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Let*s define a power series:

$$ \begin{align} p_n(x) = \sum\limits_{k=1}^n a_k x^k, \quad a_k \in \mathbb{N} \end{align} $$

Is it possible to give a certain bound $-1 \le x_b \le x < 0 $ such that

$$ \begin{align} p_n(x) &> p_{\infty}(x), \quad \text{if n is even} \\ p_n(x) &< p_{\infty}(x), \quad \text{if n is odd} \end{align} $$

although $a_n$ is kind of a random distributed sequence only bound between $a_n = n$ and $a_n = n^2$?

Edit: I would be happy also about a rough bound. F.e. let's say the best $x_b$ for a given $a_n$ sequence would be $-0.6$ but I would be able to say that $-0.25$ is a certain bound such that the conditions above are true.

Edit 2: With kind of randomly distributed I mean that I don't know the sequence. The values can jump up and down. I only know that this sequence is bounded. So, the "jumps" are bounded, too.

Clarification of the question: Is there any way to give a fixed $x_b$ only by the bounds such that the condition is true?

thinkingeye
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    The question doesn't make much sense to me. You call $p_n$ a power series, but really it's a polynomial; did you intentionally leave off the constant term? It's not clear what $x_b$ means. Perhaps you want partial sums of a power series to alternately be upper and lower bounds on the limit, for all inputs in a certain subinterval of $[-1, 0]$? I have no idea what "although $a_n$ is kind of a random distribution sequence only bound between $a_n=n$ and $a_n=n^2$?" is trying to say. The question seems too vague to be answerable. – Joshua P. Swanson Dec 28 '20 at 12:46
  • Thanks, indeed. I edited. – thinkingeye Dec 28 '20 at 18:11

1 Answers1

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I assume that $p_n(x) = \sum_{k=1}^n a_k x^k$.

As I understood the question, there exists a feasible sequence $(a_n)$ without the bound $x_b$, because the gap between the bounds for $a_n$ is too big. Namely, let $a_n=n$ when $n$ is odd and $a_n=n^2$, when $n$ is even. Given $-1<x<0$, pick an even $n$ such that $\tfrac 1n <-x$. It is easy to check that for each positive odd $k$ we have $$ a_{n+k}+xa_{n+k+1}=(n+k)+x(n+k+1)^2<0.$$ Then $$p_\infty(x)-p_n(x)= \sum_{k=n+1}^\infty a_k x^k=x^{n+1}((a_{n+1}+a_{n+2}x)+ x^2(a_{n+3}+a_{n+4}x)+\dots)>0.$$

Alex Ravsky
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    So, as I understand, your $x_b$, which depends on $n$. would be like $-\frac{1}{n}$. So there isn't any because $x$ would be go to zero as $n$ goes to infinity. There could be no other fixed $x_b$ as zero? – thinkingeye Dec 28 '20 at 18:14
  • @thinkingeye It is more correct to say that for any $-1\le x_b<x<0$ there exists an even $n$ such that the condition $p_n(x)> p_{\infty}(x)$ is violated, so there exists no $x_b$ required in the question, as I understood it. – Alex Ravsky Jan 04 '21 at 11:44