Please how draw this curve 

https://www.maplesoft.com/applications/view.aspx?SID=3589&view=html

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)

I was setting up a worksheet for my vactor calculus class and tried to show how the DifferentialDiff command from the VectorCalculus package was limited.  I was expecting to have it give undefined or something similar for all directions with the function I am using.  Unfortunately, I got 0 when the directional derivative is truly undefined and I got undefined when the directional derivative is 0.

This is the code I am using.  Unfortunately, this site will not take the worksheet or plot the function for me.
 

with(VectorCalculus):
s := min(abs(x),abs(y));
plot3d(s,x=-1..1,y=-1..1);
DirectionalDiff(min(abs(x),abs(y)),<-1,1>,[x,y],point=<0,0>);

Hi, 
I just realized inadvertently that it was possible to edit a reply even if you were not the author (mmcdara is my personal account and sand15 my professional one and I modified under the former a reply I'd made with the later).

To be sure of that I've modified the last @acer's answer here 232397-Plotting-Multiple-Vectors-On-One-3D--Graph

You could show us what you've got so far

is now

You could show us what you have got so far

(I hope @acer will excuse me)


I think this is a weakness of Mapleprimes for no one should be allowed to change what others have written.

I solved this ode with IC, and obtained a solution.  Maple odetest says my solution satisfies the ODE but not the initial conditions. But I do not see why. When I plugin manually the initial conditions into the solution, I get true.

So I am not sure what is going on.

restart;
ode:=x*diff(y(x),x)*y(x) = (x+1)*(y(x)+1);
ic:=y(1)=1;
mysol:=y(x)-ln(y(x)+1)=-ln(2)+x+ln(x);
my_sol_at_IC:=subs([y(x)=1,x=1],mysol);

#now check manually
evalb(my_sol_at_IC);
is(my_sol_at_IC);

Using odetest gives

odetest(mysol,ode);
odetest(mysol,[ode,ic]);

What Am I doing wrong? 

Maple 2021.1

Edit June 12, 2021

This is another example where odetest gives different result depending on how the solution is written !

restart;
ode:=diff(y(x),x)=2/3*(y(x)-1)^(1/2):
mysol_1:=y(x) = 1/9*x^2 - 2/9*x + 10/9:
mysol_2:=y(x) - (1/9*x^2 - 2/9*x + 10/9) = 0 :
odetest(mysol_1,ode);
odetest(mysol_2,ode);

Hi,

I have created some vectors v[1] and v[2] and a vector field..I am new at Maple and learning so I can plot each vector on a 3D plot but cannot

figure out how to plot both vectors and the vector field on the SAME graph...

Any suggestions?

Thanks

Frank

Suppose that we have something like

plots:-animate(F,[theta],theta=0..2*Pi,background=plot([cos(t)-2,sin(t),t=0..2*Pi]),
scaling=constrained,axes=none);

I want to change the range of the parameter theta to something like theta=L[i],i=1..20.
Here, L is a list of numbers. Can I?
Thank you.

mapleatha

I can't figure what is going on here. I have an expression, where when I run the program, I see inside the debugger that the result of evalb(simplify(tmp_1)) where tmp_1 is the expression, gives false

When I copy the same tmp_1 expression to new worksheet and do evalb(simplify(tmp_1)) it gives true.

I added print statements in the code to print tmp_1 and print result of simplify(tmp_1) and for some reason, it gives false when I run the program, but it gives true when I do the same thing in a work sheet, by copying tmp_1 and trying it there.

Never seen anything like this, as expression is all numbers. Here is the print out when I run the program

Here is the same thing in worksheet

restart;
tmp_1 :=-(2*ln(-2))/7 - ln(-1)/7 = ln(1) - (2*ln(-2))/7 - ln(-1)/7;
simplify(tmp_1);
evalb(simplify(tmp_1));

                      true

and if you think I made mistake somewhere, here is also screen shot from the debugger window itself

again, same thing gives true in separate worksheet, copying same exact expression from above

There seem to be something loaded when I run my program that causes this difference, but I have no idea now what it is.  This has nothing to do with the debugger loaded btw, I am just using the debugger to show the details. When I run the program, without the debugger, I get false for the above, and I was trying to find why.

Any suggestions what to try and what to check for and what could cause this? Should it not give true when I run the program also? Could it be some Digits setting changed when the program run? I do not change any system defaults of any sort when running the program. 

Maple 2020.1

edit june 12, 2021

Here is another manifestation of the same problem I just found. There is something seriously wrong for it to behave this way.

This is an expression called tmp_1 in a local variable in a proc inside a module. There are not even any local symbols in it. When I run the program, all the attempts to show it is true fail. Some give FAIL and some give false when the result should be true

When I simply copy the expression from the debugger window to an open worksheet, I get true

Here is the screen shot

Only when I did is(expand(tmp_1)) did it give true

But when I copy the expression to the worksheet, it gives true using all the other attempts:

restart;
tmp_1:=1 = 5/4*exp(0)+1/4*piecewise(0 <= 1,-1,1 < 0,exp(-2)):
is(tmp_1);
is(simplify(tmp_1));
is((rhs-lhs)(tmp_1)=0);
is(simplify((rhs-lhs)(tmp_1)=0));
evalb(simplify((rhs-lhs)(tmp_1)=0));
is(expand(tmp_1))

I can't make a MWE so far, since these problems only show up with I run my large program. I think Maple internal memory get messed up or something else is loaded that causes these stranges problems.

For now, I added expand() to the things I should try. My current code now looks like

if is(tmp_1) or is(simplify(tmp_1)) or is((rhs-lhs)(tmp_1)=0)
               or is(simplify((rhs-lhs)(tmp_1)=0)) or evalb(simplify((rhs-lhs)(tmp_1)=0))
               or is(expand(tmp_1)) then    
   ....

Maple 2021.1

Given finite points on a plane, how to calculate the Newton polygon, i.e. the lower bound of its convex hull?

The convex hull can be obtained by ConvexHull command from ComputationalGeometry package, but I don't know how to get its lower bound easily. Besides that, theoretically the Newton polygon does not depend on the full structure of convex hull, so maybe there is a more effective way to do that?

P.S. There is a method DEplot_polygon in DETools that can "generate the plot of the Newton polygon of a linear differential operator at a point“, but I not sure how it can be used to calculate the Newton polygon for general cases.