Thanks for the excellent suggestions!
Rgds
Paulo
On Sat, May 24, 2008 at 2:15 PM, Robert Dodier <robert.dodier at gmail.com>
wrote:
> On Fri, May 23, 2008 at 8:56 AM, Paulo Grahl <pgrahl at gmail.com> wrote:
>
> > The only problem is that I couldn?t find yet a good place to learn how to
> make a
> > more advanced use of it.
>
> Well, the only thing I can suggest is to take a look at the tutorials at:
> http://maxima.sourceforge.net/documentation.html
> Each tutorial handles some different topics, so in aggregate perhaps
> they cover all the bases ....
>
> > For instance, It took me a while to realize that
> > the commands `tellsimp? and `tellsimpafter? only work after
> `matchdeclare?.
> > I couldn?t find this info in Maxima?s manual or in any of the tutorials
> (I
> > might have overlooked it, though).
>
> I dumped everything I could figure out about matchdeclare
> and tellsimpafter into the reference manual descriptions.
> ? matchdeclare and ? tellsimpafter should find it.
>
> > E ('sum (h (k,j), k, 1, n));
> > => 'sum (E (h (k,j)), k, 1, n)
> > I know that E(h(k,j)) = a for k=j and E(h(k,j))=b for k<>j, where `a` and
> > `b` are constants, and j in [1,n]
> > I would like to have
> > 'sum (E (h (k,j)), k, 1, n) evaluated to
> > => a+b*(n-1).
>
> Perhaps kron_delta (Kronecker delta) is useful here.
> S.t. like subst(E(h(i, j)) = a*kron_delta(i, j) + b*(1 - kron_delta(i,
> j)), expr)
> where expr is 'sum(E(h(i, j)), i, 1, n) or something.
>
> Some further declarations are needed to reduce a sum of kron_delta.
> sum by default doesn't appear to know about kron_delta.
>
> declare (sum, linear);
> matchdeclare ([ii, jj], symbolp, [ii0, ii1], all);
> simp : false;
> tellsimp ('sum (kron_delta (ii, jj), ii, ii0, ii1), if jj=ii then ii1
> - ii0 + 1 else if ii0 <= j and j <= ii1 then 1 else 0);
> simp : true;
>
> then we can get the following:
>
> 'sum (E (h (i, j)), i, 1, n);
> foo : subst (E (h (i, j)) = a * kron_delta (i, j) + b * (1 -
> kron_delta (i, j)), %);
> => b*(n-(if 1 <= j and j <= n then 1 else 0))
> +(if 1 <= j and j <= n then 1 else 0)*a
> assume (j>=1 and j<=n);
> ''foo;
> => b*(n-1)+a
> forget (j>=1 and j<=n);
>
> Maxima makes the assumptions about a summation index
> automatically, so if the sum over i is within a sum over j we get this:
>
> 'sum ('sum (E (h (i, j)), i, 1, n), j, 1, n);
> subst (E (h (i, j)) = a * kron_delta (i, j) + b * (1 - kron_delta (i, j)),
> %);
> => b*(n^2-'sum(if 1 <= j and j <= n then 1 else 0,j,1,n))
> +('sum(if 1 <= j and j <= n then 1 else 0,j,1,n))*a
> %, nouns;
> => b*(n^2-n)+a*n
>
> This much seems OK but there are additional complications
> which are not accounted for.
>
> HTH
>
> Robert Dodier
>