On 24.08.2022 09:42, Ralf Hemmecke wrote:
>> The reason that the innermost terms (i.e., the factors in the product
>
> Still there is the question, how am I supposed to take the result apart for
> further computation. That should be explained somewhere. You cannot expect
> your
> users to know how the internals of Expression(Integer) are to be handled.
Not? ;-)
Maybe, most simple:
f.1
(13)
s - 1 p - 1 s - 1
n - 1 8 7 6
--+ ++-++ --+ ++-++
> | | > - | | [f(p ): (p - 1)f(p ) + p - 1 = 0] + 2 + 2
--+ | | --+ | | 5 5 5 5
s = 0 p = 0 s = 0 p = 0
8 7 6 5
Type: Expression(Integer)
g:=f.1::InputForm
(15)
(+
(sum
(product
(+
(sum
(* - 1
(product (rootOfRec %F (%inforec1))
(equation %F (SEGMENT 0 (+ %G - 1))))
)
(equation %G (SEGMENT 0 (+ %H - 1))))
2)
(equation %H (SEGMENT 0 (+ %I - 1))))
(equation %I (SEGMENT 0 (+ n - 1))))
2)
Type: InputForm
car g
(16) +
Type: InputForm
cdr g
(17)
(
(sum
(product
(+
(sum
(* - 1
(product (rootOfRec %F (%inforec1))
(equation %F (SEGMENT 0 (+ %G - 1))))
)
(equation %G (SEGMENT 0 (+ %H - 1))))
2)
(equation %H (SEGMENT 0 (+ %I - 1))))
(equation %I (SEGMENT 0 (+ n - 1))))
2)
Type: InputForm
>
> Ralf
>
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