home *** CD-ROM | disk | FTP | other *** search
- Path: sparky!uunet!olivea!decwrl!mips!swrinde!cs.utexas.edu!torn!watserv2.uwaterloo.ca!watdragon.uwaterloo.ca!watdragon!drclark
- From: drclark@daisy.uwaterloo.ca (David R. Clark)
- Newsgroups: sci.math.symbolic
- Subject: Re: Maple question.
- Message-ID: <DRCLARK.92Aug17085123@daisy.uwaterloo.ca>
- Date: 17 Aug 92 13:51:23 GMT
- References: <pete.714023792@marsh>
- Sender: news@watdragon.uwaterloo.ca (USENET News System)
- Organization: University of Waterloo, Waterloo, Ontario, Canada
- Lines: 33
- In-Reply-To: pete@marsh.cs.curtin.edu.au's message of Mon, 17 Aug 1992 03:56:32 GMT
-
-
- Newsgroups: sci.math.symbolic
- From: pete@marsh.cs.curtin.edu.au (Pete Staples)
- Subject: Maple question.
-
- ...
-
- I have written a procedure that uses Laplaces' method to solve integrals.
- Maple gives me an integral representation of the answer if I
- enter the limits (s=0..infinity) in the int procedure. If I evaluate
- the indefinite integral and use subs for each limit of integration
- it still doesn't evaluate the answer, eg leaves it in terms of
- erf(0), erf(x*infinity). Is there anyway that x > 0 can be assumed
- in the expression, so expressions like erf(x*infinity) can be interpreted?
-
- There are a couple of problems here. The first is that "subs" does not
- attempt to evaluate an expression after performing the substitution. The
- "eval" function can be used to force this evaluation. This will force
- "erf(0)" to 0. To compute the other endpoint you will have to use the
- "limit" function as it will handle erf(x*infinity) as x goes to infinity.
- This may well be what you want anyway and is the method used by the
- integration routines.
-
- In general, assuming x to be greater than 0 can be done with an assignment
- to the signum remember table "signum(x) := 1;" but it will not help in this
- case as erf(x*infinity) will still not evaluate.
-
- Hope this makes sense, thanks in advance..
-
- Peter.
-
-
- David Clark
-