[Newbie:] How to discard higher orders of variables?

On Wed, Jul 11, 2007 at 08:09:39PM +0200, Schirmacher, Rolf wrote:
> 
> Hello,
> 
> first, please note that I am a newbie to maxima, so perhaps the solution is
> rather simple.
> 
> My problem is that I want to discard higher order terms within an expression
> as it is often done when simplifying analytical equations in engineering.

often this is done by ratsubst if the equation is simple. I see your
example is not quite simple... 

perhaps you could post what you would manually do in this case. I am
having trouble figuring out exactly what your goal is. It is obvious
that you want to drop some terms, but not obvious exactly which terms.

You can always use the fact that subst does a strict substitution to
drop specific terms... for example

(%i25) term;
            2                   4                   2
(%o25) (beta  + 2) f + sqrt(beta  + 2) e + sqrt(beta  + 2) d
                                                   2                4         2
                                        + sqrt(beta  + 1) c + b beta  + a beta
(%i26) subst(2,beta^2+2,term);
                      4                               2                4
(%o26) 2 f + sqrt(beta  + 2) e + sqrt(2) d + sqrt(beta  + 1) c + b beta
                                                                              2
                                                                      + a beta
(%i27) subst(1,beta^2+1,%);
                         4                                4         2
(%o27)    2 f + sqrt(beta  + 2) e + sqrt(2) d + c + b beta  + a beta
(%i28) subst(0,beta^4,%);
                                                           2
(%o28)             2 f + sqrt(2) e + sqrt(2) d + c + a beta


is %o28 the result you are looking for?

> So, the remaining solution coming to my mind was to go to the taylor series:
> 
> (%i15) taylor(term,beta,0,2);
> 
> (%o15) c+sqrt(2)*d+sqrt(2)*e+2*f+((4*f+sqrt(2)*d+2*c+4*a)*beta^2)/4+...
> 
> As a second step, one would have to sort out all terms (manually?) from the
> beta^2 term, which occur in the part independent of beta (as they represent
> the beta^2 << 1 - issue). While this seems to be reasonable for the terms
> given here, it shows up to be rather involved for "real world" equations.

I'm not sure why you wouldn't just use %o15 and carry that around as
your expression, other than it looks a little ugly, it should be a
more accurate result than your result that simply drops the variation
due to the sqrt(1+...) terms. Perhaps you are trying to
reproduce/check someone elses calculation which does not use this
technique?


-- 
Daniel Lakeland
dlakelan at street-artists.org
http://www.street-artists.org/~dlakelan