading day.
What are ioBots trading signals?
Trading Signals are buying and selling trading signals generated by ioBots developed by IOB LLC, which is being transferred iob.fi DAO. Currently, ioBots is covering 20 markets, including BTC, ETH, ADA, XTZ, Gold, MSFT, and TSLA, among others. These are the same trading signals Quasar Fund uses for all trading decisions. The Signals started for 6 trading pairs in July 2019. 14 more markets were subsequently added. The best historical total return is BCH with a 601.57% profit. The worst is XTZ with a loss of 11.46%. You can find detailed Signal performance history here.
|
Let f <mo stretchy="false">( x <mo stretchy="false">) be a monic polynomial of odd de
f\left(x\right)
be a monic polynomial of odd degree. Prove that
\mathrm{\exists }A\in \mathbb{R}
f\left(A\right)<0
B\in \mathbb{R}
f\left(B\right)>0
Deduce that every polynomial of odd degree has a real root.
There are questions that answer the final part, but they do not do so by proving the first part. I am fairly sure that this involves the intermediate value theorem, but not sure how to implement it in this case.
RormFrure6h1
For the final part. If you have
f\left(A\right)<0
f\left(B\right)>0
then by the IVT every value in [f(A),f(B)] is attained by f(x) for some x between A and B, and this includes 0.
To show the existence of the A and B show that for x large one has that the sign of f(x) is the sign of the leading coefficient. And, if the degree is odd for small x one has that the sign of f(x) is the opposite sign of the leading coefficient.
-{x}^{3}+4\mathrm{sin}\left(x\right)+4{\mathrm{cos}}^{2}\left(x\right)=0
has at least two solutions. You should carefully justify each of the hypothesis of the theorem.
How do I know that at least two exist because I know you're meant to look for a change in sign.
I have just recently covered the Intermediate Value Theorem, and I wanted to practice solving problems involving this theorem. However, I encountered a problem that I am not exactly sure how to tackle (it's a question involving a periodic function).
I thought about splitting the proofs into 3 cases, but I don't think it would be applicable here?
If f is periodic with a period of 2a for some
a>0
f\left(x\right)=f\left(x+2a\right)
x\in R
. Show that if f is continous, there exists some
c\in \left[0,a\right]
f\left(c\right)=f\left(c+a\right)
On a recent problem, I received the following scenario: An object moves back and forth on a straight track. During the time interval
0\le t\le 30
minutes, the object's position,
x
, and velocity,
v
, are continuous functions; some of their values are shown in the table (which I have reproduced below).
\begin{array}{|ccc|}\hline t\mathbf{\text{ (min)}}& x\left(t\right)\mathbf{\text{ (feet)}}& v\left(t\right)\mathbf{\text{ (feet/min)}}\\ 0& \text{12}& -20\\ 10& \text{50}& \text{20}\\ 15& \text{18}& \text{3}\\ 25& \text{60}& -2\\ 30& \text{60}& \text{10}\\ \hline\end{array}
The question was, for
0<t<30
, does there exist a time t when
v\left(t\right)=-22
I tried to apply the intermediate value theorem, and concluded that the answer was "not necessarily." I reasoned that because the values of
v\left(t\right)
in the table ranged between −20 (at the beginning) and 10 (at the end), and −22 wasn't between those two values, we couldn't be sure that such a t exists.
The teacher gave me 3/4 points on the question. Their comment was that I should have considered the mean value theorem, too. They did not write anything about my intermediate value theorem analysis, but I still have grave doubts about my application of the intermediate value theorem ... and, as for the mean value theorem, I have no idea how to proceed.
Would anyone here be able to shed some light on how the intermediate value and mean value theorems could be used to determine whether there exists a t where
v\left(t\right)=-22
? Thanks so much.
f\left(a\right)<s<f\left(b\right)
f\left(a\right)
f\left(b\right)
f
be a continuous function on
\mathbb{R}
which is periodic with period
2\pi
f\left(t+2\pi \right)=f\left(t\right)
t
. Show that there exists
x\in \left[0,\pi \right]
f\left(x\right)=f\left(x+\pi \right)
I know it's an intermediate-value theorem problem. I think I have to take the difference of both sides, but not quite sure. Can anyone help?
f\left(x\right)={x}^{2}+10\mathrm{sin}\left(x\right)
c
f\left(c\right)=1000
f\left(0\right)=0
f\left(90\right)={90}^{2}+10\mathrm{sin}\left(90\right)=8100+10\ast 1=8110
f\left(c\right)=1000
I have been stuck on this Real Analysis problem for hours and am just totally clueless- I am sure it is some application of the Intermediate Value Theorem- suppose
\text{ }f:\mathbb{R}\to \mathbb{R}
is continuous at every point. Prove that the equation
\text{ }f\left(x\right)=c
cannot have exactly two solutions for every value of
\text{ }c.
|
Solve this please:- |x-1|+|x-2|+|x-3| =6 - Maths - Linear Inequalities - 10667077 | Meritnation.com
Aditi Jain. answered this
\mathrm{We} \mathrm{have}, \left|x-1\right|+\left|x-2\right|+\left|x-3\right|\ge 6 .....\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{As}, \left|x-1\right|=\left\{\begin{array}{l}x-1, x\ge 1\\ 1-x, x<1\end{array}\right\;\phantom{\rule{0ex}{0ex}}\left|x-2\right|=\left\{\begin{array}{l}x-2, x\ge 2\\ 2-x, x<2\end{array}\right\\mathrm{and}\phantom{\rule{0ex}{0ex}}\left|x-3\right|=\left\{\begin{array}{l}x-3, x\ge 3\\ 3-x, x<3\end{array}\right\\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\mathrm{Case} \mathrm{I}: \mathrm{When} x<1,\phantom{\rule{0ex}{0ex}}1-x+2-x+3-x\ge 6\phantom{\rule{0ex}{0ex}}⇒6-3x\ge 6\phantom{\rule{0ex}{0ex}}⇒3x\le 0\phantom{\rule{0ex}{0ex}}⇒x\le 0\phantom{\rule{0ex}{0ex}}\mathrm{So}, x\in \left(-\infty , 0\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Case} \mathrm{II}: \mathrm{When} 1\le x<2,\phantom{\rule{0ex}{0ex}}x-1+2-x+3-x\ge 6\phantom{\rule{0ex}{0ex}}⇒4-x\ge 6\phantom{\rule{0ex}{0ex}}⇒x\le 4-6\phantom{\rule{0ex}{0ex}}⇒x\le -2\phantom{\rule{0ex}{0ex}}\mathrm{So}, x\in \varphi \phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Case} \mathrm{III}: \mathrm{When} 2\le x<3,\phantom{\rule{0ex}{0ex}}x-1+x-2+3-x\ge 6\phantom{\rule{0ex}{0ex}}⇒x\ge 6\phantom{\rule{0ex}{0ex}}\mathrm{So}, x\mathit{\in }\varphi \phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Case} \mathrm{IV}: \mathrm{When} x\ge 3,\phantom{\rule{0ex}{0ex}}x-1+x-2+x-3\ge 6\phantom{\rule{0ex}{0ex}}⇒3x-6\ge 6\phantom{\rule{0ex}{0ex}}⇒3x\ge 12\phantom{\rule{0ex}{0ex}}⇒x\ge \frac{12}{3}\phantom{\rule{0ex}{0ex}}⇒x\ge 4\phantom{\rule{0ex}{0ex}}\mathrm{So}, x\in \left[4, \infty \right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{So}, \mathrm{from} \mathrm{all} \mathrm{the} \mathrm{four} \mathrm{cases}, \mathrm{we} \mathrm{get}\phantom{\rule{0ex}{0ex}}x\in \left(-\infty , 0\right]\cup \left[4, \infty \right)
Davinder Singh Purba answered this
|
A Potentially Useful Galactic Dark Matter Index
760 Campbell Ln. Ste. 106 #161, Bowling Green, KY 42104, USA
There have been a number of observational surprises with respect to galactic dark matter-to-visible matter ratios. These surprises confirm our continued lack of understanding of the fundamental nature of dark matter. Because of their apparent close ties with galactic gravitational entropy, at least four recent observations appear to provide the first evidence in support of Verlinde’s theory of gravity, dark energy and dark matter as emergent properties. They also appear to correlate with Roger Penrose’s gravitational entropy concept, as well as entropy defined in the Flat Space Cosmology (FSC) model. Given the observational support, two different testable versions of a “Dark Matter Index” (DMI) are introduced in this paper, and its utility is discussed in terms of potentially achieving a better understanding of the fundamental nature of dark matter.
Dark Matter, Flat Space Cosmology, Cosmology Theory, Gravitational Entropy, Black Holes, DMI
Recent observations have shown a distant exceedingly diffuse galaxy (NGC 1052-DF2) with no discernable dark matter whatsoever [1] . A second report indicates the virial volume of the Milky Way galaxy to have a dark matter-to-visible matter ratio of 23.074-to-1 [2] [3] .
Roger Penrose, in his new book [4] , introduces the concept of gravitational entropy on pages 256-258. In sharp contrast to entropy of an ideal gas, gravitational and cosmological entropy increases with the ongoing clustering of stars and galaxies. Thus, as Penrose pointed out, black holes (especially supermassive black holes) become huge repositories of total cosmic entropy over the great span of cosmic time. Furthermore, as detailed in “Clues to the Fundamental Nature of Gravity, Dark Energy and Dark Matter” [5] , there appears to be a deep connection within the Flat Space Cosmology (FSC) model between gravity and cosmic entropy. Thus, FSC appears to be the cosmological model correlate to Verlinde’s theory that gravity is an emergent property [6] [7] . The obvious implication, as discussed in the FSC “Clues” paper, is that if gravity is indeed an emergent property, dark energy and dark matter would also likely be emergent properties. If such were the case, dark energy could well be an emergent property of the cosmic system as a whole, and “dark matter” could well be an emergent property of the galactic visible matter. Recently, Verlinde’s theory as it may pertain to dark matter, was tested by Brouwer, et al. [8] . Their apparent lensing excess surface density (ESD) analysis of 33,613 galaxies provided the first observational support that Verlinde’s theory appears to be correct with respect to galactic dark matter.
Given this observational support for Verlinde’s theory, and the apparent deep connections between cosmic entropy and gravity, dark energy and dark matter in FSC, the two additional dark matter observations mentioned in the first paragraph of this paper take on a new significance. This is because they may provide evidence that dark matter observations are closely linked with galactic gravitational entropy. Could it be that the absence of apparent dark matter in the distant and exceedingly diffuse (i.e., gravitationally young) NCG 1052-DF2 galaxy is because it has an exceedingly low gravitational entropy? And could it be that the relative abundance of dark matter in the relatively compact, dense and gravitationally mature Milky Way galaxy (including of course its supermassive black hole), is because it has a high gravitational entropy? If indeed Penrose’s concept of gravitational entropy is correct, the answer would seem to be an emphatic “yes!” These additional observations of dark matter in the form of dark matter-to-visible matter ratios of zero and approximately 23-to-1, respectively, would appear to provide the second and third observations in support of Verlinde’s theory, and of FSC as an emergent gravity cosmological model. Furthermore, the recent report [9] that dark matter appears to be relatively scarce in the massive star-forming galaxies at high redshifts would appear to be the fourth observation in support of Verlinde’s theory and the FSC model presented in the “Clues” paper.
2. Recommended Dark Matter Index with Discussion
It is not always easy to quantify the amount of dark matter within a particular galaxy and its dark matter halo. Gravitational lensing, for instance, can be a hit-or-miss proposition, depending upon the alignment of more distant galaxies. However, the above recent observational results apparently in support of Verlinde’s theory suggest a possible means of indexing the dark matter of any galaxy for which the following characteristics can be measured: the galactic center intrinsic brightness IBGC; and the galactic redshift S. This author suggests that these measured characteristics could be expressed in the form of a “Dark Matter Index” (DMI) ratio according to:
DMI={I}_{BGC}/S
If Verlinde, Penrose and the FSC model are indeed correct, one would expect high DMI values to be indicative of high galactic gravitational entropy, high galactic dark matter-to-visible matter ratios, and high excess gravitational lensing powers currently attributed to dark matter. The potential usefulness of such an index could be studied by comparing galaxies in the dark matter data base for which the dark matter-to-visible matter ratio has already been measured, either by gravitational lensing or other means. Even with the more labor-intensive and alignment-dependent gravitational lensing studies of the sort reported by Brouwer, et al, a lensing DMI value (represented as DMIL below) could ultimately prove to have better correlative power than the excess surface density value alone. The proposed relation is
DM{I}_{L}=ES{D}_{DM}/S
wherein ESDDM is Brouwer’s excess surface density proposed to be attributable to dark matter and S is the galactic redshift. If correlation studies of these DMI and DMIL indices prove them to be accurate quantitative measures of galactic dark matter, this may speed up the process of truly identifying the fundamental nature of dark matter.
This paper is dedicated to Dr. Stephen Hawking and Dr. Roger Penrose for their groundbreaking work on black holes and their possible application to cosmology. Dr. Tatum also thanks Dr. Rudolph Schild of the Harvard Center for Astrophysics for his past support and encouragement.
Tatum, E.T. (2018) A Potentially Useful Galactic Dark Matter Index. Journal of Modern Physics, 9, 1564-1567. https://doi.org/10.4236/jmp.2018.98097
1. van Dokkum, P., et al. (2018) Nature, 555, 629-632. https://doi.org/10.1038/nature25767
2. Posti, L. and Helmi, A. (2018) Mass and Shape of the Milky Way’s Dark Matter Halo with Globular Clusters from Gaia and Hubble. arXiv:1805.01408v1 [astro-ph.GA].
3. Tatum, E.T. (2018) Journal of Modern Physics, 9, 1559-1563.
4. Penrose, R. (2016) Fashion Faith and Fantasy in the New Physics of the Universe. Princeton University Press, Princeton. https://doi.org/10.1515/9781400880287
7. Verlinde, E. (2016) Emergent Gravity and the Dark Universe. arXiv:1611.02269v2 [hep-th].
8. Brouwer, M.M., et al. (2016) First Test of Verlinde’s Theory of Emergent Gravity Using Weak Gravitational Lensing Measurements. Monthly Notices of the Royal Astronomical Society, 1-14. arXiv:1612.03034v2 [astro-ph.CO].
9. Genzel, R., et al. (2017) Nature, 543, 397-401. https://doi.org/10.1038/nature21685
|
(Redirected from Fuzzy search)
Finding strings that approximately match a pattern
A fuzzy Mediawiki search for "angry emoticon" has as a suggested result "andré emotions"
3 On-line versus off-line
The closeness of a match is measured in terms of the number of primitive operations necessary to convert the string into an exact match. This number is called the edit distance between the string and the pattern. The usual primitive operations are:[1]
insertion: cot → coat
deletion: coat → cot
substitution: coat → cost
These three operations may be generalized as forms of substitution by adding a NULL character (here symbolized by *) wherever a character has been deleted or inserted:
insertion: co*t → coat
deletion: coat → co*t
Some approximate matchers also treat transposition, in which the positions of two letters in the string are swapped, to be a primitive operation.[2]
transposition: cost → cots
Different approximate matchers impose different constraints. Some matchers use a single global unweighted cost, that is, the total number of primitive operations necessary to convert the match to the pattern. For example, if the pattern is coil, foil differs by one substitution, coils by one insertion, oil by one deletion, and foal by two substitutions. If all operations count as a single unit of cost and the limit is set to one, foil, coils, and oil will count as matches while foal will not.
Other matchers specify the number of operations of each type separately, while still others set a total cost but allow different weights to be assigned to different operations. Some matchers permit separate assignments of limits and weights to individual groups in the pattern.
Problem formulation and algorithms[edit]
One possible definition of the approximate string matching problem is the following: Given a pattern string
{\displaystyle P=p_{1}p_{2}...p_{m}}
and a text string
{\displaystyle T=t_{1}t_{2}\dots t_{n}}
, find a substring
{\displaystyle T_{j',j}=t_{j'}\dots t_{j}}
in T, which, of all substrings of T, has the smallest edit distance to the pattern P.
A brute-force approach would be to compute the edit distance to P for all substrings of T, and then choose the substring with the minimum distance. However, this algorithm would have the running time O(n3 m).
A better solution, which was proposed by Sellers[3], relies on dynamic programming. It uses an alternative formulation of the problem: for each position j in the text T and each position i in the pattern P, compute the minimum edit distance between the i first characters of the pattern,
{\displaystyle P_{i}}
, and any substring
{\displaystyle T_{j',j}}
of T that ends at position j.
For each position j in the text T, and each position i in the pattern P, go through all substrings of T ending at position j, and determine which one of them has the minimal edit distance to the i first characters of the pattern P. Write this minimal distance as E(i, j). After computing E(i, j) for all i and j, we can easily find a solution to the original problem: it is the substring for which E(m, j) is minimal (m being the length of the pattern P.)
Computing E(m, j) is very similar to computing the edit distance between two strings. In fact, we can use the Levenshtein distance computing algorithm for E(m, j), the only difference being that we must initialize the first row with zeros, and save the path of computation, that is, whether we used E(i − 1,j), E(i,j − 1) or E(i − 1,j − 1) in computing E(i, j).
In the array containing the E(x, y) values, we then choose the minimal value in the last row, let it be E(x2, y2), and follow the path of computation backwards, back to the row number 0. If the field we arrived at was E(0, y1), then T[y1 + 1] ... T[y2] is a substring of T with the minimal edit distance to the pattern P.
Computing the E(x, y) array takes O(mn) time with the dynamic programming algorithm, while the backwards-working phase takes O(n + m) time.
Another recent idea is the similarity join. When matching database relates to a large scale of data, the O(mn) time with the dynamic programming algorithm cannot work within a limited time. So, the idea is to reduce the number of candidate pairs, instead of computing the similarity of all pairs of strings. Widely used algorithms are based on filter-verification, hashing, Locality-sensitive hashing (LSH), Tries and other greedy and approximation algorithms. Most of them are designed to fit some framework (such as Map-Reduce) to compute concurrently.
On-line versus off-line[edit]
Traditionally, approximate string matching algorithms are classified into two categories: on-line and off-line. With on-line algorithms the pattern can be processed before searching but the text cannot. In other words, on-line techniques do searching without an index. Early algorithms for on-line approximate matching were suggested by Wagner and Fisher[4] and by Sellers[5]. Both algorithms are based on dynamic programming but solve different problems. Sellers' algorithm searches approximately for a substring in a text while the algorithm of Wagner and Fisher calculates Levenshtein distance, being appropriate for dictionary fuzzy search only.
On-line searching techniques have been repeatedly improved. Perhaps the most famous improvement is the bitap algorithm (also known as the shift-or and shift-and algorithm), which is very efficient for relatively short pattern strings. The Bitap algorithm is the heart of the Unix searching utility agrep. A review of on-line searching algorithms was done by G. Navarro.[6]
Although very fast on-line techniques exist, their performance on large data is unacceptable. Text preprocessing or indexing makes searching dramatically faster. Today, a variety of indexing algorithms have been presented. Among them are suffix trees[7], metric trees[8] and n-gram methods.[9][10] A detailed survey of indexing techniques that allows one to find an arbitrary substring in a text is given by Navarro et al.[11] A computational survey of dictionary methods (i.e., methods that permit finding all dictionary words that approximately match a search pattern) is given by Boytsov[12].
Common applications of approximate matching include spell checking.[13] With the availability of large amounts of DNA data, matching of nucleotide sequences has become an important application.[14] Approximate matching is also used in spam filtering.[15] Record linkage is a common application where records from two disparate databases are matched.
String matching cannot be used for most binary data, such as images and music. They require different algorithms, such as acoustic fingerprinting.
Regular expressions for fuzzy and non-fuzzy matching
^ Baeza-Yates, R.; Navarro, G. (June 1996). "A faster algorithm for approximate string matching". In Dan Hirchsberg; Gene Myers (eds.). Combinatorial Pattern Matching (CPM'96), LNCS 1075. Irvine, CA. pp. 1–23. CiteSeerX 10.1.1.42.1593.
^ Baeza-Yates, R.; Navarro, G. "Fast Approximate String Matching in a Dictionary" (PDF). Proc. SPIRE'98. IEEE CS Press. pp. 14–22.
^ Boytsov, Leonid (2011). "Indexing methods for approximate dictionary searching: Comparative analysis". Journal of Experimental Algorithmics. 16 (1): 1–91. doi:10.1145/1963190.1963191. S2CID 15635688.
^ Cormen, Thomas; Leiserson, Rivest (2001). Introduction to Algorithms (2nd ed.). MIT Press. pp. 364–7. ISBN 978-0-262-03293-3.
^ Galil, Zvi; Apostolico, Alberto (1997). Pattern matching algorithms. Oxford [Oxfordshire]: Oxford University Press. ISBN 978-0-19-511367-9.
^ Gusfield, Dan (1997). Algorithms on strings, trees, and sequences: computer science and computational biology. Cambridge, UK: Cambridge University Press. ISBN 978-0-521-58519-4.
^ Myers, G. (May 1999). "A fast bit-vector algorithm for approximate string matching based on dynamic programming" (PDF). Journal of the ACM. 46 (3): 395–415. doi:10.1145/316542.316550. S2CID 1158099.
^ Navarro, Gonzalo (2001). "A guided tour to approximate string matching". ACM Computing Surveys. 33 (1): 31–88. CiteSeerX 10.1.1.96.7225. doi:10.1145/375360.375365. S2CID 207551224.
^ Navarro, Gonzalo; Baeza-Yates, Ricardo; Sutinen, Erkki; Tarhio, Jorma (2001). "Indexing Methods for Approximate String Matching" (PDF). IEEE Data Engineering Bulletin. 24 (4): 19–27.
^ Sellers, Peter H. (1980). "The Theory and Computation of Evolutionary Distances: Pattern Recognition". Journal of Algorithms. 1 (4): 359–73. doi:10.1016/0196-6774(80)90016-4.
^ Skiena, Steve (1998). Algorithm Design Manual (1st ed.). Springer. ISBN 978-0-387-94860-7.
^ Ukkonen, E. (1985). "Algorithms for approximate string matching". Information and Control. 64 (1–3): 100–18. doi:10.1016/S0019-9958(85)80046-2.
^ Wagner, R.; Fischer, M. (1974). "The string-to-string correction problem". Journal of the ACM. 21: 168–73. doi:10.1145/321796.321811. S2CID 13381535.
^ Zobel, Justin; Dart, Philip (1995). "Finding approximate matches in large lexicons". Software: Practice and Experience. 25 (3): 331–345. CiteSeerX 10.1.1.14.3856. doi:10.1002/spe.4380250307. S2CID 6776819.
Efficient Similarity Query Processing Project with recent advances in approximate string matching based on an edit distance threshold.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Approximate_string_matching&oldid=1082740685"
|
Hydrochemistry and Groundwater Quality Assessment in Mafraq Province, Jordan
Hydrochemistry and Groundwater Quality Assessment in Mafraq Province, Jordan
Ali Obeidat1, Mahmoud Alawneh2
1Department of Geology, Faculty of Science, The University of Jordan, Amman, Jordan
2Ministry of Water and Irrigation, Amman, Jordan
The acceptable water quality standard varies based on the intended use of water. Therefore, the standard will be dramatically decreased while moving from Drinking, domestic, industrial and agricultural purposes. Water quality modeling software Aquachem and ArcGIS10.3 were used to analyze data sets of 12 samples from four different wells located around the Mafraq area in order to investigate its suitability for drinking. The results revealed that hydrochemistry of groundwater recorded a wide range in total dissolve solid (TDS) and the chloride is found to be most predominating. Generally, groundwater in the University and Zatari areas has high concentrations of nitrate. The major element’s data were plotted on the Piper’s diagram for working of hydrogeochemical facies. Mixed CaMgCl are the most prominent facies. The pH part of the Durov diagram reveals that groundwater in the study area is alkaline and electrical conductivity of most of samples lies in the range of drinking water standards adopted in Jordan.
Water Quality, Hydrogeological Facies, Water-Rock Interaction, Groundwater Evaluation
The Mafraq City at the North East of Jordan was selected as the study area based on the availability of data. It is the largest of the country’s four governmental provinces in terms of area. The area is located between latitudes 32.25˚ to 32.00˚ north and longitudes 36.00˚ to 36.20˚ east, as shown in the location map given in Figure 1. It is the eastern extension of the Jordanian Plateau with deserts, valleys and plains. The elevation of the province increased from very low in the extreme east at the average rainfall from 20 to 200 mm. Due to growing industry in cities and increasing urbanization, water demand is growing and will need to depend largely on the groundwater. This continued exploitation of groundwater results in declining water table and deterioration in water quality. It is crucial to evaluate the quality and quantity of groundwater for its suitable utilization.
[1] and [2] studied the geology of this area, Rocks exposed from the upper Cretaceous to lower tertiary in Mafraq, which is stratigraphically divided into three major zones namely the Muwaqqar Chalky Mar (MCM), the Umm Rijam Chalk (URC) and Wadi Shallala Chalk (WSC). The Eruptive basalt zone covers the northeastern side of the study area as it is shown in Figure 2.
Groundwater in Mafraq area in both confined and unconfined aquifers in all basins and sub-basins are generally flows from catchment boundaries to the axis of valleys and then follows the general trend of surface drainage. A similar study was carried out by [3] by investigating the groundwater quality for Azraq and Harrana basin for the same Aquifers in south eastern Jordan.
All groundwater contains inorganic and organic constituents in solution due to the interaction of the groundwater with the surrounding environment. A baseline for quality criteria, including chemical, physical, and biological properties,
Figure 1. Location of the tested wells in Mafraq and its surrounds.
Figure 2. Geology Map of the study area after bender 1975, cretaceous―green, fluvial deposits―doted, basalt lava flow―pink color.
as well as standard methods of reporting and comparing results of water analyses must be established for new development areas. The aim of the hydrochemical studies is to distinguish the geochemical properties that control the dissolved constituents in water and establish the distribution of some contaminants aquifers result from human activities. The quality of a water supply depends upon its purpose; thus, the needs of drinking, industrial and irrigation water vary widely. Representative groundwater samples from the four locations in the Mafraq Province aquifer were obtained, the wells selected on the basis of geographical distribution. Aquachem and ArcGIS10.3 software programs have been used to manage create and graphically display the water quality data.
Water Samples Preservation and Laboratory Analysis
From the studied area, the selected wells are used for drinking, irrigation and domestic use. All the water samples were collected in the period between 2010 and 2015 and it were analyzed for 11 physical and chemical water quality parameters including: pH, EC, and major cations (calcium, carbonate, magnesium sodium and potassium) and major anions (chloride, sulfate, Nitrate and bicarbonate) in laboratory of MWI using Standard Methods for the Examination of Water and analysis as per American Public Health Association [4] .
The statistical comparison of the range, mean and standard deviations (SD) of concentrations of chemical constituents of groundwater samples are given in Table 2. The chemical analyses of the groundwater showed large variations of dissolved ions concentrations with high salinity in some samples. The standard deviations for potassium and bicarbonate are greater than their averages due to the large variations for these constituents. Electrical conductivity showed large spatial variations between 280 and 3125 μS/cm, with a mean value of 1025.58 μS/cm indicating a very wide range of fresh to lightly saline water.
4.1. Water Physical Parameters
The most common physical parameters of water: pH, electrical conductivity EC and total dissolved solids (TDS) have been determined. The pH values in the study area range between 7.5 and 7.9 with mean value of 7.7.
The very pure distilled water is theoretically insulator due to the absence of any ions which act as charge carriers. So, EC distribution indicates low values of EC in the middle and northern areas of Mafraq and with gradient increases toward the south and northern-east regions.
4.2. Water Chemical Parameters
The abundance of major ions largely depends upon the nature of rocks, climatic conditions and mobility. The ion distribution is also influenced by the infinite complex surface and subsurface physicochemical environments, To asses these geochemical processes the collected water samples are chemically analyzed for the major cations (Na+, K+, Ca2+ and Mg2+) and major anions (Cl−,
{\text{SO}}_{4}^{2-}
{\text{NO}}_{3}^{-}
{\text{HCO}}_{3}^{-}
5. Graphical Presentation of Hydrochemical Data
During circulation of water in rocks and soils, ions leached out and dissolve in groundwater. The geological formations, water-rock interaction and relative mobility of ions are the prime factors influencing the geochemistry of the groundwater. Because of chemical analysis results of water in form of tables may be difficult to interpret. The graphic representations are used to discuss the water-rock interaction in the study area. Hydrogeological facies were worked out by developing Scholler, Stiff, Piper and Durov diagrams.
5.1. Schoeller
[5] Diagram is also used to present average chemical composition of Mafraq groundwater. The relative tendency of ions in mg/l shows Na > Mg > Ca > K and Cl > HCO3 > SO4. Schoeller diagram in Figure 3 below present the average composition of major cations anions of the groundwater for Mafraq province.
5.2. Trilinear Diagram (Piper Diagram)
Piper Diagrams [6] are a combination anion and cation triangles that lie on a common baseline. Diamond shape between them can be used to make a conclusion
Figure 3. Schoeller diagram showing average composition of major cations anions in mg/l and meq/l of the groundwater of Mafraq.
as to the origin of the water represented by the analysis and to characterize different water types. Piper divided waters into four basic types according to their placement near the four corners of the diamond. Water that plots at the top of the diamond is high in Ca2+ + Mg2+ and Cl− +
{\text{SO}}_{4}^{2-}
, which results in an area of permanent hardness. The water that plots near the left corner is rich in Ca2+ + Mg2+ and
{\text{HCO}}_{3}^{-}
and is the region of water of temporary hardness. Water plotted at the lower corner of the diamond is primarily composed of alkali carbonates (Na+ + K+ and
{\text{HCO}}_{3}^{-}
{\text{CO}}_{3}^{2-}
). Water lying nears the right-hand side of the diamond considered saline (Na+ + K+ and Cl− +
{\text{SO}}_{4}^{2-}
The Mafraq groundwater samples are plotted in Piper Diagram (Figure 4). The figure shows three different groups of samples that have been summarized in Table 1. Most of samples, about 58% occupy middle upper side of the diamond shape. This group indicates mixed water (chloride-calcium and chloride magnesium water type). 33.3% of samples occupy the right corner of the diamond shape, where these samples belong to sodiumcloride water type). 8.3% of the samples located in the left corner of the diamond shape where these samples belong to calcium bicarbonate type.
5.3. Durov Diagram
[7] introduce another diagram which provides more information on the hydrochemical facies by helping to identify the water types and it can display some possible geochemical processes that could help in understanding quality of groundwater and its evaluation.
The Durov Diagram for the major cations and anions was plotted using AquaChem software as given in Figure 5. The Duorv plot for Mafraq groundwater
Figure 4. Piper Diagram depicting hydrogeochemical facies of Mafraq groundwater.
Figure 5. Durov Diagram depicting geochemical processes in Mafraq groundwater.
Table 1. Hydrochemical facies carried out by Piper Diagram.
samples indicates that most of the samples are in the phase of mixing, dissolution. The pH part of the plot reveals that groundwater in study area is alkaline which is favored for drinking. The total dissolve solid for most of groundwater samples lies in the range of drinking water standards adapted in Jordan except for Znayya 3 Wells the TDS reach more than 1500 mg/L.
6. Groundwater Quality
Human welfare is directly linked to water quality and quantities, chemistry of the groundwater have been used for the evaluation of quality of water for drinking, domestic and irrigation purposes. It was estimated that about 1000 km3 of the world’s aggregated groundwater is abstracted annually.
6.1. Water Quality for Drinking Purposes
Chemical analyses define the type and significance of various chemicals present in water, the standards for water constituents used to evaluate the suitability of water for drinking and domestic purposes are more precautionary than those that would be applied to water for other purposes such as irrigations or industrial use.
The maximum permissible limits for water standard by World Health Organization [8] have been added in Table 2. The average of samples with contents their location in study area has also been given.
It is observed from this table that, out of 4 groundwater average samples, only 1 sample have shown TDS values more than maximum permissible limit of 1000 mg/l at Znayya 3. As also seen, all samples locations have nitrate concentration higher than the permissible limit. Generally, the groundwater in the most of study area is suitable for drinking purposes except for Znayya 3.
6.2. Water Quality for Irrigation Purposes
Excessive Salts may harm plant growth physically by limiting the uptake of water through modification of osmotic processes, or chemically by metabolic reactions such as those caused by toxic constituents [9] . In this study, the discussion of water quality for irrigation is mainly based on the following factors:
The relation between Na and Ca + Mg ion content affects the physical properties of soil. This relation can be expressed by sodium adsorption ratios (SAR). The higher the sodium adsorption ratio, then less suitable the water is for irrigation, that is implies that sodium in the irrigation water may re-place calcium and magnesium ions in the soil, which will cause long term damage to the soil structure. It is defined by the suitability of water irrigation is a function of both specific conductance and SAR, the relations between specific conductance and SAR values from the four sites in Mafraq province were graphically presented in Wilcox diagram in Figure 6, where the ion concentrations are expressed in equivalent per million (epm). Two Factors used to show the suitability of water
Table 2. Sample average concentration compare to [9] .
Figure 6. SAR values from the four sites in Mafraq province were graphically presented in.
for irrigation are SAR and specific conductance. These two Factors are combined to make the classifications shown below. To make the classifications shown below.
Salinity hazard dividing points are 250, 750 and 2250 μmoh, resulting in four categories: Low salinity water (C1), Medium salinity water (C2), High salinity water (C3) and Very high salinity water (C4).
The Wilcox diagram in Figure 6 shows that about 33 % of the samples fall into C4-S2 class with very high salinity and medium sodium hazard and the rest of the samples fall into C3-S1 with high salinity and low hazard. This water cannot be used on soil without special management for salinity control in well Znayya 3; plants with good salt tolerance should be planted in the Mafraq province around this area.
Mafraq province suffers from increase demand of groundwater due to rapid increase in population of the Syrian refugee. In the present study, 12 ground water samples were collected from 4 locations of Mafraq Province north of Jordan. This study recommends more detail study using geochemical models to trace the sources of high nitrate concentration in groundwater. Interpretation of hydro-chemical analysis reveals that:
The pH values range between 7.5 and 7.9 with mean value of 7.7 and EC values range from 1562 μs/cm in Znyya 3 Location to 415 μs/cm in Merhab.
Geochemistry of groundwater displays Na > Ca > Mg > K and Cl > HCO3 > SO4 trend.
The major element’s data were plotted on Piper’s diagram indicated that, three hydrogeochemical type facies were identified that 58.3% of the samples are mixed CaMgCl water type, 33.3% of them belong to the NaCl water type and 8.3% for CaHCO3 type.
The Durov plot for groundwater samples indicates that most of the samples are in the phase of mixing dissolution.
Comparisons of data with [8] standards for drinking water indicate that the groundwater in the most of study area is suitable for drinking purposes except Znayya 3.
The Wilcox diagram shows that about 67% of groundwater samples are located in C4-S1 class and about 33% of the samples fall into C4-S2 class.
Obeidat, A. and Alawneh, M. (2019) Hydrochemistry and Groundwater Quality Assessment in Mafraq Province, Jordan. Open Access Library Journal, 6: e5365. https://doi.org/10.4236/oalib.1105365
1. Bender, F. (1975) Geology of the Arabian Peninsula-Jordan. United States Geological Survey Professional Paper No. 560-1, Washington DC, 36 p.
https://doi.org/10.3133/pp560I
2. Abed, A.M. (2000) Geology of Jordan. Jordanian Geologists Association, Amman.
3. Obeidat, A.M. and Rimawi, O. (2017) Characteristics and Genesis of the Groundwater Resources Associated with Oil Shale Deposits in the Azraqand Harrana Basins, Jordan. Journal of Water Resource and Protection, 9, 121-138.
4. APHA (1995) Standard Method for the Examination of Water and Wastewater. 19th Edition, American Public Health Association, Washington DC.
5. Schoeller, H. (1960) Salinity of Groundwater, Evapotranspiration and Recharge of Aquifers. IASH Pulls., France.
6. Piper, A.M. (1944) A Graphic Procedure in the Geochemical Interpretation of Water Analysis. American Geophysical Union Transactions, 25, 914-923.
7. Durov, S.A. (1948) Natural Waters and Graphical Representation of Their Composition. Doklady Akademii Nauk, USSR, 59, 87-90.
8. WHO (2011) Gide Lines for Drinking Water Quality. Fourth Edition, Vol. 1, WHO Library Cataloguing-in-Publication Data, Gene-va.
9. Abdellatef, S.A., Ohi, A., Nabatame, T. and Taniguchi, A. (2013) The Effect of Physical and Chemical Cues on Hepatocellular Function and Morphology. Inter-national Journal of Molecular Sciences, 15, 4299-4317.
|
Balanced model truncation via Schur method - MATLAB schurmr - MathWorks United Kingdom
Balanced model truncation via Schur method
GRED = schurmr(G)
GRED = schurmr(G,order)
[GRED,redinfo] = schurmr(G,key1,value1,...)
[GRED,redinfo] = schurmr(G,order,key1,value1,...)
schurmr returns a reduced order model GRED of G and a struct array redinfo containing the error bound of the reduced model and Hankel singular values of the original system.
The error bound is computed based on Hankel singular values of G. For a stable system Hankel singular values indicate the respective state energy of the system. Hence, reduced order can be directly determined by examining the system Hankel SV's, σι.
With only one input argument G, the function will show a Hankel singular value plot of the original model and prompt for model order number to reduce.
This method guarantees an error bound on the infinity norm of the additive error ∥ G-GRED ∥∞ for well-conditioned model reduced problems [1]:
{‖G-Gred‖}_{\infty }\le 2\sum _{k+1}^{n}{\sigma }_{i}
This table describes input arguments for schurmr.
LTI model to be reduced (without any other inputs will plot its Hankel singular values and prompt for reduced order).
(Optional) an integer for the desired order of the reduced model, or optionally a vector packed with desired orders for batch runs
'MaxError' can be specified in the same fashion as an alternative for ' ORDER '. In this case, reduced order will be determined when the sum of the tails of the Hankel sv's reaches the 'MaxError'.
A real number or a vector of different errors
{Wout,Win} cell array
Optimal 1x2 cell array of LTI weights Wout (output) and Win (input); default is both identity; Weights must be invertible.
Weights on the original model input and/or output can make the model reduction algorithm focus on some frequency range of interests. But weights have to be stable, minimum phase and invertible.
LTI reduced order model. Becomes multi-dimensional array when input is a serial of different model order array.
A STRUCT array with 3 fields:
REDINFO.ErrorBound
REDINFO.StabSV
REDINFO.UnstabSV
G can be stable or unstable. G and GRED can be either continuous or discrete.
[g1, redinfo1] = schurmr(G); % display Hankel SV plot
[g2, redinfo2] = schurmr(G,20);
[g3, redinfo3] = schurmr(G,[10:2:18]);
[g4, redinfo4] = schurmr(G,'MaxError',[0.01, 0.05]);
figure(i); eval(['sigma(G,g' num2str(i) ');']);
Given a state space (A,B,C,D) of a system and k, the desired reduced order, the following steps will produce a similarity transformation to truncate the original state-space system to the kth order reduced model [16].
Find the controllability and observability Gramians P and Q.
\begin{array}{l}{V}_{A}^{T}PQ{V}_{A}=\left[\begin{array}{ccc}{\lambda }_{1}& \dots & \dots \\ 0& \dots & \dots \\ 0& 0& {\lambda }_{n}\end{array}\right]\\ {V}_{D}^{T}PQ{V}_{D}=\left[\begin{array}{ccc}\lambda n& \dots & \dots \\ 0& \dots & \dots \\ 0& 0& {\lambda }_{1}\end{array}\right]\end{array}
Find the left/right orthonormal eigen-bases of PQ associated with the kth big Hankel singular values.
{V}_{A}=\left[{V}_{R,SMALL},\stackrel{}{\overbrace{{V}_{L,BIG}\right]}}
Find the SVD of (VTL,BIG VR,BIG) = U Σ VT
{V}_{D}=\left[\stackrel{}{\overbrace{{V}_{R,BIG}}},{V}_{L,SMALL}\right]
SL,BIG = V L,BIG UΣ(1:k,1:k)–½
SR,BIG = VR,BIGVΣ(1:k,1:k)–½
\left[\begin{array}{cc}\stackrel{^}{A}& \stackrel{^}{B}\\ \stackrel{^}{C}& \stackrel{^}{D}\end{array}\right]=\left[\begin{array}{cc}{S}_{L,BIG}^{T}A{S}_{R,BIG}& {S}_{L,BIG}^{T}B\\ C{S}_{R,BIG}& D\end{array}\right]
The proof of the Schur balance truncation algorithm can be found in [2].
[1] K. Glover, “All Optimal Hankel Norm Approximation of Linear Multivariable Systems, and Their L∝- error Bounds,” Int. J. Control, vol. 39, no. 6, pp. 1145-1193, 1984.
[2] M. G. Safonov and R. Y. Chiang, “A Schur Method for Balanced Model Reduction,” IEEE Trans. on Automat. Contr., vol. 34, no. 7, July 1989, pp. 729-733.
reduce | balancmr | bstmr | ncfmr | hankelmr | hankelsv
|
discrete math functions pigeonhole princple How many multiples
functions pigeonhole princple
How many multiples of 11 are in the set
\left\{1,2,3,..,901\right\}
gigglesbuggk1co
To find the number of multiples of 11 are in the set
\left\{1,2,3,..,901\right\}
, we use the following result:
Number of multiples of number n from 1 to m is given by
m/n (integer division)
Using the above result we get,
Number of multiples of 11 are in the set
\left\{1,2,3,..,901\right\}
i.e., number of multiples of number 11 from 1 to 901 is given by
901/11 (integer division) = 81.
\left[Herem=901,n=11\right]
Therefore there are 81 multiples of 11 are in the set
\left\{1,2,3,..,901\right\}
z = -z
z=\mathrm{sin}x
\frac{{d}^{2}z}{{ dx }^{2}}=-z
a\mathrm{sin}x+b\mathrm{cos}x
\mathrm{sin}x\text{or}\mathrm{cos}x
find all values of k sor which they given matrix augmented matrix to a consistent linear system
We wish to estimate what percent of adult residents in a certain county are parents. Out of 600 adult residents sampled, 336 had kids. Based on this, construct a 90% confidence interval for the proportion p of adult residents who are parents in this county.
\text{□}
|
Generate Faces Using Ladder Variational Autoencoder with Maximum Mean Discrepancy (MMD)
Tianjin Nankai High School, Tianjin, China.
Generative Models have been shown to be extremely useful in learning features from unlabeled data. In particular, variational autoencoders are capable of modeling highly complex natural distributions such as images, while extracting natural and human-understandable features without labels. In this paper we combine two highly useful classes of models, variational ladder autoencoders, and MMD variational autoencoders, to model face images. In particular, we show that we can disentangle highly meaningful and interpretable features. Furthermore, we are able to perform arithmetic operations on faces and modify faces to add or remove high level features.
Generative Models, Ladder Variational Autoencoders, Facial Recognition
Xu, H. (2018) Generate Faces Using Ladder Variational Autoencoder with Maximum Mean Discrepancy (MMD). Intelligent Information Management, 10, 108-113. doi: 10.4236/iim.2018.104009.
Generative Models have been highly successful in a wide variety of tasks by generating new observations from an existing probability density function. These models have been highly successful in various tasks such as semi-supervised learning, missing data imputation, and generation of novel data samples.
Variational Autoencoder is a very important class of models in Generative Models [1] [2] . These models map a prior on latent variables to conditional distributions on the input space. Training by maximum likelihood is intractable, so a parametric approximate inference distribution is jointly trained, and surprisingly, jointly training the generative model for maximum likelihood, and the inference distribution to approximate the true posterior is tractable, through a “reparameterization trick” [1] . These models have been highly successful in modeling complex natural distributions such as natural images. In addition it has been observed that these models can make use of the latent space in a meaningful manner. For example, it can learn to map different regions of the latent variable space into different object classes.
Ladder Variational Autoencoders [3] have been recently proposed to further augment this ability. In particular, it is able to disentangle high level and low level features. It utilizes the assumption that high level features require deeper networks to model, so that latent variables that are connected with the input with deep neural networks learn complicated, high level features while low level features are represented by low level variables.
It has also been observed that the evidence lower bound (ELBO) used in traditional variational autoencoders suffers from uninformative latent feature problem [4] where these models tend to under-use the latent variables. Multiple methods have been proposed to alleviate this [4] [5] . In particular, [5] showed that this problem can be avoided altogether if an MMD loss is used instead of the KL divergence in the original ELBO variational autoencoders.
In this paper we combine these ideas to build a variational ladder autoencoder with MMD loss instead of KL divergence, and utilize this model to analyze of structure and hidden features of human faces. As an application we use this model to perform “arithmetic” operations on faces. For example, we can perform arithmetic operations such as: men with pale skin − men with dark skin + women with dark skin = women with pale skin. The way we do this is by performing arithmetic operations in the feature space, and transform the results back into image space. This can be potentially useful in games and virtual reality where arbitrary features can be added to a face through the above process of analogy. This further demonstrates the effectiveness of our model in learning highly meaningful latent features.
2.1. Generative Modeling and Variational Autoencoders
Generative models seek to model a distribution pdata (x) in some input space X. The model is usually a parameterized family of distribution pθ(x) trained by maximum likelihood
\mathrm{max}\text{Epdata}\left(x\right)\left[\mathrm{log}{p}_{\theta }\left(x\right)\right]\theta
Intuitively this encourages the model distribution to place probability mass where pdata is more likely.
Variational autoencoder (Kingma & Welling, 2013; Jimenez Rezende et al., 2014) is an important class of generative models. It models a probability distribution by a prior p(z) on a latent space Z, and a conditional distribution p(x|z) on. Usually p(z) is a fixed simple distribution such as white Gaussian N(0, I), and p(x|z) is parameterized by a deep network with parameters θ, so we denote it as pθ(x|z). The model distribution is defined by
{p}_{\theta }\left(x\right)=\underset{z}{\int }{p}_{\theta }\left(x|z\right)p\left(z\right)dz
However maximum likelihood training is intractable because
{p}_{\theta }\left(x\right)
requires an integration which is very difficult to compute. The solution is by jointly defining an inference distribution
{q}_{\phi }\left(z|x\right)
parameterized by φ to approximate
{p}_{\theta }\left(z|x\right)
. Jointly training both criteria give the following optimization function, called the evidence lower bound (ELBO)
\begin{array}{l}\text{LELBO}=\left(-\text{KL}{q}_{\phi }\left(z|x\right)‖{p}_{\theta }\left(z|x\right)\right)-\text{KL}\left(\text{pdata}\left(x\right)‖{p}_{\theta }\left(x\right)\right)\\ \text{}=-\text{KL}\left({q}_{\phi }\left(z|x\right)‖p\left(z\right)\right)+\text{}E{q}_{\phi }\left(z|x\right)\left[\mathrm{log}{p}_{\theta }\left(z|x\right)\right]\end{array}
where KL denotes the Kullback-Leibler divergence. Intuitively this model achieves its goal by first applying an “encoder”
{q}_{\phi }\left(z|x\right)
to the input, then “decode” the generated latent code by
{p}_{\theta }\left(x|z\right)
and compare the generated results with the original data x using the cost function
\mathrm{log}{p}_{\theta }\left(z|x\right)
2.2. Ladder Variational Autoencoder
Ladder variational autoencoders [3] add additional structure into the latent code by adding multiple layers to the model. The model is shown in Figure 1. High level latent features are connected with the input through a deep network while low level features are connected through a shallow network. The intuition is that complicated features require deeper networks to model, so that high level latent variables will be used to model the high-level features, and vice versa. This makes it possible to disentangle simple and sophisticated features.
2.3. MMD Regularization
It has been observed that the
\text{KL}\left({q}_{\phi }\left(z|x\right)‖p\left(z\right)\right)
term in ELBO criteria result in under-used latent features (Chen et al., 2016; Zhao et al., 2017a). A solution is to use the MMD (q(z), p(z)) instead, which is defined by
Figure 1. Structure of VLAE (variational ladder autoencoder). Here circles are stochastic variables and diamonds are deterministic variables.
\begin{array}{l}\text{MMD}\left(q\left(z\right),p\left(z\right)\right)=Eq\left(z\right),q\left({z}_{0}\right)\left[k\left(z,{z}_{0}\right)\right]+Ep\left(z\right),p\left({z}_{0}\right)\left[k\left(z,{z}_{0}\right)\right]\\ \text{}-2Ep\left(z\right),q\left({z}_{0}\right)\left[k\left(z,z0\right)\right]\end{array}
where k(z, z0) is a kernel function such as Gaussian.
k\left(z,{z}^{\prime }\right)={e}^{-{‖z-{z}^{\prime }‖}_{2}^{2}/{\sigma }^{2}}
Intuitively k(z, z0) measures the distance between z and z0, and Ep(z), q(z0) [k(z, z0)] measures the average distance between samples from distributions p(z) and q(z0). If two distributions are identical, then the average distance between samples from p, samples from q, and samples from p, q respectively, should all be identical, so MMD distance should be zero. This can be used to replace
\text{KL}\left({q}_{\phi }\left(z|x\right)‖p\left(z\right)\right)
in ELBO VAE to achieve better properties.
2.4. MMD Variational Ladder Autoencoder
We apply MMD regularization to Variational Ladder Autoencoders. In particular, we regularize all the latent features respectively
\begin{array}{l}\text{LMMD}-\text{VLAE}=E{q}_{\phi }\left(z|x\right)\left[\mathrm{log}{p}_{\theta }\left(x|z\right)\right]-\text{MMD}\left(p\left({z}_{0}\right),{q}_{\phi }\left({z}_{0}\right)\right)\\ \text{}-MMD\left(p\left({z}_{1}\right),{q}_{\phi }\left({z}_{1}\right)\right)\end{array}
This combines the advantage of both models and learns meaningful hierarchical features.
To verify the effective of our method we performed experiments on MNIST and CelebA [6] . We visualize the manifold learned for each dataset, and observe extremely rich disentangled features.
Samples from MNIST are shown in Figure 2. We are able to disentangle visual features such as digit width, inclination, digit identity, etc. For example, bottom layer represents style of the stroke, such as the width. Middle layer represents inclination while top layers mostly represent digit identity.
Samples from CelebA are shown in Figure 3. We are able to disentangle features such as lighting, hair style, face identity and pose.
4. Arithmetic Operations on Faces
We observed that by adding or subtracting values from latent code, we can modify
Figure 2. Training results over MNIST after 1 hour on a GTX1080Ti. Each plot is obtained by sampling one layer uniformly in the [−3, 3] range, and other layers randomly. Left: Represents stroke style and width. Middle: Represents digit inclination. Right: Represents digit identity.
Figure 3. Training results over CelebA after roughly 8 hours on a GTX1080Ti. Left: Represents lighting and white balance. Middle Left: Represents hair color, face color, and minor variations of facial feature. Middle Right: Represents face identity. Right: Represents pose and expression of the face.
Figure 4. Faces of the fourth column are acquired by subtracting the second column from first column, then by adding the third column to the first column.
certain properties of faces. In addition, we can blend multiple faces together by adding or subtracting latent codes from or to each other.
We observed convincing results from these experiments (as shown in Figure 4). The final result of fourth column has shown various arithmetical properties. For example, faces of colors and brightnesses on all images are explicitly represented by the arithmetic result: the forth images share similar colors and brightnesses with the first and the third images, while these properties differ from the second images. Moreover, more complicated features are also learned and applied, the most specific one being the facial expression.
In this paper we proposed MMD Variational Ladder Autoencoder and its applications on various tasks, especially on facial recognition and modification on the CelebA dataset. It is capable of disentangling various features of human face and also capable of modifying or blending different faces.
Possible future works might include further discussion on the accuracy and readability of its latent code, its overfitting tendency, and its application on more unlabeled datasets.
[1] Kingma, D.P. and Welling, M. (2013) Auto-Encoding Variational Bayes.
[2] Rezende, J.D., Mohamed, S. and Wierstra, D. (2014) Stochastic Backpropagation and Approximate Inference in Deep Generative Models.
[3] Zhao, S., Song, J. and Ermon, S. (2017) Learning Hierarchical Features from Generative Models.
[4] Chen, X., Kingma, D.P., Salimans, T., Duan, Y., Dhariwal, P., Schulman, J., Sutskever, I. and Abbeel, P. (2016) Variational Lossy Autoencoder.
[5] Zhao, S., Song, J. and Ermon, S. (2017) InfoVAE: Information Maximizing Variational Autoencoders.
[6] Liu, Z., Luo, P., Wang, X. and Tang, X. (2015) Deep learning face attributes in the wild.
|
Laplace transform involving step function \(\displaystyle{f{{\left({t}\right)}}}={\frac{{{\sin{{\left({2}{t}\right)}}}}}{{{e}^{{{2}{t}}}}}}+{t}{\left\lbrace\cdot\right\rbrace}{u}{\left({t}-{4}\right)}\)
Brielle James 2022-04-13 Answered
f\left(t\right)=\frac{\mathrm{sin}\left(2t\right)}{{e}^{2t}}+t·u\left(t-4\right)
srasloavfv
\text{F}\left(\text{s}\right)={L}_{t}{\left[\frac{\mathrm{sin}\left(2t\right)}{{e}^{2t}}+t\theta \left(t-4\right)\right]}_{\left(\text{s}\right)}\phantom{\rule{0.222em}{0ex}}={\int }_{0}^{\mathrm{\infty }}{e}^{-\text{s}t}\left(\frac{\mathrm{sin}\left(2t\right)}{{e}^{2t}}+t\theta \left(t-4\right)\right)dt
\theta \left(x\right)
is the Heaviside step function.
\text{F}\left(\text{s}\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-\text{s}t}\cdot \frac{\mathrm{sin}\left(2t\right)}{{e}^{2t}}dt+{\int }_{0}^{\mathrm{\infty }}t{e}^{-\text{s}t}\theta \left(t-4\right)dt=
{\int }_{0}^{\mathrm{\infty }}{e}^{-t\left(2+\text{s}\right)}\mathrm{sin}\left(2t\right)dt+{\int }_{4}^{\mathrm{\infty }}t{e}^{-st}dt
1. When
R\left[\text{s}\right]>-2
{\int }_{0}^{\mathrm{\infty }}{e}^{-t\left(2+s\right)}\mathrm{sin}\left(2t\right)dt=\frac{2}{{s}^{2}+4\left(2+s\right)}
R\left[\text{s}\right]>0
{\int }_{4}^{\mathrm{\infty }}t{e}^{-st}dt=\frac{{e}^{-4s}\left(1+4s\right)}{{s}^{2}}
lildeutsch11xq2j
You can use the formula for finding the laplace transform of secondary part-
L\left\{f\left(t\right)u\left(t-a\right)\right\}
=L\frac{\left\{f\left(t+a\right)\right\}×{e}^{-as}}{s}
For any vectors u, v and w, show that the vectors u+v, u+w and v+w form a linearly dependent set.
Calculate the laplace transform of
{t}^{2}u\left(t-2\right)
I don't know how to manipulate
{t}^{2}
in order for it to meet the form of the product between a function and a heaviside function.
Proving the double differential of
z = -z
z=\mathrm{sin}x
\frac{{d}^{2}z}{{ dx }^{2}}=-z
implies z is of the form
a\mathrm{sin}x+b\mathrm{cos}x
. Is there a proof for the same. I was trying to arrive at the desired function but couldn't understand how to get these trigonometric functions in the equations by integration. Does it require the use of taylor polynomial expansion of
\mathrm{sin}x\text{or}\mathrm{cos}x
|
Ask Answer - Linear Inequalities - Recently Asked Questions for School Students
\frac{{\left(x+1\right)}^{3}{\left(x-2\right)}^{2}}{\left(x-5\right)}\ge 0
\frac{{\left(x+1\right)}^{3}{\left(x-2\right)}^{2}}{\left(x-5\right)}>0
a\right) \frac{{\left(x-2\right)}^{2} \left(x-3\right)}{x-1} > 0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}b\right) \frac{{\left(x-2\right)}^{2} \left(x-3\right)}{x-1} \ge 0
Draw the sign scheme\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}1\right) \frac{{\left(x - 2\right)}^{2} \left(x - 3\right)}{x - 1} > 0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}2\right) \frac{{\left(x - 2\right)}^{2} \left(x - 3\right)}{x - 1} \ge 0
1. 1<\frac{3{\mathrm{x}}^{2}-7\mathrm{x}+8}{{\mathrm{x}}^{2}+1}\le 2\phantom{\rule{0ex}{0ex}}\left(1\right) \left[1,6\right] \left(2\right) \left[-1,6\right]\phantom{\rule{0ex}{0ex}}\left(3\right) \left(-1,6\right] \left(4\right) \mathrm{R}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}2. 1-{\mathrm{e}}^{\left(\frac{1}{\mathrm{x}}-1\right)} >0\phantom{\rule{0ex}{0ex}}\left(1\right) \mathrm{x}\in \left(-\infty ,0 \right)\cup \left(1,\infty \right)\phantom{\rule{0ex}{0ex}}\left(2\right) \mathrm{x}\in \left(-\infty ,0 \right)\cup \left[1,\infty \right)\phantom{\rule{0ex}{0ex}}
2nd one fast.....
2nd and 3rd one pleaseeeee
\frac{4}{x-6}-\frac{x-2}{x-3}=\frac{x+4}{x-5}-\frac{2\left(x-1\right)}{x-4}
\frac{4x+19}{x+5}<\frac{4x-17}{x-3}.
\frac{1}{x+1}-\frac{2}{{x}^{2}-x+1}<\frac{1-2x}{{x}^{3}+1}.
|
Selective Coating of Anatase and Rutile TiO2 on Carbon via Ultrasound Irradiation: Mitigating Fuel Cell Catalyst Degradation | J. Electrochem. En. Conv. Stor | ASME Digital Collection
Selective Coating of Anatase and Rutile
TiO2
on Carbon via Ultrasound Irradiation: Mitigating Fuel Cell Catalyst Degradation
P. P. George,
Department of Chemistry and Kanbar Laboratory for Nanomaterials at the Bar-Ilan University Center for Advanced Materials and Nanotechnology,
, Ramat-Gan 52900, Israel
V. G. Pol,
V. G. Pol
A. Gedanken,
A. Gedanken
A. Gabashivili,
A. Gabashivili
, 30500 Mound Road, MC 480-106-269, Warren, MI 48090-9055
A. M. Mance,
A. M. Mance
L. Feng,
Fuel Cell Activity Group
M. S. Ruthkosky
George, P. P., Pol, V. G., Gedanken, A., Gabashivili, A., Cai, M., Mance, A. M., Feng, L., and Ruthkosky, M. S. (September 11, 2008). "Selective Coating of Anatase and Rutile
TiO2
on Carbon via Ultrasound Irradiation: Mitigating Fuel Cell Catalyst Degradation." ASME. J. Fuel Cell Sci. Technol. November 2008; 5(4): 041012. https://doi.org/10.1115/1.2890105
Before mass produced automotive fuel cell technology can be made practical, the oxidative instability of the carbons currently, used as the catalyst support at the oxygen electrode, must be addressed. This article describes a method for coating carbon (Vulcan XC-72) with protective barriers of titanium dioxide (titania) and provides the results of physical characterization tests on those materials. By combining the sol-gel coating process with high intensity ultrasonication and choosing the appropriate precursor, either the anatase or rutile phases of titania can be produced. More complete coverage of the carbon was provided by the anatase process. Accelerated gas-phase oxidation tests showed that platinized, anatase-coated carbon oxidized at
1∕5
the rate of untreated carbon while the rutile phase showed a mass loss of
1∕3
that of untreated carbon.
catalysts, fuel cells, life testing, materials testing, oxidation, protective coatings, sol-gel processing, titanium compounds, ultrasonic applications
Carbon, Catalysts, Oxidation, Coatings, Coating processes, Fuel cells
Thermal Degradation of the Support in Carbon-Supported Platinum Electrocatalysts for PEM Fuel Cells
A Solid-Polymer Electrolyte Direct Methanol Fuel Cell With a Methanol-Tolerant Cathode and its Mathematical Modelling
Handbook of Fuel Cells, Fundamentals, Technology and Applications
Surface Area Loss of Supported Platinum in Polymer Electrolyte Fuel Cells
Kinetic Model of Platinum Dissolution in PEMFCs
Electrocatalytic Corrosion of Carbon Support in PEMFC Cathodes
Kaqz
In-Situ Spin Trap Electron Paramagnetic Resonance Study of Fuel Cell Processes
Characterization of Vulcan Electrochemically Oxidized Under Simulated PEM Fuel Cell Conditions
Catalyst Microstructure Examination of PEMFC Membrane Electrode Assemblies Versus Time
Sub-Stoichiometric Titanium Oxide-Supported Platinum Electrocatalyst for Polymer Electrolyte Fuel Cells
Formation of Oxidation Resistant Graphite Flakes by Ultrathin Silicone Coating
Methane Oxidation at Room Temperature and Atmospheric Pressure Activated by Light via Polytungstate Dispersed on Titania
0022-3654 B,
Template-Free Hydrothermal Synthesis of High Surface Area Nitrogen-Doped Titania Photocatalyst Active Under Visible Light
Preparation of Highly Photocatalytic Active Nano-Sized TiO2 Particles via Ultrasonic Irradiation
Effect of Hydrothermal Treatment of Amorphous Titania on the Phase Change From Anatase to Rutile During Calcination
TiO2-Coated Active Carbon Composites With Increased Photocatalytic Activity Prepared by a Properly Controlled Sol-Gel Method
Sol-Gel Synthesis and Hydrothermal Processing of Anatase and Rutile Titania Nanocrystals
Formation of TiO2 Modified Film on Carbon Steel
Iwanoto
Preparation of Efficient Titanium Oxide Photocatalysts by an Ionized Cluster Beam (ICB) Method and Their Photocatalytic Reactivities for the Purification of Water
Synthesis of Anatase TiO2 Supported on Porous Solids by Chemical Vapor Deposition
, 2003, “Titanium Dioxide Coated Activated Carbon. A Regenerative Technology for Water Recovery,” Master's thesis, University of Florida.
Yoneyyama
1st International Symposium on Environmental Issues in the Electronics/Semiconductor Industries and Electrochemical-Photochemical Methods for Pollution Abatement
Immobilization of Rutile TiO2 on Multiwalled Carbon Nanotubes
J. Mater. Comm.
Design of Unique Titanium Oxide Photocatalysts by an Advanced Metal Ion-Implantation Method and Photocatalytic Reactions Under Visible Light Irradiation
Removal and Decomposition of Malodorants by Using Titanium Dioxide Photocatalyst Supported on Fiber Activated Carbon
Titania Coating of a Microporous Carbon Surface by Molecular Adsorptio-Deposition
Preparation of TiO2 Powders With Different Morphologies by an Oxidation-Hydrothermal Combination Method
Carbon-Coated Anatase: The Role of the Carbon Layer for Photocatalytic Performance
Synthesis and Characterization of Carbon Nanotubes—TiO2 Nanocomposites
Ultrasound: Its Chemical, Physical and Biological Effects
Sonochemistry as a Tool for Preparation of Porous Metal Oxides
Preparation of Stable Porous Nickel and Cobalt Oxides Using Simple Inorganic Precursor, Instead of Alkoxides, by a Sonochemical Technique
Sonochemical Synthesis of Mesoporous Iron Oxide and Accounts of its Magnetic and Catalytic Properties
Vradman
Ultrasonically Controlled Deposition-Precipitation Co—Mo HDS Catalysts Deposited on Wide-Pore MCM Material
Sonochemical Synthesis of Mesoporous Titanium Oxide With Wormhole-Like Framework Structures
, 1992, US98520.
Preparation of Photocatalytic TiO2 Coatings of Nanosized Particles on Activated Carbon by AP-MOCVD
Quantitative Analysis of Anatase-Rutile Mixtures With an X-Ray Diffractometer
Martinez-Escandell
Rodriguez-Valero
Self-Sintering of Carbon Mesophase Powders: Effect of Extraction/Washing With Solvents
Muilenbergy
Surface and Optical Properties of Nanocrystalline Anatase Titania Films Grown by Radio Frequency Reactive Magnetron Sputtering
In Situ Electrical Conductivity Study of Titania-Supported Vanadium-Niobium Oxide Catalysts Used in the Oxidative Dehydrogenation of Propane
Hydrogen Atom Reactions in the Sonolysis of Aqueous Solutions
Sonolysis of Aqueous Surfactant Solutions: Probing the Interfacial Region of Cavitation Bubbles by Spin Trapping
A New PtRu Anode Formed by Thermal Decomposition for the Direct Method Fuel Cell
Modeling of CO Influence in PBI Electrolyte PEM Fuel Cells
|
Dynamical Behavior of a System of Second-Order Nonlinear Difference Equations
Hongmei Bao, "Dynamical Behavior of a System of Second-Order Nonlinear Difference Equations", International Journal of Differential Equations, vol. 2015, Article ID 679017, 7 pages, 2015. https://doi.org/10.1155/2015/679017
Hongmei Bao1
1Faculty of Mathematics and Physics, Huaiyin Institute of Technology, Huai’an, Jiangsu 223003, China
This paper is concerned with local stability, oscillatory character of positive solutions to the system of the two nonlinear difference equations , , where , , , and , .
Difference equation or discrete dynamical system is a diverse field which impacts almost every branch of pure and applied mathematics. Every dynamical system determines a difference equation and vice versa. Recently, there has been great interest in studying difference equation systems. One of the reasons for this is a necessity for some techniques which can be used in investigating equations arising in mathematical models describing real life situations in population biology, economics, probability theory, genetics, psychology, and so forth.
The theory of difference equations occupies a central position in applicable analysis. There is no doubt that the theory of difference equations will continue to play an important role in mathematics as a whole. Nonlinear difference equations of order greater than one are of paramount importance in applications. Such equations also appear naturally as discrete analogues and as numerical solutions of differential and delay differential equations which model various diverse phenomena in biology, ecology, physiology, physics, engineering, and economics. It is very interesting to investigate the behavior of solutions of a system of nonlinear difference equations and to discuss the local asymptotic stability of their equilibrium points.
Amleha et al. [1] investigated global stability, boundedness character, and periodic nature of the difference equation: where and .
El-Owaidy et al. [2] investigated local stability, oscillation, and boundedness character of the difference equation:
And also Stević [3] studied dynamical behavior of this difference equation. Other related difference equation readers can refer to [4–18].
Papaschinopoulos and Schinas [6] studied the system of two nonlinear difference equation:where , are positive integers.
Clark and Kulenović [9, 10] investigated the system of rational difference equations:where and the initial conditions and are arbitrary nonnegative numbers.
Our aim in this paper is to investigate local stability, oscillation, and boundedness character of positive solutions of the system of difference equations:where , and initial conditions , .
Firstly we recall some basic definitions that we need in the sequel.
is bounded and persists if there exist positive constants such that
A solution of (5) is said to be nonoscillatory about if both and are either eventually positive or eventually negative. Otherwise, it is said to be oscillatory about .
A solution of (5) is said to be nonoscillatory about equilibrium if both and are either eventually positive or eventually negative. Otherwise, it is said to be oscillatory about equilibrium .
In this section, we will prove the following results concerning system (5).
Theorem 1. The following statements are true:(i)The system (5) has a positive equilibrium point .(ii)The equilibrium point of system (5) is locally asymptotically stable if .(iii)The equilibrium point of system (5) is unstable if .(iv)The equilibrium point of system (5) is a sink or an attracting equilibrium if .
Proof. (i) Let be positive numbers such thatThen from (7) we have that the positive equilibrium point .
(ii) The linearized equation of system (5) about the equilibrium point iswhereThe characteristic equation of (8) isBy using the linearized stability theorem [11], the equilibrium point is locally asymptotically stable iff all the roots of characteristic equation (10) lie inside unit disk, in other words, iff That is,
(iii) From the proof of (ii), it is true.
(iv) By using the linearized stability theorem [11], the equilibrium point is a sink or attracting equilibrium iff It only needs . This completes the proof of the theorem.
Theorem 2. Let , and let be a solution of system (3) such that Then the following statements are true:(i), .(ii), .
Proof. Since , so and so ; then We haveThusSimilarly we haveThereforeAlsoThusBy induction, we haveThusSoThis completes the proof.
Theorem 3. Let be a positive solution of system (5) which consists of at least two semicycles. Then is oscillatory.
Proof. Consider the following two cases.
Case 1. Let ThenThereforeCase 2. LetThenThereforeThis completes the proof.
Theorem 4. Let be a positive solution of system (5). Then the following statements are true:(i), for , if .(ii), for , if .
Proof. (i) SincethusBy induction, assuming, for , we have then, for , we have Thus(ii) The proof of (ii) is similar to (i), so we omit it.
Remark 5. If , then system (5) has a unique positive equilibrium . If , then system (5) has multipositive equilibrium; however system (5) always has an equilibrium . In this paper, we only investigate the dynamical behavior of the solution to this system associated with the equilibrium . It is of further interest to study the behavior of system (5) about other equilibriums in the future.
For confirming the results of this section, we consider numerical examples, which represent different types of solutions to (5).
Example 1. When the initial conditions , , , , and , , then the solution of system (5) converges to (see Figure 1, Theorem 1).
Example 2. When the initial conditions , , , , and , , then the solution of system (5) is locally unstable (see Figure 2, Theorem 1).
Example 3. When the initial conditions , , , , and , , then , , , and (see Figure 3, Theorem 2).
Example 4. When the initial conditions , , , , and , , then and (see Figure 4, Theorem 4).
A. M. Amleha, E. A. Grovea, G. Ladasa, and D. A. Georgiou, “On the recursive sequence
{x}_{n+1}=\alpha +\left({x}_{n-1}/{x}_{n}\right)
H. M. El-Owaidy, A. M. Ahmed, and M. S. Mousa, “On asymptotic behaviour of the difference equation
{x}_{n+1}=\alpha +\left({x}_{n-1}^{p}/{x}_{n}^{p}\right)
,” Journal of Applied Mathematics & Computing, vol. 12, no. 1-2, pp. 31–37, 2003. View at: Publisher Site | Google Scholar | MathSciNet
{x}_{n+1}=\alpha +\left({x}_{n-1}^{p}/ {x}_{n}^{p}\right)
G. Papaschinopoulos, P. C. Kariakouli, and V. Hatzifilippidis, “Oscillation and asymptotic stability of a nonlinear difference equation,” Nonlinear Analysis: Theory, Methods & Applications, vol. 47, no. 7, pp. 4885–4896, 2001. View at: Publisher Site | Google Scholar
R. DeVault, G. Ladas, and S. W. Schultz, “On the recursive sequence
{x}_{n+1}=A/{x}_{n}+1/{x}_{n-2}
,” Proceedings of the American Mathematical Society, vol. 126, pp. 3257–3261, 1998. View at: Publisher Site | Google Scholar
R. DeVault, G. Ladas, and S. W. Schultz, “Necessary and sufficient conditions the boundedness of
{x}_{n+1}=A/{x}_{n}^{p}+B/{x}_{n-1}^{q}
,” Journal of Difference Equations and Applications, vol. 3, pp. 259–266, 1998. View at: Google Scholar
R. M. Abu-Saris and R. DeVault, “Global stability of
{y}_{n+1}=A+{y}_{n}/{y}_{n-k}
D. Clark, M. R. S. Kulenović, and J. F. Selgrade, “Global asymptotic behavior of a two-dimensional difference equation modelling competition,” Nonlinear Analysis: Theory, Methods & Applications, vol. 52, no. 7, pp. 1765–1776, 2003. View at: Publisher Site | Google Scholar | MathSciNet
V. L. Kocic and G. Ladas, Global Behavior of Nonlinear Difference Equations of Higher Order with Applications, Kluwer Academic Publishers Group, Dordrecht, The Netherlands, 1993. View at: Publisher Site | MathSciNet
M. R. S. Kulenonvic and G. Ladas, Dynamics of Second Order Rational Difference Equations with Open Problems and Conjectures, Chapman & Hall, CRC Press, Boca Raton, Fla, USA, 2002.
W.-T. Li and H.-R. Sun, “Dynamics of a rational difference equation,” Applied Mathematics and Computation, vol. 163, no. 2, pp. 577–591, 2005. View at: Publisher Site | Google Scholar | MathSciNet
Y.-H. Su and W.-T. Li, “Global attractivity of a higher order nonlinear difference equation,” Journal of Difference Equations and Applications, vol. 11, no. 10, pp. 947–958, 2005. View at: Publisher Site | Google Scholar | MathSciNet
L.-X. Hu and W.-T. Li, “Global stability of a rational difference equation,” Applied Mathematics and Computation, vol. 190, no. 2, pp. 1322–1327, 2007. View at: Publisher Site | Google Scholar | MathSciNet
Q. Zhang, L. Yang, and J. Liu, “Dynamics of a system of rational third-order difference equation,” Advances in Difference Equations, vol. 2012, article 136, 2012. View at: Publisher Site | Google Scholar | MathSciNet
J. Diblik, B. Iricanin, S. Stevic, and Z. Smarda, “On some symmetric systems of difference equations,” Abstract and Applied Analysis, vol. 2013, Article ID 246723, 7 pages, 2013. View at: Publisher Site | Google Scholar | MathSciNet
Q. Din, M. N. Qureshi, and A. Q. Khan, “Dynamics of a fourth-order system of rational difference equations,” Advances in Difference Equations, vol. 2012, article 215, 15 pages, 2012. View at: Publisher Site | Google Scholar | MathSciNet
Copyright © 2015 Hongmei Bao. 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.
|
iperfpow - Maple Help
Home : Support : Online Help : Mathematics : Numerical Computations : Integer Functions : iperfpow
determine if an integer is a perfect power
iperfpow( n )
iperfpow( n, 'p' )
n
n={r}^{e}
r
e
1<e
, then iperfpow( n ) returns r. If the second argument, p, is specified in the calling sequence, then it is assigned the value e.
It is an error for n to evaluate to a numeric which is not a positive integer.
This command returns FAIL if it is not able to establish that n is a perfect power of an integer.
In all other cases, the command returns unevaluated.
\mathrm{iperfpow}\left(256\right)
\textcolor[rgb]{0,0,1}{16}
\mathrm{iperfpow}\left(125,'p'\right)
\textcolor[rgb]{0,0,1}{5}
p
\textcolor[rgb]{0,0,1}{3}
\mathrm{iperfpow}\left(216,'p'\right)
\textcolor[rgb]{0,0,1}{6}
p
\textcolor[rgb]{0,0,1}{3}
\mathrm{iperfpow}\left(12\right)
\textcolor[rgb]{0,0,1}{\mathrm{FAIL}}
\mathrm{ifactor}\left(12\right)
{\left(\textcolor[rgb]{0,0,1}{2}\right)}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{3}\right)
\mathrm{iperfpow}\left(1,'p'\right)
\textcolor[rgb]{0,0,1}{1}
p
\textcolor[rgb]{0,0,1}{2}
\mathrm{iperfpow}\left(x\right)
\textcolor[rgb]{0,0,1}{\mathrm{iperfpow}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\right)
\mathrm{iperfpow}\left(-25\right)
Error, (in iperfpow) first argument must be a positive integer
|
Determine which of the following transformations are linear
Option C and D is correct
A=\left[\begin{array}{cccc}2& 3& 5& -9\\ -8& -9& -11& 21\\ 4& -3& -17& 27\end{array}\right]
A=\left[\begin{array}{c}-3\\ 2\\ 0\\ 1\end{array}\right]
A=\left[\begin{array}{ccccc}1& 2& -5& 11& -3\\ 2& 4& -5& 15& 2\\ 1& 2& 0& 4& 5\\ 3& 6& -5& 19& -2\end{array}\right]
B=\left[\begin{array}{ccccc}1& 2& 0& 4& 5\\ 0& 0& 5& -7& 8\\ 0& 0& 0& 0& -9\\ 0& 0& 0& 0& 0\end{array}\right]
A=\left[\begin{array}{ccccc}1& 5& -4& -3& 1\\ 0& 1& -2& 1& 0\\ 0& 0& 0& 0& 0\end{array}\right]
5{x}_{1}+3{x}_{2}=6
-6{x}_{1}-3{x}_{2}=-2
{A}^{-1}=\left[\begin{array}{cc}-1& -1\\ 2& \frac{5}{3}\end{array}\right]
The matrix in Hessenberg form with the help of similarity transformation and the matrix in the similarity transformations.
Find k such that the following matrix M is singular.
M=\left[\begin{array}{ccc}-1& -1& -2\\ 0& -1& -4\\ -12+k& -2& -2\end{array}\right]
k=?
If we have a matrices
A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\text{ and }{e}_{12}\left(\lambda \right)=\left[\begin{array}{cc}1& \lambda \\ 0& 1\end{array}\right]
then by doing product
A{e}_{12}\left(\lambda \right)=\left[\begin{array}{cc}a& a\lambda +b\\ c& c\lambda +d\end{array}\right]\text{ and }{e}_{12}\left(\lambda \right)A=\left[\begin{array}{cc}a+c\lambda & b+d\lambda \\ c& d\end{array}\right]
we can interpret that right multiplication by
{e}_{12}
to A gives a column-operation: add
\lambda
-times first column to the second column. In similar way, left multiplication by
{e}_{12}\left(\lambda \right)
to A gives row-operation on A.
Is there any conceptual (not computational, if any) way to see that elementary row and column operations on a matrix can be expressed as multiplication by elementary matrices on left or right, accordingly?
|
Autoregressive power spectral density estimate — covariance method - MATLAB pcov - MathWorks Australia
Covariance-Method PSD Estimate of an AR(4) Process
Covariance-Method PSD Estimate of a Multichannel Signal
Autoregressive power spectral density estimate — covariance method
pxx = pcov(x,order) returns the power spectral density (PSD) estimate, pxx, of a discrete-time signal, x, found using the covariance method. When x is a vector, it is treated as a single channel. When x is a matrix, the PSD is computed independently for each column and stored in the corresponding column of pxx. pxx is the distribution of power per unit frequency. The frequency is expressed in units of rad/sample. order is the order of the autoregressive (AR) model used to produce the PSD estimate.
pxx = pcov(x,order,nfft) uses nfft points in the discrete Fourier transform (DFT). For real x, pxx has length (nfft/2+1) if nfft is even, and (nfft+1)/2 if nfft is odd. For complex–valued x, pxx always has length nfft. If you omit nfft, or specify it as empty, then pcov uses a default DFT length of 256.
[pxx,w] = pcov(___) returns the vector of normalized angular frequencies, w, at which the PSD is estimated. w has units of radians/sample. For real-valued signals, w spans the interval [0, π] when nfft is even and [0,π) when nfft is odd. For complex–valued signals, w always spans the interval [0,2π].
[pxx,f] = pcov(___,fs) returns a frequency vector, f, in cycles per unit time. The sampling frequency, fs, is the number of samples per unit time. If the unit of time is seconds, then f is in cycles/second (Hz). For real-valued signals, f spans the interval [0,fs/2] when nfft is even and [0,fs/2) when nfft is odd. For complex-valued signals, f spans the interval [0,fs).
[pxx,w] = pcov(x,order,w) returns the two-sided AR PSD estimates at the normalized frequencies specified in the vector, w. The vector w must contain at least two elements, because otherwise the function interprets it as nfft.
[pxx,f] = pcov(x,order,f,fs) returns the two-sided AR PSD estimates at the frequencies specified in the vector, f. The vector f must contain at least two elements, because otherwise the function interprets it as nfft. The frequencies in f are in cycles per unit time. The sampling frequency, fs, is the number of samples per unit time. If the unit of time is seconds, then f is in cycles/second (Hz).
[___] = pcov(x,order,___,freqrange) returns the AR PSD estimate over the frequency range specified by freqrange. Valid options for freqrange are: 'onesided', 'twosided', or 'centered'.
[___,pxxc] = pcov(___,'ConfidenceLevel',probability) returns the probability × 100% confidence intervals for the PSD estimate in pxxc.
pcov(___) with no output arguments plots the AR PSD estimate in dB per unit frequency in the current figure window.
Create a realization of an AR(4) wide-sense stationary random process. Estimate the PSD using the covariance method. Compare the PSD estimate based on a single realization to the true PSD of the random process.
Create a realization of the AR(4) random process. Set the random number generator to the default settings for reproducible results. The realization is 1000 samples in length. Assume a sampling frequency of 1 Hz. Use pcov to estimate the PSD for a 4th-order process. Compare the PSD estimate with the true PSD.
N\left(0,1\right)
Estimate the PSD of the signal using the covariance method with a 12th-order autoregressive model. Use the default DFT length. Plot the estimate.
|
Pogrebkov A.K. Negative numbers of times of integrable hierarchies (abstract) - Geometry of Differential Equations
Pogrebkov A.K. Negative numbers of times of integrable hierarchies (abstract)
Speaker: Andrei Pogrebkov
Title: Negative numbers of times of integrable hierarchies
Time evolutions of the dressing operators of the integrable hierarchies, like Kadomtsev-Petviashvili or Davey-Stewartson, are given by linear differential operators. In the standard situation these operators result from dressing of positive powers of a
{\displaystyle \partial _{x}}
. It is natural to call these semiinfinite hierarchies of integrable (2+1)-dimensional equations as hierarchies with positive numbers of times. Here we develop hierarchies directed to negative numbers of times. Derivation of such systems, as well as of the corresponding hierarchies, is based on the commutator identities. This approach enables introduction of linear differential equations that admit lift up to nonlinear integrable ones by means of the special dressing procedure. Thus one can construct not only nonlinear equations but corresponding Lax pairs, as well. Lax operator of such evolutions coincide with the Lax operator of the "positive" hierarchy. We also derive (1+1)-dimensional reductions of equations of such hierarchies.
Retrieved from "https://gdeq.org/w/index.php?title=Pogrebkov_A.K._Negative_numbers_of_times_of_integrable_hierarchies_(abstract)&oldid=6498"
|
avalg10o 2022-03-25 Answered
Find a particular solution to the differential equation.
y{}^{″}-{y}^{\prime }-6y=\mathrm{sin}t+3\mathrm{cos}t
Please find laplace and inverse Laplace
Dumaen80p3 2022-03-24 Answered
Solve the following second order nonhomogeneous linear differential equations (where no initial values are given, find the general solution).
y{}^{″}-2{y}^{\prime }+y=-8\mathrm{cos}\left(x\right)+8\mathrm{sin}\left(x\right)
y{}^{″}-15{y}^{\prime }+56y=0
ideklaraz7xz 2022-03-24 Answered
Consider the following linear second order homogeneous differential equation wit constant coefficients and two initial conditions
\frac{{d}^{2}y\left(t\right)}{{dt}^{2}}+\frac{dy\left(t\right)}{dt}-2y\left(t\right)=0,\text{ }y\left(0\right)=-1\text{ }\frac{dy\left(0\right)}{dt}=5
Consider the following second-order nonhomogeneous linear differential equation. Use the method of undetermined coefficients to find a particular solution for each equation. Then solve each equation for real general solution.
y{}^{″}-4{y}^{\prime }+3y=\left(2-12x\right){e}^{x}+\left(60+40x\right)\left({e}^{3}x\right)
y{}^{″}+4{y}^{\prime }-y=0
ezpimpin6988ok1n 2022-03-24 Answered
Determine a unique solution of the separable differential equa- tion that satisfies the given initial condition.
\frac{{d}^{2}y}{{dt}^{2}}=\frac{dy}{dt}\frac{dy}{dt}\text{ }\left(0\right)=2,\text{ }y\left(0\right)=3
cleffavw8 2022-03-24 Answered
y{}^{″}-9{y}^{\prime }+14y=0
annlanw09y 2022-03-24 Answered
For what value of k, if any, will
y=k{e}^{-2x}+4\mathrm{cos}\left(3x\right)
be a solution to the differential equation
y+9y=26{e}^{-2x}
Consider the following second-order linear homogeneous difference equation with constant coefficients and two initial conditions:
{u}_{n+2}-6{u}_{n+1}+9{u}_{n}=0,\text{ }{u}_{0}=1,\text{ }{u}_{1}=9
Determine the sequence
{u}_{2},\text{ }{u}_{3},\text{ }{u}_{4},\text{ }{u}_{5}
. Solve the difference equation for
{u}_{n}
and use the result to check for
{u}_{5}
afasiask7xg 2022-03-23 Answered
y{}^{″}-5y=0
Aryan Salinas 2022-03-23 Answered
Solve the following second order Linear homogeneous /non-homogenous differential equation
y{}^{″}-4{y}^{\prime }-12y=3{e}^{5x}
jermandcryspza3 2022-03-23 Answered
Find the general solution y(x) of the following second order linear ODEs:
y{}^{″}-4{y}^{\prime }+3y=0
nastupnat0hh 2022-03-23 Answered
Solve the following Higher Order Non-homogenous DE using Methods of undetermined coefficients
y{}^{″}-2{y}^{\prime }+y=3x+2
monkeyman130yb 2022-03-23 Answered
y{}^{″}+9y=4\mathrm{cos}2x
Alexis Alexander 2022-03-23 Answered
y{}^{″}+5{y}^{\prime }-6y=0
Solute the equation:
\left(1-{x}^{2}\right)y{}^{″}=x{y}^{\prime }
Leroy Davidson 2022-03-23 Answered
y{}^{″}-{y}^{\prime }=\frac{{e}^{x}}{{e}^{x}+1}
Find the general solution of the give equation
3y{}^{″}-20{y}^{\prime }+12y=0
|
Lawrence Chiou, Lathan Liou, Patrick Corn, and
The covariance generalizes the concept of variance to multiple random variables. Instead of measuring the fluctuation of a single random variable, the covariance measures the fluctuation of two variables with each other.
Calculation of the Covariance
Recall that the variance is the mean squared deviation from the mean for a single random variable
X
\text{Var}(X) = E[\left(X - E[X]\right)^2].
The covariance adopts an analogous functional form.
\text{Cov}(X, Y)
of random variables
X
Y
\text{Cov}(X, Y) = E\left[(X - E[X])(Y - E[Y])\right].
Now, instead of measuring the fluctuation of a single variable, the covariance measures how two variables fluctuate together. For the covariance to be large, both
X - E[X]
Y - E[Y]
must be large at the same time or, in other words, change together.
It is generally simpler to find the covariance by taking
\begin{aligned} \text{Cov}(X, Y) &= E[XY - E[X] Y - X E[Y] + E[X] E[Y]] \\ &= \boxed{E[XY] - E[X] E[Y].} \end{aligned}
In other words, to compute the covariance, one can equivalently find
E[XY]
(in addition to the means of
X
Y
X
Y
be random variables such that
P(X = 0, Y = -1) = 1/5
P(X = 0, Y = 1) = 1/5
P(X = 1, Y = -1) = 1/2
P(X = 1, Y = 1) = 1/10
\text{Cov}(X, Y)
Similarly, one can find an expression in terms of variances:
\begin{aligned} \text{Var}(X + Y) &= E\left[(X + Y - E[X] - E[Y])^2\right] \\ &= E[\left(X - E[X]\right)^2] + E[\left(Y - E[Y]\right)^2] + 2 E\left[(X - E[X])(Y - E[Y])\right] \\ &= \boxed{\text{Var}(X) + \text{Var}(Y) + 2 \text{Cov}(X, Y).} \end{aligned}
A generalized statement of this result is as follows.
Variance of a sum. Given random variables
X_i
, each with finite variance,
\text{Var}\left( \sum_i X_i \right) = \sum_i \ \text{Var}(X_i) + 2 \sum_{i<j} \text{Cov}(X_i, X_j).
The covariance inherits many of the same properties as the inner product from linear algebra. The proof involves straightforward algebra and is left as an exercise for the reader.
Given a constant
and random variables
X
Y
Z
\text{Cov}(X, X) = \text{Var}(X) \geq 0
\text{Cov}(X, Y) = \text{Cov}(Y, X)
\text{Cov}(aX, Y) = a \text{Cov}(X, Y)
\text{Cov}(X, a) = 0
\text{Cov}(X + Y, Z) = \text{Cov}(X, Z) + \text{Cov}(Y, Z)
I, II, III, and IV I, III, and IV only I only I and II only I, II, and IV only
Given knowledge of
\text{Cov}(W, Y)
\text{Cov}(W, Z)
\text{Cov}(X, Y)
\text{Cov}(X, Z)
, which of the following can necessarily be computed?
\text{Cov}(W + X, Y + Z)
\text{Cov}(Y + Z, W + X)
\text{Cov}(W, X + Y + Z)
\text{Cov}(W, X + Y + Z)
, if it known that
W
X
X
Y
\text{Var}(X) = \sigma^2
Y = aX
\sigma
are constants. Determine
\text{Cov}(X, Y)
The inner product properties yield
\text{Cov}(X, Y) = \text{Cov}(X, aX) = \text{Cov}(aX, X) = a\text{Cov}(X, X) = a \sigma^2.
X
is a standard normal random variable and
Y = 3X
\text{Cov}(X, Y)
As a result, the Cauchy-Schwarz inequality holds for covariances.
Cauchy-Schwarz inequality. Given random variables
X
Y
\left[ \text{Cov}(X ,Y) \right]^2 \leq \text{Var}(X) \text{Var}(Y).
One of the key properties of the covariance is the fact that independent random variables have zero covariance.
Covariance of independent variables. If
X
Y
\text{Cov}(X, Y) = 0.
X
Y
E[XY] = E[X] E[Y]
\text{Cov}(X, Y) = 0
E[XY] = E[X] E[Y]
is a simple consequence of the fact that
P(X | Y) = P(X)
Dependent variables with zero covariance. However, the converse is not in general true. As a simple example, suppose that
X
is a standard normal random variable and that
Y = X^2
. Notice that knowledge of
X
Y
X
Y
are very clearly dependent. However, by symmetry it holds that
\text{Cov}(X, Y) = E[XY] - E[X] E[Y] = 0.
A simple corollary is as follows.
Variance of the sum of independent variables. Given independent random variables
X_i
\text{Var}\left( \sum_i X_i \right) = \sum_i \ \text{Var}(X_i).
X_i
are independent, it must be the case that
\text{Cov}(X_i, X_j) = 0
i \neq j
, and the result follows directly from the variance of a sum theorem.
When dealing with a large number of random variables
X_i
, it makes sense to consider a covariance matrix whose
m,n
th entry is
\text{Cov}(X_m, X_n)
\text{Cov}(X, Y) = \text{Cov}(Y, X)
, the covariance matrix is symmetric.
[1] DeGroot, Morris H. Probability and Statistics. Second edition. Addison-Wesley, 1985.
Cite as: Covariance. Brilliant.org. Retrieved from https://brilliant.org/wiki/covariance/
|
TARA: Training and Representation Alteration for AI Fairness and Domain Generalization | Neural Computation | MIT Press
Johns Hopkins University Applied Physics Laboratory Laurel, MD 20723, U.S.A. william.paul@jhuapl.edu
Armin Hadzic,
Johns Hopkins University Applied Physics Laboratory Laurel, MD 20723, U.S.A. armin.hadzic@jhuapl.edu
Johns Hopkins University Applied Physics Laboratory Laurel, MD 20723, U.S.A. neil.joshi@jhuapl.edu
Fady Alajaji,
Department of Mathematics and Statistics, Queens University, ON K7L 3N6, Canada fa@queensu.ca
Johns Hopkins University Applied Physics Laboratory Laurel, MD 20723, U.S.A.
Department of Computer Science, Johns Hopkins University, Baltimore, MD 21218, U.S.A. pburlin2@jh.edu
William Paul, Armin Hadzic, Neil Joshi, Fady Alajaji, Philippe Burlina; TARA: Training and Representation Alteration for AI Fairness and Domain Generalization. Neural Comput 2022; 34 (3): 716–753. doi: https://doi.org/10.1162/neco_a_01468
We propose a novel method for enforcing AI fairness with respect to protected or sensitive factors. This method uses a dual strategy performing training and representation alteration (TARA) for the mitigation of prominent causes of AI bias. It includes the use of representation learning alteration via adversarial independence to suppress the bias-inducing dependence of the data representation from protected factors and training set alteration via intelligent augmentation to address bias-causing data imbalance by using generative models that allow the fine control of sensitive factors related to underrepresented populations via domain adaptation and latent space manipulation. When testing our methods on image analytics, experiments demonstrate that TARA significantly or fully debiases baseline models while outperforming competing debiasing methods that have the same amount of information—for example, with (% overall accuracy, % accuracy gap)
=
(78.8, 0.5) versus the baseline method's score of (71.8, 10.5) for Eye-PACS, and (73.7, 11.8) versus (69.1, 21.7) for CelebA. Furthermore, recognizing certain limitations in current metrics used for assessing debiasing performance, we propose novel conjunctive debiasing metrics. Our experiments also demonstrate the ability of these novel metrics in assessing the Pareto efficiency of the proposed methods.
African Sculpture from the Tara Collection
Taras su tre dimensioni by Teresa Rampazzi: Documenting the Creative Process
|
Sharygin G. Quasi-differential operators on universal enveloping algebras and their applications (abstract) - Geometry of Differential Equations
Sharygin G. Quasi-differential operators on universal enveloping algebras and their applications (abstract)
Speaker: Georgy Sharygin
Title: Quasi-differential operators on universal enveloping algebras and their applications
In my talk I will describe a family of operators on the universal enveloping algebras, first of all on
{\displaystyle Ugl_{n}}
, which were first introduced by Gourevich and Saponov. We will discuss their properties, alternative definitions and relation with the algebra of differential operators on the corresponding Lie group. We shall also speculate on the possible applications of these operators to Vinberg's question to describe argument shift subalgebras in the universal enveloping algebras.
Retrieved from "https://gdeq.org/w/index.php?title=Sharygin_G._Quasi-differential_operators_on_universal_enveloping_algebras_and_their_applications_(abstract)&oldid=6474"
|
Socle of socle of module is the socle of the
slanglyn3u2 2022-04-25 Answered
Socle of socle of module is the socle of the module
Soc\left(Soc\left(M\right)\right)=Soc\left(M\right)
I write SM for the soccel of M. It is clear that
SSM\subset SM
x\in SM
. The element x may be written as a finite linear combination
x={y}_{1}+\cdots +{y}_{n}
{y}_{r}
is contained in a simple submodule
{V}_{r}
of M. But each
{V}_{r}
is also a simple submodule of SM, so in particular
{y}_{r}
is contained in SSM. Since SSM is a submodule, it follows that x is contained in it.
R=\frac{K\left[X,Y\right]}{\left(XY\right)}.F\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}P\in K\left[X,Y\right]
\left[P\right]
If H and K are subgroups of G,
|H|=20\text{ }\text{and}\text{ }|K|=32
then a possible value of
|H\cap K|
{x}^{6}-6{x}^{4}+12{x}^{2}-11
\mathbb{Q}
{M}_{2×2}\left(\mathbb{C}\right)
In an abstract algebra equation about groups, is "taking the inverse of both sides of an equation" an acceptable operation? I know you can right/left multiply equations by elements of the group, but was wondering if one can just take the inverse of both sides?
{x}^{2}=e
|
MMDCCVIII446 2022-02-15 Answered
Assume that A and B are events in a sample space and that
P\left(A\right)=.40\text{ }\text{and}\text{ }P\left(\frac{B}{A}\right)=.25
. What is the probability that both A and B occur?
Assume the random variable X has a binomial distribution with the given probability of obtaining a success. Find the following probability, given the number of trials and the probability of obtaining a success. Round your answer to four decimal places.
P(X>1), n=4, p=0.6
Cierra Woodard 2022-02-15 Answered
P\left(A\right)=0.7,P\left(B\right)=0.1,\text{and}\text{ }P\left(A\cap B\right)=0,\text{what is}\text{ }P\left(B\mid A\right)
P\left(A\mid B\right)
Joslyn Reese 2022-02-15 Answered
P\left(A\right)=0.7,P\left(B\right)=0.9\text{ }\text{and}\text{ }P\left(A\cap B\right)=0.4
P\left(B\mid A\right)?
P\left(A\mid B\right)
Reina Pennington 2022-02-15 Answered
What is the mathematical definition of conditional probability?
jkminzeszjt 2022-02-15 Answered
P\left(A\right)=0.3\text{ }\text{and}\text{ }P\left(B\right)=0.25\text{ }\text{and}P\left(A\cap B\right)=0.1
P\left(B\mid A\left(\text{complement}\right)\right)
Dangelo Le 2022-02-15 Answered
Two events A and B are such that
P\left(A\right)=0.2,P\left(B\right)=0.3,\text{and}\text{ }P\left(A\cup B\right)=0.4
P\left(A\cap B\right)
Zouridt6 2022-02-15 Answered
P\left(A\mid B\right)<P\left(A\right)
Of 100 gumdrops 73 are large, 39 of them are red, and 19 are large and red. If you choose a random gumdrop, what is the probability that it is red? If the gumdrop is red, what is the probability that it is large?
Sergio Bradshaw 2022-02-15 Answered
l2yanhatal0n 2022-02-14 Answered
An instructor randomly selects one of the pianists to play first part of the duet from a group of 5 pianists that includes 2 boys and 3 girls. What is the probability of selecting a boy for first part of duet and a girl for the second part of the duet? How the probability changes if second pianist is different?
Neil Wiley 2022-02-14 Answered
Israel Holden 2022-02-14 Answered
kaibaloveyou3cj 2022-02-14 Answered
n\left(A\cap B\right)
n\left(A\right)=7,n\left(B\right)=9,
n\left(A\cup B\right)=13
Jameson Powers 2022-02-14 Answered
When drawing two hearts from a deck without replacement, are the events independent?
Lawson Cain 2022-02-14 Answered
Among the contestants in a competition are 42 women and 28 men. If 5 winners are randomly selected, what is the probability that they are all men?
Vagotonusybk 2022-02-14 Answered
There are two boxes - one holds 11 cards numbered 1 through 11, the other 5 cards numbered 1 through 5. Two cards are randomly drawn. What is the probability of drawing 2 even cards in a row?
Misael Sweeney 2022-02-14 Answered
90 students will graduate from Lima Shawness High School this year. Out of the 90, 50 are planning to attend college. Two students are selected at to carry the flag at graduation. What is the probability that both of them are planning to attend college?
A bowl contains 7 pennies, 9 nickels, and 4 dimes. Elyse removes one coin at random from the bowl and does not replace it. She then removes a second coin at random. What is the probability that both will be dimes?
There are three urns containing black, red and green balls. urn 1 contains 4 black, 5 red and 3 green balls urn 2 contains 7 black, 4 red and 4 green balls urn three contains 6 black, 6 red and 5 green balls. An urn is chosen and ball is picked if the ball picked is green find the probability it has come from urn 3.
|
m (→Deriving Physical Quantities: bash scripted conversions from to DN to Radiances & Reflectances)
m (→Automatising Conversions: minor typos)
The conversion process ca be scripted to avoid repeating the same steps for each band separately. In bash, such a script might be as the following example. '''Note,''' however, in this example script constants, band parameters and acquisition related metadata are hard-coded!
# echo something?
{\displaystyle {\frac {W}{m^{2}*sr*nm}}}
{\displaystyle L\lambda ={\frac {10^{4}*DN\lambda }{CalCoef\lambda *Bandwidth\lambda }}}
{\displaystyle \rho _{p}={\frac {\pi *L\lambda *d^{2}}{ESUN\lambda *cos(\Theta _{S})}}}
{\displaystyle \rho }
{\displaystyle \pi }
{\displaystyle L\lambda }
{\displaystyle d}
{\displaystyle Esun}
{\displaystyle cos(\theta _{s})}
{\displaystyle {\frac {W}{m^{2}*\mu m}}}
{\displaystyle [0,255]}
{\displaystyle [0,2047]}
|
f\left(t\right)=t{e}^{-t}\mathrm{sin}\left(2t\right)
{x}^{\ast }\left(s\right)=\left(\frac{1}{\left(s+{\mu }_{1}+{\mu }_{2}\right)\left(s+{\stackrel{^}{\lambda }}_{2}\right)\left(s+{\lambda }_{1}+{\lambda }_{2}\right)}\right)
{\mu }_{1},{\mu }_{2},{\lambda }_{1},{\lambda }_{2},{\stackrel{^}{\lambda }}_{2}
x\left(t\right)={\mathcal{L}}^{-1}\left({x}^{\ast }\left(s\right)\right)
f\left(t\right)=\frac{\mathrm{sin}\left(2t\right)}{{e}^{2t}}+t·u\left(t-4\right)
\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
\frac{\mathrm{d}y}{\mathrm{d}x}+P\left(x\right)y=Q\left(x\right)
\frac{du\left(t\right)}{dt}=\left[t{u}_{2}\left(t\right);4{u}_{1}\left(t{\right)}^{3/2}\right]
\mathrm{sin}\left(-0.5\right)
For every differentiable function
f:\mathbb{R}\to \mathbb{R}
g:\mathbb{R}×\mathbb{R}\to \mathbb{R}
g\left(f\left(x\right),{f}^{\prime }\left(x\right)\right)=0
x
and for every differentiable function
h:\mathbb{R}\to \mathbb{R}
holds that
being true that for every
x\in \mathbb{R}
g\left(h\left(x\right),{h}^{\prime }\left(x\right)\right)=0
h\left(0\right)=f\left(0\right)
h\left(x\right)=f\left(x\right)
x\in \mathbb{R}
i.e every differentiable function
f
is a solution to some first order differential equation that has translation symmetry.
x\frac{dy}{dx}+y={y}^{-2}
To solve this equation, we reduce it to a linear differential equation with the substitution
v={y}^{3}
\begin{array}{rl}x{y}^{2}\frac{dy}{dx}+{y}^{3}& =1\\ \frac{dv}{dx}& =3{y}^{2}\frac{dy}{dx}\\ 3x{y}^{2}\frac{dy}{dx}+3{y}^{3}& =3\\ \frac{dv}{dx}+3v& =3\end{array}
Now we have a first order linear differential equation. To solve it, we use the integrating factor
I={e}^{\int P\left(x\right)}
. In this case, we have P(x)=3.
\begin{array}{rl}I\left(x\right)& ={e}^{3x}\\ {e}^{3x}\frac{dv}{dx}+3{e}^{3x}v& =3{e}^{3x}\\ D\left({e}^{3x}v\right)& =3{e}^{3x}\\ {e}^{3x}v& ={e}^{3x}+C\\ v& =1+{e}^{-3x}\\ {y}^{3}& =1+{e}^{-3x}\end{array}
Now to check my answer.
\begin{array}{rl}3{y}^{2}\frac{dy}{dx}& =-3C{e}^{-3x}\\ {y}^{2}\frac{dy}{dx}& =-C{e}^{-3x}\\ \frac{dy}{dx}& =-C{e}^{-3x}{y}^{-2}\\ x\frac{dy}{dx}+y& =x\left(-C{e}^{-3x}{y}^{-2}\right)+{\left(1+{e}^{-3x}\right)}^{\frac{1}{3}}\end{array}
I cannot seem to get my answer to check. Where did I go wrong?
Here is my second attempt to solve the problem:
v={y}^{3}
\begin{array}{rl}x{y}^{2}\frac{dy}{dx}+{y}^{3}& =1\\ \frac{dv}{dx}& =3{y}^{2}\frac{dy}{dx}\\ 3x{y}^{2}\frac{dy}{dx}+3{y}^{3}& =3\\ x\frac{dv}{dx}+3v& =3\\ \frac{dv}{dx}+3{x}^{-1}v& =3{x}^{-1}\end{array}
I={e}^{\int P\left(x\right)}
P\left(x\right)=3{x}^{-1}
\begin{array}{rl}I& ={e}^{3\int {x}^{-1}\phantom{\rule{thinmathspace}{0ex}}dx}={e}^{3\mathrm{ln}|x|}\\ I& =3x\\ 3x\frac{dv}{dx}+9v& =9\end{array}
Now, I want to write:
D\left(3xv\right)=9
but that is wrong. What did I do wrong?
Here is my third attempt to solve the problem. Last time, I made a mistake in finding the integrating factor.
v={y}^{3}
\begin{array}{rl}x{y}^{2}\frac{dy}{dx}+{y}^{3}& =1\\ \frac{dv}{dx}& =3{y}^{2}\frac{dy}{dx}\\ 3x{y}^{2}\frac{dy}{dx}+3{y}^{3}& =3\\ x\frac{dv}{dx}+3v& =3\\ \frac{dv}{dx}+3{x}^{-1}v& =3{x}^{-1}\end{array}
I={e}^{\int P\left(x\right)}
P\left(x\right)=3{x}^{-1}
\begin{array}{rl}I& ={e}^{3\int {x}^{-1}\phantom{\rule{thinmathspace}{0ex}}dx}={e}^{3\mathrm{ln}|x|}\\ I& ={x}^{3}\\ {x}^{3}\frac{dv}{dx}+3{x}^{2}v& =3{x}^{2}\\ D\left({x}^{3}v\right)& ={x}^{3}+C\\ {x}^{3}v& ={x}^{3}+C\\ v& =C{x}^{-3}+1\\ {y}^{3}& =C{x}^{-3}+1\end{array}
y{}^{″}+y=2t
y\left(\frac{\pi }{4}\right)=\frac{\pi }{2}
{y}^{\prime }\left(\frac{\pi }{4}\right)=2-\sqrt{2}
\frac{\mathrm{cos}t}{t}
\frac{S-4}{{S}^{2}-2S-11}
\mathrm{tan}\left(\frac{\pi }{4}+0.05\right)
dy
and
{s}^{-\frac{3}{2}}
\frac{2}{\sqrt{\pi }}\frac{\sqrt{\pi }}{2{s}^{\frac{3}{2}}}=2\sqrt{\frac{t}{\pi }}
\frac{2}{\sqrt{\pi }}
\mathrm{\Gamma }\left(\frac{3}{2}\right)
{L}^{-1}\left\{0\right\}=0
find the solution of the differential equation that satisfies the given initial condition
xy'+y=y^2 , y(1)=-1
Erik Cantu 2022-04-05 Answered
{x}^{\prime }\left(t\right)+{\int }_{\left\{0\right\}}^{t}\left(t-s\right)x\left(s\right)ds=t+\frac{1}{2}{t}^{2}+\frac{1}{24}{t}^{4}
y{}^{″}\left(t\right)+12{y}^{\prime }\left(t\right)+32y\left(t\right)=32u\left(t\right)
y\left(0\right)={y}^{\prime }\left(0\right)=0
Y\left(p\right)=x=\frac{32}{\left(p\left({p}^{2}+12p+32\right)\right)}
\frac{1}{{s}^{2}-9s+20}
{L}^{-1}\left\{\frac{1}{{\left(s-\frac{9}{2}\right)}^{2}-{\left(\frac{1}{2}\right)}^{2}}\right\}=2{L}^{-1}\left\{\frac{\frac{1}{2}}{{\left(s-\frac{9}{2}\right)}^{2}-{\left(\frac{1}{2}\right)}^{2}}\right\}
y{}^{‴}+9{y}^{\prime }=\frac{1}{f\left(x\right)}
f\left(x\right)=\mathrm{cos}11x
|
GDP - Vocabulary - Course Hero
Macroeconomics/GDP/Vocabulary
price set to a specific "constant" to adjust for inflation when calculating GDP
price at the time of calculation when calculating GDP
a number that represents the current prices of various goods and services versus their past prices of a given year
a measure of the total output of a country divided by the number of people in the country
\text{GDP per}\;\text{Capita}=\frac{\text{GDP}}{\text{Population}}
the market value of all the finished goods and services produced by a country's factors of production regardless of where the factors of production are located
good or service used in the production of other goods and services
spending on business capital, residential capital, and inventories
the total value of a country's exports minus the value of the country's imports
situation in which a nation imports more than it exports
situation in which a nation exports more than it imports
<Overview>Defining and Measuring GDP
|
discrete math functions pigeonhole principle How many numbers in
Dylan Yoder 2022-04-14 Answered
\left\{1,2,3,\dots ,346\right\}
szalbierzfytg
In the given problem we have to find how many numbers in the set
\left\{1,2,3,\dots ,346\right\}
are divisible by 5 or 7.
Denote A=
\left\{1,2,3,\dots ,346\right\}
{A}_{5}
be the subset of A containing those numbers which are divisible by 5 and
{A}_{7}
be the subset of A containing those numbers which are divisible by 7.
{A}_{5}\cap {A}_{7}
contain those numbers of A which are divisible by both 5 and 7.
And a number is divisible by both 5 and 7 if it is divisible by their least common multiple and
\text{lcm}\left(5,7\right)=35
Therefore a number is divisible by both 5 and 7 if it is divisible by 35.
{A}_{5}\cup {A}_{7}
contain those numbers of A which are divisible by both 5 or 7.
n\left({A}_{5}\right)=\left[\frac{346}{5}\right]=69
where [] is the greatest integer function.
n\left({A}_{7}\right)=\left[\frac{346}{7}\right]=49
n\left({A}_{5}\cap {A}_{7}\right)=\left[\frac{346}{35}\right]=9
n\left({A}_{5}\cup {A}_{7}\right)=n\left({A}_{5}\right)+\left({A}_{7}\right)-n\left({A}_{5}\cap {A}_{7}\right)
⇒n\left({A}_{5}\cup {A}_{7}\right)=69+49-9
⇒n\left({A}_{5}\cup {A}_{7}\right)=109
Hence there are 109 numbers in the set
\left\{1,2,3,\dots ,346\right\}
So the question I'm trying to answer looks like this:
We are interested in p, the population proportion of all people who are currently happy with their cell phone plans. In a small study done in 2012, it was found that in a sample of 150 people, there were 90 who were happy with their cell phone plans.
Find the upper confidence limit of an 82% confidence interval for p.
I know the formula is
\mathrm{estimate}±\left(\mathrm{criticalvalue}\right)\left(\mathrm{standard} \mathrm{error}\right)
How many injective functions are there from a set with 3^10 elements to a set with 2^20 elements
A system for a random amount of time X (in units of months) is given by a density ;
f\left(x\right) = \frac{1}{4}x{e}^{-\frac{x}{2}} ; x > 0\phantom{\rule{0ex}{0ex}} 0; x\le 0
(a) Find the moment generating function of X . Hence, compute the variance of X .
(b) Deduce the expression for the k th moment.
(c)Obtain the distribution function of X . Hence, compute that the probability that, 7 of such system, at least 4 will function for at least 6 units of months. State the assumptions that you make.
Given that the polynomial f(x) has 8 x-intercepts, which of the following most accurately describes the degree of the polynomial f(x)?
\left(-7,2\right)
. The function passes through the point \left(8,3\right)
f\left(x\right)=a{\left(x-h\right)}^{2}+k
|
Inductive Effect, Electromeric Effect, Resonance Effects, and Hyperconjugation | Brilliant Math & Science Wiki
Gautam Sharma, Jordan Calmes, Anandhu Raj, and
Electronic factors that influence organic reactions include the inductive effect, electromeric effect, resonance effects, and hyperconjugation. These electronic factors involve organic molecules, most of which are made from a combination of the following six elements: carbon, hydrogen, nitrogen, oxygen, phosphorus, and sulfur (known collectively as CHNOPS). Yet, the limited number of building blocks does not prevent organic compounds from taking on diverse properties in their physical characteristics and chemical reactivity. The subtle differentiation of various compounds in organic chemistry is essential for the biological functions of the molecules and creates a wide variety of reactions.
Part of this variety in organic chemistry stems from differences in electron behavior when elements other than carbon and hydrogen participate in molecular bonds. For example, the three compounds pictured above have similar formula units and structures, but react very differently from one another because of these electronic factors. Varying electronegativity can cause delocalization effects, where the electron cloud for a given bond expands to more than two atoms within the molecule.
Polarity of Organic Molecules
Examples of Electronic Effects
Partial polarity within a molecule leads to electron transfer among the atoms in a molecule, leading to different behavior than what would be expected in a non-polar version of the compound, where no sections were electron-rich or electron-deficient.
Saturated hydrocarbons are nonreactive because there is no polarity in C-C bond and practically no polarity in C-H bonds. Carbon and hydrogen are almost identical in electronegativity, so the electrons involved in a bond between the two atoms are equally attracted to each nucleus and spend roughly the same amount of time orbiting one as the other.
Electron density is evenly distributed between the two atoms in a non-polar bond, which prevents charged species from attacking or altering the bond. In contrast, charged species (electrophiles and nucleophiles) react with polar organic molecules because there is an imbalance in electron density or polarity.
Elements with higher electronegativity, including oxygen and the halide group, can change the electron density around an organic molecule and make the molecule more reactive.
Electronic effects complicate chemical reactions, and they can stabilize a molecule, make a compound less volatile, make a molecule more likely to react in a desired fashion, or affect the acidity or basicity. Understanding the factors involved in electronic imbalance is vital for understanding the underlying mechanisms of a chemical reaction, predicting the products of a reaction and predicting organic molecules' behavior.
The inductive effect is a permanent state of polarization. The electron density in a
\sigma
bond between two unlike atoms is not uniform. The electron density is more dense toward the more electronegative of the two atoms.
The inductive effect is a distance-dependent phenomenon:
C^{\delta+}-X^{\delta-}
X
above acquires a slightly negative charge
(\delta-),
and the carbon atom a slightly positive charge
(\delta+),
which means the bond is polarized:
If the electronegative atom
X
is connected to a chain of carbon atoms, then the positive charge is relayed to the other carbon atoms.
C_{1}
, with its positive
\delta
charge, exerts a pull on the electrons of
C_2
, but the pull is weaker than it is between
X
C_1
. The effect rapidly dies out and is usually not significant after the
2^\text{nd}
carbon atom, or at most the
3^\text{rd}.
The inductive effect is permanent, but relatively weak, and can be easily overshadowed by the electronic effects discussed later.
There are two categories of inductive effects: the electron-withdrawing (-I effect) and the electron-releasing (+I effect). The latter is also called the electron-donating effect. In the image above,
X
is electron-withdrawing and
Y
is electron-donating.
These relative inductive effects are measured with reference to hydrogen:
\text{NO}_2> \text{COOH}> \text{F}> \text{Cl}> \text{Br}> \text{I}> \text{OR}> \text{OH}> \text{C}_6\text{H}_5\text{(Benzene)}> \textbf{H} > \text{Me}_3\text{C}^-> \text{Me}_2\text{CH}^-> {\text{MeCH}_2}^-> {\text{CH}_3}^-.
The -I effect is seen around a more electronegative atom or group, and electron density is higher there than elsewhere in the molecule. Electron-withdrawing groups include halogen, nitro
(-\text{NO}_2),
cyano
(-\text{CN}),
(-\text{COOH}),
(-\text{COOR}),
and aryloxy
(-\text{OAr}).
The +I effect is observed among the less electronegative atoms of the molecule by electron-releasing (or electron-donating) groups. The alkyl groups are usually considered electron-releasing (or electron-donating) groups.
Sometimes, there are several correct Lewis structures for a given molecule. Ozone
(O_3)
is one example. The compound is a chain of three oxygen atoms, and minimizing the charges while giving each atom an octet of electrons requires that the central oxygen atom form a single bond with one terminal oxygen and a double bond with the other terminal oxygen.
When drawing the Lewis structure, the choice of placement for the double bond is arbitrary, and either choice is equally correct. The multiple correct ways of drawing the Lewis structure are called the resonance forms.
Based on the resonance forms, a beginning chemistry student might wonder if ozone has bonds of two different lengths, since single bonds are generally longer than double bonds. However, the ozone molecule is perfectly symmetrical, with bonds that are the same length. None of the resonance forms represent the true structure of the molecule. Rather, the negative charge of the electrons that would form a double bond are delocalized, or distributed evenly across the three oxygen atoms. The true structure is a composite, with bonds shorter than what would be expected for single bonds, but longer than the expected double bonds.
The resonance hybrid for ozone is found by identifying the multiple resonance structures for the molecule.
{ O }_{ 3 }
the two structures (I and II) shown above constitute the canonical structures or resonance structures and their hybrid (i.e. the III structure) represents the structure of
{ O }_{ 3 }
more accurately. Resonance is represented by a double-headed arrow between the resonance structures, as illustrated above.
The resonance hybrid is more stable than its canonical forms, i.e. the actual compound (hybrid) is at a lower energy state than its canonical forms. Resonance stability increases with increased number of resonance structures.
The difference in the experimental and calculated energies is the amount of energy by which the compound is stable. This difference is known as resonance energy or delocalization energy.
All resonance structures are not equivalent. The following rules help determine whether or not a resonance structure will contribute significantly to the hybrid structure.
Rules of Resonance
Rule 1: The most significant resonance contributor has the greatest number of full octets (or if applicable, expanded octets).
Rule 2: The most significant resonance contributor has the fewest atoms with formal charges.
Rule 3: If formal charges cannot be avoided, the most significant resonance contributor has the negative formal charges on the most electronegative atoms, and the positive formal charges on the least electronegative atoms.
Rule 4: The most significant resonance contributor has the greatest number of covalent bonds.
Rule 5: If a pi bond is present, the most significant resonance contributor has this pi bond between atoms of the same row of the periodic table (usually carbon pi bonded to boron, carbon, nitrogen, oxygen, or fluorine).
Rule 6: Aromatic resonance contributors are more significant than resonance contributors that are not aromatic.
The permanent polarization of a group conjugated with a
\pi
bond or a set of alternate
\pi
bonds is transmitted through the
\pi
electrons of the system, resulting in a different distribution of electrons in the unsaturated chain. This kind of electron distribution in unsaturated compounds conjugated with electron-releasing or withdrawing groups or atoms is called mesomeric effect.
As shown above, a polarity is induced in compounds due to transfer of electrons through
\pi
bonds. This effect is a consequence of resonance and is seen in compounds that contain a double bond that is separated from another double bond or a lone pair of electrons by a single bond.
The electromeric effect is an intramolecular movement of electrons from a pi bond to another atom in the molecule due to attack by a reagent. It is temporary and reversible.
There are two distinct types of electromeric effects:
(i) Positive Electromeric Effect (+E effect): In this effect the
\pi
-electrons of the multiple bond are transferred to that atom to which the reagent gets attached. For example:
(ii) Negative Electromeric Effect (-E effect): In this effect the
\pi
-electrons of the multiple bomd are transferred to that atom to which the attacking reagents do not get attached. For example:
Hyperconjugation helps explain the stability of alkyl radicals. It involves the delocalization of
\sigma
-electrons belonging to the C-H bond of the alkyl group attaching to an atom with an unshared
p
orbital. The more the hyperconjugative hydrogen, the more is the stability.
Cite as: Inductive Effect, Electromeric Effect, Resonance Effects, and Hyperconjugation. Brilliant.org. Retrieved from https://brilliant.org/wiki/inductive-effect-electromeric-efffect-resonance/
|
m (→Pan-Sharpening: Section for the HPFA algorithm)
All of the above steps can be replicated in GRASS GIS. ''A script that implements the algorithm is currently developed and hopefully published as a grass add-on.''
{\displaystyle {\frac {W}{m^{2}*sr*nm}}}
{\displaystyle L\lambda ={\frac {10^{4}*DN\lambda }{CalCoef\lambda *Bandwidth\lambda }}}
{\displaystyle \rho _{p}={\frac {\pi *L\lambda *d^{2}}{ESUN\lambda *cos(\Theta _{S})}}}
{\displaystyle \rho }
{\displaystyle \pi }
{\displaystyle L\lambda }
{\displaystyle d}
{\displaystyle Esun}
{\displaystyle cos(\theta _{s})}
{\displaystyle {\frac {W}{m^{2}*\mu m}}}
{\displaystyle [0,255]}
{\displaystyle [0,2047]}
|
Trigonometric R method | Brilliant Math & Science Wiki
Aditya Virani, Alexander Katz, and Jimin Khim contributed
The trigonometric R method is a method of rewriting a weighted sum of sines and cosines as a single instance of sine (or cosine). This allows for easier analysis in many cases, as a single instance of a basic trigonometric function is often easier to work with than multiple are.
The R method is most often used to find the extrema (maximum and minimum) of combinations of trigonometric functions, since the extrema of a basic trigonometric function are easy to work with (both sine and cosine have a minimum of -1 and a maximum of 1).
Formal Statement and Proof
Using the R Method
Applications to Extrema
The R method can be used to convert an expression of the form
a\sin\theta+b\cos\theta
into a single instance of either sine or cosine:
The sine form:
a,b,\theta\in\mathbb{R},
the following corresponds:
a\sin\theta+b\cos\theta=R\sin(\theta+\alpha),
R=\sqrt{a^2+b^2}
\tan\alpha=\frac{b}{a},
\alpha=\arctan\frac{b}{a}.
The cosine form:
a,b,\theta\in\mathbb{R},
a\sin\theta+b\cos\theta=R\cos(\theta-\alpha),
R=\sqrt{a^2+b^2}
\tan\alpha=\frac{a}{b},
\alpha=\arctan\frac{a}{b}.
First, let's prove the sine form. Let
a=R\cos\alpha
b=R\sin\alpha.
Then the following equations correspond:
\begin{aligned} a^2+b^2&=R^2(\cos^2\alpha+\sin^2\alpha)\\&=R^2\\ \tan\alpha&=\frac{\sin\alpha}{\cos\alpha}\\&=\frac{b}{a}\\ a\sin\theta+b\cos\theta&=R\sin\theta\cos\alpha+R\cos\theta\sin\alpha\\&=R\sin(\theta+\alpha).\ _\square \end{aligned}
To prove the cosine form, let
a=R\sin\alpha
b=R\cos\alpha.
\begin{aligned} a^2+b^2&=R^2(\sin^2\alpha+\cos^2\alpha)=R^2\\ \tan\alpha&=\frac{\sin\alpha}{\cos\alpha}=\frac{a}{b}\\ a\sin\theta+b\cos\theta&=R\cos\theta\cos\alpha+R\sin\theta\sin\alpha=R\cos(\theta-\alpha).\ _\square \end{aligned}
Given the condition
R>0
0<\theta<\frac{\pi}{2},
what are the appropriate values of
R
\theta
\sin\frac{\pi}{8}+\cos\frac{\pi}{8}=R\sin\theta?
R^2=1^2+1^2=2
\arctan\frac{1}{1}=\frac{\pi}{4},
\sin\frac{\pi}{8}+\cos\frac{\pi}{8}=\sqrt{2}\sin\left(\frac{\pi}{8}+\frac{\pi}{4}\right)=\sqrt{2}\sin\frac{3}{8}\pi.\ _\square
R=\sqrt{2}
\theta=\frac{3}{8}\pi.\ _\square
\sin\frac{\pi}{6}+\sqrt{3}\cos\frac{\pi}{6}
using the R method.
(1) Using the sine form:
R^2=1^2+(\sqrt{3})^2=4
\arctan\sqrt{3}=\frac{\pi}{3},
\sin\frac{\pi}{6}+\sqrt{3}\cos\frac{\pi}{6}=2\sin\left(\frac{\pi}{6}+\frac{\pi}{3}\right)=2\sin\frac{\pi}{2}=2.\ _\square
(2) Using the cosine form:
R^2=1^2+(\sqrt{3})^2=4
\arctan\frac{1}{\sqrt{3}}=\frac{\pi}{6},
\sin\frac{\pi}{6}+\sqrt{3}\cos\frac{\pi}{6}=2\cos\left(\frac{\pi}{6}-\frac{\pi}{6}\right)=2\cos0=2.\ _\square
\sqrt{3}\cos \theta-\sin \theta=\frac{1}{3}
0 < \theta < \frac{\pi}{2},
\sqrt{3}\sin \theta+\cos \theta
\frac{\sqrt{a}}{3}
a?
x
\sqrt{3}\sin x-\cos x=\sqrt{2},\quad -\pi<x<\pi.
R^2=\big(\sqrt{3}\big)^2+(-1)^2=4
\arctan\frac{-1}{\sqrt{3}}=-\frac{\pi}{6},
\begin{aligned} \sqrt{3}\sin x-\cos x=2\sin\left(x-\frac{\pi}{6}\right)&=\sqrt{2}\\ \Rightarrow \sin\left(x-\frac{\pi}{6}\right)&=\frac{\sqrt{2}}{2}. \end{aligned}
x-\frac{\pi}{6}=\theta.
-\pi<x<\pi,
-\frac{7}{6}\pi<\theta<\frac{5}{6}\pi.
In this interval the values of
\theta
\sin\theta=\frac{\sqrt{2}}{2}
\frac{\pi}{4}
\frac{3}{4}\pi.
\begin{aligned} x-\frac{\pi}{6}=\frac{\pi}{4}&\implies x=\frac{5}{12}\pi\\ x-\frac{\pi}{6}=\frac{3}{4}\pi&\implies x=\frac{11}{12}\pi. \end{aligned}
x=\frac{5}{12}\pi, \frac{11}{12}\pi.
_\square
When using the R method, there are numerous possible values of
R
\alpha,
since sine and cosine are periodic functions. Adding
2n\pi
\alpha
n
would give the same answer. Also, adding
(2n-1)\pi
\alpha
n
and changing the sign of
R
would also be equivalent to the answer. However we most often use
-\frac{\pi}{2}<\alpha<\frac{\pi}{2}
R>0
for convenience.
The R method is frequently used to find the maximum or minimum of equations in the form of
a\sin\theta+b\cos\theta.
-1\leq\sin x\leq1
-1\leq\cos x\leq1
x,
so the maximum and minimum of
a\sin\theta+b\cos\theta
R
-R,
Find the sum of the maximum and minimum values of
3\sin\left(x-\frac{\pi}{3}\right)-4\cos\left(x-\frac{\pi}{3}\right)+2.
3^2+(-4)^2=5^2,
R=5.
-5\leq3\sin\left(x-\frac{\pi}{3}\right)-4\cos\left(x-\frac{\pi}{3}\right)\leq5\\ \Rightarrow -3\leq3\sin\left(x-\frac{\pi}{3}\right)-4\cos\left(x-\frac{\pi}{3}\right)+2\leq7.
-3+7=4.
_\square
If the maximum value of
7 \cos x+6 \sin x
\sqrt{a}
a
This problem is posed by Minimario M.
Cite as: Trigonometric R method. Brilliant.org. Retrieved from https://brilliant.org/wiki/trigonometric-r-method/
|
x+1\in \mathbb{Q}\frac{x}{{x}^{3}-2}
. Explain why this is the same as finding the inverse of
\sqrt[3]{2}\in \mathbb{R}
Prove that a group of even order must have an element of order 2.
Let a,b be coprime integers. Prove that every integer
x>ab-a-b
na+mb
where n,m are non negative integers. Prove that
ab-a-b
cannot be expressed in this form.
Let F be a field and consider the ring of polynominals in two variables over F,F[x,y]. Prove that the functions sending a polyomial f(x,y) to its degree in x, its degree in y, and its total degree (i.e, the highest
i+j
{x}^{i}{y}^{i}
appears with a nonzero coefficient) all fail o be norm making F[x,y] a Euclidean domain.
In the froup
{Z}_{12}
|a|,|b|
|a+b|
a=3,b=8
Use Principle of MI to verify
n\in \mathbb{Z}
is a positive ineger then
{2}^{n}{3}^{xn}-1
divisible by 17.
(ii) For all positive integers
n\ge 5
{2}^{k}>{k}^{2}
Let F be i field with subfields K,L. Prove that there is a largest subfield of F contained in both K and L, and a smallest subfield of containung both K and L.
How many subgroups of order 4 does
{D}_{4}
Suppose G is a group, H a subgroup of G, and a and b elements of G. If
a\in bH
b\in aH
How to find the value of U(n) on abstract algebra
Example U(8) = {1,3,5,7}
An nth root of unity epsilon is an element such that
{ϵ}^{n}=1
. We say that epsilon is primitive if every nth root of unity is
{ϵ}^{k}
for some k. Show that there are primitive nth roots of unity
{ϵ}_{n}\in \mathbb{C}
for all n, and find the degree of
\mathbb{Q}\to \mathbb{Q}\left({ϵ}_{n}\right)
1\le n\le 6
Find th eminimal polynomial of
\sqrt{2}+\sqrt{3}
\mathbb{Q}
Write down a definition of a subfield. Prove that the intersection of a set of subfields of a field F is again a field
Let F be a field. Prove that there are infinitely many irreducible monic polynomials
p,q\in \mathbb{Z}
be district primes. Prove that
\mathbb{Q}\left(\sqrt{p},\sqrt{q}\right)=\mathbb{Q}\left(\sqrt{p}+\sqrt{q}\right)
\mathbb{Q}\subseteq \mathbb{Q}\left(\sqrt{p}+\sqrt{q}\right)
is a degree 4 extension.
|
To calculate:To evaluate
\left(f\ast g\right)\left(70\right)\text{ }and\text{ }\left(g\ast f\right)\left(70\right)
Professor Harsh gave a test to his college algebra class and nobody got more than 80 points (out of 100) on the test.
One problem worth 8 points had insufficient data, so nobody could solve that problem.
The professor adjusted the grades for the class by
a. Increasing everyones
To calculate: The probability that the selected student of 17-26 age group has received an "A" in the course.
Find the x- and y-intercepts of the equation.
-5x+8y=80
f\left(x\right)={\mathrm{log}}_{2}x
Trust Fund A philanthropist deposits $5000 in a trust fund that pays 7.5% interest, compounded continuously. The balance will be given to the college from which the philanthropist graduated after the money has earned interest for 50 years.
How much will the college receive?
The points needed on the final to average the points to 75 when the final carries double weight.
Grades in three teste in College Algebra are 87,59 and 73.
The final carries double weight.
The average after final is 75.
Find the x-and y-intercepts of the given equation.
g\left(x\right)=2x+4
f\left(x\right)=|x+1|
Self-Check A student scores 82, 96, 91, and 92 on four college algebra exams. What score is needed on a fifth exam for the student to earn an average grade of 90?
The points needed on the final to avetage the points to 75.
Grades in three tests in College Algebra are 87, 59 and 73.
4x-5y=12
Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. 1 gram = 1000 kilograms ___
|
Searching Practice Problems Online | Brilliant
One of the simplest questions one might have about a list is whether or not some item is in the list. For example, if you are writing an algorithm to determine if someone should have access to a members-only website, you can solicit their information and then see if it matches some item in the members list.
Suppose that you have an unsorted array of 1000 emails and you want to check if abc123@gmail.com is in the array. Using a naive algorithm where you move sequentially through the list checking if each element matches abc123@gmail.com, what is the maximum number of comparisons between emails you would ever need to make?
Remember, it is entirely possible that the email is not in the array at all.
The search algorithm described in the previous question is known as linear search. Clearly, the worst case is very bad, since you might have to go through every element in the array.
What is the best case for the number of comparisons you will need to make when searching for a specific email in an array of 1000 emails?
So, for linear search, the best case is great (just 1 comparison!) but the worst case is very bad--possibly having to compare the email with every element in the array.
What about on average for this array of length 1000? If the email is in the array (whose elements are distinct and randomly arranged), what is the average number of comparisons that you will need to make?
(Note: While we are assuming the email is in the array, the algorithm can't make that assumption; it needs to confirm the presence of the email before it stops.)
About 1 About 5 About 50 About 500 About 1000
To summarize, to determine if an element is in an array with
n
elements using linear search, the number of comparisons we need to make is
in the best case, just 1 comparison;
in the worst case,
n
comparisons;
in the average case, given that the element is actually in the array,
\frac{n+1}{2}
While the average case might seem not too bad, it’s important to think about what this means for large
n.
While 500,000 comparisons is certainly not as bad as 1,000,000, they’re both problematic in that they scale linearly with
n;
that is, with twice as much data, the algorithm will need twice as many comparisons.
This isn’t great in a world with increasingly large data sets. Can we do better?
At first glance, it feels hard to do better than linear search. If—as with an unsorted list—we know nothing about the elements and their organization within the array, then we can’t do any better than just checking all of the elements.
But what if the list is sorted?
Define a comparison as an operation which takes in one number and tells you whether it is larger than, smaller than, or equal to, another.
Suppose you are given a sorted array with 1000 elements, and you can use at most
n
comparisons to determine whether a certain number is in this array.
n
such that you will always be capable of making this determination, regardless of the values of the elements in the array?
Hint: Pick an element from the list to observe first. If the element we observe is greater than our target what does this tell us? What if it's lower? With a sorted list, knowing if an element is greater or less than the element we’re looking for can be useful!
The search described in the previous problem is called binary search.
Consider a sorted array. In short, binary search repeatedly chooses a number in the middle of the remaining possible numbers, and then determines if the desired number would lie to the left or right of this chosen number (or exactly the chosen number). In each iteration, the amount of remaining numbers is halved, making binary search very efficient. This is especially important when dealing with a very large array.
How much does the faster binary search matter? The short answer: a lot. Comparing linear search and binary search serves as a clear demonstration that the choice of algorithm matters.
Suppose you have a sorted array with 100,000,000 elements in it. Assuming the worst case for each method, what is the ratio between the number of comparisons made by linear search and the number of comparisons made by binary search?
In other words, approximately how many times could we run binary searches before making the number of comparisons required for a single linear search?
Modern data sets tend to be massive; it’s critical to store and structure data in a way that makes common questions easily answerable. Throughout this course, we’ll consider what data types and structures should be used in various situations.
So far, we’ve already seen one repeated theme: lists tend to be more useful when sorted. But we haven’t answered the question of how to sort a list, and how computationally complicated it would be to do so. We’ll start to answer this question in the next quiz!
|
's Gravesande Wikipedia
Willem Jacob 's Gravesande (1688–1742)
Experimental proof of
{\displaystyle E_{k}\propto {\begin{matrix}\end{matrix}}mv^{2}}
Pieter van Musschenbroek, Jean Allamand
Willem Jacob 's Gravesande (26 September 1688 – 28 February 1742) was a Dutch mathematician and natural philosopher, chiefly remembered for developing experimental demonstrations of the laws of classical mechanics and the first experimental measurement of kinetic energy. As professor of mathematics, astronomy, and philosophy at Leiden University, he helped to propagate Isaac Newton's ideas in Continental Europe.
3 's Gravesande's ring
Born in 's-Hertogenbosch, 's Gravesande studied law at Leiden University, where he defended a thesis on suicide and earned a doctorate in 1707. He then practised law in The Hague while also participating in intellectual discussions and cultivating his interest in the mathematical sciences. His Essai de perspective ("Essay on Perspective"), published in 1711, was praised by the influential Swiss mathematician Johann Bernoulli.[1] In The Hague, 's Gravesande also helped to establish the Journal littéraire ("Literary journal"), a learned periodical first published in 1713.[2]
In 1715, 's Gravesande visited London as part of a Dutch delegation sent to welcome the Hanoverian succession in Great Britain.[2] In London, 's Gravesande met both King George I and Isaac Newton, and was elected a Fellow of the Royal Society.[3] In 1717 he became professor of mathematics and astronomy in Leiden. From that position, he was instrumental in introducing Newton's work to the Netherlands. He also obtained the chairs of civil and military architecture in 1730 and philosophy in 1734.[2] As a philosopher, he opposed fatalists like Hobbes and Spinoza.
's Gravesande was married to Anna Sacrelaire in 1720. They had two sons, both of whom died in adolescence. In 1724, Peter the Great offered 's Gravesande a position in the new Imperial Saint Petersburg Academy of Sciences. In 1737 he received an offer from Frederick the Great to join the Prussian Academy of Sciences in Berlin. He declined both offers, opting to remain in Leiden.[2]
Gravesande's main scientific work is Physices elementa mathematica, experimentis confirmata, sive introductio ad philosophiam Newtonianam ("Mathematical Elements of Natural Philosophy, Confirmed by Experiments; or, an Introduction to Newtonian Philosophy"), published in Leiden in 1720. In that book, he laid the foundations for the teaching of Newtonian mechanics through experimental demonstrations. He presented his work before audiences that included, Voltaire and Albrecht von Haller, and Émilie du Châtelet (the translator of Newton's Principia whose later commentary incorporated 's Gravesande's 1722 experimental discovery of kinetic energy). 's Gravesande's book was soon translated into English by John Theophilus Desaguliers, curator of experiments for the Royal Society.[4]
In 1721, 's Gravesande became involved in a public controversy over whether the German inventor Johann Bessler, known as Councillor Orffyreus, had created a genuine perpetual motion machine. 's Gravesande at first argued for the feasibility of perpetual motion based on the conservation of the scalar quantity mv (mass multiplied by speed), which he erroneously believed was implied by Newtonian mechanics.[5] However, in 1722 he published the results of a series of experiments in which brass balls were dropped from varying heights onto a soft clay surface. He found that a ball with twice the speed of another would leave an indentation four times as deep, from which he concluded that the correct expression for the "live force" of a body in motion (what is modernly called its "kinetic energy") is proportional to mv2.[5]
Even though those results invalidated his original argument for the feasibility of perpetual motion, 's Gravesande continued to defend Bessler's work, claiming that Bessler might have discovered some new "active principle" of nature that allowed his wheels to keep turning. Similar views were defended at the time by Gottfried Wilhelm Leibniz, Johann Bernoulli, and others, but the modern consensus is that Bessler was perpetrating a deliberate hoax.[5] Russian Tsar Peter the Great was interested in Bessler's wheel and sought 's Gravesande's advice on the subject.[6]
's Gravesande communicated his results on the impact of falling weights to Émilie du Châtelet. Similar observations were published independently by Giovanni Poleni. The interpretation of 's Gravesande's and Poleni's results led to a controversy with Samuel Clarke and other Newtonians that became a part of the so-called "vis viva dispute" in the history of classical mechanics.[7]
's Gravesande's ring[]
Mathematical Elements of Natural Philosophy, Confirm'd by Experiments: or, An introduction to Sir Isaac Newton's philosophy (Volume I), 1747
Mathematical Elements of Natural Philosophy, Confirm'd by Experiments: or, An introduction to Sir Isaac Newton's philosophy (Volume II), 1747
9682 Gravesande, main-belt asteroid named after Willem Jacob Gravesande
^ Google Books Knight, C. (1841). Penny cyclopaedia of the Society for the Diffusion of Useful Knowledge, Volumes 21–22. Page 331. Retrieved 6 October 2009
^ a b c d Albert van Helden, "Willem Jacob 's Gravesande, 1688–1742", in A History of Science in The Netherlands, eds. K. van Berkel, A. van Helden and L. Palm, (Leiden: Brill, 1999), pp. 450–453
^ "John Theophilus Desaguliers (1683–1744): popularising Newton", Isaac Newton and Newtonianism, Whipple Library, University of Cambridge
^ a b c Jenkins, Alejandro (2013). "The mechanical career of Councillor Orffyreus, confidence man". American Journal of Physics. 81 (6): 421–427. arXiv:1301.3097. Bibcode:2013AmJPh..81..421J. doi:10.1119/1.4798617. S2CID 118678318.
^ Werrett, Simon (2010). "The Schumacher Affair: Reconfiguring Academic Expertise across Dynasties in Eighteenth-Century Russia". Osiris. 25: 104–126. doi:10.1086/657265. S2CID 145788508.
^ Iltis, Carolyn (2009). "The Leibnizian-Newtonian Debates: Natural Philosophy and Social Psychology". The British Journal for the History of Science. 6 (4): 343–377. doi:10.1017/S000708740001253X.
A. R. Hall, "'s Gravesande, Willem Jacob", in Dictionary of Scientific Biography, vol. V, (New York: 1972), pp. 509–11.
C. de Pater, "Experimental Physics", in Leiden University in the Seventeenth Century, An Exchange of Learning (Leiden: 1975), pp. 308–327.
Wikimedia Commons has media related to Willem 's Gravesande.
's Gravesande's mistaken belief in perpetuum mobile
's Gravesande's New York Public Library entry
Willem 's Gravesande at the Mathematics Genealogy Project
O'Connor, John J.; Robertson, Edmund F., "Willem 's Gravesande", MacTutor History of Mathematics archive, University of St Andrews
The Oldest Magic Lantern in the World
Presentations of Willem 's Gravesande's Lectures, Devices, Laboratory and Experiments
List of Ph.D. students of Willem 's Gravesande
|
Noise in RF Systems - MATLAB & Simulink - MathWorks Deutschland
White and Colored Noise
Noise in an RF system is generated internally by active components in the system or introduced externally like channel interference or antenna.
White noise: Noise with a flat frequency spectrum is called white noise. White noise has equal power across all frequencies of the system band width.
Colored noise: Noise with power that varies according to frequencies in an RF system bandwidth is called colored noise.
To simulate white or colored noise in RF Blockset, use the Noise block.
Thermal noise is the most common noise introduced in an RF system. This noise is generated internally by active components in the system or externally due to channel interference or antenna. Thermal noise is also known as Johnson or Nyquist noise. The equation for thermal noise is:
{P}_{N}={k}_{B}TB
B is the bandwidth.
At room temperature, the thermal noise generated by system with a band width of 1 Hz is -174 dBm.
To generate thermal noise in a RF Blockset system use Resistor and Configuration block. You can also generate thermal noise using the S-Parameters block if the S-parameter is passive.
Thermal noise floor in RF Blockset is defined by the equation:
{P}_{noise}=4{k}_{B}T{R}_{s}\Delta f
Phase noise is a short-term fluctuation in the phase of an oscillator signal. This noise introduces uncertainty in the detection of digitally modulated signals. Phase noise is defined as the ratio of power ins one-phase modulation sideband to the total signal power per unit bandwidth. It is expressed in decibels relative to the carrier power per hertz of bandwidth (dBc/Hz). To know how to model LO phase noise in an oscillator see, Model LO Phase Noise.
Noise figure value determines the degradation of signal to noise ratio of a signal as it goes through the network. Noise figure is defined by the equation:
{N}_{f}=\frac{\frac{Signal}{Noise}\text{at the input}}{\frac{Signal}{Noise}\text{at the output}}
Excessive noise figure in the system causes the noise to overwhelm the signal, making the signal unrecoverable. Noise figure is a function of frequency but it is independent of the bandwidth of the system. Noise figure is expressed in dB.
In RF Blockset, you can specify noise figure for an RF system using the Amplifier or Mixer blocks.
|
Trigonometric Substitution in Integration | Brilliant Math & Science Wiki
Trevor Arashiro, Henry Maltby, A Former Brilliant Member, and
kritarth lohomi
Utkarsh Priyam
Like other substitutions in calculus, trigonometric substitutions provide a method for evaluating an integral by reducing it to a simpler one. Trigonometric substitutions take advantage of patterns in the integrand that resemble common trigonometric relations and are most often useful for integrals of radical or rational functions that may not be simply evaluated by other methods. These substitutions are often used in conjunction with the basic trigonometric relations and occasionally product-to-sum identities as well as other integration techniques including integration by parts and
u
-substitutions.
The Half-angle Substitution
Trigonometric substitutions are a specific type of
u
-substitutions and rely heavily upon techniques developed for those.
They use the key relations
\sin^2x + \cos^2x = 1
\tan^2x + 1 = \sec^2x
\cot^2x + 1 = \csc^2x
to manipulate an integral into a simpler form. The derivatives of trigonometric functions are also necessary to determine the best way to simplify the expression.
f(x) \quad \rightarrow
\hspace{10mm}
f'(x) \quad \rightarrow
\hspace{10mm}
f'(x)
rewritten
\sin \theta
\cos\theta
\pm\sqrt{1 - \sin^2\theta}
\cos \theta
-\sin\theta
\mp\sqrt{1-\cos^2\theta}
\tan \theta
\sec^2\theta
1 + \tan^2\theta
\sec \theta
\sec\theta\tan\theta
\pm\sec\theta\sqrt{\sec^2\theta - 1}
\csc \theta
-\csc\theta\cot\theta
\mp\csc\theta\sqrt{\csc^2\theta - 1 }
\cot\theta
-\csc^2\theta
-1-\cot^2\theta
Where applicable, the sign of the square root may be determined by the value of
\theta
One of the most common and direct applications is the integration of the reciprocal of a quadratic function.
\int \frac{dx}{x^2+1}.
Note from the above table that
x = \tan\theta
\tfrac{dx}{d\theta} = 1 + x^2
. It then makes some intuitive sense to say
d\theta = \frac{dx}{1 + x^2}.
\int \frac{dx}{x^2 + 1} = \int d\theta = \int 1 \, d\theta = \theta + C,
C
is the constant of integration. Recalling
x = \tan\theta
, the "
\theta
-substitution"
\theta = \arctan x
shows that the integral is
\int \frac{dx}{x^2 + 1} = \theta + C = \arctan x + C.\ _\square
In general, tangent or cotangent substitutions help a lot with the integration of rational functions, especially those with denominators of even degree. This concept is further explored below in the section on rational functions.
Trigonometric substitutions also help integrate certain types of radical functions, especially those involving square roots of quadratic functions. In fact, this technique may provide a verification of the well-known formula for the area of a circle.
Determine the area of a circle of radius
r
y^2 + x^2 = r^2
, and it has symmetries across the
x
y
-axes. Therefore, the area of the circle is four times the area of the portion of the circle in Quadrant I, i.e. that region bounded by
y = 0
x = 0
y = \sqrt{r^2 - x^2}
. Thus, the integral in question is
4 \int_0^r \sqrt{r^2-x^2} \, dx.
The above table shows that, in Quadrant I,
x = r\sin\theta
\tfrac{dx}{d\theta} = r\sqrt{1 - x^2}
dx = r\sqrt{1 - x^2} \, d\theta.
Therefore, the "
\theta
\theta = r\arcsin x
\begin{aligned} 4 \int_0^r \sqrt{r^2 - x^2} \, dx &= 4 \int_0^{\pi/2} \sqrt{r^2 - r^2\sin^2\theta} \cdot \left(r\sqrt{1 - \sin^2\theta}\right) \, d\theta \\ &= 4r^2 \int_0^{\pi/2} 1 - \sin^2\theta \, d\theta \\ &= 4r^2 \int_0^{\pi/2} \cos^2\theta \, d\theta \\ &= 4r^2 \int_0^{\pi/2} \frac{1 + \cos(2\theta)}{2} \, d\theta \\ &= 4r^2 \left[ \frac{\theta}{2} + \frac{\sin(2\theta)}{4} \right]_0^{\pi/2} \\ &= 4r^2 \left(\frac{\pi}{4} - 0\right) \\ &= \boxed{\pi r^2}. \end{aligned}
The double angle formula, which follows from the product-to-sum formulas, was necessary to finish the calculation.
_\square
n
\cos^nx
\sin^nx
may be expressed as a sum of multiples of elements of
\{1, \, \cos x, \, \cos(2x), \, \dots, \, \cos(nx)\}
\{1, \, \sin x, \, \sin(2x), \, \dots, \, \sin(nx)\},
respectively. For instance, the double-angle formulas yield
\cos^2x = \frac{1}{2} + \frac{1}{2} \cos(2x)\ \text{ and }\ \sin^2x = \frac{1}{2} - \frac{1}{2} \cos(2x).
By expressing, a typical
u
-substitution like
u = nx
may be applied to help integrate an expression.
\displaystyle \int \sin^3x \, dx.
\begin{aligned} \sin(3x) &= \cos x \sin(2x) + \cos(2x) \sin x \\ &= 2\cos^2 x \sin x + \big(1 - 2\sin^2 x\big) \sin x \\ &= 3 \sin x - 4\sin^3 x. \end{aligned}
\sin^3 x = \frac{3}{4} \sin x - \frac{1}{4} \sin(3x)
\begin{aligned} \int \sin^3x \, dx &= \int \frac{3}{4} \sin x - \frac{1}{4} \sin(3x) \, dx \\ &= \frac{1}{12} \cos(3x) - \frac{3}{4} \cos x + C, \end{aligned}
C
_\square
\frac{1}{12} \sin 3x + \frac{3}{4} \sin 2x\\ + \frac{15}{4} \sin x + \frac{5}{2} x + C
\frac{1}{12} \cos 3x + \frac{3}{4} \cos 2x\\ - \frac{15}{4} \cos x + \frac{5}{2} x + C
\frac{1}{12} \cos 3x + \frac{3}{4} \cos 2x\\ + \frac{15}{4} \cos x + \frac{5}{2} x + C
\frac{1}{12} \sin 3x + \frac{3}{4} \sin 2x\\ - \frac{15}{4} \sin x + \frac{5}{2} x + C
\int (1 + \cos x)^3 \, dx.
C
denotes the arbitrary constant of integration.
Many of the relations between trigonometric are second-order, so the inverse relations may involve square roots. As such, integrals involving square roots may be simplified by the use of trigonometric substitutions. In particular, expressions involving square roots of quadratic functions may benefit from cosine or secant substitutions. Sine substitutions work in the same scenarios as cosine ones, and cosecant substitutions work in the same scenarios as secant ones.
Given an expression of the form
\sqrt{ax^2 + bx + c}
, consider the sign of
and
d = b^2 - 4ac
d
is positive, then a trigonometric substitution might help. For positive
a
, a secant substitution could help after completing the square; for negative
a
, a cosine substitution could help after completing the square.
(
d
is negative, then a tangent or hyperbolic trigonometric substitution might help.
)
Such a substitution may help because it can remove the radical from the expression through the usage of trigonometric identities.
\int \frac{1}{\sqrt{2x - x^2}} \, dx.
\cos\theta = x - 1
simplifies the integral:
\begin{aligned} \int \frac{1}{\sqrt{2x - x^2}} \, dx &= \int \frac{1}{\sqrt{1 - (x - 1)^2}} \, dx \\ &= \int \frac{1}{\sqrt{1 - \cos^2\theta}} \cdot (- \sin \theta) \, d\theta \\ &= \int \frac{- \sin\theta}{\sin\theta} \, d\theta \\ &= - \theta + C \\ &= - \arccos(x - 1) + C, \end{aligned}
C
_\square
\int \frac{1}{\sqrt{4x^2 - 1}} \, dx.
\sec\theta = 2x
\begin{aligned} \int \frac{1}{\sqrt{4x^2 - 1}} \, dx &= \int \frac{1}{\sqrt{\sec^2\theta - 1}} \cdot \big(\tfrac{1}{2} \sec\theta \tan\theta\big) \, d\theta \\ &= \int \frac{\sec\theta \tan\theta}{2\tan\theta} \, d\theta \\ &= \int \tfrac{1}{2} \sec\theta d\theta \\ &= \tfrac{1}{2} \ln(\tan\theta + \sec\theta) + C\\ &= \tfrac{1}{2} \ln\big(2x + \sqrt{4x^2 - 1}\big) + C, \end{aligned}
C
_\square
\frac{1}{4}
\frac{\pi}{8}
\frac{\pi}{4}
\frac{\sqrt{2}}{2} - \frac{\pi}{8}
\frac{\sqrt{2}}{8} + \frac{1}{4}
\int_0^1 x \sqrt{1 - x^4} \, dx.
\int_0^2 x \sqrt{1-(x-1)^2} \, dx = \, ?
Rational functions involving quadratic polynomials in the denominator are often susceptible to a substitution involving the tangent function, and these tangent substitutions
(
together with knowledge of
\ln)
are the cornerstone of partial fraction decomposition. For each quadratic polynomial with no real roots, the idea is to complete the square and replace the square portion with a tangent function.
The idea behind this substitution is to "cancel out" part of the denominator with the differential term
(dx
d\theta)
in order to integrate a smaller expression. When applied properly, something will cancel out, since
\tfrac{dx}{d\theta} = 1 + x^2,
x = \tan\theta
\int_0^1\frac{2}{t^2+3} \, dt.
\theta
t = \sqrt{3} \tan\theta
\begin{aligned} \int_0^1\frac{2}{t^2+3} \, dt &= \int_0^{\pi/6} \frac{2}{3\tan^2\theta + 3} \cdot \big(\sqrt{3} \sec^2\theta\big) \, d\theta \\ &= \int_0^{\pi/6} \frac{2\sqrt{3}}{3} \, d\theta \\ &= \frac{\sqrt{3}\pi}{9}. \end{aligned}
Note that the limits change in accordance with the
\theta
-substitution; that is,
\sqrt{3} \tan\tfrac{\pi}{6} = \sqrt{3} \cdot \tfrac{1}{\sqrt{3}} = 1
_\square
\frac{\pi}{4}
\frac{\pi}{e}
\frac{\pi}{2}
\frac{5}{2e}
\frac{e}{2}
\int_0^\infty \frac{1}{e^x + e^{-x}} \, dx.
Recall the half-angle formulas
\cos\left(\tfrac{\theta}{2}\right) = \sqrt{\tfrac{1}{2} (1 + \cos\theta)}\ \text{ and }\ \sin\left(\tfrac{\theta}{2}\right) = \sqrt{\tfrac{1}{2} ( 1 - \cos\theta)}.
These are helpful to know for simplifying many expressions involving trigonometric substitutions. Even more helpful is the tangent half-angle formula
\tan\left(\tfrac{\theta}{2}\right) = \frac{\sin\left(\tfrac{\theta}{2}\right)}{\cos\left(\tfrac{\theta}{2}\right)} = \sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}} = \sqrt{\frac{(1 - \cos\theta)^2}{1 - \cos^2\theta}} = \frac{1 - \cos\theta}{\sin\theta}.
The half-angle tangent substitution has the further property that, if
t = \tan\left(\tfrac{\theta}{2}\right)
\begin{array}{c}&\sin\theta=\frac{2t}{1+t^2}, &\cos\theta=\frac{1-t^2}{1+t^2}, &\tan\theta=\frac{2t}{1-t^2}.\end{array}
For many rational functions involving trigonometric functions, such a substitution can make the expression seem much more integrable.
\displaystyle \int_0^\frac{\pi}{2}\frac{dx}{2+\cos x}\, dx.
\int_0^\frac{\pi}{2}\frac{dx}{2+\cos x}\, dx = \int_0^1\frac{\frac{2dt}{1+t^2}}{2+\frac{1-t^2}{1+t^2}}=\int_0^1\frac{2}{t^2+3} \, dt.
This integral can be finished using the methods outlined in the previous section, and it was evaluated to be equal to
\frac{\pi\sqrt{3}}{9}.\ _\square
\int_0^1\frac{x^4\left(1-x^2\right)^5}{\left(1+x^2\right)^{10}}\, dx=A
\frac{1}{A}.
Half-angle Tangent Substitution
u
-Substitutions
Cite as: Trigonometric Substitution in Integration. Brilliant.org. Retrieved from https://brilliant.org/wiki/integration-u-substitution-trigonometric/
|
Thinking Probabilistically Practice Problems Online | Brilliant
This first puzzle is a classic. It's from a 1983 research study on how people think about probability.
The world is probabilistic. Even "nearly impossible" things, like winning the lottery, have some calculable probability of occurring.
Probability gives a nearly universal framework for analyzing the probabilistic world around us, with applications that stretch from games to sports to finance to engineering to medicine.
This quiz kicks off our introduction to probability. Through a series of guided exercises, we'll tour some of the most important ideas that we'll encounter later in the course. Our topics include
conditional probability, which helps account for information we know for certain, and
counting strategies and probability rules, which are used to compute probabilities.
To end the quiz, we'll see how probability can help us determine our confidence that someone committed a crime given some evidence in the case.
Whose statement is the most reasonable?
Ali: I just flipped 3 heads in a row with a fair coin. My next flip is very likely to also be heads.
Ben: I rolled a fair six-sided die 10 times and I never rolled a 6. The next roll is especially likely to be a 6 because I am "due" for one.
Cam: Usually, if it rains in Brilliantia (40 km west of where I live), it rains here a couple of hours later. It just started raining in Brilliantia, so it will probably rain here soon.
Ali Ben Cam
Probability allows for the quantification of extremely rare events. For example, suppose Tim has two ways to try to win a large sum of money:
Win at a lottery with a chance of 1 in 300 million.
Roll a fair six-sided die 20 times and roll all 6s.
Which winning event is more likely to happen?
Winning the lottery Rolling all 6s
Which of these events is the most likely to happen when flipping a fair coin?
Flip 2 or more heads when flipping 3 coins.
Flip 20 or more heads when flipping 30 coins.
Flip 200 or more heads when flipping 300 coins.
Instead of making an explicit calculation, think about how likely or unlikely each of these outcomes would be.
2 out of 3 20 out of 30 200 out of 300 They are all equally likely
Probability can help avoid logical fallacies, quantify rare events, and compare the likelihood of various outcomes. However, like any framework, it only works if the underlying assumptions are reasonable.
An incident of the misuse of probability occurred in 1999, when Sir Roy Meadow successfully convicted Sally Clark of double homicide of her two children. The first died in 1996, aged 10 weeks; the second died in 1998, aged 8 weeks. Sally Clark claimed that both of her children died of Sudden Infant Death Syndrome (SIDS), a rare condition in which an infant suddenly and inexplicably dies.
The final two problems explore Sir Roy Meadow's probabilistic argument against Sally Clark.
Sir Roy Meadow argued that in lower-risk families (e.g., a non-smoking household) like the Clarks', the probability of a single SIDS death was about 1 in 8543. He further argued that the events of two children dying of SIDS in a single family are independent, so to find the probability of two SIDS deaths you could square the probability of a single SIDS death.
What approximate probability did Sir Roy Meadow claim for the probability that both of Sally Clark's infants would have died from SIDS?
1 in 2 million 1 in 11 million 1 in 20 million 1 in 73 million
Sir Roy Meadow argued that his probability calculations implied that it was nearly certain that Sally Clark murdered both of her children. Which of the following arguments, if true, refute this?
\boxed{1}
It is not clear that SIDS (which could be caused by some genetic predisposition) is actually independent across children from the same parents.
\boxed{2}
The "a priori" probability that Sally Clark would commit a double homicide was also incredibly rare, if not more rare than double-SIDS.
Note: "A priori" refers to the probability that Sally Clark would commit such a crime before taking into account the information of the deaths, which is the same type of probability that Sir Roy Meadow calculated. You can assume the facts are true, and you are simply evaluating reasonableness of the arguments.
1 only 2 only Both Neither
In this quiz, we took a quick stroll through the world of probability. We saw how probability applies to the real world, and how it helps us form opinions and strategies based on limited information.
In the next quiz, we explore counting strategies in more detail. They'll be one of our primary tools for representing outcome probabilities with numbers.
|
Solve for x: 0=-4u-7x+4.
Rewrite the equation as -4u-7x+4=0
-4u-7x+4=0
-7x=4u-4
Divide each term in -7x=4u-4 by -7 and simplify.
P\left(x\right)=-12{x}^{2}+2136x-41000
x=r\mathrm{cos}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y=r\mathrm{sin}
\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{x}^{2}-{y}^{2}}{\sqrt{{x}^{2}+{y}^{2}}}
Requires Uploaded Supporting Analysis: Differentiate.
y={\left(3{x}^{2}+5x+1\right)}^{\frac{3}{2}}
Use the following linear regression equation to answer the questions.
{x}_{1}=1.3+3.0{x}_{2}-8.3{x}_{3}+1.6{x}_{4}
(a) Which variable is the response variable?
(b) Which number is the constant term? List the coefficients with their corresponding explanatory variables.
{x}_{2}=8,{x}_{3}=2,\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{x}_{4}=6
, what is the predicted value for
{x}_{1}
? (Use 1 decimal place)
(d) Explain how each coefficient can be thought of as a "slope" under certain conditions.
To avoid detection at customs, a traveler places 6 narcotic tablets in a bottle containing 9 vitamin tablets that are similar in appearance. If the customs official selects 3 of the tablets at random for analysis, what is the probability that the traveler will be arrested for illegal possession of narcotics
Find all the complex roots. Leave your answers in polar form with the argument in degrees. The complex fourth roots of
100+100\sqrt{3}i
True or false? If the statement is false, use one or two sentences to justify your answer.
Assuming that other factors are constant, a correlation of r=-0.95 will result in less accurate predictions than a correlation of r=+0.70.
In regression analysis, at least an actual (observed) data point is located on the regression line.
|
Flow Boiling in Minichannels Under Normal, Hyper-, and Microgravity: Local Heat Transfer Analysis Using Inverse Methods | J. Heat Transfer | ASME Digital Collection
Sébastien Luciani,
Sébastien Luciani
, Technopôle de Château-Gombert, 5, rue Enrico Fermi, 13453 Marseille, France
e-mail: sebastien.luciani@polytech.univ-mrs.fr
David Brutin,
Christophe Le Niliot,
Christophe Le Niliot
Ouamar Rahli,
Ouamar Rahli
Luciani, S., Brutin, D., Le Niliot, C., Rahli, O., and Tadrist, L. (August 8, 2008). "Flow Boiling in Minichannels Under Normal, Hyper-, and Microgravity: Local Heat Transfer Analysis Using Inverse Methods." ASME. J. Heat Transfer. October 2008; 130(10): 101502. https://doi.org/10.1115/1.2953306
Boiling in microchannels is a very efficient mode of heat transfer since high heat and mass transfer coefficients are achieved. Here, the objective is to provide basic knowledge on the systems of biphasic cooling in mini- and microchannels during hyper- and microgravity. The experimental activities are performed in the frame of the MAP Boiling project founded by ESA. Analysis using inverse methods allows us to estimate local flow boiling heat transfers in the minichannels. To observe the influence of gravity level on the fluid flow and to take data measurements, an experimental setup is designed with two identical channels: one for the visualization and the other one for the data acquisition. These two devices enable us to study the influence of gravity on the temperature and pressure measurements. The two minichannels are modeled as a rectangular rod made up of three materials: a layer of polycarbonate
(λ=0.2Wm−1K−1)
used as an insulator, a cement rod
(λ=0.83Wm−1K−1)
instrumented with 21
K
-type thermocouples, and in the middle a layer of Inconel®
(λ=10.8Wm−1K−1)
in which the minichannel is engraved. Pressure and temperature measurements are carried out simultaneously at various levels of the minichannel. Above the channel, we have a set of temperature and pressure gauges and inside the cement rods, five heating wires provide a power of
11W
K
-type thermocouple sensors enable us to acquire the temperature in various locations (
x
y
z
) of the device. With these temperatures and the knowledge of the boundary conditions, we are able to solve the problem using inverse methods and obtain local heat fluxes and local surface temperatures on several locations. The experiments are conducted with HFE-7100 as this fluid has a low boiling temperature at the cabin pressure on Board A300. We applied for each experiment a constant heat flux
(Qw=33kWm−2)
for the PF52 campaigns (Parabolic Flights). The mass flow rate varies in the range of
1<Qm<4gs−1
and the fluid saturation temperature
(Tsat)
54°C
Psat=820mbars
boiling, convection, flow visualisation, gravity waves, inverse problems, mass transfer, microchannel flow, two-phase flow, convective flow boiling, microchannel, microgravity, inverse methods, truncated S.V.D., local heat transfer coefficient
Boiling, Boundary-value problems, Flow (Dynamics), Gravity (Force), Heat flux, Heat transfer, Inverse problems, Temperature, Thermocouples, Cements (Adhesives), Heat transfer coefficients, Heat, Errors, Sensors, Heating, Boundary element methods, Microchannels
Pressure Fluctuations During Boiling in a Narrow Channel
,” HTFS Research Symposium 1998.
The Onset of Flow Instability in Uniformly Heated Horizontal Microchannels
Measurement and Prediction of Pressure Drop in Two-Phase Micro-Channels Heat Sinks
Destabilization Mechanisms and Scaling Laws of Convective Boiling in a Minichannel
Le Probème de Cauchy et les Équations aux Dérivées Partielles Linéaires Hyperboliques
Boundary Element Approach for Inverse Conduction Problems: Application to a Bidimensional Transient Numerical Experiment
Dulikravitch
Inverse Determination of Boundary Conditions and Sources in Steady Heat Conduction With Heat Generation
Application de la méthode des éléments de frontière à la résolution du problème inverse de diffusion de la chaleurmultidimensionnel: régularisation par troncature de spectre
,” Ph.D. thesis, Institut National Polytechnique de, Grenoble, France.
A Method for Multiple Steady Line Heat Sources Identification in a Diffusive System: Application to an Experimental 2D Problem
The Determination of Two Heat Sources in an Inverse Heat Conduction Problem
Two-Dimensional Inverse Heat Conduction Problem of Estimating the Time-Varying Strength of a Line Heat Source
Inverse Heat Conduction, Ill-Posed Problems
V. H. Winston & Sons
La Méthode des Eléments de Frontière Pour la Résolution de Problèmes Inverses en Conduction de la Chaleur: Applications Numériques et Expérimentales
Teulossky
Flow Boiling in Minichannels Under Normal, Hyper and Microgravity: Local Heat Transfer Analysis Using Inverse Methods
Boundary Conditions and Integrated Sensors in Microchannel Convective Heat Transfer
|
Revision as of 20:19, 17 November 2014 by NikosA (talk | contribs) (→Automatising Conversions: Added link to https://github.com/NikosAlexandris/i.quickbird.toar)
{\displaystyle {\frac {W}{m^{2}*sr*nm}}}
{\displaystyle L_{\lambda {\text{Pixel, Band}}}={\frac {K_{\text{Band}}*q_{\text{Pixel, Band}}}{\Delta \lambda _{\text{Band}}}}}
{\displaystyle L_{\lambda {\text{Pixel,Band}}}}
{\displaystyle K_{\text{Band}}}
{\displaystyle q_{\text{Pixel,Band}}}
{\displaystyle \Delta _{\lambda _{\text{Band}}}}
{\displaystyle \rho _{p}={\frac {\pi *L\lambda *d^{2}}{ESUN\lambda *cos(\Theta _{S})}}}
{\displaystyle \rho }
{\displaystyle \pi }
{\displaystyle L\lambda }
{\displaystyle d}
{\displaystyle Esun}
{\displaystyle cos(\theta _{s})}
{\displaystyle {\frac {W}{m^{2}*\mu m}}}
|
Describe the translation of figure ABCD. Use the drop-down menus to explain your answer.
Figure ABCD is translated _____ unit(s) right and ______ unit(s) up.
Consider the following linear transformation
T:{P}_{2}\to {P}_{3}
T\left(f\right)=3{x}^{2}f
That is, take the first derivative and then multiply by
3{x}^{2}
(a) Find the matrix for T with respect to the standard bases of
{P}_{n}
: that is, find
\left[T{\right]}_{ϵ}^{ϵ}
, where-
ϵ=1,x,{x}^{2},{x}^{n}
(b) Find N(T) and R(T). You can either work with polynomials or with their coordinate vectors with respect to the standard basis. Write the answers as spans of polynomials.
(c) Find the the matrix for T with respect to the alternate bases:
\left[T{\right]}_{A}^{B}
A=x-1,x,{x}^{2}+1,B={x}^{3},x,{x}^{2},1.
Prove that by setting these two expressions equal to another, the result is an identity:
d={\mathrm{cos}}^{-1}\left(\mathrm{sin}\left(L{T}_{1}\right)\mathrm{sin}\left(L{T}_{2}\right)+\mathrm{cos}\left(L{T}_{1}\right)\mathrm{cos}\left(L{T}_{2}\right)\mathrm{cos}\left({\mathrm{ln}}_{1}-{\mathrm{ln}}_{2}\right)\right)
d=2{\mathrm{sin}}^{-1}\left(\sqrt{{\mathrm{sin}}^{2}\left(\frac{L{T}_{1}-L{T}_{2}}{2}\right)+\mathrm{cos}\left(L{T}_{1}\right)\mathrm{cos}\left(L{T}_{2}\right){\mathrm{sin}}^{2}\left(\frac{{\mathrm{ln}}_{1}-{\mathrm{ln}}_{2}}{2}\right)}\right)
Whether the statement "I need to be able to graph systems of linear inequalities in order to solve linear programming problems" makes sense or not.
Prove that: If A or B is nonsingular, then AB is similar to BA
The given matrix is the augmented matrix for a system of linear equations. Give the vector form for the general solution.
\left[\begin{array}{cccc}1& 0& -1& -2\\ 0& 1& 2& 3\end{array}\right]
(i)Prove that if
{v}_{1},{v}_{2}
is linearly dependent, then are multiple of each other, that is, there exists a constant c such that
{v}_{1}=c{v}_{2}\text{ }or\text{ }{v}_{2}=c{v}_{1}
(ii)Prove that the converse of(i) is also true.That is to say, if there exists a constant c such that
{v}_{1}=c{v}_{2}\text{ }or\text{ }{v}_{2}=c{v}_{1}
, 1. then
{v}_{1},{v}_{2}
is linearly dependent.
2x+3y=8
meets the curve
2{x}^{2}+3{y}^{2}=110
at the points A and B.
Find the coordinates of A and B.
To graph: The sketch of the solution set of system of nonlinear inequality
\left[{x}^{2}\text{ }+\text{ }{y}^{2}\ge \text{ }9\right]
{x}^{2}\text{ }+\text{ }{y}^{2}\le \text{ }25
y\ge \text{ }|x|
Find the sets of points in space whose coordinates satisfy the given combinations of equation and inequalities:
y\ge {x}^{2},z\ge 0,
x\le {y}^{2},0\le z\le 2.
Interraption: To show that the system
\stackrel{˙}{r}=r\left(1-{r}^{2}\right),\stackrel{˙}{\theta }=1
\stackrel{˙}{x}=x-y-x\left({x}^{2}+{y}^{2}\right),\stackrel{˙}{y}=x+y-y\left({x}^{2}+{y}^{2}\right)
for polar to Cartesian coordinates.
A limit cycle is a closed trajectory. Isolated means that neighboring trajectories are not closed.
A limit cycle is said to be unstable or half stable, if all neighboring trajectories approach the lemin cycle.
These systems oscillate even in the absence of external periodic force.
Find the matrix of the linear mapping.
L\left(x,y,z\right)=\left(x+y+z\right)
{y}_{1}=Ay
has the given eigenvalues and eigenspace bases. Find the general solution for the system
{\lambda }_{1}=3+i⇒\left\{\left[\begin{array}{c}2i\\ i\end{array}\right]\right\},{\lambda }_{2}=3-i⇒\left\{\left[\begin{array}{c}-2i\\ -i\end{array}\right]\right\}
{R}^{3}
\alpha :=\left\{\left[\begin{array}{c}2\\ 1\\ 3\end{array}\right],\left[\begin{array}{c}-1\\ 0\\ 1\end{array}\right],\left[\begin{array}{c}3\\ 1\\ -1\end{array}\right]\right\}and\text{ }\beta :=\left\{\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right],\left[\begin{array}{c}-2\\ 3\\ 1\end{array}\right],\left[\begin{array}{c}2\\ 3\\ -1\end{array}\right]\right\}
\left[x{\right]}_{\alpha }={\left[\begin{array}{c}1\\ 2\\ -1\end{array}\right]}_{\alpha }then\text{ }find\text{ }\left[x{\right]}_{\beta }
\beta
The equivalent polar coordinates for the given rectangular coordinates.
A rectangular coordinate is given as (0, -3).
Give a full correct answer for given question 1- Let W be the set of all polynomials
a+bt+c{t}^{2}\in {P}_{2}
a+b+c=0
Show that W is a subspace of
{P}_{2},
find a basis for W, and then find dim(W) 2 - Find two different bases of
{R}^{2}
so that the coordinates of
b=\left[\begin{array}{c}5\\ 3\end{array}\right]
are both (2,1) in the coordinate system defined by these two bases
To calculate: The intercepts on the coordinate axes of the straight line with the given equation
2y-4=3x
\gamma =\left\{{t}^{2}-t+1,t+1,{t}^{2}+1\right\}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\beta =\left\{{t}^{2}+t+4,4{t}^{2}-3t+2,2{t}^{2}+3\right\}be\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}deredbasesf\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}{P}_{2}\left(R\right).
\beta \text{ coordinates into }\gamma -\text{ coordinates}
\left[\begin{array}{cccccc}1& -1& 0& -2& 0& 0\\ 0& 0& 1& 2& 0& 0\\ 0& 0& 0& 0& 1& 0\end{array}\right]
Given f(x)=6x+5, describe how the graph of g compares with the graph of f. g(x)=6(0.2x)+5
Select the correct choice below, and fill in the answer box to complete your choice.
A. The graph of g(x) is translated _ unit(s) to the left compared to the graph of f(x).
B. The graph of g(x) is translated _ unit(s) down compared to graph of f(x).
C. g(x) has a scale factor of _ compared to f(x). Because it scales the vertical direction, the graph is stretched vertically.
D. g(x) has a scale factor of _ compared to f(x). Because it scales the vertical direction, the graph is compressed vertically.
E. g(x) has a scale factor of _ compared to f(x). Because it scales the horizontal direction, the graph is stretched horizontally.
F. g(x) has a scale factor of _ compared to f(x). Because it scales the horizontal direction, the graph is compressed horizontally.
G. The graph of g(x) is translated _ unit(s) to the right compared to the graph of f(x).
H. The graph of g(x) is translated _ unit(s) up compared to graph of f(x).
|
Dave Grace Manguilimotan 2022-03-19 Answered
Assuming that, on average, 4 out of 5 planes at Mactan International Airport arrive on schedule for a particular time period, what is the probability that out of 8 planes, chosen at random:
1.1) all 8 planes arrive on schedule?
1.2) 5 arrive on schedule?
1.3) at least 6 arrive on schedule?
Toss 5 coins 20 times Simultaneously and Observe the number of heads that will occur. The Possible Values of X or The number of heads in 5 coins tossed are 0, 1,2,3,4 and 5.
Refer to the scenario and table provided below. The number of qualified voters living in the household on a randomly selected Subdivision block is described by the following probability distribution. Find the MEAN. *
Genny Gimarino 2022-03-17
Toss 5 coins 20 times simultaneously and observe the number of heads that will occur.the possible values of X or the number of heads in 5 coins tossed are 0, 1, 2, 3, 4 and 5.
Assume that when adults with smartphones are randomly selected,44%use them in meetings or classes. If7adult smartphone users are randomly selected, find the probability that exactly3of them use their smartphones in meetings or classes.
what is the probability of getting two 6's in two rolls of a balanced die?
A company prices its tornado insurance using the following assumptions:• In any calendar year, there can be at most one tornado.• In any calendar year, the probability of a tornado is 0.09.• The number of tornadoes in any calendar year is independent of the number of tornados in any other calendar year.Using the company's assumptions, calculate the probability that there are fewer than 4 tornadoes in a 21-year period.
A manufacturing machine has a 4% defect rate.If 6 items are chosen at random, what is the probability that at least one will have a defect? Round answer to 4 decimal places.
A card is drawn from a standard 52-card deck. Find the probability that the card is a queen or red.
marsbekturov 2022-03-04 Answered
On a 100 question true/false test, what is the probability of getting 0 answers right?
In a 100 question true/false test what is the probability of getting 0 right answers? I'm struggling with formulas
Dilemma1 2022-02-27 Answered
Hi,I'm unsure whether I'm oversimplifying this problem.A zoo records 94,000 visits annually from the residents of a particular region. The region has a population of 67,000 inhabitants. From this, calculate the probability that any individual within the population visits the zoo at any given month of the year.Calculation:94,000 / 67,000 = 1.41.4 is the average visit rate per resident1.4 / 12 = .12Is the probability that an individual from the population visits the zoo on any given month therefore 12%?Can anybody please provide a bit of guidance?Much appreciated.
Assume that a procedure yields a binomial distribution with a trial repeated n = 11 times. Use the binomial probability formula to find the probability of k = 9 successes given the probability p = 11/15 of success on a single trial.
In a certain region, the probability that any given day in October is wet is 0.16, independently of other days.(a) Find the probability that, in a 10-day period in October, fewer than 3 days will be wet.(b) Find the probability that the first wet day in October is 8 October.(c) For 4 randomly chosen years. find the probability that in exactly 1 of these years the first wet day in October is 8 October.
A poll is give , showing 40% are in favor of a new building project.If 6 people are chosen at random, what is the probability that exactly 3 of them favor the new building project?
On a 9 question multiple-choice test, where each question has 2 answers, what would be the probability of getting at least one question wrong?
Suppose that a family has 7 children. Also, suppose that the probability of having a girl is 12. Find the probability that the family has the following children. No girls
Anabelle Schneider 2022-02-15 Answered
P\left(A\right)=0.1,\text{ }P\left(B\right)=0.9
P\left(A\cap B\right)=0
P\left(B\mid A\right)
P\left(A\mid B\right)
Beausepoomoubor2 2022-02-15 Answered
What is the probability that the deal of a five-card hand provides exactly one ace?
|
Validator Information Terms - Harmony
Definition of validator information terms
This page introduces main terms about validator infos in the harmony-mainnet-tracker spreadsheet, you can also read some of the terms in staking dashboard.
Annual Percentage Return (APR) is the percent return of rewards for the last epoch that the validator was last elected.
This field reflects the data of the last epoch this validator elected, not the current epoch.
For example, if a validator was elected in epoch 186 with 50% APR and they were never elected again, they will still display 50% APR.
Latest Expected Return
Example Validator Profile
Latest expected return is the validator's APR for the latest epoch.
Example Validator List
The Expected Return field here is average of the validator's APR for the past 30 epochs that they were elected.
Validator uptime is calculated as:
s / b * 100
Where s is the number of blocks signed by the validator and b is the number of blocks in the epoch so far.
If a validator has never been elected, their uptime will display as None in the tracking spreadsheet. In the staking dashboard, their uptime will be 0.00%.
Example of a validate that has never been elected
Similarly to APR, uptime is also only updated if the validator is currently elected.
For example, if the same validator was elected in epoch 186 with 100% uptime and they were never elected again, they will still have 100% uptime displayed.
Epos-status
There are five types of data in this field:
eligible to be elected next epoch
not eligible to be elected next epoch
doubleSigningBanned
Epos status is based on the validator's active status. The validator's active status will be controlled either by user or system (if detected low uptime or 0 self-stake). However if the validator is never elected, as long as he set the validator status as active, he will always be shown as "eligible to be elected next epoch".
Boot-status
There are five types of the data in this field:
NotBooted: if the validator was never selected/signed blocks.
LostEPoSAuction
TurnedInactiveOrInsufficientUptime
BannedForDoubleSigning
None: when the validator is currently elected
|
Introduction to Mathematical Physics/N body problem and matter description/Other matter arrangements - Wikibooks, open books for an open world
Introduction to Mathematical Physics/N body problem and matter description/Other matter arrangements
1 Quasi crystals
5 Sand piles, orange piles
Quasi crystalsEdit
Quasi crystals ([#References|references]) discovered by Israeli physicist Shechtman in 1982 correspond to a non periodical filling of a volume by atoms or molecules. Pentagon is a forbidden form in crystallography: a volume can not be filled with elementary bricks of fifth order symmetry repartited periodically. This problem has a two dimensional twin problem: a surface can not be covered only by pentagons. English mathematician Penrose \index{Penrose paving}discovered non periodical paving of the plane by lozenges of two types (see figure ---figpenrose--- and figpenrose2).
figpenrose
File:Penrose
Penrose paving: from two types of lozenges, non periodical paving of the plane can be constructed.
figpenrose2
File:Penrose2
Penrose paving: same figure as previous one, but with only two gray levels to distinguish between the two types of lozenges.
Obtained structures are amazingly complex and it is impossible to encounter some periodicity in the paving.
secristliquides
Liquid crystalsEdit
Liquid crystals ([#References|references]), also called mesomorphic phases\index{mesomorphic phase}, are states intermediary between the cristaline perfect order and the liquid disorder. Molecules that compound liquid crystals\index{liquid crystal} have cigar--like shape. They arrange themselves in space in order to form a "fluid" state more or less ordered. Among the family of liquid crystals, several classes can be distinguished.
In the nematic \index{nematic} phase (see figure ---fignematique---), molecular axes stays parallel. There exists thus a privileged direction. Each molecule can move with respect to its neighbours, but in a fish ban way.
File:Nematique
Nematic material is similar to a fish ban: molecules have a privileged orientation and located at random points.}
fignematique
When molecules of a liquid crystal are not superposable to their image in a mirror, a torsion of nematic structure appears: phase is then called cholesteric \index{cholesteric} (see figure ---figcholesteric---).
figcholesteric
File:Cholesteric
In a cholesteric phase, molecules are arranged in layers. In each layer, molecule are oriented along a privileged direction. This direction varies from one layer to another, so that a helicoidal structure is formed.
Smectic \index{smectic} phase is more ordered : molecules are arranged in layers (see figure figsmectiqueA). The fluid character comes from the ability of layers to slide over their neighbours.
figsmectiqueA
File:SmectiqueA
To describe deformations of nematics, a field of vector is used. To describe deformations of smectics, each state is characterised by set of functions
{\displaystyle u_{i}(x,y)}
that describes the surface of the i
{\displaystyle ^{th}}
layer. Energical properties of nematic materials are presented at section secenernema.
Colloids \index{colloid} are materials finely divided and dispersed. Examples are emulsions \index{emulsion} and aerosols. Consider a molecule that has two parts: a polar head which is soluble in water and an hydrophobe tail. Such molecules are called amphiphile\index{amphiphile molecule}. Once in water, they gather into small structures called micelles\index{micelle} (see figure fighuileeau). Molecules turn their head to water and their tail to the inside of the structure.
Remark: Cells of live world are very close to micelles, and it may be that life\index{life (origin of)}\index{origin of life} appeared by the mean of micelles.
fighuileeau
File:Huileeau
Amphiphile molecules (as oil molecules) that contains a hydrophile head and an hydrophobe tail organize themselves into small structures called micelles.
Micelles can be encountered in mayonnaise ([#References|references]). Tensioactive substances allow to disolve micelles (application to soaps). Foams are similar arrangements ([#References|references]). For a mathematical introduction to soap films and minimal surfaces, check ([#References|references]).
secglassyspin
Glass state ([#References|references]) is characterized by a random distribution of molecules (see figure figglass). Glass\index{glass} is solid, that implies that movements of different constituents are small.
figglass
File:Glass
Glass is characterized by a great rigidity as crystals. However position of atoms are random.}
Phase transition between liquid state and glassy state is done progressively. A material close to glass can be made by compressing small balls together. Under pressure forces, those small balls are deformed and stick to each other. Following question arises: are remaining interstices sufficiently numerous to allow a liquid to pour trough interstices ? This pouring phenomena is a particular case of percolation phenomenon\index{percolation} ([#References|references]). Figure ---figpercol--- illustrates this phenomenom in an experiment presented in ([#References|references]).
figpercol
File:Percol
Example of percolation : black balls are conducting and white balls are insulating. Current running trough the circuit is measured as a function of proportion
{\displaystyle p}
of white balls. There exists a critical proportion
{\displaystyle p_{c}}
for which no more current can pass. Study of this critical point allows to exhibit some universal properties encountered in similar systems.
Another example is given by the vandalized grid ([#References|references]) where connections of a conducting wire is destroyed with probability
{\displaystyle p}
Spin glasses are disordered magnetic materials. A good example of spin glass is given by the alloy copper manganese, noted CuMn where manganese atoms carrying magnetic moments are dispersed at random in a copper matrix. two spins tend to orient themselves in same or in opposite direction, depending on distance between them (see figure ---figspinglass---).
figspinglass
File:Spinglass
In a spin glass, interactions between neighbour spins are at random ferromagnetic type or anti ferromagnetic type, due to the random distance between spin sites.
The resulting system is called "frustrated": there does not exist a configuration for which all interaction energies are minimal. The simplest example of frustrated system is given by a system constituted by three spins labelled 1,2 and 3 where interactions obey the following rule: energy decreases if 1 and 2 are pointing in the same direction, 2 and 3 are pointing in the same direction and 1 and 3 are pointing in opposite direction. Some properties of spin glasses are presented at section secverredespi.
Sand piles, orange pilesEdit
Physics of granular systems is of high interest and is now the subject of many researches. Those systems, as a sand pile, exhibit properties of both liquids and solids. The sand in a hour-glass doesn't pour like liquids: the speed of pouring doesn't depend on the high of sand above the hole. The formation of the sand pile down of the hole is done by internal convection and avalanche \index{avalanche} at the surface of the pile (see for instance ([#References|references])).
Retrieved from "https://en.wikibooks.org/w/index.php?title=Introduction_to_Mathematical_Physics/N_body_problem_and_matter_description/Other_matter_arrangements&oldid=3985005"
|
How can the least count of a vernier calipers be increased - Physics - Motion in one dimension - 16779363 | Meritnation.com
How can the least count of a vernier calipers be increased?
Least count of a vernier caliper = minimum measurement on main scale / total number of divisions on vernier scale. Let's derive this:
Least count (L.C) of vernier calipers is the minimum length or thickness measurable with the
Least count \left(L.C\right) =1 M.S.D -1 V.S.D
L.C = 1 M.S.D - \frac{N-1}{N}*M.S.D\phantom{\rule{0ex}{0ex}}Since the V.S.D is \frac{N-1}{N} times the main scale division
⇒L.C=1 M.S.D\left(1-\frac{N-1}{N}\right)\phantom{\rule{0ex}{0ex}}⇒L.C=\frac{1 M.S.D}{N}=\frac{S}{N}
Smaller is the magnitude of least count of a measuring instrument. more precise the measuring .
Based on the above information we can say that in order to increase the least count of vernier calipers just increase the value of one scale division by increasing the number of divisions on main scale.
In order to increase the least count of screw gauge just increase the value of one scale division by increasing the number of divisions on circular scale or linear scale.
|
Learning From MACD
MACD vs. Relative Strength
Example of MACD Crossovers
Example of Rapid Rises or Falls
The result of that calculation is the MACD line. A nine-day EMA of the MACD called the "signal line," is then plotted on top of the MACD line, which can function as a trigger for buy and sell signals. Traders may buy the security when the MACD crosses above its signal line and sell—or short—the security when the MACD crosses below the signal line. Moving average convergence divergence (MACD) indicators can be interpreted in several ways, but the more common methods are crossovers, divergences, and rapid rises/falls.
Moving average convergence divergence (MACD) is calculated by subtracting the 26-period exponential moving average (EMA) from the 12-period EMA.
\text{MACD}=\text{12-Period EMA }-\text{ 26-Period EMA}
MACD=12-Period EMA − 26-Period EMA
The MACD has a positive value (shown as the blue line in the lower chart) whenever the 12-period EMA (indicated by the red line on the price chart) is above the 26-period EMA (the blue line in the price chart) and a negative value when the 12-period EMA is below the 26-period EMA. The more distant the MACD is above or below its baseline indicates that the distance between the two EMAs is growing.
In the following chart, you can see how the two EMAs applied to the price chart correspond to the MACD (blue) crossing above or below its baseline (dashed) in the indicator below the price chart.
MACD is often displayed with a histogram (see the chart below) which graphs the distance between the MACD and its signal line. If the MACD is above the signal line, the histogram will be above the MACD’s baseline. If the MACD is below its signal line, the histogram will be below the MACD’s baseline. Traders use the MACD’s histogram to identify when bullish or bearish momentum is high.
The relative strength indicator (RSI) aims to signal whether a market is considered to be overbought or oversold in relation to recent price levels. The RSI is an oscillator that calculates average price gains and losses over a given period of time. The default time period is 14 periods with values bounded from 0 to 100.
MACD measures the relationship between two EMAs, while the RSI measures price change in relation to recent price highs and lows. These two indicators are often used together to provide analysts a more complete technical picture of a market.
These indicators both measure momentum in a market, but, because they measure different factors, they sometimes give contrary indications. For example, the RSI may show a reading above 70 for a sustained period of time, indicating a market is overextended to the buy-side in relation to recent prices, while the MACD indicates the market is still increasing in buying momentum. Either indicator may signal an upcoming trend change by showing divergence from price (price continues higher while the indicator turns lower, or vice versa).
One of the main problems with divergence is that it can often signal a possible reversal but then no actual reversal actually happens—it produces a false positive. The other problem is that divergence doesn't forecast all reversals. In other words, it predicts too many reversals that don't occur and not enough real price reversals.
"False positive" divergence often occurs when the price of an asset moves sideways, such as in a range or triangle pattern following a trend. A slowdown in the momentum—sideways movement or slow trending movement—of the price will cause the MACD to pull away from its prior extremes and gravitate toward the zero lines even in the absence of a true reversal.
As shown on the following chart, when the MACD falls below the signal line, it is a bearish signal that indicates that it may be time to sell. Conversely, when the MACD rises above the signal line, the indicator gives a bullish signal, which suggests that the price of the asset is likely to experience upward momentum. Some traders wait for a confirmed cross above the signal line before entering a position to reduce the chances of being "faked out" and entering a position too early.
Crossovers are more reliable when they conform to the prevailing trend. If the MACD crosses above its signal line following a brief correction within a longer-term uptrend, it qualifies as bullish confirmation.
When the MACD forms highs or lows that diverge from the corresponding highs and lows on the price, it is called a divergence. A bullish divergence appears when the MACD forms two rising lows that correspond with two falling lows on the price. This is a valid bullish signal when the long-term trend is still positive.
When the MACD forms a series of two falling highs that correspond with two rising highs on the price, a bearish divergence has been formed. A bearish divergence that appears during a long-term bearish trend is considered confirmation that the trend is likely to continue.
Some traders will watch for bearish divergences during long-term bullish trends because they can signal weakness in the trend. However, it is not as reliable as a bearish divergence during a bearish trend.
Image by Sabrina Jiang © Investopedia 2020
When the MACD rises or falls rapidly (the shorter-term moving average pulls away from the longer-term moving average), it is a signal that the security is overbought or oversold and will soon return to normal levels. Traders will often combine this analysis with the relative strength index (RSI) or other technical indicators to verify overbought or oversold conditions.
It is not uncommon for investors to use the MACD’s histogram the same way they may use the MACD itself. Positive or negative crossovers, divergences, and rapid rises or falls can be identified on the histogram as well. Some experience is needed before deciding which is best in any given situation because there are timing differences between signals on the MACD and its histogram.
How Do Traders Use Moving Average Convergence Divergence (MACD)?
Traders use MACD to identify changes in the direction or severity of a stock’s price trend. MACD can seem complicated at first glance, since it relies on additional statistical concepts such as the exponential moving average (EMA). But fundamentally, MACD helps traders detect when the recent momentum in a stock’s price may signal a change in its underlying trend. This can help traders decide when to enter, add to, or exit a position.
Is MACD a Leading Indicator, or a Lagging Indicator?
MACD is a lagging indicator. After all, all of the data used in MACD is based on the historical price action of the stock. Since it is based on historical data, it must necessarily “lag” the price. However, some traders use MACD histograms to predict when a change in trend will occur. For these traders, this aspect of the MACD might be viewed as a leading indicator of future trend changes.
What Is a MACD Positive Divergence?
A MACD positive divergence is a situation in which the MACD does not reach a new low, despite the fact that the price of the stock reached a new low. This is seen as a bullish trading signal—hence, the term “positive divergence.” If the opposite scenario occurs—the stock price reaching a new high, but the MACD failing to do so—this would be seen as a bearish indicator and referred to as a negative divergence.
|
Disproportionation - Wikipedia
In chemistry, disproportionation, sometimes called dismutation, is a redox reaction in which one compound of intermediate oxidation state converts to two compounds, one of higher and one of lower oxidation states.[1][2] More generally, the term can be applied to any desymmetrizing reaction of the following type: 2 A → A' + A", regardless of whether it is a redox or some other type of process.[3]
1.1 Reverse reaction
Mercury(I) chloride disproportionates upon UV-irradiation:
Phosphorous acid disproportionates upon heating to give phosphoric acid and phosphine:
3 → 3 H3PO4 + PH3
Desymmetrizing reactions are sometimes referred to as disproportionation, as illustrated by the thermal degradation of bicarbonate:
3 → CO2−
3 + H2CO3
The oxidation numbers remain constant in this acid-base reaction. This process is also called autoionization.
Another variant on disproportionation is radical disproportionation, in which two radicals form an alkane and alkene.
Reverse reactionEdit
The reverse of disproportionation, such as when a compound in an intermediate oxidation state is formed from precursors of lower and higher oxidation states, is called comproportionation, also known as synproportionation.
This was examined using tartrates by Johan Gadolin in 1788. In the Swedish version of his paper he called it 'söndring'.[4][5]
Chlorine gas reacts with dilute sodium hydroxide to form sodium chloride, sodium chlorate and water. The ionic equation for this reaction is as follows:[6]
{\displaystyle {\ce {3Cl2 + 6 OH- -> 5 Cl- + ClO3- + 3 H2O}}}
The chlorine reactant is in oxidation state 0. In the products, the chlorine in the Cl− ion has an oxidation number of −1, having been reduced, whereas the oxidation number of the chlorine in the ClO3− ion is +5, indicating that it has been oxidized.
Decompositions of numerous interhalogen compounds involve disproportionation. Bromine fluoride undergoes disproportionation reaction to form bromine trifluoride and bromine:[7]
{\displaystyle {\ce {3 BrF -> BrF3 + Br2}}}
The dismutation of superoxide free radical to hydrogen peroxide and oxygen, catalysed in living systems by the enzyme superoxide dismutase:
{\displaystyle {\ce {2 O2- + 2H+ -> H2O2 + O2}}}
The oxidation state of oxygen is −1/2 in the superoxide free radical anion, −1 in hydrogen peroxide and 0 in dioxygen.
The disproportionation of hydrogen peroxide into water and oxygen catalysed by either potassium iodide or the enzyme catalase:
{\displaystyle {\ce {2 H2O2 -> 2 H2O + O2}}}
In the Boudouard reaction, carbon monoxide disproportionates to carbon and carbon dioxide. The reaction is for example used in the HiPco method for producing carbon nanotubes, high-pressure carbon monoxide disproportionates when catalysed on the surface of an iron particle:
{\displaystyle {\ce {2 CO -> C + CO2}}}
Nitrogen has oxidation state +4 in nitrogen dioxide, but when this compound reacts with water, it forms both nitric acid and nitrous acid, where nitrogen has oxidation states +5 and +3 respectively:
{\displaystyle {\ce {2NO2 + H2O -> HNO3 + HNO2}}}
Dithionite undergoes acid hydrolysis to thiosulfate and bisulfite:[8]
{\displaystyle {\ce {2{S2O4^{2-}}+H2O->{S2O3^{2-}}+2HSO3-}}}
Dithionite also undergoes alkaline hydrolysis to sulfite and sulfide:[8]
{\displaystyle {\ce {3 Na2S2O4 + 6 NaOH -> 5 Na2SO3 + Na2S + 3 H2O}}}
Dithionate is prepared on a larger scale by oxidizing a cooled aqueous solution of sulfur dioxide with manganese dioxide:[9][citation needed]
{\displaystyle {\ce {2 MnO2 + 3 SO2 -> MnS2O6 + MnSO4}}}
In free-radical chain-growth polymerization, chain termination can occur by a disproportionation step in which a hydrogen atom is transferred from one growing chain molecule to another which produces two dead (non-growing) chains.[10]
-------CH2–C°HX + -------CH2–C°HX → -------CH=CHX + -------CH2–CH2X
In 1937, Hans Adolf Krebs, who discovered the citric acid cycle bearing his name, confirmed the anaerobic dismutation of pyruvic acid into lactic acid, acetic acid and CO2 by certain bacteria according to the global reaction:[11]
2 pyruvic acid + H2O → lactic acid + acetic acid + CO2
The dismutation of pyruvic acid in other small organic molecules (ethanol + CO2, or lactate and acetate, depending on the environmental conditions) is also an important step in fermentation reactions. Fermentation reactions can also be considered as disproportionation or dismutation biochemical reactions. Indeed, the donor and acceptor of electrons in the redox reactions supplying the chemical energy in these complex biochemical systems are the same organic molecules simultaneously acting as reductant or oxidant.
Another example of biochemical dismutation reaction is the disproportionation of acetaldehyde into ethanol and acetic acid.[12]
While in respiration electrons are transferred from substrate (electron donor) to an electron acceptor, in fermentation part of the substrate molecule itself accepts the electrons. Fermentation is therefore a type of disproportionation, and does not involve an overall change in oxidation state of the substrate. Most of the fermentative substrates are organic molecules. However, a rare type of fermentation may also involve the disproportionation of inorganic sulfur compounds in certain sulfate-reducing bacteria.[13]
^ IUPAC, Compendium of Chemical Terminology, 2nd ed. (the "Gold Book") (1997). Online corrected version: (2006–) "disproportionation". doi:10.1351/goldbook.D01799
^ a b José Jiménez Barberá; Adolf Metzger; Manfred Wolf (2000). "Sulfites, Thiosulfates, and Dithionites". Ullmann's Encyclopedia of Industrial Chemistry. Weinheim: Wiley-VCH. doi:10.1002/14356007.a25_477.
^ Cowie, J. M. G. (1991). Polymers: Chemistry & Physics of Modern Materials (2nd ed.). Blackie. p. 58. ISBN 0-216-92980-6.
^ Krebs, H.A. (1937). "LXXXVIII - Dismutation of pyruvic acid in gonoccus and staphylococcus". Biochem. J. 31 (4): 661–671. doi:10.1042/bj0310661. PMC 1266985. PMID 16746383.
^ Bak, Friedhelm; Cypionka, Heribert (1987). "A novel type of energy metabolism involving fermentation of inorganic sulphur compounds". Nature. 326 (6116): 891–892. Bibcode:1987Natur.326..891B. doi:10.1038/326891a0. PMID 22468292. S2CID 27142031.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Disproportionation&oldid=1076525448"
|
Metals and Non-Metals | Brilliant Math & Science Wiki
Sravanth C. and Skanda Prasad contributed
Chromium, a metal and sulfur, a non-metal
The periodic table can be broadly segregated into two types of elements, commonly referred to as metals and non-metals. Each of these elements have varying properties and can be found in a wide variety of places, such as bridges, buildings, roads, electric cables, cars, aircraft, mobile phones, and laptops, as well as in the oxygen we breathe and the carbon dioxide we exhale.
Over the course of this wiki, you will come to understand the physical and chemical properties of both metals and non-metals. To get the most out of this wiki please read balancing chemical reactions, chemical equilibrium, acids and bases.
Reaction of Metals with Non-metals
Let's begin by studying the physical properties of metals.
Electrical Conductivity: Almost all metals are good conductors of electricity, though they differ in their conducting power. Silver, for instance, is the best conductor of electricity, and copper
\ce{(Cu)}
and aluminium
\ce{(Al)}
are also good conductors of electricity. However, silver
\ce{(Ag)}
is highly expensive and is not used for this purpose. Instead, copper and aluminium are used. Metals conduct in a solid state.
The conductivity of metal is actually due to the electrons present in the valence shell of metal atoms which they can easily release.
\ce{(Hg)}
, on the other hand, is a poor conductor of electricity. In addition, lead
\ce{(Pb)}
is almost non-conducting. We can easily check whether a metal is a good conductor of electricity. In the case of steel, connect a steel spoon to a battery and a light bulb. If the light bulb starts glowing, then we can conclude that steel conducts electricity.
Thermal conductivity: Metals are good conductors of heat as well, which can also be easily proven. To do this, take a thin metal, a clamp, a stand, and a burner. Set up the apparatus as shown below. Heat one end of the metal sheet for a few minutes, then touch the other end of the sheet. We can observe that the end which had not been heated by the burner is also quite hot. This activity proves that metals, generally, are good conductors of heat.
Heating Iron: Thermal Conductivity[4]
Just like electrical conductivity, metals also differ in their thermal conductivity. Silver and copper are good thermal conductors, while lead and mercury are poor thermal conductors.
Malleability: Malleability is the property by which a metal can be beaten into sheets by a hammer or similar device. All of us have heard of and seen aluminium foil, which is the metal aluminium beaten into very fine, thin sheets.
A thin Sheet of gold: Malleability[5]
Ductility: Ductility is the property by which a metal is drawn into thin wires. When a metal can be drawn into wire, we say the metal is ductile. Copper and aluminium are some of the best ductile metals.
Hardness: Metals are usually hard and difficult to break, though there are exceptions. Sodium
\ce{(Na)}
and magnesium
\ce{(Mg)}
are so soft they can easily be cut with a knife. The hardness of a metal depends upon the strength of the bonds present among its atoms. Such bonds are called metallic bonds. Since the strength of the bonds differ from metal to metal, their hardness varies accordingly.
Lustre: Lustre refers to a metal's shine or brightness. Metals like silver, gold, and platinum are well known for their lustre. This property is the reason why jewelry and ornaments are often made of these metals. The lustre is present only when the metals are fresh, and once they are exposed to air, they start to lose their shine and their surface becomes dull. These metals also lose their shine when they come in contact with water vapor.
Lusture of Pyrite: Lustre[6]
Sonorosity: Metals are generally sonorous, which means they produce a ringing sound when struck.
Though the physical properties help us classify and differentiate metals from non-metals, we need better methods to classify them, and we do that by using their chemical properties. These properties work well with almost all elements and do not seem to have many exceptions. Let us explore a few of these properties.
Reaction with Oxygen: Most metals combine with oxygen in the following manner:
\ce{Metal} \ + \ \ce{Oxygen} \ \to \ \text{Metal Oxide}
We shall look at some examples of this kind of reaction.
\ce{4Na}\ + \ \ce{O}_2\ \to \ce{2Na_2O}
\ce{Note:}
\ce{(K)}
also reacts with oxygen in the same manner to form potassium oxide
\ce{(K_2O)}
Both of these metals react quite violently because they are highly reactive elements and occupy top positions in the reactivity series.
If we look at magnesium, it occupies a lower position in the reactivity series, hence, it doesn't react very vigorously. It only reacts with oxygen upon heating. It burns with a bright flame to form magnesium oxide
\ce{(Mg_2O)}
. Here is the reaction:
\ce{2Mg}\ + \ \ce{O_2}\ \mathrel{\mathop{\longrightarrow}^{\mathrm{heat}}} \ce{2MgO}
Reaction with water: The reactivity of metals with water is also linked with their positions in the reactivity series. As a result of reaction of a metal with water, metal hydroxide is formed along with the release of hydrogen gas.
The reactions take place in the following manner:
\ce{Metal} \ + \ce{Water} \ \to \text{Metal Hydroxide} + \ce{Hydrogen}
\ce{2K} \ + \ce{2H_2O}\ \to \ce{2KOH}\ + \ce{2H_2}
\ce{2Na} \ + \ce{2H_2O}\ \to \ce{2NaOH}\ + \ce{2H_2}
Note: There exists an intermediate form when the metal forms a metal oxide first. However, when the metal oxide is further made to react with water, the product formed will be a metal hydroxide.
Reaction with Acids: Generally, metals react with acids to produce hydrogen in the following manner.
\ce{Metal} \ + \ \ce{Acid}\ \rightarrow\ \ce{salt} \ + \ \ce{Hydrogen}
However, not all metals can displace hydrogen from acid to form a salt. This is because metals which are lower than hydrogen in the Reactivity Series cannot replace it. As a result, they are unable to follow the above reaction. But for now, let's see some metals which do react with acids.
\ce{2Na} \ + \ \ce{2HCl} \ \rightarrow \ \ce{2NaCl} \ + \ \ce{H2}
Here, as
\ce{Na}
is well above hydrogen in the activity series, it easily replaces hydrogen to form
\ce{NaCl}
. Let's have a look at some other examples:
\begin{aligned} \ce{Mg} \ + \ \ce{H2SO4} &\rightarrow \ \ce{MgSO4} \ + \ \ce{H2}\\ \ce{Ca} \ + \ \ce{2HCl} &\rightarrow \ \ce{CaCl2} \ + \ \ce{H2}\\ \end{aligned}
However, metals like copper, mercury, and silver cannot replace hydrogen, because they occur well below hydrogen in the reactivity series.
Reaction with Solutions of other Metals: Again, these reactions involve the reactivity series, where a metal which is higher in the series can displace a metal which is lower. Here's the general formula:
\text{Metal A} \ + \ \text{Salt Solution of Metal B} \rightarrow \text{Salt Solution of Metal A} \ + \ \text{Metal B}
provided that the reactivity of Metal A is greater that the reactivity of Metal B. These reactions help us in determining the reactivity of an element, as the displaced element will be proven to be weaker in terms of reactivity than the element which displaced it. Let us see some examples now:
One very common reaction of this kind is the
\ce{Fe-Cu}
reaction.
\ce{Fe} \ + \ce{CuSO4} \rightarrow \ce{FeSO4} \ + \ \ce{Cu}
In each of these examples, the reactions follow according to the Reactivity Series. And in each reaction, the color of the solution changes as the reaction takes place, here are a few more:
\begin{aligned} \ce{Zn} \ + \ce{CuSO4} &\rightarrow \ce{ZnSO4} \ + \ \ce{Cu}\\ \ce{Cu} \ + \ce{2AgNO3} &\rightarrow \ce{Cu(NO3)2} \ + \ \ce{2Ag}\\ \end{aligned}
Now, let us glance through some physical properties of non-metals. Physical properties alone are not good for distinguishing non-metals due to exceptions in almost every physical property. Later, we shall study in more detail chemical properties, which is the best way to differentiate between metals and non-metals.
State of existence: Non-metals usually exist in the three states of matter. However, most of them exist in gaseous form. Non-metals like nitrogen, oxygen, carbon dioxide, argon, neon, helium, krypton, chlorine, and fluorine are the ones which constitute the air in our surroundings.
Hardness: Out of all the non-metals, only solids are expected to be hard. Sulphur and phosphorus are quite soft, but diamond is very hard. Diamond is also probably the hardest substance presently known.
Diamond: An exception for Hardness [7]
Lustre: Non-metals usually have no shine since they have no loosely attached electrons which are responsible for lustre. However, there are exceptions. Out of the non-metals, Diamond and iodine have lustrous nature.
Lustrous Iodine: An exception for Lustre[8]
Electrical and Thermal Conductivity: Non-metals, in general, are quite poor conductors of heat and electricity. Graphite is an exception here.In fact, it is a very good conductor of electricity.
Malleability: Non-metals cannot be beaten into thin sheets in the way that metals can. There is a weak force of attraction. As a result, they are quite brittle. Sulphur is a brittle element. If it is hammered, it would break into pieces.
Sulfur powder[3]
Ductility: Non-metals cannot be drawn into thin wires. They would break, whether being beaten into thin sheets or drawn into wires.
Sonorosity: Non-metals are not sonorous. They produce no ringing sound when struck on their surface.
Again, to classify metals and non-metals in a more orderly fashion, we deal with their chemical properties. The chemical properties work well with almost all elements and do not seem to have many exceptions. We will discuss some of the main chemical properties of non-metals in this section.
Reaction with Oxygen: In the presence of oxygen, when non-metals are heated they react and form non-metal oxides. These oxides are generally acidic in nature. In general, they react in the following manner:
\text{Non-Metal} \ + \ \text{Oxygen} \mathrel{\mathop{\longrightarrow}^{\mathrm{heat}}} \text{Non-Metal Oxide}
Here are some examples, one of the most common one being:
\ce C \ + \ \ce{O2} \mathrel{\mathop{\longrightarrow}^{\mathrm{heat}}} \text{CO2}
\begin{aligned} \ce{S8} \ + \ \ce{8O2} &\mathrel{\mathop{\longrightarrow}^{\mathrm{heat}}} \text{8SO2}\\ \ce{P4} \ + \ \ce{5O2} &\mathrel{\mathop{\longrightarrow}^{\mathrm{heat}}} \text{P4O10}\\ \end{aligned}
Reaction with Acids: Non-metals, being electronegative, do not react with dilute acids as they cannot displace hydrogen easily. However, they react with concentrated acids when they are heated. Here's the general formula:
\text{Non-Metal} \ + \ \ce{Acid} \mathrel{\mathop{\longrightarrow}^{\mathrm{heat}}} \text{Salt} \ + \ \text{Oxide} \ + \ \text{Water}
However, the salt may not necessarily be formed in all the cases.
The following is one of the classic examples in which the salt is not formed:
\ce{S} \ + \ \ce{2H2SO4} \mathrel{\mathop{\longrightarrow}^{\mathrm{heat}}} \ce{3SO2} \ + \ \ce{2H2O}
Other common examples being:
\begin{aligned} \ce{P} \ + \ \ce{5HNO3} &\mathrel{\mathop{\longrightarrow}^{\mathrm{heat}}} \ce{H3PO4} \ + \ \ce{5NO2} \ + \ \text{2H2O}\\ \ce{2P} \ + \ \ce{5H2SO4} &\mathrel{\mathop{\longrightarrow}^{\mathrm{heat}}} \ce{2H3PO4} \ + \ \ce{5SO2} \ + \ \ce{2H2O} \end{aligned}
Reaction with Chlorine: Non-metals react with chlorine atoms upon heating to form their respective chlorides. Sulphur
\ce{(S)}
and phosphorus
\ce{(P)}
react in this way. Let's have a look at some reactions to have a better idea of this.
\text{Non-Metal} \ + \ \ce{Chlorine} \mathrel{\mathop{\longrightarrow}^{\mathrm{heat}}} \text{Metal Chloride}
\begin{aligned} \ce{2P}\ + \ce{5Cl_2}\ &\mathrel{\mathop{\longrightarrow}^{\mathrm{heat}}} \ce{2PCl_5}\\ \ce{2S}\ + \ce{Cl_2}\ &\mathrel{\mathop{\longrightarrow}^{\mathrm{heat}}} \ce{S_2Cl_2} \end{aligned}
Reaction with hydrogen: Non-metals form their respective hydrides when they react with hydrogen. These hydrides are non-electrolytic and exist as gases at room temperature. This is the general formula:
\text{Non-Metal} \ + \ \ce{Hydrogen} \mathrel{\mathop{\longrightarrow}^{\mathrm{heat}}} \text{Metal Hydride}
Hydrogen combines with sulphur at
713K
to form hydrogen sulphide in the following manner:
\ce{S}\ + \ce{H2}\ \mathrel{\mathop{\longrightarrow}^{\mathrm{heat}}} \ce{H2S}
Another common reaction is the formation of ammonia, which occurs at
773 K
in the presence of an iron catalyst and it is a reversible reaction:
\ce{N2}\ + \ce{3H2}\ \rightleftharpoons \ce{2NH3}
Reaction with Salt Solutions: This type of reaction is similar to the one we discussed in the chemical properties of metals. Here again, the reactions involve the reactivity series. A non-metal which is higher in the series can displace a non-metal from it's salt which is lower than it. Here's the general formula:
\text{Non-Metal A} \ + \ \text{Salt solution of Non-Metal B} \longrightarrow \text{Non-Metal B} \ + \ \text{Salt Solution of Non-Metal A}
provided that the reactivity of Metal A is greater than the reactivity of Metal B. Let's have a look at one very common example:
\ce{Cl2}\ + \ce{2NaBr}\ \longrightarrow \ce{2NaCl} \ + \ \ce{Br2}
Metals and non-metals do not have stable electronic configurations like the noble gas elements do. Hence, they will have a strong tendency to achieve the noble gas configuration.
Metal atoms have surplus electrons to give away in their valence shell. On the other hand, non-metals need electrons to fill their last valence shell.
From this, we can say that atoms of metals lose valence electrons while non-metal atoms accept electrons. When the metal atoms give away their valence electrons, they become positively charged or cations, while the non-metal atoms, when they lose electrons, become negatively charged, or in other words, anions. The opposing type of charges bring the ions closer due to electrostatic force of attraction between them, resulting in the formation of ionic compounds with an ionic bond between them.
The formation of common salt or table salt
\ce{(NaCl)}
is shown below. The electron from the sodium atom gets transferred to the chlorine atom. Hence, an ionic bond is formed between them.
\ce{Na}\cdot \ + \ \cdot\ce{Cl} \rightarrow \ce{Na-Cl}
Here, the chlorine atom changes to
\ce{Cl^-}
and gets the electronic configuration
\ce{2, 8, 8}
of argon
\ce{(Ar)}
. Sodium, as it changes to
\ce{Na^+}
ion, gets the electronic configuration
\ce{2, 8}
of neon
\ce{(Ne)}
When they react, they will form
\ce{NaCl}
\ce{Na^+}\ + \ce{Cl^-}\ \to \ce{NaCl}
Hence, the ionic compound sodium chloride
\ce{(NaCl)}
is formed.
[1] MTG Publications, Foundation Course for Chemistry Revised 2014 edition, MTG: Indian Subcontinent adaptation 2014.
[2] Image from https://en.m.wikipedia.org/wiki/Chromium#/media/File%3AChromiumcrystalsand1cm3cube.jpg under the creative commons attribution for reuse and modification.
[3] Image from https://en.m.wikipedia.org/wiki/Sulfur#/media/File%3ASulfur-sample.jpg under the creative commons attribution for reuse and modification.
[4] Image from https://en.m.wikipedia.org/wiki/Metal#/media/File%3AHot_metalwork.jpg under the creative Commons license for reuse and modification.
[5] Image from https://en.m.wikipedia.org/wiki/Ductility#/media/File%3AKanazawaGoldFactory.jpg under the creative Commons attribution for reuse and modification.
[6] Image from https://en.m.wikipedia.org/wiki/Lustre(mineralogy)#/media/File%3APyrite3.jpg under the creative Commons attribution for reuse and modification.
[7] Image from https://en.m.wikipedia.org/wiki/Diamond#/media/File%3ATheHopeDiamond-SIA.jpg under the creative Commons license for reuse and modification.
[8] Image from https://en.m.wikipedia.org/wiki/Nonmetal#/media/File%3AIodinecrystals.JPG under the creative Commons attribution for reuse and modification.
Cite as: Metals and Non-Metals. Brilliant.org. Retrieved from https://brilliant.org/wiki/metals-and-non-metals/
|
Jordan curve theorem - Wikipedia
Division by a closed curve of the plane into two regions
Illustration of the Jordan curve theorem. A Jordan curve (drawn in black) divides the plane into an "inside" region (light blue) and an "outside" region (pink).
In topology, the Jordan curve theorem asserts that every Jordan curve (a plane simple closed curve) divides the plane into an "interior" region bounded by the curve and an "exterior" region containing all of the nearby and far away exterior points. Every continuous path connecting a point of one region to a point of the other intersects with the curve somewhere. While the theorem seems intuitively obvious, it takes some ingenuity to prove it by elementary means. "Although the JCT is one of the best known topological theorems, there are many, even among professional mathematicians, who have never read a proof of it." (Tverberg (1980, Introduction)). More transparent proofs rely on the mathematical machinery of algebraic topology, and these lead to generalizations to higher-dimensional spaces.
The Jordan curve theorem is named after the mathematician Camille Jordan (1838–1922), who found its first proof. For decades, mathematicians generally thought that this proof was flawed and that the first rigorous proof was carried out by Oswald Veblen. However, this notion has been overturned by Thomas C. Hales and others.
1 Definitions and the statement of the Jordan theorem
2 Proof and generalizations
3 History and further proofs
Definitions and the statement of the Jordan theorem[edit]
A Jordan curve or a simple closed curve in the plane R2 is the image C of an injective continuous map of a circle into the plane, φ: S1 → R2. A Jordan arc in the plane is the image of an injective continuous map of a closed and bounded interval [a, b] into the plane. It is a plane curve that is not necessarily smooth nor algebraic.
Theorem — Let C be a Jordan curve in the plane R2. Then its complement, R2 \ C, consists of exactly two connected components. One of these components is bounded (the interior) and the other is unbounded (the exterior), and the curve C is the boundary of each component.
In contrast, the complement of a Jordan arc in the plane is connected.
Proof and generalizations[edit]
Theorem — Let X be an n-dimensional topological sphere in the (n+1)-dimensional Euclidean space Rn+1 (n > 0), i.e. the image of an injective continuous mapping of the n-sphere Sn into Rn+1. Then the complement Y of X in Rn+1 consists of exactly two connected components. One of these components is bounded (the interior) and the other is unbounded (the exterior). The set X is their common boundary.
{\displaystyle {\tilde {H}}_{q}(Y)={\begin{cases}\mathbb {Z} ,&q=n-k{\text{ or }}q=n,\\\{0\},&{\text{otherwise}}.\end{cases}}}
There is a strengthening of the Jordan curve theorem, called the Jordan–Schönflies theorem, which states that the interior and the exterior planar regions determined by a Jordan curve in R2 are homeomorphic to the interior and exterior of the unit disk. In particular, for any point P in the interior region and a point A on the Jordan curve, there exists a Jordan arc connecting P with A and, with the exception of the endpoint A, completely lying in the interior region. An alternative and equivalent formulation of the Jordan–Schönflies theorem asserts that any Jordan curve φ: S1 → R2, where S1 is viewed as the unit circle in the plane, can be extended to a homeomorphism ψ: R2 → R2 of the plane. Unlike Lebesgue's and Brouwer's generalization of the Jordan curve theorem, this statement becomes false in higher dimensions: while the exterior of the unit ball in R3 is simply connected, because it retracts onto the unit sphere, the Alexander horned sphere is a subset of R3 homeomorphic to a sphere, but so twisted in space that the unbounded component of its complement in R3 is not simply connected, and hence not homeomorphic to the exterior of the unit ball.
History and further proofs[edit]
The statement of the Jordan curve theorem may seem obvious at first, but it is a rather difficult theorem to prove. Bernard Bolzano was the first to formulate a precise conjecture, observing that it was not a self-evident statement, but that it required a proof.[citation needed] It is easy to establish this result for polygons, but the problem came in generalizing it to all kinds of badly behaved curves, which include nowhere differentiable curves, such as the Koch snowflake and other fractal curves, or even a Jordan curve of positive area constructed by Osgood (1903).
Earlier, Jordan's proof and another early proof by Charles Jean de la Vallée Poussin had already been critically analyzed and completed by Schoenflies (1924).[5]
Due to the importance of the Jordan curve theorem in low-dimensional topology and complex analysis, it received much attention from prominent mathematicians of the first half of the 20th century. Various proofs of the theorem and its generalizations were constructed by J. W. Alexander, Louis Antoine, Ludwig Bieberbach, Luitzen Brouwer, Arnaud Denjoy, Friedrich Hartogs, Béla Kerékjártó, Alfred Pringsheim, and Arthur Moritz Schoenflies.
Elementary proofs were presented by Filippov (1950) and Tverberg (1980).
The root of the difficulty is explained in Tverberg (1980) as follows. It is relatively simple to prove that the Jordan curve theorem holds for every Jordan polygon (Lemma 1), and every Jordan curve can be approximated arbitrarily well by a Jordan polygon (Lemma 2). A Jordan polygon is a polygonal chain, the boundary of a bounded connected open set, call it the open polygon, and its closure, the closed polygon. Consider the diameter
{\displaystyle \delta }
of the largest disk contained in the closed polygon. Evidently,
{\displaystyle \delta }
is positive. Using a sequence of Jordan polygons (that converge to the given Jordan curve) we have a sequence
{\displaystyle \delta _{1},\delta _{2},\dots }
presumably converging to a positive number, the diameter
{\displaystyle \delta }
of the largest disk contained in the closed region bounded by the Jordan curve. However, we have to prove that the sequence
{\displaystyle \delta _{1},\delta _{2},\dots }
does not converge to zero, using only the given Jordan curve, not the region presumably bounded by the curve. This is the point of Tverberg's Lemma 3. Roughly, the closed polygons should not thin to zero everywhere. Moreover, they should not thin to zero somewhere, which is the point of Tverberg's Lemma 4.
The first formal proof of the Jordan curve theorem was created by Hales (2007a) in the HOL Light system, in January 2005, and contained about 60,000 lines. Another rigorous 6,500-line formal proof was produced in 2005 by an international team of mathematicians using the Mizar system. Both the Mizar and the HOL Light proof rely on libraries of previously proved theorems, so these two sizes are not comparable. Nobuyuki Sakamoto and Keita Yokoyama (2007) showed that in reverse mathematics the Jordan curve theorem is equivalent to weak König's lemma over the system
{\displaystyle {\mathsf {RCA}}_{0}}
If the initial point (pa) of a ray (in red) lies outside a simple polygon (region A), the number of intersections of the ray and the polygon is even.
If the initial point (pb) of a ray lies inside the polygon (region B), the number of intersections is odd.
Main article: Point in polygon § Ray casting algorithm
In computational geometry, the Jordan curve theorem can be used for testing whether a point lies inside or outside a simple polygon.[6][7][8]
From a given point, trace a ray that does not pass through any vertex of the polygon (all rays but a finite number are convenient). Then, compute the number n of intersections of the ray with an edge of the polygon. Jordan curve theorem proof implies that the point is inside the polygon if and only if n is odd.
^ Camille Jordan (1887)
^ Oswald Veblen (1905)
^ Hales (2007b)
^ A. Schoenflies (1924). "Bemerkungen zu den Beweisen von C. Jordan und Ch. J. de la Vallée Poussin". Jahresber. Deutsch. Math.-Verein. 33: 157–160.
^ Richard Courant (1978)
^ "V. Topology". 1. Jordan curve theorem (PDF). Edinburg: University of Edinburgh. 1978. p. 267.
^ "PNPOLY - Point Inclusion in Polygon Test - WR Franklin (WRF)". wrf.ecse.rpi.edu. Retrieved 2021-07-18.
Courant, Richard (1978). "V. Topology". Written at Oxford. What is mathematics? : an elementary approach to ideas and methods. Herbert Robbins ([4th ed.] ed.). United Kingdom: Oxford University Press. p. 267. ISBN 978-0-19-502517-0. OCLC 6450129.
Filippov, A. F. (1950), "An elementary proof of Jordan's theorem" (PDF), Uspekhi Mat. Nauk (in Russian), 5 (5): 173–176
Hales, Thomas C. (2007a), "The Jordan curve theorem, formally and informally", The American Mathematical Monthly, 114 (10): 882–894, doi:10.1080/00029890.2007.11920481, ISSN 0002-9890, MR 2363054, S2CID 887392
Maehara, Ryuji (1984), "The Jordan Curve Theorem Via the Brouwer Fixed Point Theorem", The American Mathematical Monthly, 91 (10): 641–643, doi:10.2307/2323369, ISSN 0002-9890, JSTOR 2323369, MR 0769530
Osgood, William F. (1903), "A Jordan Curve of Positive Area", Transactions of the American Mathematical Society, 4 (1): 107–112, doi:10.2307/1986455, ISSN 0002-9947, JFM 34.0533.02, JSTOR 1986455
Ross, Fiona; Ross, William T. (2011), "The Jordan curve theorem is non-trivial", Journal of Mathematics and the Arts, 5 (4): 213–219, doi:10.1080/17513472.2011.634320, S2CID 3257011 . author's site
Sakamoto, Nobuyuki; Yokoyama, Keita (2007), "The Jordan curve theorem and the Schönflies theorem in weak second-order arithmetic", Archive for Mathematical Logic, 46 (5): 465–480, doi:10.1007/s00153-007-0050-6, ISSN 0933-5846, MR 2321588, S2CID 33627222
Tverberg, Helge (1980), "A proof of the Jordan curve theorem" (PDF), Bulletin of the London Mathematical Society, 12 (1): 34–38, CiteSeerX 10.1.1.374.2903, doi:10.1112/blms/12.1.34
M.I. Voitsekhovskii (2001) [1994], "Jordan theorem", Encyclopedia of Mathematics, EMS Press
Brown, R.; Antolino-Camarena, O. (2014). "Corrigendum to "Groupoids, the Phragmen-Brouwer Property, and the Jordan Curve Theorem", J. Homotopy and Related Structures 1 (2006) 175-183". arXiv:1404.0556 [math.AT].
doi:10.1007/15.40062-014-0089-0
Retrieved from "https://en.wikipedia.org/w/index.php?title=Jordan_curve_theorem&oldid=1081910309"
Theorems about curves
|
Ravi Substitution | Brilliant Math & Science Wiki
Arousse Fares, Utkarsh Gupta, Anuj Shikarkhane, and
Ravi substitution is a technique emerging as extremely useful in the field of geometric inequalities, particularly at Olympiad level. Some inequalities with the variables
a,b,c
have the side constraint that
a,b,c
are the sides of a triangle, however it may seem unclear how to use this to our advantage when solving a problem.
Ravi substitution offers a solution to this:
a,b,c
are the sides of a non-degenerate triangle if there exist three positive reals
x,y,z
a = x+y
b = y+z
c = z+x
The substitution is effective even in the most difficult problems involving the sides of a triangle. The substitution accompanied with clever manipulations and use of other basic inequalities gives beautiful solutions.
An example from IMO Shortlist 2006.
Problem (ISL 2006 A5):
a,b,c
are the sides of a triangle, prove that
\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}+\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\leq 3.
a,b,c
are the sides of a triangle, we can set
a=(q+r)^2,\quad b=(p+r)^2,\quad c=(p+q)^2.
Thus, the inequality can be rewritten as
\sum_\text{cyclic} \frac{\sqrt{2p^2+2pr+2pq-2qr}}{p} \le 6 .
Thus, is it is sufficient to show that
\begin{aligned} \sum_\text{cyclic} \frac{2p^2+2pr +2pq -2qr}{p^2} &\le 12\\ \iff \sum_\text{cyclic} \frac{pr+pq-qr}{p^2} &\le 3\\ \iff \sum_\text{cyclic} q^2r^2(pr+pq-qr) &\le 3p^2q^2r^2. \end{aligned}
This is third degree Schur's inequality for
pq,qr,rp
_\square
The complete Ravi substitution
A common way to use the Ravi substitution is by expressing x, y and z in terms of a, b and c. Meaning :
\left\{\begin{matrix} a = y + z \\ b = x + z \\ c = x+y\end{matrix}\right.
\left\{\begin{matrix} x = \frac{b+c-a}{2} \\ y = \frac{a+c-b}{2} \\ z = \frac{a+b-c}{2}\end{matrix}\right.
Memorizing the order of
a
b
c
may be quite difficult, but usually there is no need since most of the inequalities requiring Ravi are symmetric. It's however possible to use a mnemonic method to keep it in mind :
In the first part, you can notice that "
a
doesn't have
x
b
y
c
z
In the bottom part, notice as well that "
x
-a
, y has
-b
z
-c
Here is an application :
a,b,c
(a+b) (b+c) (c+a) \geqslant 8 (a+b-c) (b+c-a) (c+a-b)
We can start from both sides, but doing it from the right is more Ravi-like :
According to the Ravi substitution, we can have
x, y, z
such as :
\left\{\begin{matrix} x = \frac{b+c-a}{2} \\ y = \frac{a+c-b}{2} \\ z = \frac{a+b-c}{2}\end{matrix}\right.
We will also assume the reader knows AM-GM inequality
Therefore :
8 (a+b-c) (b+c-a) (c+a-b) = 8 \times 8 \frac{b+c-a}{2} \frac{a+c-b}{2} \frac{a+b-c}{2} = 8 \times 8 xyz = 8 \times 8 \sqrt{xy} \sqrt{yz} \sqrt{zx}
AM-GM tells us that
8 \times 8 \sqrt{xy} \sqrt{yz} \sqrt{zx} \leq 8 \times 8 \frac{x+y}{2} \frac{z+y}{2} \frac{x+z}{2}
8 \times 8 \frac{x+y}{2} \frac{z+y}{2} \frac{x+z}{2} = 8 abc = 8 \sqrt{ab} \sqrt{bc} \sqrt{ca}
again, AM-GM tells us
8 \sqrt{ab} \sqrt{bc} \sqrt{ca} \leq 8 \frac{a+b}{2} \frac{b+c}{2} \frac{c+a}{2}
8 \frac{a+b}{2} \frac{b+c}{2} \frac{c+a}{2} = (a+b) (b+c) (c+a)
and thus we have :
(a+b) (b+c) (c+a) \geqslant 8 (a+b-c) (b+c-a) (c+a-b)
Cite as: Ravi Substitution. Brilliant.org. Retrieved from https://brilliant.org/wiki/ravi-substitution/
|
Static Call Graph - Maple Help
Home : Support : Online Help : Programming : CodeTools : Program Analysis : Static Call Graph
create a static call graph from a Maple procedure
StaticCallGraph(P, opts)
Maple procedure, appliable module, string, or set or list of procedures or appliable modules
one or more options as described below
maxdepth=posint or infinity
The maximum depth of procedure calls to examine. The default is infinity, meaning the static call graph is traversed to arbitrary depth.
restrict=r, where r is a symbol or string, or a set or list of symbols or strings
Whether and how to restrict the procedures or appliable modules included as vertices of the resulting graph. If this option is not supplied, then there are no restrictions on which vertices are included. If it is supplied, then a procedure or appliable module p is included only if:
it is listed in r; or
it is a local or export of a module included in r; or
it is a local or export of a module that is, itself, a local or export of a module included in r, and so forth.
This is determined by looking at the name for such procedures or appliable modules that this code finds. Note that if a procedure p1 calls a procedure p2 that is rejected according to this criterion, then that procedure is not examined; if p2 contains any calls to a procedure p3 that would be accepted, then p3 will not be found, at least not through p2.
StaticCallGraph(P) returns a directed graph which represents a static call graph for the input P. The vertices of the graph are strings which represent procedures or appliable modules, and a directed arc exists from f to g when there is an explicit reference to g in the body of f.
If P is a string, then it is interpreted as the name of a Maple library archive, and all procedures defined in it are examined.
This command generates only simple graphs; recursive calls in a procedure body are not translated into self-loops in the resulting graph.
This command simply examines the procedure body for instances of procedure names. There are many methods of invoking a procedure in Maple (for example, by constructing the procedure's name with cat or parse) which this command makes no effort to detect.
\mathrm{with}\left(\mathrm{CodeTools}[\mathrm{ProgramAnalysis}]\right):
G≔\mathrm{StaticCallGraph}\left(\mathrm{maximize},\mathrm{maxdepth}=2\right)
\textcolor[rgb]{0,0,1}{G}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{Graph 1: a directed unweighted graph with 22 vertices and 24 arc\left(s\right)}}
\mathrm{GraphTheory}:-\mathrm{Vertices}\left(G\right)
[\textcolor[rgb]{0,0,1}{"NumericEventHandler"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"convert"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"evalf"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"has"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"hastype"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"indets"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"length"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"max"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"maximize"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"min"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"minimize"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"minimize/Handler"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"minimize/cell/check"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"nops"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"op"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"select"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"selectremove"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"seq"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"simplify"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"sort"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"subs"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"type"}]
\mathrm{GraphTheory}:-\mathrm{DrawGraph}\left(G,\mathrm{style}=\mathrm{spring}\right)
The CodeTools[ProgramAnalysis][StaticCallGraph] command was introduced in Maple 2020.
|
We show that, for any stochastic event
p
n
, there exists a measure-once one-way quantum finite automaton (1qfa) with at most
2\sqrt{6n}+25
states inducing the event
ap+b
, for constants
a>0
b\phantom{\rule{-0.166667em}{0ex}}\ge 0
a+b\le 1
. This fact is proved by designing an algorithm which constructs the desired 1qfa in polynomial time. As a consequence, we get that any periodic language of period
n
can be accepted with isolated cut point by a 1qfa with no more than
2\sqrt{6n}+26
states. Our results give added evidence of the strength of measure-once 1qfa’s with respect to classical automata.
Classification : 68Q10, 68Q19, 68Q45
Mots clés : quantum finite automata, periodic events and languages
author = {Mereghetti, Carlo and Palano, Beatrice},
title = {On the size of one-way quantum finite automata with periodic behaviors},
AU - Mereghetti, Carlo
AU - Palano, Beatrice
TI - On the size of one-way quantum finite automata with periodic behaviors
Mereghetti, Carlo; Palano, Beatrice. On the size of one-way quantum finite automata with periodic behaviors. RAIRO - Theoretical Informatics and Applications - Informatique Théorique et Applications, Tome 36 (2002) no. 3, pp. 277-291. doi : 10.1051/ita:2002014. http://www.numdam.org/articles/10.1051/ita:2002014/
[1] A. Aho, J. Hopcroft and J. Ullman, The Design and Analysis of Computer Algorithms. Addison-Wesley (1974). | MR 413592 | Zbl 0326.68005
[2] A. Ambainis, A. Ķikusts and M. Valdats, On the Class of Languages Recognizable by 1-way Quantum Finite Automata, in Proc. 18th Annual Symposium on Theoretical Aspects of Computer Science. Springer, Lecture Notes in Comput. Sci. 2010 (2001) 305-316. | MR 1890780 | Zbl 0976.68087
[3] A. Ambainis and J. Watrous, Two-way Finite Automata with Quantum and Classical States. Technical Report (1999) quant-ph/9911009. | Zbl 1061.68047
[4] A. Ambainis and R. Freivalds, 1-way Quantum Finite Automata: Strengths, Weaknesses and Generalizations, in Proc. 39th Annual Symposium on Foundations of Computer Science. IEEE Computer Society Press (1998) 332-342.
[5] A. Brodsky and N. Pippenger, Characterizations of 1-Way Quantum Finite Automata, Technical Report. Department of Computer Science, University of British Columbia, TR-99-03 (revised). | Zbl 1051.68062
[6] C. Colbourn and A. Ling, Quorums from Difference Covers. Inform. Process. Lett. 75 (2000) 9-12. | MR 1774826
[7] L. Grover, A Fast Quantum Mechanical Algorithm for Database Search, in Proc. 28th ACM Symposium on Theory of Computing (1996) 212-219. | MR 1427516 | Zbl 0922.68044
[8] J. Gruska, Quantum Computing. McGraw-Hill (1999). | MR 1978991
[9] J. Gruska, Descriptional complexity issues in quantum computing. J. Autom. Lang. Comb. 5 (2000) 191-218. | MR 1778471 | Zbl 0965.68021
[10] T. Jiang, E. Mcdowell and B. Ravikumar, The Structure and Complexity of Minimal nfa's over a Unary Alphabet. Int. J. Found. Comput. Sci. 2 (1991) 163-182. | Zbl 0746.68040
[11] A. Ķikusts, A Small 1-way Quantum Finite Automaton. Technical Report (1998) quant-ph/9810065.
[12] M. Kohn, Practical Numerical Methods: Algorithms and Programs. The Macmillan Company (1987). | MR 892962 | Zbl 0665.65002
[13] A. Kondacs and J. Watrous, On the Power of Quantum Finite State Automata, in Proc. 38th Annual Symposium on Foundations of Computer Science. IEEE Computer Society Press (1997) 66-75.
[14] M. Marcus and H. Minc, Introduction to Linear Algebra. The Macmillan Company (1965). Reprinted by Dover (1988). | MR 188221 | Zbl 0142.26801
[15] M. Marcus and H. Minc, A Survey of Matrix Theory and Matrix Inequalities. Prindle, Weber & Schmidt (1964). Reprinted by Dover (1992). | MR 1215484 | Zbl 0126.02404
[16] C. Mereghetti and B. Palano, Upper Bounds on the Size of One-way Quantum Finite Automata, in Proc. 7th Italian Conference on Theoretical Computer Science. Springer, Lecture Notes in Comput. Sci. 2202 (2001) 123-135. | MR 1915411 | Zbl 1042.68066
[17] C. Mereghetti, B. Palano and G. Pighizzini, On the Succinctness of Deterministic, Nondeterministic, Probabilistic and Quantum Finite Automata, in Pre-Proc. Descriptional Complexity of Automata, Grammars and Related Structures. Univ. Otto Von Guericke, Magdeburg, Germany (2001) 141-148. RAIRO: Theoret. Informatics Appl. (to appear). | Numdam | MR 1908867 | Zbl 1010.68068
[18] C. Mereghetti and G. Pighizzini, Two-Way Automata Simulations and Unary Languages. J. Autom. Lang. Comb. 5 (2000) 287-300. | MR 1778478 | Zbl 0965.68043
[19] C. Moore and J. Crutchfield, Quantum automata and quantum grammars. Theoret. Comput. Sci. 237 (2000) 275-306. | MR 1756213 | Zbl 0939.68037
[20] J.-E. Pin, On Languages Accepted by finite reversible automata, in Proc. 14th International Colloquium on Automata, Languages and Programming. Springer-Verlag, Lecture Notes in Comput. Sci. 267 (1987) 237-249. | MR 912712 | Zbl 0627.68069
[21] P. Shor, Algorithms for Quantum Computation: Discrete Logarithms and Factoring, in Proc. 35th Annual Symposium on Foundations of Computer Science. IEEE Computer Science Press (1994) 124-134. | MR 1489242
[22] P. Shor, Polynomial-time algorithms for prime factorization and discrete logarithms on a quantum computer. SIAM J. Comput. 26 (1997) 1484-1509. | MR 1471990 | Zbl 1005.11065
[23] B. Wichmann, A note on restricted difference bases. J. London Math. Soc. 38 (1963) 465-466. | MR 158857 | Zbl 0123.04302
|
Coach Ringdahl has 10 players on his varsity basketball team.
Coach Ringdahl has 10 players on his varsity basketball team. How many different starting line-ups (5 players) can he choose?
This is a combinations question - we don't care about the order in which we select the 5 players in the starting line-up, only that they are chosen.
{C}_{10,5}=\frac{10!}{\left(5!\right)\left(10-5\right)!}=\frac{10!}{\left(5!\right)\left(5!\right)}
And now let's evaluate this:
\frac{10!}{\left(5!\right)\left(5!\right)}=\frac{\overline{)10}×{\overline{)9}}^{3}×{\overline{)8}}^{2}×7×6×\overline{)5}!}{\left(\overline{)5}!\right)\left(\overline{)5}×\overline{)4}×\overline{)3}×\overline{)2}×1\right)}=3×2×7×6=252
10. What is the probability of these events whenwe randomly select a permutation of the 26 lowercase letters of the English alphabet?
a) The first 13 letters of the permutation are in alphabetical order.
b) a is the first letter of the permutation and z is the last letter.
c) a and z are next to each other in the permutation.
d) a and b are not next to each other in the permutation.
e) a and z are separated by at least 23 letters in the permutation.
f ) z precedes both a and b in the permutation.
The population consists of healthy, ill and hospitalized people. A healthy person remains helthy with probability 0.7 or becomes ill with probability 0.3. An ill person recovers with probability 0.6 or becomes hospitalized with probability 0.4. A hospitalized person may recover with probability 0.5 or remain hospitalized with probability 0.5. Find the number of healthy, ill and hospitalized people after a long run of the illness assuming that initially the number of healthy, ill and hospitalized people is:
{\stackrel{\to }{v}}_{A}=⟨\text{Healthy, ill, hospitalized}⟩=⟨h,\text{ }i,\text{ }o\text{ }⟩=⟨50,\text{ }25,\text{ }12⟩
{\stackrel{\to }{v}}_{B}=⟨\text{Healthy, ill, hospitalized}⟩=⟨h,\text{ }i,\text{ }o⟩=⟨87,\text{ }0,\text{ }0⟩
Sue and Ann are taking the same English class but they do not study together, so whether one passes will be Independent of whether the other passes. In other words, “Sue passes and “Ann passes” are assumed to be independent events. The probebiity that Sue passes English is 0.8 and the probablity that Ann passes English is 0.75 If Ann passes, what then is the conditional probability that Sue also passes?
Suppose the probability it will rain on an particular day is 0.6 . A man goes to the office a probability the man will take a umberalla when it is raining is 0.9 and the probability the man will take a umberalla when it is not raining is 0.2. Calculate the probability that the man carries an umberalla.
|
Preliminary Economics of Black Liquor Gasifier/Gas Turbine Cogeneration at Pulp and Paper Mills | J. Eng. Gas Turbines Power | ASME Digital Collection
Eric D. Larson,
Center for Energy and Environmental Studies, School of Engineering and Applied Science, Princeton University, Princeton, NJ 08540
Contributed by the International Gas Turbine Institute (IGTI) of THE AMERICAN SOCIETY OF MECHANICAL ENGINEERS for publication in the ASME JOURNAL OF ENGINEERING FOR GAS TURBINES AND POWER. Paper presented at the International Gas Turbine and Aeroengine Congress and Exhibition, Stockholm, Sweden, June 2–5 1998; ASME Paper 98-GT-346. Manuscript received by IGTI March 18, 1998; final revision received by the ASME Headquarters January 3, 2000. Associate Technical Editor: R. Kielb.
Larson, E. D., Consonni, S., and Kreutz, T. G. (January 3, 2000). "Preliminary Economics of Black Liquor Gasifier/Gas Turbine Cogeneration at Pulp and Paper Mills ." ASME. J. Eng. Gas Turbines Power. April 2000; 122(2): 255–261. https://doi.org/10.1115/1.483203
Black liquor, the lignin-rich byproduct of kraft pulp production, is burned in boiler/steam turbine cogeneration systems at pulp mills today to provide heat and power for onsite use. Black liquor gasification technologies under development would enable this fuel to be used in gas turbines. This paper reports preliminary economics of
100-MWe
scale integrated black-liquor gasifier/combined cycles using alternative commercially proposed gasifier designs. The economics are based on detailed full-load performance modeling and on capital, operating and maintenance costs developed in collaboration with engineers at Bechtel Corporation and Stone & Webster Engineering. Comparisons with conventional boiler/steam turbine systems are included. [S0742-4795(00)00402-6]
gas turbines, paper industry, economics
Boilers, Economics , Fuel gasification, Gas turbines, Heat, High temperature, Power stations, Project cost estimation, Pulp, Sulfate waste liquor, Combined heat and power, Oxygen, Fuels, Cogeneration systems, Kraft papers, Modeling, Steam
AFPA, 1996, Fact Sheet on 1994 Energy Use in the U.S. Pulp and Paper Industry, American Forest & Paper Association, Wash., DC.
Adams, T. N., Frederick, W. J., Hupa, M., Iisa, K., Jones, A., and Tran, H., 1997, Kraft Recovery Boilers, Tappi Press, Atlanta, GA.
Larson, E. D., Yang, W., McDonald, G., Frederick, W. J., Malcolm, E. W., McDonough, T. J., Iisa, K., Kreutz, T. G., and Brown, C., 1998, “Impacts of Integrated Gasification—Combined Cycle Recovery Technology on the Chemical Aspects of Kraft Pulping and Recovery,” Proceedings of the 1998 Int’l. Chemical Recovery Conf., Tappi Press, Atlanta, GA, pp. 1–18.
Larson, E. D., and Consonni, S., 1997, “Performance of Black Liquor Gasifier/Gas Turbine Combined Cycle Cogeneration in the Kraft Pulp and Paper Industry,” Proceedings, 3rd Biomass Conf. of the Americas, Overend and Chornet, eds., Elsevier Science Inc., Tarrytown, NY, pp. 1495–1512.
Consonni, S., 1992, “Performance Prediction of Gas/Steam Cycles for Power Generation,” Ph.D. thesis 1893-T, Mechanical and Aerospace Eng. Dept., Princeton Univ., Princeton, NJ.
Larson, E. D., Consonni, S., Berglin, N., and Kreutz, T., 1996, “Advanced Technologies for Biomass-Energy Utilization in the Pulp and Paper Industry,” report to U.S. Dept. of Energy from Center for Energy and Environmental Studies, Princeton Univ., Princeton, NJ.
Lorson, H., Schingnitz, M., White, V. F., and Dean, D. R., 1996, “Black Liquor Recovery by Pressurized Oxygen-Blown Gasification,” Proceedings, 1996 Engineering Conference, Tappi Press, Atlanta, GA, pp. 557–565.
Aghamohammadi, B., Mansour, M. N., Durai-Swamy, K., Steedman, W., Rockvam, L. N., Brown, C., and Smith, P., 1995, “Large Scale Pilot Testing of the MTCI/Thermochem Black Liquor Steam Reformer,” Proceedings, 1995 Int’l. Chemical Recovery Conf., Tappi Press, Atlanta, GA, pp. B297–B301.
Pietruszkiewicz, J., 1997, “Final Report for Costing High-Temperature Black Liquor Gasifier/Gas Turbine Cogeneration Systems,” prepared by Bechtel National, Inc. (Gaithersburg, MD) for Center for Energy & Environmental Studies, Princeton Univ., Princeton, NJ.
Gastwirth, G., 1997, “Costing Study of the Stonechem Black Liquor Gasifier/Gas Turbine Combined Cycle Cogeneration System,” prepared by Stone and Webster Engineering Corp. (New York, NY) for Center for Energy & Environmental Studies, Princeton Univ., Princeton, NJ.
|
Applied Sciences | Free Full-Text | Evaluation of Classical Mathematical Models of Tumor Growth Using an On-Lattice Agent-Based Monte Carlo Model
Review of Convective Heat Transfer Modelling in CFD Simulations of Fire-Driven Flows
Experimental and Numerical Investigation of an Innovative Method for Strengthening Cold-Formed Steel Profiles in Bending throughout Finite Element Modeling and Application of Neural Network Based on Feature Selection Method
Ruiz-Arrebola, S.
Lallena, A. M.
Evaluation of Classical Mathematical Models of Tumor Growth Using an On-Lattice Agent-Based Monte Carlo Model †
Samuel Ruiz-Arrebola
Antonio M. Lallena
Servicio de Oncología Radioterápica, Hospital Universitario Marqués de Valdecilla, E-39008 Santander, Spain
Unidad de Radiofísica, Hospital Universitario Clínico San Cecilio, E-18016 Granada, Spain
Instituto de Investigación Biosanitaria (ibs.GRANADA), Universidad de Granada, E-18016 Granada, Spain
Departamento de Radiología y Medicina Física, Universidad de Granada, E-18071 Granada, Spain
Instituto de Biopatología y Medicina Regenerativa (IBIMER), Universidad de Granada, E-18071 Granada, Spain
Departamento de Física Atómica, Molecular y Nuclear, Universidad de Granada, E-18071 Granada, Spain
This work is part of the Ph.D. thesis developed by S. Ruiz-Arrebola within the Ph.D. program “Clinical Medicine and Public Health” at the University of Granada, Spain.
Academic Editor: Alexander N. Pisarchik
Purpose: To analyze the capabilities of different classical mathematical models to describe the growth of multicellular spheroids simulated with an on-lattice agent-based Monte Carlo model that has already been validated. Methods: The exponential, Gompertz, logistic, potential, and Bertalanffy models have been fitted in different situations to volume data generated with a Monte Carlo agent-based model that simulates the spheroid growth. Two samples of pseudo-data, obtained by assuming different variability in the simulation parameters, were considered. The mathematical models were fitted to the whole growth curves and also to parts of them, thus permitting to analyze the predictive power (both prospective and retrospective) of the models. Results: The consideration of the data obtained with a larger variability of the simulation parameters increases the width of the
{\chi }^{2}
distributions obtained in the fits. The Gompertz model provided the best fits to the whole growth curves, yielding an average value of the
{\chi }^{2}
per degree of freedom of 3.2, an order of magnitude smaller than those found for the other models. Gompertz and Bertalanffy models gave a similar retrospective prediction capability. In what refers to prospective prediction power, the Gompertz model showed by far the best performance. Conclusions: The classical mathematical models that have been analyzed show poor prediction capabilities to reproduce the MTS growth data not used to fit them. Within these poor results, the Gompertz model proves to be the one that better describes the growth data simulated. The simulation of the growth of tumors or multicellular spheroids permits to have follow-up periods longer than in the usual experimental studies and with a much larger number of samples: this has permitted performing the type of analysis presented here. View Full-Text
Keywords: on-lattice agent-based models; classical tumor growth models; exponential; Gompertz; logistic; Bertalanffy; multicellular spheroids; Monte Carlo on-lattice agent-based models; classical tumor growth models; exponential; Gompertz; logistic; Bertalanffy; multicellular spheroids; Monte Carlo
Ruiz-Arrebola, S.; Guirado, D.; Villalobos, M.; Lallena, A.M. Evaluation of Classical Mathematical Models of Tumor Growth Using an On-Lattice Agent-Based Monte Carlo Model. Appl. Sci. 2021, 11, 5241. https://doi.org/10.3390/app11115241
Ruiz-Arrebola S, Guirado D, Villalobos M, Lallena AM. Evaluation of Classical Mathematical Models of Tumor Growth Using an On-Lattice Agent-Based Monte Carlo Model. Applied Sciences. 2021; 11(11):5241. https://doi.org/10.3390/app11115241
Ruiz-Arrebola, Samuel, Damián Guirado, Mercedes Villalobos, and Antonio M. Lallena. 2021. "Evaluation of Classical Mathematical Models of Tumor Growth Using an On-Lattice Agent-Based Monte Carlo Model" Applied Sciences 11, no. 11: 5241. https://doi.org/10.3390/app11115241
|
sexoagotadorogyr 2022-04-01 Answered
y{}^{″}\left(t\right)+12{y}^{\prime }\left(t\right)+32y\left(t\right)=32u\left(t\right)
y\left(0\right)={y}^{\prime }\left(0\right)=0
Y\left(p\right)=x=\frac{32}{\left(p\left({p}^{2}+12p+32\right)\right)}
Kason Palmer 2022-03-31 Answered
\frac{1}{{s}^{2}-9s+20}
{L}^{-1}\left\{\frac{1}{{\left(s-\frac{9}{2}\right)}^{2}-{\left(\frac{1}{2}\right)}^{2}}\right\}=2{L}^{-1}\left\{\frac{\frac{1}{2}}{{\left(s-\frac{9}{2}\right)}^{2}-{\left(\frac{1}{2}\right)}^{2}}\right\}
Amya Horn 2022-03-31 Answered
Solve the differential equation by using method of variation of parametrs
y{}^{‴}+9{y}^{\prime }=\frac{1}{f\left(x\right)}
f\left(x\right)=\mathrm{cos}11x
Roy Roth 2022-03-31 Answered
{\int }_{0}^{\infty }\frac{\mathrm{cos}\left(xt\right)}{1+{t}^{2}}dt
I'm supposed to solve this using Laplace Transformations.
jisu61hbke 2022-03-31 Answered
H\left(s\right)=\frac{s-5}{\left(s-3\right)\left(s-1\right)}
The inverse Laplace transform of H(s) is equal to
f\cdot g
Alvaro Baca 2022-03-31
ikramkeyslo4s 2022-03-30 Answered
I am trying to take the laplace transform of
\mathrm{cos}\left(t\right)u\left(t-\pi \right)
. Is it valid for me to treat it as
\left(\left(\mathrm{cos}\left(t\right)+\pi \right)-\pi \right)u\left(t-\pi \right)
and treat
\mathrm{cos}\left(t\right)-\pi
as f(t) and use the 2nd shifting property, or is this not the correct procedure?
Milan Manandhar 2022-03-29
wibawanyacl3q 2022-03-29 Answered
y{}^{″}+2{y}^{\prime }=3{x}^{2}+2x
Kendall Daniels 2022-03-28 Answered
{y}_{c}
{x}^{2}y{}^{″}-4x{y}^{\prime }+6y={x}^{3};\text{ }{y}_{c}={c}_{1}{x}^{2}+{c}_{2}{x}^{3}
atenienseec1p 2022-03-26 Answered
Could anyone enlighten me on how to find the Laplace Transform of
\frac{1-\mathrm{cos}\left(t\right)}{t}
Sanai Huerta 2022-03-25 Answered
Determine the Laplace Transform
{\left(t-1\right)}^{4}
Find a second-order lincar equation for which
y\left(x\right)={c}_{1}{e}^{3x}+{c}_{2}{e}^{5x}+\mathrm{sin}\left(2x\right)
is the general solution
Coradossi7xod 2022-03-25 Answered
Consider the following linear second-order homogeneous differential equation with constant coefficients and two initial conditions
\frac{{d}^{2}y\left(t\right)}{{dt}^{2}}-\frac{1}{3}\cdot \frac{dy\left(t\right)}{dt}-\frac{2}{9}\cdot y\left(t\right)=0,\text{ }y\left(0\right)=0,\text{ }\frac{dy\left(0\right)}{dt}=-3
Blackettyl2j 2022-03-25 Answered
Solve the following linear second order homogeneous differential equation
y{ }^{″}-\frac{3}{2}{y}^{\prime }+\frac{9}{16}y=0,
y\left(0\right)=1, y\text{'}\left(0\right)=-1
Pizzadililehz 2022-03-25 Answered
Find the general solution of the given differential equations
4y{}^{″}-4{y}^{\prime }-3=0
Destinee Hensley 2022-03-25 Answered
Find the general solution of the given second-order differential equations
2y{}^{″}+2{y}^{\prime }+y=0
Jackson Floyd 2022-03-25 Answered
Find ODE having the given function as its general solution.
y={x}^{2}+{c}_{1}{e}^{2x}+{c}_{2}{e}^{3x}
kembdumatxf 2022-03-25 Answered
Find the general solution to the following second order differential equations.
y{}^{″}-{6}^{\prime }+25y=0
|
Profit_model Knowpia
The profit model is the linear, deterministic algebraic model used implicitly by most cost accountants. Starting with, profit equals sales minus costs, it provides a structure for modeling cost elements such as materials, losses, multi-products, learning, depreciation etc. It provides a mutable conceptual base for spreadsheet modelers. This enables them to run deterministic simulations or 'what if' modelling to see the impact of price, cost or quantity changes on profitability.
Basic modelEdit
{\displaystyle \pi =pq-(F_{n}+wq)\,}
π is profit
p is sales price
Fn is fixed costs
w is variable costs per unit sold
q is quantity sold
For an expansion of the model see below.
The justification for wanting to express profit as an algebraic model is given by Mattessich in 1961:
'To some operations analysts the mere translation of accounting models into mathematical :terminology, without a calculus for determining an optimum, might appear to be a rather :pedestrian task. We are convinced, however, that as long as accounting methods are acceptable :to the industry the mere change to a mathematical formulation will be advantageous for :several reasons: (1) it can be considered a prerequisite for applying electronic data :processing to certain accounting problems, (2) it articulates the structure of the accounting :models and illuminates accounting methods from a new point of view, revealing many facets so :far neglected or unobserved, (3) it enables a general and hence more scientific presentation :of many accounting methods, (4) it facilitates the exploration of new areas, thereby :accelerating the advancement of accounting, (5) it leads to more sophisticated methods and :might help to lay the foundations for close cooperation of accounting with other areas of :management science.'[1]
Most of the definitions in cost accounting are in an unclear narrative form, not readily associated with other definitions of accounting calculations. For example, preparing a comparison of fixed cost variances in stock under different stock valuation methods can be confusing. Another example is modelling labour variances with learning curve corrections and stock level changes. With the absence of a basic profit model in an algebraic form, confident development of such models is difficult.
The development of spreadsheets has led to a decentralisation of financial modelling. This has often resulted in model builders lacking training in model construction. Before any professional model is built it is usually considered wise to start by developing a mathematical model for analysis. The profit model provides a general framework plus some specific examples of how such an a priori profit model might be constructed.
The presentation of a profit model in an algebraic form is not new. Mattessich's model,[1] while large, does not include many costing techniques such as learning curves and different stock valuation methods. Also, it was not presented in a form that most accountants were willing or able to read. This paper presents a more extended model analysing profit but it does not, unlike Mattessich, extend to the balance sheet model. Its form, of starting with the basic definition of profit and becoming more elaborate, may make it more accessible to accountants.
Most cost accounting textbooks [2] explain basic Cost Volume Profit modeling in an algebraic form, but then revert to an 'illustrative' [3] approach. This 'illustrative' approach uses examples or narrative to explain management accounting procedures. This format, though useful when communicating with humans, can be difficult to translate into an algebraic form, suitable for computer model building. Mepham [4] extended the algebraic, or deductive, approach to cost accounting to cover many more techniques. He develops his model to integrate with the optimizing models in operations research. The profit model comes out of Mephams work, extending it but only in a descriptive, linear form.
Model extensionsEdit
The basic profit model is sales minus costs. Sales are made up of quantity sold multiplied by their price. Costs are usually divided between Fixed costs and variable costs.
Sales revenue = pq = price × quantity sold
Cost of sales = wq = unit cost × quantity sold
Administration, selling, engineers, personnel etc. = Fn = fixed post-manufacturing overheads
Profit = π
Thus the profit can be calculated from:
{\displaystyle \pi =pq-(F_{n}+wq)\qquad \qquad (1)}
Notice that w (average unit production cost) includes the fixed and variable costs. The square brackets contain the cost of goods sold, wq not cost of good made wx where x = cost of good sold.
To show cost of good sold, the opening and closing finished goods stocks need to be included The profit model would then be:
Opening stock = go w = opening stock quantity × unit cost
Cost of stock = g1 w = closing stock quantity × unit cost
Cost of production = wx = unit production cost × quantity made:
{\displaystyle \pi =pq-[F_{n}+wx+g_{0}w-g_{1}w]\qquad \qquad (2)}
Presenting the profit calculation in this form immediately demands that some of the costs be more carefully defined.
Production costsEdit
The unit production costs (w) can be separated into fixed and variable costs:
{\displaystyle w={\frac {F_{m}+vx}{x}}\qquad \qquad (3)}
Fm = manufacturing fixed costs;
v = variable costs per unit;
x = production quantity.
The introduction of this separation of w allows for consideration of the behaviour of costs for different levels of production. A linear cost curve is assumed here, divided between the constant (F) and its slope (v). If the modeller has access to the details of non-linear cost curves then w will need to be defined by the appropriate function.
Replacing wx in (equation 2) and making F = Fn + Fm:
{\displaystyle \pi =pq-[F+vx+g0w-g1w]\qquad \qquad (4)}
Variable-cost elementsEdit
Moving on to other extensions of the basic model, the cost elements such as direct materials, direct labour and variable overheads can be included. If a non-linear function is available and thought useful such functions can be substituted for the functions used here.
The material cost of sales = m * µ * q, where
m is the amount of material in one unit of finished goods.
µ is the cost per unit of the raw material.
The labour cost of sales = l λ q, where
l is the amount of labour hours required to make one unit of finished goods
λ is the labour cost (rate) per hour.
The variable overhead cost of sales = nq where n is the variable overhead cost per unit.
This is not here subdivided between quantity per finished goods units and cost per unit.
Thus the variable cost v * q can now be elaborated into:
π = pq - [F+(mµ q + l λq + nq)] …………(equation 5)
If the production quantity is required the finished goods stock will need to be added.
In a simple case two materials can be accommodated in the model by simply adding another m * µ. In more realistic situations a matrix and a vector will be necessary (see later).
If material cost of purchases is to be used rather than material cost of production it will be necessary to adjust for material stocks. That is,
mx = md0 + mb - md1………… (equation 6)
d = material stock quantity,
0 = opening, 1 = closing,
b = quantity of material purchased
m = the amount of material in one unit of finished goods
x = quantity used in production
All depreciation rules can be stated as equations representing their curve over time. The reducing balance method provides one of the more interesting examples.
Using c = cost, t = time, L = life, s = scrap value, Fd = time based depreciation:
Depr/yr = Fd = c (s/c)(t-L)/L * [L(s/c)1/L] …………… (equation 7)
This equation is better known as the rule: Depreciation per year = Last year's written down value multiplied by a constant %
The limits are 0 < t < L, and the scrap value has to be greater than zero. (For zero use 0.1).
Remembering that time-based depreciation is a fixed cost and usage-based depreciation can be a variable cost, depreciation can easily be added into the model (equation 5).
Thus, the profit model becomes:
π = pq - [F + Fd + (mµ + lλ + n + nd)q].......... (equation 8)
where, nd = usage (as q) based depreciation and π = annual profit.
Stock valuationEdit
In the above, the value of the unit finished goods cost ‘w' was left undefined. There are numerous alternatives to how stock (w) is valued but only two will be compared here.
The marginal versus absorption costing debate, includes the question of the valuation of stock (w).
Should w = v or as (3) w = (Fm + v x)/x.
(i) Under marginal costing: w = v. Inserting in (4),
π = pq- [F + v x + g0w0 - g1 w1]
π = pq- [F + v x + g0w0 - g1 v]
This can be simplified by taking v out and noting, opening stock quantity + production - closing stock quantity = sales quantity (q) so,
π = p q - [F + v q]………….. (equation 9)
Note, v q = variable cost of goods sold.
(ii) Using full (absorption) costing Using (equation 3), where xp = planned production, x1 = period production w = (Fm + v xp)/xp = Fm/xp + v. This can be shown to result in:
π = p q - [Fn + Fm + v q + Fm/xp * (q-x1)]………..(equation 10)
Note the strange presence of 'x' in the model. Notice also that the absorption model (equation 10) is the same as the marginal costing model (equation 9) except for the end part:
F/xp * (q-x1)
This part represents the fixed costs in stock. This is better seen by remem¬bering q — x= go—g1 so it could be written
F/xp • (g0—g1)
The model form with 'q' and 'x' in place of' g0 and g1 allows profits to be calculated when only the sales and production figures are known.
A spreadsheet could be prepared for a company with increasing then decreasing levels of sales and constant production. It could have another column showing profit under increasing sales and constant production. Thus the effects of carrying fixed costs in stock can be simulated. Such modelling thus provides a very useful tool in the marginal versus full costing debate.
Modelling for lossesEdit
One way of modeling for losses is to use:
Fixed losses, (quantity) = δf,
Variable losses (%) = δv,
Material losses = mδ,
Production losses = pδ
The model, with all these losses together will look like,
π = v q – [F + µ * mδf + {mµ(1 + mδv) + lλ+n) * (1+ pδ* (q +pδf)]........ (equation 11)
Note that labour and variable overhead losses could also have been included.
Multi-productsEdit
So far the model has assumed very few products and/or cost elements. As many firms are multi-product the model they use must be able to handle this problem. Whilst the mathematics here is straightforward the accounting problems introduced are enormous: the cost allocation problem being a good example. Other examples include calculation of break-even points, productivity measures and the optimisation of limited resources. Here only the mechanics of building a multi-dimension model will be outlined.
If a firm sells two products (a and b) then the profit model (equation 9),
π = pq —(F +vq) becomes
π = (pa *qa +pb *qb) - [F + va*qa + vb *qb]
All fixed costs have been combined in F
Therefore, for multiple products
π = Σ(pq) - [F + Σ(vq)].... (equation 12)
Where Σ = the sum of. Which can be drafted as a vector or matrix in a spreadsheet
π = Σpq - [F + Σ(Σmμ +Σlλ + Σn)q]..... (equation 13)
VariancesEdit
The profit model may represent actual data (c), planned data (p)or standard data (s) which is the actual sales quantities at the planned costs.
The actual data model will be (using equation 8):
π = pc*qc - [Fc + (mµc + lλc + nc)qc]
The planned data model will be (using equation 8):
π = pp*qp - [Fp + (mµp + lλp + np)qp]
The standard data model will be (using equation 8):
π = pp*qc - [Fp + (mµp + lλp + np)qc]
Operating variances are obtained by subtracting the actual model from the standard model.
Learning Curve ModelEdit
It is possible to add non linear cost curves to the Profit model. For example, if with learning, the labour time per unit will decrease exponentially over time as more product is made, then the time per unit is:
l = r * q−b
where r = average time. b = learning rate. q = quantity.
Inserting into equation 8
π = pq - [F + (mµ + rq−bλ + n)q]
This equation is best solved by trial and error, the Newton Raphson method or graphing. Like depreciation within the model, the adjustment for learning does provide a form of non-linear sub-modelling.
Percentage Change ModelEdit
Rather than the variable be absolute amounts, they might be percentage changes. This represents a major change in approach from the model above. The model is often used in the 'now that ... (say) the cost of labour has gone up by 10%' format. If a model can be developed that only uses such percentage changes then the cost of collecting absolute quantities will be saved.[5]
The notation used below is of attaching a % sign to variables to indicate the change of that variable, for example, p% = 0.10 if the selling price is assumed to change by 10%,
Let x = q and C = contribution
Starting with the absolute form of the contribution model (equation (9) rearranged):
π + F = C = (p — v)q.
The increase in the contribution which results from an increase in p, v and/or q can be calculated thus:
C(l + C%) = [p(l+p%)- v(l + v%)]q(l+q%)
rearranging and using α = (p — v)/p,
C% = ((l+q%)/α)[p%-(l - α)v%]+q%...... (equation 18)
This model might look messy but it is very powerful. It makes very few demands on data, especially if some of the variables do not change. It is possible to develop most of the models presented above in this percentage change format.
For typical components / steps of business modeling, see the list for "Equity valuation" under Outline of finance#Discounted cash flow valuation.
^ a b Mattessich, R. (1961). 'Budgeting models and system simulation', The Accounting Review, 36(3), 384–397.
^ Drury, C. (1988). Management and Cost Accounting, London: V.N.R
^ Ijiri, Y. (1983). 'New dimensions in accounting education: computers and algorithms,' Issues In Accounting Research, 168–173.
^ Mepham, M. (1980). Accounting Models, London: Pitmans
^ Eilon, S. (1984), The Art of Reckoning: analysis of performance criteria, London: Academic Press
Girardi, Dario; Giacomello, Bruno; Gentili, Luca (2011). "Budgeting Models and System Simulation: A Dynamic Approach". SSRN Electronic Journal. doi:10.2139/ssrn.1994453.
Metcalfe M. and Powell P. (1994) Management Accounting: A Modelling Approach. Addison Wesley, Wokingham.
|
{x}^{2}+{y}^{2}-{z}^{2}=0
A=\left[1,0,1\right]
x=0
X\in P\left(U\right)
X=\left[0,1,t\right]
\alpha
\begin{array}{rl}\alpha \left(\left[0,1,t\right]\right)& =\left[B\left(\left(0,1,t\right),\left(0,1,t\right)\right)\left(1,0,1\right)-2B\left(\left(1,0,1\right),\left(0,1,t\right)\right)\left(0,1,t\right)\right]\\ & =\left[1-{t}^{2},2t,1+{t}^{2}\right]\\ \text{ or }\alpha \left(\left[0,0,1\right]\right)=\left[-1,0,1\right].\end{array}
describe maximum and minimum for the function of two variables x and y and saddle poin
Suppose that the series of a sequence an in real converges to a real number. Show that limit to infinity summation k = n + 1 = 0
{t}^{2}u\left(t-2\right)
{t}^{2}
Elliptic integrals The length of the ellipse
x=acost,y=bsint,0≤t≤2π
x=a\mathrm{cos}t,\phantom{\rule{1em}{0ex}}y=b\mathrm{sin}t,\phantom{\rule{1em}{0ex}}0\le t\le 2\pi
=4a∫π/201−e2cos2t√dt
=4a{\int }_{0}^{\pi /2}\sqrt{1-{e}^{2}{\mathrm{cos}}^{2}t}dt
where e
is the ellipse's eccentricity. The integral in this formula, called an elliptic integral, is non elementary except when e=0
or 1 a. Use the Trapezoidal Rule with n=10
to estimate the length of the ellipse when a=1
and e=1/2
. b. Use the fact that the absolute value of the second derivative of f(t)=1−e2cos2t√
f\left(t\right)=\sqrt{1-{e}^{2}{\mathrm{cos}}^{2}t}
is less than 1 to find an upper bound for the error in the estimate you obtained in part (a).
A strawberry jam company regularly receives large shipments of strawberries. For each shipment that is received, a supervisor takes a random sample of 500 strawberries to see what percent of the sample is bruised and performs a significance test. If the sample shows convincing evidence that more than 10%, percent of the entire shipment of strawberries is bruised, then they will request a new shipment of strawberries.
Let p represent the proportion of the strawberries in a shipment that are bruised.
|
\frac{dy}{dx}=y+1 Solving the above given differential equation, yields the following
\frac{dy}{dx}=y+1
y+1={e}^{x+C}
y=C{e}^{x}-1
⇒
y=-1
C=0
y=-1
timbreoizy
y=-1
is not a singular solution because it is included in the general solution
y=C{e}^{x}-1
C=0
Solutions are only called singular when they are not attainable from the general solution form. In fact, I believe all linear first order homogenous equations have no singular solutions.
y=-1
is not a singular solution, because it does not pass by any point
\left({x}_{0},\text{ }{y}_{0}\right)
such that the initial value problem
\left\{\begin{array}{l}{y}^{\prime }=y+1\\ y\left({x}_{0}\right)={y}_{0}\end{array}
has more than one solution: in point of fact, the Cauchy problem
\left\{\begin{array}{l}{y}^{\prime }=y+1\\ y\left({x}_{0}\right)=-1\end{array}
has only the one maximal solution
y=-1
, whatever
{x}_{0}
\frac{dw}{dt}
\frac{dw}{dt}
w=x\mathrm{sin}y,\text{ }x={e}^{t},\text{ }y=\pi -t
t=0
Variation of parameters method for differential equations.
x={e}^{t}
and then find the general solution for the following differential equation
2{x}^{2}y{}^{″}-6x{y}^{\prime }+8y=2x+2{x}^{2}\mathrm{ln}x
It seems a little suspicious that i can factor out 2 and x first.
{x}^{2}y{}^{″}-3x{y}^{\prime }+4y=x\left(\mathrm{ln}\left(x\right)+1\right)
and if we factor out
{x}^{2}
now we end up with
y{}^{″}-\frac{3}{x}{y}^{\prime }+\frac{4}{{x}^{2}}y=\frac{x\mathrm{ln}x+1}{x}
(1) Therefore substituting
x={e}^{t}
(1) now becomes
y{}^{″}-\frac{3}{{e}^{t}}{y}^{\prime }+\frac{4}{{e}^{2t}}y=\frac{{e}^{\mathtt{+}}1}{{e}^{t}}
. We need constant coefficients in order to use the variation of parameters method am i right?
\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)
{T}^{2}-4\mathrm{\Delta }\ge \phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\le \phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}=0
T=a+c
\mathrm{\Delta }=ad-bc=DetA
±2i
\delta
Using a change in variables, Show that the boundary value problem:
-\frac{d}{dx}\left(p\left(x\right){y}^{\prime }\right)+q\left(x\right)y=f\left(x\right),
a\le x\le b,
y\left(a\right)=\alpha ,
y\left(b\right)=\beta
Can be transformed to the form:
-\frac{d}{dw}\left(p\left(w\right){z}^{\prime }\right)+q\left(w\right)z=F\left(w\right),
0\le w\le 1,
z\left(0\right)=0,
z\left(1\right)=0
V\left(x,y\right)=x-y-y\mathrm{ln}\left(\frac{x}{y}\right)
x,y>0
and y is fixed. V is constructed to study the stability of equilibrium point
x=y
\frac{dx\left(t\right)}{dt}=y-x\left(t\right).
{y}^{\prime 2}-y{y}^{\prime }+{e}^{x}=0
How would we solve the same initial value problem but instead of
c\in \mathbb{R}
being a constant we have a function
f:{\mathbb{R}}^{n}×\left[0,t\right)\to \mathbb{R}
in its place such that the problem becomes:
\left\{\begin{array}{ll}{u}_{t}+b\cdot \left({D}_{x}u\right)+f\left(x,t\right)u=h\left(x,t\right)& \text{in }\phantom{\rule{0.167em}{0ex}}{\mathbb{R}}^{n}×\left(0,\mathrm{\infty }\right)\\ u\left(x,0\right)=g\left(x\right)& \text{on }\phantom{\rule{0.167em}{0ex}}{\mathbb{R}}^{n}×\left\{t=0\right\}\end{array}
What steps would I take to find a function u(x,t) that satisfies this? I am assuming u(x,t) will be similar to the function I found for the original problem but with some additional integrals (after playing with it for a little), but I am unsure.
I am fairly new to partial differential equations so any help will be appreciated.
|
Differential equations, solved and explained.
Differential equations questions and answers
Recent questions in Differential Equations
A particle moves along the curve
x=2{t}^{2}y={t}^{2}-4t
and z=3t-5 where t is the time.find the components of the velocity at t=1 in the direction i-3j+2k
How to solve this equation
y-4y+2y-16y=4x+1
using Method of Undetermined Coefficient, Variation of Parameters and Laplace Transformation
xydy-{y}^{2}dx=\left(x+y{\right)}^{2}{e}^{\left(}-y/x\right)
X\left(s\right)=\frac{s+1}{\left(s+2\right)\left({s}^{2}+2s+2\right)\left({s}^{2}+4\right)}
X\left(s\right)=\frac{1}{\left(2{s}^{2}+8s+20\right)\left({s}^{2}+2s+2\right)\left(s+8\right)}
X\left(s\right)=\frac{1}{{s}^{2}\left({s}^{2}+2s+5\right)\left(s+3\right)}
Find Laplace transform of the given function
t{e}^{-4t}\mathrm{sin}3t
dy/dx=15{x}^{2}{e}^{-y}
Explain First Shift Theorem & its properties?
Find the inverse laplace transform of the function
Y\left(s\right)=\frac{{e}^{-s}}{s\left(2s-1\right)}
\left(s\right)=2{t}^{2}+{\int }_{0}^{t}\mathrm{sin}\left[2\left(t-\tau \right)\right]x\left(\tau \right)d\tau
y+y={e}^{-2t}\mathrm{sin}t,y\left(0\right)=y\prime \left(0\right)=0
solution of given initial value problem with Laplace transform
Find the Laplace transform
L\left\{{u}_{3}\left(t\right)\left({t}^{2}-5t+6\right)\right\}
a\right)F\left(s\right)={e}^{-3s}\left(\frac{2}{{s}^{4}}-\frac{5}{{s}^{3}}+\frac{6}{{s}^{2}}\right)
b\right)F\left(s\right)={e}^{-3s}\left(\frac{2}{{s}^{3}}-\frac{5}{{s}^{2}}+\frac{6}{s}\right)
c\right)F\left(s\right)={e}^{-3s}\frac{2+s}{{s}^{4}}
d\right)F\left(s\right)={e}^{-3s}\frac{2+s}{{s}^{3}}
e\right)F\left(s\right)={e}^{-3s}\frac{2-11s+30{s}^{2}}{{s}^{3}}
Solve the Differential equations
2y\prime \prime +3y\prime -2y=14{x}^{2}-4x-11,y\left(0\right)=0,y\prime \left(0\right)=0
{y}^{\prime }+2y={y}^{2}
Solve the Laplace transforms
\stackrel{˙}{x}-2\stackrel{¨}{x}+x={e}^{t}t
\text{ given }t=0\text{ and }x=0\text{ and }x=1
Calculate the Laplace transform
L\left\{\mathrm{sin}\left(t-k\right)\cdot H\left(t-k\right)\right\}
Solve the following IVP using Laplace Transform
{y}^{\prime }-2y=1-t,y\left(0\right)=4
f\left(t\right)={e}^{t}\mathrm{cos}t
How to solve for the equation
{y}^{‴}+4{y}^{″}+5{y}^{\prime }+2y=4x+16
using laplace transform method given that
y\left(0\right)=0,{y}^{\prime }\left(0\right)=0,\text{and }{y}^{″}\left(0\right)=0
Find the laplace transform of the following
f\left(t\right)=t{u}_{2}\left(t\right)
F\left(s\right)=\left(\frac{1}{{s}^{2}}+\frac{2}{s}\right){e}^{-2s}
Use the Laplace transform to solve the given initial-value problem
y{}^{″}+2{y}^{\prime }+y=0,y\left(0\right)=1,{y}^{\prime }\left(0\right)=1
Speaking of differential equations, these are used not only by those students majoring in Physics because solving differential equations is also quite common in Statistics and Financial Studies. Explore the list of questions and examples of equations to get a basic idea of how it is done.
These answers below are meant to provide you with the starting points as you work with your differential equations. If you need specific help or cannot understand the rules behind the answers that are presented below, start with a simple equation and learn with the provided solutions..
|
On the Existence of Central Configurations of -Body Problems
Yueyong Jiang, Furong Zhao, "On the Existence of Central Configurations of -Body Problems", Advances in Mathematical Physics, vol. 2014, Article ID 629467, 5 pages, 2014. https://doi.org/10.1155/2014/629467
Yueyong Jiang1 and Furong Zhao 1,2
Academic Editor: Juan C. Marrero
We prove the existence of central configurations of the -body problems with Newtonian potentials in . In such configuration, masses are symmetrically located on the -axis, masses are symmetrically located on the -axis, and masses are symmetrically located on the -axis, respectively; the masses symmetrically about the origin are equal.
The Newtonian -body problems [1–3] is concerned with the motions of particles with masses and positions ; the motion is governed by Newton's second law and the Universal law: where with Newtonian potential Consider the space that is, suppose that the center of mass is fixed at the origin of the coordinate axis. Because the potential is singular when two particles have same position, it is natural to assume that the configuration avoids the collision set for some . The set is called the configuration space.
Definition 1 (see [2, 3]). A configuration is called a central configuration if there exists a constant such that The value of constant in (4) is uniquely determined by where Since the general solution of the -body problem cannot be given, great importance has been attached to search for particular solutions from the very beginning. A homographic solution is a configuration which is self-similar for all time. Central configurations and homographic solutions are linked by the Laplace theorem [3]. Collapse orbits and parabolic orbits have relations with the central configurations [2, 4–6]. So finding central configurations becomes very important.
Long and Sun [7] studied four-body central configurations with some equal masses; Albouy et al. [8] took an alternate approach to the symmetry of planar four-body convex central configurations; Llibre and Mello [9] considered triple and quadruple nested central configurations; Corbera and Llibre [10] analyzed the existence of central configurations of nested regular polyhedra. Based on the above works, we study central configuration for Newtonianc -body problems. In the -body problems, masses are symmetrically located on the -axis, masses are symmetrically located on the -axis, and masses are symmetrically located on the -axis, the masses located on the same axis and symmetrical about the origin are equal (see Figure 1 for , , and ).
In this paper, we extend the result of Corbera-Llibre by the following.
Theorem 2. For -body problem in where , , and , there is at least one central configuration such that points are located along the first axis symmetric with respect to the origin, points are located on the second axis symmetric with respect to the origin, and points are located on the third axis, also symmetric with respect to the origin. Along each axis, the masses symmetric with respect to the origin are equal.
To begin, we take a coordinate system which simplifies the analysis. The particles have positions given by where , .
The masses are given by , , and , where .
By the symmetries of the system, (4) is equivalent to the following equations: for , In order to simplify the equations, we defined the following:
, where ; ; , where ; , , where ; , where ; , where ; , where ; , where ; , , , where .
Equations (8) and (9) can be written as a linear system of the form given by The column vector is given by the variables . Since the are function of , we write the coefficient matrix as .
2.2. For , , and
Lemma 3. When , , and , there exists such that (10) has a unique solution satisfying and .
Proof. If we consider as a function of and , then is an analytic function and nonconstant. When , , and , (10) is equivalent to where It is obvious that By implicit function theorem, there exists such that (10) has a unique solution satisfying and .
The proof for , , and is done by induction. At each step, we assume we have inductively shown the existence of a central configuration symmetrically located at the -axis, and we add two particles with zero masses. Using continuity with respect to the masses of the above defined equations, we prove the existence of a central configuration when the mass of the added particles is zero and the mass of the remaining particles are changed suitably. We then verify that the Hessian of this new system of masses has no kernel and we conclude that the two new particles can have nonzero masses which still maintain a central configuration.
For , , and , we claim that there exists , such that system (10) have a unique solution , for . We have seen that the claim is true for , , and . By induction, we assume the claim is true for , , and , and we will prove it for , , and .
Assume by induction hypothesis that there exists , such that system (10) with instead of has a unique solution , where and .
Lemma 4. There exists such that , for , and is a solution of (10).
Proof. Since , we have that the first equations of (10) are satisfied when and for and . Substituting this solution into the last equations of (10), we get the equation We have that Therefore, there exists at least a value satisfying equation . This completes the proof of Lemma 4.
By using the implicit function theorem we will prove that the solution of (10) given in Lemma 4 can be continued to a solution with .
Let , we define
It is not difficult to see that the system (10) is equivalent to for . Let be the solution of system (10) given in Lemma 4. The differential of (16) with respect to the variables is where We have assumed that makes the system (10) with instead of have a unique solution, so ; therefore, . Applying the implicit function theorem, there exists a neighborhood of and unique analytic functions , for and , such that is the solution of the system (10) for all . The determinant is calculated as where is the algebraic cofactor of .
We see that , , and for do not contain the factor . If we consider as a function of , then is analytic and nonconstant. It is obvious that for . We can find sufficiently close to such that and, therefore, a solution of system (10) satisfying , for .
The proof for , , and is also done by induction. At each step, two particles are symmetrically located at the -axis. Using the similar arguments above we prove Theorem 2 for , , and . At each step, two particles are symmetrically located at the -axis. Using the similar arguments above we prove Theorem 2 for , , and .
This work is supported by Youth Found of Mianyang Normal University. The authors express their gratitude to the reviewers for their useful suggestions and constructive criticism.
R. Abraham and J. E. Marsden, Foundation of Mechanics, Addison-Wesley, Benjamin, NY, USA, 2nd edition, 1978.
D. G. Saari, Collisions, Rings and Other Newtonian N-body Problems, American Mathematical Society, Providence, RI, USA, 2005.
A. Wintner, The Analytical Foundations of Celestial Mechanics, Princeton University Press, Princeton, NJ, USA, 1941. View at: MathSciNet
D. G. Saari, “Singularities and collisions of Newtonian gravitational systems,” Archive for Rational Mechanics and Analysis, vol. 49, pp. 311–320, 1973. View at: Publisher Site | Google Scholar | MathSciNet
D. G. Saari, “On the role and properties of N body central configurations,” Celestial Mechanics and Dynamical Astronomy, vol. 21, no. 1, pp. 9–20, 1980. View at: Google Scholar
D. G. Saari and N. D. Hulkower, “On the manifolds of total collapse orbits and of completely parabolic orbits for the
n
-body problem,” Journal of Differential Equations, vol. 41, no. 1, pp. 27–43, 1981. View at: Publisher Site | Google Scholar | MathSciNet
Y. Long and S. Sun, “Four-body central configurations with some equal masses,” Archive for Rational Mechanics and Analysis, vol. 162, no. 1, pp. 25–44, 2002. View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
A. Albouy, Y. Fu, and S. Sun, “Symmetry of planar four-body convex central configurations,” Proceedings of The Royal Society of London A, vol. 464, no. 2093, pp. 1355–1365, 2008. View at: Publisher Site | Google Scholar | MathSciNet
J. Llibre and L. F. Mello, “Triple and quadruple nested central configurations for the planar
n
-body problem,” Physica D: Nonlinear Phenomena, vol. 238, no. 5, pp. 563–571, 2009. View at: Publisher Site | Google Scholar | MathSciNet
M. Corbera and J. Llibre, “On the existence of central configurations of
p
nested regular polyhedra,” Celestial Mechanics & Dynamical Astronomy, vol. 106, no. 2, pp. 197–207, 2010. View at: Publisher Site | Google Scholar | MathSciNet
Copyright © 2014 Yueyong Jiang and Furong Zhao. 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.
|
QDispersion - Maple Help
Home : Support : Online Help : Mathematics : Discrete Mathematics : Summation and Difference Equations : QDifferenceEquations : QDispersion
return the q-dispersion of two polynomials (or all the set of non-negative integers used in its definition)
QDispersion(s, t, q, x, 'maximal')
name or number used as the parameter q, usually q
(optional) indicates the q-dispersion itself must be returned rather than all the set of non-negative integers used in its definition
If two polynomials
s\left(x\right)
t\left(x\right)
, with both
s\left(0\right)
t\left(0\right)
nonzero, are given, the QDispersion(s,t,q,x,'maximal') calling sequence returns their q-dispersion, that is,
\mathrm{qdis}\left(s\left(x\right),t\left(x\right)\right)=\mathrm{max}{r|r\mathrm{in}Z,r>=0,\mathrm{deg}\left(\mathrm{gcd}\left(s\left({q}^{r}x\right),t\left(x\right)\right)\right)>=1}
if the option 'maximal' is specified. Otherwise, the QDispersion(s,t,q,x) calling sequence returns the set of all non-negative integers
r
used in the definition of the q-dispersion.
s\left(x\right)
t\left(x\right)
are as above and
k,l
are non-negative integers, then QDispersion(x^k*s,x^l*t,q,x) returns the same result as QDispersion(s,t,q,x), and similarly if the option 'maximal' is specified.
The efficient algorithm for computing the dispersion of two polynomials
\mathrm{dis}\left(s\left(x\right),t\left(x\right)\right)=\mathrm{max}{r|r\mathrm{in}Z,r>=0,\mathrm{deg}\left(\mathrm{gcd}\left(s\left(x+r\right),t\left(x\right)\right)\right)>=1}
is the algorithm by Yiu-Kwong Man and F.J.Wright. This algorithm is based on the factorization of the polynomials involved rather than on the resultant calculation as it was in earlier implementations. This algorithm is adapted for computing the q-dispersion of two polynomials.
\mathrm{with}\left(\mathrm{QDifferenceEquations}\right):
\mathrm{p1}≔\left({x}^{2}+3\right){\left(x+q\right)}^{2};
\mathrm{p2}≔\left(qx+1\right)\left({q}^{100}+{q}^{5}{x}^{2}+3\right)\left({q}^{20}x+1\right)
\textcolor[rgb]{0,0,1}{\mathrm{p1}}\textcolor[rgb]{0,0,1}{≔}\left({\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{3}\right)\textcolor[rgb]{0,0,1}{}{\left(\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{q}\right)}^{\textcolor[rgb]{0,0,1}{2}}
\textcolor[rgb]{0,0,1}{\mathrm{p2}}\textcolor[rgb]{0,0,1}{≔}\left(\textcolor[rgb]{0,0,1}{q}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\right)\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{q}}^{\textcolor[rgb]{0,0,1}{100}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{q}}^{\textcolor[rgb]{0,0,1}{5}}\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}\right)\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{q}}^{\textcolor[rgb]{0,0,1}{20}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\right)
\mathrm{QDispersion}\left(\mathrm{p1},\mathrm{p2},q,x\right)
{\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{21}}
\mathrm{QDispersion}\left(\mathrm{p1},\mathrm{p2},q,x,'\mathrm{maximal}'\right)
\textcolor[rgb]{0,0,1}{21}
q≔10;
\mathrm{QDispersion}\left(\mathrm{p1},\mathrm{p2},q,x\right)
\textcolor[rgb]{0,0,1}{q}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{10}
{\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{21}}
Khmelnov, D.E. "Improved Algorithms for Solving Difference and q-Difference Equations." Programming and Computer Software. Vol. 26 No. 2. (2000): 107-115. Translated from Programmirovanie. No. 2.
QDifferenceEquations[UniversalDenominator]
|
How can I show these polynomials are not coprime? x,\ x-1
Malachi Novak 2022-04-24 Answered
x,\text{ }x-1
x+1
{\mathbb{Z}}_{6}\left[x\right]
veceritzpzg
\left(x-1,\text{ }x+1\right)
equals the ideal
\left(2,\text{ }x-1\right)
and this is not the whole ring, otherwise
1\in \left(2,\text{ }x-1\right)
1=2f\left(x\right)+\left(x-1\right)g\left(x\right)
Now send x to 1 and get a contradiction.
G{L}_{2}\frac{\mathbb{R}}{S}{L}_{2}\left(\mathbb{R}\right)
G{L}_{2}\left(\mathbb{R}\right)
2×2
S{L}_{2}\left(\mathbb{R}\right)
2×2
S{L}_{2}\left(\mathbb{R}\right)
G{L}_{2}\left(\mathbb{R}\right)
|G{L}_{2}\frac{\mathbb{R}}{S}{L}_{2}\left(\mathbb{R}\right)|=\frac{|G{L}_{2}\left(\mathbb{R}\right)|}{|S{L}_{2}\left(\mathbb{R}\right)|},
Is it true that for a Group G with Normal Group
N:\frac{G}{N}=\frac{GN}{N}?
I think the statement is correct. But why do we have to write:
\left[G,G\right]\frac{N}{N}
here instead of just
\frac{G,G}{N}?
card\left(G\right)={p}^{2}q
p<q
{s}_{q}
{s}_{q}\in \left\{1, {p}^{2}\right\}
\underset{S\in Sy{l}_{q}\left(G\right)}{\cup }S\setminus \left\{1\right\}={\stackrel{˙}{\cup }}_{S\in Sy{l}_{q}\left(G\right)}S\setminus \left\{1\right\}
S,\text{ }T\in Sy{l}_{q}\left(G\right)
S\ne T
\mathrm{S}\setminus \left\{1\right\}\cap \mathrm{T}\setminus \left\{1\right\}=\varnothing
Two sided ideal in
{M}_{2×2}\left(\mathbb{C}\right)
There are 2 photos inserted the one is the part 1 of the problem and the other is the part 2 . The picture with the number 1 is the part 1 and the x and y axis graph pic is part 2.
Let H be a subgroup of G and
x,\text{ }y\in G
x\left(Hy\right)=\left(xH\right)y
|
The determined Wile E. Coyote is out once
The determined Wile E. Coyote is out once more to try to capture the elusive roadrunner. The coyote wears a new pair of power roller skates, which provide a constant horizontal acceleration of 15 m/s2, as shown in Figure P3.73. The coyote starts off at rest 70 m from the edge of a cliff at the instant the roadrunner zips by in the direction of the cliff. (a) If the roadrunner moves with constant speed, find the minimum speed the roadrunner must have to reach the cliff before the coyote. (b) If the cliff is 100 m above the base of a canyon, find where the coyote lands in the canyon. (Assume his skates are still in operation when he is in “flight” and that his horizontal component of acceleration remains constant at 15 m/s2.)
(a) Derive the expression for the minimum time required for the coyote to travel a distance of
\mathrm{\Delta }x
from the cliff.
The distance travelled by the coyote is,
\mathrm{\Delta }x={v}_{0x}t+\frac{1}{2}{a}_{x}{t}^{2}
{v}_{0x}
is the initial horizontal velocity and
{a}_{x}
is the horizontal acceleration of the coyote, and
t
is the time to cover the distance
\mathrm{\Delta }x
The initial velocity og the coyote is zero.
{v}_{0x}=0\text{ }m/s
0\text{ }m/s
{v}_{0x}
\mathrm{\Delta }x={v}_{0x}t+\frac{1}{2}{a}_{x}{t}^{2}
t
\mathrm{\Delta }x=\left(0\text{ }m/s\right)t+\frac{1}{2}{a}_{x}{t}^{2}
t=\sqrt{\frac{2\mathrm{\Delta }x}{{a}_{x}}}
The roadrunner is moving with a constant speed. Hence, the minimum speed to reach the cliff is as follows.
{v}_{min}=\frac{\mathrm{\Delta }x}{t}
\sqrt{\frac{2\mathrm{\Delta }x}{{a}_{x}}}
t
{v}_{min}=\frac{\mathrm{\Delta }x}{\sqrt{\frac{2\mathrm{\Delta }x}{{a}_{x}}}}
=\sqrt{\frac{{a}_{x}\mathrm{\Delta }x}{2}}
15\text{ }m/{s}^{2}
{a}_{x}
70\text{ }m
\mathrm{\Delta }x
{v}_{min}=\sqrt{\frac{\left(15\text{ }m/{s}^{2}\right)\left(70\text{ }m\right)}{2}}
=23\text{ }m/s
Hence, the required minimum speed of the roadrunner is
23\text{ }m/s
b) The initial velocity of the coyote when it goes over the cliff is equal to the horizontal velocity.
{v}_{0x}={a}_{x}t
\sqrt{\frac{2\mathrm{\Delta }x}{{a}_{x}}}
t
{v}_{0x}={a}_{x}\sqrt{\frac{2\mathrm{\Delta }x}{{a}_{x}}}
=\sqrt{2{a}_{x}\mathrm{\Delta }x}
The time taken to cover the vertical height of the cliff is,
{t}^{\prime }=\sqrt{\frac{2\mathrm{\Delta }y}{{a}_{y}}}
\mathrm{\Delta }y
is vertical heisght of the cliff.
Therefore, the horizontal displacement of the fall of the coyote is,
\mathrm{\Delta }x={v}_{0x}t+\frac{1}{2}{a}_{x}{t}^{2}
=\sqrt{2{a}_{x}\mathrm{\Delta }x}
for v_{0x} and
\sqrt{\frac{2\mathrm{\Delta }y}{{a}_{y}}}
{t}^{\prime }
\mathrm{\Delta }x=\sqrt{2{a}_{x}\mathrm{\Delta }x}\left(\sqrt{\frac{2\mathrm{\Delta }y}{{a}_{y}}}\right)+\frac{1}{2}{a}_{x}\left(\sqrt{\frac{2\mathrm{\Delta }y}{{a}_{y}}}{\right)}^{2}
15\text{ }m/{s}^{2}
{a}_{x}
70\text{ }m
\mathrm{\Delta }x
-100\text{ }m
\mathrm{\Delta }y
-9.80\text{ }m/{s}^{2}
{a}_{y}
\mathrm{\Delta }x=\sqrt{2\left(15\text{ }m/s\right)\left(70\text{ }m\right)}\left(\sqrt{\frac{2\left(-100\text{ }m\right)}{-9.80\text{ }m/{s}^{2}}}\right)+\frac{1}{2}\left(15\text{ }m/s\right)\left(\sqrt{\frac{2\left(-100\text{ }m\right)}{-9.80\text{ }m/{s}^{2}}}{\right)}^{2}
=3.6×{10}^{2}\text{ }m
Hence, the reauired dispplacement is
3.6×{10}^{2}\text{ }m
Find the x-and y-intercepts of this equation.
7x+2y=56
To calculate: The probability that a student chosen randomly from the class is not going to college if the number of students in a high school graduating class is 128, out of which 52 are on the honor roll, and out of these, 48 are going to college. The number of students who are not on the honor roll is 76,out of these 56 are going to college.
If Julia is traveling along a highway at a constant speed of 50 miles per hour, how far does she travel in two hours and 30 minutes?
Saving for college in 20 years, a father wants to accumulate $40,000 to pay for his daughters
Find the x-and y-intercepts of the graph of the equation algebraically.
3y+2.5x-3.4=0
An object moves in simple harmonic motion with period 7 minutes and amplitude 17m. At time =t0 minutes, its displacement d from rest is 0m, and initially it moves in a positive direction .Give the equation modeling the displacement
|
Zn+Cu^{+2} \to Zn^{2+}+Cu
This is a spontaneous redox reaction with a negative Gibbs free energy (
\Delta G < 0
). It can be split into two half-reactions:
Zn \to Zn^{2+}+2e^-
Cu^{2+}+2e^- \to Cu
Zinc loses 2 electrons to become the ion
Zn^{2+}
Cu^{+2}
accepts the electrons to become elemental copper. If these two half-reactions occur in two separate containers connected by a conductive wire, an electric current will form from the transfer of the electrons from one container to the other. Zinc and copper electrodes are submerged in aqueous solutions of their salts, such as
ZnSO_4
CuSO_4
, in their respective containers. These two electrodes are connected by wire and the solutions are connected by a medium or bridge which allows ions to transfer but does not allow the solutions to mix directly.
The dry cell is the most common type of battery used to power small household devices, such as flashlights, radios, and calculators. Despite its name, these cells are built in a water-based paste containing
MnO_{2}
Zn
. The chemical reactions used in a dry cell can be modified to work in acidic or alkaline solutions. Alkaline batteries are more often available commercially.
The overall reaction for a mercury cell is as follows:
Zn(s) + HgO(s) \to Hg(l) + ZnO (s)
HgO(s) +H_{2}O(l) +2e^{-} \to Hg(l) + 2OH^{-}(aq)
Corrosion is a galvanic process. When metals are exposed to air or water, many of them react with oxygen to form a metal oxide. Rust, for example, is an iron oxide
Fe_2O_3
. The tarnish of silver or the blue-green color that copper develops are other examples of corrosion. Rust in particular can be a major industrial problem, because iron is so prevalent in manufacturing and because the reddish product flakes off, exposing even more metal to be oxidized. Metal products ranging from automobiles to faucets to soda cans are often coated with a thin layer of plastic, metal oxide, or a nonreactive metal (such as chromium) in order to prevent corrosion. The anodic index between the bolts and the plate allows for galvanic corrosion. [3] Dissimilar metals can also corrode each other. This process, sometimes called galvanic corrosion occurs when two metals are in contact and have different electrical properties. The anodic index can be used to determine the voltage potential between two metals. Metals with a large difference in anodic index should be kept insulated from each other to discourage corrosion.
|
The electromagnetic fields surrounding a thin, subseabed resistive disk in response to a deep-towed, time-harmonic electric dipole antenna are investigated using a newly developed 3D Cartesian, staggered-grid modeling algorithm. We demonstrate that finite-difference and finite-volume methods for solving the governing curl-curl equation yield identical, complex-symmetric coefficient matrices for the resulting
N×N
linear system of equations. However, the finite-volume approach has an advantage in that it naturally admits quadrature integration methods for accurate representation of highly compact or exponentially varying source terms constituting the right side of the resulting linear system of equations. This linear system is solved using a coupled two-term recurrence, quasi-minimal residual algorithm that doesnot require explicit storage of the coefficient matrix, thus reducing storage costs from
22N
10N
complex, double-precision words with no decrease in computational performance. The disk model serves as a generalized representation of any number of resistive targets in the marine environment, including basaltic sills, carbonates, and stratigraphic hydrocarbon traps. We show that spatial variations in electromagnetic phase computed over the target are sensitive to the disk boundaries and depth, thus providing a useful complement to the usual amplitude-versus-offset analysis. Furthermore, we estimate through the calculation of Fréchet sensitivity kernels those regions of the 3D model which have the greatest effect on seafloor electric fields for a given source/receiver configuration. The results show that conductivity variations within the resistive disk have a stronger influence on the observed signal than do variations in the surrounding sediment conductivity at depth.
|
broggesy9 2022-04-30 Answered
Lughettitbn
Rewrite the equation as 3u-7x-9=0
3u-7x-9=0
-7x=-3u+9
Divide each term in -7x=-3u+9 by -7 and simplify.
x=\frac{3u}{7}-\frac{9}{7}
P\left(x\right)=-12{x}^{2}+2136x-41000
x=r\mathrm{cos}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y=r\mathrm{sin}
\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{x}^{2}-{y}^{2}}{\sqrt{{x}^{2}+{y}^{2}}}
There are four washing machines in an apartment complex: A, B, C, D. On any given day the probability that these machines break down is as follows:
P(A) = 0.04, P(B) = 0.01, P(C) = 0.06, P(D) = 0.01 .
Assume that the functionality of each machine is independent of that of others. What is the probability that on a given day at least one machine will be working?
Solve for G.S. / P.S. for the following differential equations using separation of variables.
{x}^{2}ydy-x{y}^{2}dx+3ydy-2xdx=0
f\left(x,y\right)=7{x}^{2}+9xy+4{y}^{3}
\frac{\partial f}{\partial x}
\frac{\partial f}{\partial y}
\frac{\partial f}{\partial x}{\mid }_{y=9}
Describe the procedure for finding critical points of a function in two independent variables.
f:{\mathbb{N}}^{2}\to \mathbb{N}
f\left(a,b\right)={a}^{b}
|
Egg Dropping Practice Problems Online | Brilliant
Consider the standard 2-egg problem. We wish to determine the critical height at which an egg breaks. That is we assume that there is some critical height
C
, such that if we drop an egg from height
h \leq C
the egg does not break where as it will always break if dropped from height
h > C
. We want to minimize the number of trials and we are only given two eggs.
Let us consider the worst case scenario. Let
W(k)
be the minimum number of trials required to identify
C
under the worst case scenario given that we have two eggs and a building a height
k
k > 0
W(k) = 18518505
Consider the classic egg drop problem. Let
S(n,k)
be the minimum number of egg droppings that will suffice to find the critical floor in an
-story building given
k
eggs.
S(k,n)
can be described as follows for an elementary function of
n
f(n)
S(k,n) = \Theta(f(n,k))
f(n,k)
is given by which of the following?
k \log n
k^2n^2
\sqrt[k]{n}
kn
Consider the standard egg dropping problem, and design an algorithm that computes the maximum number of trials required to find the critical floor in the worst case. Find the maximum number of trials required in the worst case when you have three eggs and
105
floors to check.
|
Determine the average value of F(x, y, z)
Determine the average value of F(x, y, z) = xyz throughout the cubical region D
bounded by the coordinate planes and the planes x = 2, y − 2, z = 2 in the first octant.
A Harris Interactive survey for InterContinental Hotels & Resorts asked respondents, “When traveling internationally, do you generally venture out on your own to experience culture, or stick with your tour group and itineraries?” The survey found that 23% of the respondents stick with their tour group (USA Today, January 21, 2004).
a. In a sample of six international travelers, what is the probability that two will stick with their tour group?
b. In a sample of six international travelers, what is the probability that at least two will stick with their tour group?
c. In a sample of 10 international travelers, what is the probability that none will stick with the tour group?
Assume that 60% of the students at Remmington High studied for their Psychology test. Of those that studied, 25% got an A, but only 8% of those who didn't study got an A. What is the approximate probability that someone that gets an A actually studied for the Psychology test?
translate into a variable expression. Then simplify.
a number decreased by the total of the number and twelve.
How do I prove that this ideal is not a ' ideal?
Let K be a field and
R=\frac{K\left[X,Y\right]}{\left(XY\right)}.F\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}P\in K\left[X,Y\right]
\left[P\right]
its class in R. Show that the Ideal (XY) is not a ' ideal.
You are conducting a Goodness of Fit hypothesis test for the claim that all 5 categories are equally likely to be selected. Complete the table. Report all answers correct to three decimal places.
\begin{array}{|cccc|}\hline \text{ Category }& \text{ Observed Frequency }& \text{ Expected Frequency }& \text{ (obs-exp)^2/exp }\\ \text{ A }& 15\\ \text{ B }& 19\\ \text{ C }& 23\\ \text{ D }& 18\\ \text{ E }& 5\\ \hline\end{array}
What is the chi-square test-statistic for this data?
{\chi }^{2}=
At the alpha = 0.05 level, what is the conclusion for this test?
Report all answers accurate to three decimal places.
Determine whether g(x)= x lxl is even or odd or neither function.
|
Solve for y: 0=u-5y+1.
Judith Warner 2022-04-30 Answered
gonzakunti2
Rewrite the equation as u-5y+1=0
u-5y+1=0
Move all terms not containing y to the right side of the equation.
-5y=-u-1
Divide each term in -5y=-u-1 by -5 and simplify.
y=\frac{u}{5}+\frac{1}{5}
P\left(x\right)=-12{x}^{2}+2136x-41000
x=r\mathrm{cos}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y=r\mathrm{sin}
\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{x}^{2}-{y}^{2}}{\sqrt{{x}^{2}+{y}^{2}}}
Letf\left(x\right)={x}^{5}+5{x}^{4}+6x+3.
a. Conduct a sign analysis of f"". That is, determine the sign off"" on each interval. Make sure to show how you found the boundaries (endpoints) of the intervals and how you computed the sign in each interval.
b. indicate the intervals, if any, on which fis concave up and the intervals, if any, on which fis concave down.
c. Find the x-coordinates of any inflection points of f. If there are more than one, enter them as a comma- separated list. If there are none, enter ""DNE""
The reason to find points of intersection of polar graph which may require further analysis beyond solving two equations simultaneously
Take examples of two polar equations
r=1-2\mathrm{cos}\theta
r=1
A surface is represented by the following multivariable function,
f\left(x,y\right)={x}^{3}+{y}^{3}-3x-3y+1
Calculate coordinates of stationary points.
Find all complex zeros of the polynomial function. Give exact values. List multiple zeros as necessary.
f\left(x\right)={x}^{4}-5{x}^{3}-36{x}^{2}+272x-448
All complex zeros are=?
Average value over a multivariable function using triple integrals. Find the average value of
F\left(x,y,z\right)={x}^{2}+{y}^{2}+{z}^{2}
over the cube in the first octant bounded bt the coordinate planes and the planes
x=5
y=5
z=5
|
Bernoulli's Principle (Fluids) | Brilliant Math & Science Wiki
July Thomas, Tom Finet, and Jimin Khim contributed
The air particles above the wing travel faster than those below the wing, meaning the pressure is relatively less above. This difference in pressure creates the lift force. Wikimedia
Bernoulli's principle states that the pressure of a fluid decreases when either the velocity of the fluid or the height of the fluid increases. Bernoulli's principles is integral to the design of airplane wings and ventilation systems. The actual equation itself resembles conservation of energy, however, in lieu of studying the motion of an individual particle, Bernoulli's principle generalizes for a collection of particles with a uniform density.
Bernoulli's principle actually relates pressure to two separate phenomena. One is the idea that if a fluid moves faster, the individual particles will spread out more, decreasing the pressure on the surroundings. The other is the idea that as the amount of fluid above decreases, the particles will be less compacted, thus exhibiting less apparent pressure.
In both cases, the density of the fluid provides a "default" closeness of the particles in the fluid, so density is incorporated into the principle as well.
P_A + \frac12 \rho v_A^2 + \rho gh_A = P_B + \frac12 \rho v_B^2 + \rho gh_B
Fluid flows from one pipe into a second of smaller cross-sectional area. Which pipe feels a greater pressure from the fluid? (Assume they are on level ground.)
By continuity,
A_1 v_1 = A_2 v_2
A_1 > A_2
v_1 < v_2
. Bernoulli's principle for two sections of pipe at the same height is just
P_1 + \frac12 \rho v_1^2 = P_2 + \frac12 \rho v_2^2.
Since equality must be maintained,
v_1 < v_2
P_1 > P_2
_\square
Note: Perhaps counterintuitively, the pressure is greater in the wider section with the lower fluid velocity. As an analogy, consider a traffic jam caused by road construction which takes two lanes down to one. In this scenario, the "congestion" (or pressure) is worse in the wider two-lane section, where drivers are being forced to consolidate into just one lane, than in the skinnier one-lane section later.
A chimney near the surface of the earth is used to draw the smoke out of a fireplace and carry it up and out of the building. If it is determined that a
38 \text{ Pa}
pressure difference is required to pull the smoke through a chimney, how tall in meters should the chimney be? (Assume the air around the fireplace and the top of the chimney is still.)
Since the air is still at the base and the top,
\frac12 \rho v^2 = \frac12 \rho (0)^2 = 0.
So Bernoulli's equation simplifies to
P_A + \rho g h_A = P_B + \rho g h_B.
Let the fireplace be location A and the top of the chimney be location B. The height of the chimney is then
h_B - h_A
. Solving the equation above for his relation yields
h_B - h_A = \frac{P_A-P_B}{\rho g}.
The pressure difference provided is
38 \text{ Pa}
, density of air is
1.29 {\text{ kg}^3}/\text{m}
, and the acceleration due to gravity near the surface of the earth is
9.8 \text{ m}/\text{s}^2
h_B - h_A = \frac{38}{(1.29)(9.8)} = 3.0 \text{ (m)}.
1~\mbox{atm}
1~\mbox{atm}=101,325~\mbox{Pa}
-9.8~\mbox{m/s}^2
1~\mbox{g/cm}^3
0.1~\mbox{m}
0.01~\mbox{m}
0.5~\mbox{m}
-9.8~\mbox{m/s}^2
200 \text{ m/s}
180 \text{ m/s}
\rho = 1.29 \text{ kg/m}^3
A = 2\text{ m}
In order to describe fluids in active flow, start with the conservation of matter: when fluid moves from one position to the next, it must do so in such a way that no fluid matter is destroyed.
For instance, if fluid is injected into the mouth of a tube at a rate of
J_\text{in}=1\text{ L}
per second,
1\text{ L}
of the fluid should come out the other side every second, i.e.
J_\text{in}=J_\text{out}
For a toy model, consider the device below that consists of one level section of tube
\Gamma_\text{A}
r_A
\Gamma_\text{B}
r_B
Find a relationship that connects the velocity and pressure of the fluid in either section of tube.
To start, take the system to be the fluid that's between the discs
\partial_\text{A}
\partial_\text{B}
at time zero, called
\Sigma
\Sigma
\Gamma_\text{A}
P_A
, is greater than the fluid pressure in
\Gamma_{B}
P_B
\Sigma
\Sigma_L
, will push with greater force than the fluid to the right,
\Sigma_R
\Sigma
P_A \pi r_A^2
P_B\pi r_B^2
Appealing to the work-energy principle,
W = \Delta\text{K.E.}
Here, the work is performed by the two pressures in moving
\Sigma
\Delta t
\Gamma_A
\Sigma
v_A\Delta t
\Gamma_B
\Sigma
v_B\Delta t
Hence, the net work
W
\Sigma
is given by the work done on
\Sigma
\Sigma_L
P_A \pi r_A^2 v_A \Delta t
, minus the work done by
\Sigma
\Sigma_R
P_B \pi r_B^2 v_B \Delta t
W = P_A \pi r_A^2 v_A \Delta t - P_B \pi r_B^2 v_B \Delta t.
However, the conservation condition
J_\text{in} = J_\text{out}
applies for the discs
\Gamma_A
\Gamma_B
\pi r_A^2 v_A \Delta t = \pi r_A^2 v_A \Delta t
\Delta V
that flows into
\Gamma_A
is equal to the volume of fluid that flows out from
\Gamma_B
W = \Delta V \left(P_A-P_B\right) = \frac{\Delta m}{\rho} \left(P_A-P_B\right).
\Sigma
is equal to the change in kinetic energy of the fluid. The kinetic energy of
\Sigma
\Delta m
v_A
\Gamma_A
is now travelling with velocity
v_B
\Gamma_B
\Delta \text{K.E.} = \frac12 \Delta m v_B^2 - \frac12 \Delta m v_A^2
\frac{\Delta m}{\rho} \left(P_A-P_B\right) = \Delta m \left( \frac12 v_B^2 - \frac12 v_A^2 \right)
P_A + \frac12 \rho v_A^2 = P_B + \frac12 \rho v_B^2
In this derivation, the tubes were kept at equal level for simplicity's sake. It is trivial to recalculate the relation for the case when the two tube sections are of differing heights in a gravitational field, as occurs for the plumbing system in an apartment building. In this case, the work-energy principle is given by
W = \Delta \text{K.E.} + \Delta \text{P.E.}
and thus the full Bernoulli relation is
P_A + \frac12 \rho v_A^2 + \rho gh_A = P_B + \frac12 \rho v_B^2 + \rho gh_B.
Note that the calculation did not depend in any way on the particular setup that we used (the two tubes and linker section). The same calculation applies to a tube of arbitrary shape which carries out an arbitrary trajectory through a gravitational field. Thus, the relation can be used to connect any two cross sections of a fluid's flow.
Cite as: Bernoulli's Principle (Fluids). Brilliant.org. Retrieved from https://brilliant.org/wiki/bernoullis-principle-fluids/
|
4.1 Showing G 1:1 conformal mapping
4.2 Finding the domain D
6.1 Summation a_n Convergent
6.2 Show that h is analytic
Prove there is an entire function
{\displaystyle f\!\,}
so that for any branch
{\displaystyle g\!\,}
{\displaystyle {\sqrt {z}}\!\,}
{\displaystyle \sin ^{2}(g(z))=f(z)\!\,}
{\displaystyle z\!\,}
in the domain of definition of
{\displaystyle g\!\,}
{\displaystyle \sin ^{2}(\theta )={\frac {1-\cos(2\theta )}{2}}\!\,}
{\displaystyle \cos(z)=\sum _{n=1}^{\infty }{\frac {(-1)^{n}z^{2n}}{(2n)!}}\!\,}
{\displaystyle H\!\,}
be the domain
{\displaystyle \{z:-{\frac {\pi }{2}}<\Re (z)<{\frac {\pi }{2}},\Im (z)>0\}\!\,}
{\displaystyle g=\sin(z)\!\,}
is a 1:1 conformal mapping of
{\displaystyle H\!\,}
onto a domain
{\displaystyle D\!\,}
{\displaystyle D\!\,}
Showing G 1:1 conformal mappingEdit
{\displaystyle {\begin{aligned}(1)\quad |g^{\prime }(z)|&=|\sin ^{\prime }(z)|\\&=|\cos(z)|\\&=\left|{\frac {e^{-iz}+e^{iz}}{2}}\right|\\&\geq {\frac {1}{2}}(|e^{-ix}||e^{y}|-|e^{ix}||e^{-y}|)\\&\geq {\frac {1}{2}}(e^{y}-e^{-y})\\&>0\quad {\mbox{since }}y>0\end{aligned}}}
Also, applying a trigonometric identity, we have for all
{\displaystyle z_{1},z_{2}\in H\!\,}
{\displaystyle (2)\quad \sin z_{1}-\sin z_{2}=2\sin \left({\frac {z_{1}-z_{2}}{2}}\right)\cos \left({\frac {z_{1}+z_{2}}{2}}\right)\!\,}
{\displaystyle \sin z_{1}=\sin z_{2}\!\,}
{\displaystyle \sin \left({\frac {z_{1}-z_{2}}{2}}\right)=0\!\,}
{\displaystyle \cos \left({\frac {z_{1}+z_{2}}{2}}\right)=0\!\,}
The latter cannot happen in
{\displaystyle H\!\,}
{\displaystyle |g^{\prime }(z)|=|\cos(z)|>0\!\,}
{\displaystyle \sin \left({\frac {z_{1}-z_{2}}{2}}\right)=0\!\,}
{\displaystyle z_{1}=z_{2}\!\,}
Note that the zeros of
{\displaystyle \sin(z)=\sin(x+iy)\!\,}
{\displaystyle x=k\pi ,k\in \mathbb {Z} \!\,}
. Similary the zeros of
{\displaystyle \cos(z)=\cos(x+iy)\!\,}
{\displaystyle x={\frac {\pi }{2}}+k\pi ,k\in \mathbb {Z} \!\,}
{\displaystyle (1)\!\,}
{\displaystyle (2)\!\,}
{\displaystyle g\!\,}
{\displaystyle 1:1\!\,}
Finding the domain DEdit
{\displaystyle D\!\,}
, we only need to consider the image of the boundaries.
Consider the right hand boundary,
{\displaystyle C_{3}=\{z=x+iy|x={\frac {\pi }{2}},y>0\}\!\,}
{\displaystyle {\begin{aligned}g(C_{3})&=g\left({\frac {\pi }{2}}+yi\right)\\&=\sin \left({\frac {\pi }{2}}+yi\right)\\&={\frac {e^{i({\frac {\pi }{2}}+yi)}-e^{-i({\frac {\pi }{2}}+yi})}{2i}}\\&={\frac {e^{i{\frac {\pi }{2}}}e^{-y}-e^{-i{\frac {\pi }{2}}}e^{y}}{2i}}\\&={\frac {e^{i{\frac {\pi }{2}}}e^{-y}+e^{i{\frac {\pi }{2}}}e^{y}}{2i}}\\&={\frac {e^{i{\frac {\pi }{2}}}(e^{-y}+e^{y})}{2i}}\\&={\frac {i(e^{-y}+e^{y})}{2i}}\\&={\frac {e^{-y}+e^{y}}{2}}\\\end{aligned}}}
{\displaystyle y>0\!\,}
{\displaystyle g(C_{3})=(1,\infty )\!\,}
Now, consider the left hand boundary
{\displaystyle C_{2}=\{z=x+iy|x=-{\frac {\pi }{2}},y>0\}\!\,}
{\displaystyle {\begin{aligned}g(C_{2})&=g\left(-{\frac {\pi }{2}}+yi\right)\\&=\sin \left(-{\frac {\pi }{2}}+yi\right)\\&={\frac {e^{i(-{\frac {\pi }{2}}+yi)}-e^{-i(-{\frac {\pi }{2}}+yi})}{2i}}\\&={\frac {e^{-i{\frac {\pi }{2}}}e^{-y}-e^{i{\frac {\pi }{2}}}e^{y}}{2i}}\\&={\frac {-e^{i{\frac {\pi }{2}}}e^{-y}-e^{i{\frac {\pi }{2}}}e^{y}}{2i}}\\&={\frac {e^{i{\frac {\pi }{2}}}(-e^{-y}-e^{y})}{2i}}\\&=-{\frac {i(e^{-y}+e^{y})}{2i}}\\&=-{\frac {e^{-y}+e^{y}}{2}}\\\end{aligned}}}
{\displaystyle y>0\!\,}
{\displaystyle g(C_{2})=(-\infty ,-1)\!\,}
Now consider the bottom boundary
{\displaystyle C_{1}=\{z=x+iy|-{\frac {\pi }{2}}<=x<={\frac {\pi }{2}},y=0\}\!\,}
{\displaystyle {\begin{aligned}g(C_{1})&=g(x)\\&=\sin(x)\end{aligned}}}
{\displaystyle -{\frac {\pi }{2}}<=x<={\frac {\pi }{2}}\!\,}
{\displaystyle g(C_{2})=[-1,1]\!\,}
Hence, the boundary of
{\displaystyle H\!\,}
maps to the real line. Using the test point
{\displaystyle z=i\!\,}
{\displaystyle {\begin{aligned}g(i)&=\sin(i)\\&={\frac {e^{i(i)}-e^{-i(i)}}{2i}}\\&={\frac {e^{-1}-e^{1}}{2i}}\\&={\frac {-i(e^{-1}-e^{1})}{2}}\\&={\frac {i(e^{1}-e^{-1})}{2}}\\&={\frac {i}{2}}(e-{\frac {1}{e}})\\&\in {\mbox{Upper Half Plane}}\end{aligned}}}
We then conclude
{\displaystyle D=g(H)={\mbox{Upper Half Plane}}\!\,}
Suppose that for a sequence
{\displaystyle a_{n}\in R\!\,}
{\displaystyle z,\Im (z)>0\!\,}
{\displaystyle h(z)=\sum _{n=1}^{\infty }a_{n}\sin(nz)\!\,}
is convergent. Show that
{\displaystyle h\!\,}
{\displaystyle \{\Im (z)>0\}\!\,}
and has analytic continuation to
{\displaystyle C\!\,}
Summation a_n ConvergentEdit
{\displaystyle \sum _{n=1}^{\infty }a_{n}\!\,}
is convergent. Assume for the sake of contradiction that
{\displaystyle \sum _{n=1}^{\infty }a_{n}\!\,}
is divergent i.e.
{\displaystyle \sum _{n=1}^{\infty }a_{n}=\infty \!\,}
{\displaystyle h(z)\!\,}
is convergent in the upper half plane, choose
{\displaystyle z=i\!\,}
as a testing point.
{\displaystyle {\begin{aligned}h(i)&=\sum _{n=1}^{\infty }a_{n}\sin(ni)\\&=\sum _{n=1}^{\infty }a_{n}{\frac {e^{-n}-e^{n}}{2i}}\\&=\sum _{n=1}^{\infty }a_{n}{\frac {-i(e^{-n}-e^{n})}{2}}\\&=\sum _{n=1}^{\infty }a_{n}{\frac {i(e^{n}-e^{-n})}{2}}\end{aligned}}}
{\displaystyle h(i)\!\,}
converges in the upper half plane, so does its imaginary part and real part.
{\displaystyle \Im (h(i))={\frac {1}{2}}\sum _{n=1}^{\infty }a_{n}\underbrace {(e^{n}-e^{-n})} _{E_{n}}\!\,}
{\displaystyle \{E_{n}\}\!\,}
is increasing (
{\displaystyle E_{1}<E_{2}<\ldots <E_{n}<E_{n+1}\!\,}
{\displaystyle e^{n+1}>e^{n}\!\,}
{\displaystyle e^{-n}>e^{-(n+1)}\!\,}
e.g. the gap between
{\displaystyle e^{n}\!\,}
{\displaystyle e^{-n}\!\,}
is grows as
{\displaystyle n\!\,}
grows. Hence,
{\displaystyle {\begin{aligned}\Im (h(i))&\geq {\frac {1}{2}}(e^{1}-e^{-1})\underbrace {\sum _{n=1}^{\infty }a_{n}} _{=\infty }\\\\\\\Im (h(i))&\geq \infty \end{aligned}}}
This contradicts that
{\displaystyle h(i)\!\,}
is convergent on the upper half plane.
Show that h is analyticEdit
{\displaystyle h\!\,}
is analytic, let us cite the following theorem
{\displaystyle {h_{n}}\!\,}
be a sequence of holomorphic functions on an open set
{\displaystyle U\!\,}
. Assume that for each compact subset
{\displaystyle K\!\,}
{\displaystyle U\!\,}
the sequence converges uniformly on
{\displaystyle K\!\,}
, and let the limit function be
{\displaystyle h\!\,}
{\displaystyle h\!\,}
is holomorphic.
Proof See Theorem 1.1 in Chapter V, Complex Analysis Fourth Edition, Serge Lang.
Now, define
{\displaystyle h_{n}(z)=\sum _{k=1}^{n}a_{k}\sin(kz)\!\,}
{\displaystyle K\!\,}
be a compact set of
{\displaystyle U=\{\Im {z}>0\}\!\,}
{\displaystyle \sin(kz)\!\,}
|
An qeroplane startingfrom the airpoint A flies 300 kms edut, then 359 km at 30
keletsokatty 2022-03-31
A population of values has a normal distribution with \mu =121.3
and \sigma =5.8
. You intend to draw a random sample of size n=149
Find the probability that a sample of size n=149
\begin{array}{|ccc|}\hline & \text{ High TV Violence }& \text{ Low TV Violence }\\ \text{ Yes, Physical Abuse }& 14\left(8.42\right)& 26\left(31.58\right)\\ \text{ No Physical Abuse }& 22\left(27.58\right)& 109\left(103.42\right)\\ \hline\end{array}
{\chi }^{2}=\square
. Find the Maclaurin series for f(x) = (1 + x) −3 using the definition. Find the radius of convergence.
|
Prove that n|\phi(a^n-1) in "Topics in Algebra 2nd Edition" by
Slade Higgins 2022-04-29 Answered
n\mid \varphi \left({a}^{n}-1\right)
Aut\left(G\right)
morpheus1ls1
\varphi \left({a}^{n}-1\right)
measures the number of automorphisms of
\frac{\mathbb{Z}}{\left(an-1\right)}\mathbb{Z}
There is a subgroup of order n in this group: if
\varphi
is the automorphism sending 1 to a, then
\varphi
generates a subgroup of order n. The statement follows from Lagrange's Theorem.
Using the integral test, find the positive values of p for which the series ∑∞ k=2 1/ (k(ln(k)))^p converges. Show your work and explain your reasoning.
Find k
such that the following matrix M
(1 point) A breathalyser test is used by police in an area to determine whether a driver has an excess of alcohol in their blood. The device is not totally reliable: 7 % of drivers who have not consumed an excess of alcohol give a reading from the breathalyser as being above the legal limit, while 10 % of drivers who are above the legal limit will give a reading below that level. Suppose that in fact 14 % of drivers are above the legal alcohol limit, and the police stop a driver at random. Give answers to the following to four decimal places.
What is the probability that the driver is incorrectly classified as being over the limit?
What is the probability that the driver is correctly classified as being over the limit?
Find the probability that the driver gives a breathalyser test reading that is over the limit.
Find the probability that the driver is under the legal limit, given the breathalyser reading is also below the limit.
f\left(x\right) = \frac{1}{4}x{e}^{-\frac{x}{2}} ; x > 0\phantom{\rule{0ex}{0ex}} 0; x\le 0
a−15=4a−3
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.