How to plot the (x,y,z) points on 3d plot like first point is (19.8,12,1.62) like that in order

and connect all the ponts above as a line

Kind help your help will be acknowleged

IN FIBONACCI RABBIT PROBLEM, SUPPOSE ONLY 3/4 (PROPORTION) OF FEMALES BECOME PREGNANT THEN HOW TO DETERMINE THE ANNUAL RATE OF INCREASE IN RABBITS?

restart:
with(plots):
kernelopts(version);
 

How do I setup solve to find only the real and imaginary part of Zeta to be both positive and the real part of Zeta to be the smallest positive value? For example: real part = 1.348412872 and imaginary part = 0.04230725448.

Acceleration ratio (db) and phase (deg). Convert ratio to linear and phase to radian. 
dat := <14.52407334|-162.1310124>:
A := 10^(dat[1]/20.):
phi := dat[2]*Pi/180.:
R:= 0.3036:
Characteristic equation Eqn 5
f := (Zeta,A,phi) -> cos(Zeta) - R * Zeta * sin(Zeta) - exp(-phi * I)/A:

soln := [solve(f(Zeta,A,phi), Zeta)]:
 

Tests.mw

Hi MaplePrimes team,

I am aware that Maple is not designed for CAD and I am not here for you to help me build any advanced system like the example below.

Figure 1: Trimmed surface of fuselage to create airplane windows.

Here I my problem. Even using the adaptmesh option, creating serious trimmed surfaces requires large numbers of grids like for example [3000, 3000] and even beyond, which makes a mesh unnecessary on most of the surface and especially where we don’t need high resolution.

While if we limit the display volume with the Maple’s view command (view = [x₁..x₂, y₁..y₂, z₁..z₂]), the edges are always smooth regardless of the coordinates system or the type of mesh of the surface to be treated.

And so, the reason I’m writing this message is to ask you why creating trimmed surface with plot3d surface wouldn’t use Maple view’s algorithm to get smooth edges?

Future Alternative Solution:

I also take this opportunity to ask you when would it be possible to update the Maple’s grid option such that the point space will variable in term of surface variable. For example:

plot3d(<x(s,t), y(s,t), z(s,t)>, s = s₁..s₂, t = t₁..t₂, grid = [m(s,t), n(s,t)] )

I am sure that this solution, which for me does not seem to be difficult to implement, could be an alternative to enrich the mesh where we really need it and therefore a better control. What do you think?

Thank you in advance.

Best.

Guy.

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 

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