On Wed, Sep 15, 2010 at 3:47 AM, Håkan Granath
<[email protected]> wrote:
> On Sep 15, 3:44 am, Robert Bradshaw <[email protected]>
> wrote:
>> On Tue, Sep 14, 2010 at 10:44 AM, Håkan Granath
>>
>> <[email protected]> wrote:
>> > On Sep 14, 12:16 am, Robert Bradshaw <[email protected]>
>> > wrote:
>>
>> >> Alastair correctly deduced the issue that it can't tell if the number
>> >> is less than or greater than 2, what should it do here?
>>
>> > I do not know what it should do, but what I would have expected
>> > in this case is that continued_fraction would first compute the
>> > 53 bit approximation of the input, and then return the continued
>> > fraction of that approximation.
>>
>> If this is what you want, the way to get that is to give it something
>> that has 53 bits of precision.
>
> I completely agree :-) In fact, that is what I tried to do but
> got unexpected results. Some background information:
>
> I compute some quantity a, which is a product of the result of
> some high precision (say 1000 bits) numerical computation and an
> exact algebraic factor. The number a is expected to be rational,
> so to identify it I do something like
>
>  b = a.n(1000)
>  v = continued_fraction(b)
>
> However this did not always work, because in some cases b was not
> a floating point number as expected but a symbolic expression (a
> bug in my opinion, see below). This, I found, can make the
> continued fraction computation fail in 2 ways:
>
>  1. Sometimes, since b is symbolic it is computed with the
>  default 53 bit precision which is inadequate for my purposes.
>
>  2. Sometimes v would be the empty list.
>
> Of course I can work around this by doing something like
>
>  b = a.n(1000).n(1000)
>
> but, although it works for my code, it is somewhat of a
> cludge. Hence I reported the two issues: the topic of this thread
> and the issue I found with the n() function:
>
> http://groups.google.com/group/sage-support/browse_thread/thread/b36c90f1490eac19
>
> I hope this makes sense.

Yes, that issue with n() is certainly the root of the problem.

- Robert

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