Though Prebens Answer does work. This is simple and easy to handle^^

thx

Maybe some last question. If p ist not a constant but rather a function V(x) then this procedure does not account for the non-commutivity of Dx and V(x).

@Carl Love Sorry my fault. I reinstalled maple recently and thought I have the newest version. But it's only 17.01. I just installed 17.02 and indeed the bug is gone. So sorry for the confusion.

If it's of any help: I'm using Maple 17.02 64 Bit Running on Windows 7 Home

mapleprimessimplif.mw

Download mapleprimessimplif.mw

@digerdiga 

@Carl Love 

Ok this is not really what I want. Maybe it's not possible but I try again.

restart;

f(x):=exp(-x^(2))

(((D-2)^2)(f))(x)

gives:

(-2*x*exp(-x^2)-2)^3

But I dont want the function f(x) to be pulled into the parenthesis.

In fact I want

(D^(2)-4*D+4)(f)(x)=D^(2)(f)(x)-4*D(f)(x)+4*f

@Carl Love 

Ok this is not really what I want. Maybe it's not possible but I try again.

restart;

f(x):=exp(-x^(2))

(((D-2)^2)(f))(x)

gives:

(-2*x*exp(-x^2)-2)^3

But I dont want the function f(x) to be pulled into the parenthesis.

In fact I want

(D^(2)-4*D+4)(f)(x)=D^(2)(f)(x)-4*D(f)(x)+4*f

Sorry, I apparently missed that one :-/

Sorry, I apparently missed that one :-/

you mean constants applied as function would be

2(x)=2

Now I have for example (D-p)^3(sin)(x) then the result is

(cos(x)-(p(sin))(x))^3

so obviously maple doesnt treat p as a constant. How can I let maple know that the variable p is a constant?

you mean constants applied as function would be

2(x)=2

Now I have for example (D-p)^3(sin)(x) then the result is

(cos(x)-(p(sin))(x))^3

so obviously maple doesnt treat p as a constant. How can I let maple know that the variable p is a constant?

lol^^

Thank you very much.

Either does the job. I think simplify/siderels is most convenient since its easy to remember, because one uses simplify all the time anyway. 

lol^^

Thank you very much.

Either does the job. I think simplify/siderels is most convenient since its easy to remember, because one uses simplify all the time anyway. 

Yeah but when the curve is more complex in terms of one cannot solve for the Re/Im parts I'd like to plot it implicitly.

say for example.

x^5+y^5+x+y=1

I'd like to plot this equation via implicitplots3d.

Isn't this possible?

@AxelVogt: What do you mean I missed to assign?

Could you explain with an example?

Yeah but when the curve is more complex in terms of one cannot solve for the Re/Im parts I'd like to plot it implicitly.

say for example.

x^5+y^5+x+y=1

I'd like to plot this equation via implicitplots3d.

Isn't this possible?

@AxelVogt: What do you mean I missed to assign?

Could you explain with an example?

Now what I did is:

E := (1/2)*(1/rho1^2+1/rho1^2)-1/sqrt(z1^2+rho1^2)-1/sqrt((R-z1)^2+rho1^2)-1/sqrt(z1^2+rho1^2)-1/sqrt((R-z1)^2+rho1^2)+1/R+1/sqrt((2*rho1)^2+(R-2*z1)^2)

Er1 := diff(E, rho1) = 0

Ez1 := diff(E, z1) = 0

sol := proc (r) options operator, arrow; fsolve(eval({Er1, Ez1}, R = r), {z1 = -1 .. 3, rho1 = 0 .. 3}) end proc

sol(.5)

solE := proc (s) options operator, arrow; eval(E, {op(1, sol(s)), op(2, sol(s)), R = s}) end proc

solE(.5)

plot(proc (s) options operator, arrow; solE(s) end proc, 0 .. 5, view = -.5 .. -1.2)

For me at 2.3/2.4 the plot is not continuous.

I think this is due to the fsolve command mixing up different solutions.

There is one solution which behaves like shown there in the plot for large s but also which continues less steep

e.g. if I specify rho1=0..1.02 you see the continuous function.

But how can I select the solution whose behaviour for large s you see when increasing the search bound. e.g. rho1=0..3