MaplePrimes Questions

RandC := proc(k := getSystemTimeMS())
    local r,b,g,kk;
    kk := k+ parse(convert(Time(), string)[6..18]);
    kk := (randomize(round(k*time[real]() + getSystemTimeMS() + time[real]())));    
    (randomize(round(k*time[real]() + getSystemTimeMS() + time[real]())));    
    r := rand(0..1.0)();
    kk := k+ parse(convert(Time(), string)[6..18]);
    (randomize(round(k*time[real]() + getSystemTimeMS() + time[real]())));    
    kk := (randomize(round(kk*r + k + time[real]() + getSystemTimeMS())));    
    g := rand(0..1.0)();
    kk := k+ parse(convert(Time(), string)[6..18]);
    (randomize(round(k*time[real]() + getSystemTimeMS() + time[real]())));    
    kk := (randomize(round(kk*g + k*r*time[real]() + getSystemTimeMS())));
    b := rand(0..1.0)();
    [RGB(r,g,b)];
end proc:

 

Been trying to generate random colors but nothing works because obviously maple decides to cache the results if the arguments are the same. If I call RandC with an argument that changes it will give me different values. If not it will sometimes give me different values but usually there will be long strings of the same value.

 

Hence when I generate a sequence of random colors I will almost always get a sequence of the same color or a sequence that containst the same color many times. I'd like to be able to use RandC without having to provide arguments.

(all the junk in the function was trying to get it to randomize properly as I didnt' realize maple was caching. Also Value(Time()) does not return the ms and seems to do nothing which contradicts the help and hence the parse)

 

 

This seems like a dangerous thing for maple to do. Silently caching results that depend on random generation will result in non-random results. It could skew many results.

 

each time i use this i did not have any problem but this equation not seperate any one know what is problem?

restart

with(SolveTools)

undeclare(prime)

`There is no more prime differentiation variable; all derivatives will be displayed as indexed functions`

(1)

with(PDEtools)

P := U(xi)^3*mu*C[2]*h[9]+(2*I)*(diff(U(xi), xi))*a*k*mu+4*(diff(U(xi), xi))*k*mu^3*C[2]*h[7]-4*(diff(diff(diff(U(xi), xi), xi), xi))*k^3*mu*C[2]*h[7]-U(xi)^3*mu*C[2]*h[8]+I*(diff(diff(diff(diff(U(xi), xi), xi), xi), xi))*k^4*C[2]*h[7]+I*(diff(U(xi), xi))*U(xi)^2*k*C[2]*h[9]-(6*I)*(diff(diff(U(xi), xi), xi))*k^2*mu^2*C[2]*h[7]+I*U(xi)*mu^4*C[2]*h[7]-I*(diff(U(xi), xi))*v-U(xi)*w+b*U(xi)^3-U(xi)*a*mu^2+(diff(diff(U(xi), xi), xi))*a*k^2+I*(diff(U(xi), xi))*U(xi)^2*k*C[2]*h[8]+C[1](-U(xi)^3*mu^2*h[2]+(2*I)*(diff(U(xi), xi))*U(xi)^2*k*mu*h[4]-(2*I)*(diff(U(xi), xi))*U(xi)^2*k*mu*h[5]+(diff(U(xi), xi))^2*U(xi)*k^2*h[2]-U(xi)^3*mu^2*h[5]+U(xi)^2*(diff(diff(U(xi), xi), xi))*k^2*h[5]-(4*(diff(U(xi), xi))*I)*k*mu^3*h[1]+4*(diff(diff(diff(U(xi), xi), xi), xi))*k^3*mu*h[1]*I+(2*I)*(diff(U(xi), xi))*U(xi)^2*k*mu*h[2]+h[6]*U(xi)^5-U(xi)^3*mu^2*h[4]+U(xi)^2*(diff(diff(U(xi), xi), xi))*k^2*h[4]+U(xi)*mu^4*h[1]-6*(diff(diff(U(xi), xi), xi))*k^2*mu^2*h[1]+(diff(diff(diff(diff(U(xi), xi), xi), xi), xi))*k^4*h[1]+h[3](k^2*(diff(U(xi), xi))^2+2*(0+I)*(diff(U(xi), xi))*k*mu*U(xi)-mu^2*U(xi)^2)*U(xi)) = 0

U(xi)^3*mu*C[2]*h[9]+I*(diff(U(xi), xi))*U(xi)^2*k*C[2]*h[8]+4*(diff(U(xi), xi))*k*mu^3*C[2]*h[7]-4*(diff(diff(diff(U(xi), xi), xi), xi))*k^3*mu*C[2]*h[7]-U(xi)^3*mu*C[2]*h[8]+I*(diff(U(xi), xi))*U(xi)^2*k*C[2]*h[9]-(6*I)*(diff(diff(U(xi), xi), xi))*k^2*mu^2*C[2]*h[7]+I*U(xi)*mu^4*C[2]*h[7]-I*(diff(U(xi), xi))*v+(2*I)*(diff(U(xi), xi))*a*k*mu-U(xi)*w+b*U(xi)^3-U(xi)*a*mu^2+(diff(diff(U(xi), xi), xi))*a*k^2+I*(diff(diff(diff(diff(U(xi), xi), xi), xi), xi))*k^4*C[2]*h[7]+C[1](-U(xi)^3*mu^2*h[2]+(2*I)*(diff(U(xi), xi))*U(xi)^2*k*mu*h[4]-(4*I)*(diff(U(xi), xi))*k*mu^3*h[1]+(diff(U(xi), xi))^2*U(xi)*k^2*h[2]-U(xi)^3*mu^2*h[5]+U(xi)^2*(diff(diff(U(xi), xi), xi))*k^2*h[5]+(4*I)*(diff(diff(diff(U(xi), xi), xi), xi))*k^3*mu*h[1]-(2*I)*(diff(U(xi), xi))*U(xi)^2*k*mu*h[5]+(2*I)*(diff(U(xi), xi))*U(xi)^2*k*mu*h[2]+h[6]*U(xi)^5-U(xi)^3*mu^2*h[4]+U(xi)^2*(diff(diff(U(xi), xi), xi))*k^2*h[4]+U(xi)*mu^4*h[1]-6*(diff(diff(U(xi), xi), xi))*k^2*mu^2*h[1]+(diff(diff(diff(diff(U(xi), xi), xi), xi), xi))*k^4*h[1]+h[3](k^2*(diff(U(xi), xi))^2+(2*I)*(diff(U(xi), xi))*k*mu*U(xi)-mu^2*U(xi)^2)*U(xi)) = 0

(2)

Re(P)

Re(U(xi)^3*mu*C[2]*h[9]+4*(diff(U(xi), xi))*k*mu^3*C[2]*h[7]-4*(diff(diff(diff(U(xi), xi), xi), xi))*k^3*mu*C[2]*h[7]-U(xi)^3*mu*C[2]*h[8]-U(xi)*w+b*U(xi)^3-U(xi)*a*mu^2+(diff(diff(U(xi), xi), xi))*a*k^2+C[1](-U(xi)^3*mu^2*h[2]+(2*I)*(diff(U(xi), xi))*U(xi)^2*k*mu*h[4]-(4*I)*(diff(U(xi), xi))*k*mu^3*h[1]+(diff(U(xi), xi))^2*U(xi)*k^2*h[2]-U(xi)^3*mu^2*h[5]+U(xi)^2*(diff(diff(U(xi), xi), xi))*k^2*h[5]+(4*I)*(diff(diff(diff(U(xi), xi), xi), xi))*k^3*mu*h[1]-(2*I)*(diff(U(xi), xi))*U(xi)^2*k*mu*h[5]+(2*I)*(diff(U(xi), xi))*U(xi)^2*k*mu*h[2]+h[6]*U(xi)^5-U(xi)^3*mu^2*h[4]+U(xi)^2*(diff(diff(U(xi), xi), xi))*k^2*h[4]+U(xi)*mu^4*h[1]-6*(diff(diff(U(xi), xi), xi))*k^2*mu^2*h[1]+(diff(diff(diff(diff(U(xi), xi), xi), xi), xi))*k^4*h[1]+h[3](k^2*(diff(U(xi), xi))^2+(2*I)*(diff(U(xi), xi))*k*mu*U(xi)-mu^2*U(xi)^2)*U(xi)))-Im((diff(U(xi), xi))*U(xi)^2*k*C[2]*h[8]+(diff(U(xi), xi))*U(xi)^2*k*C[2]*h[9]-6*(diff(diff(U(xi), xi), xi))*k^2*mu^2*C[2]*h[7]+U(xi)*mu^4*C[2]*h[7]-(diff(U(xi), xi))*v+2*(diff(U(xi), xi))*a*k*mu+(diff(diff(diff(diff(U(xi), xi), xi), xi), xi))*k^4*C[2]*h[7]) = 0

(3)
 

``

Download real_and_imaginary_.mw

Hi, I have plotted the bar graphs, and it's fine but I am interested in showing the value on each bar. How to implement this. As an example, I uploaded a picture for convenience.

Bar_graphs.mw

How to convert this mathematica code to Maple code? Is there any command? NN.pdf

I have encountered an issue: eq1 is not satisfied, though eq2 is satisfied for the parametric value (10). I need assistance in finding a way to ensure that both equations are satisfied simultaneously. Please provide guidance or suggest a potential approach for addressing this issue.verf_kk.mw

I have  a big problem in transformation How we can do suh transformation in  type of  procure  without use any hand work for example in physic abs|-| remove the exponential term how the maple remove that term automatically and collect all term and do my transformation this example is really hard one which is must do a lot by hand and mixed them which maybe a week take my time to get results and how i reach the results without spending that time i have a result of this equation and i am try to get but i don't know the results of this person is correct or not but i will share in here,  i did some try i will share in here too if in DEchange add U(xi) it will work and give me the other step but i need something more effective, when q^* is conjugate of q =exp(-ipsi(x,t))U(xi)

NULL

restart

with(PDEtools)

with(Physics)

with(SolveTools)

undeclare(prime)

`There is no more prime differentiation variable; all derivatives will be displayed as indexed functions`

(1)

 

 

tr := {t = tau, x = xi/k+v*tau^alpha/(k*alpha)+theta, u(x, t) = U(xi)*exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta)), u[1](x, t) = U(xi)*exp(-I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))}

{t = tau, x = xi/k+v*tau^alpha/(k*alpha)+theta, u(x, t) = U(xi)*exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta)), u[1](x, t) = U(xi)*exp(-I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))}

(2)

pde := I*(I*U(xi)*exp(I*(xi/k+v*tau^alpha/(k*alpha)-mu*tau+theta))*w-exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))*(diff(U(xi), xi))*v)+a*(diff(u(x, t), `$`(x, 2)))+b*U(xi)^2*u(x, t)+C[1](h[1]*(diff(u(x, t), `$`(x, 4)))+h[2]*(diff(u(x, t), x))^2*u[1](x, t)+h[3]*abs(diff(u(x, t), x))^2*u(x, t)+h[4]*U(xi)^2*(diff(u(x, t), `$`(x, 2)))+h[5]*u(x, t)^2*(diff(u[1](x, t), `$`(x, 2)))+h[6]*U(xi)^4*u(x, t))+I*C[2]*(h[7]*(diff(u(x, t), `$`(x, 4)))+h[8]*U(xi)^2*(diff(u(x, t), x))+h[9]*u(x, t)^2*(diff(u[1](x, t), x))) = 0

I*(I*U(xi)*exp(I*(xi/k+v*tau^alpha/(k*alpha)-mu*tau+theta))*w-exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))*(diff(U(xi), xi))*v)+a*(diff(diff(u(x, t), x), x))+b*U(xi)^2*u(x, t)+C[1](h[1]*(diff(diff(diff(diff(u(x, t), x), x), x), x))+h[2]*(diff(u(x, t), x))^2*u[1](x, t)+h[3]*abs(diff(u(x, t), x))^2*u(x, t)+h[4]*U(xi)^2*(diff(diff(u(x, t), x), x))+h[5]*u(x, t)^2*(diff(diff(u[1](x, t), x), x))+h[6]*U(xi)^4*u(x, t))+I*C[2]*(h[7]*(diff(diff(diff(diff(u(x, t), x), x), x), x))+h[8]*U(xi)^2*(diff(u(x, t), x))+h[9]*u(x, t)^2*(diff(u[1](x, t), x))) = 0

(3)

``

PDEtools:-dchange(tr, pde, [xi, tau, U, U(xi)])

I*(I*U(xi)*exp(I*(xi/k+v*tau^alpha/(k*alpha)-mu*tau+theta))*w-exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))*(diff(U(xi), xi))*v)+a*((2*I)*exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))*(diff(U(xi), xi))/k+exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))*(diff(diff(U(xi), xi), xi))-U(xi)*exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))/k^2)*k^2+b*U(xi)^3*exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))+C[1](h[1]*(-(4*I)*exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))*(diff(U(xi), xi))/k^3-6*exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))*(diff(diff(U(xi), xi), xi))/k^2+(4*I)*exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))*(diff(diff(diff(U(xi), xi), xi), xi))/k+exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))*(diff(diff(diff(diff(U(xi), xi), xi), xi), xi))+U(xi)*exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))/k^4)*k^4+h[2]*(exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))*(diff(U(xi), xi))+I*U(xi)*exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))/k)^2*k^2*U(xi)*exp(-I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))+h[3]*abs((exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))*(diff(U(xi), xi))+I*U(xi)*exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))/k)*k)^2*U(xi)*exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))+h[4]*U(xi)^2*((2*I)*exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))*(diff(U(xi), xi))/k+exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))*(diff(diff(U(xi), xi), xi))-U(xi)*exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))/k^2)*k^2+h[5]*U(xi)^2*(exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta)))^2*((diff(diff(U(xi), xi), xi))*exp(-I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))-(2*I)*(diff(U(xi), xi))*exp(-I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))/k-U(xi)*exp(-I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))/k^2)*k^2+h[6]*U(xi)^5*exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta)))+I*C[2]*(h[7]*(-(4*I)*exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))*(diff(U(xi), xi))/k^3-6*exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))*(diff(diff(U(xi), xi), xi))/k^2+(4*I)*exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))*(diff(diff(diff(U(xi), xi), xi), xi))/k+exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))*(diff(diff(diff(diff(U(xi), xi), xi), xi), xi))+U(xi)*exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))/k^4)*k^4+h[8]*U(xi)^2*(exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))*(diff(U(xi), xi))+I*U(xi)*exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))/k)*k+h[9]*U(xi)^2*(exp(I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta)))^2*((diff(U(xi), xi))*exp(-I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))-I*U(xi)*exp(-I*(xi/k+v*tau^alpha/(k*alpha)+mu*tau+theta))/k)*k) = 0

(4)
 

NULL


Download find_ODE.mw

I have a procedure that I am trying to run that would be an improvement/more sophisticated way of solving a problem that I have previously solved. When I try and run my procedure I am getting an error, and from what I gather with the error is that there are some values when inserted into my procedure that cannot be evaluated. Just for context it is a procedure that contains numerical solutions to a system of DEs and and contains inequalities. 

I would like to know is there an easy method to figure out what values are giving me this error? 

Or a follow up, is there something wrong with my procedure that is giving me this error? I have included some commentary in my workshet as well to hopefully make everything clear. 

Thanks. 

Proc_Error.mw

Hi,

I just got 2 questions regarding symbols and answers in MapleFlow (I might be doing something wrong here).

- In trying to input the integral int(F(x), x = -pi/5 .. pi/5) - this is just an example! - as a symbol,

int(F(x), x = -pi/5 .. pi/5)

MapleFlow doesn´t accept fractions or multiplications for bondaries values, unless I input "int(F(x), x = -pi/5 .. pi/5 =". Is there anyway to do it?

- is it possible to have the answer as 2*cos((3*Pi)/10), like in Maple, instead of 2.00*cos(300e-3*pi) ()symbolic expression) or 1.18 (numeric expression)?

Thanks in advance for your help.

I have some library program files written using Maple 11 (or before).

I can read and use the file programs and they have worked fine for many years.

But I would like to actually see the programming commands that are buried within with a view to modification, so, according to the Maple "Help" file, I tried the following command sequence:

***********************************************************

restart;
interface(echo=4,quiet=false);
interface(echo);
interface(quiet);
                            1, false

                               4

                             false

read "C:\\Program Files (x86)\\Maple 11\\test3.m";

**************************************************************************

But there is no echo.

So, how do I read (and modify) these files?

Thank you

How can I check if a name has been used/entered already but was not assigned to a value

The variable palette only lists assigned names.

I tried unames() but this lists all unassigned names. A 'user' option (which filters for user-assigned names) as in anames() does not seem to exist.

One of my failed attempts (in 1D-Math):

restart;
unames():
initial_unames := {%}:
new_name;
{unames()} minus initial_unames; # should ideally return a reduced set containing new_name;
has(%,new_name)

What else can be done? (I am probably overlooking something very simple.)

Я пытаюсь найти корни уравнения u = 0,6*sqrt(t)—1 + exp(-t). Команда fsolve не дает никаких результатов .

For a right triangle with two legs of the right triangle a and b, draw three circles with radius r and one ellipse as shown in the diagram (the major axis of the ellipse is parallel to the hypotenuse of the right triangle). Find a relationship between a, b, and r

I never really understood Intat. Help says 

"The intat command expresses an integral evaluated at a point; it is analogous to using the D command to express a derivative evaluated at a point."

But slope at a point is clear what it is and one can visualize it.

I do not understand what integral at single point means.

If one thinks of integration as area under the curve of the function, so what does area at single point mean? Should not integration (definite) always have lower and upper limits?

But my main question is not the above. I am sure there is a valid reason for Intat, otherwise it will not be in Maple.

But my quesiton is, in the context of solution to ode, can one replace result given using Intat by Int such that the lower limit starts from zero, and using same upper limit?

ie change Intat(...., a_ = something)  by Int( ... , a_ = 0 ... something) without changing the semantics or the correctness of the solution?

Because in  Intat, the lower limit is empty, and this always bothered me. At school the teacher says definite integration should have both lower and upper limits.

I tried it few places, and odetest verifies the solution of ode when using Intat or when using Int with lower limit zero:

ode:=diff(y(x), x) = B + C*f(a*x + b*y(x));

diff(y(x), x) = B+C*f(a*x+b*y(x))

sol_1:=Intat(1/(C*f(_a*b)*b+b*B+a),_a = (a*x+b*y(x))/b)*b-x+_C1 = 0;
sol_2:=Int(1/(C*f(_a*b)*b+b*B+a),_a = 0..(a*x+b*y(x))/b)*b-x+_C1 = 0;

Intat(1/(C*f(_a*b)*b+b*B+a), _a = (a*x+b*y(x))/b)*b-x+_C1 = 0

(Int(1/(C*f(_a*b)*b+b*B+a), _a = 0 .. (a*x+b*y(x))/b))*b-x+_C1 = 0

odetest(sol_1,ode);

0

odetest(sol_2,ode);

0

 

 

Download intat_vs_int.mw

Is there a case you know, where solution of ode which has Intat, when replaced by Int with lower limit 0, will no longer verifies the ode?  I am not able to find one so far. But may be there is.
 

Update

fyi, I found case where it makes difference. Not for odetest, but when using value. When using Int(...,tau=0..upper) vs   Intat(....,tau=upper)

value was able to find the value when using Intat, but not when using Int (for this example). 

So I think I will just stick to Intat even though both verified the ode as valid solution as it is better to be able to find value for integral if possible. I knew there must be good reason why Intat was invented.

 

ode:=diff(y(x),x)= sin(x-y(x));

diff(y(x), x) = sin(x-y(x))

sol_1:=Int(1/(1 - sin(tau)), tau = 0..x - y(x)) = x + _C1;
sol_2:=Intat(1/(1 - sin(tau)), tau = x - y(x)) = x + _C1;

Int(1/(1-sin(tau)), tau = 0 .. x-y(x)) = x+_C1

Intat(1/(1-sin(tau)), tau = x-y(x)) = x+_C1

odetest(sol_1,ode);

0

odetest(sol_2,ode);

0

value(sol_1);

int(1/(1-sin(tau)), tau = 0 .. x-y(x)) = x+c__1

value(sol_2);

-2/(tan((1/2)*x-(1/2)*y(x))-1) = x+c__1

 


 

Download int_vs_intat_v2.mw

 

@Rouben Rostamian  

Dear Sir Professor Rostamian my name is Viorel Popescu from the Polytechnic University of Bucharest if you remember in the summer of 2019 you helped me to solve the equation: rH''(r)+H'(r)+(rk^2-r^2*b^2/R^2)H(r)=0 where k, b, and R are real constants positive number, with condition H(R)=0 and H'(1/R)=R. I appreciate it very much, please I'm in a similarly embarrassing situation to beg you for an answer. I want to find the equation of audion and complete the experiment http://www.michaelvio.byethost8.com/Audion.pdf

My account in Maple Primes is the same michaelvio (35) as the email michaelvio@yahoo.com and also @gmail.com it's an experiment that I want to make for my PhD. Practically I suppose that the energy can be approx. as a series of power of frequency t from I selected severaral terms Ea := 0.00762014687*t + a*t^2 + b*t^3 + c*t^4 + d*t^5 and I guess that satisfies an equation as in the document. The case of photons is beyond my possibility, but a little help from a distinguished Professor as you should cheer me up Audion1.mw

Audion.docx

Please help! 

I need to automatically solve this system, that is, both equations simultaneously for different intervals of T that I define. For example, from T=0.1 to 4, then from 4.5 to 5 and from 7 to 8 with distinct subdivisions between them, the first varying by 0.1, the second by 0.05 and the third by 0.25. Furthermore, the index that accompanies m and G is only illustrative. Would anyone know how to help me?

Thank you for your help!

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