why isnt sqrt(x^2 + y^2) a polynomial?
mike
Thanks in advance!
Mike
Gerry
> why isnt sqrt(x^2 + y^2) a polynomial?
>
> Thanks in advance!
Which polynomial do you think it should be?
--
GM
Edson
P(x, y) = Sum(i = 0, m) Sum(j = 0, n) c_i_j x^i y^j
f(x, y) = sqrt(x^2 + y^2) = (x^2 + y^2)^(1/2) is not of this form.
f^2 is, but not f.
Edson
Axel Vogt
> why isnt sqrt(x^2 + y^2) a polynomial?
The question is a bit ill-posed: what do you mean by 'sqrt'?
About what 'objects' you are talking?
Frederick Williams
>
> Because, by definition, a polynomial in 2 variables is a function of the form
>
> P(x, y) = Sum(i = 0, m) Sum(j = 0, n) c_i_j x^i y^j
>
> f(x, y) = sqrt(x^2 + y^2) = (x^2 + y^2)^(1/2) is not of this form.
Maybe the OP wants a proof that (x^2 + y^2)^(1/2) can't be put in the
form P(x, y).
> f^2 is, but not f.
--
I can't go on, I'll go on.
Frederick Williams
There may not be a polynomial that the OP thinks sqrt(x^2 + y^2) should
be. Maybe he wants a proof that sqrt(x^2 + y^2) isn't the sum of a
finite number of terms c x^n y^m.
Axel Vogt
> Edson wrote:
>> Because, by definition, a polynomial in 2 variables is a function of the form
>>
>> P(x, y) = Sum(i = 0, m) Sum(j = 0, n) c_i_j x^i y^j
>>
>> f(x, y) = sqrt(x^2 + y^2) = (x^2 + y^2)^(1/2) is not of this form.
>
> Maybe the OP wants a proof that (x^2 + y^2)^(1/2) can't be put in the
> form P(x, y).
(x+y)^2 = x^2 + y^2 if char = 2, so P(x,y) = x+y will do.
David C. Ullrich
<mike_ne...@yahoo.com> wrote:
>why isnt sqrt(x^2 + y^2) a polynomial?
Deopends on what sort of "why" you mean.
Maybe you're just confused on what the word
"polynomial" means - if so the answer is that
a polynomial is a finite sum of terms of the
form c x^n y^m and sqrt(x^2 + y^2) is
obviously not a finite sum of such terms, it's
the square root of such a sum.
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.
>Thanks in advance!
>
>Mike
Frederick Williams
My guess is that the OP is interested in polys over R.
Dave L. Renfro
>> 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
Arturo Magidin
>
> 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
Axel Vogt
> 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 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.
W^3
<05633c05-5ee2-4408...@g19g2000yqc.googlegroups.com>,
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.
Dave L. Renfro
>> 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
Barry Schwarz
<mike_ne...@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.
--
Remove del for email
Dan Cass
A nice proof, but the assumption you make:
> Assume it is a polynomial. By definition, there must
> be a finite
> expression with integer coefficients c_i_j ...
is necessary for your proof to work.
The coefficients c_i_j could presumably be rational
or even arbitrary reals.
Butch Malahide
wrote:
Let's see. Suppose sqrt(x^2 + y^2) is a polynomial in x and y. Setting
y = 0, the function f(x) = sqrt(x^2) = |x| is a polynomial in x. Since
f(x) vanishes at x = 0 but does not change sign, 0 is at least a
double root, i.e., f(x) is divisible by x^2. But the function f(x)/x^2
= |1/x| can't be polynomial with that infinite discontinuity at x = 0.
Axel Vogt
As an algebraic variant I think one can go the following route:
Assume t^2 - (x^2+y^2) = t^2 - f^2 = (t-f(x,y))*(t+f(x,y)),
which is the question in algebraic form.
Differentiating by x gives -2*x = -2*diff(f(x,y),x)*f(x,y)
and thus f(x,y) as a polynomial in x must have degree 1.
Dito w.r.t. y.
Therefore f(x,y)= a + b*x+c*y+d*x+y. Consider 0 = l.h.s - r.h.s.
Obviously a = 0 and then looking at y^2 one has c=0 or c=-2 and
after cancelling x:
0 = (-1+2*b*d+d^2+b^2)*x +- (2*b+2*d)*y = =((b+d)^2-1)*x +- (b+d)*y.
But "(b+d)^2-1 = 0 and b+d = 0" has no solution.
Ostap Bender
wrote:
> Edson wrote:
>
> > Because, by definition, a polynomial in 2 variables is a function of the form
>
> > P(x, y) = Sum(i = 0, m) Sum(j = 0, n) c_i_j x^i y^j
>
> > f(x, y) = sqrt(x^2 + y^2) = (x^2 + y^2)^(1/2) is not of this form.
>
> Maybe the OP wants a proof that (x^2 + y^2)^(1/2) can't be put in the
> form P(x, y).
Maybe? That's CLEARLY what he wants: a proof that sqrt(x^2 + y^2)
cannot be expressed as a polynomial in x and y. Which is pretty easy
to prove by supposing that it is and then squaring both sides and
comparing terms.
Ostap Bender
When people ask such questions, they are probably dealing with real
numbers, and 'sqrt' stands for 'square root'. Can you figure out what
the expression 'square root of a positive real number' means?
Ostap Bender
Finally, a simple answer to a simple question.