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


"Bill Dubuque" <wgd@xxxxxxxxxxxxxxxxxxxx> wrote in message
news:y8zwtid4b0u.fsf@xxxxxxxxxxxxxxxxxxxxxxx
> "Dave L. Renfro" <renfr1dl@xxxxxxxxx> wrote:
>>
>> As for rationalizing denominators, here's a less trivial
>> example for you. 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
>>
>> d = -525,242,269,696.
>>
>> The numerator will be an integer-linear combination of 15
>> rationally independent square root terms.
>
> I.e. g(x) := x f(x) - d is the minimum polynomial of b over Q.
> Of course one may employ the minimum poly of any algebraic number
> w in a similar manner to rationalize w in a denominator. This is
> essentially the same as multiplying w by all of its conjugates.

Indulge me here a little, this is a hobby (not a good start, I know).

We know that sin (90 degrees) = 1, and we know that the value of sin(45
degrees) satisfies the equation 2x^2 = 1.

We further know that the value of sin(1 degree) satisfies some equation
sigma (choose(90,i)x^i) = 1 (or similar).

Therefore sin(1 degree) is algebraic, and so is sin(1+1).

We can therefore construct sin(n) or indeed sin(m/n) as algebraic.

Therefore sin maps all rational degrees to algebraic numbers.

The interesting (for me) is the fact that this doesn't occur for sin in
radians, where sin(1) seems unobtainable within algebraics.



.

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