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Re: [Axiom-math] Solving symbolic vector equations: How do I do this in


From: Zach
Subject: Re: [Axiom-math] Solving symbolic vector equations: How do I do this in Axiom?
Date: Tue, 13 May 2008 16:32:51 -0600

Bill,


> so what I would like is for axiom to solve this by
>
> a * v1 . v1 = v2 . v1
> a = (v2 . v1) / (v1 . v1)
>

??? This is not a solution to the original equation!

Hmm, the problem is that this equation is in danger of having no solution.  I guess there is an implicit assumption that this has a solution for some nonzero v2, meaning the vectors must be parallel.  My boss gave me a problem something like this and I am walking on shaky legs at the moment as I haven't tried to solve problems like these before.

I guess the correct thing to do is notice that v1 and v2 must be parallel, and then deal with the magnitudes, a=|v2|/|v1|.  Once you accept this my solution works too, as with unit vector u in the direction of v1 and v2, v1 = |v1| u and v2 = |v2| u, v1 . v2 = |v1| |v2| u . u = |v1| |v2| and v1 . v1 = |v1|^2.  Right?  Please feel free to point out any misconceptions I am fostering.

Anyway, I guess I chose too simple of an example.  But never mind that.  (The original problem was to find tuples of reals (n1, n2, m1, m2) such that n1*a1 + n2*a2 = m1*b1 + m2*b2 (where a1, a2, b1, and b2 are 2D vectors).)

> Well no, not really. Currently Axiom has no domain for symbolic
> computations with vectors.

I was afraid of that.  Over at Maxima they had no facility either.


> What is the best way of tackling these types of problems?
>

It would be possible (and quite interesting) to create a new domain in
Axiom for symbolic vector calculations.

Hmm, okay.  So, it would be a good idea to understand how this is done in polynomial symbolic manipulation before approaching such a task?

Thanks,
Zach



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