Hello everyone, 

In Maple8, I tried to plot this logistic map and an error occured (Error, (in Bifurcation) `plots` does not evaluate to a module).

What is wrong into this code?

Thank you

restart: with(plots):Warning, the name changecoords has been redefined

> Bifurcation := proc(initialpoint,xexpr,ra,rb,acc)
> local p1,hr,A,L1,i,j,phi:
> global r,L2:
> hr := unapply(xexpr,x);
> A := Vector(600):
> L1 := Vector(acc*500):
> for j from 1 to acc+1 do
> r := (ra + (j-1)*(rb-ra)/acc):
> A[1] := hr(initialpoint):
> for i from 2 to 500 do
> A[i] := evalf(hr(A[i-1])):
> end do:
> for i from 1 to 400 do
> L1[i+400*(j-1)] := [r,A[i+100]]:
> end do:
> end do:
> L2 := {seq(L1[i], i = 1..acc*400)}:
> p1 := plots:-pointplot(L2, 'symbol' = solidcircle, 'symbolsize' = 8, 'color' = blue):
> unassign('r'):
> return(p1):
> end proc:
> P1:= Bifurcation(1/2,r*x*(1-x),2.5,4,250):
>
Error, (in Bifurcation) `plots` does not evaluate to a module

Hello everyone!

Do you have any idea to solve and plot a 2-time logistic map:

x(t+ 1_\alpha)= r*x(t)*(1-x(t))  ,t=(t^1,t^2)  ?

Thank you

I want to reduce all solution of the equation sin(x)^2=1/4

restart:
sol:=solve(sin(x)^2=1/4, x, AllSolutions);

and

restart:
k:=combine((sin(x))^2);
sol:=solve(k=1/4, x, AllSolutions = true, explicit);
simplify(sol);

How can I reduce solution sol := -1/3*Pi*_B3+1/6*Pi+Pi*_Z3 ?

How can I get x= pi/6+k*pi and x= -pi/6+k*pi?

In Maple 11 we have:

> A := <a,b,c>:
> a := 1:  b := 2: c := 3:
> convert(A, list);
                                   [1, 2, 3]

In Maple 2015 we have:

> A := <a,b,c>:
> a := 1:  b := 2: c := 3:
> convert(A, list);
                                   [a, b, c]

Is that change really intended?

Hi all,

restart;#part1
epsilon:=5:Delta1:=2:Delta2:=-4: N1:=1000:
dsys :={diff(x(t),t)=-I*Delta1*x(t)+y(t)+epsilon, diff(y(t),t)=-I*Delta2*y(t)+x(t)*z(t), diff(z(t),t)=-2*(conjugate(x(t))*y(t)+conjugate(y(t))*x(t))};

res:=dsolve(dsys union {x(0)=2*I,y(0)=0,z(0)=1},numeric,output=listprocedure);

P1:=plots:-odeplot(res,[[t,Im(y(t))],[t,Re(x(t))]],0..2):

/ d
{ --- x(t) = -2 I x(t) + y(t) + 5,
\ dt

d
--- y(t) = 4 I y(t) + x(t) z(t),
dt

d ____ ____ \
--- z(t) = -2 x(t) y(t) - 2 y(t) x(t) }
dt /
tit:=sprintf("D1=%g,D2=%g",Delta1,Delta2);
"D1=2,D2=-4"
plots[odeplot](res,[[t,Im(y(t))]],0..200,axes=boxed,titlefont=[SYMBOL,14],font=[1,1,18],color=black,linestyle=1,tickmarks=[3, 4],font=[1,1,14],thickness=2,titlefont=[SYMBOL,12]);
Warning, cannot evaluate the solution further right of 90.013890, maxfun limit exceeded (see ?dsolve,maxfun for details)

when I increase the time give this msn:

Warning, cannot evaluate the solution further right of 90.013890, maxfun limit exceeded (see ?dsolve,maxfun for details)

Hello,

Today I've playied a bit with CellDecomposition from the RootFinding package and for one of the systems with which I've playied I got an error which seems to me to be a bug related.

In particular, 

with(RootFinding[Parametric]):

m := CellDecomposition([x^3-y^2 = 0, x^2+y^2-1 < 0], [x, y])

Error, (in RootFinding:-Parametric:-CellDecomposition) Segmentation Violation occurred in external routine

Did I make a mistake somewhere or Maple 2015.1 faild?

I am working on a problem involving sums in Maple and I find Maple's facilities lacking.

Specifically; I want to convert the square of a sum into a sum of squares plus the cross terms (which is a subtask of a larger problem). So I start with

sum(E[n],n=1..N)^2;

and immediately get stuck as I do not find any command that does anything with it. The expansion of this is known and easy to derive:

sum(E[n],n=1..N)^2 = sum(E[n]^2,n=1..N) + 2*sum(sum(E[m]*E[n],m=1..n-1),n=1..N);

Maple knows nothing about this relation. I have checked out packages like SumTools but have not found anything useful for this purpose. Quite some time ago I have had difficulty distributing a sum over the elements of its expression; this was answered here and involved a custom function.

Does a package exist that has these kinds of conversions, or do I need to roll my own? It seems a general enough issue that I would expect functions for this to exist.

TIA,

Mac Dude

Hi,

  I have the following input in order to obtain the dimension of an array, by means of the values has been assigned

with(ArrayTools):
with(LinearAlgebra):
f[1]:=1:
f[2]:=2:
nops(f);
Size(f);
Dimension(f);

  The result is 

---

1
[ 1 1 ]
Error, invalid input: LinearAlgebra:-Dimension expects its 1st argument, A, to be of type {Matrix, Vector} or coercible via `~Simplify`, but received f

 ---

I cannot get number "2", which is expected. Is there any way to obtain that? I looked for help and mapleprime, without any simple solution ......

Thank you very much!

Hi,

  Suppose I have an array

A[1]:=a;

A[2]:=b;

A[3]:=c;

  I would like to check if an element, e.g. "c", belong to this array. I can contrsuct a "for... end do" command to compare each element and/or construct a subroutine. 

  Is there any build-in function in maple can do that? Perferably return true/false

Thank you!

I want to compute the following solve:

C:=A, B, F, H(It has 4 unknowns):

V:=R,Y,E,I (It has 4 equations):

solve({V},{C});

By hand calculations I just found that the solve result is real ,but in maple there is no solution and no error (the maple ignores to solve this equation !!!) 

can anybody help me to find the solve soltuion ?

Thank you 

H. Kabir

hi.I want to dsolve set of nonlinear equations with one unknown parameter ...is this possible with dsolve rule.in matlab this possible with bvp4c rule..please help me for this problem.if we should another rule please attached file reform.Thanks alot12.mw

I need to solve a set of equations but changing a constant each time.

For example, x+by=0, bx-y=10 where b=10,20.

I don't want to put it in a loop because, in loop, the equations are solved repeatedly. I want Maple to solve it only once and substitute b values automatically since I want to solve big set of equations faster.

Is there any option in Maple to do so with a single command

I know how we can add a member to a mathematical set ,for example :

C := NULL;

C := C, V, B, X;

print(C)

V, B, X

but, my problem is that how we can subtract a member from the defined set ?

Thanks,

hi.how i can chose a minimum and positive answer of different answer in solve rule...

my program attached below.for example at this , the second answer should be selected as 1.965392881*10^9 ,that is the minimum and posetive among other...

thanks alot

11.mw

Dear all,

I am trying to solve the following partial differential equation (transport or advection equation) with given initial and boundary conditions:

restart: with(PDEtools):
sys := [v*diff(u(x,t), x) + diff(u(x,t), t) = 0, u(x,0) = exp(-x), u(0,t) = sin(t)];
pdsolve(sys);

But it does not work. The solution is (or should be): 

u(x, t) = exp(t*v-x)+Heaviside(t-x/v)*(sin(t-x/v)-exp(t*v-x))

I think the reason is that the interval for t (in [0, inf)) and x (in [0, 1]) is not specified. On the other hand, this works:

restart: with(PDEtools):
sys := [diff(u(x, t), t) = diff(u(x, t), x, x), u(0, t) = 0, u(1, t) = 0, u(x,0) = f(x)];
sol := pdsolve(sys);

How can I solve a PDE like the transport equation with given initial AND boundary conditions?

Thanks a lot