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Let $f(x)=a_nx^n+a_{n-1}x^{n-1}\dots+a_1x+1$. Prove that $f(x)$ is strictly positive if $$0<|x|<{1\over1+\sum_{i=1}^{n}|a_i|}.$$

Any hints on how to start?

Robert Z
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Anvit
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  • At $x=0$, the function is obviously positive because it is equal to 1. The problem is suggesting that there are no roots near 0. It is giving you a range of values for $x$ such that there are no roots of $f(x)$. Because $f(x)$ is a polynomial, it cannot become nonpositive without reaching zero.

    So, to start, assume $0 < |x| < \dfrac{1}{\displaystyle 1 + \sum_{i=1}^n |a_i|}$. Then show that $f(x)>0$.

     – SlipEternal May 01 '18 at 15:43
  • @InterstellarProbe So we need to show Absolute value of root is greater than the reciporcal of sum? – Anvit May 01 '18 at 15:55

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Let $h(x)=f(x)-1=x\sum_{i=1}^{n}a_ix^{i-1}$ then for $|x| < \dfrac{1}{1 + \sum_{i=1}^n |a_i|}$we have that $|x|<1$ and $$|h(x)|\leq |x|\sum_{i=1}^{n}|a_i|<1.$$ Therefore $$f(x)=1+h(x)\geq 1-|h(x)|>0.$$

Robert Z
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