[comp.archives] [symbolic-math] Re: wanted: public domain symbolic manipul for ibmpc

mec@sandstorm.Berkeley.EDU (Michael Chastain) (02/10/91)

Archive-name: symbolic-math/msdos/ccalc/1991-02-10
Archive: wuarchive.wustl.edu:/mirrors/msdos/calculator/ccalc.zip [128.252.135.4]
Original-posting-by: mec@sandstorm.Berkeley.EDU (Michael Chastain)
Original-subject: Re: wanted: public domain symbolic manipul for ibmpc
Reposted-by: emv@ox.com (Edward Vielmetti)

In article <2449@bnlux0.bnl.gov> kyee@bnlux0.bnl.gov (kenton yee) writes:
>i want to get a public domain symbolic manipulation program
>for an ibmpc clone with 1 megabyte ram;   TeX output or
>fancy graphics are not necessary.   

Get this program from wsmr-simtel20.army.mil or the wuarchive.wustl.edu:

    Directory PD1:<MSDOS.CALCULATOR>
 
     Filename   Type Length   Date    Description
    ==============================================
    CCALC.ZIP     B  195934  900917  The Calculus Calculator (w/graphs, more)

This is not public domain, but it is shareware.  A $25 donation
to San Francisco State University, Applied Mathematics Fund,
is requested (but not required).

I have been using this program for almost a year now and it is
just a piece of pie.  It does symbolic algebra, symbolic
differentiation, and numerical integration.  It handles lists
and has a programming facility.  It's not as nice as Mathematica,
but it doesn't cost $1000 either.

Here's an example of some Calculus Calculator input:

	f(x,n) = sin ( n * x )
	g(x,n) = sum( f(x, i), i = 1 to n ) / n
	window( -pi, +pi, -1, 1 )
	graph( g(x,10), x )

Change the second line to "sum( f(x, i) * c(i), i = 1 to n ) / n",
and you are into do-it-yourself Fourier analysis.  And, of course,
you can write functions and procedures to use CC's primitives to
make more complicated mathematical functions.

This example doesn't show off the symbolic capabilities very well,
but they're there.  Go ftp the program and try it!

Michael Chastain
mec@ocf.berkeley.edu
"He who dies with the most FRIENDS wins."
a^n + b^n != c^n, n > 2.  First, note that a^n = (c-b)*(c^n-1 + ... + b^n-1),
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