Problem: Prove that if [imath]p(x)[/imath] is a real polynomial such that [imath]p(x)\geq 0[/imath] for all real [imath]x[/imath], then [imath]p[/imath] does not have any real roots of odd multiplicity.
Proof attempt.
Clearly if [imath]p(x)\geq 0[/imath] then for any real root [imath]\alpha[/imath] must have even multiplicity, otherwise [imath]p(x)[/imath] would change sign as the value of [imath]x[/imath] changes from [imath]x>\alpha[/imath] to [imath]x<\alpha[/imath]. However I'm not sure if this reasoning proves the statement above, so I will attempt to prove it below.
Suppose [imath]p(x)=a_o+a_1x+...+a_nx^n[/imath] where [imath]a_i \in \mathbb{R}[/imath]. Then by the properties of conjugation, if [imath]z_o[/imath] is a complex root such that [imath]p(z_o)=0[/imath], then [imath]p(\bar z_o)=0[/imath]. Hence by the factor theorem [imath]p(x)[/imath] is divisible by [imath](x-z_o)[/imath] and [imath](x-\bar z_o)[/imath] such that [imath]p(x)=(x-z_o)(x-\bar z_o)q(x)[/imath] for some polynomial [imath]q(x)\neq 0[/imath]. If [imath]z_o[/imath] have multiplicty [imath]\geq 2[/imath], then [imath]\bar z_o[/imath] must be a root to [imath]q(x)[/imath], this is true for [imath]\bar z_o[/imath] also. Hence [imath]z_o[/imath] and [imath]\bar z_o[/imath] must have same multiplicty. Now suppose [imath]z_o=a+bi[/imath] where [imath]b\neq 0[/imath]. Then
[math](x-z_o)(x-\bar z_o)=x^2-(z_o+\bar z_o)x+z_o\bar z_o=x^2-2ax+a^2+b^2=(x-a)^2+b^2.[/math] Note that this expression must be positive for all [imath]x\in \mathbb{R}[/imath]. Now let [imath]x_1,...,x_m[/imath] be all the real roots for [imath]p(x)[/imath] with corresponding multiplicities [imath]k_1,k_2,...,k_m[/imath], and count all non-real roots in conjugate pairs as [imath]z_1,\bar z_1,z_2,\bar z_2,...,z_l,\bar z_l[/imath] such that [imath]deg(p)=m+2l[/imath]. Then [imath]p[/imath] can be prime-factored as
[math]p(x)=a_n(x-x_1)^{k_1}...(x-x_m)^{k_m}(x-z_1)(x-\bar z_1)...(x-z_l)(x-\bar z_l)[/math]Set [imath]z_j=a_k+ib_k[/imath] for [imath]j=1,...,l[/imath] then
[math]p(x)=a_n(x-x_1)^k_1...(x-x_m)^{k_m}(x^2-2a_1+a_1^2+b_1^2)...(x^2-2a_lx+a_l^2+b_l^2)[/math]Now suppose for contradiction that [imath]p(x)\geq 0[/imath] for all x [imath]\in \mathbb{R}[/imath] and that [imath]p[/imath] has one root of odd multiplicity. Without loss of generality suppose [imath]x_1[/imath] has multiplicity [imath]k_1=2c_1+1[/imath] for some positive integer [imath]c_1[/imath], while [imath]k_2,...,k_m[/imath] are even such that [imath]k_j=2c_j[/imath] for [imath]j=2,...m[/imath]. Then
[math]p(x)=a_n(x-x_1)^{2c_1+1}(x-x_2)^{2c_2}...(x-x_m)^{2c_m}\prod_{i=1}^{l} ((x-a_i)^2+b_i^2)[/math]
But if [imath]a_n >0[/imath], choose [imath]x=x_1-1[/imath]. Then since each factor in the polynomial is a square or a sum of squares except for [imath](x-x_1)^{2c_1+1}[/imath], there exist some [imath]x\in \mathbb{R}[/imath] such that [imath]p(x)<0[/imath]. However if [imath]a_n<0[/imath], choose [imath]x=x_1+1[/imath], then there exist some [imath]x\in \mathbb{R}[/imath] such that [imath]p(x)<0[/imath]. But we assumed that [imath]p(x)\geq 0[/imath] for all [imath]x \in \mathbb{R}[/imath] and that [imath]p[/imath] has one real root of odd multiplicity. Hence by reductio ad absurdum [imath]p(x)[/imath] does not have a real root of odd multiplicity.
Is this proof correct? I haven't done math in a while and would appreciate any feedback. Thanks.
Proof attempt.
Clearly if [imath]p(x)\geq 0[/imath] then for any real root [imath]\alpha[/imath] must have even multiplicity, otherwise [imath]p(x)[/imath] would change sign as the value of [imath]x[/imath] changes from [imath]x>\alpha[/imath] to [imath]x<\alpha[/imath]. However I'm not sure if this reasoning proves the statement above, so I will attempt to prove it below.
Suppose [imath]p(x)=a_o+a_1x+...+a_nx^n[/imath] where [imath]a_i \in \mathbb{R}[/imath]. Then by the properties of conjugation, if [imath]z_o[/imath] is a complex root such that [imath]p(z_o)=0[/imath], then [imath]p(\bar z_o)=0[/imath]. Hence by the factor theorem [imath]p(x)[/imath] is divisible by [imath](x-z_o)[/imath] and [imath](x-\bar z_o)[/imath] such that [imath]p(x)=(x-z_o)(x-\bar z_o)q(x)[/imath] for some polynomial [imath]q(x)\neq 0[/imath]. If [imath]z_o[/imath] have multiplicty [imath]\geq 2[/imath], then [imath]\bar z_o[/imath] must be a root to [imath]q(x)[/imath], this is true for [imath]\bar z_o[/imath] also. Hence [imath]z_o[/imath] and [imath]\bar z_o[/imath] must have same multiplicty. Now suppose [imath]z_o=a+bi[/imath] where [imath]b\neq 0[/imath]. Then
[math](x-z_o)(x-\bar z_o)=x^2-(z_o+\bar z_o)x+z_o\bar z_o=x^2-2ax+a^2+b^2=(x-a)^2+b^2.[/math] Note that this expression must be positive for all [imath]x\in \mathbb{R}[/imath]. Now let [imath]x_1,...,x_m[/imath] be all the real roots for [imath]p(x)[/imath] with corresponding multiplicities [imath]k_1,k_2,...,k_m[/imath], and count all non-real roots in conjugate pairs as [imath]z_1,\bar z_1,z_2,\bar z_2,...,z_l,\bar z_l[/imath] such that [imath]deg(p)=m+2l[/imath]. Then [imath]p[/imath] can be prime-factored as
[math]p(x)=a_n(x-x_1)^{k_1}...(x-x_m)^{k_m}(x-z_1)(x-\bar z_1)...(x-z_l)(x-\bar z_l)[/math]Set [imath]z_j=a_k+ib_k[/imath] for [imath]j=1,...,l[/imath] then
[math]p(x)=a_n(x-x_1)^k_1...(x-x_m)^{k_m}(x^2-2a_1+a_1^2+b_1^2)...(x^2-2a_lx+a_l^2+b_l^2)[/math]Now suppose for contradiction that [imath]p(x)\geq 0[/imath] for all x [imath]\in \mathbb{R}[/imath] and that [imath]p[/imath] has one root of odd multiplicity. Without loss of generality suppose [imath]x_1[/imath] has multiplicity [imath]k_1=2c_1+1[/imath] for some positive integer [imath]c_1[/imath], while [imath]k_2,...,k_m[/imath] are even such that [imath]k_j=2c_j[/imath] for [imath]j=2,...m[/imath]. Then
[math]p(x)=a_n(x-x_1)^{2c_1+1}(x-x_2)^{2c_2}...(x-x_m)^{2c_m}\prod_{i=1}^{l} ((x-a_i)^2+b_i^2)[/math]
But if [imath]a_n >0[/imath], choose [imath]x=x_1-1[/imath]. Then since each factor in the polynomial is a square or a sum of squares except for [imath](x-x_1)^{2c_1+1}[/imath], there exist some [imath]x\in \mathbb{R}[/imath] such that [imath]p(x)<0[/imath]. However if [imath]a_n<0[/imath], choose [imath]x=x_1+1[/imath], then there exist some [imath]x\in \mathbb{R}[/imath] such that [imath]p(x)<0[/imath]. But we assumed that [imath]p(x)\geq 0[/imath] for all [imath]x \in \mathbb{R}[/imath] and that [imath]p[/imath] has one real root of odd multiplicity. Hence by reductio ad absurdum [imath]p(x)[/imath] does not have a real root of odd multiplicity.
Is this proof correct? I haven't done math in a while and would appreciate any feedback. Thanks.
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