Brian Hulley [EMAIL PROTECTED] writes:
But how does this change the fact that y still has 1 more element
than yq? yq is after all, not a circular list.
infinity+1 = infinity
Surely this is just a mathematical convention, not reality! :-)
Not even that. Infinity isn't a number, and it
On Thu, Jun 22, 2006 at 05:44:58PM +0100, I wrote:
It works because Haskell 'data' definitions yield both an initial fixed
point (with respect to strict functions) and a terminal fixed point (with
respect to arbitrary functions), and moreover these are usually the same.
The former is
Jerzy Karczmarczuk wrote:
Brian Hulley wrote:
[EMAIL PROTECTED] wrote:
you may transform a recurrential equation yielding Y out of X:
Y[n+1] = a*X[n+1] + b*Y[n]
usually (imperatively) implemented as a loop, into a stream
definition:
...
Can you explain how this transformation was
2006/6/22, Brian Hulley [EMAIL PROTECTED]:
Jerzy Karczmarczuk wrote:
Brian Hulley wrote:
[snip]
y IS NOT a longer list than yq, since co-recursive equations without
limiting cases, apply only to *infinite* streams. Obviously, the
consumer of such a stream will generate a finite segment
minh thu wrote:
2006/6/22, Brian Hulley [EMAIL PROTECTED]:
Jerzy Karczmarczuk wrote:
Brian Hulley wrote:
[snip]
y IS NOT a longer list than yq, since co-recursive equations without
limiting cases, apply only to *infinite* streams. Obviously, the
consumer of such a stream will generate a
On 2006-06-22 at 15:16BST Brian Hulley wrote:
minh thu wrote:
y and yq are infinite...
But how does this change the fact that y still has 1 more element than yq?
yq is after all, not a circular list.
infinity+1 = infinity
I don't see why induction can't just be applied infinitely
to
On Thu, 2006-06-22 at 15:16 +0100, Brian Hulley wrote:
. . .
But how does this change the fact that y still has 1 more element than yq?
yq is after all, not a circular list.
I don't see why induction can't just be applied infinitely to prove this.
The set of all non-negative integers has
2006/6/22, Brian Hulley [EMAIL PROTECTED]:
minh thu wrote:
2006/6/22, Brian Hulley [EMAIL PROTECTED]:
Jerzy Karczmarczuk wrote:
Brian Hulley wrote:
[snip]
y IS NOT a longer list than yq, since co-recursive equations without
limiting cases, apply only to *infinite* streams. Obviously, the
Jon Fairbairn wrote:
On 2006-06-22 at 15:16BST Brian Hulley wrote:
minh thu wrote:
y and yq are infinite...
But how does this change the fact that y still has 1 more element
than yq? yq is after all, not a circular list.
infinity+1 = infinity
Surely this is just a mathematical
On 2006-06-22 at 15:45BST Brian Hulley wrote:
Jon Fairbairn wrote:
infinity+1 = infinity
Surely this is just a mathematical convention, not reality! :-)
I'm not sure how to answer that. The only equality worth
talking about on numbers (and lists) is the mathematical
one, and it's a
On Jun 22, 2006, at 10:16 AM, Brian Hulley wrote:
minh thu wrote:
2006/6/22, Brian Hulley [EMAIL PROTECTED]:
Jerzy Karczmarczuk wrote:
Brian Hulley wrote:
[snip]
y IS NOT a longer list than yq, since co-recursive equations
without
limiting cases, apply only to *infinite* streams.
On Thu, Jun 22, 2006 at 11:06:38AM -0400, Robert Dockins wrote:
aside
Every few months a discussion arises about induction and Haskell
datatypes, and I feel compelled to trot out this oft-misunderstood
fact about Haskell: 'data' declarations in Haskell introduce co-
inductive
minh thu wrote:
maybe i wrong, anyway :
induction can be used to prove a property.
we claim that the property is true for any finite i.
so what's the property that you want to prove by induction ?
you say 'by induction on the lenght of yq'.. but yq is just y (modulo
the a*xq + b*).
it's exactly
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