On 2015-01-24, john_perry_usm <[email protected]> wrote: > > >> You can do e[3,2,1,1].expand(3) to get it expanded in x0, x1, and x2. >> (more generally, x[i1,i2,...,im].expand(k) will give the expansion in >> x0,x1,...,x(k-1). >> > > I *don't* want it in the x's, though; I want it the e's. I wrote > e1=x1+x2+x3 merely to illustrate what I meant by e1. Look back at my > original example; I already know that, in three variables, m[4,2,1] > corresponds to e3*(e1^2*e2 - 2*e2^2 - e1*e3). I want *that* expansion in
> general, not the expansion into x's, nor the e[3,2,1,1]+..., which despite > being in e's is obviously different from the expansion I refer to. no, it's not different; namely, e[a1,a2,...ak] is just e(a1)*e(a2)*e(a3)*e(ak). e.g. in your case you get m[4,2,1]=e3*(e1^2*e2 - 2*e2^2 - e1*e3)=e[3,2,1,1] - 2*e[3,2,2]-e[3,3,1]. (which is only valid if you restrict to the 3st 3 vars). The different expansion you get in Sage is due to the fact that whenever k1>3 then e[k1,...]=0. Thus, no need to expand in x_i, you can just cut out all the e[]'s of this kind. You can parse the expression using .terms(), and then take .support() of each e[k1,k2,...]. E.g. e[4,1,1,1].support()[0] gives [4,1,1,1], and E.g. e[4,1,1,1].support()[0][0] gives 4. So you can filter these terms out. Perhaps there is a specail function for this, I don't know, but anyhow it is easy to DIY. HTH, Dima > > In my noggin I imagine that the algorithm has to go through the e's before > it goes into x's. I guess I should look at the code to see. > > john perry > -- You received this message because you are subscribed to the Google Groups "sage-support" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To post to this group, send email to [email protected]. Visit this group at http://groups.google.com/group/sage-support. For more options, visit https://groups.google.com/d/optout.
