MaplePrimes Questions

Hello everybody – It looks like there is a bug in the numerical evaluation of (multiple) Zeta function. Take for instance, Zeta(3, 0.5) which is approximately 8.4144. Maple gives approximaterly -96.0033. Is there a bug somewhere?

Thank you

Textbook gives this nice short implicit solution for this ode

As

But Maple does not give an implicit solution

restart;
ode:=diff(y(x),x) = 2*(2*y(x)-x)/(x+y(x));
ic:=y(0)=2;
sol:=dsolve([ode,ic],'implicit')

Removing implicit gives very complicated solution as it tries to solve for y(x).

book_sol:=(x-y(x))^2-1/2*(y(x)-2*x)^3=0;
odetest(book_sol,[ode,ic])

Any suggestion or a trick or a different approach to make Maple generate the same solution given in the textbook?

ps. manually, it is possible to obtain the same solution as book as follows

restart;
ode:=diff(y(x),x) = 2*(2*y(x)-x)/(x+y(x));
ic:=y(0)=2;
sol:=dsolve(ode,'implicit');
sol:=simplify(exp(lhs(sol))=exp(rhs(sol)));
the_constant:=solve(sol,_C1);
the_constant:=eval(the_constant,[y(x)=2,x=0]);
sol:=eval(sol,_C1=the_constant);
odetest((lhs-rhs)(sol)=0,[ode,ic])

 

edit june 14,2021

I found a direct way to get an implicit solution which is close enough to book solution. It is by using Lie symmetry method

restart;
ode:=diff(y(x),x) = 2*(2*y(x)-x)/(x+y(x));
ic:=y(0)=2;
sol:=dsolve([ode,ic], y(x),'implicit','Lie');

Another variant which gives book solution but requires one extra step

restart;
ode:=diff(y(x),x) = 2*(2*y(x)-x)/(x+y(x));
ic:=y(0)=2;
sol:=dsolve([ode,ic], y(x),'implicit','Lie',way=all,fat);
simplify(exp( (lhs-rhs)(sol)))=1

ps. corrected above now, thanks to comment below by Carl.

Maple 2021.1

 

 

 

Hi, I need to count elements in an indexed set. The below commands gives 3, as it doesn't take the powers into consideration (u[1,2,2]^3). Desired output is 5, meaning to count 1 three times.
       

with(ListTools):
>C:=x*u[]*u[1,2,2]^3*u[1,1,2];
>indx_set:=(`[]`@ op)~(indets(C, indexed));
indx_set := {[], [1, 1, 2], [1, 2, 2]}
>indx_set:=Flatten(convert(indx_set,list));
 indx_set := [1, 1, 2, 1, 2, 2]
>indx1:=[SearchAll(1,indx_set)];
[1, 2, 4]
>num:=nops(indx1);
3

Thanks

Hello there, 

Would you allow me to ask this one question?

What is the difference between 'map' and '~(element-wise operation)'?

The following worksheet snipped shows an example:



Maple Worksheet - Error

Failed to load the worksheet /maplenet/convert/Q20210612.mw .
 

I thought that both operations were the same, but seemingly they are not. 

Best Regards, 

In Kwon Park 

Download Q20210612.mw

In worksheet mode, when typing only the first 3 letters of the command piecewise, the command completion popup shows and I can hit TAB to have it automatically complete which is nice since I always makes typos tying this command

  >pie

When when I wanted to get help on the same command, I have to type 5 letters now before the popup shows to help me complete the command

 >?piece

Why is that? I would have expected the same 3 letters to be enough, as I assume Maple is smart enough to notice this.

Here is a movie to illustrate. This is Maple 2021.1 on windows 10

   

Hi!

Assume we have the following mappings:

 

f[1] := proc (x) options operator, arrow; piecewise(0 <= t and t <= 1/2, (1/2)*x(2*t), 1/2 < t and t <= 1, (1/2)*x(2*t-1)+1/2) end proc

 

f[2] := proc (x) options operator, arrow; t*x(t) end proc

 

 Note that the argument are mappings (i.e, x(t) is a continuous function defined on [0,1]). Given an array with values in {1,2}, for instance [1,1,2],and a function h (say, h:=t->t) How can we define the composition mapping f[1]f[1]f[2](h)?

In general, I want to plot the composition mapping f[i1]f[i2]...f[iN](h), i1,i2,...,iN being 1 or 2 and h a given continuous function defined on [0,1]

 

Many thanks in advance for your comments.

 

 

I have a list of objects. Each object contains many fields, one of these fields happened to be a solution to an ode.

At the end of solving the ODE, I'd like to remove all objects which contain the same solution. 

I can not do the normal trick in Maple to remove duplicates from a list, which is to convert the list to a set and let Maple automatically remove duplicates because in this case each entry in the list is an object of many fields and hence each object is always different. 

I only want to remove the objects from the list which has the same specific field.

Currently what I do is the following:   Do a first scan, and loop over all entries in the list to find the duplicates. Each time I find a duplicate solution, mark a field in the object as it is being duplicate.

Then do another scan at the end, to build a new list, which only adds those objects not marked as duplicates.

It works, but was wondering if Maple or someone might have better suggestion or more elegent way to do this.

In this MWE I am using a list and adding objects to it making it grow dynamically. I know it would be better to use Array, but for now I'd like to stick to a list, since the number of solutions is normally small. May be in the future I will change the list to Array to avoid building a list dynamically.

This example code builds a list of 5 objects

restart;
ode_solution_type:=module()
    option object;
    export sol:=NULL; #only ONE solution here.  
    export stuff;
    export mark_to_delete:=false;
end module:

SOL::list:=[]:  #where to keep the objects

for n from 1 to 5 do
    tmp:=Object(ode_solution_type);
    
    if n=3 then
        tmp:-sol:=y(x)=1;
    else
       tmp:-sol:=y(x)=0;
    fi;
    tmp:-stuff:=rand();
    SOL:=[op(SOL),tmp];
od:

In the above, I made 4 objects to have same solution which is y(x)=0 and one different. The goal is to remove all objects from the list which has duplicate solutions and keep only one copy.

for n from 1 to numelems(SOL) do
    print("sol ", n , "  is ", SOL[n]:-sol);  
od:

And this is how currently I remove the duplicates

for n from 1 to numelems(SOL) do
    for m from n+1 to numelems(SOL) do
        if is(SOL[n]:-sol=SOL[m]:-sol) and not SOL[m]:-mark_to_delete then
           print("found duplicate at ",m);
           SOL[m]:-mark_to_delete:=true;
        fi;
    od;    
od:

#now make new pass to keep the non- duplicates
new_SOL::list:=[]:
for n from 1 to numelems(SOL) do
    if not SOL[n]:-mark_to_delete then
       new_SOL:=[op(new_SOL),SOL[n]];
    fi;    
od:

checking:

for n from 1 to numelems(new_SOL) do
    print("sol ", n , "  is ", new_SOL[n]:-sol);  
od:

Does there exist a better option in Maple (while still using a list?) to remove objects in list which have the same specific field?

Could you suggest a better method?

Please how draw this curve 

Nothing special, just out of boredom.
Learned about the existence of multifocal ellipses. I stopped at the option with 5 focuses and made the corresponding equation, when the sum of the distances from each point of the ellipse to these five points is constant. If we try to get rid of the radicals, we get an equation whose graph will contain branches that do not meet the condition of the constancy of the sum of the distances from points on the curve to our 5 fixed points. In my case (as usual) Draghilev's method is used to show that each point on the curve is at a fixed distance from 5 stationary points.
Is there a mathematical way to show the same thing? I think not, maybe some other numerical method. So then: is there an easier way? It is clear that the number of focuses can be larger.
FOCI_5_EXAMPLE.mw


Hi all again,

Tried this.

some_proper_divisor_examples_8.pdf

some_proper_divisor_examples_8.mw

 

Anyone know a theorem related to this?

My guess is (p1 -1) * (p2 -1) * ... * (pb -1)  .

For p1, p2, ... pb all prime numbers.

Regards,

Matt

sometimes I get a solution to an ODE, where when I plugin the initial conditions to solve for the constant of integration, the solution by inspection can be seen to be infinity.

The constant of integration is allowed to be infinity.

But now I get no solution from solve and so my program rejects this solution and says the ode can't be solved because it can't find value for the constant of integration using this initial conditions.

I am not sure how to handle this, since Maple solve will not return an infinity as possible solution for the constant of integration. I tried different assumptions.  

This has to work for general, and not for this specific example I will show now.

Here is an ODE with IC

ode:=diff(y(x),x)=y(x)^3*sin(x);
ic:=y(0)=0;
sol:=[dsolve(ode)];

The general solution is easy to find since this is just separable. The problem is when I plugin the initial conditions. (again, this is done in a program, without looking at the screen). let consider the first solution above for now.

eq:=simplify(subs( [y(x)=0,x=0],sol[1]));

And here is the problem, when next I try to solve for the constant, Maple says no solution.

solve(eq,_C1,allsolutions=true) assuming _C1::real

No solution returned.

If Maple could return _C1=infinity, then I would plugin _C1=infinity and obtain the correct particular solution which is 

simplify(subs(_C1=infinity,sol[1])) assuming real

Which is what the smart dsolve command does

ode:=diff(y(x),x)=y(x)^3*sin(x);
ic:=y(0)=0;
sol:=dsolve([ode,ic])

I do not know how Maple found the above and what algorithm it used.  

But my not too smart dsolve program gets stuck on such problems, because it can't solve for the IC.

Is there another method or option or assumptions to use, to solve such an equation and request infinity be returned as a solution?  I tried isolate(eq,_C1) but that did not work.  

One possible option to try as a last resource is to do this

limit(eq,_C1=infinity)

And this gives 0=0

which means _C1=infinity is valid solution.  And I could do this as final attempt before giving up. Is this what Maple did? But how does one know to do this in general? Unless one tries and find out? i.e I could try limit(eq,_C1=infinity) and limit(eq,_C1=-infinity) and see if any one of these gives true (after doing  evalb or is  on the output)

Any suggestions or a better/more robust method to use for such cases? I need to use the method for any equation where infinity can be a solution for the constant of integration. It will always be one equation and one variable, and can assume all are real for now if it makes it easier.

 

 

I was wondering why dsolve do not show this one solution I obtained different way, for this ode.

Maple gives 6 solutions to the ODE. But 5 of them are signular (have no constant of integration in them). I am looking at the last solution it gives, the one with constant of integration. 

But when I solve this ODE, after simplifying it, I get different general solution than the last one Maple shows. I thought they might be equivalent, but I do not see how they could be. 

So my question is why Maple did not show the simpler solution also?  Here is the code

ode_orginal:=1/3*(-2*x^(5/2)+3*y(x)^(5/3))/x^(3/2)*diff(y(x),x)/y(x)^(5/3)+1/2*(2*x^(5/2)-3*y(x)^(5/3))/x^(5/2)/y(x)^(2/3) = 0;

This looks complicated and non-linear, but when I solve for y'(x) it gives much simpler and linear ODE (Have to convert the ODE to D before solving for diff(y(x),x), since Maple can complain otherwise).

ode:=convert(ode_orginal,D);
ode:=diff(y(x),x)=solve(ode,D(y)(x),allsolutions=true);

And solving the above gives the much simpler solution

sol:=dsolve(ode,y(x));
odetest(sol,ode);

Now when using dsolve on the original complicated lookin ODE it gives

maple_sol:=[dsolve(ode_orginal)];

if there is any hope that one of the above 6 solutions will be the same as the simpler solution, it has to be the last one, with the _C1 in it, since all the others do not have _C1 (singular solutions).

But solving for y(x) from the last one does not give the simpler solution.

So it is a new general solution for the original ode and it is correct, since odetest gives zero.

I think in the process of solving for y'(x), some solutions got lost. Even though I asked for allsolutions there?.

But my main question is: Should dsolve have also have given the simpler solution? Since that one also satisfies the original nonlinear complicated ode

odetest(sol,ode_orginal);

gives 0.

Maple 2021.1

 

ExpODE4 := Y(1)*(sum(lambda^k*t^(k*d)/factorial(k*d), k = 0 .. infinity))

Statistics[NonlinearFit](ExpODE4, X, Y, t)

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