why isnt sqrt(x^2 + y^2) a polynomial?
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From: Frederick Williams on 4 Jul 2010 05:37 Gerry wrote: > > On Jul 4, 3:07 pm, mike <mike_newsgro...(a)yahoo.com> wrote: > > why isnt sqrt(x^2 + y^2) a polynomial? > > > > Thanks in advance! > > Which polynomial do you think it should be? 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. -- I can't go on, I'll go on.
From: Axel Vogt on 4 Jul 2010 05:40 Frederick Williams 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). (x+y)^2 = x^2 + y^2 if char = 2, so P(x,y) = x+y will do.
From: David C. Ullrich on 4 Jul 2010 06:49 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? 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
From: Frederick Williams on 4 Jul 2010 09:08 Axel Vogt wrote: > > Frederick Williams 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). > > (x+y)^2 = x^2 + y^2 if char = 2, so P(x,y) = x+y will do. My guess is that the OP is interested in polys over R. -- I can't go on, I'll go on.
From: Dave L. Renfro on 4 Jul 2010 10:09 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
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