Someone over at /pol/ posted this question

Someone over at /pol/ posted this question

... 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.

Other urls found in this thread:

en.wikipedia.org/wiki/Polynomial#Definition
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sum of the squares of coordinates.

It's supposed to be R^n --> (0,\infty) but the sum of squares is zero at the origin.

please restate the question. I still dont understand.

(xy-1)^2+x^2

then no.

Is there a polynomial function P:R^n --> (0,\infty) in n variables that is surjective, i.e. onto (0,\infty)?

The image of such a function is closed in \mathbb{R}, so no

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

Yes, very good, I salute you! I thought it wasn't possible, and that seemed to be where the other thread was going.

Nice job user!

(xy-1)^2+x^2 =2 k^2 when x=k and y=1+1/k, and so it's onto.

>Nice job user!
are you from reddit?

>it's onto

Filthy plebeian.

Good. Now do it in one variable

OP said from R^n to R+.

He didn't restrict it to n>1 tho

for some n
not for all n.
you dumb fucking faggot.

This works, and extends easily to [math]n[/math] variables for [math]n >1 [/math].

Clearly no such thing.

e^x

e^x is not a polynomial

It's a polynomial of infinite order.

e^x is not a polynomial.

>infinite order

[eqn]e^x = \sum_n \frac{x^n}{n!}[/eqn]

you might want to lurk more.

Also still not a polynomial.

>composed of terms of [math]x^n[/math]
>not a polynomial
u wot m8

What the fuck is wrong with you, leave this guy alone

that's called a power series.
A polynomial specifically has a finite degree.

Polynomials by definition only have definitely many terms.

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.

and a set if a collection of objects.

that's not the definition.

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.

>and a set if a collection of objects.
That's exactly the definition of set. Where's the problem?

>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.

everyone taking the bait of this guy is a retarded newfag.

>Nowhere is it restricted to be infinite
It is if you've taken a basic course in Ring Theory