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  1. If $P(x)$ is an even polynomial such that $P(x) \in \mathbb{R}[x]$, then prove that there exists a real valued polynomial $R(x)$ such that $P(x)=R(x^2)$
  2. Would the result still be true if $P(x),R(x)$ are complex valued polynomial.

Attempt: For 1, as $P(x)$ is an even polynomial, it's degree must be even. Let $P(x)= a(x-x_1)(x-x_2) \cdots (x-x_n)$ As $P(x)=P(-x)$ we get $$(x-x_1)(x-x_2) \cdots (x-x_n)= (x+x_1)(x+x_2) \cdots (x+x_n)$$ How do I proceed?

Ellie_Wong
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1 Answers1

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Write $P(x)$ as $P(x)=E(x)+O(x)$ where $E(x)$ contains only even powers of $x$ and $O(x)$ contains only odd powers of $x.$ Since $P(x)=P(-x),$ identically, we have that $E(x)+O(x)=E(-x)+O(-x).$ But, $E(x)$ has only even powers of $x$ and $O(x)$ has only odd powers of $x.$ It is easy to see that $E(-x)=E(x),$ and $O(x)=-O(-x).$ So, we get $E(x)+O(x)=E(-x)-O(-x)=E(-x)+O(-x).$ This means that $2O(-x)=-2O(x)=0$ identically. Hence, $O$ is $0$ identically. Hence, $P(x)=E(x).$ Now, $E(x)$ contains only even powers of $x.$ So, $E(\sqrt{x})$ is a polynomial in $x.$ Take $R(x)=E(\sqrt{x}).$ Then, $R(x^2)=E(x)=P(x).$

Now, nowhere did I make the assumption that we were working in $\mathbb{R}[x].$ Indeed, the same proof works even if we consider polynomials in $\mathbb{C}[x].$ For example, take $P(x)=ix^2+2.$ In this case, $R(x)=E(\sqrt{x})=ix+2.$

aqualubix
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