Someone over at /pol/ posted this question
>>>/pol/77657495
... does there exist a surjective polynomial function R^n -> R_{>0} for some positive natural number n?
Want to try yourself? I think I might have a proof.
>>8149427
sum of the squares of coordinates.
>>8149429
It's supposed to be R^n --> (0,\infty) but the sum of squares is zero at the origin.
>>8149430
please restate the question. I still dont understand.
>>8149427
(xy-1)^2+x^2
>>8149430
then no.
>>8149435
Is there a polynomial function P:R^n --> (0,\infty) in n variables that is surjective, i.e. onto (0,\infty)?
>>8149441
The image of such a function is closed in \mathbb{R}, so no
>>8149439
To give more detail:
The polynomial is obviously non-negative, and can't be zero because this would require x=0 and xy=1 for some y.
Now let c>0. If c>=1, pick y=0 and x=sqrt(c-1).
If 0<c<1, then pick x=sqrt(c) and y=1/sqrt(c)
In either case, c=(xy-1)^2+x^2
>>8149451
>Nice job anon!
are you from reddit?
>>8149451
>it's onto
Filthy plebeian.
>>8149439
Good. Now do it in one variable
>>8150547
OP said from R^n to R+.
>>8150589
He didn't restrict it to n>1 tho
>>8150595
for some n
not for all n.
you dumb fucking faggot.
>>8150547
e^x
>>8150649
e^x is not a polynomial
>>8150694
It's a polynomial of infinite order.
>>8150709
e^x is not a polynomial.
>>8150709
>infinite order
>>8150740
you might want to lurk more.
Also still not a polynomial.
>>8150746
>composed of terms of [math]x^n[/math]
>not a polynomial
u wot m8
>>8150748
What the fuck is wrong with you, leave this guy alone
>>8150748
that's called a power series.
A polynomial specifically has a finite degree.
>>8150748
Polynomials by definition only have definitely many terms.
>>8150756
>>8150762
>>8150780
https://en.wikipedia.org/wiki/Polynomial#Definition
>A polynomial is an expression that can be built from constants and symbols called indeterminates or variables by means of addition, multiplication and exponentiation to a non-negative power.
Nowhere is the degree restricted to be finite.
>>8150800
and a set if a collection of objects.
that's not the definition.
>>8150800
The finitude of the degree is implicit in the word "addition". Addition is formally a binary operation, so building with addition can yield only finite degree.
>>8150807
>and a set if a collection of objects.
That's exactly the definition of set. Where's the problem?
>>8150813
>Addition is formally a binary operation, so building with addition can yield only finite degree.
Nothing prevents us from applying the binary operation of addition an infinite number of times.
>>8150649
>>8150709
>>8150740
>>8150748
>>8150800
>>8150831
everyone taking the bait of this guy is a retarded newfag.
>>8150800
>Nowhere is it restricted to be infinite
It is if you've taken a basic course in Ring Theory
