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From: Arturo Magidin on 4 Jul 2010 11:02
On Jul 4, 9:09 am, "Dave L. Renfro" <renfr...(a)cmich.edu> wrote:
>
> 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,

Well, you get |a|=|b|=1, at any rate... which is of course good
enough.


> and from the third term on the right side we get a = 0 or b = 0,
> so we have a contradiction.

--
Arturo Magidin
From: Axel Vogt on 4 Jul 2010 12:30
Dave L. Renfro 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.
>
(...comparing coefficients, assuming no zero divisors...)
>

Comparing coefficients also works in the general case,
char <> 2, for a polynomial ring S = A[x,y] = A[y][x]:

Assume 0 = T^2 - (x^2+y^2) has a solution p in S.

Write p = Sum( t(k)*x^k , k = 0 ... N), t(k) in A[y] and
p^2 in degree k has coefficients Sum( t(i)*t(j), i+j = k).

For k=0: t(0)^2 = y^2.

For k=1: 0 = 2*t(0)*t(1) and multiplying by t(0) gives
0 = 2*t(0)*t(1)*t(0) = 2*y^2*t(1), so t(1) = 0.

For k=2: 1 = 2*t(0)*t(2) + t(1)^2 = 2*t(0)*t(2).

That says: t(0) is invertible in A[y], hence its square.
Which is y^2.

That also works for the (formal) power series, I think.
From: W^3 on 5 Jul 2010 15:27
In article
<05633c05-5ee2-4408-907f-0451520ac64c(a)g19g2000yqc.googlegroups.com>,
"Dave L. Renfro" <renfr1dl(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

You could also argue this way: If sqrt(x^2 + y^2) is a polynomial in x
and y, then restricting to (x, 0) shows sqrt(x^2) = |x| is a
polynomial in x. Thus there exist c_0, ..., c_n in R such that |x| =
c_0 + c_1*x + ... + c_n*x^n for all x in R. This implies x = c_0 +
c_1*x + ... + c_n*x^n for x > 0, or 0 = c_0 + (c_1 - 1)*x + ... +
c_n*x^n for x > 0. But a polynomial of degree n that vanishes at more
than n points must be the zero polynomial, which happens iff all of
its coefficients are 0. Thus c_1 = 1, and all other c_k = 0. This
implies |x| = x for all x in R, contradiction.
From: Dave L. Renfro on 6 Jul 2010 09:39
Dave L. Renfro wrote (in part):

>> 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,

Arturo Magidin wrote:

> Well, you get |a|=|b|=1, at any rate... which is of course good
> enough.

Ooops . . .

Dave L. Renfro
From: Barry Schwarz on 21 Jul 2010 08:02
On Sat, 3 Jul 2010 22:07:49 -0700 (PDT), mike
<mike_newsgroups(a)yahoo.com> wrote:

>why isnt sqrt(x^2 + y^2) a polynomial?

Assume it is a polynomial. By definition, there must be a finite
expression with integer coefficients c_i_j of the form
sum(over i and j) c_i_j * x^i * y^j
which is the canonical expression for that polynomial.

Evaluate the expression when x=1 and y=1. It is just the sum of the
coefficients. Since they are all integers, the sum must be an
integer.

But the value of sqrt(1^2 + 1^2) is not an integer.

Since we have reached a contradiction, our assumption must be faulty.
Therefore, sqrt(x^2 + y^2) cannot be a polynomial.

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