Re: Can I have fries and a calculator with that?

Dave L. Renfro wrote (in part):

> We can rationalize the denominator of 1/b, where
>
> b = sqrt(2) + sqrt(10) + sqrt(12) + sqrt(56),
>
> by multiplying the numerator and denominator by f(b), where
>
> f(x) = x^15 - (640)*x^13 + (155,072)*x^11 - (18,296,832)*x^9
>
> + (1,125,983,744)*x^7 - (35,305,193,472)*x^5
>
> + (491,646,992,384)*x^3 - (1,840,594,812,928)*x.
>
> After multiplying, but before reducing, the denominator will be
>
> - 525,242,269,696.
>
> The numerator will be an integer-linear combination of 15
> rationally independent square root terms. In general, if you're
> dealing with a sum of square roots and there are k many prime
> factors of the radicands involved, then at the end you'll have
> at most (2^k) - 1 many "different" square root terms showing up.

I was so confident that all 15 square root terms would show up
when one rationalizes the denominator of

1 / [ sqrt(2) + sqrt(10) + sqrt(12) + sqrt(56) ]

that I didn't even bother to qualify my statement by saying
something like: "The numerator will be an integer-linear
combination of _at_most_ 15 rationally independent square
root terms." Well, I should have! This past weekend I
carried out the rationalization with a little help from
Scientific Workplace (MAPLE based) and it turns out that
only 8 square roots appear (i.e. 7 of the square root terms
that could have appeared have coefficients of zero).

A rationalization for this expression is:

(2117/11324) * sqrt(2) + (301/2831) * sqrt(3)

+ (681/11324) * sqrt(10) - (62/2831) * sqrt(14)

+ (425/5662) * sqrt(15) - (173/2831) * sqrt(21)

- (2117/11324) * sqrt(70) - (61/2831) * sqrt(105).


Dave L. Renfro

.

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