/sci/ - Science & Math

Previous Thread >>10536135

>>10543223
Clearly the leading coefficient is positive.
Additionally all real roots must have even multiplicity or p would change sign near it.
So p factors as:
A(Π(x-r_i)^2)(Π(x-z_i)(x-z*_i))
=Aq(x)^2 Π(x^2+|z|^2)
By fibbonaccis identity
(a^2+b^2)(c^2+d^2)
=(ac+bd)^2+(ad-bc)^2
The later product is a sum of the squares of two real polynomials and A≥0 so the first part AΠ(x-r_i)^2 is a square.
So p is a sum of two sqaures.

>>10543223
If f(x) is a polynomial function with order i such that i is an integer in I, the resulting p(x) will be the sum of any polynomial function in {1,...,k} with a order of 2j such that j = max{i_1,....,i_k}

There.

I don't think I understand the problem, because this seemed way to easy.
If each f(x) is squared, then for any real value of x, it should be positive.

>>10543362
We basically know that the sum of any polynomial is a polynomial so we just need to show that any polynomial function squared is also a polynomial then we're clear

>>10543329
Some more details.
(z* denotes complex conjugate of z.)
p factors by FTA and the complex roots are conjugates as the coefficients are real (p∈R[x] by say taylor expansion at zero, as p(x)≥0 for real x all derivatives at 0 are real.)

>>10543362
>>10543375
You got it backwards, friends.

>>10543223
Use induction on the degree of the leading term. The rest is trivial, QED

>>10543223
I was getting somewhere, but can't quite connect all the dots. It's part of Hilbert's 17th problem here https://en.wikipedia.org/wiki/Hilbert%27s_seventeenth_problem and here https://en.wikipedia.org/wiki/Positive_polynomial

P(x) must be a sum of two squares of real polynomials in one variable. So it takes the form [math](x+a)^{2}+(x+b)^{2}=2x^{2}+2ax+2bx+a^{2}+b^{2}[/math]

Then find a family of polynomials f sub such that [math]\sum_{j=1}^{k}(f_{j}(x))^{2}=[/math] shares the form for P(x).

Trying f sub j of x equals (j+x) for example gives https://www.wolframalpha.com/input/?i=sum((j%2Bx)%5E2,j,1,k) which is the correct degree, but I didn't check if coefficients jived.