Hello All 

I looking to write Maple code procedure to solve this equation 

restart;
with(PDEtools):
with(plots):
with(LinearAlgebra):
pde := diff(psi(t,x),t,t) - diff(psi(t,x),x,x) + V(x)*psi(t,x);

where V(x)->x^2;

I have to use this stencil 

u[i+1,j]=-u[i-1,j1]+(1-1/8*h^2*v[i-1])*u[i,j-1]+(1-1/8*h^2*v[i+1])*u[i,j+1]; where i is spatial step and j is time step

Thanks 

Hello,
I have a question concerning the change of the results of Maximization procedure and it will be much appreciated if you could help me.
The case is: I need to find the maximum of a function (call if f(t)) given a range of values of another function (call it g(t)).Both sets of parameters are the same. I use "Maximize" command, but I need to plot the the results on (f,g) plane afterwards. How could I change the Maximization output in order to have only objective function...

int(y^2/2 - lambda*(u*diff(y,t)-diff(u,t)-a), t);

with(Physics):

diff(y(t)^2/2 - lambda(t)*(u(t)*diff(y(t),t)-diff(u(t),t)-a), u(t)); # u -> something

diff(lambda(t),t) disappear

should be

diff(lambda(t),t)+Physics[`*`](lambda(t), diff(y(t), t))

diff(y(t)^2/2 - lambda(t)*(u(t)*diff(y(t),t)-diff(u(t),t)-a), y(t)); # y -> something

u*diff(lambda(t...

I would like to take the sin of a procedure, but it keeps giving me an error. Is it possible at all, or should I rewrite it into a matrix?

like this: sin(proc(x)...end proc)

following expression doesn't simplify to 1

conjugate(1/sqrt(I*t*lambda+m))*sqrt(lambda^2*(t^2)+m^2)*(1/sqrt(I*t*lambda+m))

Now I did use assumptions on any variable

specifically: assume(lambda>0,t>0,m>0)

Then I tried radnormal, simplify, evalc, combine(expr,radical)

But none of these work.

Hey (again) everyone. I'm back with a question similar to (but never answered on) this post: http://www.mapleprimes.com/questions/152116-Maple-Doesnt-Recognize-Function-Definition

The problem here is almost identical, except now it won't let me define a procedure. It's very bizarre. Here is my sheet:

The following equation represents the magnification (A) of an object given its distance (s1) from a closed optical system.

How can I determine the range of s1 that will provide a magnification (A) range of 2.5 to 5?

Dear maple team.

I would like to suggest something about the treatment that Maple makes of arithmetic modulo 2.
If you write the command:"x+x mod 2", the Maple output is "0", because "x+x" is always even. Perfect.
The problem I have is:
When  you write "x^2 + x mod 2", the Maple output is "x^2 + x"  instead of "0". Maple doesn't takes into account when simplifying that all the variables are idempotent modulo 2 (i.e, x*x = x)

If I write

Can maple use a Differential Operator somehow in the following sense?

Instead of expanding (D-p)^n where p is a constant and D the differential operator and n an integer and then in each term writing D^n(f) for some function can I just leave the expression as it is like

(D-p)^n and acting on everything on the right?

I have this sequence:

How can I put  this in maple, assess if it converges or not and if it is possible, calculate the limit to infinity.

Thanks

Warning, solving for expressions other than names or functions is not recommended.

is the warning I get every time I try to solve for solve(ode1 = ode2, theta/T)

Is there any way I can solve for an expression (i.eg theta/T)?

Furthermore I would like to have an answer on how I get to laplace transforms (ODE) with only "s" in the output, if I type like this:

laplace(J*(diff(theta(t), t, t...

Hi, I cannot solve the 2 implicit functions with 2 unknown (m,n) simultaneously (please see attached maple file). Note: m and n are real numbers.

May I get some advice please? Many thanks.

Is there a way to swap the names of two Arrays without creating unnecessary copies?

The scenario is as follows. I want to repeatedly apply a function to a large Array. The new value in each cell depends on seveal cells of the old Array.  So on each iteration I need to keep track of two Arrays: A_old and A_new. At the end of the iteration I would like to swap A_old and A_new efficiently. This way I will only need the memory for two Arrays and will...