[MUG] Diff remember table

[MUG] Re: Diff remember table
Author: Maple Group    Posted: 01/03/2000 19:21:46 GMT
>> From: Maple Group On Mon, 28 Feb 2000, Marc Prevosto wrote: |> I would like to impose in the remember table of the diff procedure |> diff(theta,t):=a; |> without defining theta. |> Is it possible? -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Tue, 29 Feb 2000 11:57:59 -0800 (PST) From: Robert Israel To: Subject: Diff remember table For most functions F, you could just say > F(theta,t):= a; But this doesn't work for "diff". I'm not sure why, but it looks like Maple treats "diff" in a special way. For example, although "diff" has option remember, the remember table is not changed when you say diff(theta,t) or even diff(t,t). Anyway, it looks like you could do it with a "wrapper" for the diff procedure: > olddiff:= eval(diff); > unprotect(diff); > diff:= proc() olddiff(args) end; > protect(diff); > diff(theta,t):= a; However, that is almost certainly not what you want, because while it will evaluate diff(theta,t) as a, this won't be used in more complicated expressions such as diff(theta+t,t). Instead you could try doing more with the wrapper. > restart; olddiff:= eval(diff); unprotect(diff); `diff/_theta`:= proc(e,v) if v=t then a else 0 fi end; diff:= proc(e) eval(olddiff(subs(theta=_theta(t),e),args[2..nargs]), _theta=proc() theta end); end; protect(diff); And now, for example: > diff(theta^2 + t, t); 2 theta a + 1 Robert Israel Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Wed, 01 Mar 2000 17:42:04 +0000 To: Marc Prevosto From: Stanley J Houghton Subject: Diff remember table At 17:18 01/03/2000 +0100, you wrote: >Yes of course, but I have a complicated function g of theta and t >(trig,power,exp) and I would like to compute diff(g,theta) as a function of >theta an t. If I put theta:=a*t+C and use diff it is then a difficult job >(simplify,combine,...) to obtain the function of theta. >I do not know if it's clear? > >Marc I understand. However, without seeing the function, it is difficult to suggest an approach. Function algsubs may help to reintroduce theta, using the constant C as an indicator thus: > expr:=exp(theta*t)*sin(theta+4*t+t^2); > theta:=a*t+C; #define theta in terms of a using C as an indicator constant > d:=diff(expr,t); > theta:='theta'; #revert to theta undefined > algsubs(C=theta-a*t,d); #now substitute back for theta using the indicator Please excuse the naive functional form invented for the purpose but it serves to show the approach. Alternatively, you could use: > restart; > expr:=exp(theta*t)*sin(theta+4*t+t^2); > theta:=f(t); #define theta as a function form > `diff/f`:=proc(x,t) a end; #define the differential as 'a' > d:=diff(expr,t); > theta:='theta';f(t):=theta; #now reset theta and set f(t) to evaluate to theta > d; #check the result Hope it helps Stan

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