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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 $\frac{(n+1)}{2}$

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

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ You completely changed the question with that edit. $\endgroup$ – Mathematician 42 Sep 16 '18 at 8:34
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    $\begingroup$ Why did you edit your question to something completely different? $\endgroup$ – Matheus Andrade Sep 16 '18 at 8:34
  • $\begingroup$ 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 $\endgroup$ – Marko Škorić Sep 16 '18 at 8:36
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    $\begingroup$ Why can't you ask more than $50$ questions? $\endgroup$ – Matheus Andrade Sep 16 '18 at 8:38
  • $\begingroup$ 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. $\endgroup$ – Marko Škorić Sep 16 '18 at 8:40
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Answer to the original question:

$(a)$ 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 $z \mapsto \overline{z}$.


Answer to the new question:

$(a)$ 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.

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  • $\begingroup$ how you get for b) that is linear since some arbitrary $x\in \mathbb C$ $x=\alpha_1+i\beta$ and $\alpha \in \mathbb R$ $g(\alpha x)= \alpha \alpha_1-i\alpha \beta=\alpha(\alpha \alpha_1-i\alpha \beta)=\alpha g(x)$ $\endgroup$ – Marko Škorić Sep 16 '18 at 8:54
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    $\begingroup$ @MarkoŠkorić It is not $\mathbb{C}$-linear because $\overline{i} = -i \ne i$. $\endgroup$ – mechanodroid Sep 16 '18 at 8:55
  • $\begingroup$ thanks but can you help for this new question if you have idea? $\endgroup$ – Marko Škorić Sep 16 '18 at 8:58
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    $\begingroup$ @MarkoŠkorić Have a look now. $\endgroup$ – mechanodroid Sep 16 '18 at 9:09
  • $\begingroup$ thank you so much $\endgroup$ – Marko Škorić Sep 16 '18 at 9:15

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