- #1
Gregg
- 459
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1. Let [tex] \mathbb{R}[x]_n^+ [/tex] and [tex] } \mathbb{R}[x]_n^- [/tex] denote the vector subspaces of even and odd polynomials in [tex] \mathbb{R}[x]_n[/tex]
Show [tex] \mathbb{R}[x]_n=\mathbb{R}[x]_n^+ \oplus\mathbb{R}[x]_n^- [/tex]
3. For every [tex] p^+(x) \in \mathbb{R}[x]_n^+ [/tex] [tex]\displaystyle p^+(x)=\sum_{m=0}^n a_m x^m=p^+(-x)[/tex]
So [tex] a_m = 0 [/tex] for [tex] m=2k+1, k=0,1,2,...[/tex] else [tex] a_m \in \mathbb{R} [/tex]. Similarly, [tex] a_m=0[/tex] for [tex] m=2k, k=0,1,2,...[/tex] if the function is odd.
[tex] p^+(x)=a_0+a_2x^2+a_4x^4+\cdots, a_m\in\mathbb{R} [/tex]
[tex] p^-(x)=a_1x+a_3x^3+a_5x^5+\cdots a_m\in\mathbb{R} [/tex]
[tex] p(x)=a_0+a_2x^2+a_4x^4+\cdots+a_1x+a_3x^3+a_5x^5+\cdots [/tex] for every [tex] p(x)\in \mathbb{R}[x]_n [/tex]. So every [tex] p(x) [/tex] is some [tex] p^+(x) [/tex] with some [tex] p^-(x) [/tex]. Is this enough? Is it better to find a basis for the two subspaces and show that the union of the two basis sets spans [tex] \mathbb{R}[x]_n [/tex] ?
Show [tex] \mathbb{R}[x]_n=\mathbb{R}[x]_n^+ \oplus\mathbb{R}[x]_n^- [/tex]
3. For every [tex] p^+(x) \in \mathbb{R}[x]_n^+ [/tex] [tex]\displaystyle p^+(x)=\sum_{m=0}^n a_m x^m=p^+(-x)[/tex]
So [tex] a_m = 0 [/tex] for [tex] m=2k+1, k=0,1,2,...[/tex] else [tex] a_m \in \mathbb{R} [/tex]. Similarly, [tex] a_m=0[/tex] for [tex] m=2k, k=0,1,2,...[/tex] if the function is odd.
[tex] p^+(x)=a_0+a_2x^2+a_4x^4+\cdots, a_m\in\mathbb{R} [/tex]
[tex] p^-(x)=a_1x+a_3x^3+a_5x^5+\cdots a_m\in\mathbb{R} [/tex]
[tex] p(x)=a_0+a_2x^2+a_4x^4+\cdots+a_1x+a_3x^3+a_5x^5+\cdots [/tex] for every [tex] p(x)\in \mathbb{R}[x]_n [/tex]. So every [tex] p(x) [/tex] is some [tex] p^+(x) [/tex] with some [tex] p^-(x) [/tex]. Is this enough? Is it better to find a basis for the two subspaces and show that the union of the two basis sets spans [tex] \mathbb{R}[x]_n [/tex] ?