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$$Q(x) = x^k + a_1x^{k+1}+ ... + a_nx^{k+n}$$ where $k,n$ are positive integers is a polynomial with real coefficients. I have to show that $Q(x)/x^k$ is strictly positive for all real x satisfying $$0<|x|<1/(1+\sum_{i=0}^n |a_i|)$$

Now the polynomial formed after division(which is legal since $x$ is non-zero) is the general $n$ degree polynomial $1+a_1x+...+a_nx^n.$ This approaches $1$ as $x$ approaches $0$. But how can I obtain the right hand part of the inequality? Furthermore, I tried substituting $x=1$ to obtain the right hand denominator but the it doesn't give the sum of the absolute values.

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  • $\begingroup$ Please write out your question as it means a) everyone can see it and b) it is searchable on the site. See the MathJax tutorial if you need help with formatting the maths $\endgroup$ – lioness99a May 14 '17 at 13:18
  • $\begingroup$ I'm sorry. I'll try and edit it. $\endgroup$ – Anish Bhattacharya May 14 '17 at 13:23
  • $\begingroup$ I also urge you to write "positive" instead of "+ve" as not everyone knows what the latter means. $\endgroup$ – kccu May 14 '17 at 13:40
  • $\begingroup$ Hope the edit's okay :) $\endgroup$ – Anish Bhattacharya May 14 '17 at 13:52
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We will show that $$ p(x)=1+a_1x+\ldots+a_nx^n>0\quad\text{for all }x\in X=\left[-\frac{1}{1+\sum|a_i|},\frac{1}{1+\sum|a_i|}\right]. $$ Observations:

  1. $|x|\le\frac{1}{1+\sum|a_i|}\le 1$ $\Rightarrow$ $|x|^i\le|x|$.
  2. $p(x)=1+\sum a_ix^i\ge 1-\max_{x\in X}|\sum a_ix^i|$.

We can estimate from 1 $$ |\sum a_ix^i|\le\sum|a_i|\underbrace{|x|^i}_{\le |x|}\le|x|\sum|a_i|\le\frac{1}{1+\sum|a_i|}\sum|a_i|<1,\quad\forall x\in X, $$ then $\max_{x\in X}|\sum a_ix^i|<1$ and from 2 we conclude that $p(x)>1-1=0$ in $X$.

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