Jianrong Li <lijr07 at gmail.com> writes:
> Dear all,
>
> I have some questions about summation. I am reading some codes about maxima
>
> What does " map(x->sum(sum(x,j,0,k),i,0,m), f(m+n+2*i,2*i+2*k+1-2*j)); "
> mean? Here f( , ) is a polynomial. " -> " is defined as " (arg ->
> form)::=buildq([arg,form],lambda([arg],form)) ". It is very strange since
> it seems that the summation is over x.
>
> Is " map(x->sum(sum(x,j,0,k),i,0,m), f(m+n+2*i,2*i+2*k+1-2*j)); " the same
> as " sum(sum(f(m+n+2*i,2*i+2*k+1-2*j), j, 0, k), i, 0, m) "? Maybe the they
> are not the same because I changed the codes to "
> sum(sum(f(m+n+2*i,2*i+2*k+1-2*j), j, 0, k), i, 0, m) ", then the answer is
> different.
Let's break things down a bit. I couldn't work out how to define ->
correctly, so translated your first example to
map (lambda([x], sum(sum(x,j,0,k),i,0,m)), f(m+n+2*i,2*i+2*k+1-2*j));
This probably isn't what you wanted, since
map (func, f(a,b)) => f(func(a), func(b))
Instead, you probably wanted "apply":
apply (lambda([x], sum(sum(x,j,0,k),i,0,m)),
[f(m+n+2*i,2*i+2*k+1-2*j)]);
m k
==== ====
\ \
(%o4) > > f(n + m + 2 i, 2 k - 2 j + 2 i + 1)
/ /
==== ====
i = 0 j = 0
I'm not sure what you mean by "the summation is over x".
I think your confusion about map vs. apply explains the answer to the
question in your second paragraph.
If things still aren't behaving like you expect, maybe you could put a
complete example in your email so we can try it out?
Rupert
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