Michael Goodrich wrote:
Also I know what strict means, but why are you saying that baz is strict?
Because otherwise the loop would be OK. For instance if baz were
baz x = 100 -- lazy
then the equations could be evaluated starting from
c0 = baz z0 = 100
rd0 = c0*c0*m = 100*100*m
-- etc.
On Wednesday 05 April 2006 04:51 pm, Michael Goodrich wrote:
> Oops, I just realized that you gave me the answer, namely that it won't
> find fixed points of numeric sets of equations.
>
> Pity, that would really have made Haskell useful for this kind of
> scientific computing.
See section 4 of:
Oops, I just realized that you gave me the answer, namely that it won't find fixed points of numeric sets of equations.
Pity, that would really have made Haskell useful for this kind of scientific computing.
On 4/5/06, Brandon Moore <[EMAIL PROTECTED]> wrote:
Michael Goodrich wrote:> Looks like
On 4/5/06, Roberto Zunino <[EMAIL PROTECTED]> wrote:
Michael Goodrich wrote:[snip]> r = r2+2*step*rdc> rdc = (rd2+rd1+rd0)/6> rd0 = c0*c0*m> c0 = baz(z0)> z0 = 6.378388e6-rThe equations above form a loop: each one requires the one below it, and
the last one requires the first on
Michael Goodrich wrote:
[snip]
r = r2+2*step*rdc
rdc = (rd2+rd1+rd0)/6
rd0 = c0*c0*m
c0 = baz(z0)
z0 = 6.378388e6-r
The equations above form a loop: each one requires the one below it, and
the last one requires the first one.
(And yes, baz is strict)
Regards,
Roberto Z
Michael Goodrich wrote:
Looks like my calulation involves a self referential set of definitions.
Is Haskell not able to deal with a self referential set of definitions?
I was frankly hoing it would since otherwise there is then the specter
of sequence, i.e. that I have to finesse the order i
