relate root representations

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relate root representations

Bugzilla from jakubi@df.uba.ar
I see that Axiom produces different representations for the roots of
polynomials in the form of lists, but I do not find the operations that
relates the corresponding elements of these lists.

As a concrete example, in a follow-up of the example that I have posted a few weeks
ago, consider the lists produced by:

allRootsOf(p^3 - p + 1/10)$RECLOS(FRAC INT)

[%F8,%F9,%F10]

solve((p^3 - p + 1/10=0)::EQ EXPR INT,p)


               +-----------+             +-----------+
               |      2                  |      2
              \|- 3%p0  + 4  - %p0    - \|- 3%p0  + 4  - %p0
   [p= %p0,p= --------------------,p= ----------------------]
                        2                        2
                                  Type: List Equation Expression Integer

radicalSolve(p^3 - p + 1/10=0,p)

                       +------------------+2
                       |    +-+    +-----+
           +---+       |- 3\|3  + \|- 373
      (- 3\|- 3  + 3)  |------------------  - 2
                      3|         +-+
                      \|      60\|3
  [p= -----------------------------------------,
                         +------------------+
                         |    +-+    +-----+
             +---+       |- 3\|3  + \|- 373
          (3\|- 3  + 3)  |------------------
                        3|         +-+
                        \|      60\|3
                       +------------------+2
                       |    +-+    +-----+
           +---+       |- 3\|3  + \|- 373
      (- 3\|- 3  - 3)  |------------------  + 2
                      3|         +-+
                      \|      60\|3
   p= -----------------------------------------,
                         +------------------+
                         |    +-+    +-----+
             +---+       |- 3\|3  + \|- 373
          (3\|- 3  - 3)  |------------------
                        3|         +-+
                        \|      60\|3
         +------------------+2
         |    +-+    +-----+
         |- 3\|3  + \|- 373
      3  |------------------  + 1
        3|         +-+
        \|      60\|3
   p= ---------------------------]
            +------------------+
            |    +-+    +-----+
            |- 3\|3  + \|- 373
         3  |------------------
           3|         +-+
           \|      60\|3
                                      Type: List Equation Expression Integer

So, I mean eg a command that takes %F8 as input and gives as output the
correspondig equation (or rhs) in the second or third list.

Is there such command?

Alejandro Jakubi







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Re: relate root representations

Bill Page-7
On Sat, Nov 15, 2008 at 5:23 PM, Alejandro Jakubi wrote:

>
> As a concrete example, in a follow-up of the example that I have posted
> a few weeks ago, consider the lists produced by:
>
> allRootsOf(p^3 - p + 1/10)$RECLOS(FRAC INT)
>
> [%F8,%F9,%F10]
>
> solve((p^3 - p + 1/10=0)::EQ EXPR INT,p)
>
>
>              +-----------+             +-----------+
>              |      2                  |      2
>             \|- 3%p0  + 4  - %p0    - \|- 3%p0  + 4  - %p0
>  [p= %p0,p= --------------------,p= ----------------------]
>                       2                        2
>                                 Type: List Equation Expression Integer
>
> radicalSolve(p^3 - p + 1/10=0,p)
>
> So, I mean eg a command that takes %F8 as input and gives as output the
> correspondig equation (or rhs) in the second or third list.
>
> Is there such command?
>

No there is no such command since RECLOS does not actually solve for
the roots exactly. It only finds the intervals within which each
unique real root lies. But in a previous email I suggested a simple
routine that did a lookup of the exact root based on computing it's
floating point value. Didn't you like that approach?

Regards,
Bill Page.


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Re: relate root representations

Bugzilla from jakubi@df.uba.ar
On Sun, 16 Nov 2008, Bill Page wrote:

> No there is no such command since RECLOS does not actually solve for
> the roots exactly. It only finds the intervals within which each
> unique real root lies. But in a previous email I suggested a simple
> routine that did a lookup of the exact root based on computing it's
> floating point value. Didn't you like that approach?

Not very much as it requires matching via their respective numeric
approximations. Of course, if there is no other option, it is
better something than nothing.

But my question was about all these three representations (is there any
more?). What about relating between the other two representations, which
are purely algebraic?

Alejandro Jakubi


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