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Geometrical Optics | Brilliant Math & Science Wiki
Satyajit Ghosh, Skanda Prasad, Sravanth C., and
Light can behave in many ways like a wave or a packet of photons which travel in a straight line exhibiting the property of rectilinear propagation. Geometrical optics deals with the propagation of light in a straight line and phenomena such as reflection, refraction, polarization, etc. A ray of light gives the direction of propagation of light. In the absence of an obstacle, the rays advance in a straight line without changing direction. When light meets a surface separating two transparent media, reflection and refraction occur and the light rays bend.
Reflection at Smooth Surfaces
Shifting of Image and Object
A light ray is reflected by a smooth surface in accordance with the 2 laws of reflection:
The incident ray, the reflected ray, and the normal to the reflecting surface are coplanar. The images formed by a plane mirror are equidistant from the distance between the object and the mirror.
An object is placed in front of the mirror. If the mirror is moved away from the object through a distance of
x,
by how much distance will the image move?
The answer will be added shortly.
Our final answer is
2x.\ _\square
A ray of light falls on a plane mirror. Show that if the mirror is tilted through an angle of
\theta,
then the reflected ray tilts through an angle of 2
\theta.
The laws of reflection are the same for plane and curved surfaces. A normal can be drawn from any point of the curved surface by first drawing the tangent plane from that point and then drawing the line perpendicular to that plane. Angles of incidence and reflection are defined from this normal. The angle of incidence is equal to angle of reflection.
A spherical mirror is a part cut from a hollow sphere. The spherical mirrors are generally constructed from glass. One surface of the glass is silvered. If reflection takes place at the convex surface, it is called a convex mirror and if it takes place on the concave surface, it is called a concave mirror.
Terms used in spherical mirrors:
Center of curvature (C): it is the center of the sphere from which the mirror is made.
Focus (f): it is the point at which the rays meet (either virtually or physically). Although we have not verified, from further exploration we will understand that
R=2f.
Pole (P): it is the surface of the mirror on which light falls.
Principle axis (PA): it is the imaginary line which is perpendicular to the pole.
Paraxial rays: These are rays parallel and close to the principal axis.
u:
This is the distance between the object and the pole and is sometimes referred to as the object distance.
v:
This is the distance between the pole and the image and is sometimes referred to as the image distance.
(0,0)
C,
and the principal axis is the
x
Cartesian Sign Conventions
1) The pole is taken to be the origin.
2) The direction in which the light moves is taken as the positive direction.
3) The height above the principal axis is taken as positive while the height below it is taken as negative.
The main formula used for solving problems on mirrors is
\frac { 1 }{ v } +\frac { 1 }{ u } =\frac { 1 }{ f }.
Note: While solving problems, do not forget to apply sign conventions and make sure you always take the object distance,
u,
as negative only. See the examples below which will make you comfortable with solving the problems based on the mirror formula.
If the distance between the object and the mirror is
10\text{ cm}
and its focal length is
5\text{ cm}
given it is a convex mirror, find the distance between the mirror and the object.
u = -10\text{ cm}
f = 5\text{ cm}.
Applying the mirror formula, we get
\begin{aligned} \frac { 1 }{ v } +\frac { 1 }{ u } &=\frac { 1 }{ f } \\ \frac { 1 }{ v } +\frac { 1 }{ -10 } &=\frac { 1 }{ 5 } \\ \frac { 1 }{ v } &=\frac { 3 }{ 10 } \\ \Rightarrow v&=3.33\text{ cm}. \ _\square \end{aligned}
The magnification of a mirror is given by
\frac { { h }_{ i } }{ { h }_{ o } }
{ h }_{ i }
is height of the image and
{ h }_{ o }
is height of the object.
Magnification is also given by
-\frac { v }{ u }
Again we should remember not to forget applying the sign conventions.
If magnification is
-ve,
it means the formed image is inverted.
If the image formed by a spherical mirror is half the size of the object and it is virtual, then find the distance between the mirror and the image given that the object's distance is
16\text{ cm}.
m=\frac { { h }_{ i } }{ { h }_{ o } } =\frac { 1 }{ 2 } =\frac { -v }{ u }.
u=-2v=-16 \text{ cm} \Rightarrow v=8\text{ cm}. \ _\square
When light passes from one medium to another and undergoes a change in speed and direction, it is called refraction. The light does reflect but continues on its path with a change in its direction.
When light moves from a rarer to denser medium, i.e. moves from a light to heavy medium, the light tends to bend towards the normal while the opposite happens when light travels from a denser to rarer medium.
For the concept of refraction, one important thing to know is to learn about Snell's law.
This law tells us that
{ \mu } _{ 1 } \sin i = { \mu }_{ 2 } \sin r,
{ \mu }_{ 1 }
is the medium's refractive index from which light travels and
{ \mu }_{ 2 }
is the medium's refractive index to which light goes. The angle of incidence is
i
and the angle of refraction is
r.
Generally, we deal with a situation where light goes from air to another medium and since the refractive index of air is 1, we get the equation
\sin i = { \mu } \sin r.
\frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f }.
Again it is important to use the sign conventions, or the answers are going to be wrong.
Most importantly, the sign conventions are going to remain the same as that of the mirrors, but here we have two focal lengths rather than one in the case of mirrors. So the focal length of concave is
-\text{ve}
by convention and that of convex is
+\text{ve}.
If the image formed by a convex lens is
20\text{ cm}
from the lens and the radius of curvature is
30\text{ cm},
then find the distance of the object from the lens.
R=30\text{ cm} \implies f=15\text{ cm}
. Using the lens formula, we have
\frac { 1 }{ +20 } -\frac { 1 }{ u } =\frac { 1 }{ 15 }
. Remember not to put the negative sign of
u
before solving it:
\begin{aligned} \frac { 1 }{ 20 } -\frac { 1 }{ 15 } &=\frac { 1 }{ u } \\ \frac { 3-4 }{ 60 } &=\frac { 1 }{ u } \\ \frac { -1 }{ 60 } &=\frac { 1 }{ u } \\ u &= -60\text{ cm}. \ _\square \end{aligned}
The lens formula we used above was a result of using a biconvex or biconcave lens. These lenses have equal radii of curvature and equal foci. However, the same cannot be used for a lens having different radii of curvature. So a new method of finding the focus,
u,
v
was developed, called the lens maker formula.
\frac { 1 }{ F } =\left(\frac { { \mu }_{ 2 } }{ { \mu }_{ 1 } } -1\right)\left(\frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right).
F
is the resultant focal length,
{ \mu }_{ 2 }
is the refractive index of the object in which light enters, and
{ \mu }_{ 1 }
is the refractive index from which light travels to reach the object. Normally
{ \mu }_{ 1 }
is ignored when light travels from air because it has a refractive index of
1,
{ \mu }_{ 1 }
1.
{ R }_{ 1 }
is the radius of curvature of the
1^\text{st}
lens, and
{ R }_{ 2 }
{2}^\text{nd}
We know that the earth is surrounded by a layer of air called the atmosphere. The density of air is not the same everywhere. The refractive index of air depends on its density, i.e. the higher the density of air, the greater its refractive index. The changes in refractive index can give rise to many phenomena.
For example, a rising current of hot air makes the objects viewed through it flicker and sway. You might have seen that on a summer day; hot air seems to be rising above the surface of the roads. It also happens when you see above a fire.
Early sunrise and late sunset
Consider an oblique ray from a heavenly body such as the sun or a star. While travelling through the atmosphere, it continuously moves into regions of higher refractive index. So, it continuously bends towards the normal, resulting in a path similar to the figure given below.
Since we see the object, that is the sun now, in the direction of the ray incident on the eye, the heavenly body appears higher than its actual position.
Now, consider a situation when the sun is just below the horizon. Rays of light coming from it get bent such that they seem to be coming from above the horizon (to the observer). Thus, when the sun's position is just below the horizon, the sun is visible to us. So, at sunrise, we see the sun before it actually comes to the horizon. At sunset, we see it even after it has dipped below the horizon. The fact is that it increases daylight by 4 minutes everyday (2 minutes at sunrise and 2 minutes at sunset).
You might also have noticed that the sun looks oval/flattened at sunrise and sunset. The rays from the lower regions of the sun travel a bit greater distance than those rays which are from the middle and upper regions. So, they bend more. As a result, the image of the lower region gets shifted upwards more than that of the upper region. This makes the sun appear like a flattened circle or an oval.
Similarly, twinkling of stars is also a result of atmospheric refraction.
Now, we can define atmospheric refraction as follows:
Atmospheric refraction is a phenomenon, where light/any electromagnetic wave deviates from its rectilinear path because of passing through atmosphere where there are variations in the air density and refractive index.
When light travels from a denser to rarer medium with an angle greater than the critical angle, the ray of light does not deviate in its path or does not refract, but it undergoes a reflection known as total internal reflection.
The critical angle differs from medium to medium.
\mu =\frac { 1 }{ \sin{ \theta }_{ c } }.
This is very useful as it is used in fiber glasses where total internal reflection helps in fast movement of wavelengths.
n_{\text{air}} = 1.00
n_{\text{glass}} = 1.50,
n_{\text{coating}} = 1.46
When viewing an object from different media the refraction of light often creates a shifting of the image. Because of this, for example, when viewing a pencil through a thick glass slab there is a shift of the image.
This shift of image is given by
\mu =\frac {(\text{real height})}{(\text{apparent height})}.
This is useful when viewing an object from water where the light bends only once.
However, when viewing an object from a glass slab, the light rays bend twice: once when going to the glass and again when coming out from the glass. So here we use
\triangle x=t\left( 1-\frac { 1 }{ \mu } \right),
\triangle x
is the shift in the image and
t
is the thickness of the glass slab.
Q1: Prove that only two real images can be formed when the distance between the screen and the image is
4f
in the case of a convex lens.
Details and Assumptions: The image should only form on the screen and the lens can be placed anywhere in between the object and the screen, so that no virtual image should form.
Hint: Use the information
v+u=4f
to form a quadratic equation. A quadratic equation has 2 roots, which implies it will have two such positions of
u
v.
Q2: The speed of an object travelling towards a concave mirror is
4 \text{ m/s}
and the image distance is
10\text{ cm},
while the object distance is
20\text{ cm}
. Find the speed of the image.
1\text{ m/s}
Hint: Use differentiation for this question.
4.00\text{ cm}
tall light bulb is placed a distance of
45.7\text{ cm}
from a double convex lens having a focal length of
15.2\text{ cm}.
(a) the image distance
(b) the image size.
a+b,
speed frequency wavelength all of the above none of the above
The scattering of light depends inversely upon the
4^\text{th}
\text{\_\_\_\_\_\_\_\_}.
Cite as: Geometrical Optics. Brilliant.org. Retrieved from https://brilliant.org/wiki/geometrical-optics/
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(Not recommended) Solve parabolic PDE problem - MATLAB parabolic - MathWorks United Kingdom
Parabolic Equation Using Legacy Syntax
Parabolic Problem Using Matrix Coefficients
Reducing Parabolic Equations to Elliptic Equations
(Not recommended) Solve parabolic PDE problem
parabolic is not recommended. Use solvepde instead.
u = parabolic(u0,tlist,model,c,a,f,d)
u = parabolic(u0,tlist,b,p,e,t,c,a,f,d)
u = parabolic(u0,tlist,Kc,Fc,B,ud,M)
u = parabolic(___,rtol)
u = parabolic(___,rtol,atol)
u = parabolic(___,'Stats','off')
Solves PDE problems of the type
d\frac{\partial u}{\partial t}-\nabla \cdot \left(c\nabla u\right)+au=f
on a 2-D or 3-D region Ω, or the system PDE problem
d\frac{\partial u}{\partial t}-\nabla \cdot \left(c\otimes \nabla u\right)+au=f
The variables c, a, f, and d can depend on position, time, and the solution u and its gradient.
u = parabolic(u0,tlist,model,c,a,f,d) produces the solution to the FEM formulation of the scalar PDE problem
d\frac{\partial u}{\partial t}-\nabla \cdot \left(c\nabla u\right)+au=f
d\frac{\partial u}{\partial t}-\nabla \cdot \left(c\otimes \nabla u\right)+au=f
with geometry, mesh, and boundary conditions specified in model, and with initial value u0. The variables c, a, f, and d in the equation correspond to the function coefficients c, a, f, and d respectively.
u = parabolic(u0,tlist,b,p,e,t,c,a,f,d) solves the problem using boundary conditions b and finite element mesh specified in [p,e,t].
u = parabolic(u0,tlist,Kc,Fc,B,ud,M) solves the problem based on finite element matrices that encode the equation, mesh, and boundary conditions.
u = parabolic(___,rtol) and u = parabolic(___,rtol,atol), for any of the previous input arguments, modify the solution process by passing to the ODE solver a relative tolerance rtol, and optionally an absolute tolerance atol.
u = parabolic(___,'Stats','off'), for any of the previous input arguments, turns off the display of internal ODE solver statistics during the solution process.
Solve the parabolic equation
\frac{\partial u}{\partial t}=\Delta u
on the square domain specified by squareg.
Create a PDE model and import the geometry.
Set Dirichlet boundary conditions
u=0
on all edges.
Set the initial condition to have
u\left(0\right)=1
{x}^{2}+{y}^{2}\le 0.{4}^{2}
u\left(0\right)=0
elsewhere.
u0 = zeros(size(p,2),1);
ix = find(sqrt(p(1,:).^2 + p(2,:).^2) <= 0.4);
u0(ix) = ones(size(ix));
Set solution times to be from 0 to 0.1 with step size 0.005.
Plot the initial condition, the solution at the final time, and two intermediate solutions.
title('t = 0')
pdeplot(model,'XYData',u(:,11))
pdeplot(model,'XYData',u(:,end))
title('t = 0.1')
\frac{\partial u}{\partial t}=\Delta u
on the square domain specified by squareg, using a geometry function to specify the geometry, a boundary function to specify the boundary conditions, and using initmesh to create the finite element mesh.
Specify the geometry as @squareg and plot the geometry.
u=0
on all edges. The squareb1 function specifies these boundary conditions.
b = @squareb1;
[p,e,t] = initmesh(g,'Hmax',0.02);
u\left(0\right)=1
{x}^{2}+{y}^{2}\le 0.{4}^{2}
u\left(0\right)=0
u = parabolic(u0,tlist,b,p,e,t,c,a,f,d);
pdeplot(p,e,t,'XYData',u(:,1));
pdeplot(p,e,t,'XYData',u(:,5))
pdeplot(p,e,t,'XYData',u(:,11))
pdeplot(p,e,t,'XYData',u(:,end))
Create finite element matrices that encode a parabolic problem, and solve the problem.
The problem is the evolution of temperature in a conducting block. The block is a rectangular slab.
handl = pdegplot(model,'FaceLabels','on');
handl(1).FaceAlpha = 0.5;
Faces 1, 4, and 6 of the slab are kept at 0 degrees. The other faces are insulated. Include the boundary condition on faces 1, 4, and 6. You do not need to include the boundary condition on the other faces because the default condition is insulated.
The initial temperature distribution in the block has the form
{u}_{0}=1{0}^{-3}xyz.
x = p(1,:);
y = p(2,:);
z = p(3,:);
u0 = x.*y.*z*1e-3;
The parabolic equation in toolbox syntax is
d\frac{\partial u}{\partial t}-\nabla \cdot \left(c\nabla u\right)+au=f.
Suppose the thermal conductivity of the block leads to a
coefficient value of 1. The values of the other coefficients in this problem are
d=1
a=0
f=0
Create the finite element matrices that encode the problem.
[Kc,Fc,B,ud] = assempde(model,c,a,f);
[~,M,~] = assema(model,0,d,f);
Solve the problem at time steps of 1 for times ranging from 0 to 40.
tlist = linspace(0,40,41);
u = parabolic(u0,tlist,Kc,Fc,B,ud,M);
72 function evaluations
71 solutions of linear systems
Plot the solution on the outside of the block at times 0, 10, 25, and 40. Ensure that the coloring is the same for all plots.
title 't = 0'
pdeplot3D(model,'ColorMapData',u(:,11))
title 't = 10'
vector | character vector | character array | string scalar | string vector
Initial condition, specified as a scalar, vector of nodal values, character vector, character array, string scalar, or string vector. The initial condition is the value of the solution u at the initial time, specified as a column vector of values at the nodes. The nodes are either p in the [p,e,t] data structure, or are model.Mesh.Nodes.
If the initial condition is a constant scalar v, specify u0 as v.
If there are Np nodes in the mesh, and N equations in the system of PDEs, specify u0 as a column vector of Np*N elements, where the first Np elements correspond to the first component of the solution u, the second Np elements correspond to the second component of the solution u, etc.
Give a text expression of a function, such as 'x.^2 + 5*cos(x.*y)'. If you have a system of N > 1 equations, give a text array such as
char('x.^2 + 5*cos(x.*y)',...
'tanh(x.*y)./(1+z.^2)')
Example: x.^2+5*cos(y.*x)
Solution times, specified as a real vector. The solver returns the solution to the PDE at the solution times.
c — PDE coefficient
scalar | matrix | character vector | character array | string scalar | string vector | coefficient function
PDE coefficient, specified as a scalar, matrix, character vector, character array, string scalar, string vector, or coefficient function. c represents the c coefficient in the scalar PDE
d\frac{\partial u}{\partial t}-\nabla \cdot \left(c\nabla u\right)+au=f
or in the system of PDEs
d\frac{\partial u}{\partial t}-\nabla \cdot \left(c\otimes \nabla u\right)+au=f
Example: 'cosh(x+y.^2)'
a — PDE coefficient
PDE coefficient, specified as a scalar, matrix, character vector, character array, string scalar, string vector, or coefficient function. a represents the a coefficient in the scalar PDE
d\frac{\partial u}{\partial t}-\nabla \cdot \left(c\nabla u\right)+au=f
d\frac{\partial u}{\partial t}-\nabla \cdot \left(c\otimes \nabla u\right)+au=f
Example: 2*eye(3)
f — PDE coefficient
PDE coefficient, specified as a scalar, matrix, character vector, character array, string scalar, string vector, or coefficient function. f represents the f coefficient in the scalar PDE
d\frac{\partial u}{\partial t}-\nabla \cdot \left(c\nabla u\right)+au=f
d\frac{\partial u}{\partial t}-\nabla \cdot \left(c\otimes \nabla u\right)+au=f
Example: char('sin(x)';'cos(y)';'tan(z)')
d — PDE coefficient
PDE coefficient, specified as a scalar, matrix, character vector, character array, string scalar, string vector, or coefficient function. d represents the d coefficient in the scalar PDE
d\frac{\partial u}{\partial t}-\nabla \cdot \left(c\nabla u\right)+au=f
d\frac{\partial u}{\partial t}-\nabla \cdot \left(c\otimes \nabla u\right)+au=f
b — Boundary conditions
boundary matrix | boundary file
Boundary conditions, specified as a boundary matrix or boundary file. Pass a boundary file as a function handle or as a file name. A boundary matrix is generally an export from the PDE Modeler app.
Example: b = 'circleb1', b = "circleb1", or b = @circleb1
Kc — Stiffness matrix
Stiffness matrix, specified as a sparse matrix or as a full matrix. See Elliptic Equations. Typically, Kc is the output of assempde.
Fc — Load vector
Load vector, specified as a vector. See Elliptic Equations. Typically, Fc is the output of assempde.
B — Dirichlet nullspace
Dirichlet nullspace, returned as a sparse matrix. See Algorithms. Typically, B is the output of assempde.
ud — Dirichlet vector
Dirichlet vector, returned as a vector. See Algorithms. Typically, ud is the output of assempde.
Mass matrix. specified as a sparse matrix or a full matrix. See Elliptic Equations.
To obtain the input matrices for pdeeig, hyperbolic or parabolic, run both assema and assempde:
Create the M matrix using assema with d, not a, as the argument before f.
rtol — Relative tolerance for ODE solver
1e-3 (default) | positive real
Relative tolerance for ODE solver, specified as a positive real.
atol — Absolute tolerance for ODE solver
Absolute tolerance for ODE solver, specified as a positive real.
PDE solution, returned as a matrix. The matrix is Np*N-by-T, where Np is the number of nodes in the mesh, N is the number of equations in the PDE (N = 1 for a scalar PDE), and T is the number of solution times, meaning the length of tlist. The solution matrix has the following structure.
The first Np elements of each column in u represent the solution of equation 1, then next Np elements represent the solution of equation 2, etc. The solution u is the value at the corresponding node in the mesh.
Column i of u represents the solution at time tlist(i).
To obtain the solution at an arbitrary point in the geometry, use pdeInterpolant.
To plot the solution, use pdeplot for 2-D geometry, or see 3-D Solution and Gradient Plots with MATLAB Functions.
parabolic internally calls assema, assemb, and assempde to create finite element matrices corresponding to the problem. It calls ode15s to solve the resulting system of ordinary differential equations.
Partial Differential Equation Toolbox™ solves equations of the form
m\frac{{\partial }^{2}u}{\partial {t}^{2}}+d\frac{\partial u}{\partial t}-\nabla ·\left(c\nabla u\right)+au=f
When the m coefficient is 0, but d is not, the documentation refers to the equation as parabolic, whether or not it is mathematically in parabolic form.
A parabolic problem is to solve the equation
d\frac{\partial u}{\partial t}-\nabla \cdot \left(c\nabla u\right)+au=f\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{in }\Omega
u(x,0) = u0(x) for x∊Ω
where x represents a 2-D or 3-D point and there are boundary conditions of the same kind as for the elliptic equation on ∂Ω.
The heat equation reads
\rho C\frac{\partial u}{\partial t}-\nabla \text{\hspace{0.17em}}·\text{\hspace{0.17em}}\left(k\nabla u\right)+h\left(u-{u}_{\infty }\right)=f
in the presence of distributed heat loss to the surroundings. ρ is the density, C is the thermal capacity, k is the thermal conductivity, h is the film coefficient, u∞ is the ambient temperature, and f is the heat source.
For time-independent coefficients, the steady-state solution of the equation is the solution to the standard elliptic equation
–∇ · (c∇u) + au = f.
Assuming a mesh on Ω and t ≥ 0, expand the solution to the PDE (as a function of x) in the Finite Element Method basis:
u\left(x,t\right)=\sum _{i}{U}_{i}\left(t\right){\varphi }_{i}\left(x\right)
Plugging the expansion into the PDE, multiplying with a test function ϕj, integrating over Ω, and applying Green's formula and the boundary conditions yield
\begin{array}{l}\sum _{i}\underset{\Omega }{\int }d{\varphi }_{j}{\varphi }_{i}\text{\hspace{0.17em}}\frac{d{U}_{i}\left(t\right)}{dt}\text{\hspace{0.17em}}dx+\sum _{i}\left(\underset{\Omega }{\int }\left(\nabla {\varphi }_{j}\cdot \left(c\nabla {\varphi }_{i}\right)+a{\varphi }_{j}{\varphi }_{i}\right)\text{\hspace{0.17em}}dx+\underset{\partial \Omega }{\int }q{\varphi }_{j}{\varphi }_{i}\text{\hspace{0.17em}}ds\right){U}_{i}\left(t\right)\\ \text{ }\text{ }\text{ }\text{ }=\underset{\Omega }{\int }f{\varphi }_{j}\text{\hspace{0.17em}}dx+\underset{\partial \Omega }{\int }g{\varphi }_{j}\text{\hspace{0.17em}}ds\text{ }\forall j\end{array}
In matrix notation, we have to solve the linear, large and sparse ODE system
M\frac{dU}{dt}+KU=F
This method is traditionally called method of lines semidiscretization.
Solving the ODE with the initial value
Ui(0) = u0(xi)
yields the solution to the PDE at each node xi and time t. Note that K and F are the stiffness matrix and the right-hand side of the elliptic problem
–∇ · (c∇u) + au = f in Ω
with the original boundary conditions, while M is just the mass matrix of the problem
–∇ · (0∇u) + du = 0 in Ω.
When the Dirichlet conditions are time dependent, F contains contributions from time derivatives of h and r. These derivatives are evaluated by finite differences of the user-specified data.
The ODE system is ill conditioned. Explicit time integrators are forced by stability requirements to very short time steps while implicit solvers can be expensive since they solve an elliptic problem at every time step. The numerical integration of the ODE system is performed by the MATLAB® ODE Suite functions, which are efficient for this class of problems. The time step is controlled to satisfy a tolerance on the error, and factorizations of coefficient matrices are performed only when necessary. When coefficients are time dependent, the necessity of reevaluating and refactorizing the matrices each time step may still make the solution time consuming, although parabolic reevaluates only that which varies with time. In certain cases a time-dependent Dirichlet matrix h(t) may cause the error control to fail, even if the problem is mathematically sound and the solution u(t) is smooth. This can happen because the ODE integrator looks only at the reduced solution v with u = Bv + ud. As h changes, the pivoting scheme employed for numerical stability may change the elimination order from one step to the next. This means that B, v, and ud all change discontinuously, although u itself does not.
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Self-similarly expanding networks to curve shortening flow
Schnürer, Oliver C. ; Schulze, Felix
We consider a network in the Euclidean plane that consists of three distinct half-lines with common start points. From that network as initial condition, there exists a network that consists of three curves that all start at one point, where they form
120
degree angles, and expands homothetically under curve shortening flow. We also prove uniqueness of these networks.
Classification : 53C44, 35Q51, 74K30, 74N20
author = {Schn\"urer, Oliver C. and Schulze, Felix},
title = {Self-similarly expanding networks to curve shortening flow},
AU - Schnürer, Oliver C.
AU - Schulze, Felix
TI - Self-similarly expanding networks to curve shortening flow
Schnürer, Oliver C.; Schulze, Felix. Self-similarly expanding networks to curve shortening flow. Annali della Scuola Normale Superiore di Pisa - Classe di Scienze, Série 5, Tome 6 (2007) no. 4, pp. 511-528. http://www.numdam.org/item/ASNSP_2007_5_6_4_511_0/
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[3] K. Ecker and G. Huisken, Mean curvature evolution of entire graphs, Ann. of Math. 130 (1989), 453-471. | MR 1025164 | Zbl 0696.53036
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[5] M. E. Gage and R. S. Hamilton, The heat equation shrinking convex plane curves, J. Differential Geom. 23 (1986), 69-96. | MR 840401 | Zbl 0621.53001
[6] M. A. Grayson, The heat equation shrinks embedded plane curves to points, J. Differential Geom. 26 (1987), 285-314. | MR 906392 | Zbl 0667.53001
[7] G. Huisken, A distance comparison principle for evolving curves, Asian J. Math. 2 (1998), 127-133. | MR 1656553 | Zbl 0931.53032
[8] T. Ilmanen, “Lectures on Mean Curvature Flow and Related Equations”, 1998, available from http://www.math.ethz.ch/
\sim
ilmanen/
[9] C. Mantegazza, M. Novaga and V. M. Tortorelli, Motion by curvature of planar networks, Ann. Scuola Norm. Sup. Pisa Cl. Sci. (5) 3 (2004), 235-324. | Numdam | MR 2075985 | Zbl 1170.53313
[10] R. Mazzeo and M. Sáez, Self similar expanding solutions of the planar network flow, arXiv:0704.3113v1 [math.DG]. | MR 3060453 | Zbl 1280.53062
[11] N. Stavrou, Selfsimilar solutions to the mean curvature flow, J. Reine Angew. Math. 499 (1998), 189-198. | MR 1631112 | Zbl 0895.53039
[12] B. Von Querenburg, “Mengentheoretische Topologie”, Springer-Verlag, Berlin, 1973, Hochschultext. | MR 467641 | Zbl 0431.54001
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Jug - Detailed Documentation - Maker Protocol Technical Docs
Accumulation of Stability Fees for Collateral Types
Contract Name: Jug
The primary function of the Jug smart contract is to accumulate stability fees for a particular collateral type whenever its drip() method is called. This effectively updates the accumulated debt for all Vaults of that collateral type as well as the total accumulated debt as tracked by the Vat (global) and the amount of Dai surplus (represented as the amount of Dai owned by the Vow).
Ilk : contains two uint256 values—duty, the collateral-specific risk premium, and rho, the timestamp of the last fee update
VatLike : mock contract to make Vat interfaces callable from code without an explicit dependency on the Vat contract itself
wards : mapping(address => uint) that indicates which addresses may call administrative functions
ilks : mapping (bytes32 => Ilk) that stores an Ilk struct for each collateral type
vat : a VatLike that points the the system's Vat contract
vow : the address of the Vow contract
base : a uint256 that specifies a fee applying to all collateral types
rely/deny : add or remove authorized users (via modifications to the wards mapping)
init(bytes32) : start stability fee collection for a particular collateral type
file(bytes32, bytes32, uint) : set duty for a particular collateral type
file(bytes32, data) : set the base value
file(bytes32, address) : set the vow value
drip(bytes32) : collect stability fees for a given collateral type
drip(bytes32 ilk) performs stability fee collection for a specific collateral type when it is called (note that it is a public function and may be called by anyone). drip does essentially three things:
calculates the change in the rate parameter for the collateral type specified by ilk based on the time elapsed since the last update and the current instantaneous rate (base + duty);
calls Vat.fold to update the collateral's rate, total tracked debt, and Vow surplus;
updates ilks[ilk].rho to be equal to the current timestamp.
The change in the rate is calculated as:
\Delta rate = (base+duty)^{now-rho} \cdot rate- rate
where "now" represents the current time, "rate" is Vat.ilks[ilk].rate, "base" is Jug.base, "rho" is Jug.ilks[ilk].rho, and "duty" is Jug.ilks[ilk].duty. The function reverts if any sub-calculation results in under- or overflow. Refer to the Vat documentation for more detail on fold.
rpow(uint x, uint n, uint b), used for exponentiation in drip, is a fixed-point arithmetic function that raises x to the power n. It is implemented in Solidity assembly as a repeated squaring algorithm. x and the returned value are to be interpreted as fixed-point integers with scaling factor b. For example, if b == 100, this specifies two decimal digits of precision and the normal decimal value 2.1 would be represented as 210; rpow(210, 2, 100) returns 441 (the two-decimal digit fixed-point representation of 2.1^2 = 4.41). In the current implementation, 10^27 is passed for b, making x and the rpow result both of type ray in standard MCD fixed-point terminology. rpow's formal invariants include "no overflow" as well as constraints on gas usage.
Parameters Can Only Be Set By Governance
Jug stores some sensitive parameters, particularly the base rate and collateral-specific risk premiums that determine the overall stability fee rate for each collateral type. Its built-in authorization mechanisms need to allow only authorized MakerDAO governance contracts/actors to set these values. See "Failure Modes" for a description of what can go wrong if parameters are set to unsafe values.
Ilk Initialization
init(bytes32 ilk) must called when a new collateral is added (setting duty via file() is not sufficient)—otherwise rho will be uninitialized and fees will accumulate based on a start date of January 1st, 1970 (start of Unix epoch).
base + Ilk.duty imbalance in drip()
A call to drip(bytes32 ilk)will add the base rate to the Ilk.duty rate. The rate is a calculated compounded rate, so rate(base + duty) != rate(base) + rate(duty). This means that if base is set, the duty will need to be set factoring the existing compounding factor in base, otherwise the result will be outside of the rate tolerance. Updates to the base value will require all of the ilks to be updated as well.
If drip() is called very infrequently for some collateral types (due, for example, to low overall system usage or extremely stable collateral types that have essentially zero liquidation risk), then the system will fail to collect fees on Vaults opened and closed between drip() calls. As the system achieves scale, this becomes less of a concern, as both Keepers and MKR holders are have an incentive to regularly call drip (the former to trigger liquidation auctions, the latter to ensure that surplus accumulates to decrease MKR supply); however, a hypothetical asset with very low volatility yet high risk premium might still see infrequent drip calls at scale (there is not at present a real-world example of this—the most realistic possibility is base being large, elevating rates for all collateral types).
Malicious or Careless Parameter Setting
Various parameters of Jug may be set to values that damage the system. While this can occur by accident, the greatest concern is malicious attacks, especially by an entity that somehow becomes authorized to make calls directly to Jug's administrative methods, bypassing governance. Setting duty (for at least one ilk) or base too low can lead to Dai oversupply; setting either one too high can trigger excess liquidations and therefore unjust loss of collateral. Setting a value for vow other than the true Vow's address can cause surplus to be lost or stolen.
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Prime Numbers Warmup Practice Problems Online | Brilliant
What is the largest even prime number?
2 There are no even prime numbers There are infinitely many even prime numbers It is not known whether there are finitely many or infinitely many even prime numbers
A teacher would like to have their students work together in small groups of at least 2. If there are 23 students and each student can only be in one group, is it possible to divide the students into at least 2 groups such that all the groups have the same number of students?
8 can be written as the sum of 2 prime numbers:
8 = 3 + 5
Which of the following numbers cannot be written as the sum of 2 prime numbers?
20 22 24 26 They can all be written as the sum of 2 prime numbers
3, 5, and 7 are all prime. Is there an
n > 3
n,
n + 2,
n + 4
are all prime?
Is there a positive integer
n
6n + 15
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Determine whether the geometric series is convergent or divergent. 10-4+1.6-0.64+.... If it's
10-4+1.6-0.64+\dots .
If it's convergent find its sum.
Juan Spiller
S=10-4+1.60.64+\dots
Given that S is a geometric series,
The known fact is that a geometric series
a+ar+a{r}^{2}+\dots
is convergent
⇔|r|<1
By comparing the given series with its general form,
a=10
r=\frac{-4}{10}=-0.4
|r|=|-0.4|<1
This implies that the series
10-4+1.6-0.64+\dots
a+ar+a{r}^{2}+\dots
- is equal to
\frac{a}{1-r}\left(r<1\right)
10-4+1.6-0.64±\dots =\left(10+1.6+0.256+\dots \right)-\left(4+0.64+\dots \right)
=\left(\frac{10}{1-0.16}\right)-\left(\frac{4}{1-0.16}\right)\left[\begin{array}{c}\because a=10,r=0.16\text{ }in\text{ }first\text{ }geometric\text{ }series\\ a=4,r=0,16in\text{ }second\text{ }geometric\text{ }series\end{array}\right]
=\left(\frac{10-4}{1-0.16}\right)
=\left(\frac{6}{0.84}\right)
=7.1428571428571429
Identify whether the series is an arithmetic series (AS), geometric series (GS) or not one of the difined types ( NASGS ) and determine the sum (for NASGS, write NO SUM). Write your Solution.
1+\frac{3}{2}+\frac{9}{4}+...+\frac{81}{16}
Consider two series
\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2}\pi nZ\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{\pi }^{2}}n
. Which of them converges?
Q1. Does a series
\sum _{\mathrm{\infty }}^{n=1}bn
converge if
bn\to 0
? Justify your answer by at least two examples?
Let P(k) be a statement that
\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+...+\frac{1}{k\cdot \left(k+1\right)}=
for: The basis step to prove
P\left(k\right)
is that at
k=1,?
for:Show that
P\left(1\right)
is true by completing the basis step proof. Left side of
P\left(k\right)
and Right side of
P\left(k\right)
for: Identify the inductive hypothesis used to prove
P\left(k\right)
for: Identify the inductive step used to prove
P\left(k+1\right).
1+\frac{1}{2\sqrt{2}}+\frac{1}{3\sqrt{3}}+\frac{1}{4\sqrt{4}}+\frac{1}{5\sqrt{5}}+\dots
Missing number in the series 9, ____, 6561, 43046721 is: 81 25 62 31 18
Use rationialization to simplify the given expression.
|
Linear regression model for incremental learning - MATLAB - MathWorks Switzerland
\ell \left[y,f\left(x\right)\right]=\mathrm{max}\left[0,|y-f\left(x\right)|-\epsilon \right]
\ell \left[y,f\left(x\right)\right]=\frac{1}{2}{\left[y-f\left(x\right)\right]}^{2}
f\left(x\right)=x\beta +b.
\ell \left[y,f\left(x\right)\right]=\mathrm{max}\left[0,|y-f\left(x\right)|-\epsilon \right]
\ell \left[y,f\left(x\right)\right]=\frac{1}{2}{\left[y-f\left(x\right)\right]}^{2}
{\gamma }_{t}=\frac{{\gamma }_{0}}{{\left(1+\lambda {\gamma }_{0}t\right)}^{c}}.
{\beta }_{1}
{\beta }_{1}
{\beta }_{1}
{\beta }_{10}
{\beta }_{10}
{\beta }_{10}
{\beta }_{313}
{\beta }_{313}
{\beta }_{313}
{x}_{j}^{\ast }=\frac{{x}_{j}-{\mu }_{j}^{\ast }}{{\sigma }_{j}^{\ast }}.
{\mu }_{j}^{\ast }=\frac{1}{\sum _{k}{w}_{k}}\sum _{k}{w}_{k}{x}_{jk}.
{\left({\sigma }_{j}^{\ast }\right)}^{2}=\frac{1}{\sum _{k}{w}_{k}}\sum _{k}{w}_{k}{\left({x}_{jk}-{\mu }_{j}^{\ast }\right)}^{2}.
|
Solve the following initial value problem for the function y(x). y'=2xy^2;\
Solve the following initial value problem for the function y(x).
{y}^{\prime }=2x{y}^{2};\text{ }y\left(1\right)=\frac{1}{2}
{y}^{\prime }=2x{y}^{2}
⇒\frac{dy}{dx}=2x{y}^{2}
⇒\int \frac{dy}{{y}^{2}}=\int 2xdx
=-{y}^{-1}={x}^{2}+c
⇒\frac{-1}{y}={x}^{2}+c
y\left(1\right)=\frac{1}{2}
⇒-2=1+c
⇒c=-3
\therefore \frac{-1}{y}={x}^{2}-3
⇒y=\frac{1}{3}-{x}^{2}
- is the solution
Solve the separable equation
\frac{dy\left(x\right)}{dx}=2xy{\left(x\right)}^{2}
, such that y(1) = 1/2:
y{\left(x\right)}^{2}
\frac{\frac{dy\left(x\right)}{dx}}{y}{\left(x\right)}^{2}=2x
Integrate both sides with respect to
x
\int \frac{\frac{dy\left(x\right)}{dx}}{y}{\left(x\right)}^{2}dx=\int 2xdx
-\frac{1}{y}\left(x\right)={x}^{2}+{c}_{1}
{c}_{1}
y\left(x\right)
y\left(x\right)=-\frac{1}{{x}^{2}+{c}_{1}}
{c}_{1}
y\left(1\right)=\frac{1}{2}\to y\left(x\right)=-\frac{1}{{x}^{2}+{c}_{1}}
-\frac{1}{{c}_{1}+1}=\frac{1}{2}
{c}_{1}=-3
{c}_{1}=-3\to y\left(x\right)=-\frac{1}{{x}^{2}+{c}_{1}}
y\left(x\right)=\frac{1}{-{x}^{2}+3}
Thank you, both answers are good!
If y(x)=uv, then y
f\left(x\right)=\frac{3{x}^{3}-4}{2{x}^{2}+3}
Two sides of a triangle have lengths a and b, and the angle between them is 0. What value of 0 will maximize the triangle's area? [Hint:
A=\left(\frac{1}{2}\right)ab\mathrm{sin}0
-5{x}^{2}
f\left(x\right)={e}^{x}\left({x}^{3}-3{x}^{2}+6x-6\right)
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Sound Synthesis Theory/Additive Synthesis - Wikibooks, open books for an open world
1.2 Constructing common harmonic waveforms in additive synthesis
1.3 Additive resynthesis
Additive Synthesis[edit | edit source]
As previously discussed in Section 1, sine waves can be considered the building blocks of sound. In fact, it was shown in the 19th Century by the mathematician Joseph Fourier that any periodic function can be expressed as a series of sinusoids of varying frequencies and amplitudes. This concept of constructing a complex sound out of sinusoidal terms is the basis for additive synthesis, sometimes called Fourier synthesis for the aforementioned reason. In addition to this, the concepts of additive synthesis have also existed since the introduction of the organ, where different pipes of varying pitch are combined to create a sound or timbre.
Figure 6.1. Additive synthesis block diagram.
A simple block diagram of the additive form may appear like in Fig. 6.1, which has a simplified mathematical form based on the Fourier series:
{\displaystyle f(t)=a_{0}+\sum _{n=1}^{\infty }a_{n}\sin(f_{n}t)}
{\displaystyle a_{0}}
is an offset value for the whole function (typically 0),
{\displaystyle a_{n}}
are the amplitude weightings for each sine term, and
{\displaystyle f_{n}}
is the frequency multiplier value. With hundreds of terms each with their own individual frequency and amplitude weightings, we can design and specify some incredibly complex sounds, especially if we can modulate the parameters over time. One of the key features of natural sounds is that they have a dynamic frequency response that does not remain fixed. However, a popular approach to the additive synthesis system is to use frequencies that are integer multiples of the fundamental frequency, which is known as harmonic additive synthesis. For example, if the first oscillator's frequency,
{\displaystyle f_{1}}
represents the fundamental frequency of the sound at 100 Hz, then the second oscillator's frequency would be
{\displaystyle f2=2f_{1}}
, and the third
{\displaystyle f3=3f_{1}}
and so on. This series of sine waves produces an even "harmonic" sound that can be described as "musical". Oscillator frequency relationships that are not integer related, on the other hand, are called "inharmonic" and tend to be noisier and take on the characteristics of bells or other percussive sounds.
Constructing common harmonic waveforms in additive synthesis[edit | edit source]
Figure 6.2. The first four terms of a square wave constructed from sinusoidal components (partials).
If we know the amplitude weightings and frequency components of the first
{\displaystyle x}
sinusoidal components or partials of a complex waveform, we can reconstruct that waveform using an additive system with
{\displaystyle x}
oscillators. The popular waveforms square, sawtooth and triangle are harmonic waveforms because the constituent sinusoidal components all have frequencies that are integer multiples of the fundamental. The property that distinguishes them in this form is that they all have unique amplitude weightings for each sinusoid. Fig. 6.2 demonstrates the appearance of the time-domain waveform as a set of sines at unique amplitude weightings are added together; in this case the form begins to approximate a square wave, with the accuracy increasing with each added partial. Note that to construct a square wave we only include odd numbered harmonics- the amplitude weightings for
{\displaystyle a_{2}}
{\displaystyle a_{4}}
{\displaystyle a_{6}}
etc. are 0. Below is a table that demonstrates the partial amplitude weightings of the common waveshapes:
{\displaystyle a_{1}}
{\displaystyle a_{2}}
{\displaystyle a_{3}}
{\displaystyle a_{4}}
{\displaystyle a_{5}}
{\displaystyle a_{6}}
{\displaystyle a_{7}}
{\displaystyle a_{8}}
{\displaystyle a_{9}}
Sine 1 0 0 0 0 0 0 0 0
Square 1 0 1/3 0 1/5 0 1/7 0 1/9
{\displaystyle 1/x}
{\displaystyle x}
Triangle 1 0 -1/9 0 1/25 0 -1/49 0 1/81
{\displaystyle 1/x^{2}}
{\displaystyle x}
, alternating + and -.
Sawtooth 1 1/2 1/3 1/4 1/5 1/6 1/7 1/8 1/9
{\displaystyle 1/x}
A conclusion you may draw from Fig. 6.2 and the table is that it requires a large amount of frequency partials to create a waveform that closely approximates the idealised mathematical forms of the waveforms introduced in Section 5. For this reason, it should be apparent that additive synthesis techniques are perhaps not the best method for producing these forms. The strengths of additive synthesis lie in the fact that we can exert control over every partial component of our sound, which can produce some very intricate and wonderful results. With the constant modification of the frequency and amplitude values of each oscillator, the possibilities are endless. Some examples of ways to control the weightings and frequencies of each component oscillator are illustrated:
Manual control. The user controls a bank of oscillators with an external control device (typically MIDI), tweaking the values in real time. More than one person can join in and change / alter the timbre to their whims.
External data. Digital information from another source is taken and converted into appropriate frequency and amplitude values. The varying data source will then effectively be in 'control' of the timbral outcomes. Composers have been known to use data from natural sources or pieces derived from interesting geometric, aleatoric and mathematical models.
Recursive data. Given a source set of values and a set of algorithmic rules, the control parameters reference the previous value entered into the system to determine the result of the next one. Users may wish to "interfere" with the system to set the process on a new path. See Markov chains.
There is, however, the major consideration of computational power: complex sounds may require many oscillators all operating at once which will put major demand on the system in question.
Additive resynthesis[edit | edit source]
In Section 1 it was mentioned that just as it is possible to construct waveforms using additive techniques, we can analyse and deconstruct waveforms as well. It is possible to analyse the frequency partials of a recorded sound and then resynthesize a representation of the sound using a series of sinusoidal partials. By calculating the frequency and amplitude weighting of partials in the frequency domain (typically using a Fast Fourier transform), an additive resynthesis system can construct an equally weighted sinusoid at the same frequency for each partial. Older techniques rely on banks of filters to separate each sinusoid; their varying amplitudes are used as control functions for a new set of oscillators under the user's control. Because the sound is represented by a bank of oscillators inside the system, a user can make adjustments to the frequency and amplitude of any set of partials. The sound can be 'reshaped' - by alterations made to timbre or the overall amplitude envelope, for example. A harmonic sound could be restructured to sound inharmonic, and vice versa.
Retrieved from "https://en.wikibooks.org/w/index.php?title=Sound_Synthesis_Theory/Additive_Synthesis&oldid=3583941"
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Presentation_of_a_group Knowpia
{\displaystyle \langle S\mid R\rangle .}
{\displaystyle \langle a\mid a^{n}=1\rangle ,}
{\displaystyle \langle a\mid a^{n}\rangle ,}
{\displaystyle s_{1}^{a_{1}}s_{2}^{a_{2}}\cdots s_{n}^{a_{n}}}
rfrf = 1,
r8 = 1, or
f2 = 1.
If we then let N be the subgroup of F generated by all conjugates x−1Rx of R, then it follows by definition that every element of N is a finite product x1−1r1x1 ... xm−1rm xm of members of such conjugates. It follows that each element of N, when considered as a product in D8, will also evaluate to 1; and thus that N is a normal subgroup of F. Thus D8 is isomorphic to the quotient group F/N. We then say that D8 has presentation
{\displaystyle \langle r,f\mid r^{8}=1,f^{2}=1,(rf)^{2}=1\rangle .}
Here the set of generators is S = {r, f }, and the set of relations is R = {r 8 = 1, f 2 = 1, (rf )2 = 1}. We often see R abbreviated, giving the presentation
{\displaystyle \langle r,f\mid r^{8}=f^{2}=(rf)^{2}=1\rangle .}
An even shorter form drops the equality and identity signs, to list just the set of relators, which is {r 8, f 2, (rf )2}. Doing this gives the presentation
{\displaystyle \langle r,f\mid r^{8},f^{2},(rf)^{2}\rangle .}
Although the notation ⟨S | R⟩ used in this article for a presentation is now the most common, earlier writers used different variations on the same format. Such notations include the following:[citation needed]
⟨S | R⟩
(S | R)
{S; R}
⟨S; R⟩
Let S be a set and let FS be the free group on S. Let R be a set of words on S, so R naturally gives a subset of
{\displaystyle F_{S}}
. To form a group with presentation
{\displaystyle \langle S\mid R\rangle }
, take the quotient of
{\displaystyle F_{S}}
by the smallest normal subgroup that contains each element of R. (This subgroup is called the normal closure N of R in
{\displaystyle F_{S}}
.) The group
{\displaystyle \langle S\mid R\rangle }
{\displaystyle \langle S\mid R\rangle =F_{S}/N.}
The elements of S are called the generators of
{\displaystyle \langle S\mid R\rangle }
and the elements of R are called the relators. A group G is said to have the presentation
{\displaystyle \langle S\mid R\rangle }
if G is isomorphic to
{\displaystyle \langle S\mid R\rangle }
It is a common practice to write relators in the form
{\displaystyle x=y}
where x and y are words on S. What this means is that
{\displaystyle y^{-1}x\in R}
. This has the intuitive meaning that the images of x and y are supposed to be equal in the quotient group. Thus, for example, rn in the list of relators is equivalent with
{\displaystyle r^{n}=1}
For a finite group G, it is possible to build a presentation of G from the group multiplication table, as follows. Take S to be the set elements
{\displaystyle g_{i}}
of G and R to be all words of the form
{\displaystyle g_{i}g_{j}g_{k}^{-1}}
{\displaystyle g_{i}g_{j}=g_{k}}
The definition of group presentation may alternatively be recast in terms of equivalence classes of words on the alphabet
{\displaystyle S\cup S^{-1}}
. In this perspective, we declare two words to be equivalent if it is possible to get from one to the other by a sequence of moves, where each move consists of adding or removing a consecutive pair
{\displaystyle xx^{-1}}
{\displaystyle x^{-1}x}
for some x in S, or by adding or removing a consecutive copy of a relator. The group elements are the equivalence classes, and the group operation is concatenation.[1]
Finitely presented groupsEdit
Recursively presented groupsEdit
the free group on S
{\displaystyle \langle S\mid \varnothing \rangle }
Cn, the cyclic group of order n
{\displaystyle \langle a\mid a^{n}\rangle }
Dn, the dihedral group of order 2n
{\displaystyle \langle r,f\mid r^{n},f^{2},(rf)^{2}\rangle }
D∞, the infinite dihedral group
{\displaystyle \langle r,f\mid f^{2},(rf)^{2}\rangle }
Dicn, the dicyclic group
{\displaystyle \langle r,f\mid r^{2n},r^{n}=f^{2},frf^{-1}=r^{-1}\rangle }
Z × Z
{\displaystyle \langle x,y\mid xy=yx\rangle }
Z/mZ × Z/nZ
{\displaystyle \langle x,y\mid x^{m},y^{n},xy=yx\rangle }
the free abelian group on S
{\displaystyle \langle S\mid R\rangle }
Sn, the symmetric group on n symbols generators:
{\displaystyle \sigma _{1},\ldots ,\sigma _{n-1}}
{\displaystyle \sigma _{i}^{2}=1}
{\displaystyle \sigma _{i}\sigma _{j}=\sigma _{j}\sigma _{i}{\mbox{ if }}j\neq i\pm 1}
{\displaystyle \sigma _{i}\sigma _{i+1}\sigma _{i}=\sigma _{i+1}\sigma _{i}\sigma _{i+1}\ }
{\displaystyle {(\sigma _{i}\sigma _{i+1}})^{3}=1\ }
{\displaystyle \sigma _{i}^{2}=1}
Here σi is the permutation that swaps the ith element with the i+1st one. The product σiσi+1 is a 3-cycle on the set {i, i+1, i+2}.
Bn, the braid groups generators:
{\displaystyle \sigma _{1},\ldots ,\sigma _{n-1}}
{\displaystyle \sigma _{i}\sigma _{j}=\sigma _{j}\sigma _{i}{\mbox{ if }}j\neq i\pm 1}
{\displaystyle \sigma _{i}\sigma _{i+1}\sigma _{i}=\sigma _{i+1}\sigma _{i}\sigma _{i+1}\ }
Note the similarity with the symmetric group; the only difference is the removal of the relation
{\displaystyle \sigma _{i}^{2}=1}
V4 ≅ D2, the Klein 4 group
{\displaystyle \langle s,t\mid s^{2},t^{2},(st)^{2}\rangle }
T ≅ A4, the tetrahedral group
{\displaystyle \langle s,t\mid s^{2},t^{3},(st)^{3}\rangle }
O ≅ S4, the octahedral group
{\displaystyle \langle s,t\mid s^{2},t^{3},(st)^{4}\rangle }
I ≅ A5, the icosahedral group
{\displaystyle \langle s,t\mid s^{2},t^{3},(st)^{5}\rangle }
Q8, the quaternion group
{\displaystyle \langle i,j\mid i^{4},jij=i,iji=j\rangle \,}
{\displaystyle \langle a,b\mid aba=bab,(aba)^{4}\rangle }
GL(2, Z)
{\displaystyle \langle a,b,j\mid aba=bab,(aba)^{4},j^{2},(ja)^{2},(jb)^{2}\rangle }
PSL(2, Z), the modular group
{\displaystyle \langle a,b\mid a^{2},b^{3}\rangle }
{\displaystyle \langle x,y,z\mid z=xyx^{-1}y^{-1},xz=zx,yz=zy\rangle }
BS(m, n), the Baumslag–Solitar groups
{\displaystyle \langle a,b\mid a^{n}=ba^{m}b^{-1}\rangle }
{\displaystyle \langle a,b\mid a^{2},b^{3},(ab)^{13},[a,b]^{5},[a,bab]^{4},((ab)^{4}ab^{-1})^{6}\rangle }
[a, b] is the commutator
An example of a finitely generated group that is not finitely presented is the wreath product
{\displaystyle \mathbf {Z} \wr \mathbf {Z} }
To see this, given a group G, consider the free group FG on G. By the universal property of free groups, there exists a unique group homomorphism φ : FG → G whose restriction to G is the identity map. Let K be the kernel of this homomorphism. Then K is normal in FG, therefore is equal to its normal closure, so ⟨G | K⟩ = FG/K. Since the identity map is surjective, φ is also surjective, so by the First Isomorphism Theorem, ⟨G | K⟩ ≅ im(φ) = G. This presentation may be highly inefficient if both G and K are much larger than necessary.
Novikov–Boone theoremEdit
The negative solution to the word problem for groups states that there is a finite presentation ⟨S | R⟩ for which there is no algorithm which, given two words u, v, decides whether u and v describe the same element in the group. This was shown by Pyotr Novikov in 1955[4] and a different proof was obtained by William Boone in 1958.[5]
Suppose G has presentation ⟨S | R⟩ and H has presentation ⟨T | Q⟩ with S and T being disjoint. Then
the free product G ∗ H has presentation ⟨S, T | R, Q⟩ and
the direct product G × H has presentation ⟨S, T | R, Q, [S, T]⟩, where [S, T] means that every element from S commutes with every element from T (cf. commutator).
The deficiency of a finite presentation ⟨S | R⟩ is just |S| − |R| and the deficiency of a finitely presented group G, denoted def(G), is the maximum of the deficiency over all presentations of G. The deficiency of a finite group is non-positive. The Schur multiplicator of a finite group G can be generated by −def(G) generators, and G is efficient if this number is required.[6]
Geometric group theoryEdit
^ a b c Peifer, David (1997). "An Introduction to Combinatorial Group Theory and the Word Problem". Mathematics Magazine. 70 (1): 3–10. doi:10.1080/0025570X.1997.11996491.
^ Sir William Rowan Hamilton (1856). "Memorandum respecting a new System of Roots of Unity" (PDF). Philosophical Magazine. 12: 446.
^ Stillwell, John (2002). Mathematics and its history. Springer. p. 374. ISBN 978-0-387-95336-6.
^ Novikov, Pyotr S. (1955), "On the algorithmic unsolvability of the word problem in group theory", Proceedings of the Steklov Institute of Mathematics (in Russian), 44: 1–143, Zbl 0068.01301
^ Boone, William W. (1958), "The word problem" (PDF), Proceedings of the National Academy of Sciences, 44 (10): 1061–1065, Bibcode:1958PNAS...44.1061B, doi:10.1073/pnas.44.10.1061, PMC 528693, PMID 16590307, Zbl 0086.24701
^ Johnson, D.L.; Robertson, E.L. (1979). "Finite groups of deficiency zero". In Wall, C.T.C. (ed.). Homological Group Theory. London Mathematical Society Lecture Note Series. Vol. 36. Cambridge University Press. pp. 275–289. ISBN 0-521-22729-1. Zbl 0423.20029.
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Risk Adjusted (deprecated) - Idle Finance
This strategy allows you to automatically rebalance the best risk/yield allocation
The risk-adjusted allocation strategy provides a way to earn the best rate at the lowest risk level. The risk-management algorithm takes account of the total assets within a pool, incorporates underlying protocol rate functions and levels of supply and demand, skimming protocols with a bad score/rate mix, and finally determining an allocation that achieves the highest risk-return score possible after the rebalance happens.
It has been developed in collaboration with DeFiScore, a framework for quantifying risk in permissionless lending pools. DeFiScore is a single, consistently comparable value for measuring protocol risk, based on factors including smart contract risk, collateralization, and liquidity. The model outputs a 0–10 score that represents the level of risk on a specific lending protocol (where 10 is the upper bound = lowest risk, and 0 is the lower bound = highest risk).
You can read more about the risk assessment model here.
With this strategy, we are trying to find the right balance between risk and returns. We are weighting score and apr based on k parameter. This can be modeled as follows:
max\ q(x) = \sum_{i=0}^{n} \frac{x_i}{tot} * (\frac{\frac{nextRate_i(x_i)}{maxNextRate} + k * \frac{nextScore_i(x_i)}{maxNextScore}}{k + 1})
where n is the number of lending protocols used, x_i is the amount (in underlying) allocated in protocol i , nextRate(x_i) is a function which returns the new APR for protocol i after supplying x_i ,nextScore(x_i) is a function which returns the new Score for protocol i after supplying x_iamount of underlying, maxNextRate is the highest rate of all implemented protocols after supplying x_i amount, same for maxNextScore with regard to the score, tot is total amount to rebalance, finally k is a coefficient for expressing weights of score and apr (k = 1 means equally weighted, currently k = 2 so score weights twice the APR).
tot=\sum_{i=0}^{n} x_i
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Xcommutator - Maple Help
Home : Support : Online Help : Mathematics : Differential Equations : DEtools : Lie Symmetry Method : Commands for ODEs : Xcommutator
calculate the commutator of two generators of one-parameter Lie groups
Xcommutator(X1, X2, k, y(x), ODE)
lists of the coefficients of the symmetry generators (infinitesimals) as in [xi, eta]
(optional) extension of the generators entering the commutator
(optional) right hand side is used to replace the highest derivative in the result; required if dynamical symmetries are given
The Xcommutator command receives two generators of one-parameter Lie groups, either in the form of a pair of infinitesimals [xi, eta] or in the form of differential operators, and the dependent variable y(x), and returns the commutator of these generators.
If k is given, the k extension of the generators X1 and X2 is calculated at first, and the command returns the commutator of these extended generators (that is, another extended generator; see eta_k and infgen ).
If the given generators are in the form of a list containing the infinitesimals, the result is returned as a list; otherwise, if X1 and X2 are given as differential operators (mappings) then the result is returned as a mapping (see examples).
This function is part of the DEtools package, and so it can be used in the form Xcommutator(..) only after executing the command with(DEtools). However, it can always be accessed through the long form of the command by using DEtools[Xcommutator](..).
\mathrm{with}\left(\mathrm{DEtools}\right):
\mathrm{X1}≔[x,-y]
\textcolor[rgb]{0,0,1}{\mathrm{X1}}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{y}]
\mathrm{X2}≔[ax,cy+{y}^{2}]
\textcolor[rgb]{0,0,1}{\mathrm{X2}}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{c}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{2}}]
The commutator of X1 with X2
\mathrm{Xcommutator}\left(\mathrm{X1},\mathrm{X2},y\left(x\right)\right)
[\textcolor[rgb]{0,0,1}{\mathrm{_ξ}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{_η}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{2}}]
The commutator of the third extensions of X1 and X2 involves four elements.
\mathrm{Xcommutator}\left(\mathrm{X1},\mathrm{X2},y\left(x\right),3\right)
[\textcolor[rgb]{0,0,1}{\mathrm{_ξ}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{_η}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{\mathrm{_η}}}_{\textcolor[rgb]{0,0,1}{1}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{_y1}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{\mathrm{_η}}}_{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{_y1}}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{_y2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{\mathrm{_η}}}_{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{6}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{_y1}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{_y2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{_y3}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}]
The generators G1 and G2 associated to the lists X1 and X2
\mathrm{X1}
[\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{y}]
\mathrm{G1}≔\mathrm{infgen}\left(\mathrm{X1},y\left(x\right)\right)
\textcolor[rgb]{0,0,1}{\mathrm{G1}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{_F1}}\textcolor[rgb]{0,0,1}{→}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{}\left(\frac{\textcolor[rgb]{0,0,1}{∂}}{\textcolor[rgb]{0,0,1}{∂}\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{_F1}}\right)\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{}\left(\frac{\textcolor[rgb]{0,0,1}{∂}}{\textcolor[rgb]{0,0,1}{∂}\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{_F1}}\right)
\mathrm{X2}
[\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{c}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{2}}]
\mathrm{G2}≔\mathrm{infgen}\left(\mathrm{X2},y\left(x\right)\right)
\textcolor[rgb]{0,0,1}{\mathrm{G2}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{_F1}}\textcolor[rgb]{0,0,1}{→}\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{}\left(\frac{\textcolor[rgb]{0,0,1}{∂}}{\textcolor[rgb]{0,0,1}{∂}\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{_F1}}\right)\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{c}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{2}}\right)\textcolor[rgb]{0,0,1}{}\left(\frac{\textcolor[rgb]{0,0,1}{∂}}{\textcolor[rgb]{0,0,1}{∂}\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{_F1}}\right)
If G1 and G2 are differential operators (mappings), the commutator of G1 and G2 is returned as a mapping as well:
\mathrm{Xcommutator}\left(\mathrm{G1},\mathrm{G2},y\left(x\right)\right)
\textcolor[rgb]{0,0,1}{\mathrm{_F2}}\textcolor[rgb]{0,0,1}{→}\textcolor[rgb]{0,0,1}{-}\left(\frac{\textcolor[rgb]{0,0,1}{∂}}{\textcolor[rgb]{0,0,1}{∂}\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{_F2}}\right)\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{2}}
The commutator of the third extensions of G1 and G2
\mathrm{Xcommutator}\left(\mathrm{G1},\mathrm{G2},y\left(x\right),3\right)
\textcolor[rgb]{0,0,1}{\mathrm{_F2}}\textcolor[rgb]{0,0,1}{→}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\left(\frac{\textcolor[rgb]{0,0,1}{∂}}{\textcolor[rgb]{0,0,1}{∂}\textcolor[rgb]{0,0,1}{\mathrm{_y1}}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{_F2}}\right)\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{_y1}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{\mathrm{_y1}}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{_y2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}\right)\textcolor[rgb]{0,0,1}{}\left(\frac{\textcolor[rgb]{0,0,1}{∂}}{\textcolor[rgb]{0,0,1}{∂}\textcolor[rgb]{0,0,1}{\mathrm{_y2}}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{_F2}}\right)\textcolor[rgb]{0,0,1}{+}\left(\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{6}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{_y1}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{_y2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{_y3}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}\right)\textcolor[rgb]{0,0,1}{}\left(\frac{\textcolor[rgb]{0,0,1}{∂}}{\textcolor[rgb]{0,0,1}{∂}\textcolor[rgb]{0,0,1}{\mathrm{_y3}}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{_F2}}\right)\textcolor[rgb]{0,0,1}{-}\left(\frac{\textcolor[rgb]{0,0,1}{∂}}{\textcolor[rgb]{0,0,1}{∂}\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{_F2}}\right)\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{2}}
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Dimension (vector space) - Wikipedia
Number of vectors in any basis of the vector space
For every vector space there exists a basis,[a] and all bases of a vector space have equal cardinality;[b] as a result, the dimension of a vector space is uniquely defined. We say
{\displaystyle V}
is finite-dimensional if the dimension of
{\displaystyle V}
is finite, and infinite-dimensional if its dimension is infinite.
The dimension of the vector space
{\displaystyle V}
{\displaystyle F}
{\displaystyle \dim _{F}(V)}
{\displaystyle [V:F],}
read "dimension of
{\displaystyle V}
{\displaystyle F}
". When
{\displaystyle F}
can be inferred from context,
{\displaystyle \dim(V)}
is typically written.
{\displaystyle \mathbb {R} ^{3}}
{\displaystyle \left\{{\begin{pmatrix}1\\0\\0\end{pmatrix}},{\begin{pmatrix}0\\1\\0\end{pmatrix}},{\begin{pmatrix}0\\0\\1\end{pmatrix}}\right\}}
as a standard basis, and therefore
{\displaystyle \dim _{\mathbb {R} }(\mathbb {R} ^{3})=3.}
{\displaystyle \dim _{\mathbb {R} }(\mathbb {R} ^{n})=n,}
and even more generally,
{\displaystyle \dim _{F}(F^{n})=n}
{\displaystyle F.}
{\displaystyle \mathbb {C} }
are both a real and complex vector space; we have
{\displaystyle \dim _{\mathbb {R} }(\mathbb {C} )=2}
{\displaystyle \dim _{\mathbb {C} }(\mathbb {C} )=1.}
So the dimension depends on the base field.
The only vector space with dimension
{\displaystyle 0}
{\displaystyle \{0\},}
the vector space consisting only of its zero element.
{\displaystyle W}
{\displaystyle V}
{\displaystyle \dim(W)\leq \dim(V).}
To show that two finite-dimensional vector spaces are equal, the following criterion can be used: if
{\displaystyle V}
is a finite-dimensional vector space and
{\displaystyle W}
{\displaystyle V}
{\displaystyle \dim(W)=\dim(V),}
{\displaystyle W=V.}
{\displaystyle \mathbb {R} ^{n}}
has the standard basis
{\displaystyle \left\{e_{1},\ldots ,e_{n}\right\},}
{\displaystyle e_{i}}
{\displaystyle i}
-th column of the corresponding identity matrix. Therefore,
{\displaystyle \mathbb {R} ^{n}}
{\displaystyle n.}
Any two finite dimensional vector spaces over
{\displaystyle F}
with the same dimension are isomorphic. Any bijective map between their bases can be uniquely extended to a bijective linear map between the vector spaces. If
{\displaystyle B}
is some set, a vector space with dimension
{\displaystyle |B|}
{\displaystyle F}
can be constructed as follows: take the set
{\displaystyle F(B)}
{\displaystyle f:B\to F}
{\displaystyle f(b)=0}
{\displaystyle b}
{\displaystyle B.}
These functions can be added and multiplied with elements of
{\displaystyle F}
to obtain the desired
{\displaystyle F}
{\displaystyle F/K}
is a field extension, then
{\displaystyle F}
is in particular a vector space over
{\displaystyle K.}
Furthermore, every
{\displaystyle F}
{\displaystyle V}
{\displaystyle K}
-vector space. The dimensions are related by the formula
{\displaystyle \dim _{K}(V)=\dim _{K}(F)\dim _{F}(V).}
In particular, every complex vector space of dimensio{\displaystyle n}
{\displaystyle 2n.}
Some formulae relate the dimension of a vector space with the cardinality of the base field and the cardinality of the space itself. If
{\displaystyle V}
is a vector space over a field
{\displaystyle F}
then and if the dimension of
{\displaystyle V}
{\displaystyle \dim V,}
If dim
{\displaystyle V}
is finite then
{\displaystyle |V|=|F|^{\dim V}.}
{\displaystyle V}
is infinite then
{\displaystyle |V|=\max(|F|,\dim V).}
The dimension of a vector space may alternatively be characterized as the trace of the identity operator. For instance,
{\displaystyle \operatorname {tr} \ \operatorname {id} _{\mathbb {R} ^{2}}=\operatorname {tr} \left({\begin{smallmatrix}1&0\\0&1\end{smallmatrix}}\right)=1+1=2.}
This appears to be a circular definition, but it allows useful generalizations.
Firstly, it allows for a definition of a notion of dimension when one has a trace but no natural sense of basis. For example, one may have an algebra
{\displaystyle A}
with maps
{\displaystyle \eta :K\to A}
(the inclusion of scalars, called the unit) and a map
{\displaystyle \epsilon :A\to K}
(corresponding to trace, called the counit). The composition
{\displaystyle \epsilon \circ \eta :K\to K}
is a scalar (being a linear operator on a 1-dimensional space) corresponds to "trace of identity", and gives a notion of dimension for an abstract algebra. In practice, in bialgebras, this map is required to be the identity, which can be obtained by normalizing the counit by dividing by dimension (
{\displaystyle \epsilon :=\textstyle {\frac {1}{n}}\operatorname {tr} }
), so in these cases the normalizing constant corresponds to dimension.
A subtler generalization is to consider the trace of a family of operators as a kind of "twisted" dimension. This occurs significantly in representation theory, where the character of a representation is the trace of the representation, hence a scalar-valued function on a group
{\displaystyle \chi :G\to K,}
whose value on the identity
{\displaystyle 1\in G}
is the dimension of the representation, as a representation sends the identity in the group to the identity matrix:
{\displaystyle \chi (1_{G})=\operatorname {tr} \ I_{V}=\dim V.}
The other values
{\displaystyle \chi (g)}
of the character can be viewed as "twisted" dimensions, and find analogs or generalizations of statements about dimensions to statements about characters or representations. A sophisticated example of this occurs in the theory of monstrous moonshine: the
{\displaystyle j}
-invariant is the graded dimension of an infinite-dimensional graded representation of the monster group, and replacing the dimension with the character gives the McKay–Thompson series for each element of the Monster group.[3]
Krull dimension – In mathematics, dimension of a ring
Matroid rank – Maximum size of an independent set of the matroid
Rank (linear algebra) – Dimension of the column space of a matrix
^ if one assumes the axiom of choice
^ see dimension theorem for vector spaces
^ Itzkov, Mikhail (2009). Tensor Algebra and Tensor Analysis for Engineers: With Applications to Continuum Mechanics. Springer. p. 4. ISBN 978-3-540-93906-1.
^ Gannon, Terry (2006), Moonshine beyond the Monster: The Bridge Connecting Algebra, Modular Forms and Physics, ISBN 0-521-83531-3
Axler, Sheldon (2015). Linear Algebra Done Right. Undergraduate Texts in Mathematics (3rd ed.). Springer. ISBN 978-3-319-11079-0.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Dimension_(vector_space)&oldid=1075564704"
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Let V, W, and Z be vector spaces, and let
T:V\to W
U:W\to Z
be linear.
If UT is onto, prove that U is onto.Must T also be onto?
Let UT is onto.
It is needed to prove that U is onto.
z\in Z
then for some
x\in V
UT\left(x\right)=z
T\left(x\right)\in W
So U is onto.
But T may not be onto.
T:R\to {R}^{2}
T\left(x\right)=\left(x,y\right)
U:{R}^{2}\to R
U\left(x.y\right)=0
Therefore, T may not be onto.
\left(1,3,0\right),\left(-2,0,2\right),\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left(-1,3,-1\right)
r\left(t\right)=<{t}^{2},\frac{2}{3}{t}^{3},t>
<4,-\frac{16}{3},-2>
{\mathbb{R}}^{4}
W=\left\{⟨4s-t,s,t,s⟩\mid s,t\in \mathbb{R}\right\}
\left\{⟨0,1,4,1⟩,⟨1,1,3,1⟩\right\}
{\mathbb{R}}^{4}
{\mathbb{R}}^{n}
Find the rate of change of f at P in the direction of the vector u.
f\left(x,y,z\right)={y}^{2}{e}^{x}yz,\text{ }P\left(0,1,-1\right),\text{ }u=<\frac{3}{13},\text{ }\frac{4}{13},\text{ }\frac{12}{13}>
{R}^{p}
\left(x+y\right)=norm\left(x\right)+norm\left(y\right)
x=cy\text{ }or\text{ }y=cx\text{ }with\text{ }c>0
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Describe in words the surface whose equation is given.\phi =\pi
Describe in words the surface whose equation is given.\phi =\pi /3
\varphi =\frac{\pi }{3}
\varphi =\frac{\pi }{3}
\mathrm{cos}\varphi =\mathrm{cos}\left(\frac{\pi }{3}\right)=\frac{1}{2}
x=\rho \mathrm{sin}\varphi \mathrm{cos}0
{\mathrm{cos}}^{2}\varphi =\frac{1}{4}
y=\rho \mathrm{sin}\varphi \mathrm{sin}0
{\rho }^{2}{\mathrm{cos}}^{2}\varphi =\frac{1}{4}{\rho }^{2}
z=\rho \mathrm{cos}\varphi
{z}^{2}=\frac{1}{4}\left({x}^{2}+{y}^{2}+{z}^{2}\right)
{x}^{2}+{y}^{2}+{z}^{2}={\rho }^{2}
4{z}^{2}={x}^{2}+{y}^{2}+{z}^{2}
3{z}^{2}={x}^{2}+{y}^{2}
3{z}^{2}={x}^{2}+{y}^{2}
is a double cone, however, since the original equation
\varphi =\frac{\pi }{3}>0
A=\left[\begin{array}{cc}3& 1\\ 1& 1\\ 1& 4\end{array}\right],b\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]
\stackrel{―}{x}=
Determine whether the given problem is an equation or an expression. If it is an equation, then solve. If it is an expression, then simplify. 1/2 (2x - 1) - 1/3 (5x + 2) = 3
\left(\frac{5}{6}\right)x-\left(\frac{5}{12}\right)x+\left(\frac{7}{24}\right)x
Establish the formula
ab+\left(\frac{a-b}{2}{\right)}^{2}=\left(\frac{a+b}{2}{\right)}^{2}
Write the logarithmic equation as an exponential equation.
\frac{{\mathrm{log}}_{5}1}{625}=-4
State whether the equation or system of equation is linear.
{x}^{2}+{y}^{2}+{z}^{2}=4
Find a relationship between x and y such that (x, y) is equidistant (the same distance) from the two points. (-1/2, -4), (7/2, 5/4)
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v is a set of ordered pairs (a, b) of
v is a set of ordered pairs (a, b) of real numbers. Sum and scalar multiplication are defined by: (a, b) + (c, d) = (a + c, b + d) k (a, b) = (kb, ka) (attention in this part)
v is a set of ordered pairs (a, b) of real numbers. Sum and scalar multiplication are defined by: (a, b) + (c, d) = (a + c, b + d) k (a, b) = (kb, ka) (attention in this part) show that V is not linear space.
If V is a linear space, we must have that lv=v,
for all v from V. Take for example v=(0,1)
Then lv=1(0,1)=(1*1,1*0)=(1,0)=/v.
Thus V is not linear space.
\left(1,3,0\right),\left(-2,0,2\right),\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left(-1,3,-1\right)
r\left(t\right)=<{t}^{2},\frac{2}{3}{t}^{3},t>
<4,-\frac{16}{3},-2>
A vector is first rotated by
{90}^{\circ }
along x-axis and then scaled up by 5 times is equal to
\left(15,-10,20\right)
. What was the original vector
Find the coordinate vector of w relative to the basis
S=\left\{{u}_{1},{u}_{2}\right\}\text{ }for\text{ }{\mathbb{R}}^{\mathbb{2}}\text{ }{u}_{1}=\left(1,-1\right),{u}_{2}=\left(1,1\right);w=\left(1,0\right)
Write the vector v in the form ai + bj, given its magnitude ∥v∥∥v∥ and the angle
\alpha
it makes with the positive x-axis.
\parallel v\parallel =8,\alpha =45\ast \parallel v\parallel =8,\alpha =45
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The system of equation \displaystyle{b}{e}{g}\in{\left\lbrace{c}{a}{s}{e}{s}\right\rbrace}\frac{n}{{2}}\backslash-\frac{{{n}+{1}}}{{2}}{e}{n}{d}{\left\lbrace{c}{a}{s}{e}{s}\right\rbrace} by graphing method
The system of equation {(2x + y = 1),(4x +2y = 3):} by graphing method
\left\{\begin{array}{l}2x+y=1\\ 4x+2y=3\end{array}
\left\{\begin{array}{l}2x+y=1\\ 4x+2y=3\end{array}
The first equation is given below.
2x+y=1\dots \left(1\right)
The second equation is given below
4x+2y=3\dots \left(2\right)
Write the equation (1) in slope-intercept standard form as follows:
2x+y=1
y=1-2x
Compare the equation
y=-2x+1
with the slope-intercept form
y=mx+c
m=-2
c=1
Therefore, the value of slope m is -2 and c is 1.
Thus, the value of y-intercept (0, c) is (0, 1).
Subtitute 0 for y in equation
y=-2x+1
to obtain the value of x-intercept.
0=-2x+1
-1=-2x
x=\frac{1}{2}
Therefore, the x-intercept is
\left(\frac{1}{2},0\right).
Join the points (0, 1) and
\left(\frac{1}{2},0\right)
to make a line on the graph as shown below in Figure 1:
4x+2y=3
2y=-4x+3
y=-\frac{-4}{2}x+\frac{3}{2}
y=-2x+\frac{3}{2}
y=-2x+\frac{3}{2}
y=mx+c
m=-2
c=\frac{3}{2}
Therefore, the value of slope m is -2 and c is
\frac{3}{2}
Thus, the value of y-intercept (0, c) is
\le \left(\frac{{x}^{2}}{{y}^{3}}\right)
Substitute 0 for y in equation
y=-2x+\frac{3}{2}
0=-2x+\frac{3}{2}
-\frac{3}{2}=-2x
x=\frac{3}{4}
Thus, the x-intercept is
\left(0,\frac{3}{4}\right)
Join the points
\left(0,\frac{3}{2}\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left(\frac{3}{4},0\right)
to make a line on the same graph of Figure 1 then the updated graph is shown below in Figure 2.
From Figure 2, it is observed that there is no point of intersection of lines.
Therefore, the lines never meet and are parallel to each so the system has no solution in the systems of linear equations.
Thus, the system is inconsistent and the system has no solution as shown below in Figure 2.
Provide notes on how triple integrals defined in cylindrical and spherical coordinates and the reason to prefer one of these coordinate systems to working in rectangular coordinates.
Explain the difference between Alternating Direction Method of Multipliers(ADMM) and coordinate descent (CD) .
To Sketch: The points on three - axis Cartesian coordinate system.
Point A:
\left[\begin{array}{ccc}-6& \text{ }2& \text{ }1\end{array}\right]
\left[\begin{array}{ccc}-6& \text{ }2& \text{ }1\end{array}\right]
\left[\begin{array}{ccc}5& \text{ }-3& \text{ }-6\end{array}\right]
y\ge {x}^{2},z\ge 0,
x\le {y}^{2},0\le z\le 2.
class council determined that its profit from the upcoming homecoming dance is directly related to the ticket price for the dance. Looking at past dances, the council determined that the profit pp can be modeled by the function
p\left(t\right)=-12{t}^{2}+480t+30
, where tt represents the price of each ticket. What should be the price of a ticket to the homecoming dance to maximize the council's profit? Price
1. Show that sup
\left\{1-\frac{1}{n}:n\in N\right\}=1\left\{1-\frac{1}{n}:n\in N\right\}=\frac{1}{2}
S\phantom{\rule{0.222em}{0ex}}=\left\{\frac{1}{n}-\frac{1}{m}:n,m\in N\right\}S\phantom{\rule{0.222em}{0ex}}=\left\{\frac{1}{n}-\frac{1}{m}:n,m\in N\right\},f\in d\in fS\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\supset S.end\left\{tab\underset{―}{a}r\right\}
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Form (a) the coefficient matrix and (b) the augmented matrix for the system of linear equa
Form (a) the coefficient matrix and (b) the augmented matrix for the system of linear equations. \left{\begin{array}{c} 7 x-5 y=11 x-y=-5 \end{array}\
\left\{\begin{array}{l}7x-5y=11\\ x-y=-5\end{array}
\left[\begin{array}{cc}7& -5\\ 1& -1\end{array}\right]
sugmented matrix:
\left[\begin{array}{cccc}7& -5& |& 11\\ 1& -1& |& -5\end{array}\right]
\left(1,3,0\right),\left(-2,0,2\right),\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left(-1,3,-1\right)
r\left(t\right)=<{t}^{2},\frac{2}{3}{t}^{3},t>
<4,-\frac{16}{3},-2>
Find a unit vector that has the same direction as the given vector. 8i-j+4k
Use the figure shawn 10 identify each a. vertical sngles b. adjacent angles c. linear pairs d. supplementary angles e. complementary angles
<1,3>\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}<-2,4>
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Estimation of initial elastic properties of 2D homogenised masonry model based on tensor scale indices - Brozovsky Maluchova 2015a - Scipedia
Tensor scale
J. Brozovsky and M. Maluchova, Estimation of initial elastic properties of 2D homogenised masonry model based on tensor scale indices, Perspectives in Science (2016). Vol. 7 URL https://www.scipedia.com/public/Brozovsky_Maluchova_2015a
Masonry ; Numerical analysis ; Tensor scale ; Finite element method
There are numerous masonry structures across the Europe. A lot of them have important cultural or historical value. A repair or a rehabilitation of such structure usually requires a very good knowledge about its structural function and bearing capacity (Cajka et al., 2012 ). This information can be usually obtained by a combination of in situ tests and numerical simulations. Numerical simulations are often very important because in situ tests on the historical masonry structures have to be limited as much as possible in order to protect original structures and materials (Witzany and Zigler, 2015 ).
Numerical simulations of structural behaviour can be conducted with use of numerical methods, in the most cases the finite element method is used (Zucchini and Lourenco, 2004 ). The computation can be both elastic (Witzany et al., 2014 ) or in-elastic (Portioli et al., 2012 ). One of the main problems is an acquirement of proper input data for the computations because masonry is a highly inhomogeneous material (Matysek and Witkowski, 2015 ). There have been numerous works that studied the masonry properties in certain regions and time periods, for example, see Matysek and Witkowski (2015) . Many researches have been conducting different types of laboratory experiment to simulate behaviour of masonry structures in certain conditions in order to obtain data for numerical modelling (Cajka et al., 2012 , Cajka et al., 2014 , Witzany et al., 2014 and Witzany and Zigler, 2015 ). The results of above mentioned works have been resulted for many types of models and approaches, the work of Sykora et al. (2013) , can be mentioned as an example of numerical model for estimation of compressive strength of masonry.
The further problem is masonry anisotropy which is caused by its structure: it can be created from regular or irregular brick or stones which can have various shape and location. The structure of the masonry has an important influence on local masonry mechanical properties. It causes serious problems even for linear elastic analysis of masonry structures because precise modelling of all bricks (so-called micromodel approach) is usually very time-consuming or even impossible. For this reason numerical models with a homogenised material are used. It this case a micromodel with several materials is replaced by homogeneous model with one material which has to represent the material behaviour. An orthotropic material is often assumed (Cajka et al., 2014 ).
The homogenised material properties can be obtained from laboratory tests of numerical modelling of a relatively small piece of masonry (Brozovsky and Pankaj, 2007 ) but such approach assumes than structures of masonry does not differ too much within the structure. This assumption is correct for testing samples (which usually have regular shape) and for experimental structures but it can be incorrect for real buildings with arches, windows and with non-trivial shapes. An example of such structure is illustrated in Fig. 1 . To simplify the process of numerical model preparation it is proposed to use available information of masonry structure (2D photos or X-ray pictures, for example) as a basis for automatic computation of orthotropic mechanical properties.
Example of masonry structure with complex geometry.
The proposed work is one of the first steps of an effort to create an automatic tool for computation of homogenised orthotropic properties which have to be based on a structure of the masonry. The current effort concentrates on 2D models but it is possible to extend it to the 3D later.
The approach consists of two main steps. The first one is an obtaining of 2D computer image of masonry structure. It may be a digital picture or X-ray scan. The second step is an analysis of the picture in order to describe the masonry structure in selected points. This step should result in a ready to use 2D finite element method (FEM) model with homogenised orthotropic material properties. The article deals with an approach required for the second step.
Measurement of material anisotropy
Brozovsky and Pankaj (2007) used the tensor scale indices (introduced by Saha and Wehrli, 2004 , for analysis of human bone and then extended for other uses by Saha, 2005 and Genau et al., 2009 ) for an estimation of material properties of material with complicated internal structure. The paper by Brozovsky and Pankaj (2007) compared the results of tensor scale analysis with finite element simulations of numerical with both original and homogenised material properties. It was shown that the tensor scale can be used for such tasks but some form of correlation between tensor scale indices and the mechanical properties has to be established for individual types of structures. The tensor scale can be directly used to find the orientations of material axes for homogenised orthotropic material, though.
The tensor scale in a material point is defined by sizes of axes of the largest possible ellipse that has centre in that point and that can be fitted in the material of the same type. An example of such ellipse is illustrated in Fig. 2 . There are three main parameters of the ellipse: two axes sizes and the angle between x coordinate axis and the a axis of the ellipse.
Tensor scale parameters (a , b ) in a material point.
There are several possible approaches to compute the tensor scale. The authors of this paper use the original approach proposed by Saha and Wehrli (2004) . The computation is based on use of certain number of measurements of distances between the studied point and the nearest material point of different material type. The angle a between the all testing lines is usually the same. The procedure is illustrated in Fig. 3 . A more detailed overview of the procedure is available in the paper by Brozovsky and Pankaj (2007) .
Analysis of space with one material type.
The original approach was designed for measurement of homogeneous materials that include empty spaces. It is use for cases which involve more than one material requires certain modifications.
As it was already stated before it is possible to use the approach to compute orientation of orthotropic material axes. The ellipse orientation angle can be directly used. An estimation of mechanical properties (Young moduli and Poisson ratios) can requires further steps.
Estimation of mechanical properties based on tensor scale
On the basis of numerical studies conducted by authors of this paper it can be stated that only the tensor scale data of the masonry are required for analysis. The analysis of mechanical properties requires masonry model of certain size. Their size will be illustrated below.
The obtained global tensor scale parameters (a , b ) on such area can be used to define a ratio between Young moduli of homogenised orthotropic material as it is shown in Eq. (1) .
{\displaystyle {\frac {a}{b}}={\frac {E_{1}}{E_{2}}}}
However, in arbitrary case it is not possible to define the values of E1 or E2 just from tensor scale data and the initial E parameters of mortar and bricks. There are of course additional data that have to be included to the computation: volume ratio between mortar and bricks and their mass ratio. It is possible to use neural network approach to define the relations for certain cases but it is out of the scope of this paper. There are also problems with determination of Poisson ratios.
It should be also mentioned that the a /b ratio obtained in Eq. (1) can differ up to 20% from a ratio of Young moduli which can be obtained from finite analysis of the same sample.
There is proposed a simpler approach to obtain values of orthotropic material properties which uses the finite element model of the masonry sample. The finite element model can be subjected to simulated compression and shear tests than can be used to provide Young moduli and Poisson ratios in defined directions. The selection of those directions should utilise the ellipse angle obtained from the tensor scale computation.
The required finite element simulations are illustrated in Fig. 4 . The material properties can be obtained from formulas of linear elasticity which use reactions and x and y deformations from the FEM computations.
Schemes of FEM simulations of simple material tests.
It is proposed to use such computation to compute a correction factor q for the a /b values that can be computed in other parts of the analysed structure (2) .
{\displaystyle q={\frac {aE_{2}}{bE_{1}}}}
The custom computer codes in the C language have been written for discussed calculations. A general in-house written finite element code uFEM was used to conduct the FEM computations. Fig. 5 shows the user interfaces for the tensor scale evaluating code and for the finite element method code.
Tensor scale code (left) and FEM code (right) evaluating the same model.
A simple example is provided to illustrate the proposed approach. The basic masonry sample is shown in Fig. 6 where are also shown the larger sizes of the samples. Several levels of larger samples with the same basic structure have been created.
The basic sample (left) and larger samples.
The analysis of these samples was conducted to show how bit the sample have to be to obtain the results for the a /b ratio.
The a /b ratios for the models are given in Table 1 . The largest model is not shown in Fig. 6 due to its size. The Size parameter in Table 1 describes the number of basic units which were used to create the model. The basic unit is shown in Fig. 6 (a picture at the left). The computed angle of the tensor scale axes (the ellipse orientation) is not shown as it was obviously 0 rad in this case. The expected 0 rad value was also obtained from the computations.
Table 1. Comparison of tensor scale values for different sample sizes.
1 2 × 2 0.131559
3 12 × 12 0.146460
It can be concluded that for the given case the model of 4 × 4 basic units is enough for the tensor scale analysis.
A study to determine a necessary number of sample lines (see Fig. 3 ) was also conducted. For simple cases with rectangular bricks it can be shown that 8 sample lines are enough. Larger number of sample lines (12 or more) gives comparable results but it requires greater computational time. In cases with more irregular brick shapes it is required to use at least 12 sample lines.
The initial material properties are given in Table 2 . The selected computed mechanical properties are given in Table 3 . It can be noted that there is slight difference between a /b ratio (0.146) and the computed E1 /E2 ratio (0.145).
Table 2. Input data for materials.
10 1 GPa
Table 3. Obtained data for homogenised orthotropic material.
Young modulus E1
The paper discussed a proposed procedure to simplify the computation of homogenised linear elastic orthotropic material properties for finite element modelling of masonry. The proposed procedure is based on tensor scale anisotropy indices which was originally introduced by Saha and Wehrli (2004) . The proposed procedure complements the tensor scale indices by finite element analysis of a masonry sample model which is used to obtain a correction coefficient for the a /b tensor scale data and to obtain the actual sizes of Young moduli of the homogenised material. The obtained data can be used to set the orthotropic properties of other parts of the numerical model: the orientation angle is directly used and the mechanical values are based on the a /b data for the particular part of the model and on the q coefficient which was obtained from the numerical analysis of a masonry sample. The proposed procedure can be used for estimating of initial elastic properties of the material. There are ongoing works that are aimed to allow to use the procedure for estimation of inelastic material parameters.
This outcome has been achieved with the financial support of the Ministry of Education, Youth and Sports of the Czech Republic with use of the institutional support of conceptual development of research in 2015.
Brozovsky and Pankaj, 2007 J. Brozovsky, P. Pankaj; Towards modeling of a trabecular bone; Comput. Struct., 85 (9) (2007), pp. 512–517
Cajka et al., 2012 R. Cajka, P. Mateckova, M. Stara, M. Janulikova; Testing of pre-stressed masonry corner for tri-axial stress–strain analysis, Life-Cycle and Sustainability of Civil Infrastructure Systems; Proceedings of the 3rd International Symposium on Life-Cycle Civil Engineering, IALCCE 2012 (2012), pp. 1955–1958
Cajka et al., 2014 R. Cajka, M. Kozielova, K. Burkovic, L. Mynarzova; Strengthening of masonry structures on the undermined area by prestressing; Acta Montan. Slovaca, 19 (2) (2014), pp. 95–104
Genau et al., 2009 A. Genau, P. Voorhees, K. Thornton; The morphology of topologically complex interfaces; Scripta Mater., 60 (2009), pp. 301–304
Matysek and Witkowski, 2015 P. Matysek, M. Witkowski; A comparative study on the compressive strength of bricks from different historical periods; Int. J. Archit. Herit. (2015) http://dx.doi.org/10.1080/15583058.2013.855838
Portioli et al., 2012 F. Portioli, L. Cascini, M. D’Aniello, R.A. Landolfo; Rigid block model with cracking units for limit analysis of masonry walls subject to in-plane loads; Proceedings of the Eleventh International Conference on Computational Structures Technology, Civil-Comp Press (2012), p. 117
Saha and Wehrli, 2004 P.K. Saha, F.W. Wehrli; A robust method for measuring trabecular bone orientation anisotropy at in vivo resolution using tensor scale’; Pattern Recogn., 37 (2004), pp. 1935–1944
Saha, 2005 P.K. Saha; Tensor scale: a local morphometric parameter with applications to computer vision and image processing; Comput. Vision Image Underst., 99 (2005), pp. 384–413
Sykora et al., 2013 M. Sykora, T. Cejka, M. Holicky, J. Witzany; Probabilistic model for compressive strength of historic masonry; European Safety and Reliability Conference, ESREL 2013 (2013), pp. 2645–2652
Witzany et al., 2014 J. Witzany, T. Cejka, R. Zigler; Failure mechanism of compressed short brick masonry columns confined with FRP strips; Constr. Build. Mater., 63 (2014), pp. 180–188
Witzany and Zigler, 2015 J. Witzany, R. Zigler; Failure mechanism of compressed reinforced and non-reinforced stone columns; Materiaux et Constructions, 48 (5) (2015), pp. 1603–1613
Zucchini and Lourenco, 2004 A. Zucchini, P.B. Lourenco; A coupled homogenisation-damage model for masonry cracking; Comput. Struct., 82 (2004), pp. 917–929
Masonry • Numerical analysis • Tensor scale • Finite element method
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Analysis of Laminar Falling Film Condensation Over a Vertical Plate With an Accelerating Vapor Flow | J. Fluids Eng. | ASME Digital Collection
A.-R. A. Khaled,
Department of Thermal Engineering and Desalination Technology,
e-mail: akhaled@kau.edu.sa
Abdulhaiy M. Radhwan,
Abdulhaiy M. Radhwan
S. A. Al-Muaikel
Khaled, A. A., Radhwan, A. M., and Al-Muaikel, S. A. (June 23, 2009). "Analysis of Laminar Falling Film Condensation Over a Vertical Plate With an Accelerating Vapor Flow." ASME. J. Fluids Eng. July 2009; 131(7): 071304. https://doi.org/10.1115/1.3155992
Laminar falling film condensations over a vertical plate with an accelerating vapor flow is analyzed in this work in the presence of condensate suction or slip effects at the plate surface. The following assumptions are made: (i) laminar condensate flow having constant properties, (ii) pure vapor with a uniform saturation temperature in the vapor region, and (iii) the shear stress at the liquid/vapor interface is negligible. The appropriate fundamental governing partial differential equations for the condensate and vapor flows (continuity, momentum, and energy equations) for the above case are identified, nondimensionalized, and transformed using nonsimilarity transformation. The transformed equations were solved using numerical, iterative, and implicit finite-difference methods. It is shown that the freestream striking angle has insignificant influence on the condensation mass and heat transfer rates, except when slip condition is present and at relatively small
Grl/Re2
values. Moreover, it is shown that increasing the values of the dimensionless suction parameter
(VS)
results to an increase in dimensionless mass of condensate
(Γ(L)/(μl Re))
and Nusselt number
(Nu(L)/Re1/2)
. Thus, it results in an increase in condensation mass and heat transfer rates. Finally, it is found that the condensation and heat transfer rates increase as Jakob number, slip parameter, and saturation temperature increase. Finally, the results of this work not only enrich the literature of condensation but also provide additional methods for saving thermal energy.
film condensation, film flow, finite difference methods, heat transfer, iterative methods, mass transfer, slip flow, condensation, mass transfer, buoyancy driven flow, laminar, gas-liquid, heat transfer
Condensation, Condensed matter, Film condensation, Flow (Dynamics), Heat transfer, Suction, Temperature, Vapors, Vertical plates, Finite difference methods, Mass transfer
Heat Transfer in Condensation—Effect of Heat Capacity Condensate
Heat Transfer Theory of Film Condensation of Saturated Vapor at Rest
Heat Transfer by Film Condensation
The Two-Phase Boundary Layer in Laminar Film Condensation
An Analytical Study of Laminar Film Condensation Part I—Flat Plates
Heat Transmission During Condensation of Superheated Steam
Heat Transfer Coefficients in Surface Condensers
Heat Transfer During Film Condensation of a Liquid Metal Vapor
The Condensation of Superheated Steam
Proceedings of a Conference Held at the National Engineering Laboratory
, East Kilbride, Glasgow, pp.
A Survey of the Condensation of Superheated Steam
Laminar Film Condensation on Plane and Axisymmetric Bodies in Non-Uniform Gravity
Mixed Convection Laminar Film Condensation on an Inclined Tube
Free Forced Convection Laminar Film Condensation on Horizontal Elliptical Tubes
Mixed Convection Laminar Film Condensation on a Horizontal Elliptical Tube With Uniform Surface Heat Flux
Exit Condition, Gravity, and Surface-Tension Effects on Stability and Noise-Sensitivity Issues for Steady Condensing Flows Inside Tubes and Channels
Internal Condensing Flows Inside a Vertical Pipe—Experimental/Computational Investigations of the Effects of Specified and Unspecified (Free) Conditions at Exit
Non-Linear Stability of the Classical Nusselt Problem of Film Condensation and Wave-Effects
Direct Computational Simulations for Internal Condensing Flows and Results on Attainability/Stability of Steady Solutions, Their Intrinsic Waviness, and Their Noise-Sensitivity
Effect of Gravity, Shear and Surface Tension in Internal Condensation Flows—Results From Direct Computational Simulations
Royale des Sciences de I'Institut de France
Mixed-Convection Laminar Film Condensation on an Inclined Elliptical Tube
Combined Free and Forced Convection Film Condensation on a Finite-Size Horizontal Wavy Plate
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Head - Maple Help
Home : Support : Online Help : Connectivity : Web Features : Web Page Access : URL package : Head
test specified URL for validity
Head( url, options )
a list of one or more of content, headers, and code, or one of the symbols content, headers, or code. This determines what will be returned by Head. The output will be an expression sequence with the corresponding values. The default is headers. This option is only valid with http and https schemes.
The Head command tests the URL specified in url for validity using an HTTP request with the HEAD method. Head requires the URL schemes to be one of http or https.
As specified in RFC 2616, "[t]he HEAD method is identical to GET except that the server must not return a message-body in the response. The metainformation contained in the HTTP headers in response to a HEAD request should be identical to the information sent in response to a GET request."
The output option is a list of one or more of content, headers, and code or one of the symbols content, headers, or code.
\mathrm{URL}:-\mathrm{Head}\left("http://www.maplesoft.com"\right)
\textcolor[rgb]{0,0,1}{table}\textcolor[rgb]{0,0,1}{}\left([\textcolor[rgb]{0,0,1}{"Content-Length"}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{"151"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"Content-Type"}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{"text/html; charset=utf-8"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"X-Powered-By"}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{"ASP.NET"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"Location"}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{"https://www.maplesoft.com/formdata"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"X-Frame-Options"}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{"sameorigin"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"Server"}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{"Microsoft-IIS/8.5"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"Date"}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{"Tue, 09 Mar 2021 02:27:17 GMT"}]\right)
\mathrm{URL}:-\mathrm{Head}\left("https://www.maplesoft.com",\mathrm{output}=\mathrm{content}\right)
\textcolor[rgb]{0,0,1}{""}
\mathrm{URL}:-\mathrm{Head}\left("http://www.maplesoft.com",\mathrm{proxy}="http://myproxy:3421",\mathrm{proxyuser}="graymond",\mathrm{proxypassword}="apple"\right)
The URL[Head] command was introduced in Maple 2017.
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Use Laplace transform to solve the following initial value problem:y"-2y'+5y=1+t
Use Laplace transform to solve the following initial value problem:y"-2y'+5y=1+t , y(0)=0 , y'(0)=4
{y}^{″}+5y=1+t,y\left(0\right)=0,{y}^{\prime }\left(0\right)=4
\frac{7}{25}{e}^{t}\mathrm{cos}\left(2t\right)+\frac{21}{10}{e}^{t}\mathrm{sin}\left(2t\right)
\frac{7}{2}+\frac{t}{5}-\frac{t}{25}{e}^{t}\mathrm{cos}\left(2t\right)+\frac{21}{10}{e}^{t}\mathrm{sin}\left(2t\right)
\frac{7}{25}+\frac{t}{5}
\frac{7}{25}+\frac{t}{5}-\frac{7}{25}{e}^{t}\mathrm{cos}\left(2t\right)+\frac{21}{10}{e}^{t}\mathrm{sin}\left(2t\right)
\frac{7}{25}+\frac{t}{5}-\frac{7}{5}{e}^{t}\mathrm{cos}\left(2t\right)+\frac{21}{10}{e}^{t}\mathrm{sin}\left(2t\right)
{y}^{\prime }=|x|,y\left(-1\right)=2
Find the Laplace Transform of the function
f\left(t\right)={e}^{at}
A baseball diamond is a square with side 90 ft. A batter hits the ball and runs toward first base with a speed of 24 ft/s. At what rate is his distance from second base decreasing when he is halfway to first base?
L\left\{9+6t\right\}
L\left\{k\right\}=\frac{k}{s}
L\left\{t\right\}=\frac{1}{{s}^{2}}
y5{y}^{\prime }-6y=0
2\cdot \left(\frac{dy}{dx}\right)+2y=0
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Gases - Course Hero
A gas is a form of matter without definite shape or volume. In an ideal gas, there are no intermolecular forces between the gas atoms or molecules, and the molecules do not take up space. The ideal gas law explains and predicts the behavior of a hypothetical ideal gas. Laws derived from the ideal gas law allow for calculations regarding the behavior of gases. These laws are based on kinetic molecular theory, which is a model that describes the macroscopic behavior of gases based on their microscopic components. This theory also explains how gases effuse through small holes in solids and diffuse from areas of high concentration to areas of low concentration. Dalton's law of partial pressures states that the pressures of each individual gas in a mixture add up to the total pressure of the mixture. This can be used to calculate stoichiometric reactions involving gases. Although the ideal gas law is commonly used for gases at low pressures, at higher pressures, real gases do not behave like ideal gases. At these pressures, the van der Waals equation can be used to correct for these changes to the ideal gas law.
Gases lack definite shape or volume. Gas pressure is measured in a number of scientific units, including atmospheres (atm), bars, and kilopascal (kPa).
PV=nRT
Stoichiometry of gases relies on the partial pressure of gases—the pressure of one gas component in a mixture of gases.
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The series is, \sum_{n=9}^\infty \frac{1}{n(n-1)}
\sum _{n=9}^{\mathrm{\infty }}\frac{1}{n\left(n-1\right)}
\sum _{n=9}^{\mathrm{\infty }}\frac{1}{n\left(n-1\right)}=\sum _{n=9}^{\mathrm{\infty }}\left(\frac{1}{n-1}-\frac{1}{n}\right)
=\left(\frac{1}{8}-\frac{1}{9}\right)+\left(\frac{1}{9}-\frac{1}{10}\right)+\left(\frac{1}{10}-\frac{1}{11}\right)+\dots
=\frac{1}{9}+\left(-\frac{1}{9}+\frac{1}{9}\right)+\left(-\frac{1}{10}+\frac{1}{10}\right)+\dots
=\frac{1}{8}
Fasaniu
\frac{1}{n\left(n-1\right)}=\frac{A}{n}+\frac{B}{n-1}
B=lim\left(n-1\right)×\frac{1}{n\left(n-1\right)}=-1
A=limn×\frac{1}{n\left(n-1\right)}=-1
This is why you get
\frac{1}{n\left(n-1\right)}=\frac{1}{n-1}-\frac{1}{n}
and then (keep the first term for the lower index and the last with the bigger)
{S}_{N}=\sum _{n=9}^{N}\frac{1}{n-1}-\frac{1}{n}=\frac{1}{8}-\frac{1}{N}
\frac{1}{n\left(n-1\right)}=\frac{A}{n}+\frac{B}{n-1}=\frac{A\left(n-1\right)+Bn}{n\left(n-1\right)}=\frac{\left(A+B\right)n-A}{n\left(n+1\right)}
A+B=0\to A=-B
-A=1\to A=-1,\text{ }B=1
\frac{1}{n\left(n-1\right)}=\frac{1}{n-1}-\frac{1}{n}
Now write out some terms:
\left(\frac{1}{8}-\frac{1}{9}\right)+\left(\frac{1}{9}\right)+\left(\frac{1}{9}-\frac{1}{10}\right)+\left(\frac{1}{10}-\frac{1}{11}\right)+...
It's clear that the
-\frac{1}{n}
is canceling the next term. So it should be clear that this gives:
\sum _{9}^{N}\left(\frac{1}{n-1}-\frac{1}{n}\right)=\frac{1}{8}-\frac{1}{N}
Now take the limit as N goers to infinity:
\underset{N\to \mathrm{\infty }}{lim}\sum _{9}^{N}\left(\frac{1}{n-1}-\frac{1}{N}\right)=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{8}-\frac{1}{N}=\frac{1}{8}
\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}=e
Suppose you take a dose of m mg of a particular medication once per day. Assume f equals the fraction of the medication that remains in your blood one day later. Just after taking another dose of medication on the second day, the amount of medication in your blood equals the sum of the second dose and the fraction of the first dose remaining in your blood, which is m+mf. Continuing in this fashion, the amount of medication in your blood just after your nth does is
{A}_{n}=m+mf+\dots +m{f}^{n-1}
. For the given values off and m, calculate
{A}_{5},{A}_{10},{A}_{30}
\underset{n\to \mathrm{\infty }}{lim}{A}_{n}
. Interpret the meaning of the limit
\underset{n\to \mathrm{\infty }}{lim}{A}_{n}
f=0.25,
m=200mg.
Sum of series:
\sum _{n=1}^{\mathrm{\infty }}\frac{1}{4{n}^{2}-1}
\sum _{n=0}^{\mathrm{\infty }}\frac{2{n}^{7}+{n}^{6}+{n}^{5}+2{n}^{2}}{n!}
\sum _{x=1}^{\mathrm{\infty }}\frac{1}{{8}^{n}}
f\left(x\right)=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{{\left(x+k\right)}^{2}}
What is the idea to sum this series of powers?
\sum _{n=1}^{+\mathrm{\infty }}\left({n}^{2}+n+1\right){x}^{n-1}
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1Department of Civil Engineering, Faculty of Engineering, Kandahar University, Kandahar, Afghanistan.
Rahmat, M. and Mizokami, S. (2019) Evaluation of Disparities between Transportation Needs and Trip Patterns of Men and Women in Kandahar City, Afghanistan. Journal of Transportation Technologies, 9, 78-94. doi: 10.4236/jtts.2019.91005.
{V}_{j}={\beta }_{1}\ast {X}_{j1}+{\beta }_{2}\ast {X}_{j2}+{\beta }_{13}\ast {X}_{j3}+\cdots +{\epsilon }_{j}
{U}_{j}={V}_{j}+{\epsilon }_{j}
{V}_{j}
{\beta }_{1},{\beta }_{2},\cdots ,{\beta }_{3}
{X}_{j1},{X}_{j2},{X}_{j3}
{\epsilon }_{j}
\mathrm{Pr}\left(i:C\right)=\frac{{V}_{i}}{{\sum }_{C}{V}_{j}}
\mathrm{Pr}\left(i:C\right)
{V}_{j}
[1] Riverson, J., Kunieda, M., Roberts, P., et al. (2005) An Overview of Women’s Transport Issues in Developing Countries. Transportation Research Board, Annual Meeting CD-ROM.
[2] Romer, H.R., Poulsen, H., Oldrup, H.H., et al. (2007) Gender Mainstreaming European Transport Research and Policies, Building the Knowledge Base and Mapping Good Practices. (TRANSGEN), University of Copenhagen.
[3] Hasson, Y. and Polevoy, M. (2011) Gender Equality Initiatives in Transportation Policy: A Review of the Literature. Women’s Budget Forum.
[4] Turner, J. (2012) Urban Mass Transit, Gender Planning Protocols and Social Sustainability: The Case of Jakarta. Research in Transportation Economics, 34, 48-53.
[5] Sarmiento, S. (1998) Household, Gender and Travel in Women’s Travel Issues. Proceedings from the Second Nation Conference, Office of Highway Information Management, 97-024.
[6] Claudia, N. and Lenz, B. (2005) Gender Differences in Travel Patterns: Role of Employment Status and Household Structure. Research on Women’s Issues in Transportation, Report of a Conference, 2, 114-123.
[7] Peters, D. (2002) Gender and Transport in Less Developed Countries: A Background Paper for the Expert Workshop. Gender Perspectives for Earth Summit 2002: Energy, Transport Information for Decision-Making, Federal Ministry for the Environment, Nature Protection and Nuclear Safety; Heinrich Boell Foundation, Berlin.
[8] Kudat, A. and Cernea, M. (1996) Strengthening Ashgabat’s Urban Transport System, “Social Assessments for Better Development: Case Studies in Russia and Central Asia”. The World Bank, Washington DC, 165-187.
[9] Buehler, R. and Pucher, J. (2010) Demand for Public Transport in Germany and the USA: An Analysis of Rider Characteristics. Transport Reviews, 32, 541-567.
[10] Duchene, C. (2011) Gender and Transport, Discussion Paper. The International Transport Forum on Transport for Society, Leipzig, 11.
[11] Lambrick, M., Rainero, L., Andrew, C., et al. (2010) Recognise the Barriers to Women’s Safe, Efficient and Affordable Travel. UN-Women.
http://www.endvawnow.org/en/articles/286-recognise-the-barriers-to-womens-safe-efficient-andafordable-travel.html
[12] Sham, R., Omar, N. and Amat, D.W. (2013) Women and Crime in Central Business District. Asian Behavioral Studies, 3, 1-12.
[13] Sarkar, M.S. and Partheeban, P. (2011) Abandon All Hope, Ye Who Enter Here, Women’s Issues in Transportation. Transport Research Board, 2, 74-84.
[14] Needle, J.R. and Cobb, M. (1997) Improving Transit Security: Synthesis of Transit Practice 21. Transit Cooperative Research Program. National Academy Press, Washington DC.
[15] Sideris, A.L. and Fink, C. (2008) Addressing Women’s Fear of Victimization in Transportation Settings, a Survey of U.S. Transit Agencies. Urban Affairs Review, 44, 554-587.
[16] Henderson, S. and Associates, R.H. (2015) Women and Transport: Moving Forward. The Scottish Executive Central Research Unit.
[17] Bhatt, A., Menon, R. and Khan, A. (2015) Women’s Safety in Public Transport: A Pilot Initiative in Bhopal. The WRI Ross Center for Sustainable Cities, Bhopal.
[18] Zelinka, A. and Brennan, D. (2001) SafeScape: Creating Safer, More Livable Communities through Planning and Design. American Planning Association Planner’s Press, Washington DC.
[19] Babinard, J. (2011) World Bank Gender Transport Surveys: An Overview. Transport Notes Series, No. TRN 43. World Bank, Washington DC.
[20] Golden, S. (2008) Gender Mainstreaming in Transport for London. Transport for London.
[21] Dahlström, K. (2007) National Travel Survey RES 2005-2006. Swedish Institute for Transport and Communications Analysis (SIKA).
[22] CSO-Afghanistan (2017) Statistical Year Book of Afghanistan 2016-17. Central Statistic Organization, Kabul.
[23] Biljana, R.P. and Jovi, J. (2014) Women and Transportation Demands in Rural Serbia. Rural Studies, 36, 207-218.
[24] Dobbs, L. (2005) Wedded to the Car: Women, Employment and the Importance of Private Transport. Transport Policy, 12, 266-278.
[25] Macdonal, H.I. (1999) Women’s Employment and Commuting: Explaining the links. Planning Literature, 13, 267-283.
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Let C be the curve described by the parametric equations
prelimaf1 2021-12-04 Answered
x\left(t\right)=t,y\left(t\right)={t}^{3}
a) sketch a graph of C.
c) find the velocity vector and the speed of f(t) at t=1.
kayleeveez7
x\left(t\right)=t,y\left(t\right)={t}^{3}
To sketch the curve let convert it into Cartesian form
Put\text{ }t=x\text{ }in\text{ }y={t}^{3}
⇒y={x}^{3}
Graph of curve Cis as shown
vector-valued function f(t) associated with this parameterization of the curve C.
⇒f\left(t\right)={t}^{3}
c) the velocity vector and the speed of f(t) at t=1.
⇒{f}^{\prime }\left(t\right)=3{t}^{2}
Put\text{ }t=1
⇒{f}^{\prime }\left(t\right)=3{t}^{2}
⇒{f}^{\prime }\left(t\right)=3×{1}^{2}
⇒{f}^{\prime }\left(t\right)=3
\left(x,\text{ }y\right)=
\begin{array}{|cc|}\hline \text{Point}& \text{Parallel to}v=\\ \left(0,\text{ }0,\text{ }0\right)& <<8,\text{ }1,\text{ }4>>\\ \hline\end{array}
(a) parametric equations
(b) symmetric equations
8x=y=4z
4x=y=8z
\frac{x}{8}=y=\frac{z}{4}
\frac{x}{4}=y=\frac{z}{8}
Determine whether the graphs of the polar equation are symmetric with respect to the x-axis, the y -axis, or the origin.
r=1+\mathrm{cos}0
\left(x,\text{ }y\right)
\begin{array}{|cccccccccc|}\hline x& & 2& 4& 6& 8& 10& 12& 14& 16\\ y& 0.08& 0.12& 0.18& 0.25& 0.36& 0.52& 0.73& 1.06\\ \hline\end{array}
\left(x,\text{ }\mathrm{ln}y\right)
\left(\mathrm{ln}x,\text{ }\mathrm{ln}y\right)
What is the polar form of (5, -3)?
Find a set of parametric equations of the line with the given characteristics. The line passes through the point
\left(-4,\text{ }5,\text{ }2\right)
and is parallel to the xy-plane and the yz-plane
{x}^{2}+{y}^{2}=25
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\lim_{x \to 0} \frac{x(1+a \cos x)-b \sin x}{x^3}=1, how to
\underset{x\to 0}{lim}\frac{x\left(1+a\mathrm{cos}x\right)-b\mathrm{sin}x}{{x}^{3}}=1
, how to find the constants a,b?
If you can use Taylor series, the numerator is
x\left(1+a-\frac{a{x}^{2}}{2}\right)-b\left(x-\frac{{x}^{3}}{6}\right)+o\left({x}^{3}\right)=\left(1+a-b\right)x+\left(b-3a\right)\frac{{x}^{3}}{6}+o\left({x}^{3}\right)
1+a-b=0,b-3a=6
a=-\frac{52}{,}b=-\frac{32}{}
kaluitagf
Apply LHopitals rule again:
\underset{x\to 0}{lim}\frac{x\left(1+a\mathrm{cos}x\right)-b\mathrm{sin}x}{{x}^{3}}\stackrel{LHR}{=}\underset{x\to 0}{lim}\frac{1+\left(a-b\right)\mathrm{cos}x-ax\mathrm{sin}x}{3{x}^{2}}
\stackrel{LHR}{=}\underset{x\to 0}{lim}\frac{\left(b-2a\right)\mathrm{sin}x-ax\mathrm{cos}x}{6x}
and recall that
\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{x}=1
\mathrm{sin}x+\mathrm{sin}y=a\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{cos}x+\mathrm{cos}y=b
\mathrm{tan}\left(x-\frac{y}{2}\right)
\frac{\mathrm{csc}x+\mathrm{cot}x}{\mathrm{tan}x+\mathrm{sin}x}=\mathrm{cot}x\mathrm{csc}x
\int \mathrm{csc}\left(x-\frac{\pi }{3}\right)\mathrm{csc}\left(x-\frac{\pi }{6}\right)dx
What I Tried- I tried dividing both numerator and denominator by
\mathrm{sin}\frac{\pi }{6}
but couldnt
Use a calculator to find the solution of teh equation, rounded to six decimal places.
{5}^{2x+1}={3}^{4x-1}
How do I evaluate
\underset{z\to 1}{lim}\left(1-z\right)\mathrm{tan}\frac{\pi z}{2}
I tried using the identity,
\mathrm{tan}\frac{x}{2}=\frac{1-\mathrm{cos}x}{\mathrm{sin}x}
to simplify this to:
\underset{z\to 1}{lim}\left(1-z\right)\frac{\mathrm{sin}\pi z}{1+\mathrm{cos}\pi z}
\frac{1-\mathrm{sin}x}{1+\mathrm{sin}x}=4
\mathrm{tan}\frac{x}{2}
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Evaluate the integrals \int_0^{\ln9}e^\theta(e^\theta-1)^{1/2}d\theta
{\int }_{0}^{\mathrm{ln}9}{e}^{\theta }{\left({e}^{\theta }-1\right)}^{\frac{1}{2}}d\theta
wheezym
{\int }_{0}^{\mathrm{ln}9}{e}^{\theta }{\left({e}^{\theta }-1\right)}^{\frac{1}{2}}d\theta
{e}^{\theta }-1==t
on differentiating with respect to t we get
{e}^{\theta }d\theta =dt
d\theta =\frac{dt}{{e}^{\theta }}
now the limits
t={e}^{\theta }-1
t={e}^{\mathrm{ln}9}-1
t=9-1
t=8
t={e}^{0}-1
t=1-1
t=0
now the definite integral
{\int }_{0}^{8}{e}^{\theta }{t}^{\frac{1}{2}}\frac{dt}{{e}^{\theta }}
{\int }_{0}^{8}{t}^{\frac{1}{2}}dt
⇒{\left[\frac{{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_{0}^{8}
⇒\frac{2}{3}{\left[{t}^{\frac{3}{2}}\right]}_{0}^{8}
⇒\frac{2}{3}\left[{8}^{\frac{3}{2}-0}\right]
⇒\frac{2}{3}\left[{2}^{3×\frac{3}{2}}\right]
⇒\frac{{2}^{\frac{11}{2}}}{3}=\frac{32×{2}^{\frac{1}{2}}}{3}
{\int }_{0}^{\mathrm{ln}9}{e}^{\theta }{\left({e}^{\theta }-1\right)}^{\frac{1}{2}}d\theta =\frac{32×{2}^{\frac{1}{2}}}{3}
x={t}^{2},y=2t,0\le t\le 5
\left({x}^{2}+2xy-4{y}^{2}\right)dx-\left({x}^{2}-8xy-4{y}^{2}\right)dy=0
{\int }_{0}^{1}\sqrt{x}\left(\sqrt{x}+1\right)dx
Use the Integral Test to determine whether the series is convergent or divergent.
\sum _{n=1}^{\mathrm{\infty }}\frac{n}{{n}^{2}+1}
Calculate the iterated integral
{\int }_{0}^{1}{\int }_{0}^{1}4xy\sqrt{{x}^{2}+{y}^{2}}dxdy
{\int }_{b}^{a}{x}^{7}dx
Evaluating triple integrals: Each of the triple integrals that follow represents the volume of a solid. Sketch the solid and evaluate the integral.
{\int }_{-2}^{2}{\int }_{0}^{\sqrt{4-{x}^{2}}}{\int }_{0}^{y}dzdydx
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Independent Researcher, Al-Hokail, Khobar, KSA.
Abstract: Introduction: Intraoperative blood loss remains as a concern for all surgeons. Proper estimation of intraoperative blood loss is critical and can be challenging, especially if the blood is mixed with other fluids such as tumescent fluid in liposuction cases. In such cases, proper estimation of intraoperative blood loss will lead to fewer mistakes in fluid resuscitation. In this article, Tallquist Haemoglobin Scale was tried to estimate intraoperative blood loss in liposuction. Objectives: Proper estimation of intraoperative blood loss in liposuction cases. Method: Tallquist Haemoglobin Scale will be tried to estimate the approximate intraoperative blood loss in liposuction cases using a mathematical formula that considers total fluid loss, patient’s preoperative haemoglobin and the reading from Tallquist kit. Results: Tallquist Haemoglobin Scale can be considered as a valid method for proper estimation of intraoperative blood loss in liposuction cases, the thing that will lead to correct fluid resuscitation and fewer complications. Conclusion: Proper estimation of intraoperative blood loss leads to fewer mistakes in fluid resuscitation and fewer related complications of under or overcorrection. Tallquist Haemoglobin scale is a trusted, cheap and fast method for proper estimation of intraoperative blood loss in liposuction cases.
Keywords: Tallquist, Haemoglobin, Scale, Blood Loss, Super Wet, Liposuction, Tumescent, Fluid, Resuscitation
Intraoperative blood loss remains as a concern for all surgeons. Proper estimation of intraoperative blood loss is critical and can be challenging, especially if the blood is mixed with other fluids such as tumescent fluid in liposuction cases. In such cases, proper estimation of intraoperative blood loss will lead to fewer mistakes in fluid resuscitation and protecting patients from under or over fluid balance correction.
To understand fluid balance in liposuction cases, two definitions should be covered:
1) Intraoperative fluid ratio 8: defined as the volume of super-wet solution and intraoperative intravenous fluid divided by the aspiration volume.
Trott et al., stated that it should be 2.1 for small-volume (aspirate < 4000 cc) and 1.4 for large-volume (aspirate > 4000 cc) [1] .
Rohrich et al., revised the ratio to be 1.8 for the small-volume reductions (aspirate < 5000 cc) and 1.2 for the large-volume reductions (aspirate > 5000 cc) [2] .
2) Final residual fluid volume: all intakes and outputs should be meticulously measured as follows: residual fluid volume = (total fluid volume in) − (total volume out); total fluid volume in = (total fluid volume of wetting solution used) + (total volume of IV fluid used) + (volume of bupivacaine/steroid solution); and total volume out = [total volume of aspirated wetting solution (which is approximately 30% of the total aspirated volume in most cases)] + urine output [3] .
The average residual volume by Wang et al. [3] was around 110 mL/kg, and was 120 mL/kg in the report of Commons et al. [4] .
IV fluid infused intraoperatively should equal:
1) Fasting deficit: hourly maintenance*number of fasting hours.
2) Hourly maintenance: body weight + 40 ml/hour.
3) Estimated fluid loss from surgical trauma: 4 - 6 mL/kg/hr. (liposuction was considered as a moderate surgical trauma) [5] .
In 1996, Trott et al. suggested guidelines for intraoperative fluid resuscitation. Patients having less than 4000 cc of lipoaspirate removed received maintenance fluid only, whereas those having more than 4000 cc of lipoaspirate removed received maintenance plus 0.25 cc of intravenous fluid for each milliliter aspirated over 4000 cc [1] .
To limit the degree of fluid overloading and possibility of pulmonary edema, Rohrich et al. modified the replacement fluid delivery at 0.25 cc of intravenous fluid for each milliliter aspirated over 5000 cc [2] .
Although blood loss in super wet liposuction is supposed to be 1% of the aspirate [6] , it is possible to exceed that percentage in some occasions e.g. technical issues, coagulopathy, etc. I recommend replacing the estimated blood loss with colloids in 1:1 ratio [7] , unless one of the blood transfusion indications is met.
As a fast review of blood transfusion indications, World Health Organization (WHO) defined them, in adults, as one of the following:
1) Perioperative transfusion: 8 g/dL for patient undergoing major surgery or experiencing GIT bleeding.
2) Acute blood loss: 30% of volume of blood.
considering that One unit of whole blood/packed red blood cells can increase haemoglobin by 1g/dL in an adult or haematocrit by 3% (haemoglobin of unit must be >75%) [8] .
National Institute for Health and Care Excellence (NICE) guidelines mention that when using a restrictive red blood cell transfusion threshold, consider a threshold of 70 g/liter and a haemoglobin concentration target of 70 - 90 g/liter after transfusion.
Restrictive red blood cell transfusion thresholds are for patients who need red blood cell transfusions and who do not:
l have major haemorrhage or
l have acute coronary syndrome or
l need regular blood transfusions for chronic anaemia [9] .
In literature, several methods of estimating haemoglobin can be found. Direct cyanmethaemoglobin method has been the gold standard for haemoglobin estimation but other methods like haemoglobin color scale, Sahli technique, Lovibond-Drabkin technique, Tallquist technique, copper-sulfate method, HemoCue and automated haematology analyzers are also available [10] .
In my opinion, two points had to be considered before choosing the haemoglobin estimation method:
1) The method should be manual, simple, fast without need for power supply.
2) The method should be able to detect very low concentrations of haemoglobin as we are trying to estimate the fraction/percentage of blood in a fluid not to estimate haemoglobin concentration in a proper blood sample.
For both of the considerations above, digital devices are not the best choice for the task. I chose Tallquist haemoglobin scale as a method (Figure 1).
The use of the Tallquist method in assessing anemia is of, relatively, high validity and diagnostic accuracy. Tallquist method should be an effective method to detect mild to moderate anemia in a reliable manner comparable to other standard methods such as the haematocrit and haemoglobin cyanide methods [12] .
1) After finishing liposuction, super wet in our case, the aspirated fluid volume (VF) is to be calculated, 400 cc in our example here (Figure 2).
2) Take a drop of the aspirated fluid and apply it on one filter paper, comes with Tallquist kit, wait till glistening disappears and compare it, before it completely dries, with the different color grades on the scale. The closest in our example was 4.7 g/dl (fluid haemoglobin concentration) (HbF) (Figure 3). Percentages on the scale indicate the ratio of the sample haemoglobin to a standardized sample of 15.6 g/dl haemoglobin. Do not use percentages in calculations not to get mixed with Hematocrit values.
3) Go back to the patient’s preoperative investigations and get the haemoglobin value (HbP) (Table 1) (13.6 g/dl in our case).
Figure 1. Tallquist haemoglobin scale [11] .
Figure 2. The aspirated fluid volume (VF) (400 cc in our example).
Figure 3. Take a drop of the aspirated fluid and apply it on one filter paper, comes with Tallquist kit, wait till glistening disappears and compare it, before it completely dries, with the different color grades on the scale. The closest in our example was 4.7 g/dl (HbF).
Table 1. Example patient’s preoperative investigations: Haemoglobin concentration (HbP).
4) Calculate the volume of intraoperative blood loss (VL) through applying the classic concentration/volume formula (C1V1 = C2V2)
\begin{array}{l}\text{Preoperativepatient}’\text{s haemoglibin concentration}\left(\text{HbP}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\times \text{Volume of intraoperative blood loss}\left(\text{VL}\right)\\ =\text{Aspirated fluid haemoglobin concentration}\left(\text{HbF}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\times \text{Volume of Total aspirated fluid}\left(\text{VF}\right)\end{array}
{\text{V}}_{\text{L}}{\text{=Hb}}_{\text{F}}\ast {\text{V}}_{\text{F}}/{\text{Hb}}_{\text{P}}=4.7\ast 400/13.6=138\text{cc}
Tallquist Haemoglobin Scale can be considered as a valid method for proper estimation of intraoperative blood loss in liposuction cases using a mathematical formula that considers total fluid loss, patient’s preoperative haemoglobin and the reading from Tallquist kit, the thing that will lead to correct fluid resuscitation and fewer complications.
Proper estimation of intraoperative blood loss will lead to fewer mistakes in fluid resuscitation, using crystalloids, colloids or blood, and protecting patients from under or over fluid balance correction. Application of the mentioned method can be expanded to be used in other surgeries with further blood loss. Tallquist Haemoglobin Scale can be considered as a valid method for proper estimation of intraoperative blood loss in liposuction cases. A classic mathematical formula that addresses volume and concentration of both pure patient’s blood and the aspirated fluid can be applied to detect the approximate intraoperative blood loss. Blood loss from the postoperative oozing should be considered as a next target for analysis.
Cite this paper: Alsayed, A. (2020) Using Tallquist Haemoglobin Scale for Estimating Intraoperative Blood Loss in Liposuction. Modern Plastic Surgery, 10, 17-22. doi: 10.4236/mps.2020.102003.
[1] Trott, S.A., Beran, S.J., Rohrich, R.J., Kenkel, J.M., Adams Jr., W.P. and Klein, K.W. (1998) Safety Considerations and Fluid Resuscitation in Liposuction. An Analysis of 53 Consecutive Patients. Plastic and Reconstructive Surgery, 102, 2220-2229.
[2] Rohrich, R., Leedy, J., Swamy, R., Brown, S.A. and Coleman, J. (2006) Fluid Resuscitation in Liposuction: A Retrospective Review of 89 Consecutive Patients. Plastic and Reconstructive Surgery, 117, 431-435.
[3] Wang, G., Cao, W. and Zhao, T. (2018) Fluid Management in Extensive Liposuction. Medicine, 97, e12655.
[4] Commons, G.W., Halperin, B. and Chang, C.C. (2001) Large-Volume Liposuction: A Review of 631 Consecutive Cases over 12 Years. Plastic and Reconstructive Surgery, 108, 1753-1763.
[5] Intraoperative Fluid Dosing in Adult Patients-MDCalc.
https://www.mdcalc.com/intraoperative-fluid-dosing-adult-patients#evidence
[6] Granados-Tinajero, S., Buenrostro-Vásquez, C., Cárdenas-Maytorena, C. and Contreras-López, M. (2019) Anesthesia Management for Large-Volume Liposuction. Anesthesia Topics for Plastic and Reconstructive Surgery, 4, 71-90.
[7] Broadstone, R. (1999) Fluid Therapy and Newer Blood Products. Veterinary Clinics of North America: Small Animal Practice, 29, 611-628.
[8] World Health Organization (2020) Clinical Transfusion Practice Guidelines for Medical Interns.
https://www.who.int/bloodsafety/transfusion_services/ClinicalTransfusionPracticeGuidelinesfor
MedicalInternsBangladesh.pdf
[9] NICE (2015) Blood Transfusion, NICE Guideline [NG24], Guidance: Recommendations: Red Blood Cells: Thresholds and Targets (1.2.2). National Institute for Health and Care Excellence (NICE).
[10] Srivastava, T., Negandhi, H., Neogi, S.B., Sharma, J. and Saxena, R. (2014) Methods for Haemoglobin Estimation: A Review of “What Works”. Journal of Hematology & Transfusion, 2, 1028.
[11] Stemfinity.com. (2020). Tallquist Haemoglobin Scale. Kemtec Science. STEMfinity.
https://www.stemfinity.com/Kemtec-Science-Tallquist-Haemoglobin-Scale
[12] Wasiu Olalekan, A. and Olufemi Emmanuel, A. (2016) How Valid Is the Tallquist Method in Screening Pregnant Women with Anemia in Poor Rural Settings of Southwestern Nigeria? Medical Journal of the Islamic Republic of Iran, 30, 389.
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Find the Laplace transform for displaystyle f{{left({t}right)}}={2}{u}{left({t}-{3}right)}{{cos}^{2}{2}}{t}-{6}{e}^{{{2}{t}+{7}}}delta{left({t}+{3}right)}
Find the Laplace transform for f(t)=2u(t-u)cos^2 2t-6e^(2t+7) delta(t+3)
Find the Laplace transform for
f\left(t\right)=2u\left(t-3\right){\mathrm{cos}}^{2}2t-6{e}^{2t+7}\delta \left(t+3\right)
Calculate the Laplace transform:
{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f\left(t\right){e}^{-st}dt={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\left(2u\left(t-3\right){\mathrm{cos}}^{2}2t-6{e}^{2t+7}\delta \left(t+3\right)\right){e}^{-st}dt
=2{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}u\left(t-3\right){\mathrm{cos}}^{2}2t\cdot {e}^{-st}dt-6{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{2t+7}\delta \left(t+3\right){e}^{-st}dt
=2{\int }_{3}^{\mathrm{\infty }}{e}^{-st}{\mathrm{cos}}^{2}2tdt-6{e}^{3s+1}
={\int }_{3}^{\mathrm{\infty }}{e}^{-st}\left(\mathrm{cos}4t+1\right)dt-6{e}^{3s+1}
={\int }_{3}^{\mathrm{\infty }}{e}^{-st}\mathrm{cos}4tdt+{\int }_{3}^{\mathrm{\infty }}{e}^{-st}dt-6{e}^{3s+1}
={\int }_{3}^{\mathrm{\infty }}{e}^{-st}\mathrm{cos}4tdt+\frac{1-{e}^{-3s}}{s}-6{e}^{3s+1}
={\int }_{3}^{\mathrm{\infty }}{e}^{-st}\frac{{e}^{4jt}+{e}^{-4jt}}{2}dt+\frac{1-{e}^{-3s}}{s}-6{e}^{3s+1}
={\int }_{3}^{\mathrm{\infty }}\frac{{e}^{-st+4jt}+{e}^{-st-4jt}}{2}dt+\frac{1-{e}^{-3s}}{s}-6{e}^{3s+1}
\theta
1.00×{10}^{-3}
Y\left(s\right)=\frac{8}{s}+\frac{1}{s-3}\cdot {e}^{-2s}
Using the definition of continuity, explain why the beatiful piece-wise function
f\left(x\right)=\left\{\begin{array}{ll}\mathrm{cos}\left(x\right),& for\text{ }x<0\\ 1,& for\text{ }x=0\\ 1-{x}^{2},& for\text{ }x>0\end{array}
is continuous at x=0. Show work to support your explanation.
Prove that for
n\ge 2,2\cdot \left(\begin{array}{c}n\\ 2\end{array}\right)+\left(\begin{array}{c}n\\ 1\end{array}\right)={n}^{2}
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Thermal Decomposition of Diphenyl Tetroxane in Chlorobenzene Solution
Thermal Decomposition of Diphenyl Tetroxane in Chlorobenzene Solution
Alexander G. Bordon, Andrea N. Pila, Mariela I. Profeta, María J. Jorge, Lilian C. Jorge, Jorge M. Romero, Nelly L. Jorge
Laboratorio de Investigación en Tecnología Ambiental, Facultad de Ciencias Exactas, Naturales y Agrimensura de la Universidad Nacional del Nordeste, Corrientes, Argentina
The thermal decomposition of Cyclic Diperoxide of Benzaldehyde 3,6-diphenyl-1,2,4,5-tetroxane, (DFT) in chlorobenzene solution in the studied temperature range (130˚C - 166˚C) satisfactorily satisfies a first order law up to 60% conversions of diperoxide. DFT would decompose through a mechanism in stages and initiated by the homolytic breakdown of one of the peroxidic bonds of the molecule, with the formation of the corresponding intermediate biradical. The concentration studied was very low, so that the effects of secondary reactions of decomposition induced by free radicals originated in the reaction medium can be considered minimal or negligible. The activation parameters for the unimolecular thermal decomposition reaction of the DFT are ΔH# = 30.52 ± 0.3 kcal∙mol−1 and ΔS# = −6.38 ± 0.6 cal∙mol−1 K−1. The support for a step-by-step mechanism instead of a process concerted is made by comparison with the theoretically calculated activation energy for the thermal decomposition of 1,2,4,5-tetroxane.
DFT, Thermal Decomposition, Chlorobenzene, Mechanism
Peroxides have been widely used in several areas of chemistry due to their high reactivity [1] . Due to their specific properties as initiators of free radical reaction they are widely used in synthetic chemistry [2] , especially in polymerization processes [3] . At present, it has been discovered that tetroxanes possess an impressive antimalarial activity [4] and herbicide activity. The association between the functional group of peroxide and the antimalarial activity allows the development of new antimalarial drugs [4] [5] [6] . Unfortunately, several members of this composite class are very unstable; only a small number is easy to handle with adequate security and are available as commercial products.
The thermal decomposition of tetraderivates of 1,2,3,4-tetroxanes have been studied in solution and gas phase. The mechanism of thermal decomposition was evaluated in solution [7] [8] [9] .
Unimolecular thermal decomposition can happen by two different kinds:
1) A stepwise homolysis initiated by rupture of peroxidic bond with intermediate biradical formation
2) A concerted mechanism where peroxidic bond breaking and peroxidic bond making may occur simultaneously.
Both processes in the thermolysis of tetroxanes give ketone or aldehyde and molecular oxygen as reaction products.
In previous theoretical studies, it was determined that in the gas phase the mechanism of the thermal decomposition reaction for 1,2,4,5-tetroxane is a mechanism in stages and not concerted [10] .
In our research group, the kinetics and reaction mechanism of the 3,6-diderivatives of tetroxane in different solvents including 3,6-diphenyl-1,2,4,5-te-troxane were investigated the influence of solvents with different physicochemical characteristics [11] [12] [13] [14] .
This work has been performed to understand the kinetics and the mechanism thermal decomposition of 3,6-difenil-1,2,4,5-Tetroxane, (DFT) in solution of chlorobenzene These kinetic studies are intended to analyze the solvent effect on the parameters governing the reaction. Steric effects, inductive, mesomeric and stereoelectronic of the substituents on peroxidic ring are considered in relation by the force of peroxidic bond.
DFT was prepared by drop wise addition of 7.2 mL benzaldehyde (Fluka) in 20 mL ethanol solution to a vigorously stirred and cooled (−20˚C) mixture of 56% hydrogen peroxide and sulfuric acid. After stirring for 2 hs, a microcrystalline white solid was obtained.
It was repeatedly washed with water and further purified by recrystallizing from methanol. The DFT purity was checked by GC. The structure and conformation was confirmed by NMR, obtaining the same results as in previous synthesis [15] .
2.2. Kinetics Methods
Pyrex glass tubes (8 mm i.d., 10 cm length) filled with the appropriated volume of DFT in chlorobenzene solution, with added n-octane (internal standard) were thoroughly degassed in the vacuum line at −196˚C and then sealed with a flame torch.
These Pyrex glass tubes were submerged in a thermostated silicone oil bath (±0.1˚C) to temperatures between 130˚C and 166˚C. There were withdrawn after selected times and cooling rapidly to 0˚C to stop the reaction
It was conserved to −18˚C up till its analysis with GC.
2.3. Analysis of Remaining Reagent
The remaining DFT was evaluated by Gas Chromatographer AGILENT 7890A, with FID detector, using nitrogen like carrier gas and quantitative method of n-octane (internal standard). The capillary column used was HP5-Agilent (stationary phase: (5%-Phenyl)-methylpolysiloxane, length 25 m, film 0.33 µm, id. 0.2 mm)
Injector temperature 150˚C,
Detector temperature 300˚C.
Oven with progameted temperature:
Ti = 50˚C initial, Ti = 1 min.
Ramp of 20˚/min, FT = 220˚C and Tt = 15 min.
The measurement of the areas of the chromatographic peaks was made using the EZ Crom software.
2.4. Calculations Methods
Rate constant values were calculated by a mean square data treatment (r > 0.998) and the parameters values of activation were obtained by computational method employing the Eyring and Arrhenius equations and the corresponding error limits with a method executed with a computational treatment.
The thermal decomposition of the DFT in solution of chlorobenzene in the range studied, as shown in Table 1 (Figure 1) meet a first order kinetic law up to 60% conversion of the DFT.
The influence of temperature on the rate constant values of DFT (kexp) in chlorobenzene solution corresponding to the unimolecular decomposition reaction, can be represented by the Arrhenius equation, where the value of the
Table 1. Rate constant values for DFT thermal decomposition reaction in chlorobenzene solution at different temperatures.
aErrors calculated as in ref. [11] .
Figure 1. Kinetic of DFT thermal decomposition reaction in chlorobenzene solution at different temperatures.
activation energy is expressed in kJ∙mol−1 and the errors correspond to the standard deviations derived from a treatment of the kinetic data by a method of minimum squares, r2(n)a. Equation (1).
\mathrm{ln}k\left({s}^{-1}\right)=\left(28.25\pm 0.8\right)-\left(31.92\pm 4.0\right)/RT
The linearity of the Arrhenius equation (r2 = 0.987) over a relatively wide temperature range (40˚C) suggests that the calculated activation parameters (activation energy and pre-exponential factor), for the DFT reaction in chlorobenzene correspond to a simple process represented by the homolytic breakdown of the O-O bond, (Scheme 1).
This intermediate biradical in the condition experimental , can undergo subsequent reactions involving C-O bond breaks, justifying the appearance of benzaldehyde with a 50% yield, or C-C cleavages giving rise to phenyl radicals, which then react with the solvent extracting hydrogen to form a biphenyl and two phenyl radicals are bound to give biphenyl (Scheme 2).
The products of the reaction were confirmed by mass spectrometry with HP5890-Series II plus chromatograph.
The values of the activation parameters (ΔH# = 30.52 ± 0.3 kcal∙mol−1 and ΔS# = −6.38 ± 0.6 cal∙mol−1 K−1) corresponding to the thermolysis of DFT in chlorobenzene solution were determined using Eyring’s equation [16] , whose graphical representation (Figure 2) is practically linear (r2 = 0.984) over a wide range of
Scheme 1. Homolytic rupture of O-O bond.
Scheme 2. Biphenyl formation.
Figure 2. Graphical representation of the Eyring equation for the thermal decomposition of the DFT in chlorobenzene solution.
temperatures (ΔT = 40˚C). The values of the enthalpy variation are similar to those corresponding to the homolytic decomposition of other cyclic diperoxides already studied.
This suggests that the determinant stage of the reaction rate in the thermolysis of the DFT is the rupture of the peroxidic bond with formation of the intermediate biradical.
The variation of the activation entropy results in a negative value. This is explained because the DFT is a disubstituted diperoxide that has a low steric hindrance, which results in a higher degree of solvation of the activated complex, causing a decrease in the activation entropy.
The principal products generated in the decomposition of DFT in chlorobenzene solution were benzaldehyde and molecular oxygen (Scheme 3).
Analysis of the Results Obtained in the Thermolysis of the DFT in Chlorobenzene and Other Solvents Previously Studied
If we try to understand the influence that the solvent can have on the kinetics of the decomposition reaction, we will have to analyze the kinetic parameters in different solvents of the DFT (Table 2).
Scheme 3. Formation of benzaldehyde and molecular oxygen.
Table 2. Activation parameters of the thermal decomposition of DFT in different solvents.
According to the transition state theory, solvents can modify the activation parameters of the thermolysis reactions according to the degree of solvation of the reactants and activated complex.
The rate constant of the reaction is determined by the speed with which the activated complex passes the energy barrier in the direction of product formation. If the reaction is carried out in solution, the DFT can be solvated with different scope, according to the solvation power of the solvent used.
These differences can accelerate or retard the reaction, varying the reaction rate.
The relationship between entropy and the enthalpy of activation of a reaction is not easy to analyze. However, when evaluating the data obtained in this work with the facts above, it can be seen that a change in the value of the enthalpy (ΔH0#) is accompanied by a change in the same sense of the activation entropy (ΔS0#) [17] [18] . This phenomenon is known as Compensation Effect or Isokinetic Ratio. [17] [18] (Figure 3).
The linearity observed in this representation shows the compensation effect for the DFT in all the solvents studied.
Thermal decompositions reaction of DFT in chlorobenzene solution, follow first order kinetic laws up to ca. 60% diperoxide conversion at temperature from 130.0˚C to 166.0˚C and an 1.00 × 10−3 mol∙L−1 initial concentration.
Figure 3. Representation of the “Isokinetic Ratio” according to the Leffler treatment for the unimolecular thermolysis of DFT in 1-Metilcellosolve; 2-Methanol; 3-Tetra-hydrofuran; 4-Toluene; 5-Chlorobenzene; 6-Benzene.
The reaction products and the activation parameter values contribute to postulate the mechanism for the thermolysis of DFT in chlorobenzene. The thermolysis follows the same mechanism of decomposition for other tetroxanes, which begins with the homolytic rupture of the peroxydic bond, C-C leading to the formation of an intermediate biradical and then C-O bond ruptures giving benzaldehyde and oxygen molecule as final products.
The concentration at which the thermolysis reaction of the DFT was studied was very low, whereby the effects of secondary decomposition reactions induced by free radicals originated in the reaction medium can be considered minimal or negligible.
In chlorobenzene, the thermolysis of the DFT follows the same reaction pathway as in other solvents studied.
Bordon, A.G., Pila, A.N., Profeta, M.I., Jorge, M.J., Jorge, L.C., Romero, J.M. and Jorge, N.L. (2019) Thermal Decomposition of Diphenyl Tetroxane in Chlorobenzene Solution. International Journal of Organic Chemistry, 9, 1-9. https://doi.org/10.4236/ijoc.2019.91001
1. Jorge, N.L. and Castro, E.A. (2009) Structure and Properties. Trends in Organic Chemistry, 13, 65-74.
2. Smith, M.B. and March, J. (2007) March’s Advances Organic Chemistry. 6th Edition, Wiley & Sons, New Jersey.
3. Acuna, P. and Morales, G. (2011) Síntesis de poliestireno de alto impacto (hips) a partir de diferentes iniciadores multifuncionales: efecto de la estructura y del contenido de oxígeno activo del iniciador. Rev. Iberoam. Polim., 12, 160-168.
4. Dong, Y., McCulloug, K.J., Wittlin S., Chollet, J. and Vennerstrom J.L. (2010) The Structure and Antimalarial Activity of Dispiro-1,2,4,5-Tetraoxanes Derived from (+)-Dihydrocarvone. Bioorganic & Medicinal Chemistry Letters, 20, 6359-6361. https://doi.org/10.1016/j.bmcl.2010.09.113
5. Atheaya, H., Khan, S.I., Mamgaina, R. and Rawata, D.S. (2008) Synthesis, Thermal Stability, Antimalarial Activity of Symmetrically and Asymmetrically Substituted Tetraoxanes. Bioorganic & Medicinal Chemistry Letters, 18, 1446-1452. https://doi.org/10.1016/j.bmcl.2007.12.069
6. Creek, D.J., Ryan, E., Charman, W.N., Chiu, F.C., Prankerd, R.J., Vennerstrom, J.L. and Charman, S.A. (2009) Stability of Peroxide Antimalarials in the Presence of Human Hemoglobin. Antimicrobial Agents and Chemotherapy, 53, 3496-3500. https://doi.org/10.1128/AAC.00363-09
7. Oxley, J.C., Smith, J.L. and Chen, H. (2002) Decomposition of a Multiperoxidic Compound: Triacetone Triperoxide (TATP). Propellants, Explosives, Pyrotechnics, 27, 209-216. https://doi.org/10.1002/1521-4087(200209)27:4<209::AID-PREP209>3.0.CO;2-J
8. Leiva, L.C.A., Jorge, N.L., Romero, J.M., Cafferata, L.F.R., Gómez Vara, M.E. and Castro, E.A. (2008) Decomposition of the Acetone Cyclic Diperoxide in Octanol Solution. Journal of the Argentine Chemical Society, 96, 110-122.
9. Profeta, M.I., Romero, J.M., Leiva, L.C.A., Jorge, N.L., Gómez Vara, M.E. and Castro, E.A. (2011) Solvent Effect of Oxygen in the Thermolisys Decomposition of the Acetone Diperoxide. International Journal of Chemoinformatics and Chemical Engineering, 1, 96-102. https://doi.org/10.4018/ijcce.2011010107
10. Jorge, N.L., Romero, J.M., Grand, A. and Hernández-Laguna, A. (2012) Gas Phase Thermolysis Reaction of Formaldehyde Diperoxide. Kinetic Study and Theoretical Mechanisms. Chemical Physics, 39, 37-45. https://doi.org/10.1016/j.chemphys.2011.11.019
11. Jorge, N.L., Hernandez-Laguna, A. and Castro, E.A. (2013) Some Recent Developments on the Synthesis, Chemical Reactivity, and Theoretical Studies of Tetroxanes. International Journal of Chemoinformatics and Chemical Engineering, 3, 48-73. https://doi.org/10.4018/ijcce.2013010105
12. Reguera, M.B., Frette, S.G., Romero, J.M., Jorge, N.L. and Castro, E.A. (2012) Synthesis and Thermical Decomposition Reaction of 3,6-Dibutanoic-1,2,4,5-Tetroxane in Solution. Bentham Science, 4, 1-4.
13. Pila, A.N., Profeta, M.I., Romero, J.M., Jorge, N.L. and Castro, E.A. (2012) Kinetics and Mechanism of the Thermal Decomposition Reaction of 3,6-Diphenyl-1,2,3,5- Tetroxane in Solution. International Journal of Chemical Modeling, 4, 405-411.
14. Bordón, A.G., Profeta, M.I., Romero, J.M. and Jorge, N.L. (2015) Thermal Decomposition of Benzaldehyde Diperoxide in Isopropyl Alcohol, Effect of Solvent Polarity. Asian Journal of Science and Technology, 6, 1928-1932.
15. Jorge, N.L., Gómez Vara, M.E., Castro, E.A., Autino, J.C. and Cafferata, L.F.R. (1999) Experimental and Theoretical Study of trans-3,6-diphenyl 1,2,4,5-Tetroxane Molecule. Journal of Molecular Structure: THEOCHEM, 459, 29-35. https://doi.org/10.1016/S0166-1280(98)00254-1
16. Huyberechts, S., Halleux, A. and Kruys, P. (1955) Une application de Calcule. Statistique a le Cinétique Chimique. Bulletin des Sociétés Chimiques Belges, 64, 203-209. https://doi.org/10.1002/bscb.19550640502
17. Exner, O. (1972) Statistic of the Enthalpy-Entropy Relationship. I. The Special Case. Collection of Czechoslovak Chemical Communications, 27, 1425-1444. https://doi.org/10.1135/cccc19721425
18. Leffler, J.E. (1955) The Enthalpy-Entropy Relationship and Its Implications for Organic Chemistry. The Journal of Organic Chemistry, 20, 1202-1231. https://doi.org/10.1021/jo01126a009
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Equilibrium constant - Simple English Wikipedia, the free encyclopedia
Equilibrium constant is a mathematical quantity which expresses the relationship between products and reactants of a reaction at equilibrium with respect to a specific unit. In other words, equilibrium constant is the reaction quotient of a chemical reaction at chemical equilibrium. The equilibrium constant can help us to understand whether the reaction tends to have a higher concentration of products or reactants at equilibrium. We can also use to Equilibrium constant determine if the reaction is already at equilibrium.
There are several different types of equilibrium constants that provide relationships between the products and the reactants of a chemical reaction at chemical equilibrium in terms of different units. Dissociation constants can be used as an example.
1 Equilibrium constant of a reaction
2 Relation between '"`UNIQ--postMath-00000008-QINU`"' and '"`UNIQ--postMath-00000009-QINU`"'
Equilibrium constant of a reactionEdit
For a general chemical equilibrium
{\displaystyle \alpha A+\beta B...\rightleftharpoons \sigma S+\tau T...}
the equilibrium constant can be defined by[1]
{\displaystyle K={\frac {{\{S\}}^{\sigma }{\{T\}}^{\tau }...}{{\{A\}}^{\alpha }{\{B\}}^{\beta }...}}}
where {A} is the activity of the chemical species A, etc. (activity is a dimensionless quantity). It is conventional to put the activities of the products in the numerator and those of the reactants in the denominator.
For equilibria in solution, activity is the product of concentration and activity coefficient. Most chemists determine equilibrium constants in a solution with a high ionic strength. In high strength solutions, the quotient of activity coefficients changes very little. So, the equilibrium constant is defined as a concentration quotient:
{\displaystyle K_{c}={\frac {{[S]}^{\sigma }{[T]}^{\tau }...}{{[A]}^{\alpha }{[B]}^{\beta }...}}}
However, the value of Kc will depend on the ionic strength. (The square brackets mean the concentration of A, B and so on.)
This is a simple idea. In an equilibrium, atoms can combine or break apart because the reaction can work in both directions. For the reaction to work, all of the parts must be present to combine. This is more likely to happen if the reactants have a higher concentration. So, the concentrations of all the necessary pieces are multiplied together to get the probability that they will be in the same place for the reaction. (If the reaction requires two molecules of a particular compound, then the concentration of that compound is squared.) Going the other way, all of the concentrations of those necessary pieces are multiplied together to get the probability that they will be in the same place to react in the opposite direction. The ratio between those two numbers represents how popular each side of the reaction will be when equilibrium is reached. An equilibrium constant of 1 means that both sides are equally popular. Chemists perform experiments to measure the equilibrium constant of various reactions.
There is a relation between Gibbs free energy (
{\displaystyle \Delta G}
) and Equilibrium constant which is,
{\displaystyle \Delta G=-RT\ln K}
{\displaystyle R}
is the universal gas constant and
{\displaystyle T}
{\displaystyle K_{p}}
{\displaystyle K_{c}}
From ideal gas law, we know that,[2]
{\displaystyle PV=nRT\,}
{\displaystyle {\frac {n}{V}}={\frac {P}{RT}}}
So, concentration (as, concentration
{\displaystyle C={\frac {n}{V}}}
{\displaystyle C={\frac {P}{RT}}}
{\displaystyle P}
{\displaystyle V}
{\displaystyle n}
{\displaystyle R}
{\displaystyle T}
is the temperature. So,
{\displaystyle {\frac {[AB]}{[A][B]}}={\frac {\frac {P_{AB}}{RT}}{{\frac {P_{A}}{RT}}{\frac {P_{B}}{RT}}}}}
{\displaystyle K_{c}={\frac {P_{AB}}{{P_{A}}{P_{B}}}}\times {RT}^{1+1-1}}
{\displaystyle P_{X}}
is the partial pressure of
{\displaystyle X}
{\displaystyle {\frac {P_{AB}}{{P_{A}}{P_{B}}}}=K_{p}}
{\displaystyle K_{c}{(RT)}^{-1-1+1}=K_{p}}
{\displaystyle K_{p}}
is the equilibrium constant in terms of partial pressure.
In this same process,
{\displaystyle \alpha A+\beta B...\rightleftharpoons \sigma S+\tau T...}
For the reaction above,
{\displaystyle K_{c}{(RT)}^{-\alpha -\beta ...+\sigma +\tau ...}=K_{p}={\frac {{p_{\mathrm {S} }}^{\sigma }{p_{\mathrm {T} }}^{\tau }...}{{p_{\mathrm {A} }}^{\alpha }{p_{\mathrm {B} }}^{\beta }...}}}
So, the relation between
{\displaystyle K_{c}}
{\displaystyle K_{p}}
{\displaystyle K_{c}{(RT)}^{\Delta n}=K_{p}}
{\displaystyle \Delta n}
the number of moles of gas on the product side minus the number of moles of gas on the reactant side in the balanced reaction
Science Aid: Equilibrium Constants Explanation of Kc and Kp for High School level
{\displaystyle K_{eq}}
Derivation intuition[permanent dead link]
↑ "Equation of State". 11 August 2014.
Retrieved from "https://simple.wikipedia.org/w/index.php?title=Equilibrium_constant&oldid=7758816"
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group(deprecated)/permrep - Maple Help
Home : Support : Online Help : group(deprecated)/permrep
find a permutation representation of a group
permrep(sbgrl)
subgroup of a group given by generators and relations (i.e. a subgrel)
Important: The group package has been deprecated. Use the superseding command GroupTheory[PermutationGroup] instead.
This function finds all the right cosets of the given subgroup in the given group, assigns integers consecutively to these cosets, constructs a permutation on these coset numbers for each group generator, and returns the permutation group generated by these permutations. Thus the permutation group will be a homomorphic image of (but not necessarily isomorphic to) the original group. A permgroup is returned whose generators are named the same as the original group generators.
The command with(group,permrep) allows the use of the abbreviated form of this command.
\mathrm{with}\left(\mathrm{group}\right):
g≔\mathrm{grelgroup}\left({x,y},{[x,x,y,x,y,y,y],[y,y,x,y,x,x,x]}\right):
\mathrm{sg}≔\mathrm{subgrel}\left({y=[y]},g\right):
\mathrm{permrep}\left(\mathrm{sg}\right)
\textcolor[rgb]{0,0,1}{\mathrm{permgroup}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{8}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{=}[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{7}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{6}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}]]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{=}[[\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{8}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{6}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{7}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}]]}\right)
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Using the Extended Power Rule Find the following derivatives. \frac{d}{dx}(\fra
Using the Extended Power Rule Find the following derivatives. \frac{d}{dx}(\frac{9}{x^{5}})
Using the Extended Power Rule Find the following derivatives.
\frac{d}{dx}\left(\frac{9}{{x}^{5}}\right)
\frac{d}{dx}\left(\frac{9}{{x}^{5}}\right)=\frac{d}{dx}\left(9{x}^{-5}\right)
Use the power rule of derivative which states
\frac{d}{dx}\left({x}^{n}\right)=n{x}^{n-1}
\frac{d}{dx}\left(9{x}^{-5}\right)=9\left(-5\right){x}^{-6}
=-45{x}^{-6}
=\frac{-45}{{x}^{6}}
\frac{2+6x+5{x}^{2}+7{x}^{3}}{{x}^{2}}
Find all the second-order partial derivatives of the functions
g\left(x,y\right)={x}^{2}y+\mathrm{cos}y+y\mathrm{sin}x
Find the derivatives of the function.
y=\frac{5x-7}{2x+1}
The derivative is y'=
f\left(x\right)=5x-3\mathrm{ln}x
Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.All continuous functions have derivatives.
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Pulsating_white_dwarf Knowpia
A pulsating white dwarf is a white dwarf star whose luminosity varies due to non-radial gravity wave pulsations within itself. Known types of pulsating white dwarfs include DAV, or ZZ Ceti, stars, with hydrogen-dominated atmospheres and the spectral type DA;[1] DBV, or V777 Her, stars, with helium-dominated atmospheres and the spectral type DB;[2] and GW Vir stars, with atmospheres dominated by helium, carbon, and oxygen, and the spectral type PG 1159. (Some authors also include non-PG 1159 stars in the class of GW Vir stars.) GW Vir stars may be subdivided into DOV and PNNV stars;[3][4] they are not, strictly speaking, white dwarfs but pre-white dwarfs which have not yet reached the white dwarf region on the Hertzsprung-Russell diagram.[5][6] A subtype of DQV stars, with carbon-dominated atmospheres, has also been proposed,[7] and in May 2012, the first extremely low mass variable (ELMV) white dwarf was reported.[8]
These variables all exhibit small (1%–30%) variations in light output, arising from a superposition of vibrational modes with periods of hundreds to thousands of seconds. Observation of these variations gives asteroseismological evidence about the interiors of white dwarfs.[9]
Types of pulsating white dwarf[4][7][8][10]
DQV DQ spectral type; hot, carbon-dominated atmosphere
ELMV DA spectral type;
{\displaystyle \lesssim 0.2M_{\odot }}
DAV starsEdit
Early calculations suggested that white dwarfs should vary with periods around 10 seconds, but searches in the 1960s failed to observe this.[11][12] The first variable white dwarf found was HL Tau 76; in 1965 and 1966, Arlo U. Landolt observed it to vary with a period of approximately 12.5 minutes.[13] The reason for this period being longer than predicted is that the variability of HL Tau 76, like that of the other pulsating variable white dwarfs known, arises from non-radial gravity wave pulsations.[14] In 1970, another white dwarf, Ross 548, was found to have the same type of variability as HL Tau 76;[15] in 1972, it was given the variable star designation ZZ Ceti.[16] The name ZZ Ceti also refers to this class of pulsating variable white dwarfs, which, as it consists of white dwarfs with hydrogen atmospheres, is also called DAV.[17] These stars have periods between 30 seconds and 25 minutes and are found in a rather narrow range of effective temperatures between about 12,500 and 11,100 K.[18] The measurement of the rate of change of period with time for the gravity wave pulsations in ZZ Ceti stars is a direct measurement of the cooling timescale for a DA white dwarf, which in turn can give an independent measurement of the age of the galactic disk.[19]
DBV starsEdit
A white-light light curve for GD 358, adapted from Winget et al. (1982)[20]
In 1982, calculations by Don Winget and his coworkers suggested that helium-atmosphere DB white dwarfs with surface temperatures around 19,000 K should also pulsate.[21] Winget then searched for such stars and found that GD 358 was a variable DB, or DBV, white dwarf.[20] This was the first prediction of a class of variable stars before their observation.[22] In 1985, this star was given the designation V777 Her, which is also another name for this class of variable stars.[2][23] These stars have effective temperatures around 25,000K.[24]
GW Vir starsEdit
A light curve for GW Virginis, adapted from Winget et al. (1985)[25]
The third known class of pulsating variable white dwarfs is the GW Vir stars, sometimes subdivided into DOV and PNNV stars. Their prototype is PG 1159-035.[5] This star (also the prototype for the class of PG 1159 stars) was observed to vary in 1979,[26] and was given the variable star designation GW Vir in 1985,[23] giving its name to the class. These stars are not, strictly speaking, white dwarfs; rather, they are stars which are in a position on the Hertzsprung-Russell diagram between the asymptotic giant branch and the white dwarf region. They may be called pre-white dwarfs.[5][6] They are hot, with surface temperatures between 75,000 K and 200,000 K, and have atmospheres dominated by helium, carbon, and oxygen. They may have relatively low surface gravities (log g ≤ 6.5.)[27] It is believed that these stars will eventually cool and become DO white dwarfs.[5]
The periods of the vibrational modes of GW Vir stars range from about 300 to about 5,000 seconds.[27] How pulsations are excited in GW Vir stars was first studied in the 1980s[28] but remained puzzling for almost twenty years.[29] From the beginning, the excitation mechanism was thought to be caused by the so-called κ-mechanism associated with ionized carbon and oxygen in the envelope below the photosphere, but it was thought this mechanism would not function if helium was present in the envelope. However, it now appears that instability can exist even in the presence of helium.[30]
DQV starsEdit
A new class of white dwarfs, with spectral type DQ and hot, carbon-dominated atmospheres, has recently been discovered by Patrick Dufour, James Liebert and their coworkers.[31] Theoretically, such white dwarfs should pulsate at temperatures where their atmospheres are partially ionized. Observations made at McDonald Observatory suggest that SDSS J142625.71+575218.3 is such a white dwarf; if so, it would be the first member of a new, DQV, class, of pulsating white dwarfs. However, it is also possible that it is a white dwarf binary system with a carbon-oxygen accretion disk.[7]
^ Koester & Chanmugam 1990, p. 891–895.
^ a b Murdin, Paul, ed. (2001). Encyclopedia of Astronomy and Astrophysics. Bristol: Nature Publishing Group. p. 3525. ISBN 978-0-333-75088-9.
^ Nagel, T.; Werner, K. (1 November 2004). "Detection of non-radial g-mode pulsations in the newly discovered PG 1159 star HE 1429-1209". Astronomy and Astrophysics. 426 (2): L45–L48. arXiv:astro-ph/0409243. Bibcode:2004A&A...426L..45N. doi:10.1051/0004-6361:200400079. ISSN 0004-6361. S2CID 9481357. §1.
^ a b Quirion, Fontaine & Brassard 2007, §1.1, 1.2.
^ a b c d Quirion, Fontaine & Brassard 2007, §1.1.
^ a b O'Brien, M. S. (1 April 2000). "The Extent and Cause of the Pre-White Dwarf Instability Strip". The Astrophysical Journal. 532 (2): 1078–1088. arXiv:astro-ph/9910495. Bibcode:2000ApJ...532.1078O. doi:10.1086/308613. ISSN 0004-637X. S2CID 115958740.
^ a b c Montgomery, M. H.; Williams, Kurtis A.; Winget, D. E.; Dufour, Patrick; DeGennaro, Steven; Liebert, James (2008). "SDSS J142625.71+575218.3: A Prototype for a New Class of Variable White Dwarf". The Astrophysical Journal Letters. 678 (1): L51. arXiv:0803.2646. Bibcode:2008ApJ...678L..51M. doi:10.1086/588286. ISSN 1538-4357. S2CID 15385909.
^ a b Hermes, J. J.; Montgomery, M. H.; Winget, D. E.; Brown, Warren R.; Kilic, Mukremin; Kenyon, Scott J. (1 May 2012). "SDSS J184037.78+642312.3: The First Pulsating Extremely Low Mass White Dwarf". The Astrophysical Journal Letters. 750 (2): L28. arXiv:1204.1338. Bibcode:2012ApJ...750L..28H. doi:10.1088/2041-8205/750/2/L28. ISSN 0004-637X. S2CID 119188878.
^ Winget, D. E. (1998). "Asteroseismology of white dwarf stars". Journal of Physics: Condensed Matter. 10 (49): 11247–11261. Bibcode:1998JPCM...1011247W. doi:10.1088/0953-8984/10/49/014. ISSN 0953-8984.
^ Association Française des Observateurs d'Etoiles Variables. "ZZ Ceti variables". Centre de Données astronomiques de Strasbourg. Archived from the original on 2007-02-05. Retrieved 2007-06-06.
^ Koester & Chanmugam 1990, § 7.1.1.
^ Lawrence, George M.; Ostriker, Jeremiah P.; Hesser, James E. (1 June 1967). "Ultrashort-Period Stellar Oscillations. I. Results from White Dwarfs, Old Novae, Central Stars of Planetary Nebulae, 3c 273, and Scorpius XR-1". The Astrophysical Journal Letters. 148: L161–L163. Bibcode:1967ApJ...148L.161L. doi:10.1086/180037. ISSN 0004-637X.
^ Landolt, Arlo U. (1 July 1968). "A New Short-Period Blue Variable". The Astrophysical Journal. 153: 151–164. Bibcode:1968ApJ...153..151L. doi:10.1086/149645. ISSN 0004-637X.
^ Koester & Chanmugam 1990, § 7.
^ Lasker, Barry M.; Hesser, James E. (1 February 1971). "High-Frequency Stellar Oscillations.VI. R548, a Periodically Variable White Dwarf". The Astrophysical Journal Letters. 163: L89–L93. Bibcode:1971ApJ...163L..89L. doi:10.1086/180673. ISSN 0004-637X.
^ Kukarkin, B. V.; Kholopov, P. N.; Kukarkina, N. P.; Perova, N. B. (1 September 1972). "58th Name-List of Variable Stars". Information Bulletin on Variable Stars. 717: 1. Bibcode:1972IBVS..717....1K. ISSN 0374-0676.
^ Koester & Chanmugam 1990, pp. 891, 895.
^ Bergeron, P.; Fontaine, G.; Billères, M.; Boudreault, S.; Green, E. M. (2004). "On the Purity of the ZZ Ceti Instability Strip: Discovery of More Pulsating DA White Dwarfs on the Basis of Optical Spectroscopy". The Astrophysical Journal. 600 (1): 404–8. arXiv:astro-ph/0309483. Bibcode:2004ApJ...600..404B. doi:10.1086/379808. ISSN 0004-637X. S2CID 16636294.
^ Kepler, S. O.; Vauclair, G.; Nather, R. E.; Winget, D. E.; Robinson, E. L. (1989). G117-B15A - How is it evolving?. IAU Colloq. 114: White Dwarfs. Vol. 328. pp. 341–345. doi:10.1007/3-540-51031-1_344.
^ a b Winget, D. E.; Robinson, E. L.; Nather, R. D.; Fontaine, G. (1 November 1982). "Photometric observations of GD 358 - DB white dwarfs do pulsate". The Astrophysical Journal Letters. 262: L11–L15. Bibcode:1982ApJ...262L..11W. doi:10.1086/183902. ISSN 0004-637X.
^ Winget, D. E.; van Horn, H. M.; Tassoul, M.; Fontaine, G.; Hansen, C. J.; Carroll, B. W. (1 January 1982). "Hydrogen-driving and the blue edge of compositionally stratified ZZ Ceti star models". The Astrophysical Journal Letters. 252: L67. Bibcode:1982ApJ...252L..65W. doi:10.1086/183721. ISSN 0004-637X.
^ Kawaler, Steven D.; Novikov, I. D.; Srinivasan, G. (1997). Meynet, G.; Schaerer (eds.). Stellar remnants. Saas-Fee advanced course 25 lecture notes. Berlin: Springer. p. 89. ISBN 978-3-540-61520-0. Lecture notes for Saas-Fee advanced course number 25.
^ a b Kholopov, P. N.; Samus, N. N.; Kazarovets, E. V.; Perova, N. B. (1 March 1985). "The 67th Name-List of Variable Stars". Information Bulletin on Variable Stars. 2681: 1. Bibcode:1985IBVS.2681....1K. ISSN 0374-0676.
^ Koester & Chanmugam 1990, p. 895.
^ Winget, D. E.; Kepler, S. O.; Robinson, E. L.; Nather, R. E.; O’Donoghue, Darragh (May 1985). "A measurement of secular evolution in the pre-white dwarf star PG 1159-035". The Astrophysical Journal. 292: 606–613. doi:10.1086/163193. Retrieved 6 May 2022.
^ McGraw, J. T.; Liebert, James; Starrfield, S. G.; Green, R. (1979). PG1159-035: A new, hot, non-DA pulsating degenerate. IAU Colloq. 53: White Dwarfs and Variable Degenerate Stars. pp. 377–381. Bibcode:1979wdvd.coll..377M.
^ a b Quirion, Fontaine & Brassard 2007, Table 1.
^ Cox, Arthur N. (1 March 2003). "A Pulsation Mechanism for GW Virginis Variables". The Astrophysical Journal. 585 (2): 975–982. Bibcode:2003ApJ...585..975C. doi:10.1086/346228. ISSN 0004-637X.
^ Cox, A. N. (1 May 2002). An Instability Mechanism for GW Vir Variables. Bulletin of the American Astronomical Society. Vol. 200. p. 85.07. Bibcode:2002AAS...200.8507C.
^ Córsico, A. H.; Althaus, L. G.; Miller Bertolami, M. M. (1 October 2006). "New nonadiabatic pulsation computations on full PG 1159 evolutionary models: the theoretical GW Virginis instability strip revisited". Astronomy and Astrophysics. 458 (1): 259–267. arXiv:astro-ph/0607012. Bibcode:2006A&A...458..259C. doi:10.1051/0004-6361:20065423. ISSN 0004-6361. S2CID 16700443. §1.
^ Dufour, P.; Liebert, James; Fontaine, G.; Behara, N. (November 2007). "White dwarf stars with carbon atmospheres". Nature. 450 (7169): 522–524. arXiv:0711.3227. Bibcode:2007Natur.450..522D. doi:10.1038/nature06318. ISSN 0028-0836. PMID 18033290. S2CID 4398697.
Koester, D.; Chanmugam, G. (1990). "REVIEW: Physics of white dwarf stars" (PDF). Reports on Progress in Physics. 53 (7): 837–915. Bibcode:1990RPPh...53..837K. doi:10.1088/0034-4885/53/7/001. ISSN 0034-4885.
Quirion, P.O.; Fontaine, G.; Brassard, P. (1 July 2007). "Mapping the Instability Domains of GW Vir Stars in the Effective Temperature-Surface Gravity Diagram". The Astrophysical Journal Supplement Series. 171 (1): 219–248. Bibcode:2007ApJS..171..219Q. doi:10.1086/513870. ISSN 0067-0049.
Variable White Dwarf Data Tables, Paul A. Bradley, 22 March 2005 version. Accessed online June 7, 2007.
A Progress Report on the Empirical Determination of the ZZ Ceti Instability Strip, A. Gianninas, P. Bergeron, and G. Fontaine, arXiv:astro-ph/0612043.
Asteroseismology of white dwarf stars, D. E. Winget, Journal of Physics: Condensed Matter 10, #49 (December 14, 1998), pp. 11247–11261. DOI 10.1088/0953-8984/10/49/014.
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NDF_STYPE
The routine sets a new full type for an NDF array component, causing its storage type to be changed. If the component’s values are defined, they will be converted from from the old type to the new one. If they are undefined, then no conversion will be necessary. Subsequent enquiries will reflect the new type. Conversion may be performed between any types supported by the NDF_ routines, including from a non-complex type to a complex type (and vice versa).
CALL NDF_STYPE( FTYPE, INDF, COMP, STATUS )
\ast
\ast
The new full type specification for the NDF component (e.g. ’_REAL’ or ’COMPLEX_INTEGER’).
\ast
\ast
Name of the array component whose type is to be set: ’DATA’ or ’VARIANCE’.
The routine may only be used to change the type of a component of a base NDF. If it is called for an NDF which is not a base NDF, then it will return without action. No error will result.
A comma-separated list of component names may also be supplied, in which case the type of each component will be set to the same value in turn.
An error will result if a component being modified, or any part of it, is currently mapped for access (e.g. through another identifier).
If the type of a component is to be changed without its values being retained, then a call to NDF_RESET should be made beforehand. This will avoid the cost of converting all the values.
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Home : Support : Online Help : Mathematics : Algebra : Expression Manipulation : Simplifying : logarithms
simplify expressions involving logarithms
simplify(expr, ln)
literal name; ln
The simplify/ln function is used to simplify logarithmic expressions. It applies the following simplifications whenever it can determine that the appropriate conditions hold:
\mathrm{ln}\left({a}^{b}\right)
b\mathrm{ln}\left(a\right)
\mathrm{signum}\left(a\right)=1
b
\mathrm{ln}\left({a}^{b}\right)
b\mathrm{ln}\left(-a\right)
\mathrm{signum}\left(a\right)=-1
b
\mathrm{ln}\left({a}^{b}\right)
b\mathrm{ln}\left(a\right)
\mathrm{signum}\left(a\right)=-1
b
\mathrm{ln}\left({a}^{b}\right)
\frac{b\mathrm{ln}\left({a}^{2}\right)}{2}
a
b
\mathrm{ln}\left({a}^{b}\right)
b\mathrm{ln}\left(a\right)
a
b
\mathrm{ln}\left(xy\right)
\mathrm{ln}\left(x\right)+\mathrm{ln}\left(y\right)
0<x
\mathrm{signum}\left(y\right)
is unknown
\mathrm{ln}\left(xy\right)
\mathrm{ln}\left(-x\right)+\mathrm{ln}\left(-y\right)
\mathrm{signum}\left(x\right)=-1
\mathrm{ln}\left(xy\right)
\mathrm{ln}\left(-x\right)+\mathrm{ln}\left(y\right)+I\mathrm{\pi }
\mathrm{signum}\left(x\right)=-1
\mathrm{signum}\left(y\right)=1
\mathrm{ln}\left({ⅇ}^{x}\right)
x
x
\mathrm{ln}\left(\mathrm{LambertW}\left(x\right)\right)
\mathrm{ln}\left(x\right)-\mathrm{LambertW}\left(x\right)
x
In the case of an integer argument to ln, the integer is factored and the logarithm is returned as a sum of logarithms.
In the case of a sum of terms as the argument to ln, the integer content of the sum is factored out and the logarithm is returned as a sum of two logarithms.
Making the appropriate assumptions on the names in the expression to be simplified (see assume) provides simplify with enough information to apply the above identities correctly.
It is inappropriate to apply the above identities in these cases since nothing is known about n, x, and y:
\mathrm{simplify}\left(\mathrm{ln}\left({x}^{3}\right),\mathrm{ln}\right)
\textcolor[rgb]{0,0,1}{\mathrm{ln}}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{3}}\right)
\mathrm{simplify}\left(\mathrm{ln}\left(xy\right)\right)
\textcolor[rgb]{0,0,1}{\mathrm{ln}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{y}\right)
\mathrm{simplify}\left(\mathrm{ln}\left(\mathrm{exp}\left(x\right)\right),\mathrm{ln}\right)
\textcolor[rgb]{0,0,1}{\mathrm{ln}}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{x}}\right)
\mathrm{simplify}\left(\mathrm{ln}\left({y}^{n}\right),\mathrm{ln}\right)
\textcolor[rgb]{0,0,1}{\mathrm{ln}}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{n}}\right)
However, by making appropriate assumptions on the variables, enough information is provided to correctly apply the identities:
\mathrm{assume}\left(n,\mathrm{even}\right)
\mathrm{assume}\left(x,\mathrm{real}\right)
\mathrm{assume}\left(y<0\right)
\mathrm{simplify}\left(\mathrm{ln}\left({x}^{3}\right),\mathrm{ln}\right)
\textcolor[rgb]{0,0,1}{\mathrm{ln}}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{\mathrm{x~}}}^{\textcolor[rgb]{0,0,1}{3}}\right)
\mathrm{simplify}\left(\mathrm{ln}\left(xy\right)\right)
\textcolor[rgb]{0,0,1}{\mathrm{ln}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{y~}}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{ln}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{x~}}\right)
\mathrm{simplify}\left(\mathrm{ln}\left(\mathrm{exp}\left(x\right)\right)\right)
\textcolor[rgb]{0,0,1}{\mathrm{x~}}
\mathrm{simplify}\left(\mathrm{ln}\left({y}^{3}\right),\mathrm{ln}\right)
\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{ln}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{y~}}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{I}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{\pi }}
\mathrm{simplify}\left(\mathrm{ln}\left({y}^{n}\right),\mathrm{ln}\right)
\textcolor[rgb]{0,0,1}{\mathrm{n~}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{ln}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{\mathrm{y~}}\right)
Simplifications involving integer factors:
\mathrm{simplify}\left(\mathrm{ln}\left(-40a+15b\right),\mathrm{ln}\right)
\textcolor[rgb]{0,0,1}{\mathrm{ln}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{5}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{ln}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{8}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{b}\right)
\mathrm{simplify}\left(\mathrm{ln}\left(345366\right),\mathrm{ln}\right)
\textcolor[rgb]{0,0,1}{\mathrm{ln}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{2}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{ln}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{3}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{ln}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{7}\right)\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{\mathrm{ln}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{2741}\right)
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Evaluate the definite integral. \int_{1}^{2}\frac{e^{\frac{1}{x^{4}}}}{x^{5}}dx
{\int }_{1}^{2}\frac{{e}^{\frac{1}{{x}^{4}}}}{{x}^{5}}dx
To evaluate the definite integral.
{\int }_{1}^{2}\left(\frac{{e}^{\frac{1}{{x}^{4}}}}{{x}^{5}}\right)dx
Integrate the function with respect to x.
{\int }_{1}^{2}\left(\frac{{e}^{\frac{1}{{x}^{4}}}}{{x}^{5}}\right)dx
\frac{1}{{x}^{4}}=u
so, differentiate with respect to x.
\frac{d}{dx}\left(\frac{1}{{x}^{4}}\right)=\frac{d}{dx}\left(u\right)
\frac{-4}{{x}^{5}}=\frac{du}{dx}
\left(\frac{1}{{x}^{5}}\right)dx=\frac{1}{-4}du
[put in the given integral]
So, the integral become,
{\int }_{x=1}^{2}\left(\frac{{e}^{\frac{1}{{x}^{4}}}}{{x}^{5}}\right)dx={\int }_{x=1}^{2}\left(\frac{{e}^{u}}{-4}\right)du
\left[put\left(\frac{1}{{x}^{5}}\right)dx=\left(\frac{1}{{x}^{-4}}\right)du,\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}u=\frac{1}{{x}^{4}}\right]
=-\frac{1}{4}{\left[{e}^{u}\right]}_{x=1}^{2}
\left[\because \int {e}^{x}dx={e}^{x}\right]
=-\frac{1}{4}{\left[{e}^{\frac{1}{{x}^{4}}}\right]}_{x=1}^{2}
\left[substitute\text{ }u=\frac{1}{{x}^{4}}\right]
=-\frac{1}{4}\left[{e}^{\frac{1}{{2}^{\left(4\right)}}}-{e}^{\frac{1}{{1}^{\left(4\right)}}}\right]
limacarp4
u=\frac{1}{{x}^{4}}
du=-\frac{4}{{x}^{5}}dx
\int \frac{{e}^{\frac{1}{{x}^{4}}}}{{x}^{5}}dx=\int {e}^{u}\left(-\frac{1}{4}du\right)
\int \frac{{e}^{\frac{1}{{x}^{4}}}}{{x}^{5}}dx=-\frac{1}{4}\int {e}^{u}du
\int {e}^{u}du={e}^{u}+C
\int \frac{{e}^{\frac{1}{{x}^{4}}}}{{x}^{5}}dx=-\frac{1}{4}{e}^{u}+C
\int \frac{{e}^{\frac{1}{{x}^{4}}}}{{x}^{5}}dx=-\frac{1}{4}{e}^{\frac{1}{{x}^{4}}}+C
{\int }_{1}^{2}\frac{{e}^{\frac{1}{{x}^{4}}}}{{x}^{5}}dx={\left[-\frac{1}{4}{e}^{\frac{1}{{x}^{4}}}\right]}_{1}^{2}
=-\frac{1}{4}{e}^{\frac{1}{{2}^{4}}}-\left(-\frac{1}{4}{e}^{\frac{1}{{1}^{4}}}\right)
=-\frac{1}{4}{e}^{\frac{1}{16}}+\frac{1}{4}e
=\frac{1}{4}\left(-{e}^{\frac{1}{16}}+e\right)
\approx 0.413
We have to calculate
{\int }_{1}^{2}\frac{{e}^{\frac{1}{{x}^{4}}}}{{x}^{5}}dx
Let us first calculate indefinite integral.
\int \frac{{e}^{\frac{1}{{4}^{4}}}}{{x}^{5}}dx=-\frac{1}{4}\int {e}^{u}du
u=\frac{1}{{x}^{4}}
dx=-\frac{{x}^{5}}{4}du
-\frac{1}{4}\int {e}^{u}du=-\frac{{e}^{u}}{4}=-\frac{{e}^{\frac{1}{{x}^{4}}}}{4}
u=\frac{1}{{x}^{4}}
\int \frac{{e}^{\frac{1}{{x}^{4}}}}{{x}^{5}}dx=-\frac{{e}^{\frac{1}{{4}^{5}}}}{4}+C
{\int }_{1}^{2}\frac{{e}^{\frac{1}{{x}^{4}}}}{{x}^{5}}dx=-\left[\frac{{e}^{\frac{1}{{4}^{5}}}}{4}{\right]}_{1}^{2}
=\frac{e}{4}-\frac{\sqrt[16]{e}}{4}
How do I solve the following?
\underset{x\to 0}{lim}{\int }_{0}^{1}\mathrm{cos}\left(\frac{1}{xt}\right)dt
Evaluate the line integral, where C is the given curve.C
{y}^{3}ds,C:x={t}^{3},y=t,0\le t\le 5
Show the steps needed to find the given solutions.
\int \frac{\mathrm{arctan}c}{1+{x}^{2}}dx
Ry{e}^{-xy}dA,R=\left[0,2\right]×\left[0,3\right]
Which is the easiest way to evaluate
{\int }_{0}^{\frac{\pi }{2}}\left(\sqrt{\mathrm{tan}x}+\sqrt{\mathrm{cot}x}\right)
\int \frac{3}{4}t\mathrm{cos}\left(5t\right)dt
Evaluate the line integral, where C is the given curve.
\int y3\text{ }ds,\text{ }C÷x=t3,\text{ }y=t,\text{ }0?\text{ }t?\text{ }3
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I want to solve for x 2^{\sin^4 x-\cos^2 x}-2^{\cos^4 x-\sin^2 x}=\cos
Mabel Breault 2021-12-30 Answered
I want to solve for x
{2}^{{\mathrm{sin}}^{4}x-{\mathrm{cos}}^{2}x}-{2}^{{\mathrm{cos}}^{4}x-{\mathrm{sin}}^{2}x}=\mathrm{cos}2x
u={\mathrm{sin}}^{2}\left(x\right)\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }v={\mathrm{cos}}^{2}\left(x\right)
{2}^{{u}^{2}-v}-{2}^{{v}^{2}-u}=v-u
and u+v=1. Thus
{2}^{{u}^{2}+u-1}+u={2}^{{v}^{2}+v-1}+v
f\left(u\right)={2}^{{u}^{2}+u-1}
,then we are looking for a
u\in \left[0,1\right]
such that f(u)=f(1-u). However
{f}^{\prime }\left(u\right)=\mathrm{ln}\left(2\right)\left(2u+1\right){2}^{{u}^{2}+u-1}+1>0
u\in \left[0,1\right]
. Hence f is injective on [0,1] and thus f(u)=f(1-u) if and only if u=1-u. Thus
{\mathrm{sin}}^{2}\left(x\right)=u=\frac{12}{}
The left side of the equation is greater than 0 if and only if the right side is less than 0, and vice versa.
\left[{\mathrm{sin}}^{4}x-{\mathrm{cos}}^{2}x\right]-\left[{\mathrm{cos}}^{4}x-{\mathrm{sin}}^{2}x\right]=-\left({\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x\right)\left({\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x+1\right)=-2\mathrm{cos}2x
Therefore all possible solutions correspond to zeros of the right hand side (which are also automatically zeroes of the left hand side). This gives
x=±\frac{\pi }{4}+2\pi \mathbb{Z},±\frac{3\pi }{4}+2\pi \mathbb{Z}
{\mathrm{cos}}^{2}x=a
{\mathrm{sin}}^{4}x-{\mathrm{cos}}^{2}x=\left(1-a{\right)}^{2}-a={a}^{2}-3a+1
{\mathrm{cos}}^{4}x-{\mathrm{sin}}^{2}x={a}^{2}-\left(1-a\right)={a}^{2}+a-1
{2}^{{a}^{2}-3a+1}-{2}^{{a}^{2}+a-1}=2a-1
{2}^{{a}^{2}-3a+1}\left(1-{2}^{4a-2}\right)=2a-1
a-1>0,\text{ }\text{ }4a-2>0⇒{2}^{4a-2}>{2}^{0}=1⇒
the left hand side is < 0
Similarly, if 2a-1 <0 the right hand side is > 0
If 2a-1=0, both sides become 0
⇒2{\mathrm{cos}}^{2}x-1=0⇔\mathrm{cos}2x=0⇒2x=\left(2n+1\right)\frac{\pi }{2}
where n is any integer
\mathrm{sin}x+\mathrm{sin}y=a\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{cos}x+\mathrm{cos}y=b
\mathrm{tan}\left(x-\frac{y}{2}\right)
Determining the period of
\frac{\mathrm{sin}2x}{\mathrm{cos}3x}
I would like to compute the period of this function which is a fraction of two trigonometric functions.
\frac{\mathrm{sin}2x}{\mathrm{cos}3x}
Is there a theorem for this? what trick to use to easily find the period? I started by reducing the fraction but Im
{\int }_{0}^{2\pi }\left(\mathrm{cos}\left(t{\right)}^{2n}\right)=\left(\genfrac{}{}{0}{}{2n}{n}\right)\frac{2\pi }{{2}^{2n}}
using the following result
{\int }_{\gamma }\left(z+\frac{1}{z}{\right)}^{2n}\frac{dz}{z}=\left(\genfrac{}{}{0}{}{2n}{n}\right)2\pi i
{\int }_{0}^{2\pi }\left(\mathrm{cos}\left(t{\right)}^{2n}\right)=\left(\genfrac{}{}{0}{}{2n}{n}\right)\frac{2\pi }{{2}^{2n}}
\mathrm{cos}360+\mathrm{cos}234+\mathrm{cos}162+\mathrm{cos}18=
Note - numbers are in degree.
I've been trying to figure out why this equation is satisfied:
\frac{1-\left(\mathrm{cos}\left(x\right){\right)}^{3}}{{x}^{2}}=\frac{2\cdot \left(\mathrm{sin}\left(\frac{x}{2}\right){\right)}^{2}}{{x}^{2}}\cdot \left(1+\mathrm{cos}\left(x\right)+\left(\mathrm{cos}\left(x\right){\right)}^{2}\right)
but I can't find the proper trigonometric changes in order to change from one to another. I know that the sine comes from the double angle formula, but I obtain slightly different results in other parts and it's never the same as the formula above.
4\mathrm{sin}\left(x\right)+7\mathrm{cos}\left(x\right)=6
0\le x\le {360}^{\circ }
I put the equation into the form
a\mathrm{sin}\left(x\right)+b\mathrm{cos}\left(x\right)=R\mathrm{sin}\left(x+a\right)
, but after determining that
R\mathrm{cos}\left(a\right)=4,\text{ }R\mathrm{sin}\left(a\right)=7
R\mathrm{sin}\left(x+a\right)=6
, I don't know how to proceed.
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Board Paper Solutions for ICSE Class 10 MATHS Board Paper 2020 MeritNation
Board Paper of Class 10 2020 Math - Solutions
Board Paper of Class 10 2020 Math MeritNation - Solutions
{x}^{2}-7x+3=0
(b) Given
\mathrm{A}=\left[\begin{array}{cc}x& 3\\ y& 3\end{array}\right]
{A}^{2}=3I
, where I is the identity matrix of order 2, find x and y.
(c) Using ruler and compass construct a triangle ABC where AB = 3 cm, BC = 4 cm and
\angle \mathrm{ABC}=90°
. Hence construct a circle circumscribing the triangle ABC. Measure and write down the radius of the circle. VIEW SOLUTION
(a) Use factor theorem to factories
6{x}^{3}+17{x}^{2}+4x-12
(b) Solve the following inequation and represent the solution set the number line
\frac{3x}{5}+2<x+4\le \frac{x}{2}+5, x\in \mathbf{R}
(c) Draw a Histogram for the give data, using a graph paper:
Weekly Wages(in Rs) Number of People
(a) In what ratio is the line joining P(5, 3) and Q(-5, 3) divided by the y-axis? Also find coordinates of the point of intersection.
(b) Prove that:
\frac{\mathrm{sin}A}{1+\mathrm{cot}A}-\frac{\mathrm{cos}A}{1+\mathrm{tan}A}=\mathrm{sin}A-\mathrm{cos}A
(c) In the figure given bellow,
\mathrm{O}
is the center of the circle and
\mathrm{AB}
\mathrm{AC}=\mathrm{BD}
\angle \mathrm{AOC}=72°
(i) ∠ABC (ii) ∠BAD (iii) ∠ABD
(a) A solid spherical ball of radius 6 cm is melted and recast into 64 identical spherical marbles. Find the radius of each marble,
(b) Each of the letters of the word `AUTHORIZES’ is written on identical circular discs and put in a bag. They are well shuffled. If a disc is drawn at random from the bag, what is the probability that the letter is:
(i) a vowel (ii) one of the first 9 letters of the English alphabet which appears in the given word (iii) one of the last 9 letters of the English alphabet which appears in the in word?
Medicines costing ₹ 950, GST @ 5% A pair of shoes costing ₹ 3000, GST @ 18% A Laptop bag costing ₹ 1000 with a discount of 30%, GST @, 18%, (i) Calculate the total amount of GST paid. (ii) The total bill amount including GST paid by Mr. Bedi, VIEW SOLUTION
(a) A company with 500 shares of nominal value 120 declares an annual dividend ut 131 15%. Calculate:
(ii) annual income of Mr. Sharma who holds 80 shares of the company If the return percent of Mr. Sharma from his shares is 10%, find the Market value of each share.
(b) The mean of the following data is 16 Calculate the value of f
Number of Students 3 7 f 9 6
(c) The 4th, 6th and the last term of a geometric progression are 10, 40 and 640 respectively. If the common ratio is positive, find the first term, common ratio number of terms of the series. VIEW SOLUTION
(a) From the top of a cliff, the angle of depression of the top and bottom of a tower are observed to be
45°
60°
respectively. If the height of the tower is 20 m. Find: (i) the height of the cliff (ii) the distance between the cliff and the tower.
(b) In the given figure AB = 9cm, PA = 7.5 and PC = 5Cm. Chords AD and BC intersect at P.
(i) Prove that ∆ PAB ~ ∆PCD
(ii) Find the length of CD
(iii) Find area of ∆PAB: area of ∆PCD
A=\left[\begin{array}{cc}3& 0\\ 5& 1\end{array}\right], B=\left[\begin{array}{cc}-4& 2\\ 1& 0\end{array}\right]
{A}^{2}-2AB+{B}^{2}
In the given figure TP and TQ are two tangents to the circle with center O, touching at A and C respectively, If
\angle \mathrm{BCQ}=55°
\angle \mathrm{BAP}=60°
, find: (i)
\angle \mathrm{OBA}
\angle \mathrm{OBC}
\angle \mathrm{AOC}
\angle \mathrm{ATC}
(b) Using properties of find x : y, given:
\frac{{x}^{2}+2x}{2x+4}=\frac{{y}^{2}+3y}{3y+9}
(c) Find the value of 'p' if the lines,
5x-3y+2=0
6x-py+7=0
are perpendicular to each other. Hence find the equation of a line passing through
\left(-2, -1\right)
6x-py+7=0
(a) What must be added to the polynomial
2{x}^{3}-3{x}^{2}-8x
, so that it leaves a remainder 10 when divided by 2x + 1?
(b) Mr. Sonu has a recurring deposit account and deposits ₹ 750 per month for 2 years If he gets ₹ 19125 at the time of maturity, find the rate of interest.
(c) Use graph paper for this question. Take 1 cm = 1 unit on both x and y axes
(i) Plot the following points on your graph sheets: A(-4, 0), B(-3, 2), C(0, 4) D(4, 1) and E(7, 3)
(ii) Reflected the point B,C,D and E on the x-axis and name then as B,C,D and E respectively.
(iii) Join the points A,B,C,D,E,E,D,C,B and A in order.
(iv) Name the close figure farmed VIEW SOLUTION
x=\frac{\sqrt{2a+1}+\sqrt{2a-1}}{\sqrt{2a+1}-\sqrt{2a-1}}
{x}^{2}-4ax+1=0
(b) 40 student enter for a game of shot-put competition. The distance thrown (in meters) is recorded below:
Distance (In m) 12-13 13-14 14-15 15-16 16-17 17-18 18-19
Use a graph paper to draw an ogive for the above distribution. Use a scale of 2 cm = 1 m on one axis and 2 cm = 5 students on the other axis. Hence using your graph find: (i) the median (ii) Upper Quartile (iii) number of student who cover a distance which is above
16\frac{1}{2} \mathrm{m}
(c) Use ruler and compass for this question. Construct a circle of radius 4.5 cm. Draw a chord. AB = 6cm.
(ii) Join AD and find the locus of points which are equidistant from AD and AB. Murk the point where it meets the circle as C.
(iii) Join and CD. Measure and write down the length of side CD of the quadrilateral ABCD. VIEW SOLUTION
(a) A model of a high rise building is made to a scale of 1: 50. (i) If the height of the model is 0.8 m, find the height of the actual building. (ii) If the floor area of a flat in the building is 20 m2, find the floor area of that in the model.
(b) From a solid wooden cylinder of height 28 cm and diameter 6 cm, two conical cavities are hollowed out. The diameters of the cones are also of 6 cm and height 10.5 cm. Taking 22 7 π= find the volume of the remaining solid.
{\left(\frac{1-\mathrm{tan}\theta }{1-\mathrm{cot}\theta }\right)}^{2}={\mathrm{tan}}^{2}\theta
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19 Q19 A cyclist paddling at a speed of 10 m/s on a level road takes a sharp - Physics - Work Energy And Power - 12003739 | Meritnation.com
Q19. A cyclist paddling at a speed of 10 m/s on a level road takes a sharp circular turn of radius 10 m without reducing the speed. The angle made by cyclist with vertical is
\left(1\right) \frac{\mathrm{\pi }}{4} \left(2\right) \frac{\mathrm{\pi }}{3}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(3\right) \frac{\mathrm{\pi }}{6} \left(4\right) \frac{\mathrm{\pi }}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}
Abid Hussain answered this
1st one is correct
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Use Laplace transforms to solve the initial value problems. y"+y'=\delta(t)-\delta(t-3),
Use Laplace transforms to solve the initial value problems. y"+y'=\delta(t)-\delta(t-3), y(0)=1 , y'(0)=0
Use Laplace transforms to solve the initial value problems.
y{y}^{\prime }=\delta \left(t\right)-\delta \left(t-3\right),y\left(0\right)=1,{y}^{\prime }\left(0\right)=0
L\left\{\delta \left(t-a\right)\right\}={e}^{-as}
{L}^{-1}\left\{\frac{1}{s-a}\right\}={e}^{at}
{L}^{-1}\left\{\frac{{e}^{-as}}{s}\right\}=u\left(t-a\right)
y{y}^{\prime }=\delta \left(t\right)-\delta \left(t-3\right),y\left(0\right)=1,{y}^{\prime }\left(0\right)=0
Taking laplace transformation both sides:
L\left\{y"\right\}+L\left\{{y}^{\prime }\right\}=L\left\{\delta \left(t\right)\right\}-L\left\{\delta \left(t-3\right)\right\}
\left[{s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)\right]+\left[sy\left(s\right)-y\left(0\right)\right]=1-{e}^{-3s}
\left({s}^{2}y\left(s\right)-s\right)+\left(sy\left(s\right)-1\right)=1-{e}^{-3s}
y\left(s\right)\left[{s}^{2}+s\right]-s-1=1-{e}^{-3s}
y\left(s\right)\left[{s}^{2}+s\right]=2-{e}^{-3s}+s
y\left(s\right)=\frac{2+s-{e}^{-3s}}{s\left(s+1\right)}
y\left(s\right)=\frac{2+s}{s\left(s+1\right)}-\frac{{e}^{-3s}}{s\left(s+1\right)}
y\left(s\right)=\frac{2}{s}-\frac{1}{s+1}-\frac{{e}^{-3s}}{s}+\frac{{e}^{-3s}}{s+1}
Now taking inverse laplace transformation both sides:
{L}^{-1}\left\{y\left(s\right)\right\}=2{L}^{-1}\left\{\frac{1}{s}\right\}-{L}^{-1}\left\{\frac{1}{s+1}\right\}-{L}^{-1}\left\{\frac{{e}^{-3s}}{s}\right\}+{L}^{-1}\left\{\frac{{e}^{-3s}}{s+1}\right\}
y\left(t\right)=2×1-{e}^{-t}-u\left(t-3\right)+{e}^{3-t}u\left(t-3\right)
\frac{1}{2a}\left(\mathrm{sin}at+at\mathrm{cos}at\right)
Sketch the Fourier transforms of the following functions assuming
|{\omega }_{1}|>|omeg{a}_{2}|
a\right)f\left(t\right)=\left[\mathrm{cos}\left({\omega }_{1}t\right)+i\mathrm{sin}\left({\omega }_{1}t\right)\right]-\left[\mathrm{cos}\left({\omega }_{2}t\right)+i\mathrm{sin}\left({\omega }_{2}t\right)\right]
b\right)f\left(t\right)=\left[\mathrm{cos}\left({\omega }_{0}t\right)\right]{e}^{-\frac{t}{{T}_{2}}}
f\left(t\right)={e}^{t}
g\left(t\right)={e}^{-2t}
0\le t<\mathrm{\infty }
compute in two different ways:
a) By directly evaluating the integral in the defination of
f\cdot g
b) By computing
{L}^{-1}\left\{F\left(s\right)G\left(s\right)\right\}\text{ where }F\left(s\right)=L\left\{f\left(t\right)\right\}\text{ and }G\left(s\right)=L\left\{g\left(t\right)\right\}
\text{Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by }
\left(F\left(s\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt
\text{where we assume s is a positive real number. For example, to find the Laplace transform of}
f\left(t\right)={e}^{-t}\text{ , the following improper integral is evaluated using integration by parts:}
F\left(s\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}{e}^{-t}dt={\int }_{0}^{\mathrm{\infty }}{e}^{-\left(s+1\right)t}dt=\frac{1}{s+1}
\text{ Verify the following Laplace transforms, where u is a real number. }
f\left(t\right)=t\to F\left(s\right)=\frac{1}{{s}^{2}}
Discuss is there a solution or not (why) of Laplace transformation of
L\left(\frac{1}{t}\right)
Find the laplace transform of the following function
f\left(t\right)=t{u}_{2}\left(t\right)
\frac{Y\left(s\right)}{U\left(s\right)}=\frac{50}{{\left(s+7\right)}^{2}+25}
f\left(t\right)=10{e}^{-7t}\mathrm{sin}\left(5t\right)
f\left(t\right)=10{e}^{7t}\mathrm{sin}\left(5t\right)
f\left(t\right)=50{e}^{-7t}\mathrm{sin}\left(5t\right)
f\left(t\right)=2{e}^{-7t}\mathrm{sin}\left(5t\right)
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I have to make a crossword puzzle for mathsfrom chapters linear equations in two variables,introduction to Euclidsgeometry, lines and angles - Maths - Introduction to Euclid\s Geometry - 9447437 | Meritnation.com
I have to make a crossword puzzle for mathsfrom chapters linear equations in two variables,introduction to Euclidsgeometry, lines and angles .plzzzz help
\mathrm{The} \mathrm{crossword} \mathrm{on} \mathrm{the} \mathrm{mathematics} \mathrm{chapter} "\mathrm{Linear} \mathrm{Equations} \mathrm{in} \mathrm{two} \phantom{\rule{0ex}{0ex}}\mathrm{variables}" \mathrm{is} \mathrm{shown} \mathrm{below}:\phantom{\rule{0ex}{0ex}}
\mathrm{The} \mathrm{clues} \mathrm{to} \mathrm{fill} \mathrm{the} \mathrm{options} \mathrm{in} \mathrm{above} \mathrm{crossword} \mathrm{are} \mathrm{given} \mathrm{below}:\phantom{\rule{0ex}{0ex}} \mathbf{ACROSS}\phantom{\rule{0ex}{0ex}}1. \mathrm{x}=0 \mathrm{is} \mathrm{the} \mathrm{_________} \mathrm{of} \mathrm{the} \mathrm{y}-\mathrm{axis}\phantom{\rule{0ex}{0ex}}3. \mathrm{an} \mathrm{equation} \mathrm{of} \mathrm{the} \mathrm{type} \mathrm{y}=\mathrm{mx} \mathrm{represents} \mathrm{a} \mathrm{straight} \mathrm{line} \mathrm{passing} \phantom{\rule{0ex}{0ex}}\mathrm{through} \mathrm{the} \mathrm{_________}\phantom{\rule{0ex}{0ex}}6. \mathrm{the} \mathrm{graph} \mathrm{of} \mathrm{x}=\mathrm{a} \mathrm{is} \mathrm{a} \mathrm{staight} \mathrm{line} \mathrm{_______} \mathrm{to} \mathrm{the} \mathrm{y}-\mathrm{axis}\phantom{\rule{0ex}{0ex}} \mathbf{DOWN}\phantom{\rule{0ex}{0ex}}2. \mathrm{a} \mathrm{linear} \mathrm{equation} \mathrm{in} \mathrm{two} \mathrm{variables} \mathrm{has} \mathrm{_______} \mathrm{many} \mathrm{solutions}\phantom{\rule{0ex}{0ex}}4. \mathrm{the________} \mathrm{of} \mathrm{every} \mathrm{linear} \mathrm{equation} \mathrm{in} \mathrm{two} \mathrm{variables} \mathrm{is} \mathrm{a} \mathrm{staright} \mathrm{line}.\phantom{\rule{0ex}{0ex}}5 \mathrm{an} \mathrm{equation} \mathrm{of} \mathrm{the} \mathrm{form} \mathrm{ax}+\mathrm{by}+\mathrm{c}=\mathrm{________}
\mathrm{The} \mathrm{answer} \mathrm{to} \mathrm{the} \mathrm{above} \mathrm{crossword} \mathrm{is} \mathrm{shown} \mathrm{below}:\phantom{\rule{0ex}{0ex}}
\mathbf{Similarly}\mathbf{ }\mathbf{by}\mathbf{ }\mathbf{taking}\mathbf{ }\mathbf{appropriate}\mathbf{ }\mathbf{words}\mathbf{ }\mathbf{from}\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{other}\mathbf{ }\mathbf{chapters}\mathbf{,}\mathbf{ }\mathbf{their}\phantom{\rule{0ex}{0ex}}\mathbf{corresponding}\mathbf{ }\mathbf{crosswords}\mathbf{ }\mathbf{can}\mathbf{ }\mathbf{be}\mathbf{ }\mathbf{formed}\mathbf{.}
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Coproximinality in the Space of Bounded Functions
Eyad Abu-Sirhan, Zuhier Altawallbeh, "Coproximinality in the Space of Bounded Functions", International Journal of Mathematics and Mathematical Sciences, vol. 2014, Article ID 196391, 5 pages, 2014. https://doi.org/10.1155/2014/196391
Eyad Abu-Sirhan 1 and Zuhier Altawallbeh 1
1Department of Mathematics, Tafila Technical University, Tafila 6610, Jordan
As a counterpart to best approximation in Banach spaces, the best coapproximation was introduced by Franchetti and Furi (1972). In this paper, we shall consider the relation between coproximinality of a nonempty subset of a Banach space and of in .
Let be a nonempty subset of a Banach space . We say that is proximinal in if each there corresponds at least to one point such that
Recently, another kind of approximation from has been introduced by Franchetti and Furi [1] who have considered those elements (if any) satisfying
Such an element is called best coapproximant of in . The set of all such elements, satisfying above inequality, is denoted by . The subset is called coproximinal in if is nonempty for any . If is singleton for any , then is called coChebyshev; see [2–4].
It is clear that is convex if is convex and closed. The kernal of is the set defined by
Many results in the theory of best coapproximation have appeared since Franchetti and Furri’s paper, 1972. These results are largely concerned with the question of existence and uniqueness of best coapproximation (see for example [5–8]). Let and denote the Banach spaces of all bounded (resp., continuous) functions from a topological space into a Banach space and let be a closed subset of . It should be remarked that if is a compact space, then is a subspace of . In this paper, we discuss the coproximinality of and in and , respectively. For uniqueness and existence of best coapproximation in , see [7].
Definition 1. Let be a coproximinal subset of a Banach space . A map which associates with each element in one of its best coapproximation in (i.e., for all ) is called a coproximity map.
Remark 2. Let be a coproximinal subset of a Banach space . We state some basic properties of a proximity map .(1)If is coChebyshev, [5], then(a), for any .(b), for any scalar and . (i.e., is homogeneous).(c), for any and .(2)If is a subspace of , then(a), for any and .(b), for all . (set and in (a) and take into account that ).(c) is continuous at .(d)If is linear, then is continuous.
Theorem 3 (see [3]). If is coproximinal hyperplane or -dimensional subspace of a Banach space , then has a continuous coproximity map.
Lemma 4 (see [3]). Let be a coproximinal subspace of . Then the following are equivalent.(1) has linear coproximity map.(2) contains a closed subspace such that . Moreover, if (2) holds, then the linear coproximity map for can be defined by , , and .
Definition 5. Let be a Banach space. Consider the following.(a)A subspace of is called one complemented in if there is a closed subspace such that and the projection is contractive.(b)A linear projection is called an -projection if for all . A closed subspace of is called an -summand if it is the range of an -projection, [9].(c)A linear projection is called an -projection if for all . A closed subspace of is called an -summand if it is the range of an -projection.
Clearly every -summand is one complemented.
Theorem 6. Let be a subspace of a Banach space . Then the following are equivalent.(1) is one complemented in .(2) is coproximinal in and has linear coproximity map.
Proof. : Let , a contractive projection, and . Then ,
Hence is coproximinal in . Now let . Then for any , . Thus and has linear coproximity map by Lemma 4.
: If has a linear coproximity map, then contains a closed subspace such that . Moreover, , is a linear coproximity map for and so is the projection of along by Lemma 4. To show that is contractive, if , then for all , and so which means that is contractive.
Corollary 7. If is -summand, then is coproximinal and has continuous coproximity map.
For more information about coproximinal sets, optimal sets, and contractive sets, the reader is referred to [10–18].
2. Coproximinality in and
Let be a compact Hausdorff space and be a Banach space. We denote and to be the Banach space of all bounded ( resp., continuous) functions from into .
Lemma 8 (see [19]). Let be a continuous map of a Banach space into a closed subset of . If is compact Hausdorff space, then the map defined by is continuous from to .
Lemma 9. Let be a closed subset of a Banach space . If there is a continuous coproximity map of onto , then is coproximinal in and in fact it has a continuous coproximity map.
Proof. Let be a continuous coproximity map for . Define by , then is continuous by Lemma 8. Let , then for all ,
Hence for all and is a continuous coproximity map from onto .
Lemma 10. Let be a closed subset of a Banach space . If is coproximinal in , then is coproximinal in . Moreover, if has a continuous coproximity map, then has a continuous coproximity map.
Proof. For , define by for all . The map . By assumption, there exists such that for all . In particular, for all . For any fixed , for all and so is a best coapproximation of . Now let be a continuous coproximity map. For any fixed , define by . Then is a proximity map. Suppose that in . Then . and
Hence is continuous.
Theorem 11. Let be a closed subset of a Banach space . Then the following are equivalent.(1) is coproximinal in and has a continuous coproximity map.(2) is coproximinal in and has a continuous coproximity map.
Proof. The proof follows from Lemmas 9 and 10.
Theorem 12. Let be a closed subset of a Banach space and let be a topological space. Then the following are equivalent.(1) is coproximinal in .(2) is coproximinal in .
Proof. . Let . Since is coproximinal in , then is nonempty for any . By axiom of choice, we may define a function such that for all . Since for all and is a best coapproximation of in .
. For , define by for all . Since is coproximinal in , the map . By assumption, there exists such that for all . In particular, for all . For any fixed , for all and is a best coapproximation of .
We recall that.
Definition 13. Let be a set valued mapping, taking points of a topological space into the family of all subsets of a topological space . The mapping is said to be lower semicontinuous if, for each open set in , the set is open.
Theorem 14 (see [19, Michael Selection Theorem]). Let be a lower semicontinuous of a paracompact space into the family of nonvoid closed convex subsets of a Banach space . Then has a continuous selection; that is, there exists a continuous map such that for all .
Definition 15. A sequence of sets in a topological space is convergent and has the limit if and only if where is the set of those elements which are limits of points in for infinitely many and is the set of those elements which are limits of points in for cofinitely many (for all but finitely many) .
Lemma 16. Let be a compact Hausdorff space, a Banach space, and a compact subset of . For , we define by
Then is lower semicontinuous.
Proof. Let be an open set in . We need to prove that is open. For this purpose, let be a sequence in and as , so for all which means that Thus for all , and is a sequence of subsets of . We shall show that as . First, we show that Let . Then there is a sequence of points and a subsequence of such that and as . By the definition of , we get Taking and , and since is continuous, we have so . This proves that . Now we shall show that . Let . In this case, if as then for infinitely many , and, so, We may assume that, without loss of generality, as , for some , since is compact. Taking , we have that so and we have as a conclusion the follwing result: Thus, . Finally, we know that is a closed set and the members of the sequence of sets are all subsets of for all . Now, to complete the proof, we still need to show that To do so, let . By the definition of , there is a sequence of points and a subsequence of the sequence such that Since is closed, we get . This proves that . This means that , and so . Thus is open set and so is lower semicontinuous.
Theorem 17. Let be a compact convex subset of a Banach space and be a compact Hausdorff space. Then the following are equivalent.(1) is coproximinal in .(2) is coproximinal in .
Proof. . The proof is similar to that given in Lemma 10.
. Let and as defined in Lemma 16. Then is lower semicontinuous by Lemma 16. Since is closed, nonvoid, and convex, then is closed, nonvoid, and convex. By Michael Selection Theorem, has a continuous selection . Then and
Hence is a best coapproximation of in .
Corollary 18. Let be a compact Hausdorff space and let be a reflexive Banach space. Then (the closed unit ball of ) is coproximinal in if and only if is coproximinal in .
Proof. Since is reflexive, then is compact by Alaoglu Theorem. As is closed and convex subset of , the result follows from Theorem 17.
Problem 19. If is a compact Hausdorff space and is a Banach space, when is coproximinal in .
Finally the author would like to thank the referees for their valuable advice.
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Copyright © 2014 Eyad Abu-Sirhan and Zuhier Altawallbeh. 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.
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Predictive driver controller to track longitudinal speed and lateral path - Simulink - MathWorks France
y=\frac{{K}_{ff}}{{v}_{nom}}{v}_{ref}+\frac{{K}_{p}{e}_{ref}}{{v}_{nom}}+\int \left(\frac{{K}_{i}{e}_{ref}}{{v}_{nom}}+{K}_{aw}{e}_{out}\right)dt+{K}_{g}\theta
y=\frac{{K}_{ff}\left(v\right)}{{v}_{nom}}{v}_{ref}+\frac{{K}_{p}\left(v\right){e}_{ref}}{{v}_{nom}}+\int \left(\frac{{K}_{i}\left(v\right){e}_{ref}}{{v}_{nom}}+{K}_{aw}{e}_{out}\right){e}_{ref}dt+{K}_{g}\left(v\right)\theta
\begin{array}{l}\text{where:}\\ \\ {e}_{ref}={v}_{ref}-v\\ {e}_{out}={y}_{sat}-y\\ \\ {y}_{sat}=\left\{\begin{array}{cc}-1& y<-1\\ y& -1\le y\le 1\\ 1& 1<y\end{array}\end{array}
H\left(s\right)=\frac{1}{{\tau }_{err}s+1}\text{ for }{\tau }_{err}>0
\begin{array}{l}{y}_{acc}=\left\{\begin{array}{cc}0& {y}_{sat}<0\\ {y}_{sat}& 0\le {y}_{sat}\le 1\\ 1& 1<{y}_{sat}\end{array}\\ \\ {y}_{dec}=\left\{\begin{array}{cc}0& {y}_{sat}>0\\ -{y}_{sat}& -1\le {y}_{sat}\le 0\\ 1& {y}_{sat}<-1\end{array}\end{array}
\begin{array}{l}{x}_{1}=U\\ {\stackrel{˙}{x}}_{1}={x}_{2}=\frac{{K}_{pt}}{m}+ vr-g\text{sin}\left(\gamma \right)+{F}_{r}{x}_{1}\\ \stackrel{˙}{y}=v+U\psi \\ \stackrel{˙}{v}=\left[-\frac{2\left({C}_{\alpha F}+{C}_{\alpha R}\right)}{mU}\right]v+\left[\frac{2\left(b{C}_{\alpha R}-a{C}_{\alpha F}\right)}{mU}-U\right]r+\left(\frac{2{C}_{\alpha F}}{m}\right){\delta }_{F}\\ \stackrel{˙}{r}=\left[\frac{2\left(b{C}_{\alpha R}-a{C}_{\alpha F}\right)}{IU}\right]v+\left[-\frac{2\left({a}^{2}{C}_{\alpha F}+{b}^{2}{C}_{\alpha R}\right)}{IU}\right]r+\left(\frac{2a{C}_{\alpha F}}{I}\right){\delta }_{F}\\ \stackrel{˙}{\psi }=r\end{array}
\begin{array}{l}\stackrel{˙}{x}=Fx+gu\\ \\ \text{where:}\\ \\ x=\left[\begin{array}{c}\begin{array}{c}{x}_{1}\\ {x}_{2}\\ y\\ v\\ r\end{array}\\ \psi \end{array}\right]\\ \\ F=\left[\begin{array}{cccccc}0& 1& 0& 0& 0& 0\\ \frac{{F}_{r}}{m}& 0& 0& 0& v& 0\\ 0& 0& 0& 1& 0& U\\ 0& 0& 0& -\frac{2\left({C}_{\alpha F}+{C}_{\alpha R}\right)}{mU}& \frac{2\left(b{C}_{\alpha R}-a{C}_{\alpha F}\right)}{mU}-U& 0\\ 0& 0& 0& \frac{2\left(b{C}_{\alpha R}-a{C}_{\alpha F}\right)}{IU}& -\frac{2\left({a}^{2}{C}_{\alpha F}+{b}^{2}{C}_{\alpha R}\right)}{IU}& 0\\ 0& 0& 0& 0& 1& 0\end{array}\right]\\ \\ g=\left[\begin{array}{c}\begin{array}{cc}0& 0\\ \frac{{K}_{pt}}{m}& 0\\ 0& 0\\ 0& \frac{2{C}_{\alpha F}}{m}\\ 0& \frac{2a{C}_{\alpha F}}{I}\\ 0& 0\end{array}\end{array}\right]\\ \\ u=\left[\begin{array}{c}\begin{array}{c}\overline{u}\\ {\delta }_{F}\end{array}\end{array}\right] \\ \\ \overline{u}=u- \frac{{m}^{2}}{{K}_{pt}}g\text{sin}\left(\gamma \right)\end{array}
\begin{array}{l}{a}^{*}=\left({T}^{*}\right){m}^{T}\left[I+\sum _{n=1}^{\infty }\frac{{F}^{n}{\left({T}^{*}\right)}^{n}}{\left(n+1\right)!}\right]g\\ {b}^{*}={m}^{T}\left[I+\sum _{n=1}^{\infty }\frac{{F}^{n}{\left({T}^{*}\right)}^{n}}{n!}\right]\\ {m}^{T}=\left[\begin{array}{cccccc}1& 1& 1& 0& 0& 0\end{array}\right]\end{array}
J=\frac{1}{T}{\int }_{t}^{t+T}{\left[f\left(\eta \right)-y\left(\eta \right)\right]}^{2}d\eta
\frac{dJ}{du}=0
{u}^{o}\left(t\right)=u\left(t\right)+\frac{e\left(t+{T}^{*}\right)}{{a}^{*}}
{T}^{*}=\frac{L}{U}
H\left(s\right)={e}^{-s\tau }
\underset{0}{\overset{t}{\int }}{e}_{ref}{}^{2}dt
\mathrm{max}\left({e}_{ref}\left(t\right)\right)
\mathrm{min}\left({e}_{ref}\left(t\right)\right)
\underset{0}{\overset{t}{\int }}{e}_{ref}{}^{2}dt
\mathrm{max}\left({e}_{ref}\left(t\right)\right)
\mathrm{min}\left({e}_{ref}\left(t\right)\right)
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If I have a such equation to solve: \sin 3x=\frac12
If I have a such equation to solve:\sin 3x=\frac12
guringpw 2021-12-26 Answered
If I have a such equation to solve:
\mathrm{sin}3x=\frac{1}{2}
\left[0,2\pi \right]\text{ }\text{ },\mathrm{sin}\frac{\pi }{6}=\mathrm{sin}\frac{5\pi }{6}=\frac{12}{}
\mathrm{sin}\left(3x\right)=\mathrm{sin}\frac{\pi }{6}=\mathrm{sin}\frac{5\pi }{6}
Also sine has periodicity of
2\pi
\mathrm{sin}\left(2n\pi +\theta \right)=\mathrm{sin}\theta
\mathrm{sin}\left(3x\right)=\mathrm{sin}\frac{\pi }{6}=\mathrm{sin}\left(2n\pi +\frac{\pi }{6}\right)⇒3x=2n\pi +\frac{\pi }{6}
\mathrm{sin}\left(3x\right)=\mathrm{sin}\frac{5\pi }{6}=\mathrm{sin}\left(2n\pi +\frac{5\pi }{6}\right)⇒3x=2n\pi +\frac{5\pi }{6}
\mathrm{sin}x=\mathrm{sin}\alpha ⇒x=n\pi +{\left(-1\right)}^{n}\alpha
n=0,±±1,±2,±3,\dots
So here
\mathrm{sin}3x=\mathrm{sin}\frac{\pi }{6}⇒3x=n\pi +{\left(-1\right)}^{n}\frac{\pi }{3}⇒x=\frac{n\pi }{3}+{\left(-1\right)}^{n}\frac{\pi }{9},\text{ }\text{ }\text{ }n=0,±1,±3,\dots
\mathrm{sin}\left(x\right)
is periodic, as in it has a period of
2\pi
. Therefore, it will intersect the same y-value every period, or every
2n\pi
where n is an integer. By scaling
\theta
by 3, you are actually scaling the period of
\mathrm{sin}\theta
by 13. Thus the new period is
\frac{2\pi }{3}
. You can see this in action with your solution you gave, which was
\mathrm{sin}\theta =\frac{1}{2}
\theta ={\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)
Here's the kicker:
\mathrm{sin}\theta
intersects 12 twice every period, so there are two solutions. By inspecting the unit circle, find why this must be true. The two solutions are
3\theta =\frac{\pi }{6}
3\theta =\frac{5\pi }{6}
But we want to know the solutions for the entire domain
\left(-\mathrm{\infty },\mathrm{\infty }\right)
, so we want
3\theta =\frac{\pi }{6}+2n\pi
3\theta =\frac{5\pi }{6}+2n\pi
And now by dividing by 3, you can see why your period is scaled by
\frac{1}{3}
(as well as the solution):
\theta =\frac{\pi }{18}+\frac{2n\pi }{3}
\theta =\frac{5\pi }{18}+\frac{2n\pi }{3}
An elevator (mass 4850 kg) is to be designed so that themaximum acceleration is 0.0600g. What is the maximum and the minimum forces the motor should exert on the supportingcable?
I already know how to find the maximum force (just multiplyingthe maximum acceleration by the mass), but I've got no clue as tohow to find the minimum acceleration when no minimum accelerationhas been found.
A skier of mass 70 kg is pulled up a slope by a motor driven cable. (a) How much work is required to pull him 60m up a 30 degree slope (assumed frictionless) at a constant speed of 2.0 m/s ?(b) What power must a motor have to perform this task?
A research submarine has a 20-cm-diameter window 8.0 cm thick. The manufacturer says the window can withstand forces up to
1.0×{10}^{6}
N. What is the submarine's maximum safe depth in seawater? The pressure inside the submarine is maintaned at 1.0 atm
A morsel of food with a mass of 4.2 g is injected into theesophagus with an initial speed of
2.5cm/s
. On the way down to thestomach, the walls of the esophagus exert an upward resistive forceof
2.7×{10}^{-3}N
on the morsel. If the esophagus is 20.0 cm long,with what speed does the morsel of food enter the stomach? I need some help setting this problem up.
A piano has been pushed to the top of the ramp at the back of a moving van. The workers think it is safe, but as they walk away, it begins to roll down the ramp. If the back of the truck is 1.0 m above the ground and the ramp is inclined at
{20}^{\circ }
, how much time do the workers have to get to the piano before it reaches the bottom of the ramp?
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What is the antiderivative of \sec(x)?
\mathrm{sec}\left(x\right)
\int \mathrm{sec}xdx
\frac{\mathrm{sec}\left(x\right)+\mathrm{tan}\left(x\right)}{\mathrm{sec}\left(x\right)+\mathrm{tan}\left(x\right)}
\int \frac{\mathrm{sec}\left(x\right)\left(\mathrm{sec}\left(x\right)+\mathrm{tan}\left(x\right)\right)}{\mathrm{sec}\left(x\right)+\mathrm{tan}\left(x\right)}dx=
=\int \frac{{\mathrm{sec}}^{2}\left(x\right)+\mathrm{sec}\left(x\right)\mathrm{tan}\left(x\right)}{\mathrm{sec}\left(x\right)+\mathrm{tan}\left(x\right)}dx=
=\int \frac{1}{u}du=\mathrm{ln}|u|+C=\mathrm{ln}|\mathrm{sec}\left(x\right)+\mathrm{tan}\left(x\right)|+C
Is there any other easier explanation?
f\left(x\right)={x}^{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}g\left(x\right)=\sqrt{x-5}
, to find:
\left(fog\right)\left(x\right)
2. Domain of
\left(fog\right)\left(x\right)
\left(gof\right)\left(x\right)
\left(gof\right)\left(x\right)
f\left(x\right)={x}^{2},\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}x\le 4\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}f\left(x\right)=m+b,\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}x>4
I need a review or step by step explain how these two problems work.
\mathrm{cos}\left(14\pi -0\right)
Determine the limit as x approaches the given x-coordinate and the continuity of the function at the that x-coordinate.
\underset{x\to 15\frac{\pi }{8}}{lim}\mathrm{cos}\left(4x-\mathrm{cos}\left(4x\right)\right)
f\left(x\right)={x}^{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}g\left(x\right)=\sqrt{x-5}
1. (fog)(x)
2. Domain of (fog)(x)
3. (gof)(x)
4. Domain of (gof)(x)
f\left(x\right)=-\mathrm{sin}\left(8x\right)+6\mathrm{cos}\left(4x\right)-8x
p=\mathrm{sin}\theta \mathrm{sin}\varphi
Find the arc length of the astroid
x\left(t\right)={\mathrm{cos}}^{3}t,\text{ }y\left(t\right)={\mathrm{sin}}^{3}t
Calculate the arc length of
\frac{1}{4}
of the astroid
\left(0\le t\le \frac{\pi }{2}\right)
|
Find the derivative of the following functions y=2^{3+\sin x}
y={2}^{3+\mathrm{sin}x}
Consider the provided function,
y={2}^{3+\mathrm{sin}x}
Find the derivative of the above functions.
First, we taking log both the sides.
\mathrm{ln}y=\mathrm{ln}\left({2}^{3+\mathrm{sin}x}\right)
Apply the log property
\mathrm{ln}\left({m}^{n}\right)=n\text{ }\mathrm{ln}\left(m\right)
\mathrm{ln}y=\mathrm{ln}\left({2}^{3+\mathrm{sin}x}\right)
\mathrm{ln}y=\left(3+\mathrm{sin}x\right)\mathrm{ln}\left(2\right)
Now, the above equation differentiate with respect to x.
Apply the common derivative rule
\frac{d}{dx}\left(\mathrm{ln}x\right)=\frac{1}{x}\cdot 1\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }\frac{d}{dx}\left(\mathrm{sin}x\right)=\mathrm{cos}x
\frac{d}{dx}\left(\mathrm{ln}y\right)=\frac{d}{dx}\left[\left(3+\mathrm{sin}x\right)\mathrm{ln}\left(2\right)\right]
\frac{1}{y}\frac{dy}{dx}=\mathrm{ln}\left(2\right)\frac{d}{dx}\left[\left(3+\mathrm{sin}x\right)\right]
\frac{1}{y}\frac{dy}{dx}=\mathrm{ln}\left(2\right)\cdot \left(0+\mathrm{cos}x\right)
\frac{1}{y}\frac{dy}{dx}=\mathrm{ln}\left(2\right)\cdot \mathrm{cos}x
\frac{dy}{dx}=y\cdot \mathrm{ln}\left(2\right)\cdot \mathrm{cos}x
Now, in the equation (1) we put the value of y in the above derivative.
\frac{dy}{dx}=\left({2}^{3+\mathrm{sin}x}\right)\cdot \mathrm{ln}\left(2\right)\cdot \mathrm{cos}x
=\mathrm{ln}\left(2\right)\left({2}^{3+\mathrm{sin}x}\right)\mathrm{cos}x
f\left(x\right)={e}^{2x+7}
g\left(x\right)=\sqrt{3x+1}
\frac{\mathrm{ln}\left(x\right)}{{e}^{x}}
Use formulas (1) and (2) and the power rule to find the derivatives of
f\left(x\right)=\frac{{x}^{2}}{3}
Derivatives of vector-valued functions Differentiate the following function.
r\left(t\right)=<4,3\mathrm{cos}2t,2\mathrm{sin}3t>
f\left(0\right)={\left(\frac{\mathrm{sin}0}{1+\mathrm{cos}0}\right)}^{2}
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Find derivatives for the functions. Assume a, b, c, and k are constants. g(x)=5x^{4}
Find derivatives for the functions. Assume a, b, c, and k are constants.g(x)=5x^{4}
g\left(x\right)=5{x}^{4}
{g}^{\prime }\left(x\right)={\left(5{x}^{4}\right)}^{\prime }
=5{\left({x}^{4}\right)}^{\prime }
=5\stackrel{˙}{4}{x}^{3}
=20{x}^{3}
f\left(x\right)={e}^{x}\mathrm{ln}|x|
Calculate the second-order partial derivatives off
\left(x,y\right)={x}^{3}+{y}^{2}{e}^{x}
On a frictionless, horizontal air table, puck A (with mass 0.250 kg) is moving toward puck B (with mass 0.350 kg), which is initially at rest. After the collision, puck A has a velocity of 0.120 m/s to the left, and puck B has a velocity of 0.650 m/s to the right.
(a) What was the speed of puck A before the collision?
(b) Calculate the change in the total kinetic energy of the system that occurs during the collision.
Find the derivatives of the given functions.
w\left(x\right)=\mathrm{sec}x\mathrm{tan}\left({x}^{2}-1\right)
z={e}^{xy}
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Momentum: Level 1-2 Challenges Practice Problems Online | Brilliant
Two billiard balls of equal mass are rolled in such a way that their velocities have the same magnitude but are in opposite directions.
Which of the following is an impossible outcome for the velocities of the billiard balls after a collision?
Note: Collision may or may not be elastic.
A B C All scenarios are possible
You went to a park in winter to practice ice skating on a frozen lake, but because you're a beginner, you asked your friend to give you some impulse to start.
He applied a force of 150 Newtons to you during 1.2 seconds, after which you began moving with a constant velocity.
Another person is skating toward you with a velocity of 5 m/s. Unable to stop, you both collide and both stop moving.
Suppose that you have a mass of 60kg. Find the mass (in kilograms) of the person who collided with you.
Details and assumptions: Because the lake is frozen, the icy surface causes negligible friction.
A cricket ball of mass 150g moving with a speed of 126km/h hits at the middle of the bat, held firmly at its position by the batsman. The ball moves straight back to the bowler after hitting the bat. Assuming that collision between ball and bat is completely elastic and the two remain in contact for 0.001s, what is the magnitude of the average force that the batsman had to apply in order to hold the bat firmly in its place?
1.05 x 10^4 N 21 N 2.1 x 10^4 N 10.5 N
m
is moving along the
x
-axis with speed
v
when it collides with a body of mass
2m
initially at rest. After the collision, the first body comes to rest and the second body splits into two pieces of equal mass that move with equal velocities at equal angles off the
x
0 < \theta < 90^\circ
Which of the following statements correctly describes the speeds of the two pieces?
Each piece moves with speed v. Each piece moves with speed less than v/2. Each piece moves with speed v/2. Each piece moves with speed greater than v/2.
Mr. Andrew excites his students with his home-made cannon demonstration. The 1.27kg cannon is loaded with a 54 grams cricket ball and placed on the ground. Mr. Andrew fires the cannonball horizontally and determines that the cannon recoiled backwards with a speed of 7.8m/s with help of a photogate measurement and asks his students to determine the speed of the ball. So, what is the speed of the ball in m/s?
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Filter disturbances through vector error-correction (VEC) model - MATLAB filter - MathWorks Deutschland
Filter Disturbances Through VEC Model
Filter Multiple Disturbance Paths
Filter disturbances through vector error-correction (VEC) model
Y = filter(Mdl,Z)
Y = filter(Mdl,Z,Name,Value)
[Y,E] = filter(___)
Y = filter(Mdl,Z) returns the multivariate response series Y, which results from filtering the underlying multivariate disturbance series Z. The Z series are associated with the model innovations process through the fully specified VEC(p – 1) model Mdl.
Y = filter(Mdl,Z,Name,Value) uses additional options specified by one or more name-value arguments. For example, 'X',X,'Scale',false specifies X as exogenous predictor data for the regression component and refraining from scaling the disturbances by the lower triangular Cholesky factor of the model innovations covariance matrix.
[Y,E] = filter(___) returns the multivariate model innovations series E using any of the input arguments in the previous syntaxes.
Consider a VEC model for the following seven macroeconomic series. Then, fit the model to the data and filter disturbances through the fitted model.
Estimate the model using the entire data set and the default options. By default, estimate uses the first p = 2 observations as presample data.
Generate a numobs-by-7 series of random Gaussian distributed values, where numobs is the number of observations in the data minus p.
numobs = size(FRED,1) - Mdl.P;
Z = randn(numobs,Mdl.NumSeries);
To simulate responses, filter the disturbances through the estimated model. Specify the first p = 2 observations as presample data.
Y = filter(EstMdl,Z,'Y0',FRED{1:2,:});
Y is a 238-by-7 matrix of simulated responses. Columns correspond to the variable names in EstMdl.SeriesNames.
plot(FRED.Time(3:end),[FRED.GDP(3:end) Y(:,1)]);
legend('Simulation','True','Location','Best')
plot(FRED.Time(3:end),[FRED.GDPDEF(3:end) Y(:,2)]);
plot(FRED.Time(3:end),[FRED.COE(3:end) Y(:,3)]);
plot(FRED.Time(3:end),[FRED.HOANBS(3:end) Y(:,4)]);
plot(FRED.Time(3:end),[FRED.FEDFUNDS(3:end) Y(:,5)]);
plot(FRED.Time(3:end),[FRED.PCEC(3:end) Y(:,6)]);
plot(FRED.Time(3:end),[FRED.GPDI(3:end) Y(:,7)]);
\begin{array}{rcl}\Delta {y}_{t}& =& c+A{B}^{\prime }{y}_{t-1}+{\Phi }_{1}\Delta {y}_{t-1}+{\epsilon }_{t}\\ & =& \\ & =& \left[\begin{array}{c}-1\\ -3\\ -30\end{array}\right]+\left[\begin{array}{cc}-0.3& 0.3\\ -0.2& 0.1\\ -1& 0\end{array}\right]\left[\begin{array}{ccc}0.1& -0.2& 0.2\\ -0.7& 0.5& 0.2\end{array}\right]{y}_{t-1}+\left[\begin{array}{ccc}0& 0.1& 0.2\\ 0.2& -0.2& 0\\ 0.7& -0.2& 0.3\end{array}\right]\Delta {y}_{t-1}+{\epsilon }_{t}.\end{array}
\Sigma =\left[\begin{array}{ccc}1.3& 0.4& 1.6\\ 0.4& 0.6& 0.7\\ 1.6& 0.7& 5\end{array}\right].
Generate 1000 paths of 100 observations from a 3-D Gaussian distribution. numobs is the number of observations in the data without any missing values.
Z = randn(numobs,Mdl.NumSeries,numpaths);
Filter the disturbances through the estimated model. Return the innovations (scaled disturbances).
[Y,E] = filter(Mdl,Z);
Y and E are 100-by-3-by-1000 matrices of filtered responses and scaled disturbances, respectively.
For each time point, compute the mean vector of the filtered responses among all paths.
MeanFilt = mean(Y,3);
MeanFilt is a 100-by-3 matrix containing the average of the filtered responses at each time point.
Plot the filtered responses and their averages.
plot(squeeze(Y(:,j,:)),'Color',[0.8,0.8,0.8])
plot(MeanFilt(:,j));
Z — Underlying multivariate disturbance series
Underlying multivariate disturbance series associated with the model innovations process, specified as a numobs-by-numseries numeric matrix or a numobs-by-numseries-by-numpaths numeric array.
numobs is the sample size. numseries is the number of disturbance series (Mdl.NumSeries). numpaths is the number of disturbance paths.
Rows correspond to sampling times, and the last row contains the latest set of disturbances.
Columns correspond to individual disturbance series for response variables.
Pages correspond to separate, independent paths. For a numeric matrix, Z is a single numseries-dimensional path of disturbance series. For a 3-D array, each page of Z represents a separate numseries-dimensional path. Among all pages, disturbances in corresponding rows occur at the same time.
The 'Scale' name-value pair argument specifies whether to scale the disturbances before filter filters them through Mdl. For more details, see Scale.
Example: 'Scale',false,'X',X does not scale Z by the lower triangular Cholesky factor of the model covariance matrix before filtering, and uses the matrix X as predictor data in the regression component.
Presample responses that provide initial values for the model Mdl, specified as the comma-separated pair consisting of 'Y0' and a numpreobs-by-numseries numeric matrix or a numpreobs-by-numseries-by-numprepaths numeric array.
Rows correspond to presample observations, and the last row contains the latest presample observation. Y0 must have at least Mdl.P rows. If you supply more rows than necessary, filter uses the latest Mdl.P observations only.
If Y0 is a matrix, then filter applies it to each path (page) in Y. Therefore, all paths in Y derive from common initial conditions.
Otherwise, filter applies Y0(:,:,j) to Y(:,:,j). Y0 must have at least numpaths pages, and filter uses only the first numpaths pages.
By default, filter sets any necessary presample observations.
For stationary VAR processes without regression components, filter uses the unconditional mean
\mu ={\Phi }^{-1}\left(L\right)c.
For nonstationary processes or models containing a regression component, filter sets presample observations to an array composed of zeros.
Rows correspond to observations, and the last row contains the latest observation. X must have at least as many observations as Z. If you supply more rows than necessary, filter uses only the latest observations. filter does not use the regression component in the presample period.
filter applies X to each path (page) in Z; that is, X represents one path of observed predictors.
By default, filter excludes the regression component, regardless of its presence in Mdl.
Scale — Flag indicating whether to scale disturbances
Flag indicating whether to scale disturbances by the lower triangular Cholesky factor of the model covariance matrix, specified as the comma-separated pair consisting of 'Scale' and true or false.
For each page j = 1,...,numpaths, filter filters the numobs-by-numseries matrix of innovations E(:,:,j) through the VAR(p) model Mdl, according to these conditions.
If Scale is true, then E(:,:,j) = L*Z(:,:,j) and L = chol(Mdl.Covariance,'lower').
If Scale is false, then E(:,:,j) = Z(:,:,j).
Example: 'Scale',false
NaN values in Z, Y0, and X indicate missing values. filter removes missing values from the data by list-wise deletion.
If Z is a 3-D array, then filter horizontally concatenates the pages of Z to form a numobs-by-numpaths*numseries matrix.
If a regression component is present, then filter horizontally concatenates X to Z to form a numobs-by-(numpaths*numseries + numpreds) matrix. filter assumes that the last rows of each series occur at the same time.
filter removes any row that contains at least one NaN from the concatenated data.
filter applies steps 1 and 3 to the presample paths in Y0.
This process ensures that the filtered responses and innovations of each path are the same size and are based on the same observation times. In the case of missing observations, the results obtained from multiple paths of Z can differ from the results obtained from each path individually.
Y — Filtered multivariate response series
Filtered multivariate response series, returned as a numobs-by-numseries numeric matrix or a numobs-by-numseries-by-numpaths numeric array. Y represents the continuation of the presample responses in Y0.
E — Multivariate model innovations series
Multivariate model innovations series, returned as a numobs-by-numseries numeric matrix or a numobs-by-numseries-by-numpaths numeric array. For details on the value of E, see Scale.
filter computes Y and E using this process for each page j in Z.
If Scale is true, then E(:,:,j) = L*Z(:,:,j), where L = chol(Mdl.Covariance,'lower'). Otherwise, E(:,:,j) = Z(:,:,j). Set et = E(:,:,j).
Y(:,:,j) is yt in this system of equations.
\Delta {y}_{t}={\stackrel{^}{\Phi }}^{-1}\left(L\right)\left(\stackrel{^}{c}+\stackrel{^}{d}t+\stackrel{^}{A}\stackrel{^}{B}\prime {y}_{t-1}+\stackrel{^}{\beta }{x}_{t}+{e}_{t}\right).
For variable definitions, see Vector Error-Correction Model.
filter generalizes simulate. Both functions filter a disturbance series through a model to produce responses and innovations. However, whereas simulate generates a series of mean-zero, unit-variance, independent Gaussian disturbances Z to form innovations E = L*Z, filter enables you to supply disturbances from any distribution.
filter uses this process to determine the time origin t0 of models that include linear time trends.
Otherwise, filter sets t0 to size(Y0,1) – Mdl.P. Therefore, the times in the trend component are t = t0 + 1, t0 + 2,..., t0 + numobs, where numobs is the effective sample size (size(Y,1) after filter removes missing values). This convention is consistent with the default behavior of model estimation in which estimate removes the first Mdl.P responses, reducing the effective sample size. Although filter explicitly uses the first Mdl.P presample responses in Y0 to initialize the model, the total number of observations in Y0 and Y (excluding missing values) determines t0.
estimate | simulate | infer
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Data types and features - Upsolver
This page provides an overview of the data types supported by different outputs. It also indicates whether or not the output supports upserts, aggregations, and SQL.
When creating an output, these specific data types are supported. Additionally, upserts, aggregations, and SQL are supported for most of the outputs.
^{6}
^{6}
DECIMAL(18,6) See: Note 6
^{6}
^{6}
See: Note 1 and Note 6
^{6}
See: Note 6
^{6}
^{6}
The type is selected based on the output types. You can cast fields to get any types supported by the output format.
You cannot choose the destination data type as it is fetched from the original database.
Supported by definition. If a Primary Key (PK) is configured in the database, we will only have the latest record per PK.
Only aggregations by key are supported for lookup tables.
DOUBLE, BIGINT and Upsolver NUMBER data types are limited to 64 bits.
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Friedman number - Wikipedia
A Friedman number is an integer, which represented in a given numeral system, is the result of a non-trivial expression using all its own digits in combination with any of the four basic arithmetic operators (+, −, ×, ÷), additive inverses, parentheses, exponentiation, and concatenation. Here, non-trivial means that at least one operation besides concatenation is used. Leading zeros cannot be used, since that would also result in trivial Friedman numbers, such as 024 = 20 + 4. For example, 347 is a Friedman number in the decimal numeral system, since 347 = 73 + 4. The decimal Friedman numbers are:
25, 121, 125, 126, 127, 128, 153, 216, 289, 343, 347, 625, 688, 736, 1022, 1024, 1206, 1255, 1260, 1285, 1296, 1395, 1435, 1503, 1530, 1792, 1827, 2048, 2187, 2349, 2500, 2501, 2502, 2503, 2504, 2505, 2506, 2507, 2508, 2509, 2592, 2737, 2916, ... (sequence A036057 in the OEIS).
Friedman numbers are named after Erich Friedman, a now-retired mathematics professor at Stetson University, located in DeLand, Florida.
A Friedman prime is a Friedman number that is also prime. The decimal Friedman primes are:
127, 347, 2503, 12101, 12107, 12109, 15629, 15641, 15661, 15667, 15679, 16381, 16447, 16759, 16879, 19739, 21943, 27653, 28547, 28559, 29527, 29531, 32771, 32783, 35933, 36457, 39313, 39343, 43691, 45361, 46619, 46633, 46643, 46649, 46663, 46691, 48751, 48757, 49277, 58921, 59051, 59053, 59263, 59273, 64513, 74353, 74897, 78163, 83357, ... (sequence A112419 in the OEIS).
1 Results in base 10
2 Finding 2-digit Friedman numbers
3.2 Duodecimal
4 Using Roman numerals
Results in base 10[edit]
The expressions of the first few Friedman numbers are:
number expression number expression number expression number expression
25 52 127 27−1 289 (8+9)2 688 8×86
121 112 128 28−1 343 (3+4)3 736 36+7
125 51+2 153 3×51 347 73+4 1022 210−2
126 6×21 216 62+1 625 56−2 1024 (4−2)10
A nice Friedman number is a Friedman number where the digits in the expression can be arranged to be in the same order as in the number itself. For example, we can arrange 127 = 27 − 1 as 127 = −1 + 27. The first nice Friedman numbers are:
127, 343, 736, 1285, 2187, 2502, 2592, 2737, 3125, 3685, 3864, 3972, 4096, 6455, 11264, 11664, 12850, 13825, 14641, 15552, 15585, 15612, 15613, 15617, 15618, 15621, 15622, 15623, 15624, 15626, 15632, 15633, 15642, 15645, 15655, 15656, 15662, 15667, 15688, 16377, 16384, 16447, 16875, 17536, 18432, 19453, 19683, 19739 (sequence A080035 in the OEIS).
A nice Friedman prime is a nice Friedman number that's also prime. The first nice Friedman primes are:
127, 15667, 16447, 19739, 28559, 32771, 39343, 46633, 46663, 117619, 117643, 117763, 125003, 131071, 137791, 147419, 156253, 156257, 156259, 229373, 248839, 262139, 262147, 279967, 294829, 295247, 326617, 466553, 466561, 466567, 585643, 592763, 649529, 728993, 759359, 786433, 937577 (sequence A252483 in the OEIS).
Friedman's website shows around 100 zeroless pandigital Friedman numbers as of April 2020[update]. Two of them are: 123456789 = ((86 + 2 × 7)5 − 91) / 34, and 987654321 = (8 × (97 + 6/2)5 + 1) / 34. Only one of them is nice: 268435179 = −268 + 4(3×5 − 17) − 9.
Michael Brand proved that the density of Friedman numbers among the naturals is 1,[1] which is to say that the probability of a number chosen randomly and uniformly between 1 and n to be a Friedman number tends to 1 as n tends to infinity. This result extends to Friedman numbers under any base of representation. He also proved that the same is true also for binary, ternary and quaternary nice Friedman numbers.[2] The case of base-10 nice Friedman numbers is still open.
Vampire numbers are a subset of Friedman numbers where the only operation is a multiplication of two numbers with the same number of digits, for example 1260 = 21 × 60.
Finding 2-digit Friedman numbers[edit]
There usually are fewer 2-digit Friedman numbers than 3-digit and more in any given base, but the 2-digit ones are easier to find. If we represent a 2-digit number as mb + n, where b is the base and m, n are integers from 0 to b−1, we need only check each possible combination of m and n against the equalities mb + n = mn, and mb + n = nm to see which ones are true. We need not concern ourselves with m + n or m × n, since these will always be smaller than mb + n when n < b. The same clearly holds for m − n and m / n.
General results[edit]
{\displaystyle b=mk-m}
{\displaystyle b^{2}+mb+k=(mk-m+m)b+k=mbk+k=k(mb+1)}
is a Friedman number (written in base
{\displaystyle b}
as 1mk = k × m1).[3]
{\displaystyle b>2}
{\displaystyle {(b^{n}+1)}^{2}=b^{2n}+2{b^{n}}+1}
{\displaystyle b}
as 100...00200...001 = 100..0012, with
{\displaystyle n-1}
zeroes between each nonzero number).[3]
{\displaystyle b={\frac {k(k-1)}{2}}}
{\displaystyle 2b+k=2\left({\frac {k(k-1)}{2}}\right)+k=k^{2}-k+k=k^{2}}
{\displaystyle b}
as 2k = k2). From the observation that all numbers of the form 2k × b2n can be written as k000...0002 with n 0's, we can find sequences of consecutive Friedman numbers which are arbitrarily long. For example, for
{\displaystyle k=5}
, or in base 10, 250068 = 5002 + 68, from which we can easily deduce the range of consecutive Friedman numbers from 250000 to 250099 in base 10.[3]
Repdigit Friedman numbers:
The smallest repdigit in base 8 that is a Friedman number is 33 = 33.
The smallest repdigit in base 10 that is thought to be a Friedman number is 99999999 = (9 + 9/9)9−9/9 − 9/9.[3]
It has been proven that repdigits with at least 22 digits are nice Friedman numbers.[3]
There are an infinite number of prime Friedman numbers in all bases, because for base
{\displaystyle 2\leq b\leq 6}
{\displaystyle n\times 10^{1111}+11111111=n\times 10^{1111}+10^{1000}-1+0+0}
{\displaystyle n\times 10^{102}+1101221=n\times 10^{102}+2^{101}+0+0}
{\displaystyle n\times 10^{20}+310233=n\times 10^{20}+33^{3}+0}
{\displaystyle n\times 10^{13}+2443111=n\times 10^{4+4}+(2\times 3)^{11}}
{\displaystyle n\times 10^{13}+25352411=n\times 10^{2\times 5-1}+(5+2)^{(3+4)}}
{\displaystyle 7\leq b\leq 10}
{\displaystyle n\times 10^{60}+164351=n\times 10^{60}+(10+4-3)^{5}+0+0+\ldots }
in base 7,
{\displaystyle n\times 10^{60}+163251=n\times 10^{60}+(10+3-2)^{5}+0+0+\ldots }
{\displaystyle n\times 10^{60}+162151=n\times 10^{60}+(10+2-1)^{5}+0+0+\ldots }
{\displaystyle n\times 10^{60}+161051=n\times 10^{60}+(10+1-0)^{5}+0+0+\ldots }
in base 10,
and for base
{\displaystyle b>10}
{\displaystyle n\times 10^{50}+{\text{15AA51}}=n\times 10^{50}+(10+{\text{A}}/{\text{A}})^{5}+0+0+\ldots }
are Friedman numbers for all
{\displaystyle n}
. The numbers of this form are an arithmetic sequence
{\displaystyle pn+q}
{\displaystyle p}
{\displaystyle q}
are relatively prime regardless of base as
{\displaystyle b}
{\displaystyle b+1}
are always relatively prime, and therefore, by Dirichlet's theorem on arithmetic progressions, the sequence contains an infinite number of primes.
Duodecimal[edit]
In base 12, the Friedman numbers less than 1000 are:
441 (4+1)4
Using Roman numerals[edit]
In a trivial sense, all Roman numerals with more than one symbol are Friedman numbers. The expression is created by simply inserting + signs into the numeral, and occasionally the − sign with slight rearrangement of the order of the symbols.
Some research into Roman numeral Friedman numbers for which the expression uses some of the other operators has been done. The first such nice Roman numeral Friedman number discovered was 8, since VIII = (V - I) × II. Other such nontrivial examples have been found.
The difficulty of finding nontrivial Friedman numbers in Roman numerals increases not with the size of the number (as is the case with positional notation numbering systems) but with the numbers of symbols it has. For example, it is much tougher to figure out whether 147 (CXLVII) is a Friedman number in Roman numerals than it is to make the same determination for 1001 (MI). With Roman numerals, one can at least derive quite a few Friedman expressions from any new expression one discovers. Since 8 is a nice nontrivial nice Roman numeral Friedman number, it follows that any number ending in VIII is also such a Friedman number.
^ Michael Brand, "Friedman numbers have density 1", Discrete Applied Mathematics, 161(16–17), Nov. 2013, pp. 2389-2395.
^ Michael Brand, "On the Density of Nice Friedmans", Oct 2013, https://arxiv.org/abs/1310.2390.
^ a b c d e "Math Magic".
Home page for Friedman numbers
Friedman numbers The On-Line Encyclopedia of Integer Sequences
Friedman numbers have density 1 Discrete Applied Mathematics, Vol 161, Issues 16–17, Nov 2013, pp 2389–2395
Pretty wild narcissistic numbers - numbers that pwn
Retrieved from "https://en.wikipedia.org/w/index.php?title=Friedman_number&oldid=1080135134"
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Write the solutions or, if there is no solution, say
Write the solutions or, if there is no solution, say the system is inconsistent. [ 1 0 0 -1 0 1 0 3 0 0 1 4 0 0 0 0 ]
\left[\begin{array}{cccc}1& 0& 0& -1\\ 0& 1& 0& 3\\ 0& 0& 1& 4\\ 0& 0& 0& 0\end{array}\right]
Bach variable is represented by a leading 1 in the corresponding column. The last row represents an equation that is always true (0=0), 20... the system is consistent and has exactly one solution.
(No parameters) Interpreting row by row as equations.
{P}_{5}
{P}_{5}
V={R}_{3}
\left({x}_{1},{x}_{2},{x}_{3}\right)
{x}_{1}-6{x}_{2}+{x}_{3}=5
V={R}^{n}
{C}^{2}\left(I\right)
{P}_{n}
{P}_{n}
V={M}_{n}\left(R\right)
Write the vector form of the general solution of the given system of linear equations.
-{x}_{1}+2{x}_{3}-5{x}_{4}+{x}_{5}-{x}_{6}=0
{6}^{3}×{6}^{-2}
\left[\begin{array}{ccccc}1& 0& -1& |& 1\\ 0& 1& 2& |& 1\end{array}\right]
x1+2x2−x3+x4=0
{x}_{1},{x}_{2}\dots
\left[1,2,0,2,-1,3\right]
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Demodulate orthogonal frequency division modulated data - Simulink - MathWorks Switzerland
OFDM Demodulator Baseband
Demodulate orthogonal frequency division modulated data
OFDM, in Digital Baseband sublibrary of Modulation
The Orthogonal Frequency Division Modulation (OFDM) Demodulator Baseband block demodulates an OFDM input signal. The block accepts a single input and has one or two output ports, depending on the status of Pilot output port.
Pilot Output Port
Pilot Carrier Indices
false N/A NCPTotal+NFFT×Nsym-by-Nr Ndata-by-Nsym-by-Nr N/A
true 2-D Npilot-by-Nsym-by-Nr
3-D Npilot-by-Nsym-by-Nt-by-Nr
Nsym represents the number of symbols as determined by Number of OFDM symbols.
Nr represents the number of receive antennas as determined by Number of receive antennas.
Npilot represents the number of pilot symbols determined by the second dimension in the Pilot subcarrier indices array.
Nt represents the number of transmit antennas. This parameter is derived from the third dimension of the Pilot subcarrier indices array.
Specify the FFT length, which is equivalent to the number of subcarriers. The length of the FFT, NFFT, must be greater than or equal to 8.
Assign the number of subcarriers to the left, NleftG, and right, NrightG, guard bands. The input is a 2-by-1 vector. The number of subcarriers must fall within [0,NFFT/2 − 1].
Remove DC carrier
Select to remove the DC subcarrier.
Select to separate the data from the pilot signal and output the demodulated pilot signal.
Specify the pilot subcarrier indices. This field is available only when the Pilot output port check box is selected. You can assign the indices can be assigned to the same or different subcarriers for each symbol. Similarly, the pilot carrier indices can differ across multiple transmit antennas. Depending on the desired level of control for index assignments, the dimensions of the indices’ array vary from 1 to 3. Valid pilot indices fall in the range
\left[{N}_{\text{leftG}}+1,\text{\hspace{0.17em}}{N}_{\text{FFT}}/2\right]\cup \left[{N}_{\text{FFT}}/2+2,\text{\hspace{0.17em}}{N}_{\text{FFT}}-{N}_{\text{rightG}}\right],
where the index value cannot exceed the number of subcarriers. If the number of transmit antennas is greater than one, ensure that the indices per symbol are mutually distinct across antennas to minimize interference.
Specify the length of the cyclic prefix. If you specify a scalar, the prefix length is the same for all symbols through all antennas. If you specify a row vector of length Nsym, the prefix length can vary across symbols but remains the same length through all antennas.
Specify the number of OFDM symbols, Nsym, in the time-frequency grid.
Specify the number of receive antennas, Nr, as a positive integer such that Nr ≤ 64.
Select the simulation type from these choices:
This block implements the algorithm, inputs, and outputs described in the OFDM Demodulator System object reference page. The object properties correspond to the block parameters.
[1] Dahlman, E., S. Parkvall, and J. Skold. 4G LTE/LTE-Advanced for Mobile Broadband.London: Elsevier Ltd., 2011.
[2] Andrews, J. G., A. Ghosh, and R. Muhamed. Fundamentals of WiMAX.Upper Saddle River, NJ: Prentice Hall, 2007.
QPSK Demodulator Baseband | Rectangular QAM Demodulator Baseband | OFDM Modulator Baseband
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How do I evaluate the power series \sum_{n=1}^\infty\frac{n}{9^n} without using the formula
kloseyq 2022-01-02 Answered
How do I evaluate the power series
\sum _{n=1}^{\mathrm{\infty }}\frac{n}{{9}^{n}}
without using the formula for infinite geometric series?
Check with induction that the partial sum is
\sum _{k=1}^{N}\frac{k}{{9}^{k}}=\frac{9-{9}^{-N}\left(8N+9\right)}{64}
Evaluate the limit when N goes to infinity and you get the result.
x\frac{d}{dx}\frac{1}{1-x}=\sum _{i=0}^{\mathrm{\infty }}i{x}^{i}
. So differenciating you get
x\frac{d}{dx}\frac{1}{1-x}=\frac{x}{{\left(1-x\right)}^{2}}
x=\frac{1}{9}
you get required answer.
A=\sum _{n\ge 1}n{9}^{-n},B=\sum _{n\ge 2}n{9}^{-n},\text{ }C=\sum _{n\ge 1}{9}^{-n},\text{ }D=\sum _{n\ge 2}{9}^{-n}
. You have the closed system
A=B+\frac{1}{9},\text{ }9B-A=C,\text{ }C=D+\frac{1}{9},\text{ }9D=C
which gives the value of A. Of course, as a byproduct of the computation one also gets the value of the geometric series C.
f\left(x\right)=\sum _{i=0}^{\mathrm{\infty }}\frac{{x}^{i\text{mod}\left(k-1\right)}}{i!}
\sum _{n=1}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{n+1}}{n\left(n+1\right)}=2\mathrm{ln}2-1
{e}^{x}
Use the Limit Comparison Test to prove convergence or divergence of the infinite series.
\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\sqrt{n}+\mathrm{ln}n}
\sum _{n,k}\frac{1}{\left(\begin{array}{c}n\\ k\end{array}\right)}
where the summation ranges over all positive integers n, k with
1<k<n-1
\left(n+1\right)\left(n-2\right){a}_{n}=n\left({n}^{2}-n-1\right){a}_{n-1}-{\left(n-1\right)}^{3}{a}_{n-2}
{a}_{2}={a}_{3}=1
Approximating powers Compute the coefficients for the Taylor series for the following functions about the given point a, and then use the first four terms of the series to approximate the given number.
f\left(x\right)=\frac{1}{\sqrt{x}}
a=4
\frac{1}{\sqrt{3}}
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Apply memoryless nonlinearity to complex baseband signal - Simulink - MathWorks France
{F}_{\text{AMAM}}\left(u\right)=\frac{{\alpha }_{\text{AMAM}}×u}{1+{\beta }_{\text{AMAM}}×{u}^{2}}\text{,}
{F}_{\text{AMPM}}\left(u\right)=\frac{{\alpha }_{\text{AMPM}}×{u}^{2}}{1+{\beta }_{\text{AMPM}}×{u}^{2}}\text{,}
{F}_{\text{AMAM}}\left(u\right)=\frac{{x}_{1}{u}^{{x}_{2}}}{1+{x}_{3}{u}^{{x}_{2}}}+{x}_{4}u\text{,}
{F}_{\text{AMPM}}\left(u\right)=\frac{{y}_{1}{u}^{{y}_{2}}}{1+{y}_{3}{u}^{{y}_{2}}}+{y}_{4}u\text{,}
{F}_{\text{AMAM}}\left(u\right)=\frac{{g}_{\text{lin}}×u}{{\left(1+{\left(\frac{{g}_{\text{lin}}×u}{{O}_{\text{sat}}}\right)}^{\text{2}S}\right)}^{\text{1/2}S}}\text{,}
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Replace the polar equations with equivalent Cartesian equations. Then describe o
Replace the polar equations with equivalent Cartesian equations. Then describe or identify the graph.
{r}^{2}=-6r\mathrm{sin}\theta
{r}^{2}=-6r\mathrm{sin}\theta
The Cartesian equation is
{x}^{2}+{y}^{2}=-6y\text{ }\text{ }\text{ }\left[\because x=r\mathrm{cos}\theta ,y=r\mathrm{sin}\theta \text{ }\mathrm{&}\text{ }r=\sqrt{{x}^{2}+{y}^{2}}\right]
⇒{x}^{2}+{y}^{2}+6y=0
⇒{x}^{2}+{y}^{2}+6y+9=9
⇒{x}^{2}+{\left(y+3\right)}^{2}=9
⇒{\left(x-0\right)}^{2}+{\left(y-\left(-3\right)\right)}^{2}={3}^{2}
Now its the form of of
⇒{\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={r}^{2}
Hence it represents a circle with centre at (h,k)(0,-3)& radius 3
\left(x,\text{ }y\right)=
What is (-2, 9) in polar coordinates?
What is the polar form of (42, 77)?
r\left(t\right)=\frac{1}{t+1}i+\frac{t}{2}j-3tk
Find the area of the region that lies inside both curves.
r=\sqrt{3}\mathrm{cos}\theta ,r=\mathrm{sin}\theta
f\left(x\right)={\left({x}^{2}–9\right)}^{2}
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group(deprecated)/core - Maple Help
Home : Support : Online Help : group(deprecated)/core
find the core of a subgroup of a permutation group
core(sg, pg)
group in which the core is to be found
subgroup of pg
Important: The group package has been deprecated. Use the superseding command GroupTheory[Core] instead.
This function finds the largest normal subgroup of the permutation group pg contained in sg. To speed up the computation, it is required that sg be a subgroup of pg. The result is returned as an unevaluated permgroup call.
The command with(group,core) allows the use of the abbreviated form of this command.
\mathrm{with}\left(\mathrm{group}\right):
\mathrm{pg}≔\mathrm{permgroup}\left(7,{[[1,2]],[[1,2,3,4,5,6,7]]}\right):
\mathrm{sg}≔\mathrm{permgroup}\left(7,{[[1,2,3]],[[3,4,5,6,7]]}\right):
\mathrm{core}\left(\mathrm{sg},\mathrm{pg}\right)
\textcolor[rgb]{0,0,1}{\mathrm{permgroup}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{7}\textcolor[rgb]{0,0,1}{,}{[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}]]\textcolor[rgb]{0,0,1}{,}[[\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{6}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{7}]]}\right)
group[issubgroup]
group[NormalClosure]
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Grain is loaded onto a train initially coasting at the speed
v_0 \text{ m/s}
by dropping it vertically from a stationary silo at the rate
\alpha_m \text{ kg/s}
. How fast is the train moving when it exits from under the silo (in m/s)?
The train is
l = 200
m long, and has an empty weight of
m_0 = 500
v_0=45
\alpha_m = 20
v_0 \text{ m/s}
\alpha_m \text{ kg/s}
. How long does it take for the train to pass the silo (in seconds)?
l = 200
m_0 = 500
v_0=45
\alpha_m = 20
Bouncy balls are stacked on top of each other with decreasing mass as shown In the figure above. They are configured such that the one on the bottom is at a height of
H
above the ground. The figure above shown for
n=5
, but we can increase the number of ball to our liking.
After the balls are released, and the first ball touches the ground, each ball bounces off its neighboring ball. If
v_{n}
is the speed of the
nth
ball right after it bounces of its adjacent ball. What is the ratio of the speed of the
20th
ball right after it bounces to the
5th
ball right after it bounces.
Simply put what is
\large{\frac { { v }_{ 20 } }{ { v }_{ 5 } }}
All the collisions that take place are elastic collisions.
{ M }_{ 1 }\gg { M }_{ 2 }\gg { M }_{ 3 }\gg ........{ M }_{ n }
Suppose you want to throw a super ball so that it bounces back and forth, retracing the same trajectory, as shown above. If you throw the ball horizontally with velocity
\SI[per-mode=symbol]{1}{\meter\per\second},
what spin
\omega
\si[per-mode=symbol]{\radian\per\second}
) must you give it so that it does what you want?
The collision with the ground is perfectly elastic.
The ball does not slip while it contacts the ground.
The radius of the ball is
\SI{3}{\centi\meter}.
A 100 m rope of line density
\lambda=0.1
kg/m is suspended vertically so that the bottom of the rope is just touching a scale. It is then released and falls onto the scale, which perceives a time dependent weight.
What is the reading on the scale in Newtons at the moment half the rope lies on the scale?
g=9.8
^{2}
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Disturbance Attenuation Using a dc Motor for Radial Force Actuation in a Rotordynamic System | J. Dyn. Sys., Meas., Control. | ASME Digital Collection
Rainer Leuschke,
Rainer Leuschke Research Assistant
Brian C. Fabien Professor
Leuschke, R., and Fabien, B. C. (January 18, 2007). "Disturbance Attenuation Using a dc Motor for Radial Force Actuation in a Rotordynamic System." ASME. J. Dyn. Sys., Meas., Control. November 2007; 129(6): 804–812. https://doi.org/10.1115/1.2789471
This paper presents a technique for attenuating the external disturbances acting on the rotor of a prototype flywheel energy storage device. The approach uses a three-phase axial flux brushless dc motor to simultaneously produce a torque and a radial force. This is accomplished by using two phases of the motor for torque generation, and one phase to produce the radial force. The paper develops a set of equations that can be used to predict the forces generated by the motor coils. These equations are used to implement a feedback control system to suppress the effects of external excitations. The nonlinear controller requires the velocity measurements and the angular displacement of the flywheel. The controller essentially adds damping to the system, and the constant feedback gains solve an optimization problem that involves a
H∞
bound on the disturbance attenuation. The experimental results clearly demonstrate that the dc motor can be used to suppress unwanted radial vibrations due to external disturbances.
angular velocity control, DC motors, energy storage, feedback, nonlinear control systems, rotors, flywheel energy storage system, rotor dynamics, vibration control, dc motor actuator, H2∕H∞ control
Control equipment, Engines, Flywheels, Motors, Rotors, Torque, Actuators, Damping, Energy storage, Sensors, Magnetic bearings, Feedback
Stability and Damped Critical Speeds of Rotor-Bearing Systems
Vibration of Larger Turbo-Rotors in Fluid-Film Bearings on an Elastic Foundation
, Number 12, pp.
Stienmier
Analysis and Control of a Flywheel Energy Storage System With a Hybrid Magnetic Bearing
Optimal Control of a Flywheel Energy Storage System
Vibration Control of a Flexible Rotor∕Magnetic Bearing System Subject to Direct Forcing and Base Motion Disturbances
Proceedings of the Institution of Mechanical-Engineers, Part C, Journal of Mechanical Engineering Science
Vibration Control of Flexible Rotor by Inclination Control Magnetic Bearings With Axial Self-Bearing Motor
Imashima
Levitation and Torque Control of Internal Permanent Magnet Type Bearingless Motor
Design and Analysis of Permanent Magnet-Type Bearingless Motors
Closed-Loop Performance of a Six Degree-Of-Freedom Precision Magnetic Actuator
Modeling and Control of a New Horizontal-Shaft Hybrid-Type Magnetic Bearing
Application and Study of Optimal Control Theory in Control System of Active Magnetic Bearings
Imbalance Estimation for Speed-Varying Rigid Rotors Using Time-Varying Observer
Structure of Magnetic Bearing Control System for Compensating Unbalance Force
Dynamics of Controlled Mechanical Systems, IUTAM∕IFAC Symposium Zurich∕Switzerland 1998
Constrained Output Feedback Control of Flexible Rotor-Bearing Systems
Motor Integrated Actuation for a Flywheel Energy Storage System
,” Ph.D. thesis, University of Washington, Seattle, WA.
Output Feedback Stabilizing Control With a Bounds on Disturbance Attenuation
A New Static Output Feedback Approach to the Suboptimal Mixed H2∕H∞ Problem
http://abs-5.me.washington.edu/flywheel/flywheel.movhttp://abs-5.me.washington.edu/flywheel/flywheel.mov
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Find the tangent line(s) to the parametric curve given by
Find the tangent line(s) to the parametric curve given by x=t^5-4t^3 y=t^2
x={t}^{5}-4{t}^{3}
y={t}^{2}
x={t}^{5}-4{t}^{3}
y={t}^{2}
m=\frac{dy}{dx}=\frac{2t}{5{t}^{4}-12{t}^{2}}
⇒\frac{2}{5{t}^{3}-12t}\left[\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\right]
line of tangent
Y-{y}_{1}=m\left(x-{x}_{1}\right)
Y-{t}^{2}=\frac{2}{5}{t}^{3}-12t\left(x-{t}^{5}+4{t}^{3}\right)
\left(x,\text{ }y\right)=
Find the values of x for which the series coverges.(Enter your answer using interval notation.)
\sum _{\eta =0}^{\mathrm{\infty }}\frac{{\left(x-1\right)}^{\eta }}{{2}^{\eta }}
Find the sum of the series fot those values of x.
Express the plane
z=x
in cylindrical and spherical coordinates.
a) cykindrical
z=r\mathrm{cos}\left(\theta \right)
b) spherical coordinates
\theta =\mathrm{arcsin}\left(\mathrm{cot}\left(\varphi \right)\right)
Differentiate the vector-valued function
r\left(t\right)=⟨\mathrm{sin}\left({t}^{2}\right),4{t}^{2}-t⟩
\left(2,\text{ }-1\right)
\left(1,\text{ }3\right)
Find the lengths of the sides of the triangle PQR. Is it a right triangle? Is it an isosceles triangle? P(2, -1, 0), Q(4, 1, 1), R(4, -5, 4)
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Optical flow or optic flow is the pattern of apparent motion of objects, surfaces, and edges in a visual scene caused by the relative motion between an observer and a scene.[1][2] Optical flow can also be defined as the distribution of apparent velocities of movement of brightness pattern in an image.[3] The concept of optical flow was introduced by the American psychologist James J. Gibson in the 1940s to describe the visual stimulus provided to animals moving through the world.[4] Gibson stressed the importance of optic flow for affordance perception, the ability to discern possibilities for action within the environment. Followers of Gibson and his ecological approach to psychology have further demonstrated the role of the optical flow stimulus for the perception of movement by the observer in the world; perception of the shape, distance and movement of objects in the world; and the control of locomotion.[5]
The optic flow experienced by a rotating observer (in this case a fly). The direction and magnitude of optic flow at each location is represented by the direction and length of each arrow.
The term optical flow is also used by roboticists, encompassing related techniques from image processing and control of navigation including motion detection, object segmentation, time-to-contact information, focus of expansion calculations, luminance, motion compensated encoding, and stereo disparity measurement.[6][7]
1.1 Methods for determination
3 Optical flow sensor
Sequences of ordered images allow the estimation of motion as either instantaneous image velocities or discrete image displacements.[7] Fleet and Weiss provide a tutorial introduction to gradient based optical flow.[8] John L. Barron, David J. Fleet, and Steven Beauchemin provide a performance analysis of a number of optical flow techniques. It emphasizes the accuracy and density of measurements.[9]
The optical flow methods try to calculate the motion between two image frames which are taken at times
{\displaystyle t}
{\displaystyle t+\Delta t}
at every voxel position. These methods are called differential since they are based on local Taylor series approximations of the image signal; that is, they use partial derivatives with respect to the spatial and temporal coordinates.
For a (2D + t)-dimensional case (3D or n-D cases are similar) a voxel at location
{\displaystyle (x,y,t)}
with intensity
{\displaystyle I(x,y,t)}
will have moved by
{\displaystyle \Delta x}
{\displaystyle \Delta y}
{\displaystyle \Delta t}
between the two image frames, and the following brightness constancy constraint can be given:
{\displaystyle I(x,y,t)=I(x+\Delta x,y+\Delta y,t+\Delta t)}
Assuming the movement to be small, the image constraint at
{\displaystyle I(x,y,t)}
with Taylor series can be developed to get:
{\displaystyle I(x+\Delta x,y+\Delta y,t+\Delta t)=I(x,y,t)+{\frac {\partial I}{\partial x}}\,\Delta x+{\frac {\partial I}{\partial y}}\,\Delta y+{\frac {\partial I}{\partial t}}\,\Delta t+{}}
higher-order terms
By truncating the higher order terms (which performs a linearization) it follows that:
{\displaystyle {\frac {\partial I}{\partial x}}\Delta x+{\frac {\partial I}{\partial y}}\Delta y+{\frac {\partial I}{\partial t}}\Delta t=0}
or, dividing by
{\displaystyle \Delta t}
{\displaystyle {\frac {\partial I}{\partial x}}{\frac {\Delta x}{\Delta t}}+{\frac {\partial I}{\partial y}}{\frac {\Delta y}{\Delta t}}+{\frac {\partial I}{\partial t}}{\frac {\Delta t}{\Delta t}}=0}
{\displaystyle {\frac {\partial I}{\partial x}}V_{x}+{\frac {\partial I}{\partial y}}V_{y}+{\frac {\partial I}{\partial t}}=0}
{\displaystyle V_{x},V_{y}}
{\displaystyle x}
{\displaystyle y}
components of the velocity or optical flow of
{\displaystyle I(x,y,t)}
{\displaystyle {\tfrac {\partial I}{\partial x}}}
{\displaystyle {\tfrac {\partial I}{\partial y}}}
{\displaystyle {\tfrac {\partial I}{\partial t}}}
are the derivatives of the image at
{\displaystyle (x,y,t)}
in the corresponding directions.
{\displaystyle I_{x}}
{\displaystyle I_{y}}
{\displaystyle I_{t}}
can be written for the derivatives in the following.
{\displaystyle I_{x}V_{x}+I_{y}V_{y}=-I_{t}}
{\displaystyle \nabla I\cdot {\vec {V}}=-I_{t}}
This is an equation in two unknowns and cannot be solved as such. This is known as the aperture problem of the optical flow algorithms. To find the optical flow another set of equations is needed, given by some additional constraint. All optical flow methods introduce additional conditions for estimating the actual flow.
Methods for determinationEdit
Phase correlation – inverse of normalized cross-power spectrum
Block-based methods – minimizing sum of squared differences or sum of absolute differences, or maximizing normalized cross-correlation
Differential methods of estimating optical flow, based on partial derivatives of the image signal and/or the sought flow field and higher-order partial derivatives, such as:
Lucas–Kanade method – regarding image patches and an affine model for the flow field[10]
Horn–Schunck method – optimizing a functional based on residuals from the brightness constancy constraint, and a particular regularization term expressing the expected smoothness of the flow field[10]
Buxton–Buxton method – based on a model of the motion of edges in image sequences[11]
Black–Jepson method – coarse optical flow via correlation[7]
General variational methods – a range of modifications/extensions of Horn–Schunck, using other data terms and other smoothness terms.
Discrete optimization methods – the search space is quantized, and then image matching is addressed through label assignment at every pixel, such that the corresponding deformation minimizes the distance between the source and the target image.[12] The optimal solution is often recovered through Max-flow min-cut theorem algorithms, linear programming or belief propagation methods.
Many of these, in addition to the current state-of-the-art algorithms are evaluated on the Middlebury Benchmark Dataset.[13][14] Other popular benchmark datasets are KITTI and Sintel.
Motion estimation and video compression have developed as a major aspect of optical flow research. While the optical flow field is superficially similar to a dense motion field derived from the techniques of motion estimation, optical flow is the study of not only the determination of the optical flow field itself, but also of its use in estimating the three-dimensional nature and structure of the scene, as well as the 3D motion of objects and the observer relative to the scene, most of them using the image Jacobian.[15]
Optical flow was used by robotics researchers in many areas such as: object detection and tracking, image dominant plane extraction, movement detection, robot navigation and visual odometry.[6] Optical flow information has been recognized as being useful for controlling micro air vehicles.[16]
The application of optical flow includes the problem of inferring not only the motion of the observer and objects in the scene, but also the structure of objects and the environment. Since awareness of motion and the generation of mental maps of the structure of our environment are critical components of animal (and human) vision, the conversion of this innate ability to a computer capability is similarly crucial in the field of machine vision.[17]
Consider a five-frame clip of a ball moving from the bottom left of a field of vision, to the top right. Motion estimation techniques can determine that on a two dimensional plane the ball is moving up and to the right and vectors describing this motion can be extracted from the sequence of frames. For the purposes of video compression (e.g., MPEG), the sequence is now described as well as it needs to be. However, in the field of machine vision, the question of whether the ball is moving to the right or if the observer is moving to the left is unknowable yet critical information. Not even if a static, patterned background were present in the five frames, could we confidently state that the ball was moving to the right, because the pattern might have an infinite distance to the observer.
Optical flow sensorEdit
Not to be confused with Optical flowmeter.
An optical flow sensor is a vision sensor capable of measuring optical flow or visual motion and outputting a measurement based on optical flow. Various configurations of optical flow sensors exist. One configuration is an image sensor chip connected to a processor programmed to run an optical flow algorithm. Another configuration uses a vision chip, which is an integrated circuit having both the image sensor and the processor on the same die, allowing for a compact implementation.[18][19] An example of this is a generic optical mouse sensor used in an optical mouse. In some cases the processing circuitry may be implemented using analog or mixed-signal circuits to enable fast optical flow computation using minimal current consumption.
One area of contemporary research is the use of neuromorphic engineering techniques to implement circuits that respond to optical flow, and thus may be appropriate for use in an optical flow sensor.[20] Such circuits may draw inspiration from biological neural circuitry that similarly responds to optical flow.
Optical flow sensors are used extensively in computer optical mice, as the main sensing component for measuring the motion of the mouse across a surface.
Optical flow sensors are also being used in robotics applications, primarily where there is a need to measure visual motion or relative motion between the robot and other objects in the vicinity of the robot. The use of optical flow sensors in unmanned aerial vehicles (UAVs), for stability and obstacle avoidance, is also an area of current research.[21]
^ Burton, Andrew; Radford, John (1978). Thinking in Perspective: Critical Essays in the Study of Thought Processes. Routledge. ISBN 978-0-416-85840-2.
^ Warren, David H.; Strelow, Edward R. (1985). Electronic Spatial Sensing for the Blind: Contributions from Perception. Springer. ISBN 978-90-247-2689-9.
^ Horn, Berthold K.P.; Schunck, Brian G. (August 1981). "Determining optical flow" (PDF). Artificial Intelligence. 17 (1–3): 185–203. doi:10.1016/0004-3702(81)90024-2. hdl:1721.1/6337.
^ Gibson, J.J. (1950). The Perception of the Visual World. Houghton Mifflin.
^ Royden, C. S.; Moore, K. D. (2012). "Use of speed cues in the detection of moving objects by moving observers". Vision Research. 59: 17–24. doi:10.1016/j.visres.2012.02.006. PMID 22406544. S2CID 52847487.
^ a b Aires, Kelson R. T.; Santana, Andre M.; Medeiros, Adelardo A. D. (2008). Optical Flow Using Color Information (PDF). ACM New York, NY, USA. ISBN 978-1-59593-753-7.
^ a b c Beauchemin, S. S.; Barron, J. L. (1995). "The computation of optical flow". ACM Computing Surveys. ACM New York, USA. 27 (3): 433–466. doi:10.1145/212094.212141. S2CID 1334552.
^ Fleet, David J.; Weiss, Yair (2006). "Optical Flow Estimation" (PDF). In Paragios, Nikos; Chen, Yunmei; Faugeras, Olivier D. (eds.). Handbook of Mathematical Models in Computer Vision. Springer. pp. 237–257. ISBN 978-0-387-26371-7.
^ Barron, John L.; Fleet, David J. & Beauchemin, Steven (1994). "Performance of optical flow techniques" (PDF). International Journal of Computer Vision. 12: 43–77. CiteSeerX 10.1.1.173.481. doi:10.1007/bf01420984. S2CID 1290100.
^ a b Zhang, G.; Chanson, H. (2018). "Application of Local Optical Flow Methods to High-Velocity Free-surface Flows: Validation and Application to Stepped Chutes" (PDF). Experimental Thermal and Fluid Science. 90: 186–199. doi:10.1016/j.expthermflusci.2017.09.010.
^ Glyn W. Humphreys and Vicki Bruce (1989). Visual Cognition. Psychology Press. ISBN 978-0-86377-124-8.
^ B. Glocker; N. Komodakis; G. Tziritas; N. Navab; N. Paragios (2008). Dense Image Registration through MRFs and Efficient Linear Programming (PDF). Medical Image Analysis Journal.
^ Baker, Simon; Scharstein, Daniel; Lewis, J. P.; Roth, Stefan; Black, Michael J.; Szeliski, Richard (March 2011). "A Database and Evaluation Methodology for Optical Flow". International Journal of Computer Vision. 92 (1): 1–31. doi:10.1007/s11263-010-0390-2. ISSN 0920-5691. S2CID 316800.
^ Baker, Simon; Scharstein, Daniel; Lewis, J. P.; Roth, Stefan; Black, Michael J.; Szeliski, Richard. "Optical Flow". vision.middlebury.edu. Retrieved 2019-10-18.
^ Corke, Peter (8 May 2017). "The Image Jacobian". QUT Robot Academy.
^ Barrows, G. L.; Chahl, J. S.; Srinivasan, M. V. (2003). "Biologically inspired visual sensing and flight control". Aeronautical Journal. 107 (1069): 159–268. doi:10.1017/S0001924000011891 (inactive 28 February 2022) – via Cambridge University Press. {{cite journal}}: CS1 maint: DOI inactive as of February 2022 (link)
^ Brown, Christopher M. (1987). Advances in Computer Vision. Lawrence Erlbaum Associates. ISBN 978-0-89859-648-9.
^ Moini, Alireza (2000). Vision Chips. Boston, MA: Springer US. ISBN 9781461552673. OCLC 851803922.
^ Mead, Carver (1989). Analog VLSI and neural systems. Reading, Mass.: Addison-Wesley. ISBN 0201059924. OCLC 17954003.
^ Stocker, Alan A. (2006). Analog VLSI circuits for the perception of visual motion. Chichester, England: John Wiley & Sons. ISBN 0470034882. OCLC 71521689.
^ Floreano, Dario; Zufferey, Jean-Christophe; Srinivasan, Mandyam V.; Ellington, Charlie, eds. (2009). Flying insects and robots. Heidelberg: Springer. ISBN 9783540893936. OCLC 495477442.
Wikimedia Commons has media related to Optic flow.
Finding Optic Flow
Art of Optical Flow article on fxguide.com (using optical flow in visual effects)
Optical flow evaluation and ground truth sequences.
Middlebury Optical flow evaluation and ground truth sequences.
mrf-registration.net - Optical flow estimation through MRF
The French Aerospace Lab: GPU implementation of a Lucas-Kanade based optical flow
CUDA Implementation by CUVI (CUDA Vision & Imaging Library)
Horn and Schunck Optical Flow: Online demo and source code of the Horn and Schunck method
TV-L1 Optical Flow: Online demo and source code of the Zach et al. method
Robust Optical Flow: Online demo and source code of the Brox et al. method
Retrieved from "https://en.wikipedia.org/w/index.php?title=Optical_flow&oldid=1075679216"
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Q Calculate the time in which Rs 5,000 would become Rs 5,500 at an interest rate of - Maths - Playing with Numbers - 11882197 | Meritnation.com
Q. Calculate the time in which Rs. 5,000 would become Rs. 5,500 at an interest rate of 8% per annum.
Q. In what time will money deposited in a bank double itself if rate of interest is
12\frac{1}{2}
% p.a.
Q. A sum of Rs. 4000 was borrowed on April 1 and repaid on Aug. 25 of the same year at an interest of
7\frac{1}{2}
% . What was the interest paid ?
Kindly refer the links for answer similar to your queries:
https://www.meritnation.com/ask-answer/question/calculate-the-time-of-which-rs-5-000would-become-rs-5-500at/practical-geometry/1582271
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40 CFR § 63.11 - Control device and work practice requirements. | CFR | US Law | LII / Legal Information Institute
40 CFR § 63.11 - Control device and work practice requirements.
(b) Flares.
{H}_{T}=K\sum _{i=1}^{n}{C}_{i}{H}_{i}
1.740×{10}^{-7}\left(\frac{1}{\mathrm{ppmv}}\right)\left(\frac{\text{g-mole}}{\mathrm{scm}}\right)\left(\frac{\mathrm{MJ}}{\mathrm{kcal}}\right)
{E}_{\mathrm{dic}}-\left({E}_{\mathrm{sds}}\right)\sum _{i=1}^{k}{x}_{i}
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Impact of Climate and Bioclimate Factors on Apricot (Prunus armeniaca L.) Yield to Increase Economy and Achieve Maintaining Food Security of Palestine ()
1Department of Plant Production and Protection, Faculty of Agriculture, Al-Quds Open University, Hebron, Palestine.
Apricot (Prunus armeniaca L.) is one of the most important export crops in Hebron of Palestine. In this work, we analyzed the mean monthly temperature and precipitation using data from one weather station of the Palestine Meteoro-logical Department, recorded in the period from 1993-2014, with the same years of plant production (rain-fed) from the Palestinian Central Bureau of Statistics (PCBS). Statistical analysis included a bioclimatic analysis of Palestinian meteorological station for the period previous by using bioclimatic classification of the Earth of Rivas Martinez Salvador, with regard to simple continentality index, compensated thermicity index, annual ombrothermic index, water deficit and soil water reserve. In concluded, climate and bioclimatic factors play a key role in apricot production, when we analyzed of variance (ANOVA), with a standard coefficients (95% confidence interval), reveals significant differences, in case of the bioclimate factors as simple thermicity index and climate factors as precipitation, whereas there are no significant differences between the apricot yield and the reset of climate, bioclimate factors. However, the optimum of the plant production is achieved with values of simple thermicity index between (14-18), annual ombrothermic index (2.5-4.5), compensated thermicity index (250-450), precipitation more than 750 mm, mean monthly temperature (15.4℃ - 20℃), and inframediterranean, thermomediterranean and mesomediterranean of thermotype in Hebron of Palestine.
Palestine, Hebron, Bioclimate, Biology, Climate, Yield
Ighbareyeh, J. and Carmona, E. (2017) Impact of Climate and Bioclimate Factors on Apricot (Prunus armeniaca L.) Yield to Increase Economy and Achieve Maintaining Food Security of Palestine. Open Access Library Journal, 4, 1-13. doi: 10.4236/oalib.1104119.
r=\frac{{\displaystyle {\sum }_{i=1}^{n}\left({X}_{i}-\stackrel{¯}{X}\right)\left({Y}_{i}-\stackrel{¯}{Y}\right)}}{\left(n-1\right){S}_{X}{S}_{Y}}
|
Genomic region with a linear time effect. Fitted linear time effect for the genomic region of gene AT1G62500 on the forward strand of chromosome 1. The different replicates are indicated by ∘, + and
▵
, while the different days are represented by different colors: black (D8), red (D9), green (D10), blue (D11), cyan (D12) and magenta (D13). The grey rectangles indicate the detected regions showing a significant linear time effect, while the black line corresponds with the coefficient function of the linear effect. The negative sign of the coefficients implies a decreasing effect over time. More specifically, the effect at probe t is
{\stackrel{̂}{\beta }}_{\mathbf{1}}\left(t\right)×\mathrm{time}
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Macroeconomics Models - Vocabulary - Course Hero
Macroeconomics/Macroeconomics Models/Vocabulary
the total individual and household purchases of consumer goods and services
the relationship between consumption and disposable income
\text{C}=a+(\text{MPC}\times Y_d)
the change in consumer spending that occurs in response to an incremental change in disposable income
the change in consumer savings that occurs in response to an incremental change in disposable income
an economic law that states when an individual produces a product or service, they get paid for that work, and are then able to use that income to demand other goods and services
the factor by which gains in total output are greater than the change in spending that caused it
\text{The}\;\text{Spending}\;\text{Multiplier}=\frac1{1-\;\text{MPC}}
use of expansionary fiscal or monetary policy to jumpstart a sluggish or depressed economy or kickstart economic growth
Expansionary Monetary PolicyExpansionary Fiscal Policy
the effect on the economy from changes in tax policy
\text{Tax Multiplier}=\frac{-\text{MPC}}{(1-\text{MPC})}
unplanned inventory accumulation
an excess of unsold goods caused by an unplanned event
unplanned negative inventory
a shortage of stored materials or goods caused by increased demand
<Overview>Macroeconomic Viewpoints
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Code-excited_linear_prediction Knowpia
Code-excited linear prediction (CELP) is a linear predictive speech coding algorithm originally proposed by Manfred R. Schroeder and Bishnu S. Atal in 1985. At the time, it provided significantly better quality than existing low bit-rate algorithms, such as residual-excited linear prediction (RELP) and linear predictive coding (LPC) vocoders (e.g., FS-1015). Along with its variants, such as algebraic CELP, relaxed CELP, low-delay CELP and vector sum excited linear prediction, it is currently the most widely used speech coding algorithm[citation needed]. It is also used in MPEG-4 Audio speech coding. CELP is commonly used as a generic term for a class of algorithms and not for a particular codec.
The CELP algorithm is based on four main ideas:
Using the source-filter model of speech production through linear prediction (LP) (see the textbook "speech coding algorithm");
Using an adaptive and a fixed codebook as the input (excitation) of the LP model;
Performing a search in closed-loop in a "perceptually weighted domain".
Applying vector quantization (VQ)
The original algorithm as simulated in 1983 by Schroeder and Atal required 150 seconds to encode 1 second of speech when run on a Cray-1 supercomputer. Since then, more efficient ways of implementing the codebooks and improvements in computing capabilities have made it possible to run the algorithm in embedded devices, such as mobile phones.
CELP decoderEdit
Figure 1: CELP decoder
Before exploring the complex encoding process of CELP we introduce the decoder here. Figure 1 describes a generic CELP decoder. The excitation is produced by summing the contributions from fixed (a.k.a. stochastic or innovation) and adaptive (a.k.a. pitch) codebooks:
{\displaystyle e[n]=e_{f}[n]+e_{a}[n]\,}
{\displaystyle e_{f}[n]}
is the fixed (a.k.a. stochastic or innovation) codebook contribution and
{\displaystyle e_{a}[n]}
is the adaptive (pitch) codebook contribution. The fixed codebook is a vector quantization dictionary that is (implicitly or explicitly) hard-coded into the codec. This codebook can be algebraic (ACELP) or be stored explicitly (e.g. Speex). The entries in the adaptive codebook consist of delayed versions of the excitation. This makes it possible to efficiently code periodic signals, such as voiced sounds.
The filter that shapes the excitation has an all-pole model of the form
{\displaystyle 1/A(z)}
{\displaystyle A(z)}
is called the prediction filter and is obtained using linear prediction (Levinson–Durbin algorithm). An all-pole filter is used because it is a good representation of the human vocal tract and because it is easy to compute.
CELP encoderEdit
The main principle behind CELP is called analysis-by-synthesis (AbS) and means that the encoding (analysis) is performed by perceptually optimizing the decoded (synthesis) signal in a closed loop. In theory, the best CELP stream would be produced by trying all possible bit combinations and selecting the one that produces the best-sounding decoded signal. This is obviously not possible in practice for two reasons: the required complexity is beyond any currently available hardware and the “best sounding” selection criterion implies a human listener.
In order to achieve real-time encoding using limited computing resources, the CELP search is broken down into smaller, more manageable, sequential searches using a simple perceptual weighting function. Typically, the encoding is performed in the following order:
Linear prediction coefficients (LPC) are computed and quantized, usually as line spectral pairs (LSPs).
The adaptive (pitch) codebook is searched and its contribution removed.
The fixed (innovation) codebook is searched.
Noise weightingEdit
Most (if not all) modern audio codecs attempt to shape the coding noise so that it appears mostly in the frequency regions where the ear cannot detect it. For example, the ear is more tolerant to noise in parts of the spectrum that are louder and vice versa. That's why instead of minimizing the simple quadratic error, CELP minimizes the error for the perceptually weighted domain. The weighting filter W(z) is typically derived from the LPC filter by the use of bandwidth expansion:
{\displaystyle W(z)={\frac {A(z/\gamma _{1})}{A(z/\gamma _{2})}}}
{\displaystyle \gamma _{1}>\gamma _{2}}
MPEG-4 Part 3 (CELP as an MPEG-4 Audio Object Type)
G.728 – Coding of speech at 16 kbit/s using low-delay code excited linear prediction
G.718 – uses CELP for the lower two layers for the band (50–6400 Hz) in a two-stage coding structure
G.729.1 – uses CELP coding for the lower band (50–4000 Hz) in a three-stage coding structure
CELT is a related audio codec that borrows some ideas from CELP.
B.S. Atal, "The History of Linear Prediction," IEEE Signal Processing Magazine, vol. 23, no. 2, March 2006, pp. 154–161.
M. R. Schroeder and B. S. Atal, "Code-excited linear prediction (CELP): high-quality speech at very low bit rates," in Proceedings of the IEEE International Conference on Acoustics, Speech, and Signal Processing (ICASSP), vol. 10, pp. 937–940, 1985.
This article is based on a paper presented at Linux.Conf.Au
Some parts based on the Speex codec manual
reference implementations of CELP 1016A (CELP 3.2a) and LPC 10e.
Selected readingsEdit
Speech Processing: Theory of LPC Analysis and Synthesis
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The Assessment of Residual Stress Effects on Ductile Tearing Using Continuum Damage Mechanics | J. Pressure Vessel Technol. | ASME Digital Collection
Andrew H. Sherry,
Andrew H. Sherry
, Sackville Street, Manchester, M60 1QD, UK
e-mail: a.sherry@manchester.ac.uk
Mark A. Wilkes,
Mark A. Wilkes
, Birchwood Park, Risley, Warrington WA3 6GA, UK
e-mail: mark.wilkes@sercoassurance.com
John K. Sharples,
John K. Sharples
e-mail: john.sharples@sercoassurance.com
Peter J. Budden
, Barnett Way, Gloucester GL4 3RS, UK
e-mail: peter.budden@british-energy.com
J. Pressure Vessel Technol. Nov 2008, 130(4): 041212 (8 pages)
Sherry, A. H., Wilkes, M. A., Sharples, J. K., and Budden, P. J. (October 14, 2008). "The Assessment of Residual Stress Effects on Ductile Tearing Using Continuum Damage Mechanics." ASME. J. Pressure Vessel Technol. November 2008; 130(4): 041212. https://doi.org/10.1115/1.2967876
This paper presents the results of a numerical study undertaken to assess the influence of residual stresses on the ductile tearing behavior of a high strength low toughness aluminum alloy. The Gurson–Tvergaard model was calibrated against conventional fracture toughness data using parameters relating to void nucleation, growth, and coalescence. The calibrated model was used to predict the load versus ductile tearing behavior of a series of full-scale and quarter-scale wide-plate tests. These center-cracked tension tests included specimens that contained a self-balancing residual stress field that was tensile in the region of the through-wall crack. Analyses of the full-scale wide-plate tests indicated that the model provides a good prediction of the load versus the ductile tearing behavior up to approximately
3mm
of stable tearing. The influence of residual stress on the load versus the crack growth behavior was accurately simulated. Predictions of the load versus the crack growth behavior of full-scale wide-plate tests for crack extensions greater than
3mm
and of the quarter-scale tests were low in terms of predicted load at a given amount of tearing. This was considered to result from (i) the “valid” calibration range in terms of specimen thickness and crack extension, (ii) the development of shear lips, and (iii) the differences in the micromechanism of ductile void formation under plane strain and under plane stress conditions.
cracks, ductile fracture, internal stresses, plates (structures), ductile fracture, residual stresses, fracture mechanics, damage mechanics, Gurson–Tvergaard model
Ductile fracture, Fracture (Materials), Stress, Calibration, Residual stresses, Fracture toughness, Damage mechanics, Plane strain
A View on Ductile-Fracture Modeling
Mirzaee-Sisan
Interaction of Residual Stress With Mechanical Loading in a Ferritic Steel
Residual Stress Effects on Cleavage Fracture
,” ASME Paper No. 6, pp.
Proceedings of ASME Pressure Vessel and Piping Conference
Residual Stress and Constraint Effects on Fracture in the Transition Temperature Regime
,” ASME Paper No. PVP200861475.
Experimental Validation of R6 Treatment of Residual Stresses
A Constraint-Based Failure Assessment Diagram for Fracture Assessment
Use of Shallow-Cracked SENB Specimens to Obtain Upper Shelf Low Constraint Fracture Toughness Data
,” Serco Assurance Report No. SA/SIS/18507/R001 Issue 1.
Application of Local Approach Modelling of Constraint Contributions to the Ductile to Brittle Transition
Continuum Theory of Ductile Rupture by Voice Nucleation and Growth: Part I—Yield Criteria and Flow Rules for Porous Ductile Materials
Influence of Voids on Shear Band Instabilities Under Plain Strain Conditions
2001 (with amendments in 2007), R6 Revision 4, “
2000, ABAQUS Version 6.1 User’s Manual, Hibbit, Karlsson and Sorensen, Inc.
Elastic and Plastic Fracture: Metals, Polymers, Ceramics, Composites, Biological Materials
Mixed Mode Ductile Fracture: A Literature Review
,” Royal Institute of Technology, Department of Solid Mechanics, Stockholm.
A Note on Ductile Fracture Criteria in Metal Forming
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Such a formula can be made either true or false based on the values assigned to its propositional variables. The double turnstile notation
{\displaystyle \vDash S}
is used to indicate that S is a tautology. Tautology is sometimes symbolized by "Vpq", and contradiction by "Opq". The tee symbol
{\displaystyle \top }
is sometimes used to denote an arbitrary tautology, with the dual symbol
{\displaystyle \bot }
(falsum) representing an arbitrary contradiction; in any symbolism, a tautology may be substituted for the truth value "true", as symbolized, for instance, by "1".[1]
4 Verifying tautologies
5 Tautological implication
7 Semantic completeness and soundness
8 Efficient verification and the Boolean satisfiability problem
9 Tautologies versus validities in first-order logic
10.1 Normal forms
10.2 Related logical topics
In 1800, Immanuel Kant wrote in his book Logic:
Propositional logic begins with propositional variables, atomic units that represent concrete propositions. A formula consists of propositional variables connected by logical connectives, built up in such a way that the truth of the overall formula can be deduced from the truth or falsity of each variable. A valuation is a function that assigns each propositional variable to either T (for truth) or F (for falsity). So by using the propositional variables A and B, the binary connectives
{\displaystyle \lor }
{\displaystyle \land }
representing disjunction and conjunction respectively, and the unary connective
{\displaystyle \lnot }
representing negation, the following formula can be obtained:
{\displaystyle (A\land B)\lor (\lnot A)\lor (\lnot B)}
A valuation here must assign to each of A and B either T or F. But no matter how this assignment is made, the overall formula will come out true. For if the first conjunction
{\displaystyle (A\land B)}
is not satisfied by a particular valuation, then one of A and B is assigned F, which will make one of the following disjunct to be assigned T.
This article or section may be written in a style that is too abstract to be readily understandable by general audiences. Please improve it by defining technical terminology, and by adding examples. (May 2020)
{\displaystyle (A\lor \lnot A)}
("A or not A"), the law of excluded middle. This formula has only one propositional variable, A. Any valuation for this formula must, by definition, assign A one of the truth values true or false, and assign
{\displaystyle \lnot }
A the other truth value. For instance, "The cat is black or the cat is not black".
{\displaystyle (A\to B)\Leftrightarrow (\lnot B\to \lnot A)}
("if A implies B, then not-B implies not-A", and vice versa), which expresses the law of contraposition. For instance, "If it's a book, it is blue; if it's not blue, it's not a book."
{\displaystyle ((\lnot A\to B)\land (\lnot A\to \lnot B))\to A}
("if not-A implies both B and its negation not-B, then not-A must be false, then A must be true"), which is the principle known as reductio ad absurdum. For instance, "If it's not blue, it's a book, if it's not blue, it's also not a book, so it is blue."
{\displaystyle \lnot (A\land B)\Leftrightarrow (\lnot A\lor \lnot B)}
("if not both A and B, then not-A or not-B", and vice versa), which is known as De Morgan's law. "If it's either not a book or it's not blue, it's either not a book, or it's not blue, or neither."
{\displaystyle ((A\to B)\land (B\to C))\to (A\to C)}
("if A implies B and B implies C, then A implies C"), which is the principle known as syllogism. "If it's a book, then it's blue, if it's blue, it's on that shelf. Hence, if it's a book, it's on that shelf."
{\displaystyle ((A\lor B)\land (A\to C)\land (B\to C))\to C}
("if at least one of A or B is true, and each implies C, then C must be true as well"), which is the principle known as proof by cases. "Books and blue things are on that shelf. If it's either a book or it's blue, it's on that shelf."
{\displaystyle (A\lor B)\to (A\lor B)}
is a tautology, but not a minimal one, because it is an instantiation of
{\displaystyle C\to C}
Verifying tautologiesEdit
For example, consider the formula
{\displaystyle ((A\land B)\to C)\Leftrightarrow (A\to (B\to C)).}
{\displaystyle A}
{\displaystyle B}
{\displaystyle C}
{\displaystyle A\land B}
{\displaystyle (A\land B)\to C}
{\displaystyle B\to C}
{\displaystyle A\to (B\to C)}
{\displaystyle ((A\land B)\to C)\Leftrightarrow (A\to (B\to C))}
Tautological implicationEdit
Main article: Tautological consequence
A formula R is said to tautologically imply a formula S if every valuation that causes R to be true also causes S to be true. This situation is denoted
{\displaystyle R\models S}
. It is equivalent to the formula
{\displaystyle R\to S}
being a tautology (Kleene 1967 p. 27).
{\displaystyle S}
{\displaystyle A\land (B\lor \lnot B)}
{\displaystyle S}
is not a tautology, because any valuation that makes
{\displaystyle A}
false will make
{\displaystyle S}
false. But any valuation that makes
{\displaystyle A}
true will make
{\displaystyle S}
true, because
{\displaystyle B\lor \lnot B}
is a tautology. Let
{\displaystyle R}
be the formula
{\displaystyle A\land C}
{\displaystyle R\models S}
, because any valuation satisfying
{\displaystyle R}
{\displaystyle A}
true—and thus makes
{\displaystyle S}
It follows from the definition that if a formula
{\displaystyle R}
is a contradiction, then
{\displaystyle R}
tautologically implies every formula, because there is no truth valuation that causes
{\displaystyle R}
to be true, and so the definition of tautological implication is trivially satisfied. Similarly, if
{\displaystyle S}
is a tautology, then
{\displaystyle S}
is tautologically implied by every formula.
Main article: Substitution instance
For example, let S be the tautology
{\displaystyle (A\land B)\lor \lnot A\lor \lnot B}
Let SA be
{\displaystyle C\lor D}
and let SB be
{\displaystyle C\to E}
It follows from the substitution rule that the sentence
{\displaystyle ((C\lor D)\land (C\to E))\lor \lnot (C\lor D)\lor \lnot (C\to E)}
Semantic completeness and soundnessEdit
Efficient verification and the Boolean satisfiability problemEdit
The problem of determining whether there is any valuation that makes a formula true is the Boolean satisfiability problem; the problem of checking tautologies is equivalent to this problem, because verifying that a sentence S is a tautology is equivalent to verifying that there is no valuation satisfying
{\displaystyle \lnot S}
. It is known that the Boolean satisfiability problem is NP complete, and widely believed that there is no polynomial-time algorithm that can perform it. Consequently, tautology is co-NP-complete. Current research focuses on finding algorithms that perform well on special classes of formulas, or terminate quickly on average even though some inputs may cause them to take much longer.
Tautologies versus validities in first-order logicEdit
A tautology in first-order logic is a sentence that can be obtained by taking a tautology of propositional logic and uniformly replacing each propositional variable by a first-order formula (one formula per propositional variable). For example, because
{\displaystyle A\lor \lnot A}
is a tautology of propositional logic,
{\displaystyle (\forall x(x=x))\lor (\lnot \forall x(x=x))}
is a tautology in first order logic. Similarly, in a first-order language with a unary relation symbols R,S,T, the following sentence is a tautology:
{\displaystyle (((\exists xRx)\land \lnot (\exists xSx))\to \forall xTx)\Leftrightarrow ((\exists xRx)\to ((\lnot \exists xSx)\to \forall xTx)).}
It is obtained by replacing
{\displaystyle A}
{\displaystyle \exists xRx}
{\displaystyle B}
{\displaystyle \lnot \exists xSx}
{\displaystyle C}
{\displaystyle \forall xTx}
in the propositional tautology
{\displaystyle ((A\land B)\to C)\Leftrightarrow (A\to (B\to C))}
{\displaystyle (\forall xRx)\to \lnot \exists x\lnot Rx}
is true in any first-order interpretation, but it corresponds to the propositional sentence
{\displaystyle A\to B}
which is not a tautology of propositional logic.
Normal formsEdit
Related logical topicsEdit
^ Weisstein, Eric W. "Tautology". mathworld.wolfram.com. Retrieved 2020-08-14.
^ a b "tautology | Definition & Facts". Encyclopedia Britannica. Retrieved 2020-08-14.
^ "New Members". Naval Engineers Journal. 114 (1): 17–18. January 2002. doi:10.1111/j.1559-3584.2002.tb00103.x. ISSN 0028-1425.
Bocheński, J. M. (1959) Précis of Mathematical Logic, translated from the French and German editions by Otto Bird, Dordrecht, South Holland: D. Reidel.
Enderton, H. B. (2002) A Mathematical Introduction to Logic, Harcourt/Academic Press, ISBN 0-12-238452-0.
Kleene, S. C. (1967) Mathematical Logic, reprinted 2002, Dover Publications, ISBN 0-486-42533-9.
Reichenbach, H. (1947). Elements of Symbolic Logic, reprinted 1980, Dover, ISBN 0-486-24004-5
Wittgenstein, L. (1921). "Logisch-philosophiche Abhandlung", Annalen der Naturphilosophie (Leipzig), v. 14, pp. 185–262, reprinted in English translation as Tractatus logico-philosophicus, New York City and London, 1922.
"Tautology", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
Retrieved from "https://en.wikipedia.org/w/index.php?title=Tautology_(logic)&oldid=1082990624"
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In this article, we learn about control flow during the intermediate code generation phase of the compiler.
Short-circuit code.
Translating boolean expressions.
How to avoid redundant gotos.
Jumping code.
Translating if-else and while statements involves translating boolean expressions.While programming, we use boolean expressions to compute logical values and alter the flow of control.
The use of boolean expressions is determined by the syntactic context, e.g, an expression that follows the keyword if alters the flow of control, while an expression on the right side is used to denote a logical value.
Such syntactic contexts are specified in several ways, we can use two different non-terminals, use inherited attributes, or set a flag during parsing.
We can also build a syntax tree then invoke different procedures for the two different uses of boolean expressions.
Boolean expressions comprise of boolean operators denoted by && - (AND), || - (OR), ! - (NOT).
Relational expressions are in the form of
{\mathrm{E}}_{1}
{\mathrm{E}}_{2}
{\mathrm{E}}_{1}
{\mathrm{E}}_{1}
are arithmetic expressions.
We will consider boolean expressions generated using the following grammar.
\to
B || B | B && B | !B | (B) | E rel E | true | false
The attribute rel.op is used to indicate the operator represented by rel among the 6 comparison operators <, <=, =, !=, > or >=.
It is conventional to assume that || and && are left-associative and || has the lowest precedence, then && and finally !.
Given an expression
{\mathrm{B}}_{1}
{\mathrm{B}}_{2}
, if we determine that
{\mathrm{B}}_{1}
is true then we can conclude the entire expression if true without evaluating
{\mathrm{B}}_{2}
. Similarly, given
{\mathrm{B}}_{1}
{\mathrm{B}}_{s}
{\mathrm{B}}_{1}
is false, the entire expression is false.
Short-circuit code
In a short-circuit code, boolean operators such as &&, ||, and ! translate into jumps. Instead, if the operators themselves appear in the code, the value of the boolean expression is represented by a position in the code sequence. An example is shown below;
if(x < 100 || x > 200 && x != y) x = 0;
The statement is translated into the following code segment;
if x < 100 goto L2
ifFalse x > 200 goto L1
ifFalse x != y goto L1
L2: x = 0
In the above translation, the boolean expression is true if the control reaches label 2. If the expression is false, control goes to L1 skipping L2 and the assignment x = 0.
In this section, we consider the translation of boolean expressions into three-address code in the context of statements generated by the following grammar;
{\mathrm{S}}_{1}
{\mathrm{S}}_{1}
{\mathrm{S}}_{2}
S → while(B)
{\mathrm{S}}_{1}
In the productions, non-terminal B represents a boolean expression while non-terminal S represents a statement.
The translation of if(B)
{\mathrm{S}}_{1}
consists of B.code followed by
{\mathrm{S}}_{1}
.code as can be seen in the image below;
The if-else translation
The while translation
Inside B.code we have jumps based on the value of B. If the value is true, control flows to the first instruction of
{\mathrm{S}}_{1}
.code otherwise B is false, therefore control flows to the instruction that follows
{\mathrm{S}}_{1}
.code.
Inherited attributes are used to manage labels for B.code and S.code. With a boolean expression B, two labels B.true and B.false are associated with each other if B is false.
We use statement S to associate the inherited attribute S.next that denotes a label for the instruction immediately following the code for S.
In other cases, the instruction that follows S.code is a jump to another label L.
A jump to L from S.code is avoided using S.next.
We have the following Syntax-directed definition that produces a three-address code for boolean expressions in the context of if, if-else and while statements.
Generating three-address code for booleans;
We have a program that consists of a statement that is generated by P → S. Semantic rules associated with this production initialize S.next to a new label.
P.code will consist of S.code which is followed by a new label S.next. assign in the production S → assign. This acts as a placeholder for the assignment statement.
In translating S → if(B)
{\mathrm{S}}_{1}
, semantic rules as shown in image (4) create a new label B.true then attach it to the first three-address instruction that is generated for
{\mathrm{S}}_{1}
By setting B.false to S.next, we make sure that control skips code for
{\mathrm{S}}_{1}
if B results in a false.
Translating if-else statement that is S → if(B)
{\mathrm{S}}_{1}
{\mathrm{S}}_{2}
, the code for the boolean expression B has jumps out of it to the first instruction of the code for
{\mathrm{S}}_{1}
if B is true and to the first instruction of the code for
{\mathrm{S}}_{2}
, if B is false as can be seen from image (2).
Furthermore, control flows from
{\mathrm{S}}_{1}
{\mathrm{S}}_{2}
to the three-address instruction that immediately follows code for S. S.next is its label which is also an inherited attribute.
S.next is an explicit goto that appears after code for
{\mathrm{S}}_{1}
to skip code for
{\mathrm{S}}_{2}
{\mathrm{S}}_{2}
, we don't need any goto since
{\mathrm{S}}_{2}
.next will be similar to S.next.
For translating S → while(B)
{\mathrm{S}}_{1}
, we form its code from B.code and
{\mathrm{S}}_{1}
.code. This is seen from image (3).
The local variable begin holds a new label that is attached to the first instruction for this while statement, which is also the first instruction for B.
A variable is preferred to an attribute since begin is local to the semantic rules for this production.
Inherited label S.next marks the instruction where control must flow to, that is, when B is false, therefore, B.false is set to S.next. B.true which is a new label is attached to the first instruction for
{\mathrm{S}}_{1}
. Code for B generates a jump to this label, that is, if B is true.
Following the code for
{\mathrm{S}}_{1}
, we place the instruction goto begin. This causes a jump back to the start of the code for boolean expressions. Keep in mind that
{\mathrm{S}}_{1}
.next is currently set to the label begin, therefore, jumps within
{\mathrm{S}}_{1}
.code can go directly to begin.
Code for S →
{\mathrm{S}}_{1}
{\mathrm{S}}_{2}
consists of code for
{\mathrm{S}}_{1}
that is followed by code for
{\mathrm{S}}_{2}
. Semantic rules manage labels, the first instruction after
{\mathrm{S}}_{1}
is the beginning of the code for
{\mathrm{S}}_{2}
, and the instruction after code
{\mathrm{S}}_{2}
code, is the instruction after code for S.
Control-flow translation of boolean expressions.
Semantic rules for boolean expressions in image (5) complement semantic rules for image (4). As we can see from the code layout in images (1-3), a boolean B is translated into three-address instructions that evaluate B using conditional and unconditional jumps to one of two labels, these are B.true and B.false, the former is B is true and the latter otherwise.
Now, we have the following 4th production for B →
{\mathrm{E}}_{1}
{\mathrm{E}}_{2}
, is translated into a comparison three-address instruction with jumps to appropriate locations.
For example, if we have B with the form of a < b. It translates to;
if a < b goto B.true
goto B.false
For the other productions, we translate them as follows;
If B is in the form of
{\mathrm{B}}_{1}
{\mathrm{B}}_{2}
, if B is true, we know that B is true, therefore
{\mathrm{B}}_{1}
.true is similar to B.true. If
{\mathrm{B}}_{1}
is false, B2 must be evaluated. Therefore we make,
{\mathrm{B}}_{1}
.false the label of the first instruction in the code for
{\mathrm{B}}_{2}
. True and false exits for
{\mathrm{B}}_{2}
are similar to true and false exits of B.
Translations of
{\mathrm{B}}_{1}
{\mathrm{B}}_{2}
No code is needed for expression B in the form of !
{\mathrm{B}}_{1}
. Here we just interchange true and false exits of B to get those for
{\mathrm{B}}_{1}
The constants true and false translate into jumps to B.true and B.false.
From the if statement from section 3 (Short-Circuit code), the comparison x > 200 translates into the following;
if x > 200 goto L4
Consider the following instruction;
The above ifFalse instruction takes advantage of the natural flow of instructions from one to the next in sequence and therefore control simply falls through to L4 if x > 200 and thus a jump is avoided.
From the code layouts from images 1, 2 and 3, the statement
{\mathrm{S}}_{1}
follows code for the boolean expression B. Using a special label fall which states, 'don't generate any jump', we adapt semantic rules from images 4 and 5 so as to allow control to fall through from the code for B to the code for
{\mathrm{S}}_{1}
The new rules for S → if(B)
{\mathrm{S}}_{1}
from image 4 set B.true to the label fall;
B.true = fall
B.false =
{\mathrm{S}}_{1}
.next = S.next
S.code = B.code ||
{\mathrm{S}}_{1}
In a similar way, rules for if-else and while statements set B.true to fall.
Now, we adapt semantic rules for boolean expressions which allows control to fall through when possible. Given the following rules for B →
{\mathrm{E}}_{1}
{\mathrm{E}}_{2}
We generate two instructions as shown in image 5, if both B.true and B.false are explicit labels, meaning neither equals fall., Otherwise, if B.true is explicit then B.false must be fall and thus they generate an if instruction that lets control fall through if the condition is false.
Conversely, if B.false is explicit, then they generate an ifFalse instruction. In the last case, both B.true and B.false are fall, here no jump is generated.
We have the following semantic rules for B → B1 || B2;
Here the meaning of the label fall for B is different from the meaning for
{\mathrm{B}}_{1}
. If B.true is fall, that is, control falls through B, if B evaluates to true. Although B evaluates to true if
{\mathrm{B}}_{1}
also evaluates to true,
{\mathrm{B}}_{1}
.true must make sure that control jumps over the code for
{\mathrm{B}}_{2}
to get to the next instruction after B.
{\mathrm{B}}_{1}
evaluates to false, B's truth-value is determined by the value of
{\mathrm{B}}_{2}
and therefore rules from image 7 make sure that
{\mathrm{B}}_{1}
.false corresponds to control falling through from
{\mathrm{B}}_{1}
to the code for
{\mathrm{B}}_{2}
Also note that the semantic rules for B →
{\mathrm{B}}_{1}
{\mathrm{B}}_{2}
will be similar to those of image 7.
Boolean values and jumping code.
Apart from using boolean expressions to change the flow of control in statements, boolean expressions can also be evaluated for their values.
For example, x = true or x = a > b.
An effective way of handling both these roles of boolean expressions is to construct a syntax tree for expressions using one of the following approaches.
Use two passes. Here we build a complete syntax tree for the input the traverse the tree using depth-first, computing translations specified by semantic rules.
Use a single pass for statements and two passes for expressions. Here we translate E in while (E)
{\mathrm{S}}_{1}
{\mathrm{S}}_{1}
is examined. The translation of
{\mathrm{S}}_{1}
would then be done by constructing the tree and traversing it.
We have the following grammar that has a single nonterminal E for expressions;
S → id = E; | if(E) S | while(E)S | S S
E → E || E | E && E | E rel E | E + E | (E) | id | true | false
The non-terminal E governs the flow of control in S → while(E)
{\mathrm{S}}_{1}
, it also denotes a value S → id = E; and E → E + E.
We handle the two roles of expressions by using separate code generation functions. We assume that an attribute E.n denotes the syntax tree node for an expression E and that nodes are objects.
We have the method jump that generates the jumping code at an expression node and the method rvalue that generates code for computing the value of the node into a temporary.
When E appears in S → while(E)
{\mathrm{S}}_{1}
, the method jump is called at the node E.n. jump is implemented based on the rules for boolean expressions as can be seen from image 5.
By calling E.n.jump(t, f), jumping code is generated, here t represents a label for for the first instruction of
{\mathrm{S}}_{1}
.code and f represents the label for S.next.
When E appears in S → id = E;, the method rvalue is called at E.n node. If E is of the form
{\mathrm{E}}_{1}
{\mathrm{E}}_{2}
, E.n.rvalue generates code. If E is of the form
{\mathrm{E}}_{1}
{\mathrm{E}}_{2}
, we will first generate the jumping code for E then assign true or false to a new temporary t at the true and false exits from the jumping code.
While programming, we use boolean expressions for two reasons, to alter the flow of control or to compute logical values. The use of boolean expressions is determined by the syntactic context used.
In short-circuit code, boolean operators such as &&, ||, and ! are translated into jumps.
Compilers Principles, Techniques, & Tools - Alfred V. Aho Monica S. Lam
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Concatenation and Kleen closure operations.
Closure under regular operations
A language is a set of strings from an a finite or infinite alphabet. Here we discuss three simple but important operations used on languages, these are union, concatenation and kleen closure.
Others include, intersection and difference.
There are three operations which are performed on languages namely: union, concatenation and kleen closure.
Let A and B be two languages over a similar alphabet:
The union of A and B is defined as:
A ∪ B = {w : w ∈ A or w ∈ B}
The concatenation of the two is defined as:
AB = {ww′ : w ∈ A and w′ ∈ B} where AB is the set of all strings obtained by taking an arbitrary string w in A and an arbitrary string w′ in B then putting them together such that the former is to the left of the latter.
The kleen closure of A is defined as:
A* = {
{\mathrm{u}}_{\mathrm{1}}
{\mathrm{u}}_{\mathrm{2}}
{\mathrm{u}}_{\mathrm{k}}
: k ≥ 0 and
{\mathrm{u}}_{\mathrm{i}}
∈ A for all i = 1, 2, ..., k}
Where A* is obtained by taking an infinite number of strings in A and putting them together.
Note that k cannot be zero, in this case it will correspond to an empty string ϵ and therefore ϵ ∈ A*.
Let A = {0, 01} and B = {1, 10}.
union: A ∪ B= {0, 01, 1, 10}
concatenation: AB = {01, 010, 011, 0110}
kleen closure: A* = {ϵ, 0, 01, 00, 001, 010, 0101, 000, 0001, 00101, ...}
If Σ = {0, 1}, Σ* is the set of all binary strings including an empty string.
Alternative kleen closure definition for a language A,
{\mathrm{A}}^{0}
= {ϵ}.
For k ≥ 1,
{\mathrm{A}}^{k}
{\mathrm{AA}}^{k-1}
{\mathrm{A}}^{k}
is a concatenation of the languages A and
{\mathrm{A}}^{k-1}
We define A* as follows:
A* =
\bigcup _{\mathrm{k}=0}^{\infty }{\mathrm{A}}^{\mathrm{k}}
Theorem 1: The set of regular languages is closed under the union operation, that is, if A and B are regular languages over a similar alphabet Σ, A ∪ B is a regular language.
Proof: A and B are regular languages, therefore there are finite automata M1 = (Q1, Σ, δ1, q1, F1) and M2 = (Q2, Σ, δ2, q2, F2) that accept A and B respectively.
The next step is to construct a finite automaton M that accepts A ∪ B so as to prove A ∪ B is regular.
M must have the following property:
for every string w ∈ Σ*, M accepts w ⇔ M1 accepts w or M2 accepts w.
Assuming M could do the following:
Starting in the start state q1 of M1, M runs M1 on w.
After reading w, M1 is in the F1 state, then w ∈ A and therfore w ∈ A ∪ B and so M accepts w.
If after reading w, M1 is in the state other than F1, w ∉ A and M runs M2 on w, starting in the start state q2 of M2. If after reading w, M2 is in F2 state, we know w ∈ A ∪ B and so M accepts w. Otherwise w ∉ A ∪ B and M rejects w.
This idea won't work since the finite automaton M reads the input string once.
A better approach is to run M1 and M2 simultaneously.
First, we define the set Q of states of M as the cartesian product Q1 x Q2. If M is in the state (r1, r2), this indicates that;
If M1 reads the input string to this point, it will be in r1 state.
If M2 reads the input string to the same point, it will be in r2 state.
An therefore we have the finite automaton M = (Q,Σ, δ, q, F) where:
Q = Q1 x Q2 = {(r1, r2) : r1 ∈ Q1 and r2 ∈ Q2}, note that |Q| = |Q1| x |Q2| and this is finite.
Σ is the alphabet of A and B, here we assume the both A and B are languages over the same alphabet.
M's start state q is equal to (q1, q2), i.e, q = (q1, q2).
Set F of accept states of M is: F = {(r1, r2) : r1 ∈ F1 or r2 ∈ F2} = (F1 x Q2) ∪ (Q1 x F2).
The transition function δ : Q x Σ → Q is:
δ((r1, r2), a) = (δ1(r1, a), δ2(r2, a)), for all r1 ∈ Q1, r2 ∈ Q2, a ∈ Σ
We complete the proof by showing that this finite automaton M accepts the language A ∪ B. This is clear from what we seen from above.
To give a formal proof we use extended transition functions
\stackrel{-}{\delta 1}
\stackrel{-}{\delta 2}
M accepts w ⇔ M1 accepts w or M2 accepts w.
That is, M accepts w ⇔
\stackrel{-}{\delta 1}
(q1, w) ∈ F1 or
\stackrel{-}{\delta 2}
(q2, w) ∈ F2.
And in terms of the extended transition function
\stackrel{-}{\delta }
of the transition function δ of M, it becomes:
\stackrel{-}{\delta }
((q1, q2),w) ∈ F ⇔
\stackrel{-}{\delta 1}
\stackrel{-}{\delta 2}
(q2, w) ∈ F2
By applying the extended transition function definition we can see that
\stackrel{-}{\delta }
((q1, q2), w) = (
\stackrel{-}{\delta 1}
(q1, w),
\stackrel{-}{\delta 2}
(q2, w)).
The above implies that the former is true and thus M accepts the language A ∪ B.
Theorem 2: Regular languages are closed under concatenation and kleen closure operations and we shall see in the proof below:
Proof: Let A and B be two languages and M1 and M2 be finite automata that accept both languages respectively.
Now to construct a finite automaton M that accepts the concatenation of AB.
Let u be an input string, M has to decide whether it can be broken into two strings w and w′ such that
w ∈ A and w′ ∈ B.
In other words, M has to decide whether or not u can be broken down into two substrings such that the first is accepted by M1 and the second is accepted by M2.
The complexity comes in where by M has to make the decision by scanning the string only once.
If u ∈ AB, then M decides if it can be broken and where to break u into two substrings such that the first substring is in the A and the second is in B, similarly if u ∉ AB, M has to decide that the string u cannot be broken into substrings all this is done in a single scan of the input string.
It is even more difficult to prove A* is a regular language if A is regular since for this proof we need a finite automaton that when given an arbitrary input string u, decides whether it can be broken into substrings such that each substring is in A.
The issue is that the finite automaton u ∈ A* has to determine the number of substrings and where to break and do this in a single scan of u.
As previously stated, if A and B are regular languages, both AB and A* are also regular.
To prove this claim we use a more general type of finite automaton, Non-deterministic finite automaton which we have discussed in the prerequisite article.
With this automaton we can easily prove the theorem that the class of regular languages is closed under the concatenation and kleen closure operations.
Closure under regular operations.
We have the following theorem from the prerequisite article.
Theorem 3: Let A be a language. Then A is regular if and only if there exists a nondeterministic finite automaton that accepts A.
Here we will use the concept of NFA from the prerequisite article together with the above theorem to to prove that regular languages are after all closed under regular operations.
An alternative proof of theorem 1:
Theorem 1.1: The set of regular languages is closed under the union operation, that is, if A1 and A2 are regular languages over a similar alphabet Σ, then A1 ∪ A2 is also a regular language.
Proof: Since A1 is regular, by theorem 3: and NFA M1 = (Q1,Σ, δ1, q1, F1) such that A1 = (LM1).
Similarly, there exists a NFA M2 = (Q2,Σ, δ2, q2, F2) such that A2 = L(M2).
We also assume that Q1 ∩ Q2 = ∅ because otherwise we can give new names to states Q1 and Q2.
From the two NFAs we construct a NFA M = (Q,Σ, δ, q0, F) such that L(M) = A1 ∪ A2
NFA M is defined as:
Q ={q0} ∪ Q1 ∪ Q2 where q0 is a new state.
q0 is the start state of M.
δ : Q × Σϵ → P(Q) where for any r ∈ Q and a ∈ Σϵ
*δ(r, a) = {
\left\{\begin{array}{ll}\mathrm{\delta }1\left(\mathrm{r},\mathrm{a}\right)& \mathrm{if} \mathrm{r}\in \mathrm{Q}1\\ \mathrm{\delta }2\left(\mathrm{r},\mathrm{a}\right)& \mathrm{if} \mathrm{r}\in \mathrm{Q}2\\ \left\{\mathrm{q}1,\mathrm{q}2\right\}& \mathrm{if} \mathrm{r}=\mathrm{q}0 \mathrm{and} \mathrm{a}=\mathrm{ϵ}\\ \varnothing & \mathrm{if} \mathrm{r}=\mathrm{q}0 \mathrm{and} \mathrm{a}\ne \mathrm{ϵ}\end{array}\right\
The set of regular languages is closed under the concatenation operation, that is, if A1 and A2 are regular languages over a similar alphabet Σ, then A1A2 is also a regular language.
Proof: Let M1 = (Q1,Σ, δ1, q1, F1) be an NFA such that A1 = L(M1). Similarly, let M2 = (Q2,Σ, δ2, q2, F2) be an NFA such that *A2 = L(M2. From above proof of theorem 1.1 we assume that Q1 ∩ Q2 = ∅ and therefore construct an NFA M = (Q,Σ, δ, q0, F) such that L(M) = A1A2.
NFA M is defined as follows,
Q = Q1 ∪ Q2.
δ : Q × Σϵ → P(Q), where for any r ∈ Q and a ∈ Σϵ
δ(r, a) = {
\left\{\begin{array}{ll}\mathrm{\delta }1\left(\mathrm{r},\mathrm{a}\right)& \mathrm{if} \mathrm{r}\in \mathrm{Q}1 \mathrm{and} \mathrm{r}\notin {\mathrm{F}}_{1}\\ \mathrm{\delta }1\left(\mathrm{r},\mathrm{a}\right)& \mathrm{if} \mathrm{r}\in {\mathrm{F}}_{1} \mathrm{and} \mathrm{a}\ne \mathrm{ϵ}\\ \mathrm{\delta }1\left(\mathrm{r},\mathrm{a}\right)\cup \left\{\mathrm{q}2\right\}& \mathrm{if} \mathrm{r}={\mathrm{F}}_{1} \mathrm{and} \mathrm{a}=\mathrm{ϵ}\\ \mathrm{\delta }2\left(\mathrm{r},\mathrm{a}\right)& \mathrm{if} \mathrm{r}=\mathrm{Q}2\end{array}\right\
Theorem 5: The set of regular languages is closed under the kleen closure operation, that is, If A is a regular language, the A* is also a regular language.
Proof: Let Σ be the alphabet of A and N = (Q1,Σ, δ1, q1, F1) be a NFA such that A = L(N). We now construct a NFA M = (Q,Σ, δ, q0, F) such that L(M) = A*.
We define it as follows,
Q = {q0} ∪ Q1 where q0 is a new state.
F = {q0} ∪ F1, (considering that ϵ ∈ A* has to be an accept state)
δ : Q × Σϵ → P(Q), where for any r ∈ Q and a Σϵ
\left\{\begin{array}{ll}\mathrm{\delta }1\left(\mathrm{r},\mathrm{a}\right)& \mathrm{if} \mathrm{r}\in \mathrm{Q}1 \mathrm{and} \mathrm{r}\notin {\mathrm{F}}_{1}\\ \mathrm{\delta }1\left(\mathrm{r},\mathrm{a}\right)& \mathrm{if} \mathrm{r}\in {\mathrm{F}}_{1} \mathrm{and} \mathrm{a}\ne \mathrm{ϵ}\\ \mathrm{\delta }1\left(\mathrm{r},\mathrm{a}\right)\cup \left\{\mathrm{q}1\right\}& \mathrm{if} \mathrm{r}={\mathrm{F}}_{1} \mathrm{and} \mathrm{a}=\mathrm{ϵ}\\ \left\{\mathrm{q}1\right\}& \mathrm{if} \mathrm{r}=\mathrm{q}0 \mathrm{and} \mathrm{a}=\mathrm{ϵ}\\ \varnothing & \mathrm{if} \mathrm{r}=\mathrm{q}0 \mathrm{and} \mathrm{a}\ne \mathrm{ϵ}\end{array}\right\
Theorem 6: The set of regular languages is closed under the complement and intersection operations:
If a language A is over an alphabet Σ, the complement
\stackrel{-}{\mathrm{A}}
= {w ∈ Σ* : w ∉ A} is also a regular language.
If A1 and A2 are regular languages over a similar alphabet Σ the intersection A1 ∩ A2 = {w ∈ Σ*: w ∈ A1 and w ∈ A2} is also a regular language.
We have discussed the union of two languages for example A and B which produce a set of strings either in A or B or both is written as:
Their concatenation is denoted by AB = {ww′ : w ∈ A and w′ ∈ B} where AB is the set of all strings obtained by taking an arbitrary string w in A and an arbitrary string w′ in B then putting them together such that the former is to the left of the latter.
A kleen closure refers to zero or more occurrences of symbols in a string, it also includes an empty string ϵ which has a length of zero.
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Subroutines | Brilliant Math & Science Wiki
Red Diamond, Agnishom Chattopadhyay, A Former Brilliant Member, and
Vasanth M M
A function is a set of code statements that are executed in sequence to affect some outcome. In the most general sense, a function can be almost anything, e.g. the square root operation
x \mapsto \sqrt{x}
, an instruction to draw the Mona Lisa with a given color palette, a call to align the residue sequence of homologous proteins across evolutionarily distant organisms, an instruction to update the status on your Facebook profile, a command to update the balance on a customer's savings account, etc.
Immediately we can see a division in the example functions listed above: the first three take an argument (a number, a color palette, amino acid sequences) and return an output (a number, a painting, a sequence alignment), while the last two take an argument (text string, a transaction) and make a change to a variable that lives at some other place in the codebase (a side effect).
The first kind of function (termed a pure function) takes input and gives an output in a way that's completely self-contained, whereas the second kind takes an input, performs an operation that changes things in the outside world, possibly using other things from the outside world. Given the same inputs, a pure function will return the same answer every single time.
It is easy (relatively) to reason about programs which are built using pure functions, and in fact there exist entire programming languages which were built to embrace this style (Haskell, Lisp, OCaml, F#, etc.). While side effects are necessary for a program that "does something" (i.e. user interfaces), they're more difficult to reason through, and should be avoided when possible.
A pure function is a map from a set of inputs to an output
f : I \to O
such that every element in
I
has one and only one image in
O
I
O
are sets of data, other functions, etc.
For example, we can define a function that takes two numbers (
x
n
) and returns
x
n
Implemented in OCaml, the power function (for positive reals x and n) is given by
let power x n = x ** n ;;
This function maps
f : \mathbb{R}^+ \times \mathbb{R}^+ \to \mathbb{R}^+
with no side effects.
We might want a function, prime, that takes a given integer and determines whether or not it is prime. A particularly inefficient algorithm for achieving this is a brute force test for divisibility by all numbers less than or equal to our integer.
Brute force primality in Python
def prime(candidate):
for number in range(2, candidate):
if candidate % number == 0:
f : \mathbb{N} \to \{\text{True},\text{False}\}
with no side effects. If
x \mod n = 0
n \in [0, N-1]
, our function breaks out of the for loop, returns False, and terminates the function. If it makes it through the list without finding any divisors, the loop finishes without returning anything, and the return True statement is executed.
Suppose the global variable a is initially set to 7,
and we have the two functions add_a, which returns the value x + a, and square_a, which reassigns the square of a to a.
def add_a(x):
def square_a():
If we call add_a(3) without calling square_a, we'll get 10, every time we call it. However, if we call square_a first, add_a(3) will return 52, and if we call square_a twice, add_a(3) will return 2404. This is a consequence of the fact that add_a depends on the global variable a that is not an explicit argument of add_a. This makes the function susceptible to changes in a that are outside the scope of add_a. If we write a program consisting of calls to add_a and square_a, the result depends crucially on the sequence in which we called the functions.
Thus, in order to reason about the output of add_a, we need to know the full history of our environment, specifically, how many times square_a has been called. The flagrant use of global variables can lead to situations where debugging requires a full-blown forensic analysis. We can largely avoid these problems by making our function dependencies explicit, thus isolating the task of debugging to single functions, and promoting a more modular style of programming.
In writing a function, it is important to stay clear about what variables are accessible at different levels of code. If we try to reference at the top level, a variable that is defined deeper in the code, or expect a certain change to have been made to our variable, we'll run into problems.
For example, consider the function kronecker(n) below which is meant to return the Kronecker tensor
\delta
n
. (In computing, the Kronecker tensor
\delta
n
is simply the identity matrix of order
n \times n
i=j
\delta_{ij} = 1
\delta_{ij} = 0
def kronecker(n):
tensor = []
def calculate(i, j):
calculate(i, j)
row.append(var)
tensor.append(row)
If we call kronecker(3), we expect to get
However, if we call kronecker(3) as implemented above, we find that we have a scope error.
What went wrong? In each loop over the j index, we calculate a new tensor element var (using the function calculate(i, j)) and append it to our newly formed list row. At the end of each loop over i, row is appended to the list tensor, to accumulate our results.
The problem is that we expect var to be defined at the level of the j loop after we define it in calculate(i, j). However, var is defined at the deepest level of our code, and so, won't be defined at a higher level. The usual convention is that variables are available to code at the same level at which they're defined, and to all levels below that, but not at levels above their definition (see diagram below).
One way to fix our kronecker function would be
Functions need not be encapsulations of useful snippets of code, whose sequence of events is explicitly outlined, e.g. "loop over this list of numbers, and print the square of each number." Functions can also be defined as a set of rules such that any given input maps to a single output. In this way, we can carefully define a computation in our code, but leave the explicit computation up to the computer. This method is called recursion, and works by declaring a set of defined base cases in terms of which all other arguments to the function can be computed.
For a much abused, but clear example, let us consider the integer factorial function. One way to define
n!
f(n) = n! = n \times (n-1) \times \ldots 3 \times 2 \times 1
, so that for instance,
7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1
This is a fine definition, but it is tedious to write out, especially as our arguments get bigger. However, there is a clear pattern in the definition of
n!
7!
as written above, we see that we can also write this as
7 \times 6!
. This insight allows us to define the integer factorial function as follows
f(n) = \begin{cases} n\times f(n-1) & \text{ if \$n > 1\$} \\ 1 & \text{ if \$n = 1\$} \\ \end{cases}
To see how this simplifies our code, let us implement both factorials in Python.
Implemented without recursion, we have the following code to calculate
n!
for number in range(1, n + 1):
with recursion we have
Without using recursion, we have to make use of some unsavory programming practices. For one, we have used a sort of global variable, in result, whose value gets updated with every run through the loop. We also have to specify a range which opens us up to a potential fencepost (off-by-one) error.
In the recursive case, we stipulate only the essentials: the base case, and the recursive definition. Here, any mistake is glaringly obvious because it will come from the mathematical definition, not from a subtle detail of our implementation. Using recursion, we are more likely to end up with clean code.
You have in your wallet $300, which you want to spend completely. You decide to spend all of it by buying food from a fancy restaurant with the following menu:
tofu scramble: $1
brunch combo: $20
saffron infused peach tea: $50
truffles: $100
caviar: $200
O
be the number of different ways that you can spend exactly $300. What are the last 3 digits of
O
Here we have seen the benefit of recursion on organization and avoiding mistakes. However, the structure of the factorial calculation is quite simple, just a string of numbers that we multiply together. Let's try to calculate the Fibonacci numbers, a problem with a slightly richer structure.
As we did with the integer factorial function, let's write a recursive definition for the
n
th Fibonacci number:
f(n) = \begin{cases} f(n-1) + f(n-2) & \text{ if } n \geq 2 \\ 1 & \text{ if } n \in\{0, 1\} \\ \end{cases}
For example, to find the third Fibonacci number, this definition yields
\begin{aligned} f(3) &= f(2) + f(1) \\ &= \left[f(1) + f(0)\right] + f(1) \\ &= 1 + 1 + 1 \\ &= 3 \end{aligned}
For the fourth, we have
\begin{aligned} f(4) &= f(3) + f(2) \\ &= \left[f(2) + f(1)\right] + \left[f(1)+f(0)\right] \\ &= \left[\left[f(1) + f(0)\right] + f(1)\right] + \left[f(1)+f(0)\right] \\ &= 1+1+1+1+1 \\ &= 5 \end{aligned}
We can implement this function in Python as follows
if number in [0, 1]:
return fib(number - 2) + fib(number - 1)
This code works fine for computing fib(n) when n is small, however, it is cripplingly slow as n grows. From the cases we tried above, we can see that the computation branches out quickly.
Let's inspect the diagram for
f(5)
below. At the first branch, we have
f(5)\to f(4)+f(3)
f(4)
branches, it leads to computation of
f(3)
f(2)
. However, we already needed to compute
f(3)
from the first branching. Once we compute
f(3)
a single time, it makes little sense to do it again.
As we compute bigger and bigger Fibonacci numbers, we'll find more and more of these cases, where the tree repeats computations that have already been performed (how many numbers must we compute in total?). Unless we cut these branches from the tree, we are going to waste a lot of processor time repeating things we've already done.
The Ackermann function is a computable function which grows very, very quickly as its inputs grow. For example, while
A(1,2),
A (2,2),
A(3,2)
4,7,
29,
A(4,2) \approx 2 \times 10^{19728}
The Ackermann function can be defined as follows:
A(m,n) = \begin{cases} n+1 & \mbox{if } m = 0 \\ A(m-1, 1) & \mbox{if } m > 0 \mbox{ and } n = 0 \\ A(m-1, A(m, n-1)) & \mbox{if } m > 0 \mbox{ and } n > 0. \end{cases}
A(3,6)?
As we noted, the recursive Fibonacci definition we implemented above leads to a lot of redundant computation. We can save time by remembering the answer to computations after the first time we perform them. This way, fib(n) will be computed a maximum of one time for each value of n. This method of halting computation of previously obtained values is known as "memoization" and can often be employed to obtain significant speedups.
Memoization of the recursive Fibonacci function
First let us make a lookup table called fib_history, and change our code so that fib consults this table before committing to a new computation, i.e. it will only execute a new branch if the parent of that branch has not been computed before.
fib_history = {1:1, 0:1};
if n not in fib_history:
fib_history[n] = fib_memo(n - 2) + fib_memo(n - 1)
return fib_history[n]
Here we see that fib_memo asks, politely, whether the value of fib_memo(n) is yet in fib_history, and only if that answer is no, will it enter the recursion. In this way, we only ever evaluate fib_memo a maximum of n times in determining fib_memo(n).
Revisiting the tree representation of our computation allows us to visualize the cost savings:
``fib_history accumulates knowledge over time, so we can even use the results from previous calls to benefit future calls. For example, if we've called fib_memo(67) in the past, we don't need to make 68 calls to find fib_memo(68) as nearly all the work has already been performed.
Below, we plot the performance of fib against fib_memo for moderate values of n. It is clear that fib's runtime grows exponentially with n whereas fib_memo has essentially constant cost for this range of n.
It is common to find functions tend to have the same value for some of their arguments, each time they're called. Or, you might have a function whose behavior you'd like to be able to modify, if needed, but which in most cases will run without the need for adjustments. In both of these cases, it is convenient to have default values for function arguments.
An optional argument helps true love last
Suppose you're writing an app to automatically answer your text messages when you're busy. Also suppose there exists some method by which an incoming text message can trigger functions, namely one you've called send_text. The default return of send_text() is "Sorry, can't talk now, in a meeting.".
def send_text(number):
message = "Sorry, can't talk now, in a meeting."
sms.message(number, message)
This works fine most of the time, but can get you in trouble with your significant other, who might think you're exploring other options and ignoring them. In this case, it is convenient to call the function with an optional argument is_honeybear to indicate that your significant other is the recipient. In this case the call send_text(is_honeybear = True), prompts the app to append the message so that it reads "Sorry, can't talk now, in a meeting. I love you so much!".
def send_text(number, is_honeybear=False):
alt_message = " I love you so much!"
to_send = message + (alt_message if is_honeybear else "")
In the first case, where somebody other than the significant other is calling, send_text uses the default value for is_honeybear which is False.
In practice, default values tend to reveal themselves as you begin to use a function. In the design phase, you usually make the function and arguments as simple as possible, but after use in the codebase, it becomes clear that sometimes you need to override certain values.
As a concrete example, suppose you've written a function that reads in a tabular dataset (a spreadsheet with column names), and by default, expects columns to be separated by tabs:
with file(filename) as fp:
return csv.reader(fp, delimiter='\t')
If we call read_data on a dataset like the file sample_data.txt
Product Amount UnitPrice
wagon 2 30.00
shovel 1 10.00
fertilizer 22 1.50
It will work fine, because the argument for the separator tells pandas.read_csv to expect columns separated by tabs, as we find in the file.
However, you start receiving similarly formatted spreadsheets from a new client to analyze, but instead of separating the columns by tabs, they're separated by commas. Because the tasks are so similar we don't want to write a separate function to handle comma-separated and tab-separated data.
We can fix this by giving read_data an optional argument for the separator, separator. For convenience, can set the default value to the most common argument we expect to use in importing datasets, the tab separator \t.
def read_data(filename, separator='\t'):
return csv.reader(fp, delimiter=separator)
Thus, whenever we call read_data on tab separated data, we can just provide the filename. When we need to specify the separator to be something different, commas for instance, we can call read_data('sample_data.txt', separator=',').
Cite as: Subroutines. Brilliant.org. Retrieved from https://brilliant.org/wiki/subroutines/
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Residuals of fitted generalized linear mixed-effects model - MATLAB - MathWorks Deutschland
Residuals of fitted generalized linear mixed-effects model
r = residuals(glme)
r = residuals(glme,Name,Value)
r = residuals(glme) returns the raw conditional residuals from a fitted generalized linear mixed-effects model glme.
r = residuals(glme,Name,Value) returns the residuals using additional options specified by one or more Name,Value pair arguments. For example, you can specify to return Pearson residuals for the model.
Conditional residuals include contributions from both fixed- and random-effects predictors. Marginal residuals include contribution from only fixed effects. To obtain marginal residual values, residuals computes the conditional mean of the response with the empirical Bayes predictor vector of random effects, b, set to 0.
Residual type, specified as the comma-separated pair consisting of 'ResidualType' and one of the following.
{r}_{ci}={y}_{i}-{g}^{-1}\left({x}_{i}^{T}\stackrel{^}{\beta }+{z}_{i}^{T}\stackrel{^}{b}+{\delta }_{i}\right)
{r}_{mi}={y}_{i}-{g}^{-1}\left({x}_{i}^{T}\stackrel{^}{\beta }+{\delta }_{i}\right)
{r}_{ci}^{pearson}=\frac{{r}_{ci}}{\sqrt{\frac{\stackrel{^}{{\sigma }^{2}}}{{w}_{i}}{v}_{i}\left({\mu }_{i}\left(\stackrel{^}{\beta },\stackrel{^}{b}\right)\right)}}
{r}_{mi}^{pearson}=\frac{{r}_{mi}}{\sqrt{\frac{\stackrel{^}{{\sigma }^{2}}}{{w}_{i}}{v}_{i}\left({\mu }_{i}\left(\stackrel{^}{\beta },0\right)\right)}}
\stackrel{^}{\beta }
\stackrel{^}{b}
Residuals of the fitted generalized linear mixed-effects model glme returned as an n-by-1 vector, where n is the number of observations.
{\text{defects}}_{ij}\sim \text{Poisson}\left({\mu }_{ij}\right)
\mathrm{log}\left({\mu }_{ij}\right)={\beta }_{0}+{\beta }_{1}{\text{newprocess}}_{ij}+{\beta }_{2}{\text{time}\text{_}\text{dev}}_{ij}+{\beta }_{3}{\text{temp}\text{_}\text{dev}}_{ij}+{\beta }_{4}{\text{supplier}\text{_}\text{C}}_{ij}+{\beta }_{5}{\text{supplier}\text{_}\text{B}}_{ij}+{b}_{i},
{\text{defects}}_{ij}
i
j
{\mu }_{ij}
i
i=1,2,...,20
j
j=1,2,...,5
{\text{newprocess}}_{ij}
{\text{time}\text{_}\text{dev}}_{ij}
{\text{temp}\text{_}\text{dev}}_{ij}
i
j
{\text{newprocess}}_{ij}
i
j
{\text{supplier}\text{_}\text{C}}_{ij}
{\text{supplier}\text{_}\text{B}}_{ij}
i
j
{b}_{i}\sim N\left(0,{\sigma }_{b}^{2}\right)
i
Generate the conditional Pearson residuals and the conditional fitted values from the model.
r = residuals(glme,'ResidualType','Pearson');
Display the first ten rows of the Pearson residuals.
r(1:10)
Plot the Pearson residuals versus the fitted values, to check for signs of nonconstant variance among the residuals (heteroscedasticity).
scatter(mufit,r)
GeneralizedLinearMixedModel | fitted | response | designMatrix
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Describe in words the surface whose equation is given. 0=\pi
Describe in words the surface whose equation is given. 0=\pi /4
0=\frac{\pi }{4}
Consider the equation we are given. Since angle is fixed, it means that only two dimensions can vary. These are r and z. This means that the surface is a vertical half plane since r > 0 and z can be anything. Result: The describer surface is a half-plane.
A=\left[\begin{array}{cc}3& 1\\ 1& 1\\ 1& 4\end{array}\right],b\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]
\stackrel{―}{x}=
Set up an equation and solve each problem. Explain how you would solve the equation
3\left(x-1\right)\left(x+2\right)=0
and also how you would solve the equation
x\left(x-1\right)\left(x+2\right)=0
Determine whether the given problem is an equation or an expression. If it is an equation, then solve. If it is an expression, then simplify. 2n/5 - n/6 = - 7/10
\mathrm{sin}2x-\mathrm{sin}x=0
\left[0,2\pi \right]
Determine whether the given problem is an equation or an expression. If it is an equation, then solve. If it is an expression, then simplify.
n+\frac{\left(2n-3\right)}{9}-2=\frac{\left(2n+1\right)}{3}
y '(x)=-y+2
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Linked List | Brilliant Math & Science Wiki
Kristian Takvam, Thaddeus Abiy, Alex Chumbley, and
Sun mi Kang
Linked lists are linear data structures that hold data in individual objects called nodes. These nodes hold both the data and a reference to the next node in the list.
Linked lists are often used because of their efficient insertion and deletion. They can be used to implement stacks, queues, and other abstract data types.
Properties of Linked Lists
Sample Java Implementation
Iteration and Recursion on Linked Lists
You can visualize a linked list using the following image:
Each node contains a value--in this case, an integer--and a reference (also known as a pointer) to the next node. The last node, in this case 2, points to a null node. This means the list is at its end.
However, linked lists do have some drawbacks. Unlike arrays, linked lists aren't fast at finding the
n^\text{th}
item. To find a node at position
, you have to start the search at the first node in the linked list, following the path of references
n
times. Also, because linked lists are inherently sequential in the forward direction, operations like backward traversal--visiting every node starting from the end and ending in the front--is especially cumbersome.
Additionally, linked lists use more storage than the array due to their property of referencing the next node in the linked list.
Finally, unlike an array whose values are all stored in contiguous memory, a linked list's nodes are at arbitrary, possibly far apart locations in memory. This means that the CPU can't effectively cache the contents of a linked list nearly as well as an array resulting in poor performance. This is the main reason why ring buffers are used to implement queues instead of linked lists in high-performance applications where middle insertion and deletion functionality isn't needed (e.g. network drivers).
Linked lists have most of their benefit when it comes to the insertion and deletion of nodes in the list. Unlike the dynamic array, insertion and deletion at any part of the list takes constant time.
However, unlike dynamic arrays, accessing the data in these nodes takes linear time because of the need to search through the entire list via pointers. It's also important to note that there is no way of optimizing search in linked lists. In the array, we could at least keep the array sorted. However, since we don't know how long the linked list is, there is no way of performing a binary search:
\begin{array}{c}&&\text{Insertion - O(1),} &\text{Deletion - O(1),} \\ &\text{Indexing - O(n),} &\text{Search - O(n)}.\end{array}
Linked lists hold two main pieces of information (the value and pointer) per node. This means that the amount of data stored increases linearly with the number of nodes in the list. Therefore, the space complexity of the linked list is linear:
\begin{array}{c}&&\text{Space - O(n)} \end{array}.
Example code in Java:
public Node(int v , Node n){
The code above describes a node object; these types of objects will usually contain a value (stored in value) and a pointer next which points to another node somewhere in memory. This can be represented as a list with one element in it by writing
Node one = new Node(6, null)
Variable one points to a node that contains the value 6. The next field of the cell is null, which could indicate the end of the list. This can be represented as a list with two elements in it by writing
Node two = new Node(4, one);
Variable two now points to a cell with a value of 4. The next field of that node points to a node with a value of 6. The next field of the second points to a null object. So, the list is
4 \rightarrow 6
. A 3 element list is created like this:
Node three = new Node(5, two)
Now, it's
5 \rightarrow 4 \rightarrow 6
Note: In this implementation, we have to pass in the pointers to the next node in the list. We do this for simplicity, but usually code has to be written to handle the insertion and deletion of nodes in the list.
It is not necessary for us to create a local variable whenever we need to add a new item to a linked list. A linked list can also be constructed by reading from an array.
public static Node createArray(int[] array){
Node list = null;
list = new Node(i, list)
A linked list can be printed either by using recursion or using a loop (iteration). Here is an iterative way of printing a linked list:
public static void printIteratively(Node cell) {
System.out.println(cell.value);
cell = cell.next;
We can write the same method recursively:
public static void printRecursively(Node cell) {
printRecursively(cell.next);
Write a method that determines the length of a given linked list recursively.
Using our discussed method of iteration, the method can be written as follows:
public static void length(Node cell) {
while (cell != null){
Now write a method that determines the length of a linked list recursively.
We can solve this by considering the base case (when P is null) and the recursive case
\text{length}(n) = \text{length}(n-1) + 1.
public static int length(Node cell) {
return 1 + length(cell.next);
Most methods on linked lists can be implemented both recursively and iteratively. But recursive solutions are more convenient when dealing with linked lists in general. Consider, for example, the following method that takes a linked list as an input and reverses it. For the recursive version, we again identify the base case ( when the list is empty) and the recursive case:
\text{reverse}( Node ) = \text{reverse}(Node.next) + Node.value
Node reverse(Node cell) {
else if(cell.next == null)
Node second_segment = reverse(cell.next);
//cell.next points to the last item in the second segment
cell.next.next = cell;
//the next of cell.next now points to cell
cell.next = null;
//cell is now the last item in the list
return second_segment;
So far we have only talked about singly linked lists, lists in which pointers go from each cell to the next in one direction. It is however possible to create a doubly linked lists where pointers go both ways.
This allows us to traverse the list in both directions, and it also makes operations such as deletion easier. Because we have pointers to both the nodes before and after the node we'd like to delete, we don't need any more information beyond our target node. In a singly linked list, we also need the pointer to the node we'd like to delete.
Node (E v, Node n, Node p) {
value=v; next=n; prev=p;
Node () { // used only to create header Node in a list
// that's initially empty
next=prev=this;
private Node header = new Node();
return header.next == header;
private E delete(Node p) {
It is now more convenient to traverse the linked list and we can move both backwards and forwards using the prev and next fields, respectively. The design of this type node allows flexibility of storing any data type as the linked list data.
Implement a method to add a node to the end of a linked list:
public void addLast(int s){
Cell p = new Cell(s , header , header.prev);
header.prev.next = p;
header.prev = p;
Cite as: Linked List. Brilliant.org. Retrieved from https://brilliant.org/wiki/linked-lists/
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Applying Differentiation Rules to Exponential Functions | Brilliant Math & Science Wiki
Mahindra Jain, Aditya Virani, Yuliya Skripchenko, and
Thiago Baldissarelli
We have learned that the derivative of an exponential function is given by the following formulas:
\begin{array}{c}&y=e^x &\implies y'=e^x\\ y=e^{f(x)} &\implies y'=e^{f(x)}f'(x)\\ y=a^x &\implies y'=a^x(\ln a)\\ y=a^{f(x)} &\implies y'=a^{f(x)}(\ln a)\cdot f'(x), \end{array}
\frac{dy}{dx}
y'
\frac{df(x)}{dx}
f'(x),
a
is a random base such that
a>0
a\ne1.
When differentiating complex exponential functions, just stick to the formulas above along with the differentiation rules that we have learned earlier. To remind, here is a list of some differentiation rules:
\begin{array}{c}&y=cf(x) &\implies y'=cf'(x)\\ y=f(x)+g(x) &\implies y'=f'(x)+g'(x)\\ y=f(x)g(x) &\implies y'=f'(x)g(x)+f(x)g'(x)\\ y=\frac{f(x)}{g(x)} &\implies y'=\frac{g(x)f'(x)-f(x)g'(x)}{\big(g(x)\big)^2}\\ y=g\big(f(x)\big) &\implies y'=g'\big(f(x)\big)\cdot f'(x). \end{array}
Another rule that might be handy is one that we have learned when studying logarithms in algebra. Recall that
a^{\log_b c}=c^{\log_b a}.
Now let's try differentiating complex exponential functions using these rules.
y=3^{x^3+1}?
y=a^{f(x)},
y'=a^{f(x)}(\ln a)\cdot f'(x).
y=3^{x^3+1}
in this problem, we have
\begin{aligned} y'&=3^{x^3+1} (\ln 3) \cdot \left(x^3+1\right)'\\ &=3^{x^3+1} (\ln 3) \cdot 3x^2\\ &=3^{x^3+2}x^2 \cdot \ln 3. \ _\square \end{aligned}
y=5^{\sin x}?
y=a^{f(x)},
y'=a^{f(x)}(\ln a)\cdot f'(x).
y=5^{\sin x}
\begin{aligned} y'&=5^{\sin x}(\ln 5) \cdot (\sin x)'\\ &=5^{\sin x}(\ln 5) \cdot (\cos x)\\ &=5^{\sin x} (\cos x) \cdot (\ln 5). \ _\square \end{aligned}
f(x)=5^{\cos3x }?
This problem can be solved using the same method as example #2. However, here is another way to look at it.
Converting the base to
e
e^{\ln a^x} =a^x
\begin{aligned} f(x)&=5^{\cos3x }\\&=e^{\ln5^{\cos3x}}\\&=e^{(\cos3x)\cdot(\ln5)}\\\\ \Rightarrow f'(x)&=-3\sin3x\cdot\ln5\cdot e^{(\cos3x)\cdot(\ln5)}.\ _\square \end{aligned}
y=xe^{\cos x}?
\begin{aligned} y'&=(x)'e^{\cos x}+x\left(e^{\cos x}\right)'\\ &=1\cdot e^{\cos x}+x \cdot e^{\cos x}\cdot(\cos x)'\\ &= e^{\cos x}+x\cdot e^{\cos x}\cdot(-\sin x)\\ &= e^{\cos x}(1-x\sin x).\ _\square \end{aligned}
y=e^{3 x}\sin 2x?
\begin{aligned} y'&=\left(e^{3 x}\right)'\sin 2x+e^{3 x}(\sin 2x)'\\ &=3e^{3 x}\sin 2x+e^{3 x}\cdot 2\cos 2x\\ &=e^{3x}(3\sin 2x+2\cos 2x).\ _\square \end{aligned}
\displaystyle{y=\frac{e^x-e^{-x}}{e^x+e^{-x}}?}
\begin{aligned} y'&=\frac{\left(e^x-e^{-x}\right)'\left(e^x+e^{-x}\right)-\left(e^x-e^{-x}\right)\left(e^x+e^{-x}\right)'}{\left(e^x+e^{-x}\right)^2}\\\\ &=\frac{\left(e^x+e^{-x}\right)\left(e^x+e^{-x}\right)-\left(e^x-e^{-x}\right)\left(e^x-e^{-x}\right)}{\left(e^x+e^{-x}\right)^2}\\\\ &=\frac{\left(e^x+e^{-x}\right)^2-\left(e^x-e^{-x}\right)^2}{\left(e^x+e^{-x}\right)^2}\\\\ &=\frac{4}{\left(e^x+e^{-x}\right)^2}.\ _\square \end{aligned}
Cite as: Applying Differentiation Rules to Exponential Functions. Brilliant.org. Retrieved from https://brilliant.org/wiki/applying-differentiation-rules-to-exponential/
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Sketch the parametric curve for the following set of parametric
Sketch the parametric curve for the following set of parametric equations. Clearly indicate direction of motion 1)x=5cost 2)y=2sint 3)0>=t>=2pi
Sketch the parametric curve for the following set of parametric equations. Clearly indicate direction of motion
x=5\mathrm{cos}t
y=2\mathrm{sin}t
0\ge t\ge 2\pi
To sketch the parametric curve for the following set of parametric equations you need
\left(x,\text{ }y\right)=
For the given rectangular equation, write an equivalent polar equation.
{x}^{2}+{y}^{2}-10x=0
State the Fundamental Theorem for Line Integrals
Find all points (if any) of horizontal and vertical tangency to the curve:
x=1-t.y={t}^{2}
Compute the derivative of the vector-valued function
r\left(t\right)=⟨\mathrm{sin}2t,e{t}^{2}⟩.
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Let W be the set of all vectors of the form shown, where a, b, and c r
Let W be the set of all vectors of the form shown, where a, b, and c represent arbitrary real numbers. In each case, either find a set S of vectors that spans W or give an example to show that W is not a vector space.
\left[\begin{array}{c}4a+3b\\ 0\\ a+b+c\\ c-2a\end{array}\right]
Since any vector w in W may be written as
w=a\left[\begin{array}{c}4\\ 0\\ 1\\ -2\end{array}\right]+b\left[\begin{array}{c}3\\ 0\\ 1\\ 0\end{array}\right]+c\left[\begin{array}{c}0\\ 0\\ 1\\ 1\end{array}\right]
S=\left\{\left[\begin{array}{c}4\\ 0\\ 1\\ -2\end{array}\right],\left[\begin{array}{c}3\\ 0\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 0\\ 1\\ 1\end{array}\right]\right\}
is a set that spans W.
S=\left\{\left[\begin{array}{c}4\\ 0\\ 1\\ -2\end{array}\right],\left[\begin{array}{c}3\\ 0\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 0\\ 1\\ 1\end{array}\right]\right\}
r\left(t\right)=<{t}^{2},\frac{2}{3}{t}^{3},t>
<4,-\frac{16}{3},-2>
\left(1,3,0\right),\left(-2,0,2\right),\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left(-1,3,-1\right)
|u\cdot v|
and determine whether
u\cdot v
is directed into the screen or out of the screen.
|u\cdot v|=??
Label the following statements as being true or false.
(a) There exists a linear operator T with no T-invariant subspace.
(b) If T is a linear operator on a finite-dimensional vector space V, and W is a T-invariant subspace of V, then the characteristic polynomial of Tw divides the characteristic polynomial of T.
(c) Let T be a linear operator on a finite-dimensional vector space V, and let x and y be elements of V. If W is the T-cyclic subspace generated by x, W
(a) find the projection of u onto v and (b) find the vector component of u orthogonal to v. u = ⟨6, 7⟩, v = ⟨1,4⟩
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Time reversibility - Wikipedia
A mathematical or physical process is time-reversible if the dynamics of the process remain well-defined when the sequence of time-states is reversed.
A deterministic process is time-reversible if the time-reversed process satisfies the same dynamic equations as the original process; in other words, the equations are invariant or symmetrical under a change in the sign of time. A stochastic process is reversible if the statistical properties of the process are the same as the statistical properties for time-reversed data from the same process.
4 Waves and optics
In mathematics, a dynamical system is time-reversible if the forward evolution is one-to-one, so that for every state there exists a transformation (an involution) π which gives a one-to-one mapping between the time-reversed evolution of any one state and the forward-time evolution of another corresponding state, given by the operator equation:
{\displaystyle U_{-t}=\pi \,U_{t}\,\pi }
Any time-independent structures (e.g. critical points or attractors) which the dynamics give rise to must therefore either be self-symmetrical or have symmetrical images under the involution π.
In physics, the laws of motion of classical mechanics exhibit time reversibility, as long as the operator π reverses the conjugate momenta of all the particles of the system, i.e.
{\displaystyle \mathbf {p} \rightarrow \mathbf {-p} }
(T-symmetry).
In quantum mechanical systems, however, the weak nuclear force is not invariant under T-symmetry alone; if weak interactions are present, reversible dynamics are still possible, but only if the operator π also reverses the signs of all the charges and the parity of the spatial co-ordinates (C-symmetry and P-symmetry). This reversibility of several linked properties is known as CPT symmetry.
Thermodynamic processes can be reversible or irreversible, depending on the change in entropy during the process. Note, however, that the fundamental laws that underlie the thermodynamic processes are all time-reversible (classical laws of motion and laws of electrodynamics),[1] which means that on the microscopic level, if one were to keep track of all the particles and all the degrees of freedom, the many-body system processes are all reversible; However, such analysis is beyond the capability of any human being (or artificial intelligence), and the macroscopic properties (like entropy and temperature) of many-body system are only defined from the statistics of the ensembles. When we talk about such macroscopic properties in thermodynamics, in certain cases, we can see irreversibility in the time evolution of these quantities on a statistical level. Indeed, the second law of thermodynamics predicates that the entropy of the entire universe must not decrease, not because the probability of that is zero, but because it's so unlikely that it's a statistical impossibility for all practical considerations (see Crooks fluctuation theorem).
A stochastic process is time-reversible if the joint probabilities of the forward and reverse state sequences are the same for all sets of time increments { τs }, for s = 1, ..., k for any k:[2]
{\displaystyle p(x_{t},x_{t+\tau _{1}},x_{t+\tau _{2}},\ldots ,x_{t+\tau _{k}})=p(x_{t'},x_{t'-\tau _{1}},x_{t'-\tau _{2}},\ldots ,x_{t'-\tau _{k}})}
A univariate stationary Gaussian process is time-reversible. Markov processes can only be reversible if their stationary distributions have the property of detailed balance:
{\displaystyle p(x_{t}=i,x_{t+1}=j)=\,p(x_{t}=j,x_{t+1}=i)}
Kolmogorov's criterion defines the condition for a Markov chain or continuous-time Markov chain to be time-reversible.
Time reversal of numerous classes of stochastic processes has been studied, including Lévy processes,[3] stochastic networks (Kelly's lemma),[4] birth and death processes,[5] Markov chains,[6] and piecewise deterministic Markov processes.[7]
Waves and opticsEdit
Time reversal method works based on the linear reciprocity of the wave equation, which states that the time reversed solution of a wave equation is also a solution to the wave equation since standard wave equations only contain even derivatives of the unknown variables.[8] Thus, the wave equation is symmetrical under time reversal, so the time reversal of any valid solution is also a solution. This means that a wave's path through space is valid when travelled in either direction.
Time reversal signal processing[9] is a process in which this property is used to reverse a received signal; this signal is then re-emitted and a temporal compression occurs, resulting in a reverse of the initial excitation waveform being played at the initial source.
^ David Albert on Time and Chance
^ Tong (1990), Section 4.4
^ Jacod, J.; Protter, P. (1988). "Time Reversal on Levy Processes". The Annals of Probability. 16 (2): 620. doi:10.1214/aop/1176991776. JSTOR 2243828.
^ Kelly, F. P. (1976). "Networks of Queues". Advances in Applied Probability. 8 (2): 416–432. doi:10.2307/1425912. JSTOR 1425912.
^ Tanaka, H. (1989). "Time Reversal of Random Walks in One-Dimension". Tokyo Journal of Mathematics. 12: 159–174. doi:10.3836/tjm/1270133555.
^ Norris, J. R. (1998). Markov Chains. Cambridge University Press. ISBN 978-0521633963.
^ Löpker, A.; Palmowski, Z. (2013). "On time reversal of piecewise deterministic Markov processes". Electronic Journal of Probability. 18. arXiv:1110.3813. doi:10.1214/EJP.v18-1958.
^ Parvasi, Seyed Mohammad; Ho, Siu Chun Michael; Kong, Qingzhao; Mousavi, Reza; Song, Gangbing (19 July 2016). "Real time bolt preload monitoring using piezoceramic transducers and time reversal technique—a numerical study with experimental verification". Smart Materials and Structures. 25 (8): 085015. Bibcode:2016SMaS...25h5015P. doi:10.1088/0964-1726/25/8/085015. ISSN 0964-1726.
^ Anderson, B. E., M. Griffa, C. Larmat, T.J. Ulrich, and P.A. Johnson, “Time reversal,” Acoust. Today, 4 (1), 5-16 (2008). https://acousticstoday.org/time-reversal-brian-e-anderson/
Isham, V. (1991) "Modelling stochastic phenomena". In: Stochastic Theory and Modelling, Hinkley, DV., Reid, N., Snell, E.J. (Eds). Chapman and Hall. ISBN 978-0-412-30590-0.
Tong, H. (1990) Non-linear Time Series: A Dynamical System Approach. Oxford UP. ISBN 0-19-852300-9
Retrieved from "https://en.wikipedia.org/w/index.php?title=Time_reversibility&oldid=1083631916"
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If the hypotenuse of a 45°-45°-90° triangle has a length
\surd 2
A 45°-45°-90° triangle is a right isosceles triangle. If x is the length of each leg, we can use the Pythagorean Theorem to write:
{x}^{2}+{x}^{2}={\left(\surd 2\right)}^{2}
2{x}^{2}=2
{x}^{2}=1
\stackrel{―}{AB}
\mathrm{△}ABC
\mathrm{△}ABC
\mathrm{△}ABC
\theta
\mathrm{cot}\left(\theta \right)=1
1:2
4\mathrm{cos}45°-2\mathrm{sin}45°
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This article is about mechanical gears. For other uses, see Gear (disambiguation).
For the gear-like device used to drive a roller chain, see Sprocket.
"Geared" redirects here. For the video game, see Geared (video game).
3 Comparison with drive mechanisms
4.1 External versus internal gears
4.3.1 Skew gears
4.4 Double helical
4.6 Spiral bevels
4.10 Non-circular
4.12 Epicyclic gear train
4.12.1 Sun and planet
4.13 Harmonic gear
4.14 Cage gear
4.15 Cycloidal gear
4.16 Magnetic gear
5.2 Helical gear
5.4 Tooth contact
5.5 Tooth thickness
7 Shifting of gears
9 Gear materials
10 Standard pitches and the module system
12 Gear model in modern physics
13 Gear mechanism in natural world
Comparison with drive mechanismsEdit
External versus internal gearsEdit
HelicalEdit
Skew gearsEdit
{\displaystyle E=\beta _{1}+\beta _{2}}
for gears of the same handedness,
{\displaystyle E=\beta _{1}-\beta _{2}}
for gears of opposite handedness,
{\displaystyle \beta }
is the helix angle for the gear. The crossed configuration is less mechanically sound because there is only a point contact between the gears, whereas in the parallel configuration there is a line contact.[24]
Double helicalEdit
Spiral bevelsEdit
The cylindrical gear tooth profile corresponds to an involute, but the bevel gear tooth profile to an octoid. All traditional bevel gear generators (like Gleason, Klingelnberg, Heidenreich & Harbeck, WMW Modul) manufacture bevel gears with an octoidal tooth profile. For 5-axis milled bevel gear sets it is important to choose the same calculation / layout like the conventional manufacturing method. Simplified calculated bevel gears on the basis of an equivalent cylindrical gear in normal section with an involute tooth form show a deviant tooth form with reduced tooth strength by 10-28% without offset and 45% with offset [Diss. Hünecke, TU Dresden]. Furthermore, the "involute bevel gear sets" cause more noise.
HypoidEdit
Crown gearEdit
WormEdit
Non-circularEdit
Rack and pinionEdit
Epicyclic gear trainEdit
Sun and planetEdit
Harmonic gearEdit
Cage gearEdit
Cycloidal gearEdit
Magnetic gearEdit
A predefined diametral position on the gear where the circular tooth thickness, pressure angle and helix angles are defined. The standard pitch diameter is a design dimension and cannot be measured, but is a location where other measurements are made. Its value is based on the number of teeth (N), the normal module (mn; or normal diametral pitch, Pd), and the helix angle (
{\displaystyle \psi }
{\displaystyle d={\frac {Nm_{n}}{\cos \psi }}}
in metric units or
{\displaystyle d={\frac {N}{P_{d}\cos \psi }}}
in imperial units.[34]
{\displaystyle m={\frac {p}{\pi }}}
{\displaystyle m={\frac {25.4}{DP}}}
in conventional metric units.
{\displaystyle a={\frac {m}{2}}(z_{1}+z_{2})}
{\displaystyle d_{w}={\frac {2a}{u+1}}={\frac {2a}{{\frac {z_{2}}{z_{1}}}+1}}.}
Pressure angle,
{\displaystyle \theta }
Outside diameter,
{\displaystyle D_{o}}
Radial distance from the pitch surface to the outermost point of the tooth.
{\displaystyle a={\frac {1}{2}}(D_{o}-D)}
Radial distance from the depth of the tooth trough to the pitch surface.
{\displaystyle b={\frac {1}{2}}(D-{\text{root diameter}})}
Whole depth,
{\displaystyle h_{t}}
{\displaystyle DP={\frac {N}{d}}={\frac {\pi }{p}}}
{\displaystyle DP={\frac {25.4}{m}}}
Base pitch, normal pitch,
{\displaystyle p_{b}}
Helix angle,
{\displaystyle \psi }
Normal circular pitch,
{\displaystyle p_{n}}
Circular pitch in the plane of rotation of the gear. Sometimes just called "circular pitch".
{\displaystyle p_{n}=p\cos(\psi )}
Worm gearEdit
Lead angle,
{\displaystyle \lambda }
Pitch diameter,
{\displaystyle d_{w}}
Tooth contactEdit
{\displaystyle \epsilon _{\gamma }=\epsilon _{\alpha }+\epsilon _{\beta }}
{\displaystyle m_{\rm {t}}=m_{\rm {p}}+m_{\rm {F}}}
{\displaystyle m_{\rm {o}}={\sqrt {m_{\rm {p}}^{2}+m_{\rm {F}}^{2}}}}
Tooth thicknessEdit
{\displaystyle P_{\rm {d}}={\frac {N}{d}}={\frac {25.4}{m}}={\frac {\pi }{p}}}
{\displaystyle P_{\rm {nd}}={\frac {P_{\rm {d}}}{\cos \psi }}}
{\displaystyle \tau ={\frac {360}{z}}}
{\displaystyle {\frac {2\pi }{z}}}
Shifting of gearsEdit
Tooth profileEdit
Gear materialsEdit
Numerous nonferrous alloys, cast irons, powder-metallurgy and plastics can be used in the manufacture of gears. However, steels are most commonly used because of their high strength-to-weight ratio and low cost. Plastic is commonly used where cost or weight is a concern. A properly designed plastic gear can replace steel in many cases because it has many desirable properties, including dirt tolerance, low speed meshing, the ability to "skip" quite well[37] and the ability to be made with materials that don't need additional lubrication. Manufacturers have used plastic gears to reduce costs in consumer items including copy machines, optical storage devices, cheap dynamos, consumer audio equipment, servo motors, and printers. Another advantage of the use of plastics, formerly (such as in the 1980s), was the reduction of repair costs for certain expensive machines. In cases of severe jamming (as of the paper in a printer), the plastic gear teeth would be torn free of their substrate, allowing the drive mechanism to then spin freely (instead of damaging itself by straining against the jam). This use of "sacrificial" gear teeth avoided destroying the much more expensive motor and related parts. This method has been superseded, in more recent designs, by the use of clutches and torque- or current-limited motors.
Standard pitches and the module systemEdit
Gear model in modern physicsEdit
Gear mechanism in natural worldEdit
Retrieved from "https://en.wikipedia.org/w/index.php?title=Gear&oldid=1089244922"
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Write an equations in indicated coordinate system. Also write a
Write an equations in indicated coordinate system. Also write a parametric equation and a vector equation for each of the following equations.1) frac{x-1}{2}=frac{y-5}{3}=frac{z-2}{4}
Write an equations in indicated coordinate system. Also write a parametric equation and a vector equation for each of the following equations.
\frac{x-1}{2}=\frac{y-5}{3}=\frac{z-2}{4}
cylindrical coordinates, sherical coordinates
3x+4y+5z=6
{x}^{2}+{y}^{2}+{z}^{2}=100
{x}^{2}+{y}^{2}=100
Equations in indicated coordinate system is
\left(x,\text{ }y\right)=
What is the polar form of (-2, 10)?
3x+4y-12=0
\begin{array}{|cc|}\hline \text{Point}& \text{Parallel to}\\ \left(-1,\text{ }0,\text{ }8\right)& v=3i+4j-8k\\ \hline\end{array}
\frac{x+1}{3}=\frac{y}{4}=\frac{8-z}{8}
The equations of two functions are
y=-21x+9
y=-24x+8
. Which function is changing more quickly? How do you know?
Anellipse in te plane
x=2
\frac{{y}^{2}}{9}+\frac{{z}^{2}}{16}=1
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Noncovalent Bonds - Course Hero
Cell Biology/Interactions Among Atoms/Noncovalent Bonds
Noncovalent bonds, such as hydrogen bonds, van der Waals forces, ionic bonds, and hydrophobic bonds, play an important role in forming the structures of biological molecules.
Hydrogen (H) bonds are part of a class of intermolecular interactions known as noncovalent bonds. A noncovalent bond is any relatively weak intermolecular attraction that does not involve the sharing of electrons. Noncovalent bonds include hydrogen bonds, van der Waals forces, ionic bonds, and hydrophobic bonds.
In a water molecule (H2O), which contains a bond between two H atoms and one oxygen (O) atom, the van der Waals attraction, which is a weak, noncovalent attractive force between two distinct molecules resulting from instantaneous dipoles, causes the molecule to bend such that the O atom is on one side, the H atoms are on the other, and the two
{\rm {H{-}O}}
bonds are at an angle of 104.5°. Since oxygen is highly electronegative, the oxygen atom draws the electrons toward its nucleus and away from the hydrogen nucleus, and
{\rm {H{-}O}}
forms a polar covalent bond. With oxygen on one "side" of the molecule and hydrogen on the other, the H2O molecule has a net dipole moment, which is a vector quantity that defines the extent of the charge on either side of a polar covalent bond, with the direction that points from the positive side of the bond toward the negative side, and the entire molecule is polar.
In water the partially positively charged hydrogen poles are attracted to the partially negatively charged oxygen poles of other water molecules, and as water molecules move closer together, they arrange themselves so as to maximize the intermolecular
{\rm {H{-}O}}
interactions. This intermolecular attraction is an example of a hydrogen bond. A hydrogen bond is a weak bond that results from an attraction between a positively charged hydrogen in one molecule and a negatively charged atom in another. Hydrogen bonds can form with any molecule that contains a positively charged dipole. Since both nitrogen and oxygen are more electronegative than hydrogen, all
{\rm {O{-}H}}
{\rm {N{-}H}}
bonds are polar, and H atoms in these bonds can form hydrogen bonds.
{\rm {C{-}H}}
bonds, in contrast, are nonpolar, and the H atoms do not form hydrogen bonds.
Hydrophobic bonds are another type of noncovalent bond. To be hydrophobic is having a weak or no affinity to water. A hydrophobic bond is a force that pushes nonpolar molecules together in an aqueous environment. When a nonpolar compound, such as an oil, is mixed with water, the oil molecules aggregate. This is not because the nonpolar molecules are attracted to each other; rather, it is because the nonpolar molecules are repelled by the polar water molecules.
Hydrogen and Hydrophobic Bonds
Although hydrogen bonds and the other noncovalent bonds are weaker than covalent bonds, they are responsible for many physical properties in water and other biological molecules. It is because of hydrogen bonds, for example, that water exists in liquid form on Earth. Nonpolar molecules of similar molecular weight have much lower freezing and boiling points, but the attractive force of hydrogen bonds holds water molecules together at higher temperatures. Hydrogen bonds also play a critical role in the structures of large biological molecules, such as DNA and proteins. Without the stabilizing force of hydrogen bonds, molecules of DNA, RNA, and proteins would fall apart and be unable to function in cells.
<Isotopes and Research>Carbon Bonds
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Quantum Teleportation | Brilliant Math & Science Wiki
Matt DeCross and Satyabrata Dash contributed
In quantum teleportation, the properties of quantum entanglement are used to send a spin state (qubit) between observers without physically moving the involved particle. The particles themselves are not really teleported, but the state of one particle is destroyed on one side and extracted on the other side, so the information that the state encodes is communicated. The process is not instantaneous, because information must be communicated classically between observers as part of the process. The usefulness of quantum teleportation lies in its ability to send quantum information arbitrarily far distances without exposing quantum states to thermal decoherence from the environment or other adverse effects.
Although quantum teleportation can in principle be used to actually teleport macroscopic objects (in the sense that two objects in exactly the same quantum state are identical), the number of entangled states necessary to accomplish this is well outside anything physically achievable, since maintaining such a massive number of entangled states without decohering is a difficult problem. Quantum teleportation, is, however, vital to the operation of quantum computers, in which manipulation of quantum information is of paramount importance. Quantum teleportation may eventually assist in the development of a "quantum internet" that would function by transporting information between local quantum computers using quantum teleportation [1].
Process of Quantum Teleportation
Mathematics of Quantum Teleportation
Below is a sketch of an algorithm for teleporting quantum information. Suppose Alice has state C, which she wants to send to Bob. To achieve this, Alice and Bob should follow the sequence of steps:
1) Generate an entangled pair of electrons with spin states A and B, in a particular Bell state:
|\Phi_0\rangle = \frac{1}{\sqrt{2}} (|\uparrow\rangle_A \otimes |\uparrow\rangle_B + |\downarrow\rangle_A \otimes |\downarrow\rangle_B).
Separate the entangled electrons, sending A to Alice and B to Bob.
2) Alice measures the "Bell state" (described below) of A and C, entangling A and C.
3) Alice sends the result of her measurement to Bob via some classical method of communication.
4) Bob measures the spin of state B along an axis determined by Alice's measurement
Since step 3 involves communicating via some classical method, the information in the entangled state must respect causality. Relativity is not violated because the information cannot be communicated faster than the classical communication in step 3 can be performed, which is sub-lightspeed.
The idea of quantum teleportation, which can be seen in the mathematics below, is that Alice's measurement disentangles A and B and entangles A and C. Depending on what particular entangled state Alice sees, Bob will know exactly how B was disentangled, and can manipulate B to take the state that C had originally. Thus the state C was "teleported" from Alice to Bob, who now has a state that looks identical to how C originally looked. It is important to note that state C is not preserved in the processes: the no-cloning and no-deletion theorems of quantum mechanics prevent quantum information from being perfectly replicated or destroyed. Bob receives a state that looks like C did originally, but Alice no longer has the original state C in the end, since it is now in an entangled state with A.
Schematic of quantum teleportation algorithm. Entangled state AB is split so that particle A goes to Alice and B to Bob. Alice then entangles A and C, disentangling B and sending Bob the information he needs to recover the state of C on the other end. Based on [2].
1 only 1 and 2 1, 2, 3, and 4 1 and 4 1 and 3 1, 3, and 4 2 only
Which of the following is true of quantum teleportation?
1) Quantum information is transferred between states
2) The teleported particle is physically transferred between locations
3) A quantum state is cloned between observers
4) Quantum information is permanently removed from the system
As a review, recall the Pauli matrices:
\sigma_0 = I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \qquad \sigma_1 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad \sigma_2 = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \qquad \sigma_3 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.
The spin operators along each axis are defined as
\frac{\hbar}{2}
times each of
\sigma_1, \sigma_2, \sigma_3
x,y,z
These Pauli matrices are used to construct Bell states, an orthonormal basis of entangled states for the tensor product space of spin-
\frac12
particles:
\begin{aligned} |\Phi_0\rangle &= I \otimes \sigma_0 |\Phi_0\rangle = \frac{1}{\sqrt{2}} (|\uparrow\rangle \otimes |\uparrow\rangle + |\downarrow\rangle \otimes |\downarrow\rangle) \\ |\Phi_1\rangle &= I \otimes \sigma_1 |\Phi_0\rangle =\frac{1}{\sqrt{2}} (|\uparrow\rangle \otimes |\downarrow\rangle + |\downarrow\rangle \otimes |\uparrow\rangle) \\ |\Phi_2\rangle &=I \otimes \sigma_2 |\Phi_0\rangle = \frac{i}{\sqrt{2}} (|\uparrow\rangle \otimes |\downarrow\rangle - |\downarrow\rangle \otimes |\uparrow\rangle) \\ |\Phi_3\rangle &=I \otimes \sigma_3 |\Phi_0\rangle = \frac{1}{\sqrt{2}} (|\uparrow\rangle \otimes |\uparrow\rangle - |\downarrow\rangle \otimes |\downarrow\rangle) \end{aligned}
Measurements that project tensor products of spin states onto the Bell basis are called Bell measurements.
\frac{1}{\sqrt{2}} (\vert\Phi_1\rangle + \vert\Phi_2\rangle)
\frac{1}{\sqrt{2}} (\vert\Phi_1\rangle - \vert\Phi_2\rangle)
\frac{1}{\sqrt{2}} (\vert\Phi_1\rangle - i\vert\Phi_2\rangle)
\frac{1}{\sqrt{2}} (\vert\Phi_1\rangle + i\vert\Phi_2\rangle)
Which of the following correctly rewrites the state
|\uparrow\rangle \otimes |\downarrow\rangle
in the Bell basis?
1
\frac14
\frac12
0
In the product of spin states
\frac{1}{\sqrt{2}} (|\uparrow\rangle\otimes|\uparrow\rangle + |\uparrow\rangle\otimes|\downarrow\rangle )
, what is the probability that a Bell measurement will find the Bell state
|\Phi_1\rangle
Now, follow the algorithm sketched in the previous section. Suppose Alice starts with state C, which she wants to send Bob. State C can be written in the most general form:
|\Psi\rangle_C = c_1 |\uparrow\rangle_C + c_2|\downarrow\rangle_C,
c_1
c_2
normalized complex constants.
1) Generate an entangled pair of electrons A and B in the Bell state:
|\Phi_0\rangle_{AB} = \frac{1}{\sqrt{2}} (|\uparrow\rangle_A \otimes |\uparrow\rangle_B + |\downarrow\rangle_A \otimes |\downarrow\rangle_B).
The state of the full system of three particles is therefore
|\Psi\rangle_{ABC} = |\Phi_0\rangle_{AB} \otimes |\Psi\rangle_C
. This is a product state between entangled pair AB and non-entangled C.
2) Alice measures the Bell state of AC, entangling A and C while disentangling B. The process of measuring the Bell state projects a non-entangled state into an entangled state, since all four Bell states are entangled.
Expanding Alice's full original state, she starts with:
\begin{aligned} |\Psi\rangle_{ABC} &=\frac{1}{\sqrt{2}} (|\uparrow\rangle_A \otimes |\uparrow\rangle_B + |\downarrow\rangle_A \otimes |\downarrow\rangle_B) \otimes (c_1 |\uparrow\rangle_C + c_2 |\downarrow\rangle_C) \end{aligned}
Multiplying out the states and changing to the Bell basis of A and C, this state can be rewritten:
\begin{aligned} |\Psi\rangle_{ABC} &= \frac12 |\Phi_0\rangle_{AC} \otimes |\Psi\rangle_B + \frac12 |\Phi_1\rangle_{AC} \otimes \sigma_1 |\Psi\rangle_B + \frac12 |\Phi_2\rangle_{AC} \otimes \sigma_2 |\Psi\rangle_B + \frac12 |\Phi_3\rangle_{AC} \otimes \sigma_3 |\Psi\rangle_B \\ &= \sum_{i=0}^3 \frac12 |\Phi_i\rangle_{AC} \otimes \sigma_i |\Psi\rangle_B \end{aligned}
When Alice measures the Bell state of A and C, she will find one of
|\Phi_0\rangle_{AC}, |\Phi_1\rangle_{AC}, |\Phi_2\rangle_{AC}, |\Phi_3\rangle_{AC}
, each with probability
\frac14
. Whichever
|\Phi_i\rangle_{AC}
she measures, the state of particle B will be
\sigma_i |\Psi\rangle_B
after measurement.
3) To send Bob the state of particle C, therefore, Alice does not need to send Bob the possibly infinite amount of information contained in the coefficients
c_1
c_2
which may be real numbers out to arbitrary precision. She needs only to send the integer
i
of the Bell state of A and C, which is a maximum of two bits of information. Alice can send this information to Bob in whatever classical way she likes.
4) Bob receives the integer
i
from Alice that labels the Bell state
|\Phi_i\rangle_{AC}
that she measured. After Alice's measurement, the overall state of the system is:
|\Psi\rangle_{ABC} = |\Phi_i\rangle_{AC} \otimes \sigma_i |\Psi\rangle_B .
Bob therefore applies
\sigma_i
to the disentangled
|\Psi\rangle_B
state on his end, by measuring the spin along axis
i
\sigma_i^2 = I
i
, Bob is left with the overall state:
|\Psi\rangle_{ABC} = |\Phi_i\rangle_{AC} \otimes |\Psi\rangle_B
Bob has therefore changed the spin state of particle B to:
|\Psi\rangle_B = c_1 |\uparrow\rangle_B + c_2|\downarrow\rangle_B,
which is identical to the original state of particle C that Alice wanted to send. The information in state C has been "teleported" to Bob's state: the final spin state of B looks like C's original state. Note, however, that the particles involved never change between observers: Alice always has A and C, and Bob always has B.
Pirandola, S., & Braunstein, S. Physics: Unite to build a quantum Internet. Retrieved from http://www.nature.com/news/physics-unite-to-build-a-quantum-internet-1.19716
Debenben, . quantum teleportation diagram. Retrieved from https://commons.wikimedia.org/w/index.php?curid=34503176
Cite as: Quantum Teleportation. Brilliant.org. Retrieved from https://brilliant.org/wiki/quantum-teleportation/
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Type expressions.
Storage layout for local names.
Sequences and declarations.
Records and classes fields.
Applications of types are grouped under checking and translation. First type checking is used to reason about the behavior of a program at runtime by using logical rules.
In applications of translation, given the type of a name, the compiler can determine the storage needed for the name during runtime.
In this article, we learn about types and storage layouts for names declared within classes or procedures. The storage procedure call or an object is allocated during runtime when the procedure is called or when the object is created.
Types have a structure we will represent using type expressions, a type expression can either be formed by applying a type constructor operator to a type expression or can be a basic type.
Basic types are determined by the language that is being checked.
We have an array of type int[2][3] that is read as array of 2 arrays of 3 integers each. It is written as a type expression as follows; array(2, array(3, integer))
We represent it as a tree as shown below (1);
The following definitions are used for type expressions;
A basic type expression. These include, boolean, char, integer, float and void.
A type name is a type expression.
A type expression is formed by applying an array type constructor to a number and type expression.
A record is a data structure with named fields. A type expression is formed by applying the record type constructor to the fields' names and their types.
A type expression is formed by using the type constructor → for function types. That is, we write s → t for 'function from type s to type t'.
If s and t are type expressions, their cartesian product s x t is a type expression.
Type expressions may contain variables whose values are type expressions.
Type-checking rules are of the form; 'if two type expressions are equal then return a certain type, else return an error'.
When similar names are used for type expressions and other subsequent type expressions, ambiguities arise.
The problem is whether a name in a type expression represents itself or represents an abbreviation for another type expression.
When representing type expressions using graphs, we say that two types are structurally equivalent if and only if either of the conditions is true;
They have the same basic type.
They are formed by applying a similar constructor to structurally equivalent types.
One is a type name that denotes the other.
If we treat type names as standing for themselves, then the first two conditions in the definition above lead to the name equivalence of type expressions.
We learn about types and declarations using simplified grammar that declares a single name at a time.
We have the following grammar; (2)
The non-terminal D generates a sequence of declarations. Non-terminal T generates basic types, array, or record types.
The non-terminal B generates a basic type either an int or float.
The non-terminal C generates strings of zero or more integers, each surrounded by brackets.
An array type consists of the basic type specified by B followed by array components specified by a non-terminal C. A record type is a sequence of declarations for fields of the record surrounded by curly braces.
Storage layout for local names
Given a type name, we can determine the amount of storage that is needed for the name at runtime. During compile time we use these amounts to assign each name a relative address.
Type and relative addresses are stored in a symbol table entry for the name. Varying-length data such as strings or data whose size cannot be determined until runtime such as dynamic arrays is handled by keeping a fixed amount of storage for a pointer to data.
Assuming that storage is in blocks of contiguous bytes whereby a byte is the smallest unit of addressable memory. Multibyte objects are stored in consecutive bytes and given the address of the first byte.
We have the following SDT(Syntax Directed Translation) that computes types and their widths for basic and array types. (3)
The above SDT uses synthesized attributes type and width for each non-terminal and two variables t and w to pass type and width information from B node in a parse tree to a node fo the production C
\to ϵ
In an SDD(Syntax-Directed Definition), t and w would be inherited attributes for C.
The body of the T-production consists of a non-terminal B, an action, and a non-terminal C that appears on the next line. The action between B and C sets t to B-type and w to B.width.
If B
\to
int, B.type is set to an integer and B.width is set to 4 which is the width of an integer.
Similarly, if B
\to
float, B.type is float and B.width is 8, which is the width of a float.
Productions for C determine whether T generates a basic type or an array type. If C
\to ϵ
, then t is C.type and w is C.width.
Otherwise, C specifies an array component. The action for C
\to
[num]
{\mathrm{C}}_{1}
forms C.type by applying the type constructor array to the operands num.value and
{\mathrm{C}}_{1}
.type. For example, the resulting tree structure for applying an array can be seen from the first image.
To obtain the width of an array we multiply the width of an element by the number of elements in an array. If addresses of consecutive integers differ by 4, then the address calculations for an array of integers include multiplications by 4.
These multiplications give opportunities for optimization and therefore the front end needs to make them explicit.
Sequences and declarations
In programming languages such as C and Java, declarations in a single group are processed as a group. These declarations can be distributed within a Java procedure but can still be processed when the procedure s analyzed.
We can use a variable to track the next available relative address.
The following translation scheme(4) deals with a sequence of declarations in the form of T id where T generates a type as shown in image (3).
Before the first declaration is considered, a variable offset that tracks the next available relative address is set to 0.
The value of offset is incremented by the width of the type of x, x is a new name entered into the symbol table with its relative address set to the current value of offset.
The semantic action, within the production D
\to
T id,
{\mathrm{D}}_{1}
creates a symbol table entry by executing top.put(id.lexeme, T.type, offset). top denotes the current symbol table.
top.put creates a symbol table entry for id.lexeme with T.type and relative address offset in its data area.
The initialization of offset in image (4) is more evident in the first production that appears as;
\to
{offset = 0;} D
Non-terminals generating
ϵ
referred to as marker non-terminals are used to rewrite productions so that all actions appear at the right ends.
By using a maker non-terminal M, the above production is restated as;
\to
M D
*M
\to
ϵ
{ offset = 0; }
Records and classes fields
The translation of declarations in image (4) carries over to fields in records and classes. Record types are added to the grammar in image (3) by adding the following production;
\to
record '{' D '}'.
The fields in the record type are specified by a sequence of declarations that are generated by D.
The approach used in image (4) can be used to determine the types and relative addresses of fields as long as we are careful about the following;
Field names within a record must be distinct, that is, a name can only appear once in declarations generated by D.
The offset or relative address for a field name is relative to the data area for that record.
Graphs are useful for representing type expressions.
Applications of types can be under checking or translation. Checking reasons about the behavior of a program at runtime by using logical rules.
For translation, given the type of a name, the compiler can determine the storage needed for the name during runtime.
Basics of Compiler Design - Torben Mogensen
Compilers Principles, Techniques, & Tools - Alfred V. Aho Monica
Control flow in Intermediate Code Generation
During the construction of the front end of a compiler, we implement statements using control flow. We translate the statements by inheriting a label next that marks the first instruction after the code for this statement.
In this article, we learn about evaluation orders fpr syntax-directed definitions. Dependency graphs are used to determine an evaluation order for instances of attributes in a parse tree.
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\int \frac{\mathrm{sin}\left(\mathrm{ln}x\right)}{4x}dx
\int \frac{\mathrm{sin}\left(\mathrm{ln}\left(x\right)\right)}{4x}dx
We have to evaluate it.
\int \frac{\mathrm{sin}\left(\mathrm{ln}\left(x\right)\right)}{4x}dx
u=\mathrm{ln}\left(x\right)
differentiate equation w.r.t x we get
du=\frac{1}{x}dx,⇒dx=xdu
⇒\int \frac{\mathrm{sin}\left(\mathrm{ln}\left(x\right)\right)}{4x}dx=\frac{1}{4}\int \mathrm{sin}\left(u\right)du
⇒\int \frac{\mathrm{sin}\left(\mathrm{ln}\left(x\right)\right)}{4x}dx=\frac{1}{4}\left(-\mathrm{cos}\left(u\right)\right)+C
, where C is arbitrary constant
u=\mathrm{ln}\left(x\right)
in above equation, we get
⇒\int \frac{\mathrm{sin}\left(\mathrm{ln}\left(x\right)\right)}{4x}dx=\frac{1}{4}\left(-\mathrm{cos}\left(u\right)\right)+C
\int \frac{\mathrm{sin}\left(\mathrm{ln}\left(x\right)\right)}{4x}dx=\frac{\mathrm{cos}\left(\mathrm{ln}\left(x\right)\right)}{4}+C
\left({x}^{2}+2xy-4{y}^{2}\right)dx-\left({x}^{2}-8xy-4{y}^{2}\right)dy=0
x={t}^{2},y=2t,0\le t\le 5
The graph of f is shown. Evaluate each integral by interpreting it in terms of areas. integral.
{\int }_{0}^{2}f\left(x\right)dx
{\int }_{1}^{4}\left(\frac{x}{8}+\frac{1}{2x}\right)dx
f\left(x,y\right)={e}^{y}\sqrt{x+{e}^{y}},R=\left[0,4\right]×\left[0,1\right]
Evaluate the line integral along the path C given by
x=2t,\text{ }y=10t,
0\le t\le 1\int cxydx+ydy
How do you find the integration of
\mathrm{log}x
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Which of the following polar coordinate pairs does not represent
Which of the following polar coordinate pairs does not represent the point with rectangular coordinates (-2,-2)?(A)(2sqrt2.−135°)(B)(2sqrt2. 225°)
Which of the following polar coordinate pairs does not represent the point with rectangular coordinates (-2,-2)?
\left(2\sqrt{2}.-135°\right)
\left(2\sqrt{2}.225°\right)
\left(-2\sqrt{2}.-315°\right)
\left(-2\sqrt{2}.45°\right)
\left(-2\sqrt{2}.135°\right)
A polar coordinate
\left(r;\theta \right)
is given, then
\left(x;y\right)=\left(r×\mathrm{cos}\theta ;r×\mathrm{sin}\theta \right)
\left(2\sqrt{2}\mathrm{cos}\left(-{135}^{\circ }\right),2\sqrt{2}\mathrm{sin}\left(-{135}^{\circ }\right)\right)=\left(2\sqrt{2}×-\frac{\sqrt{2}}{2};2\sqrt{2}×-\frac{\sqrt{2}}{2}\right)=\left(-2;-2\right)
\left(2\sqrt{2}\mathrm{cos}{225}^{\circ },2\sqrt{2}\mathrm{sin}{225}^{\circ }\right)=\left(2\sqrt{2}×-\frac{\sqrt{2}}{2};2\sqrt{2}×-\frac{\sqrt{2}}{2}\right)=\left(-2;-2\right)
\left(-2\sqrt{2}\mathrm{cos}\left(-{315}^{\circ }\right),-2\sqrt{2}\mathrm{sin}\left(-{315}^{\circ }\right)\right)=\left(-2\sqrt{2}×\frac{\sqrt{2}}{2};-2\sqrt{2}×\frac{\sqrt{2}}{2}\right)=\left(-2;-2\right)
\left(-2\sqrt{2}\mathrm{cos}{45}^{\circ },-2\sqrt{2}\mathrm{sin}{45}^{\circ }\right)=\left(-2\sqrt{2}×\frac{\sqrt{2}}{2};-2\sqrt{2}×\frac{\sqrt{2}}{2}\right)=\left(-2;-2\right)
\left(-2\sqrt{2}\mathrm{cos}{135}^{\circ },-2\sqrt{2}\mathrm{sin}{135}^{\circ }\right)=\left(-2\sqrt{2}×-\frac{\sqrt{2}}{2};-2\sqrt{2}×\frac{\sqrt{2}}{2}\right)=\left(2;-2\right)
Choice (E) does not represent
\left(-2;-2\right)
\left(x,\text{ }y\right)=
x\left(t\right)=t,y\left(t\right)={t}^{3}
The exit velocity of a baseball (its velocity as it leaves the bat) is 128 feat per second, in the direction of 30 above horizontal and directly cowards the monster 30 foot tall center field wall. If the ball was hit 4 feet above ground level, find
(i) Find polar coordinates
\left(r,\theta \right)
of the point, where
r>0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}0\le \theta <2\pi
(ii) Find polar coordinates
\left(r,\theta \right)
of the point, where r<0 and
0\le \theta <2\pi
\left(3\sqrt{3},3\right)
Determine the interval(s) on which the vector-valued function continuous.
r\left(t\right)=ti+\sqrt{t+1}j+\left({t}^{2}+1\right)k
A\left(2,\text{ }1,\text{ }1\right),\text{ }B\left(1,\text{ }2,\text{ }1\right)
C\left(1,\text{ }1,\text{ }2\right)
be three given points. Find
i) parametric and symmetric equations of the line through the point A and parallel to the line through the points B and C;
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Calculating Distance Practice Problems Online | Brilliant
When speed changes over time, it is difficult to determine the distance traveled. But we can get a good approximation by dividing the time interval into a large number of pieces and assuming the speed remains constant over each individual period. Then, in each piece, we multiply the change in time by the speed at the start of that interval and sum the products.
We’ll call dividing into
n
intervals an
-approximation. So the above shows a 5-approximation of the distance traveled.
A car accelerates with constant acceleration from 0 mph to 60 mph over a 3-hour period.
If we divide this time interval into three intervals (i.e. using a 3-approximation), what are the speeds at the start of the intervals?
0 mph, 30 mph, and 60 mph 0 mph, 20 mph, and 40 mph 20 mph, 40 mph, and 60 mph 15 mph, 30 mph, and 45 mph
The same car still accelerates at a constant rate from 0 mph to 60 mph over a 3-hour period.
d_1
be a 3-approximation for the distance traveled (i.e. divide the 3-hour period into three smaller time intervals and assume constant speed on them); similarly, let
d_2
be a 60-approximation (i.e. divide into 60 smaller time intervals) for the distance traveled over the whole 3-hour period.
d_1
d_2?
(Note we are still taking the speed at the start of each interval.)
60 and 88.5 60 and 90 88.5 and 88.5 90 and 90
Now suppose there is a function that tells us the speed of the car at any given time: in other words,
f(t)
represents the speed of the car at time
t
. The time interval
[a,b]
is split into 4 smaller intervals:
In terms of
, what are the speeds at the beginning of these smaller intervals?
1. \, \, \, \begin{cases}\ &f(a),\ f\left(a + \frac{b-a}{4}\right),\\ & f\left(a + 2 \cdot \frac{b-a}{4}\right),\ f\left(a + 3 \cdot \frac{b-a}{4}\right)\end{cases}
2.\, \, \, \begin{cases} & f(a),\ f\left(a + \frac{b-a}{3}\right),\\ & f\left(a + 2 \cdot \frac{b-a}{3}\right),\ f(b) \end{cases}
3.\, \, \, \begin{cases} & f\left(a + \frac{b-a}{4}\right),\ f\left(a + 2 \cdot \frac{b-a}{4}\right),\\ & f\left(a + 3 \cdot \frac{b-a}{4}\right),\ f(b) \end{cases}
4.\, \, \, \begin{cases} & f\left(a + \frac{b-a}{5}\right),\ f\left(a + 2 \cdot \frac{b-a}{5}\right),\\ & f\left(a + 3 \cdot \frac{b-a}{5}\right),\ f\left(a + 4 \cdot \frac{b-a}{5}\right) \end{cases}
Again the car’s speed at time
t
f(t)
. Over the time interval
[a,b]
, what is the 100-approximation (i.e. splitting the time into 100 intervals) to the distance traveled?
\sum_{i=0}^{99}f\left(a + i \cdot \frac{b-a}{100}\right)
\frac{b-a}{100}\sum_{i=0}^{99}f\left(a + i \cdot \frac{b-a}{100}\right)
\sum_{i=1}^{100}f\left(a + i \cdot \frac{b-a}{100}\right)
\frac{b-a}{100}\sum_{i=1}^{100}f\left(a + i \cdot \frac{b-a}{100}\right)
In the last problem, we saw that the 100-approximation for distance travelled over the time period
t \in [a,b]
\frac{b-a}{100}\sum_{i=0}^{99}f\left(a + i \cdot \frac{b-a}{100}\right).
Generalizing, the
n
-approximation is similarly
\frac{b-a}{n}\sum_{i=0}^{n-1}f\left(a + i \cdot \frac{b-a}{n}\right).
This is called a Riemann sum, which for now we’ll write as
\text{RS}(f, a, b, n)
(
meaning the
n
-approximation of the distance travelled over the time interval
[a,b]
with speed function
f).
The idea is that the "true" distance should be a limit of these approximations as
n
In the coming quizzes, we’ll see how the Riemann sum appears not just with distance but in more and more seemingly unrelated places. Then we’ll build up tools to easily evaluate all of these sums!
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Shelter measurements - FIThydrowiki
Figure 1: Juvenile salmon in the upper layers of river bed sediment (Forseth et al., 2014).
Figure 2: Example showing hiding depth of 2cm (first marker covered) on a tube (Forseth et al., 2014).
Developed by: A. G. Finstad
The technique was developed by Finstad to estimate the shelter availability for juvenile salmon (Finstad et al. 2007a). In this stage of their life cycle, the salmon seeks shelter in the free spaces of the upper layers of the riverbed sediment (Figure 1). The existence and size of such openings to be used as shelter is measured with the described technique. It is relatively easy to be carried out in the field. The method is based on the idea to mimic the small fish by using a plastic tube, which has the approximate diameter of such a fish.
For salmon, Finstad used tubes with five different diameters (5, 10, 13, 16, 22 mm). As a result of their experiments Forseth et al. (2014) suggest the use of a 13mm tube to conduct the tests in an area of about 0.25m2. The tube is marked according to the three different shelter categories at lengths of 2, 5 and 10 cm to be able to detect certain possible hiding depths. Within the fixed framed area one is aiming to find every hole or opening where one can stick the tube into. The length of the tube (defined by the three rings) which is not visible anymore, defines the potential hiding depth (Figure 2).
The post-processing of the data is quite simple and described in the following section.
The method was developed in and for Norwegian rivers with large grain sizes. It is currently (also in FIThydro) under investigation if, when using it in rivers with a smaller D50, i.e. smaller grain sizes, this relation still applies or another relation becomes relevant.
It should further be applied also to other fish species than the salmon. Therefor additional tubes with diameters of 5 mm and 8 mm are used within the project`s measurement campaigns. These results might then be linked to the D5 and D10 grain diameter of the sediment (Szabo-Meszaros et al 2016).
For the measurement the following material is needed:
One tube with a diameter of 13 mm and marks (rings) at 2, 5 and 10 cm
One frame (metal or wood) to define the size of the area to be counted on
Something to take notes (Laptop, paper, etc.)
The hiding depth thresholds are defined as 2, 5 and 10 cm and often named as class S1 (2 cm – 5 cm), class S2 (5 cm – 10 cm) and class S3 (> 10 cm). It is recommended that the same person is carrying out the counting, to minimize the human based systematic error as much as possible.
When all values are collected in the field, the counted numbers for the different classes / hiding depth per spot are weighted using the following empirical formula described in Forseth et al (2014):
{\displaystyle S1+S2*2+S3*3}
Based on the following matrix the shelter availability can be defined:
Table 1: shelter availability depending on the number of shelter spots available.
Values for weighted shelter
Complete or partial migration barrier removal
Mitigating rapid, short-term variations in flow (hydro-peaking operations)
Sensory, behavioural barriers (electricity, light, sound, air-water curtains)
Finstad, A.G., Einum, S., Forseth, T., Ugedal, O., 2007a. Shelter availability affects behaviour, size-dependent and mean growth of juvenile Atlantic salmon. Freshwater Biology 52, 1710–1718. https://doi.org/10.1111/j.1365-2427.2007.01799.x
Finstad, A.G., Forseth, T., Ugedal, O., Naesje, T.F., 2007b. Metabolic rate, behaviour and winter performance in juvenile Atlantic salmon. Functional Ecology 21, 905–912. https://doi.org/10.1111/j.1365-2435.2007.01291.x
Forseth, T., Harby, A., Ugedal, O., Pulg, U., Fjeldstad, H.-P., Robertsen, G., Barlaup, B.T., Alfredsen, K., Sundt, H., Salveit, S.J., Skoglund, H., Kvingedal, E., Sundt-Hansen, L.E.B., Finstad, A., Einum, S., Arnekleiv, J.V., 2014. Handbook for environmental design in regulated salmon rivers (NINA temahefte No. 53). Norsk institutt for naturforskning, Trondheim, Norway.
Szabo-Meszaros, Marcel, Rüther, N., Alfredsen, K., 2016. Correlation between the shelter of juvenile salmonids and bed substrate, in: Wieprecht, S., Haun, S., Weber, K., Noack, M., Terheiden, K. (Eds.), River Sedimentation: Proceedings of the 13th International Symposium on River Sedimentation (Stuttgart, Germany, 19-22 September, 2016).
Valdimarsson, S.K., Metcalfe, N.B., 1998. Shelter selection in juvenile Atlantic salmon, or why do salmon seek shelter in winter? Journal of Fish Biology 52, 42–49. https://doi.org/10.1111/j.1095-8649.1998.tb01551.x
Retrieved from "https://www.fithydro.wiki/index.php?title=Shelter_measurements&oldid=7271"
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what does (e) mean in this 1 1258999e+15 - Maths - Statistics - 10260551 | Meritnation.com
\mathrm{It} \mathrm{is} \mathrm{an} \mathrm{irrational} \mathrm{number} \mathrm{just} \mathrm{like} \mathrm{\pi } \mathrm{which} \mathrm{is} \mathrm{approximately} \mathrm{equal} \mathrm{to} 2.718\phantom{\rule{0ex}{0ex}}\mathrm{The} \mathrm{value} \mathrm{of} \mathrm{the} \mathrm{types} 1.1258999\mathrm{e}+15 \mathrm{is} \mathrm{generally} \mathrm{obtained} \mathrm{in} \mathrm{calulations} \mathrm{of} \mathrm{very} \mathrm{large} \mathrm{numbers} \mathrm{in} \mathrm{calculator} \mathrm{or} \mathrm{excel} \mathrm{sheets}\phantom{\rule{0ex}{0ex}}\mathrm{It} \mathrm{is} \mathrm{simply} \mathrm{the} \mathrm{approximate} \mathrm{value} \mathrm{of} \mathrm{the} \mathrm{scientific}\left(\mathrm{or} \mathrm{exponential}\right) \mathrm{notation} \mathrm{of} \mathrm{the} \mathrm{result}\phantom{\rule{0ex}{0ex}}\mathrm{Here}, 1.1258999\mathrm{e}+15=1.1258999×{10}^{15}\phantom{\rule{0ex}{0ex}}\mathrm{So}, \mathrm{here} \mathrm{e} \mathrm{simply} \mathrm{stands} \mathrm{for} \mathrm{the} \mathrm{exponential} \mathrm{notation}
Noun: e
The base of the natural system of logarithms; approximately equal to 2.718282...
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Find the length of the curve. r(t)=<8t, t^2, \frac{1}{12}
r\left(t\right)=<8t,{t}^{2},\frac{1}{12}{t}^{3}>,0\le t\le 1
The curve will be
\frac{dw}{dt}
\frac{dw}{dt}
w=x\mathrm{sin}y,\text{ }x={e}^{t},\text{ }y=\pi -t
t=0
\left({x}^{2}-{y}^{2}+2x-y\right)dx+\left({x}^{2}-{y}^{2}+x-2y\right)dy=0
\frac{dx}{dt}=f\left(x\right)g\left(t\right)
x\left(0\right)={x}_{0}
Can 2 different ODE's have the same set of solutions?
If I have two differents linear ODE's:
{x}^{\text{'}\text{'}}+p\left(t\right){x}^{\text{'}}+q\left(t\right)x=f\left(t\right).\phantom{\rule{1em}{0ex}}p,q\in C\left(I,\mathrm{\infty }\right).
{x}^{\text{'}\text{'}}+j\left(t\right){x}^{\text{'}}+g\left(t\right)x=h\left(t\right).\phantom{\rule{1em}{0ex}}j,g\in C\left(I,\mathrm{\infty }\right).
Coul they have exactly the same set of solutions?
And if we have two different non-linear ODE's could they?
Help needed solving logistic differential equation with initial conditions
\frac{dP}{dt}=P\left(10-2P\right)
with initial conditions of
P\left(0\right)=1.
{x}^{\prime }=x\mathrm{sin}\left(\frac{\pi }{x}\right)
{x}^{\prime }=\left\{\begin{array}{l}x\mathrm{sin}\left(\frac{\pi }{x}\right)\text{ if }x\ne 0\\ 0\text{ else}\end{array}
Understanding how to apply dominant balance method
ϵy{}^{″}-{x}^{2}{y}^{\prime }-y=0
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Pouer (pheesics) - Wikipedia
In physics, pouer (seembol: P) is the rate that wirk is duin or energy is transmitted, or the amoont o energy needit or uised for a gien unit o time. As a rate o chynge o wirk duin or the energy o a subseestem, pouer is:
{\displaystyle P={\frac {W}{t}}\,}
whaur P is pouer, W is wirk an t is time.
The average pouer (aften juist cried "pouer" whan the context maks it clear) is the average amoont o wirk duin or energy transferred per unit time. The instantaneous pouer is than the limitin vailyie o the average pouer as the time interval Δt approaches zero.
{\displaystyle P=\lim _{\Delta t\rightarrow 0}{\frac {\Delta W}{\Delta t}}=\lim _{\Delta t\rightarrow 0}P_{\mathrm {avg} }\,}
Whan the rate o energy transfer or wirk is constant, aw o this can be simplifee'd tae
{\displaystyle P={\frac {W}{t}}={\frac {E}{t}}}
whaur W an E are, respectively, the work duin or energy transferred in time t (for ordinar measured in seicont).
Taen frae "https://sco.wikipedia.org/w/index.php?title=Pouer_(pheesics)&oldid=809645"
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Alert - Maple Help
Home : Support : Online Help : Programming : Maplets : Examples : Alert
display an alert dialog Maplet application
Alert(msg, opts)
The Alert(msg) calling sequence displays a Maplet application that contains a message. The user can click OK or Cancel. The value returned is true or FAIL, respectively.
The Alert sample Maplet worksheet describes how to write a Maplet application that behaves similarly to this routine by using the Maplets[Elements] package.
Specifies the Maplet application title. By default, the title is Alert.
\mathrm{with}\left(\mathrm{Maplets}[\mathrm{Examples}]\right):
\mathrm{Alert}\left("Invalid input!"\right)
Alert Sample Maplet
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ThreadSafetyCheck - Maple Help
Home : Support : Online Help : Programming : Multithreaded Programming : ThreadSafetyCheck
analyze a procedure or module for potential thread safety issues
ThreadSafetyCheck(p)
procedure or module
The ThreadSafetyCheck command analyzes the given procedure or module and warns about specific thread-safety issues.
Currently warnings are limited to the use of global variables and the use of lexically scoped local variables that are seen in assignment statements.
The return value consists of a sequence of two integers counting the number of procedures flagged with thread-safety issues, followed by the total number of procedures analyzed.
\mathrm{with}\left(\mathrm{CodeTools}\right):
p := proc(a) global x; x*a; end;
\textcolor[rgb]{0,0,1}{p}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathbf{proc}}\left(\textcolor[rgb]{0,0,1}{a}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\textcolor[rgb]{0,0,1}{\mathbf{global}}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{;}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{*}\textcolor[rgb]{0,0,1}{a}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\textcolor[rgb]{0,0,1}{\mathbf{end proc}}
[\mathrm{ThreadSafetyCheck}\left(p\right)]
[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}]
m := module() export p1, p2; local a; p1 := proc(n) a := n; end; p2 := proc(n) a^n; end; end module;
\textcolor[rgb]{0,0,1}{m}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathbf{module}}\left(\textcolor[rgb]{0,0,1}{}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\textcolor[rgb]{0,0,1}{\mathbf{local}}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\textcolor[rgb]{0,0,1}{a}\textcolor[rgb]{0,0,1}{;}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\textcolor[rgb]{0,0,1}{\mathbf{export}}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\textcolor[rgb]{0,0,1}{\mathrm{p1}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{p2}}\textcolor[rgb]{0,0,1}{;}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\textcolor[rgb]{0,0,1}{}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\textcolor[rgb]{0,0,1}{\mathbf{end module}}
[\mathrm{ThreadSafetyCheck}\left(m\right)]
[\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}]
The CodeTools[ThreadSafetyCheck] command was introduced in Maple 2016.
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Simulation and Prediction at the Command Line - MATLAB & Simulink - MathWorks Benelux
Simulation and Prediction Commands
Initial Conditions in Simulation and Prediction
Simulate a Continuous-Time State-Space Model
Simulate Model Output with Noise
If you estimated a linear model from detrended data and want to simulate or predict the output at the original operation conditions, use retrend to add trend data back into the simulated or predicted output.
Determine how closely the simulated model response matches the measured output signal.
Plots simulated or predicted output of one or more models on top of the measured output. You should use an independent validation data set as input to the model.
To plot five-step-ahead predicted output of the model mod against the validation data data, use the following command:
compare(data,mod,5)
Omitting the third argument assumes an infinite horizon and results in the comparison of the simulated response to the input data.
Simulate and plot the model output only.
To simulate the response of the model model using input data data, use the following command:
sim(model,data)
Predict and plot the model output only.
To perform one-step-ahead prediction of the response for the model model and input data data, use the following command:
predict(model,data,1)
Use the following syntax to compute k-step-ahead prediction of the output signal using model m:
yhat = predict(m,[y u],k)
predict computes the prediction results only over the time range of data. It does not forecast results beyond the available data range.
Forecast a time series into the future.
To forecast the value of a time series in an arbitrary number of steps into the future, use the following command:
forecast(model,past_data,K)
Here, model is a time series model, past_data is a record of the observed values of the time series, and K is the forecasting horizon.
The process of computing simulated and predicted responses over a time range starts by using the initial conditions to compute the first few output values. The sim, forecast, and predict commands provide options and default settings for handling initial conditions.
Simulation: Default initial conditions are zero for all model types except the idnlgrey model, whose default initial conditions are the internal model initial states (model property x0). You can specify other initial conditions using the InitialCondition simulation option. For more information on simulation options, see simOptions.
Use the compare command to validate models by simulation because its algorithm estimates the initial states of a model to optimize the model fit to a given data set. You can also use compare to return the estimated initial conditions for follow-on simulation and comparison with the same data set. These initial conditions can be in the form of an initial state vector (state-space models) or an initialCondition object (transfer function or polynomial models.)
If you are using sim to validate the quality of the identified model, you need to use the input signal from the validation data set and also account for initial condition effects. The simulated and the measured responses differ in the first few samples if the validation data set output contains initial condition effects that are not captured when you simulate the model. To minimize this difference, estimate the initial state values or the initialCondition model from the data using either findstates (state-space models) or compare (all LTI models) and specify these initial states using the InitialCondition simulation option (see simOptions). For example, compute the initial states that optimize the fit of the model m to the output data in z:
% Estimate the initial states
X0est = findstates(m,z);
% Simulate the response using estimated initial states
sim(m,z.InputData,opt)
For an example of obtaining and using initialCondition models, see Apply Initial Conditions when Simulating Identified Linear Models.
Prediction: Default initial conditions depend on the type of model. You can specify other initial conditions using the InitialCondition option (see predictOptions). For example, compute the initial states that optimize the 1-step-ahead predicted response of the model m to the output data z:
opt = predictOptions('InitialCondition','estimate');
[Yp,IC] = predict(m,z,1,opt);
This command returns the estimated initial conditions as the output argument IC. For information about other ways to specify initial states, see the predictOptions reference page.
This example shows how to simulate a continuous-time state-space model using a random binary input u and a sample time of 0.1 s.
Consider the following state-space model:
\begin{array}{l}\underset{}{\overset{˙}{x}}=\left[\begin{array}{cc}-1& 1\\ -0.5& 0\end{array}\right]x+\left[\begin{array}{c}1\\ 0.5\end{array}\right]u+\left[\begin{array}{c}0.5\\ 0.5\end{array}\right]e\\ y=\left[\begin{array}{cc}1& 0\end{array}\right]x+e\end{array}
where e is Gaussian white noise with variance 7.
Create a continuous-time state-space model.
A = [-1 1; -0.5 0];
B = [1;0.5];
K = [0.5;0.5];
% Ts = 0 indicates continuous time
model_ss = idss(A,B,C,D,K,'Ts',0,'NoiseVariance',7);
Create a random binary input.
u = idinput(400,'rbs',[0 0.3]);
Create an iddata object with empty output to represent just the input signal.
data = iddata([],u);
data.ts = 0.1;
Simulate the output using the model
y = sim(model_ss,data,opt);
This example shows how you can create input data and a model, and then use the data and the model to simulate output data.
In this example, you create the following ARMAX model with Gaussian noise e:
\begin{array}{l}y\left(t\right)-1.5y\left(t-1\right)+0.7y\left(t-2\right)=\\ u\left(t-1\right)+0.5u\left(t-2\right)+e\left(t\right)-e\left(t-1\right)+0.2e\left(t-1\right)\end{array}
Then, you simulate output data with random binary input u.
Create an ARMAX model.
m_armax = idpoly([1 -1.5 0.7],[0 1 0.5],[1 -1 0.2]);
Simulate the output data.
y = sim(m_armax,u,opt);
The 'AddNoise' option specifies to include in the simulation the Gaussian noise e present in the model. Set this option to false (default behavior) to simulate the noise-free response to the input u , which is equivalent to setting e to zero.
sim | predict | forecast | compare
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Khinchin's constant - Wikipedia
In number theory, Aleksandr Yakovlevich Khinchin proved that for almost all real numbers x, coefficients ai of the continued fraction expansion of x have a finite geometric mean that is independent of the value of x and is known as Khinchin's constant.
{\displaystyle x=a_{0}+{\cfrac {1}{a_{1}+{\cfrac {1}{a_{2}+{\cfrac {1}{a_{3}+{\cfrac {1}{\ddots }}}}}}}}\;}
it is almost always true that
{\displaystyle \lim _{n\rightarrow \infty }\left(a_{1}a_{2}...a_{n}\right)^{1/n}=K_{0}}
{\displaystyle K_{0}}
is Khinchin's constant
{\displaystyle K_{0}=\prod _{r=1}^{\infty }{\left(1+{1 \over r(r+2)}\right)}^{\log _{2}r}\approx 2.6854520010\dots }
{\displaystyle \prod }
denoting the product over all sequence terms).
Although almost all numbers satisfy this property, it has not been proven for any real number not specifically constructed for the purpose. Among the numbers x whose continued fraction expansions are known not to have this property are rational numbers, roots of quadratic equations (including the golden ratio Φ and the square roots of integers), and the base of the natural logarithm e.
Khinchin is sometimes spelled Khintchine (the French transliteration of Russian Хинчин) in older mathematical literature.
2 Series expressions
3 Hölder means
The proof presented here was arranged by Czesław Ryll-Nardzewski[1] and is much simpler than Khinchin's original proof which did not use ergodic theory.
Since the first coefficient a0 of the continued fraction of x plays no role in Khinchin's theorem and since the rational numbers have Lebesgue measure zero, we are reduced to the study of irrational numbers in the unit interval, i.e., those in
{\displaystyle I=[0,1]\setminus \mathbb {Q} }
. These numbers are in bijection with infinite continued fractions of the form [0; a1, a2, ...], which we simply write [a1, a2, ...], where a1, a2, ... are positive integers. Define a transformation T:I → I by
{\displaystyle T([a_{1},a_{2},\dots ])=[a_{2},a_{3},\dots ].\,}
The transformation T is called the Gauss–Kuzmin–Wirsing operator. For every Borel subset E of I, we also define the Gauss–Kuzmin measure of E
{\displaystyle \mu (E)={\frac {1}{\log 2}}\int _{E}{\frac {dx}{1+x}}.}
Then μ is a probability measure on the σ-algebra of Borel subsets of I. The measure μ is equivalent to the Lebesgue measure on I, but it has the additional property that the transformation T preserves the measure μ. Moreover, it can be proved that T is an ergodic transformation of the measurable space I endowed with the probability measure μ (this is the hard part of the proof). The ergodic theorem then says that for any μ-integrable function f on I, the average value of
{\displaystyle f\left(T^{k}x\right)}
is the same for almost all
{\displaystyle x}
{\displaystyle \lim _{n\to \infty }{\frac {1}{n}}\sum _{k=0}^{n-1}(f\circ T^{k})(x)=\int _{I}fd\mu \quad {\text{for }}\mu {\text{-almost all }}x\in I.}
Applying this to the function defined by f([a1, a2, ...]) = log(a1), we obtain that
{\displaystyle \lim _{n\to \infty }{\frac {1}{n}}\sum _{k=1}^{n}\log(a_{k})=\int _{I}f\,d\mu =\sum _{r=1}^{\infty }\log(r){\frac {\log {\bigl (}1+{\frac {1}{r(r+2)}}{\bigr )}}{\log 2}}}
for almost all [a1, a2, ...] in I as n → ∞.
Taking the exponential on both sides, we obtain to the left the geometric mean of the first n coefficients of the continued fraction, and to the right Khinchin's constant.
Series expressions[edit]
Khinchin's constant may be expressed as a rational zeta series in the form[2]
{\displaystyle \log K_{0}={\frac {1}{\log 2}}\sum _{n=1}^{\infty }{\frac {\zeta (2n)-1}{n}}\sum _{k=1}^{2n-1}{\frac {(-1)^{k+1}}{k}}}
or, by peeling off terms in the series,
{\displaystyle \log K_{0}={\frac {1}{\log 2}}\left[-\sum _{k=2}^{N}\log \left({\frac {k-1}{k}}\right)\log \left({\frac {k+1}{k}}\right)+\sum _{n=1}^{\infty }{\frac {\zeta (2n,N+1)}{n}}\sum _{k=1}^{2n-1}{\frac {(-1)^{k+1}}{k}}\right]}
where N is an integer, held fixed, and ζ(s, n) is the complex Hurwitz zeta function. Both series are strongly convergent, as ζ(n) − 1 approaches zero quickly for large n. An expansion may also be given in terms of the dilogarithm:
{\displaystyle \log K_{0}=\log 2+{\frac {1}{\log 2}}\left[{\mbox{Li}}_{2}\left({\frac {-1}{2}}\right)+{\frac {1}{2}}\sum _{k=2}^{\infty }(-1)^{k}{\mbox{Li}}_{2}\left({\frac {4}{k^{2}}}\right)\right].}
Hölder means[edit]
The Khinchin constant can be viewed as the first in a series of the Hölder means of the terms of continued fractions. Given an arbitrary series {an}, the Hölder mean of order p of the series is given by
{\displaystyle K_{p}=\lim _{n\to \infty }\left[{\frac {1}{n}}\sum _{k=1}^{n}a_{k}^{p}\right]^{1/p}.}
When the {an} are the terms of a continued fraction expansion, the constants are given by
{\displaystyle K_{p}=\left[\sum _{k=1}^{\infty }-k^{p}\log _{2}\left(1-{\frac {1}{(k+1)^{2}}}\right)\right]^{1/p}.}
This is obtained by taking the p-th mean in conjunction with the Gauss–Kuzmin distribution. For example, the harmonic mean (p = −1) of the terms of a continued fraction is almost always
{\displaystyle K_{-1}=1.74540566240\dots }
The value for K0 is obtained in the limit of p → 0.
{\displaystyle \lim _{n\rightarrow \infty }(\pi _{1}\pi _{2}...\pi _{n})^{1/n}}
seems to tend to Khinchin's constant.
π, the Euler–Mascheroni constant γ, and Khinchin's constant itself, based on numerical evidence,[3][4] are thought to be among the numbers whose geometric mean of the coefficients ai in their continued fraction expansion tends to Khinchin's constant. However, none of these limits have been rigorously established.
It is not known whether Khinchin's constant is a rational, algebraic irrational or transcendental number.[5]
Lochs' theorem
^ Ryll-Nardzewski, Czesław (1951), "On the ergodic theorems II (Ergodic theory of continued fractions)", Studia Mathematica, 12: 74–79, doi:10.4064/sm-12-1-74-79
^ Bailey, Borwein & Crandall, 1997. In that paper, a slightly non-standard definition is used for the Hurwitz zeta function.
^ Weisstein, Eric W. "Euler-Mascheroni Constant Continued Fraction". mathworld.wolfram.com. Retrieved 2020-03-23.
^ Weisstein, Eric W. "Pi Continued Fraction". mathworld.wolfram.com. Retrieved 2020-03-23.
^ Weisstein, Eric W. "Khinchin's constant". MathWorld.
David H. Bailey; Jonathan M. Borwein; Richard E. Crandall (1995). "On the Khinchine constant" (PDF). Mathematics of Computation. 66 (217): 417–432. doi:10.1090/s0025-5718-97-00800-4.
Jonathan M. Borwein; David M. Bradley; Richard E. Crandall (2000). "Computational Strategies for the Riemann Zeta Function" (PDF). J. Comput. Appl. Math. 121 (1–2): 11. doi:10.1016/s0377-0427(00)00336-8.
Thomas Wieting (2007). "A Khinchin Sequence". Proceedings of the American Mathematical Society. 136 (3): 815–824. doi:10.1090/S0002-9939-07-09202-7.
Aleksandr Ya. Khinchin (1997). Continued Fractions. New York: Dover Publications.
Wikimedia Commons has media related to Khinchin's constant.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Khinchin%27s_constant&oldid=1064597746"
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Multiplication - Wikipedia
Find sources: "Multiplication" – news · newspapers · books · scholar · JSTOR (April 2012) (Learn how and when to remove this template message)
{\displaystyle a\times b=\underbrace {b+\cdots +b} _{a{\text{ times}}}}
For example, 4 multiplied by 3, often written as
{\displaystyle 3\times 4}
and spoken as "3 times 4", can be calculated by adding 3 copies of 4 together:
{\displaystyle 3\times 4=4+4+4=12}
{\displaystyle 4\times 3=3+3+3+3=12}
2.1 Historical algorithms
2.1.1 Egyptians
2.2.1 Grid method
2.3 Computer algorithms
3 Products of measurements
4 Product of a sequence
4.1 Capital pi notation
4.1.1 Properties of capital pi notation
7 Multiplication with set theory
8 Multiplication in group theory
9 Multiplication of different kinds of numbers
10 Exponentiation
In arithmetic, multiplication is often written using the multiplication sign (either × or
{\displaystyle \times }
) between the terms (that is, in infix notation).[2] For example,
{\displaystyle 2\times 3=6}
("two times three equals six")
{\displaystyle 3\times 4=12}
{\displaystyle 2\times 3\times 5=6\times 5=30}
{\displaystyle 2\times 2\times 2\times 2\times 2=32}
Historical algorithmsEdit
Grid methodEdit
Computer algorithmsEdit
The classical method of multiplying two n-digit numbers requires n2 digit multiplications. Multiplication algorithms have been designed that reduce the computation time considerably when multiplying large numbers. Methods based on the discrete Fourier transform reduce the computational complexity to O(n log n log log n). In 2016, the factor log log n was replaced by a function that increases much slower, though still not constant.[15] In March 2019, David Harvey and Joris van der Hoeven submitted a paper presenting an integer multiplication algorithm with a complexity of
{\displaystyle O(n\log n).}
[16] The algorithm, also based on the fast Fourier transform, is conjectured to be asymptotically optimal.[17] The algorithm is not practically useful, as it only becomes faster for multiplying extremely large numbers (having more than 2172912 bits).[18]
Products of measurementsEdit
Product of a sequenceEdit
Capital pi notationEdit
The product of a sequence of factors can be written with the product symbol, which derives from the capital letter
{\displaystyle \textstyle \prod }
(pi) in the Greek alphabet (much like the same way the capital letter
{\displaystyle \textstyle \sum }
(sigma) is used in the context of summation).[19][20] Unicode position U+220F ∏ contains a glyph for denoting such a product, distinct from U+03A0 Π , the letter. The meaning of this notation is given by:
{\displaystyle \prod _{i=1}^{4}i=1\cdot 2\cdot 3\cdot 4,}
{\displaystyle \prod _{i=1}^{4}i=24.}
{\displaystyle \prod _{i=1}^{6}i=1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6=720.}
{\displaystyle \prod _{i=m}^{n}x_{i}=x_{m}\cdot x_{m+1}\cdot x_{m+2}\cdot \,\,\cdots \,\,\cdot x_{n-1}\cdot x_{n},}
Properties of capital pi notationEdit
{\displaystyle \prod _{i=1}^{n}x_{i}=x_{1}\cdot x_{2}\cdot \ldots \cdot x_{n}.}
{\displaystyle \prod _{i=1}^{n}x=x\cdot x\cdot \ldots \cdot x=x^{n}.}
{\displaystyle \prod _{i=1}^{n}{x_{i}y_{i}}=\left(\prod _{i=1}^{n}x_{i}\right)\left(\prod _{i=1}^{n}y_{i}\right)}
{\displaystyle \left(\prod _{i=1}^{n}x_{i}\right)^{a}=\prod _{i=1}^{n}x_{i}^{a}}
if a is a nonnegative integer, or if all
{\displaystyle x_{i}}
are positive real numbers, and
{\displaystyle \prod _{i=1}^{n}x^{a_{i}}=x^{\sum _{i=1}^{n}a_{i}}}
{\displaystyle a_{i}}
are nonnegative integers, or if x is a positive real number.
Infinite productsEdit
{\displaystyle \prod _{i=m}^{\infty }x_{i}=\lim _{n\to \infty }\prod _{i=m}^{n}x_{i}.}
{\displaystyle \prod _{i=-\infty }^{\infty }x_{i}=\left(\lim _{m\to -\infty }\prod _{i=m}^{0}x_{i}\right)\cdot \left(\lim _{n\to \infty }\prod _{i=1}^{n}x_{i}\right),}
{\displaystyle x\cdot y=y\cdot x.}
{\displaystyle (x\cdot y)\cdot z=x\cdot (y\cdot z)}
{\displaystyle x\cdot (y+z)=x\cdot y+x\cdot z}
{\displaystyle x\cdot 1=x}
{\displaystyle x\cdot 0=0}
{\displaystyle (-1)\cdot x=(-x)}
{\displaystyle (-x)+x=0}
{\displaystyle (-1)\cdot (-1)=1}
Every number x, except 0, has a multiplicative inverse,
{\displaystyle {\frac {1}{x}}}
{\displaystyle x\cdot \left({\frac {1}{x}}\right)=1}
{\displaystyle x\times 0=0}
{\displaystyle x\times S(y)=(x\times y)+x}
{\displaystyle x\times 1=x\times S(0)=(x\times 0)+x=0+x=x.}
{\displaystyle (x_{p},\,x_{m})\times (y_{p},\,y_{m})=(x_{p}\times y_{p}+x_{m}\times y_{m},\;x_{p}\times y_{m}+x_{m}\times y_{p}).}
{\displaystyle (0,1)\times (0,1)=(0\times 0+1\times 1,\,0\times 1+1\times 0)=(1,0).}
Multiplication with set theoryEdit
Multiplication in group theoryEdit
Multiplication in group theory is typically notated either by a dot or by juxtaposition (the omission of an operation symbol between elements). So multiplying element a by element b could be notated as a
{\displaystyle \cdot }
b or ab. When referring to a group via the indication of the set and operation, the dot is used. For example, our first example could be indicated by
{\displaystyle \left(\mathbb {Q} /\{0\},\,\cdot \right)}
Multiplication of different kinds of numbersEdit
{\displaystyle N\times M}
is the sum of N copies of M when N and M are positive whole numbers. This gives the number of things in an array N wide and M high. Generalization to negative numbers can be done by
{\displaystyle N\times (-M)=(-N)\times M=-(N\times M)}
{\displaystyle (-N)\times (-M)=N\times M}
Generalization to fractions
{\displaystyle {\frac {A}{B}}\times {\frac {C}{D}}}
is by multiplying the numerators and denominators respectively:
{\displaystyle {\frac {A}{B}}\times {\frac {C}{D}}={\frac {(A\times C)}{(B\times D)}}}
. This gives the area of a rectangle
{\displaystyle {\frac {A}{B}}}
high and
{\displaystyle {\frac {C}{D}}}
wide, and is the same as the number of things in an array when the rational numbers happen to be whole numbers.[21][22]
Considering complex numbers
{\displaystyle z_{1}}
{\displaystyle z_{2}}
as ordered pairs of real numbers
{\displaystyle (a_{1},b_{1})}
{\displaystyle (a_{2},b_{2})}
{\displaystyle z_{1}\times z_{2}}
{\displaystyle (a_{1}\times a_{2}-b_{1}\times b_{2},a_{1}\times b_{2}+a_{2}\times b_{1})}
. This is the same as for reals
{\displaystyle a_{1}\times a_{2}}
when the imaginary parts
{\displaystyle b_{1}}
{\displaystyle b_{2}}
are zero.[citation needed]
Equivalently, denoting
{\displaystyle {\sqrt {-1}}}
{\displaystyle i}
{\displaystyle z_{1}\times z_{2}=(a_{1}+b_{1}i)(a_{2}+b_{2}i)=(a_{1}\times a_{2})+(a_{1}\times b_{2}i)+(b_{1}\times a_{2}i)+(b_{1}\times b_{2}i^{2})=(a_{1}a_{2}-b_{1}b_{2})+(a_{1}b_{2}+b_{1}a_{2})i.}
Alternatively, in trigonometric form, if
{\displaystyle z_{1}=r_{1}(\cos \phi _{1}+i\sin \phi _{1}),z_{2}=r_{2}(\cos \phi _{2}+i\sin \phi _{2})}
{\textstyle z_{1}z_{2}=r_{1}r_{2}(\cos(\phi _{1}+\phi _{2})+i\sin(\phi _{1}+\phi _{2})).}
Often division,
{\displaystyle {\frac {x}{y}}}
, is the same as multiplication by an inverse,
{\displaystyle x\left({\frac {1}{y}}\right)}
. Multiplication for some types of "numbers" may have corresponding division, without inverses; in an integral domain x may have no inverse "
{\displaystyle {\frac {1}{x}}}
" but
{\displaystyle {\frac {x}{y}}}
may be defined. In a division ring there are inverses, but
{\displaystyle {\frac {x}{y}}}
may be ambiguous in non-commutative rings since
{\displaystyle x\left({\frac {1}{y}}\right)}
need not be the same as
{\displaystyle \left({\frac {1}{y}}\right)x}
{\displaystyle a^{n}=\underbrace {a\times a\times \cdots \times a} _{n}}
^ Angell, David. "ORDERING COMPLEX NUMBERS... NOT*" (PDF). web.maths.unsw.edu.au. Retrieved 29 December 2021. {{cite web}}: CS1 maint: url-status (link)
Boyer, Carl B. (revised by Merzbach, Uta C.) (1991). History of Mathematics. John Wiley and Sons, Inc. ISBN 978-0-471-54397-8. {{cite book}}: CS1 maint: multiple names: authors list (link)
Retrieved from "https://en.wikipedia.org/w/index.php?title=Multiplication&oldid=1087449671"
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Magnetic Fields: Level 2-4 Challenges Practice Problems Online | Brilliant
An electric current of 1 Ampere is flowing along a horizontal wire. At
x=0
the wire splits into a circle of radius 0.05 m and then comes back together at
x=0.1
m. What is the magnitude in Tesla of the magnetic field in the middle of this loop of wire?
Consider a standard cyclotron accelerator in which two semi-circular regions (dees) are connected to an AC voltage source which provides an electric field
E
and a uniform vertical (out of the page) magnetic
B
field inside cyclotron which is perpendicular to the electric field. The electric field switches such that is always points toward the dee the particle is not in.
A proton is released from rest such that it starts rotating in the cyclotron at radius
R
and finally comes out from the slits of the cyclotron. The distance between the two semi-circular regions is
d
. Find the maximum number of turns proton take before coming out from slit's.
{ { m }_{ p }=1.6\times { 10 }^{ -27 }\text{ kg}\\ { q }_{ p }=1.6\times { 10 }^{ -19 } C\\ B={ 10 }^{ -4 }\text{ T}\\ R=6\text{ m}\\ E=10\frac{\text{V}}{\text{m}}\\ d=10\text{ cm}\\ }
The charged particles from solar eruptions hit Earth mainly around the North and South poles (and cause auroras). This is because our Earth is similar to a big magnet. The Earth generates a magnetic field and this magnetic field funnels the charged particles towards the poles. In order to see an example of this funneling, we can think of the following problem:
Consider each magnetic pole separately (so there's only one pole in the problem). The magnetic field near a single pole is
\vec{B} = k\vec{r}/r^3=k\hat{r}/r^2
\vec{r}
is the radial vector point of the pole and the point of interest and
\hat{r}=\vec{r}/r
is the radial unit vector. The sign of
k
is opposite for the North and South poles. If there's an electric charge moving in that magnetic field, its trajectory is on a surface of a cone, i.e. a big funnel. Find the vertex angle (the angle between the axis and a line on the surface of the cone) in degrees with these given initial conditions: the distance between the charge and the pole is
r = 1~\mbox{m}
and the velocity vector of the charge is
v = 2~\mbox{m/s}
perpendicular to the line connecting the pole and the charge. We will consider a north pole and so let
k = 3~\mbox{T}\cdot\mbox{m}^2
. The charge of the particle is
q = 4~\mbox{C}
m = 5~\mbox{kg}
Hint: For anyone who has not taken electromagnetism yet, the force on the charge particle is given by the Lorentz force law:
\vec{F}=q\vec{v} \times \vec {B}
A non-relativistic proton beam passes without deflection through a region where there are two transverse, mutually perpendicular, electric (
E=120 ~\textrm{kV/m}
B=50 ~\textrm{mT}
) fields. Then the beam strikes a grounded target. Find the force in Newtons with which the beam acts on the target if the beam current is
I=0.8 ~ \textrm{mA}
Assume that the collisions with the target are inelastic.
The proton's mass and charge are:
m_{p}=1.67 \times 10^{-27}\textrm{ kg}
e= 1.6 \times 10^{-19} C.
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