MaplePrimes Questions

For some reason machine thinks that I am trying to add a matrix to an equal sign.

Is it italic when copied and pasted?  Is it bold when copied from maple 8?  I just ahve not been able to work it out.

The only way that I can think of doing it is by multiplying by a tetrad.  Even then it does not work well see my worksheet:  The Dirac Equation in Robertson-Walker spacetime.

Factorise

   

into your Maple worksheet and copy and paste the Maple output into the answer box below.

I have a recursive function that spits out recursive formulates "evaluated inertly" so bascially it produces a formula simply like a symbolic formula(doesn't reduce the addition of values though).

 

What I would like is to color code each recursive step

e.g.,

F := proc(n)
    local i,k:
    if n <= 0 then return 1; end if:
    F(n-2) %+ F(n-1):
end proc:

F(5);

value(F(5));
 

so I would like to cover each part of the recursion differently so it is easier to see, maybe even use the previous colors so one can visualize the nestings of the recursion. (e.g., F(n-2) takes the previous formals color and tweaks it towards the red and and F(n-1) takes the porevious formulas color and tweaks it towards green)
(this would then pass an RGB value that(default, say blue) to the functions above).

 

Any way to accomplish this?

 

 

 

 

Enter the value of cos^-1(0.62) to 15 significant figures in the box below.

by cos^-1(0.62) I mean arccos(0.62) or inverse of cos(0.62)

 

For the following system, the parameters D_1,D_2,D_3,D_4,D_5,D_6,S are all positive, How can I get all the  equilibrium  and their stability.

diff(x1,t)=fS;diff(x2,t)=fV;diff(x3,t)=fC;diff(x4,t)=fR;

fS := -D2*x1*x2-D1*x1-x1*x3+S; fV := D2*x1*x2-D4*x2+x1*x3; fC := -D6*x3*x4-D5*x3+x2; fR := D6*x3*x4-x4

I want to set up an RSS or IFTTT so that I get a notification on my phone whenever there's something new in the Active Conversations stack. My phone uses Google Android. I'd prefer to not need to rely on email/gmail for the notification; something more endogenous to Android would be preferable. I've never used RSS or IFTTT, so detailed instructions are preferable.

Hello everyone

First day using Maple and I've been given the following task: 

Design a function rangetolist(intrange) which expects an expression intrange of the type .., i.e., of the form m..n as its argument and converts it to the list
[m,m+1,...n] Also test what happens if m is larger than n. Hint: This readily works using seq
.

I can't quite seem to get it working, online resources haven't been too helpful in resolving the issue unfortunately. 

Here is my attempt 

rangetolist := (x::range) -> seq(i = op(1, x), i .. op(nops(x), x), 1);
 
which gives me this error:  Error, (in rangetolist) unable to execute seq
 
Very grateful for any constructive input, thanks :)

 

 

Hi , I have a system of equations and initial or boundary conditions. My.question how to solve these equations and their conditions together? And I can’t use dsolve because my result system doesn’t contain any derivative. In addition I couldn’t use fsolve because I don’t know how to collect the system and the conditions to solve together as a one system , and some examples contains equations morethan number of variables.... please can you advice me a solution 

thanks 

Why won't simplify give the obvious result and Why does testeq give a Fail result in this case?

 

with(RealDomain);

assume(0<f,0<g);

simplify(f*sqrt(g/f));

testeq(f*sqrt(g/f)=sqrt(f)*sqrt(g))

Hi everyone,

I have 4 expression, each containing several parameters (3, I think).

I need to find the range of the values of the parameters for which one of the expressions is at least as great as any of the other expressions.

So the input of the program/function would be the expression and which one I want to be the (weakly) greatest, and the  output should be something like: [parameter a is a member of the set [0.1,0.5] and parameter b is a member of the set [1.2, 2.4] and parameters c is negative] (meaning if these are true, then the given exoression is the (weakly) greatest of all the given expressions.

 

Is there a way to do this in Maple?

 

(For the curious: I have a game-theoretic problem, where the payoff functions include several parameters. I am trying to find the best response functions to find the Nash.equilibria of the game for the different range of parameters)

 

Thank you in advance,

JTamas

I have a formula that essentially is a m-order sum over a totally ordered set(the integers).

 

sum_{n_1=1)^{n_0} ... m times ... sum_{n_m=1}^{n_{m-1}}

 

It may be a little confusing but essentially it is just summing over all monotonic lidyd of size m of [1..n_0]

e.g., if n_0 = 4 and m = 3 then

1 2 3

1 2 4

1 3 4

2 3 4

 

Does maple have a convient way to do this? Ideally some function like osum(..., i=1..n_0, m)

where i is a multi-index(just the ordered list here so I could access i[1] the smallest, value, i[2] is the next largest, etc).

 

How can i solve below equation?

diff(v,x,x)+diff(v,x)*cot(x)-v(cot2(x)+nu)=a2Q/D

that Q is function of x

Q=Q[x]


 

Analysis of the semiclassical (SC) momentum rate equations

Plotting the ICs and BCs and examining sensitivity to the Re and Im forces

MRB: 24/2/2020, 27/2/2020, 2/3/2020.

We examine solution of the SC version of the momentum rate equations, in which O`&hbar;`^2 terms for u(x, t) are removed. A high level of sensitivity to ICs and BCs makes solution finding difficult.

restart;

with(PDETools): with(CodeTools):with(plots):

We set up the initial conditions:

ICu := {u(x, 0) = .1*sin(2*Pi*x)}; ICv := {v(x, 0) = .2*sin(Pi*x)};

{u(x, 0) = .1*sin(2*Pi*x)}

 

{v(x, 0) = .2*sin(Pi*x)}

(1)

plot([0.1*sin(2*Pi*x),0.2*sin(Pi*x)],x = 0..2, title="ICs:\n u(x,0) (red), v(x,0) (blue)",color=[red,blue],gridlines=true);  

 

The above initial conditions represent a positive velocity field v(x, 0) (blue) and a colliding momentum field u(x, t)(red).

 

Here are the BCs

BCu := {u(0,t) = 0.5*(1-cos(2*Pi*t))};

{u(0, t) = .5-.5*cos(2*Pi*t)}

(2)

BCv := {v(0,t) = 0.5*sin(2*Pi*t),v(2,t)=-0.5*sin(2*Pi*t)};  

{v(0, t) = .5*sin(2*Pi*t), v(2, t) = -.5*sin(2*Pi*t)}

(3)

plot([0.5*(1-cos(2*Pi*t)),0.5*sin(2*Pi*t),-0.5*sin(2*Pi*t)],t=0..1,color=[red,blue,blue],linestyle=[dash,dash,dot],title="BCs:\n u(0,t) (red-dash),\n v(0,t) (blue-dash), v(1,t) (blue-dot)",gridlines=true);

 

 

We can now set up the PDEs for the semiclassical case.

hBar:= 1:m:= 1:Fu:= 0.2:Fv:= 0.1:#1.0,0.2

pdeu := diff(u(x,t),t)+u(x,t)/m*(diff(u(x,t),x)) = Fu;

diff(u(x, t), t)+u(x, t)*(diff(u(x, t), x)) = .2

(4)

pdev := diff(v(x,t),t)+u(x,t)/m*(diff(v(x,t),x))-hBar*(diff(u(x,t),x$2))/(2*m)+v(x,t)*(diff(u(x,t),x))/m = Fv;  

diff(v(x, t), t)+u(x, t)*(diff(v(x, t), x))-(1/2)*(diff(diff(u(x, t), x), x))+v(x, t)*(diff(u(x, t), x)) = .1

(5)

ICu:={u(x,0) = 0.1*sin(2*Pi*x)};  

{u(x, 0) = .1*sin(2*Pi*x)}

(6)

ICv:={v(x,0) = 0.2*sin(Pi*x/2)};  

{v(x, 0) = .2*sin((1/2)*Pi*x)}

(7)

IC := ICu union ICv;  

{u(x, 0) = .1*sin(2*Pi*x), v(x, 0) = .2*sin((1/2)*Pi*x)}

(8)

BCu := {u(0,t) = 0.5*(1-cos(2*Pi*t)), D[1](u)(2,t) = 0.1*cos(2*Pi*t)};

{u(0, t) = .5-.5*cos(2*Pi*t), (D[1](u))(2, t) = .1*cos(2*Pi*t)}

(9)

BCv := {v(0,t) = 0.2*(1-cos(2*Pi*t))};  

{v(0, t) = .2-.2*cos(2*Pi*t)}

(10)

BC := BCu union BCv;  

{u(0, t) = .5-.5*cos(2*Pi*t), v(0, t) = .2-.2*cos(2*Pi*t), (D[1](u))(2, t) = .1*cos(2*Pi*t)}

(11)

We now set up the PDE solver:

pds := pdsolve({pdeu,pdev},{BC[],IC[]},time = t,range = 0..2,numeric);#'numeric' solution

_m2592591229440

(12)

Cp:=pds:-animate({[u, color = red, linestyle = dash],[v,color = blue,linestyle = dash]},t = 30,frames = 400,numpoints = 400,title="Semiclassical momentum equations solution for Re and Im momenta u(x,t) (red) and v(x,t) (blue) \n under respective constant positive forces [0.2, 0.1] \n with sinusoidal boundary conditions at x = 0, 1 and sinusoidal initial conditions: \n time = %f ", gridlines = true,linestyle=solid):Cp;

Error, (in pdsolve/numeric/animate) unable to compute solution for t>HFloat(0.0):
Newton iteration is not converging

 

Cp

(13)

Observations on the quantum case:

The classical equation for u(x, t) is independent of the equation for v(x, t).  u(x, t) (red) is a solution of the classical Burgers equation subject to a force 0.2, but u(x, t) is NOT influenced by v(x, t).  On the otherhand, v(x, t) (blue) is a solution of the quantum dynamics equation subject to force 0.1 and is influenced by u(x, t).   This one way causality (u " implies v")  is a feature of the semiclassical case, and it emphasises the controlling influence of the classical u(x, t), which modulates the quantum solution for v(x, t).  Causally, we have u" implies v".

 

The initial conditions are of low momentum amplitude:"+/-"0.1 for the classical u(x, 0) (red) field and`&+-`(0).2 for v(x, 0) (blue)  but their influence is soon washed out by the boundary conditions "u(0,t) ~1, v(0,t)~0.5" and "v(1,t)~0.5" that drive the momentum dynamics.

 

The temporal frequency of the boundary condition on the v-field is twice that of the classical u-field. This is evident in the above blue transient plot. Moreover, the">=0" boundary condition on the classical u-momentum (red), drives that field in the positive direction, initially overtaking the quantum v(x, t) field, as consistent with the applied forces [0.2, 0.1]. NULLAlthough initially of greater amplitude than the classical u(x, t)field, the v(x, t) momentum field is asymptotically of the same amplitude as the u(x, t) field, but has greater spatial and temporal frequency, owing to the boundary conditions.

 

Referring to the semiclassical momentum rate equations, we note that the classical field u(x, t) (red) modulates the quantum momentum rate equation for v(x, t).

``

 

 

 


 

Download SC-plots.mw

I am having difficulty getting solutions to a pair of PDEs.  Would anyone like to cast an eye over the attached file, incase I am missing something.

Thanks

Melvin

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