What is the minimum value of $n$ for a nonnegative polynomial with degree $d$?

  • Context: MHB 
  • Thread starter Thread starter melese
  • Start date Start date
  • Tags Tags
    Polynomials
Click For Summary
SUMMARY

The minimum value of \( n \) for a nonnegative polynomial \( f(x) \) of degree \( d \) can be determined through the decomposition of \( f(x) \) into a sum of squares of polynomials. Specifically, it is established that \( n = 1 \) when \( d = 0 \), \( n = 2 \) when \( d = 2 \), and further analysis suggests that the relationship between \( n \) and \( d \) follows a pattern related to linear independence in polynomial spaces. This topic is closely tied to Hilbert's seventeenth problem, which addresses the representation of nonnegative polynomials as sums of squares.

PREREQUISITES
  • Understanding of polynomial functions and their degrees
  • Familiarity with linear algebra concepts, particularly vector spaces
  • Knowledge of polynomial decomposition techniques
  • Awareness of Hilbert's seventeenth problem and its implications
NEXT STEPS
  • Research the implications of Hilbert's seventeenth problem on polynomial representation
  • Study the properties of sums of squares in polynomial algebra
  • Explore linear independence in polynomial spaces and its applications
  • Examine examples of polynomial decomposition for various degrees
USEFUL FOR

Mathematicians, particularly those focused on algebra and polynomial theory, students studying advanced algebra concepts, and researchers interested in the applications of Hilbert's problems in modern mathematics.

melese
Messages
19
Reaction score
0
(HUN,1979) Prove the following statement: If a polynomial $f(x)$ with real coefficients takes only nonnegative values, then there exists a positive integer $n$ and polynomials $g_1(x),g_2(x),...,g_n(x)$ such that $f(x)=g_1(x)^2+g_2(x)^2+\cdots+g_n(x)^2$.

A related question of my own, but I don't have/know the answer:
If $\deg({f})=d$, then what is the smallest possible value of $n$.
For example: I know that it's $2$, when $d=2$ and $1$ when $d=0$.

መለሰ
 
Last edited by a moderator:
Mathematics news on Phys.org
melese said:
(HUN,1979) Prove the following statement: If a polynomial $f(x)$ with real coefficients takes only nonnegative values, then there exists a positive integer $n$ and polynomials $g_1(x),g_2(x),...,g_n(x)$ such that $f(x)=g_1(x)^2+g_2(x)^2+\cdots+g_n(x)^2$.

A related question of my own, but I don't have/know the answer:
If $\deg({f})=d$, then what is the smallest possible value of $n$.
For example: I know that it's $2$, when $d=2$ and $1$ when $d=0$.

መለሰ

I thought of this in terms of linear algebra and vector spaces. Consider the set

\begin{matrix}
g_0(x)=\sqrt{c_0} & g_0^2(x)=c_0 \\
g_1(x)=(1+c_1x) & g_1^2(x)=1+2c_1x+c_1^2x^2\\
g_2(x)=\sqrt{c_2}x & g_2(x)=c_1x^2\\
g_3(x)=(x+\sqrt{c_3}x^2) & g_3(x)=x^2+c_3x^3+c_3^2x^4\\
g_4(x)=\sqrt{c_4}x^2) & g_4(x)=c_4x^4 \\
g_5(x)=(x^2+\sqrt{c_5}x^3 & g_5(x)=x^4+c_5x^5+c_5^2x^6 \\
\vdots & \vdots \\
\end{matrix}

This gives a matrix that looks like this

\begin{bmatrix}
c_0 & 0 & 0 & 0 & 0 & \dots \\
1 & 2c_1 & c_1^2 &0 & 0 & \dots \\
0 & 0 & c_2 & 0 & 0 & \dots \\
0 & 0& 1 & 2c_3 & c_3^2 & \dots \\
\vdots &\vdots &\vdots &\vdots &\vdots & \ddots \\
\end{bmatrix}

As long as the number of rows $$ k$$ is odd the set will be linearly independent and thus span polynomial space of degree $$k-1$$

Just to be clear now we can pick [math]c_0,c_1,c_2,...[/math] so that we can get any polynomail

For example if $$f(x)=3x^2-8x+10$$, then

We solve
\begin{matrix}

c_0+1=10 \\
2c_1=-8 \\
c_1^2+c_2=3
\end{matrix}

This gives
$$c_0=9, \quad c_1=-4, \quad c_2=-13$$

and $$9g_0^2-4g_1^2-13g_2=9+1+2(-4)x+(-4)^2x-13x^2=3x^2-8x+10$$

I think this will help settle your other question as well.
 
Last edited:

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K