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Re: Why isn't Cancel more or less automatic?

• To: mathgroup at smc.vnet.net
• Subject: [mg35441] Re: Why isn't Cancel more or less automatic?
• From: "John Doty" <jpd at space.mit.edu>
• Date: Fri, 12 Jul 2002 04:29:10 -0400 (EDT)
• Organization: MIT Center for Space Research
• References: <ag6eh2$bo5$1@smc.vnet.net> <agbejo$jcb$1@smc.vnet.net> <ageghu$p7b$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

In article <ageghu$p7b$1 at smc.vnet.net>, "AES" <siegman at stanford.edu>
wrote:

> If you really wanted to be sure you got all the system degrees of
> freedom, however, wouldn't you just look at the roots of the determinant
> of the system, without going on to the transfer function for one
> particular input/output pair, which might indeed suffer from this
> problem.

Well, sure, you could do that. The advantage of the transfer function is
you can ask it other questions, and it's a little less abstract (which
makes it less likely you'll make a mistake). It's easy to type in a bunch
of equations, KCL and Ohm's law generalizations, and grind them through
Solve[] to get the transfer function. Your determinant is lurking in
there, of course, but it's a bit of extra work to reveal it.

In general, it comes down to what 0/0 means to you. In numerical
calculation, singularities must be avoided or removed, but in this sort of
symbolic work they are the keys to what's going on. Is a pole with a zero
residue still a pole? I think so (is an empty glass still a glass?). If
you see such a thing in an ideal calculation, you ought to be prepared to
see the signature of a pole with small but nonzero residue in the physical
process you're modeling.

-- 
|John Doty		"You can't confuse me, that's my job."
|Home: jpd at w-d.org
|Work: jpd at space.mit.edu