side-effects using declare(var,constant) in Maxima - maxima

In maxima, is the following behavior intended?
First example:
(%i1) declare(a,constant);
(%o1) done
(%i2) constantp(a);
(%o2) true
(%i3) square(a):=a^2;
define: in definition of square, found bad argument a
-- an error. To debug this try: debugmode(true);
(%i4) load("linearalgebra.mac");
define: in definition of dotproduct, found bad argument a
-- an error. To debug this try: debugmode(true);
Second example:
(%i1) a:5;
(%o1) 5
(%i2) constantp(a);
(%o2) true
(%i3) square(a):=a^2;
2
(%o3) square(a) := a
(%i4) square(a);
(%o4) 25
Third example:
(%i1) declare(a,scalar);
(%o1) done
(%i2) mat_f(a,b):=a.b - b.a;
(%o2) mat_f(a, b) := a . b - b . a
(%i4) mat_f(matrix([1,2],[3,4]),matrix([3,4],[1,2]));
[ - 10 - 14 ]
(%o4) [ ]
[ 6 10 ]
It seems like declare(a,constant) has a global effect which to me seems strange in maxima. The second and third example work exactly how I would expect it.
Also are there any similar cases where something like this happens in maxima?

Maxima has a very weak notion of scope. Essentially all symbols are in the same scope, so when you make a declaration about a, it is about all instances of a, even the ones which are function arguments.
Maxima is actually a very old program and this is one of those aspects which has never been updated. There is discussion about giving Maxima a stronger notion of scope, but that will take some time.

Related

Wrong limit with Maxima CAS

This very simple limit:
(%i1) limit((z^(2*n)-1)/(z^2-1),z,-1);
(%o1) infinity
goes catastrophically wrong. We can think that a reasonable solution will be
(%i1) declare(n,integer);
(%o1) done
(%i2) limit((z^(2*n)-1)/(z^2-1),z,-1);
(%o2) 0
And wrong again! BTW all (well, actually I haven't checked all, :-)) the particular cases are correct:
(%i1) limit((z^(2*99)-1)/(z^2-1),z,-1);
(%o1) 99
More mystery: the function is even and the other limit
(%i1) limit((z^(2*n)-1)/(z^2-1),z,1);
(%o1) n
Is OK! (even without the declare) Why?

differentiation dependencies

I want to understand how dependencies in Maxima for differentiation work
for iterated cases.
I tried here:
(%i1) depends([f],[x,y]);
(%o1) [f(x,y)]
(%i2) depends([g],[x,y]);
(%o2) [g(x,y)]
(%i3) depends([x,y],[ε]);
(%o3) [x(ε),y(ε)]
(%i4) diff(g,ε);
(%o4) (g[y])*(y[ε])+(g[x])*(x[ε])
(%i5) h(x,y):=f(x,y)+g(x,y);
(%o5) h(x,y):=f(x,y)+g(x,y)
(%i6) diff(h(x,y),ε);
(%o6) g(x,y)[ε]+f(x,y)[ε]
(%i7) diff(h,ε);
(%o7) 0
In (%o4) I get the total derivative with respect to \epsilon. Whereas in (%o6) the derivatives of x and y with respect to \epsilon are not shown. Why?
And can I make Maxima to show these derivatives in the result?
Dependencies declared by depends are only recognized for symbolic, undefined functions. The dependency is associated with the function name (a symbol).
A function with an actual definition, as defined by := or define, is not recognized. The body of the function could contain any combination of other functions, so the only way to know on which other functions the function depends is to evaluate the function body. That is what is happening when you write diff(h(x, y), ε).

Does it matter if I use new to initialize a defstruct in maxima?

Initializing a defstruct using new or without new, seems to do the same, is there any difference?
(%i1) defstruct(foo(a,b));
(%o1) [foo(a, b)]
(%i2) f: foo(1,2);
(%o2) foo(a = 1, b = 2)
(%i3) f#a;
(%o3) 1
(%i4) f_new: new(foo(1,2));
(%o4) foo(a = 1, b = 2)
(%i5) f_new#a;
(%o5) 1
(%i6)
Without using new, the code seems a bit shorter and easier, but I'm not sure if some code will break if I use this pattern.
I think both ways are OK. The only difference so far as I know is that new(foo(...)) will ensure that there are the correct number of initial values. For example new(foo(1)) and new(foo(1, 2, 3)) will give errors, with foo as defined above. Just writing foo(1) or foo(1, 2, 3) doesn't trigger an error (maybe it should).

maxima - matchdeclare won't accept first argument to be a variable

Consider the following statements:
(%i1) matchdeclare([a,b], constantp);
(%o1) done
(%i2) defmatch(match, a*x+b);
(%o2) match
(%i3) match(2*x+3);
(%o3) [b = 3, a = 2]
I want to generalise this pattern in a function. But then it doesn't seem to work anymore:
(%i1) matchForm(test, form, constants) := block(
matchdeclare(constants, constantp),
defmatch(match, form),
match(test)
);
(%o1) matchForm(test, form, constants) :=
block(matchdeclare(constants, constantp), defmatch(match, form), match(test))
(%i2) test: 2*x+3;
(%o2) 2 x + 3
(%i3) form: a*x+b;
(%o3) a x + b
(%i4) constants: [a,b];
(%o4) [a, b]
(%i5) matchForm(test, form, constants);
defmatch: evaluation of atomic pattern yields: a x + b
(%o5) false
Debugging a bit, a bit it seems like the issue is that matchdeclare doesn't accept the argument to be a variable. Is ther any way to make a function like I try to make in maxima?
I haven't tried this, but: perhaps you can get the effect you want via
apply (matchdeclare, [constants, 'constantp]),
which will evaluate constants before calling matchdeclare.

Maxima: Assign value to a variable, by addressing it in a matrix

I had a complex problem which is solved. Now i would love to automate it. In doing so, I do address a position in a matrix. This positions does contain a variable. I want to assign a value to this variable, by its position in the matrix. Here is a simple example of what I want to do:
(%i1) M:[a,b,c];
(%o1) [a,b,c]
(%i2) M[1];
(%o2) a
(%i3) ev(M[1]):3;
(%o3) error: assignment: cannot assign to ev(M[1])
(%i4) float(a);
(%o4) a
I would love the variable "a" to be 3 now. But ev() is clearly not the right function for this.
My question is: is what i want even possible? does anyone know how to do? I looked the whole day and I am really depressed now :-(
I think what you want is the :: operator, which evaluates its left-hand side and assigns to that.
(%i1) M : [a, b, c] $
(%i2) M[1] :: 123 $
(%i3) a;
(%o3) 123
To replace a in M with the value 3:
M: M, a=3;
Which is, of course, just a shorter version of writing:
M: ev(M, a=3);
Or simply use ev(M, a=3) instead of M to use M with a=3 once without changing M in memory.
To set the 1st element in M to 3:
M[1]: 3
To replace the variable stored in M[1] in all of M:
M: ev(M, M[1]=3);
I hope one of those answers your question..

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