Hi everybody,

I have written a module (let's say MyModule) that  I use as a package in a worksheet (with(MyModule)).

At some point in my worksheet I call the procedure MyProc which is part of MyModule.
I find it not to work as expected. So I modify it within MyModule and generate again the archive which contains this module.

Because my worksheet does a lot of things before the call to MyProc, I would like to test quickly the above modifications.
The idea is to do :
unwith(MyModule):  # packages() no longer exhibits its name
with(MyModule);      # to load the corrected one

Unfortunately, contrary to what happens with a "native" package,  the command unwith(MyModule) is ineffective:

• once done showstat(MyProc) still displays the content of the procedure, and running  the command MyProc(...) proves that it still "exists" in the worksheet
   
• forcing a reload of MyModule ( with(MyModule) ) and acanning again MyProc ( showstat(MyProc) ) reveals the code MyProc had when MyModule has been loaded for the first time.

Is it possible to "free" a user package through the "unwith" command ?

Hope to read you soon, TIA


PS : to be clearer

A worksheet contains  the definition of N procedures, plus the one of MyModule, and ends with the commands to generate an archive file named MyModule.mla.
MyModule is defined that way

MyModule := module()
option package

export Proc1 := eval(:-Proc1),
           ......
           MyProc := eval(:-MyProc),
           ......
           ProcN := eval(:-ProcN):
end module
 

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wave_eq_3.mws

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Thank you ~

Hi,

I have problem with plot of the function ( See attachment)
 

Download PlotProblème.mw

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Hi I've been trying to solve these set of PDEs below and have been at it for very long

My codes are below

I tried solving the first equation by using:

sys_ode := 2*(diff(T(eta), eta, eta, eta))+T(eta)*(diff(T(eta), eta, eta)) = 0

ics := T(0) = 0, (D(T))(0) = 1, (D(T))(20) = 0

Digits := 10

sol1 := dsolve({ics, sys_ode}, numeric, output = operator)

{q(eta) = rhs(sol1[2](eta)), w(eta) = rhs(sol1[3](eta))}

Then inputting into the subsequent two equations:

PDE1 := eval([2/P . (diff(g(x, eta), eta, eta))+q(eta)*(diff(g(x, eta), eta))-g(x, eta)*w(eta) = 2*x*w(eta)*(diff(g(x, eta), x)), 2/S . (diff(phi(x, eta), eta, eta))+q(eta)*(diff(phi(x, eta), eta)) = 2*x*w(eta)*(diff(phi(x, eta), x))])

subBC1 := -phi(x, 0)*exp(g(x, 0)*sqrt(x)/(1+varepsilon*g(x, 0)*sqrt(x)))

subBC2 := alpha . ((phi(x, 0))(sqrt(x)))(exp(g(x, 0)*sqrt(x)/(1+varepsilon*g(x, 0)*sqrt(x))))

BC := {diff(g(0, eta), eta, eta) = 0, diff(phi(0, eta), eta, eta) = 0, g(0, eta) = 0, g(x, 20) = 0, phi(0, eta) = 1, phi(x, 20) = 1, (D[2](g))(x, 0) = subBC1, (D[2](phi))(x, 0) = subBC2}

P = 1

S = 1

pds := pdsolve(PDE, {BC}, numeric, spacestep = .25)

but always end up with :Error, (in pdsolve/numeric/process_PDEs) number of dependent variables and number of PDE must be the same

I know my BC conditions might probably have some major errors too but i really cant proceed on cos i always end up with this same error. I really hope anyone would be able to help me on this 

help please

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LinearAlgebra[Rank]

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Thanks

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Thanks

factor_problem.mw

I want to factor the following polynomial:

Teller := 2*i1^4*i2*i3+2*i1^4*i2*i4+2*i1^4*i2*i5+2*i1^4*i3*i4+2*i1^4*i3*i5+2*i1^4*i4*i5+4*i1^3*i2^2*i3+4*i1^3*i2^2*i4+4*i1^3*i2^2*i5+4*i1^3*i2*i3^2+6*i1^3*i2*i3*i4+6*i1^3*i2*i3*i5+4*i1^3*i2*i4^2+6*i1^3*i2*i4*i5+4*i1^3*i2*i5^2+4*i1^3*i3^2*i4+4*i1^3*i3^2*i5+4*i1^3*i3*i4^2+6*i1^3*i3*i4*i5+4*i1^3*i3*i5^2+4*i1^3*i4^2*i5+4*i1^3*i4*i5^2+2*i1^2*i2^3*i3+2*i1^2*i2^3*i4+2*i1^2*i2^3*i5+4*i1^2*i2^2*i3^2+6*i1^2*i2^2*i3*i4+6*i1^2*i2^2*i3*i5+4*i1^2*i2^2*i4^2+6*i1^2*i2^2*i4*i5+4*i1^2*i2^2*i5^2+2*i1^2*i2*i3^3+6*i1^2*i2*i3^2*i4+6*i1^2*i2*i3^2*i5+6*i1^2*i2*i3*i4^2+24*i1^2*i2*i3*i4*i5+6*i1^2*i2*i3*i5^2+2*i1^2*i2*i4^3+6*i1^2*i2*i4^2*i5+6*i1^2*i2*i4*i5^2+2*i1^2*i2*i5^3+2*i1^2*i3^3*i4+2*i1^2*i3^3*i5+4*i1^2*i3^2*i4^2+6*i1^2*i3^2*i4*i5+4*i1^2*i3^2*i5^2+2*i1^2*i3*i4^3+6*i1^2*i3*i4^2*i5+6*i1^2*i3*i4*i5^2+2*i1^2*i3*i5^3+2*i1^2*i4^3*i5+4*i1^2*i4^2*i5^2+2*i1^2*i4*i5^3+2*i1*i2^3*i3*i4+2*i1*i2^3*i3*i5+2*i1*i2^3*i4*i5+4*i1*i2^2*i3^2*i4+4*i1*i2^2*i3^2*i5+4*i1*i2^2*i3*i4^2+6*i1*i2^2*i3*i4*i5+4*i1*i2^2*i3*i5^2+4*i1*i2^2*i4^2*i5+4*i1*i2^2*i4*i5^2+2*i1*i2*i3^3*i4+2*i1*i2*i3^3*i5+4*i1*i2*i3^2*i4^2+6*i1*i2*i3^2*i4*i5+4*i1*i2*i3^2*i5^2+2*i1*i2*i3*i4^3+6*i1*i2*i3*i4^2*i5+6*i1*i2*i3*i4*i5^2+2*i1*i2*i3*i5^3+2*i1*i2*i4^3*i5+4*i1*i2*i4^2*i5^2+2*i1*i2*i4*i5^3+2*i1*i3^3*i4*i5+4*i1*i3^2*i4^2*i5+4*i1*i3^2*i4*i5^2+2*i1*i3*i4^3*i5+4*i1*i3*i4^2*i5^2+2*i1*i3*i4*i5^3+4*i1^3*i2*i3+4*i1^3*i2*i4+4*i1^3*i2*i5+4*i1^3*i3*i4+4*i1^3*i3*i5+4*i1^3*i4*i5+8*i1^2*i2^2*i3+8*i1^2*i2^2*i4+8*i1^2*i2^2*i5+8*i1^2*i2*i3^2+12*i1^2*i2*i3*i4+12*i1^2*i2*i3*i5+8*i1^2*i2*i4^2+12*i1^2*i2*i4*i5+8*i1^2*i2*i5^2+8*i1^2*i3^2*i4+8*i1^2*i3^2*i5+8*i1^2*i3*i4^2+12*i1^2*i3*i4*i5+8*i1^2*i3*i5^2+8*i1^2*i4^2*i5+8*i1^2*i4*i5^2+4*i1*i2^3*i3+4*i1*i2^3*i4+4*i1*i2^3*i5+8*i1*i2^2*i3^2+12*i1*i2^2*i3*i4+12*i1*i2^2*i3*i5+8*i1*i2^2*i4^2+12*i1*i2^2*i4*i5+8*i1*i2^2*i5^2+4*i1*i2*i3^3+12*i1*i2*i3^2*i4+12*i1*i2*i3^2*i5+12*i1*i2*i3*i4^2+48*i1*i2*i3*i4*i5+12*i1*i2*i3*i5^2+4*i1*i2*i4^3+12*i1*i2*i4^2*i5+12*i1*i2*i4*i5^2+4*i1*i2*i5^3+4*i1*i3^3*i4+4*i1*i3^3*i5+8*i1*i3^2*i4^2+12*i1*i3^2*i4*i5+8*i1*i3^2*i5^2+4*i1*i3*i4^3+12*i1*i3*i4^2*i5+12*i1*i3*i4*i5^2+4*i1*i3*i5^3+4*i1*i4^3*i5+8*i1*i4^2*i5^2+4*i1*i4*i5^3+4*i2^3*i3*i4+4*i2^3*i3*i5+4*i2^3*i4*i5+8*i2^2*i3^2*i4+8*i2^2*i3^2*i5+8*i2^2*i3*i4^2+12*i2^2*i3*i4*i5+8*i2^2*i3*i5^2+8*i2^2*i4^2*i5+8*i2^2*i4*i5^2+4*i2*i3^3*i4+4*i2*i3^3*i5+8*i2*i3^2*i4^2+12*i2*i3^2*i4*i5+8*i2*i3^2*i5^2+4*i2*i3*i4^3+12*i2*i3*i4^2*i5+12*i2*i3*i4*i5^2+4*i2*i3*i5^3+4*i2*i4^3*i5+8*i2*i4^2*i5^2+4*i2*i4*i5^3+4*i3^3*i4*i5+8*i3^2*i4^2*i5+8*i3^2*i4*i5^2+4*i3*i4^3*i5+8*i3*i4^2*i5^2+4*i3*i4*i5^3+i1^4+3*i1^3*i2+3*i1^3*i3+3*i1^3*i4+3*i1^3*i5+3*i1^2*i2^2+6*i1^2*i2*i3+6*i1^2*i2*i4+6*i1^2*i2*i5+3*i1^2*i3^2+6*i1^2*i3*i4+6*i1^2*i3*i5+3*i1^2*i4^2+6*i1^2*i4*i5+3*i1^2*i5^2+i1*i2^3+3*i1*i2^2*i3+3*i1*i2^2*i4+3*i1*i2^2*i5+3*i1*i2*i3^2+10*i1*i2*i3*i4+10*i1*i2*i3*i5+3*i1*i2*i4^2+10*i1*i2*i4*i5+3*i1*i2*i5^2+i1*i3^3+3*i1*i3^2*i4+3*i1*i3^2*i5+3*i1*i3*i4^2+10*i1*i3*i4*i5+3*i1*i3*i5^2+i1*i4^3+3*i1*i4^2*i5+3*i1*i4*i5^2+i1*i5^3+4*i2^2*i3*i4+4*i2^2*i3*i5+4*i2^2*i4*i5+4*i2*i3^2*i4+4*i2*i3^2*i5+4*i2*i3*i4^2+4*i2*i3*i5^2+4*i2*i4^2*i5+4*i2*i4*i5^2+4*i3^2*i4*i5+4*i3*i4^2*i5+4*i3*i4*i5^2

What is the best strategy using Maple(latest version)? In a previous, less complicated example, the polynomial could be not be factored in a single expression, but I was succesfull to factor it in multiple factors.

kind regards,

Harry Garst