Find $n\in N$ such that exist base in space $P_n$ [closed]

Find for which $n\in \mathbb N$ exist a base of space $\{p_0,p_1,p_2,p_3,\cdots,p_n\}$ of space $P_n$ such that:

a) none of polynomials does not have degree two

b) every polynomial have even degree

c) every polynomial have odd degree

I do not uderstand what he want , for a) dimension is n, for b and c dimension is $\frac{(n+1)}{2}$

closed as unclear what you're asking by Mathematician 42, Gibbs, Delta-u, José Carlos Santos, CesareoSep 16 '18 at 18:59

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• You completely changed the question with that edit. – Mathematician 42 Sep 16 '18 at 8:34
• Why did you edit your question to something completely different? – Matheus Andrade Sep 16 '18 at 8:34
• Sorry I just want to check can I ask question because I can not ask more then 50 question, but thank you for help but this is real question, sorry for confusion – Marko Škorić Sep 16 '18 at 8:36
• Why can't you ask more than $50$ questions? – Matheus Andrade Sep 16 '18 at 8:38
• i do not know I can not ask more then 50 question for 30 days, and I study for exam now I need to delete a some question and check every day can I ask one question, I ask for help and they said that they can not help me, you can ask only 50 question and that is it. – Marko Škorić Sep 16 '18 at 8:40

$(a)$ Your solution is correct, but it is not defined for $y = 0$. You can add something like:

$$f(x,y) = \begin{cases} \frac{x^2}y, &\text{ if }y \ne 0 \\ 0, &\text{ if } y = 0\end{cases}$$

$(b)$ Consider complex conjugation $z \mapsto \overline{z}$.

$(a)$ If $\deg p_i = 2, \forall i$ then $P_n = \operatorname{span}\{p_0, \ldots, p_n\} \subseteq P_2$ so the only candidate is $n = 2$. Indeed, $\{1+x^2, x+x^2, x^2\}$ is a basis for $P_2$.
$(b)$ If all $p_i$ are of even degree, then clearly $n$ cannot be odd because then $\deg p_i \le n-1, \forall i$ so $P_n = \operatorname{span}\{p_0, \ldots, p_n\} \subseteq P_{n-1}$. If $n$ is even then $$\{1+x^n, x+x^n, \ldots, x^{n-1} + x^n, x^n\}$$ is a basis for $P_n$ with $\deg p_i = n$ which is even.
$(c)$ Similarly we see that $n$ must be odd.
• how you get for b) that is linear since some arbitrary $x\in \mathbb C$ $x=\alpha_1+i\beta$ and $\alpha \in \mathbb R$ $g(\alpha x)= \alpha \alpha_1-i\alpha \beta=\alpha(\alpha \alpha_1-i\alpha \beta)=\alpha g(x)$ – Marko Škorić Sep 16 '18 at 8:54
• @MarkoŠkorić It is not $\mathbb{C}$-linear because $\overline{i} = -i \ne i$. – mechanodroid Sep 16 '18 at 8:55