MaplePrimes Questions

I have downloaded Maple and install the trial version , i entered the Purchase code  and every thing was done, I did not get any errors , however, while i am trying to open Maple, when I open the software , I see this window which i have uploaded below , and when i click on Active i eneter my purchase code again , and it says it has acitivated and then when i click ok , it exit , what is the issue ?

can you help me ?

Hi.

1- I want to solve the following equation:
eq:=diff(theta(t),t)=-(A[1]*Dirac(t-T[1])+A[2]*Dirac(t-T[2]));

with the boundary conditions:

bcon := theta(0)=-v[1],theta(1)=v[3]:

but the code

dsolve({eq,bcon},theta(t)) assuming T[1]<T[2];
and also

dsolve({eq,bcon},theta(t), method=laplace) assuming T[1]<T[2];

do not work. 

The solution must be:

theta(t)=piecewise(0 <= t and t < T[1], -v[1], T[1] < t and t < T[2], 0, T[2] < t and t <= 1, v[3]);

2-Furthemore, solving the ode:

restart;
eq2 := diff(phi(t),t)=piecewise(0 <= t and t < T[1], -v[1], T[1] < t and t < T[2], 0, T[2] < t and t <= 1, v[3]);

with the similar bc:

bcon2 := phi(0)=-v[2],phi(1)=v[4];

is another problem which I am trying to solve it. but again the code

dsolve({eq2,bcon2},phi(t)) assuming T[1]<T[2];

does not work!! while the solution must be:

phi(t)=piecewise(0 <= t and t < T[1], v[1]*(t-T[1]), T[1] < t and t < T[2], 0, T[2] < t and t <= 1, v[3]*(t-T[2]));

Could you please help me to solve these two problems??

I really appreciate any help you can provide.

 

Write a recursive procedure proc2 with option remember, to compute the function 
                             "f[n]"

 recursively defined by 
                           "f[0] = 0"

,   
                        "f[1] = -7*x^2"

,  
       "f[n] = -3*f[n - 2]^2 + (n + 21)*cos(2*f[n - 1])"

  for n>1.
Evaluate the first derivative of 
f[18];
 at  "x = 2*Pi"

 numerically by the two procedures, and compare running times.

Okay, so for this problem, here is what I have:

>proc2:=proc(n)
option remember;
if n=0 then 0
elif n=1 then -7*x^2;
elif n>1 then -3*f^2[n-2]+(n+21)*cos(2*f[n-1]);
end if;
end proc:

>proc2(18)

and I get this as my result: -3*f^2[16] + 39*cos(2*f[17])

which is true but what can I do to let the proc2 solve it all the way til they get n=0 or 1? 

and how can I plug in 2*pi in to x?

Please help! Thank you so much for your time!

 Let a matrix be like below

[[0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0.]
 [1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0.]
 [0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0.]
 [0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0.]
 [0. 0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 1. 1. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 0. 1. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 1. 1. 0. 0. 0.
  0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 0. 1. 0. 0.
  0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 1. 0.
  0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 1.
  0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0.
  1. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1.
  0. 1. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  1. 0. 1. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 1. 0. 1. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 1. 0. 1.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 1. 0.]]
Adjacency Matrix of Mol4 that is 12,14-Nonacosanedione is
[[0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0. 0. 0.]
 [1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0. 0. 0.]
 [0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 1. 1. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 0. 1. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 1. 1. 0. 0. 0.
  0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 0. 1. 0. 0.
  0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 1. 0.
  0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 1.
  0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0.
  1. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1.
  0. 1. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  1. 0. 1. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 1. 0. 1. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 1. 0. 1. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 1. 0. 1. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 1. 0. 1.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
  0. 0. 0. 0. 0. 1. 0.]]

 

in some places we may have ". " or we have a "\n" or newline after "." we need to convert it to a regular adjacency matrix which can be recogonized by the graph function of maple

Is their anyway we can get the smiles file of given IUPAC name 

Smiles with explicit hydrogen

And draw its chemical graph 

Hi seniors, I have solved some questions of derivative, intgration and limits by using built in command. I am interested to you show full step of each questions. Could any one plase help  me in this regards

Help_derivative_and_limit.mw

This is the problem: My object is getting large with many private methods.

In non-OOP setup, I could make different modules A,B,C and put relevent methods inside each module.

Now I find I can not do this inside the object. All methods have to be flat and at same level. So I lose the benefit of using different name space for different methods.

It will be easier to explain with a simple example. Lets say I have this now

restart;

A:=module() 
    option object; 
    local name::string:="";  

    export ModuleCopy::static:= proc( self, proto, name::string, $)          
         self:-name := name;
    end proc;

    export process_1::static:=proc(_self,$)
       _self:-process_2();
    end proc;

    local process_2::static:=proc(_self,$)
       print("in process_2 ",_self:-name);
    end proc;

end module; 

The above works by having all methods at same level. I can now do 

o:=Object(A,"me");
o:-process_1()

                  "in process_2 ", "me"

But now as the number of private methods increases I want to put a name space around collection of these for better orgnization, but keep all methods logically as part of the object as before.

I change the above to be

restart;

A:=module() 
    option object; 
    local name::string:="";  

    export ModuleCopy::static:= proc( self, proto, name::string, $)          
         self:-name := name;
    end proc;

    export process_1::static:=proc(_self,$)
       o:-B:-process_2();
    end proc;
    
    #method process_2 is now inside a module. 
    local B::static:=module()
       export process_2::static:=proc(_self,$)
           print("in B:-process_2 ",_self:-name);
       end proc;
    end module;

end module; 

Now

o:=Object(A,"me");
o:-process_1()

Gives an error

Error, (in unknown) invalid input: process_2 uses a 1st argument, _self, which is missing

I tried many different combinations, but nothing works.

So right now I have all methods flat. All at same level. But I do not like this setup. Since I lose the nice modular orgnization I had with using different modules with different methods inside different modules.  

Is there a way to have a module inside an object and call its methods from inside the object as if these method were part of the object private method with only differece is adding the module name in between?

So instead of _self:-foo()  I just change it to  _self:-B:-foo() and have both behave the same way?

In C++, one can use what is called a friend class to do this. 

Please I will like to know if there exists any command to isolate the known regular mathematical functions from an expression. For example, given the equation 

I need a means of isolating (ex & sin(v)) from the above expression

Thanks

How to get a plot where a unit cube [0,1]X[0,1]X[0,1] in R^3 is intersected by a plane $x+z=1$. Please give me a code

 

where p, q, r, a are positive constants and phi(0) be a initial condition

I'm not sure if this can even be done, but if it's possible, how do I do it:

I have a python script called "Blue.py", the path should simply be "Blue.py", since my worksheet and the python script are saved in the same place.

I want to somehow import two arrays and an equation from maple to said python script, use them in a procedure there, and then import the output of said procedure, a 2d array, backwards to maple.

I don't really care what format the equation and the arrays take in python (the current version assumes types "np.array" for the arrays and type procedure for the equation, but I can always adjust it), but the type of the output of Blue.py should be of the form 2d Array (for example [[0,1],[8,-2],[3,3]] ).

Is it possible to shuffle input between maple worksheets and python scripts like that, and if so, how? 

The uploaded worksheet is a failed attempt at this proof.

Please either refer me to or show me a valid proof.

  *** I have tried to upload my worksheet using the green arrow but the insert link operation of the upload failed ***

cc

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Error occurred during PDF generation. Please refresh the page and try again

int(int(x^(alpha + 1)*y^(beta + 1)*exp(-(x^2 + y^2)/(4*Pi))/sqrt(x^2 + y^2), x = -R .. R), y = -R..R) assuming x>0, y>0, R>0;

I am trying to evaluate the above symbolic integral but Maple returns the same input in Symbolic form without evaluating it at all. Is there something wrong with my approach?

Hi there!

I'm attempting to develop a custom library in MapleSim (a Modelica solver engine) that can use convolution integrals to model the hydrodynamic behaviour of floating bodies. The convolution integral is mathematically represented as follows:

a general form of a convolution integral, spanning from negative to positive infinity in time

And the equation I'm particularly interested in solving is as follows:

Cummins equation to determine the motion of a floating body in waves

I'm trying to solve this convolution integral in Modelica using a model that can be imported to MapleSim. So far, I've had no luck in implementing this convolution for continuous functions symbolically. I have used a numerical approach on 2 arrays using the following approach in Modelica:

// Modelica function
function convIntegral
  input Real simTime;  // Simulation time
  input Real f[:];     // Kernel function array
  input Real g[:];     // Second function array

  output Real h[(2*simTime) - 1];    // Output the convolution integral in the form of an array
  
  // Define the algorithm to numerically compute the convolution integral of 2 arrays
  algorithm
    // Initialize the output array with zeroes
    for i in 1:((2*simTime) - 1) loop
      h[i] := 0;
    end for;
    
    // Iterate over the simulation time
    // Recursively increment the convolution array with the pointwise product of the 2 functions
    for i in 1:simTime loop
      for j in 1:simTime loop
        h[i+j-1] := (f[i] * g[j]) + h[i+j-1];
      end for;
    end for;
end convIntegral;
// End of function to compute the convolution integral

This works perfectly for discrete samples and I have verified it with output from Matlab's inbuilt function

conv(A,B)  % For 2 arrays A and B

However, I would like to implement this on 2 continuous functions and this numerical approach does not work since MapleSim does not support conversion between discrete and continuous signals.

I understand that convolution is essentially an operation between two functions, where we time-flip one of the functions (kernel) and then slide it across the other function while measuring the bounded area and outputting that as the result of the convolution. I include this image from Wikipedia that sums up convolution: (Not including links as they mark questions as spam)

I've tried implementing this in Modelica using the following code:

// Model to perform convolution on 2 continuous functions
model ConvolutionalIntegral

  extends Modelica.Blocks.Icons.Block;

  // Define model variables
  Real kernelFunction = (e ^ (-0.5 * time)) * cos(0.5 * time);  // I've taken an example of a function I might use
  Real kernelFunctionFlipped = (e ^ (-0.5 * (T_sim - time))) * cos(0.5 * (T_sim - time));  // I've flipped the kernel function about the vertical axis by converting the (time) variable to (T_sim - time) where (T_sim) is a variable storing the simulation duration
  Real secondFunction;  // The other function for the convolution
  Real convolutionIntegralOutput;  // Function to store the output
  
  equation
    // Convolution implementation
    der(convolutionIntegralOutput) = kernelFunctionFlipped * secondFunction;

    // Final equation to solve
    der(secondFunction) + convolutionIntegralOutput = 0;
    // An example of a differential equation I'd like to solve involving the convolution integral
    
end ConvolutionIntegral;

I had hoped that this would yield the output of the convolution since I'm essentially multiplying the time-flipped kernel and the other function and then integrating them over time. However, the output does not provide the expected result and it appears that Modelica interprets my code to mean that I'm integrating the pointwise product of these 2 functions over time instead of sliding the kernel over the other function.

I'd appreciate it if you could take a look at my code and my approach to solving the convolution integral symbolically, and point out where I'm making a mistake and what a possible fix might be.

Thank you!

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