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fnavarro

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How to run an executable program from Ma...

fnavarro 20 Maple 2021
programming linux define_external
January 20 2023
2 0

Dear All,

I have an executable program which I have generated with Fortran. Is it possible to run such a program from Maple? It sounds a bit weird but that would simplify my computation, since I would not need to use scripts in Linux. Thank you very much.

Cheers.

Problem with sum in linked functions...

fnavarro 20 Maple 2020
sum
February 21 2022
2 1

I am trying to define a sum which is obtained from linked functions (which also contain the sum command). 

The problem is: if I modify the first function p and rerun the worksheet (except fot the restart command), then the changes do not affect the result. Does anyone know why? Thank you very much.
 

I start with this

> 

restart
> 

p := proc (i, j) options operator, arrow; sum('cat(A, _, i, j, k)', 'k' = 1 .. 3) end proc

proc (i, j) options operator, arrow; sum('cat(A, _, i, j, k)', 'k' = 1 .. 3) end proc

 (1)
> 

q := proc (i) options operator, arrow; sum('p(i, j)', 'j' = 1 .. 3) end proc

proc (i) options operator, arrow; sum('p(i, j)', 'j' = 1 .. 3) end proc

 (2)
> 

r := sum('q(i)', 'i' = 1 .. 3)

A_331+A_332+A_333+A_321+A_322+A_323+A_311+A_312+A_313+A_231+A_232+A_233+A_221+A_222+A_223+A_211+A_212+A_213+A_133+A_123+A_131+A_132+A_121+A_122+A_111+A_112+A_113

 (3)

If I modify p (I make it twice its initial value), then this does not affect the value of q and r

> 

p := proc (i, j) options operator, arrow; 2*(sum('cat(A, _, i, j, k)', 'k' = 1 .. 3)) end proc

proc (i, j) options operator, arrow; 2*(sum('cat(A, _, i, j, k)', 'k' = 1 .. 3)) end proc

 (4)
> 

q := proc (i) options operator, arrow; sum('p(i, j)', 'j' = 1 .. 3) end proc

proc (i) options operator, arrow; sum('p(i, j)', 'j' = 1 .. 3) end proc

 (5)
> 

r := sum('q(i)', 'i' = 1 .. 3)

A_331+A_332+A_333+A_321+A_322+A_323+A_311+A_312+A_313+A_231+A_232+A_233+A_221+A_222+A_223+A_211+A_212+A_213+A_133+A_123+A_131+A_132+A_121+A_122+A_111+A_112+A_113

 (6)

Why?
Even if I  unassign p, q and r, this does not work, either.

> 

unassign('p', 'q', 'r')
> 

p := proc (i, j) options operator, arrow; 2*(sum('cat(A, _, i, j, k)', 'k' = 1 .. 3)) end proc

proc (i, j) options operator, arrow; 2*(sum('cat(A, _, i, j, k)', 'k' = 1 .. 3)) end proc

 (7)
> 

q := proc (i) options operator, arrow; sum('p(i, j)', 'j' = 1 .. 3) end proc

proc (i) options operator, arrow; sum('p(i, j)', 'j' = 1 .. 3) end proc

 (8)
> 

r := sum('q(i)', 'i' = 1 .. 3)

A_331+A_332+A_333+A_321+A_322+A_323+A_311+A_312+A_313+A_231+A_232+A_233+A_221+A_222+A_223+A_211+A_212+A_213+A_133+A_123+A_131+A_132+A_121+A_122+A_111+A_112+A_113

 (9)

Only if I restart, then I can modify p and change the rest of values

> 

restart
> 

p := proc (i, j) options operator, arrow; 2*(sum('cat(A, _, i, j, k)', 'k' = 1 .. 3)) end proc

proc (i, j) options operator, arrow; 2*(sum('cat(A, _, i, j, k)', 'k' = 1 .. 3)) end proc

 (10)
> 

q := proc (i) options operator, arrow; sum('p(i, j)', 'j' = 1 .. 3) end proc

proc (i) options operator, arrow; sum('p(i, j)', 'j' = 1 .. 3) end proc

 (11)
> 

r := sum('q(i)', 'i' = 1 .. 3)

2*A_332+2*A_333+2*A_323+2*A_331+2*A_313+2*A_321+2*A_322+2*A_311+2*A_312+2*A_231+2*A_232+2*A_233+2*A_221+2*A_222+2*A_223+2*A_212+2*A_213+2*A_211+2*A_131+2*A_132+2*A_133+2*A_121+2*A_122+2*A_123+2*A_111+2*A_112+2*A_113

 (12)
> 

``
> 

``


 

Download problem_with_sums.mw

(Apparent) Problem between seq/sum and a...

fnavarro 20 Maple 2020
seq alias
April 12 2021
3 3

Hello,

My name is Francisco Navarro. I have a problem when using diff and alias, together with seq or sum.

A very simplified worksheet of what I mean would be

restart

alias(a1 = a1(r), a2 = a2(r), a3 = a3(r))

array1 := [a1, a2, a3]

array2 := [seq(cat(a, i), i = 1 .. 3)]

array1 - array2

This gives [0,0,0] so they are the same thing but, under differentiation, they do not behave the same way (so they are NOT exactly the same object for Maple)

diff(array1, r)

diff(array2, r)

which are different. Any idea of what is wrong with this?

I need to create long arrays with funcions that depend on several values and then differentiate those arrays. But clearly this approach does not work. Thank you for your answer.

Francisco

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