Re: Make the denominator rational



Rainer Rosenthal wrote:

jhnrmsdn@xxxxxxxxxxx schrieb:
Thomas Mautsch wrote:

1
----------------------
(1/4) (1/4) 1/2
2 + 3 + 2

[ ... ]
But by Newton's [?] theorem these polynomials
can be expressed as integral polynomials in the
so-called elementary symmetric polynomials, which
are the rational integer coefficients of the monic
quartic of which x is a root.

Do have an /elegant/ solution? I think this is
what Thomas is asking for.

In contrast to William Elliot's postings I fail
to make a solution following your advice.
How to do it with Maple for example?

Well I thought my solution was moderately elegant
in its way, although admittedly calculating those
symmetric functions would be a pain; but here's
what you'll hopefully agree is a big improvement:

Start with three 4th roots a, b, c (It's easier
and more reliable to work with symbols, and of
course more general.)

As before, multiply top and bottom by the three
complementary factors, and note that:

(a + b + c).(a + b - c).(c - a + b)(c + a - b) =

((a + b)^2 - c^2).(c^2 - (a - b)^2) =

c^2.((a + b)^2 + (a - b)^2) - (a^2 - b^2)^2 - c^4 =

2.(a^2.b^2 + a^2.c^2 + b^2.c^2) - (a^4 + b^4 + c^4) =

2.(a^2 + c^2).(b^2 + c^2) - (a^4 + b^4 + 3.c^4)

Then multiplying top and bottom by b^2 - c^2
gives the denominator in the following form
for rational integers p, q:

p.(a^2 + c^2) + q.(b^2 - c^2)

which equals:

p.a^2 + q.b^2 + (p - q).c^2

and then we can multiply by three complementary
factors of _this_ (as above) to obtain a rational
integral denominator.


Cheers

John R Ramsden

.

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