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I have eddited the Question, I hope the Statement is now true.

I want to show that every such polynomial has a zeropoint, If I can prove the Statement above, I can use the indermediate value Theorem.

What I have worked out so far is that

$P(x)$, can be rewritten in the form of:

$P(x)= a_0x^{2k+1}+a_1x^{i_1}+a_2x^{i_2}+...a_jx^{i_j}+C,j\in\mathbb{N}_0$ and $2k+1>i_1>…>i_j$

$= (a_0x^{2k+1-i_j}+a_1x^{i_1-i_j}+...+a_j)x^{i_j}+C$

$= ((a_0x^{2k+1-i_j-(i_{j-1}-i_j)}+...a_{j-1})x^{i_{j-1}-i_j}+a_j)x^{i_j}+C$

$=((a_0x^{2k+1-i_{j-1}}+...a_{j-1})x^{i_{j-1}-i_j}+a_j)x^{i_j}+C$

$...$

$=(…(a_0x^{2k+1-i_1}+a_1)x^{i_2-i_1}+a_2)…)x^{i_{j-1}-i_j}+a_j)x^{i_j}+C$

How can I continue, do you have any suggestions?

I had the idea to pick

$\max\limits_{0 \leq s \leq j}|a_s|$ as my $x$

I somehow think that $P(x)$ would either be positive or negative dependant whether I pick $+x$ or $-x$. If that goes in the right direction I need somebody to tell me how to proceed.

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  • $\begingroup$ You want to use this to prove the intermediate value theorem? $\endgroup$ – Mindlack Jan 11 at 16:05
  • $\begingroup$ Your statement is imprecise. The polynomial $P(x)=x$ has odd degree $1$ but does not have a positive or a negative zero. $\endgroup$ – MPW Jan 11 at 16:05
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    $\begingroup$ The claim in your title is not true. What are you actually trying to prove? $\endgroup$ – user3482749 Jan 11 at 16:10
  • $\begingroup$ No what I want to show is that every polynomial of uneven Degree has a zeropoint. If I can show that for every such polynomial there exists a positive and a negative value i.e. $\exists_{z,z'\in\mathbb{R}}:P(z)>0,P(z')<0$ I can make use of the intemediate value Theorem to say that $\exists x$ in the intervall $[z,z']$ or $([z',z])$: $P(x)=0$. Therefor I assume without loss of generality that $P(x)>0,\forall x\in \mathbb{R}$ and try to invoke a contradiction by constructing a $x$ with negative value. $\endgroup$ – RM777 Jan 11 at 16:11
  • $\begingroup$ look at the limits. (if $P \in \mathbb{R}[X]$) it should give you the answer $\endgroup$ – Alexis Jan 11 at 16:15
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Assume WLOG that $a_{2n+1}\gt0$. We have $p(x)=a_{2n+1}x^{2n+1}+\dots+a_0$. But, as you take the limit as $x$ approaches $\pm\infty$, the leading term dominates. That is, $\lim_{x\to\pm\infty}p(x)=lim_{x\to\pm\infty}x^{2n+1}(a_{2n+1}+\frac{a_{2n}}{x}+\frac{a_{2n-1}}{x^2} +\dots+\frac{a_0}{x^{2n+1}})=\lim_{n\to\pm\infty}a_{2n+1}x^{2n+1}=\pm\infty$.

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  • $\begingroup$ Why can you Claim that so easily don't you have to give an explicit $a$ such that $\forall_{r\in\mathbb{R}}\exists_{a\in\mathbb{R}}\forall_{x\in\mathbb{R}}:x\in(a,+\infty)\Rightarrow p(x)>r$ $\endgroup$ – RM777 Jan 11 at 19:32
  • $\begingroup$ But no worries I could solve and understand the Problem now. $\endgroup$ – RM777 Jan 11 at 19:40
  • $\begingroup$ Well that's fairly easy to do; given that the expression in parentheses approaches $1$. That is, take $x$ such that both $x^{2n+1}\gt\frac r{a_{2n+1}}$ and the expression in parentheses is sufficiently close to $1$. There are some details to consider, but it clearly works out. $\endgroup$ – Chris Custer Jan 11 at 19:45
  • $\begingroup$ Sorry I thought I had it but I still don't understand it why does the expression in the parentheses approach 1? Take for example $\frac{a_0}{x^{2n+1}}$ it is true if I take a big $x$ then the term goes as close to 0 as I want but I if I multiply this value by$ x^{2n+1}$ then I have$ a_0 $again which does not have to be $1$ $\endgroup$ – RM777 Jan 11 at 20:05
  • $\begingroup$ Yeah, typo on my part. The expression in parentheses approaches $a_{2n+1}$. And you have given the reason why: the other terms all go to zero. $\endgroup$ – Chris Custer Jan 11 at 20:08

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