Polynomials (1 Viewer)

[seanieg89]

Prove that a polynomial with real coefficients for all real if and only if is expressible as the sum of two squares of real polynomials.

 

[nrlwinner]

Can you please rephrase the question because I can't see the question in that line.

 

[seanieg89]

Prove that a real polynomial is non-negative everywhere if and only if it is expressible as the sum of two squares of polynomials.

eg 2x^2 + 2x + 1 = x^2 + (x+1)^2, and the LHS is always non-negative.

Don't see how it can be phrased much clearer than that...

 

[Trebla]

Isn't that trivial? A square of a polynomial is always non-negative therefore the sum of squares of polynomials is clearly also non-negative.

 

[fixing]

If the gay ass polynomial can be written as a sum of two squares it can't have a nigga negative numba as you can't get nigga negatives when you add two squares.

You do your own fucken proof from that. Fucken lol you're doing a Maths degree and you're asking school kids for help.

 

[scroates]

nowhere as easy as everyone is making it out to be. Think you guys missed the if and only if part that complicates things

 

[seanieg89]

scroates is correct, you are all missing the 'if and only if'. The fact that the sum of two squares of polynomials is non-negative is indeed trivial, but the fact that every non-negative polynomial is expressible thus is not. Hint: Try induction in some form.

And this is not my uni work, I already have a proof. I simply post some of the more difficult questions i consider setting for my 4U students on BOS to gauge how hard they are.

 

[shaon0]

scroates is correct, you are all missing the 'if and only if'. The fact that the sum of two squares of polynomials is non-negative is indeed trivial, but the fact that every non-negative polynomial is expressible thus is not. Hint: Try induction in some form.

And this is not my uni work, I already have a proof. I simply post some of the more difficult questions i consider setting for my 4U students on BOS to gauge how hard they are.

Trebla may have been thinking the 'converse' was trivial ie. If P is sum of two real poly's then P>=0.

 

[seanieg89]

a further hint:

You can make use of the following factorisation

 

[Trebla]

I've 'found' the solution, but will not post it up as it is not my solution haha

 

[Affinity]

a further hint:

You can make use of the following factorisation

well this takes care of the hard part, the rest is just tidying things