I don't know how you can avoid using DEPENDS, but I do know how you can use
TELLSIMP in a function to do what you apparently want to do:
(C1) foo(x):=(depends(x,t),apply('tellsimp,['diff(x,t),dx_dt]));
dx
(D1) foo(x) := (DEPENDS(x, t), APPLY('TELLSIMP, [--, dx_dt]))
dt
(C2) foo(x1);
(D2) [DERIVATIVERULE1, SIMPDERIV]
(C3) diff(x1,t);
(D3) dx_dt
Viktor
-----Original Message-----
From: maxima-admin@math.utexas.edu [mailto:maxima-admin at math] On
Behalf Of Andrei Zorine
Sent: Thursday, May 06, 2004 1:48 PM
To: maxima
Subject: tellsimp inside functuons
Hello,
I want to write a function, which amongst all does what is done below.
(C1) depends(x,t);
(D1) [x(t)]
(C2) tellsimp('diff(x,t),dx/dt);
(D2) [DERIVATIVERULE1, SIMPDERIV]
(C3) diff(x,t,2);
(D3) 0
(C4) diff(x,t);
dx
(D4) --
dt
(C5) :lisp $d4
((MTIMES SIMP) ((MEXPT SIMP) $dt -1) $dx)
So, the function should look like
f(x):=( ... do this and that, then tellsimp to substitute diff(x,t) with
dx/dt, then do more and more...)
My intention is to be able to do f(x1), f(x2), etc., and the have
diff(x1,t,2)=0 , diff(x1,t)=dx1/dt, etc.
the problem is either f(x1) results in tellsimp('diff(x,t),...).
If I write a macro like buldq([x],'tellsimp('diff(x,t),dx/dt)), then
tellsimp gets not evaluated :(
How do I define patterns with differentiation of a function's argument?
I have to say depends(x,t) to emulate mma's Dt call. Can I define
simplification of 'diff(x,t) without saying depends(...) in the above
framework?
--
Andrei Zorine
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