Polynomial proof question.

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Problem.

Let [imath]f(x)[/imath] be a polynomial with real coefficients such that [imath]f(x)>0[/imath] for all real [imath]x[/imath]. Prove that there are real polynomials [imath]p(x)[/imath] and [imath]q(x)[/imath] such that [imath]f(x)=p^2(x)+q^2(x)[/imath]

My solution attempt:

The polynomial has no real roots since [imath]f(x)>0[/imath] for all real x. Suppose [imath]f(x)[/imath] has corresponding complex roots and count them as pairs

[math]z_1,\bar{z_1},...,z_l,\bar{z_1}[/math] And clearly [imath]deg f=2l[/imath] in this case. Now suppose [imath]z_j=a_j+ib_j[/imath] for [imath]j=1,...,l.[/imath] Then by the fundamental theorem of algebra
[imath]f(x)[/imath]
[math]f(x)=a_n(x-z_1)(x-\bar{z_1})\cdot\cdot\cdot(x-z_l)(x-\bar{z_l})[/math][math]f(x)=a_n(x^2-(z_1+\bar{z_1})x+z_1\bar{z_1})\cdot\cdot\cdot(x^2-(z_l+\bar{z_l})x+z_l\bar{z_l})[/math][math]f(x)=a_n(x^2-2a_1x+a_1^2+b_1^2)\cdot\cdot\cdot(x^2-2a_lx+a_l^2+b_l^2)[/math]
Hence [imath]f(x)[/imath] can be prime factored into real second-degree polynomials of the form

[math]f(x)=a_n\prod_{j=1}^{l}(x-a_j)^2+b_j^2[/math]
And we notice that each term is always positive because it's a sum of squares, therefore [imath]f(x)>0[/imath] provided [imath]a_n>0[/imath]. However, there remains to prove that the expansion of this expression can be written as [imath]p^2(x)+q^2(x)[/imath] and I'm not sure how to continue.

 

[Aion]

Problem.

Let [imath]f(x)[/imath] be a polynomial with real coefficients such that [imath]f(x)>0[/imath] for all real [imath]x[/imath]. Prove that there are real polynomials [imath]p(x)[/imath] and [imath]q(x)[/imath] such that [imath]f(x)=p^2(x)+q^2(x)[/imath]

My solution attempt:

The polynomial has no real roots since [imath]f(x)>0[/imath] for all real x. Suppose [imath]f(x)[/imath] has corresponding complex roots and count them as pairs

[math]z_1,\bar{z_1},...,z_l,\bar{z_1}[/math] And clearly [imath]deg f=2l[/imath] in this case. Now suppose [imath]z_j=a_j+ib_j[/imath] for [imath]j=1,...,l.[/imath] Then by the fundamental theorem of algebra
[imath]f(x)[/imath]
[math]f(x)=a_n(x-z_1)(x-\bar{z_1})\cdot\cdot\cdot(x-z_l)(x-\bar{z_l})[/math][math]f(x)=a_n(x^2-(z_1+\bar{z_1})x+z_1\bar{z_1})\cdot\cdot\cdot(x^2-(z_l+\bar{z_l})x+z_l\bar{z_l})[/math][math]f(x)=a_n(x^2-2a_1x+a_1^2+b_1^2)\cdot\cdot\cdot(x^2-2a_lx+a_l^2+b_l^2)[/math]
Hence [imath]f(x)[/imath] can be prime factored into real second-degree polynomials of the form

[math]f(x)=a_n\prod_{j=1}^{l}(x-a_j)^2+b_j^2[/math]
And we notice that each term is always positive because it's a sum of squares, therefore [imath]f(x)>0[/imath] provided [imath]a_n>0[/imath]. However, there remains to prove that the expansion of this expression can be written as [imath]p^2(x)+q^2(x)[/imath] and I'm not sure how to continue.

One observation is that each term in the product has the form
[math](x-a_j)^2+b_j^2=(x-a_j+ib_j)(x-a_j-ib_j)=\lvert x-(a_j+ib_j) \rvert^2[/math]which is the magnitude squared of the complex number [imath]x-a_j-ib_j[/imath]. Hence each factor [imath](x-a_j)^2+b_j^2[/imath] can be rewritten using the modulus of a complex number.

 

[Aion]

One observation is that each term in the product has the form
[math](x-a_j)^2+b_j^2=(x-a_j+ib_j)(x-a_j-ib_j)=\lvert x-(a_j+ib_j) \rvert^2[/math]which is the magnitude squared of the complex number [imath]x-a_j-ib_j[/imath]

The constant [imath]a_n[/imath] is included to ensure scaling of [imath]f(x)[/imath] and each factor [imath]x-(a_j+ib_j)[/imath] is a linear term in [imath]x[/imath], involving the complex roots [imath]a_j+ib_j[/imath].

 

[Aion]

The constant [imath]a_n[/imath] is included to ensure scaling of [imath]f(x)[/imath] and each factor [imath]x-(a_j+ib_j)[/imath] is a linear term in [imath]x[/imath], involving the complex roots [imath]a_j+ib_j[/imath].

Thus the product can be rewritten as

[math]f(x)=a_n\prod_{j=1}^{l}\rvert x-(a_j+ib_j) \lvert^2[/math]Now, define a complex-valued polynomial [imath]P(x)[/imath] whose modulus squared gives [imath]f(x)[/imath].

Let
[math]P(x)=a_n\prod_{j=1}^{l}(x-(a_j+ib_j))[/math]and
[math]\bar{P(x)}=a_n\prod_{j=1}^{l}(x-(a_j-ib_j))[/math]When [imath]P(x)[/imath] and [imath]\bar{P(x)}[/imath] are multiplied, the total product includes both [imath]x-(a_j+ib_j)[/imath] and [imath](x-(a_j-ib_j))[/imath] for each [imath]j[/imath]:

[math]P(x)\cdot\bar{P(x)}=a_n^2\prod_{j=1}^{l}(x-(a_j+ib_j))(x-(a_j-ib_j)[/math]And the full product becomes
[math]P(x)\cdot\bar{P(x)}=a_n^2\prod_{j=1}^{l}((x-a_j)^2+b_j^2)[/math]
This matches the original function [imath]f(x)[/imath] up to a constant [imath]a_n^2[/imath] with [imath]2l[/imath] terms.

We can conclude that [math]f(x) = \rvert P(x)\lvert^2=P(x)\bar{P(x)}[/math] (up to a constant)

The complex polynomial [imath]P(x)[/imath] can be written in terms of its real and imaginary parts as [math]P(x)=p(x)+iq(x)[/math] where [imath]p(x)[/imath] and [imath]q(x)[/imath] are real polynomials as the real and imaginary parts of [imath]P(x)[/imath]. Note that [imath]\bar{P(x)}=p(x)-iq(x)[/imath], hence

[math]f(x)=\rvert P(x)\lvert^2=p^2(x)+q^2(x)[/math] as required.