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From: Ostap Bender on 22 Jul 2010 05:20
On Jul 4, 7:09 am, "Dave L. Renfro" <renfr...(a)cmich.edu> wrote:
> Mike wrote:
> >> why isnt sqrt(x^2 + y^2) a polynomial?
>
> David C. Ullrich wrote (in part):
>
> > Or maybe you want a _proof_ that this function
> > is not such a sum. If so, one proof is that any
> > polynomial is obviously continuously differentiable,
> > while sqrt(x^2 y^2) is not differentiable at the origin.
>
> I was curious how hard it would be to prove this using
> "high school algebra" methods, and came up with the following.
>
> Assume sqrt(x^2 + y^2) is a polynomial. Let n be the degree
> of this polynomial. Since the degree of
>
> sqrt(x^2 + y^2) * sqrt(x^2 + y^2) = x^2 + y^2
>
> is 2, it follows that n + n = 2, or n = 1.
>
> Therefore, there exist constants a, b, c such that
>
> sqrt(x^2 + y^2) = ax + by + c.
>
> Now square both sides and equate corresponding coefficients
> (a quadratic-in-x-and-y polynomial that is identically equal
> to zero on a planar set containing more than 4 points must be
> the zero polynomial . . .):
>
> x^2 + y^2 = (a^2)(x^2) + (b^2)(y^2) + c^2 + (2ab)(xy) + (2ac)x +
> (2bc)y
>
> Since c = 0, we get
>
> x^2 + y^2 = (a^2)(x^2) + (b^2)(y^2) + (2ab)(xy)
>
> From the first two terms on the right side we get a = b = 1,
> and from the third term on the right side we get a = 0 or b = 0,
> so we have a contradiction.
>
> Dave L. Renfro

Finally, a simple answer to a simple question.
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