Hello,
> solve returns a list like [x = ..., x = ...] to represent multiple solutions;
> that's why a[1, 1] = ... shows up in the value assigned to a[1, 1].
> You could put a[i, i - j] : map(rhs, solve(...)) and likewise for K[i].
> But it seems like a helpful consequence of the present code that
> a[i, j] = ... shows up in the value --- otherwise you would have to
> piece that together from arrayinfo(a) (to get the indices) and
> listarray(a) (to get the values).
Yes, when I run my code with n=2 and then check the output I get
(%i3) go(2);
(%o3) done
(%i4) listarray(a);
(%o4)
[[-1/48],[(120*a[2,2]+6*K[1]*a[1,1]-K[1])/48],[-(60*a[1,1]-1)/2880]]
(%i5) listarray(K);
(%o5) [1,[-6*a[1,1]],[-6*a[2,1]]]
So here is all the information needed to evaluate each of the terms in K
and a. I do not know, and haven't found but searching, a method to
substitute the values into the expressions.
I am looking for
a=[-1/48, -139/512,1/1280]
K=[1,-1/8, 417,256]
Any help much appreciated!
Thank you for your time,
Zoho